P O WER ELE CTRONICS
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By the same author • Electrical Machinery . .. • Generalized Theory of Electrical Machines
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KHAN l~A PUBLISHERS 2-B, Natb Market, Nai Sarak Delhi-1l00C6
il pOWER ELECTRONICS i
Dr. P. S. Binibhra ' '"
Ph. D., M.E. (Hons.), F.l.E. (India), M.I.S. T.E. Ex-Dean, Ex-Prof. and Head of Electrical & Electronics Engg. Dept.
Thapar Institute of Engineering and Techn ology
PATlAld\-147004
KHANNA PUBLISHE S
2-B Nath Market, Nai Sal'ak Delhi-ll0006 Phone s : 2391 23 80 ; 2722 41 79 • Fa;{
J IN BOO
DE POT
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1''1 011.23.116 01 / 02 / 03 . 011·863 7232
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Published by " Romesh Chander Khanna
for KHANNA PUBLISHERS
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Delhi-ll0006,
I ISBN No. : 81-7409-215-3
All RWht, Reserved [This book or part thereof cannot be translated or reproduced in any form (except for review or criticism) without the 'written .permission of the Authors and the Publishers] ,';
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Fourth Edition: 20Q6
Fourth Reprint : 2007. . '. ~
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Price,' Rs. 225.00
Text Composition by " , Soft Serve Computers Krishna Nagar, Delhi-51
Printed at " Moh an Lal Prfntars Shahdara, Delh i-32
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In
Loving Memory . . . of M y Late Parents
PREFACE
Power electronics blend s the three m ajor areas of E lectr ical En gin eering-power, electronics and control. Under controlled power conciitions, loa d performs better. So ther e has always been a popular demand to have power modula tors. It is power 'electronics th at has made possible the availability of a wide variety of controlled power converters. Power electronics has really r evlut;"(oniLed the art of power conversion and its control. The advent of power semiconduct or device, 't hyristor', in 1957 has been the most exciting breakthrough, because its launch gave a boost to the art of power conversion and its control, and t ook this art to its fore-front. As a result of technological evolution, many more semiconductor devices such as triacs, asymm etrical thyristors, gate turn off thyristors, power MOSFETs, insulated gate bipolar transistors, SITs, SITHs and MOS-controlled thyristors are n ow available. The use of these semicon ductor devices has perva ded the industrial applications r elating to the field of Electrical, Electr onics, Instrumentation and Control Engineering. In other words , power-electronic components find their use in low as well as high-power applications . The purpose of this book is to provide a good understanding of the power-electronic components and t he behaviour of power-electronic converters by presenting systematically all important aspects of semiconduct or devices amI the common type of electric-power controller s. The book begins with the study of salient features of power diodes, power transistors, MOS-controlled thyristor, silicon controlled rectifier an d other memb ers of thyr ' stor f am ily. Th en their applications in the differ ent t yp es of power-conv ert er configurations are presen t ed in a lucid det ail. In other w ords , this book fo llow s the bottom-down approach (device characteristics first and t hen their applications). Maj or part of the book is intended to serve as an intr odu ctory course in' pow er~el ectronics t o t he undergradu at e students of Electrical, Electronics, Instrumen tation and Control disciplines. It is presumed that the reader is familiar with the basics of elementary electronics and circuit theory. The material presented here can be covered in one sem est e with the amiss ' on of some topics . The instructor, after browsing through the b ook for some tim e, can plan t he course contents and its sequence without loss of continuity. The book contains thirteen chapt ers . Chapter 1 gives an overview of merits and demerits of power -electronic controllers and br' efly discusses the t opics covered in this book . This chapter also touches upon the significance of power· electr onics. Chapter 2 des cribes th e characteristics of power diodes, powe transistors and MOS-con tr olled thyristors. In C apter 3 are pr esented diode characteristics, rectifiers, performance par amet ers and fil ers. Chapter 4 explains the characteri tics of thyristors in detail and of Triacs, GTOs etc. Thyristor commuta j on t echniques are given in Chapter 5. In C~pt er 6, the prin ciples of conve sian from ac to de involving singl e-phase as well as three-phase converters are presented. Chapt ers 7 to 10 pertain to the treatment of de chopper s, inverters, ac voltage controllers an d cycloconverter s re·s pectively. While Chapter 11 gives study of several applications of power electronics, Chapter 12 discusses electric drives. P ower facto improvement an th e m eth ods o reactive power compensation are detailed in Chapte 13. A large number of ilJu trative diagrams and a wide variety of worked example add t o the clarity of th e subject matter. The material given in this boo is class-room tested. In the appendices ; Fo rier analysis, Laplace transform and a large numb er of objective-type qu es tions r elating to C ap te s 2 t o 13 are given.
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The material added in the present edition includes: (i) performance parameters of uncontrolled and controlled rectifiers and filters, (ii) structural modifications in thyristors and GTOs to make than more efficient, (iii) SIT and SITH and (iv) Chapter 13 on power factor improvement. Some topics have been r~-written to make the presentation more lucid. Many more illustrative examples to reinforce the understanding of the subject matter are also included. Objective-type questions are thoroughly updated. It is hoped that the book in its present form will serve the purpose for the courses on power electronics of all Indian as well as foreign universities. . The author is grateful to all those students who had interacted with the author, in the c1ass~room or outside, during the teaching of this subject. This interaction has greatly influenced the author's style of teaching and writing to a large extent and every effort has ,gone into making the subject matter presentation as easily comprehensible as possible . . Discussion with several instructors has also been of immense help and inspiration. The author is beholden to all of them. The author, however, regrets he cannot name them all, for it is a voluminous task. Finally, the author expresses his gratitude to his wife for her perennial encouragement, understanding and patience during the preparation of this book. The author would also thank his two sons a.'ld daughter-in-law for consistently boosting the author's morale, much needed during the revision of the book. Suggestions leading to the improvement of the book will be gratefully acknowledged.
Dr. P.S. Bimbhra
CONTENTS
1.
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INTRODUCTION ••••••••.••.• .• ..
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1.1. 1.2 . 1.3. 1.4. 1.5.
Concept of Power Elect r onics . . . . ' .' . . Applications of Power Electronics . . .. . . Advantages and Disadvantages of Power-electronic Converters. Power Electronic Systems . . . . . . . Power Semiconductor Devices . . . . . 1.6. Types of Power Electronic Converters 1. 7. Power Electronic Modules . . . . . . .
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4
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POWER SEMICONDUCTOR DIODES AND TRANSISTORS
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2.4.
2. 5.
2.6.
2.7.
2.8 . 2.9. 2.10.
3..
2.1.1. Depletion Layer .. .. . Basic Structure of Power Diodes Characteristics of Power Diodes 2.3.1. Diode]- V Characteristics . 2.3.2. Diode Reverse Recovery Characteristics Types of Power Diodes . . . . . 2.4.1. General-purpose Diodes 2.4.2. Fast-recovery Diodes 2.4.3. Schottky Diodes . . . . . Power Transistors . . . .. .. . 2.5.1. Bipolar J unction Transistors . 2.5.1.1. Steady-state Characteristics. 2.5.1.2. BJT Switching Performance. 2.5.1.3. Safe OperatingArea . . Power MOSFETS . . . . .. . . . 2.6. 1. Pmosfet Characteristics . . . . . 2.6.2. Pmosfet Applications . . . . . . 2.6.3. Co~parison of PMOSFET With BJT Insulated Gate Bipolar Transistor (IGBT) 2.7.1. Basic Structure . . 2.7.2. Equivalent Circuit . 2.7.3. Working . . . . . . 2.7.4. Lat.ch-up in IGBT . 2.7.5. IGBT Characteristics 2.7.6. Switching Characteristics 2.7.7. Application oflGBTh . . . 2.7.8. Comparison oflGBT With MOSFET Static Induct ion Transistor (SIT) MOS-controlled Thyristor (MCT) New Semiconducting Materials.
DIODE CIRCUITS AND
C"tIFIERS. . • • •
3. 1. Diode Circuits With de Source 3.1.1. B..esi:Jti"ve LoA 3.1.2. RC Load . ·
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2.1. The p-n Junction . . . .. . . . .
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3.2. 3. 3. 3.4. 3.5. 3.6. 3.7.
3.8.
3.9.
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3.1.3. RI Load ..
3.1.4. LC Load .
3.1.5. RIC Load.
Freewheeling Diodes
Diode and L Circuit
Recovery of Trapped Energy.
Single-phase Diode Rectifiers " 3.5.1. Single-phase' Half~wave Rectifier Zener Diodes .. . . . . . . . . . . .. .
Performance Parameters . . ~ .. . . . .
3.7.1. Input Performance Parameters . ~
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Comparis()nof Single-phase Diode Rectifiers 3.8.1. Single-phase Half-wave Rectifier ..
3.8.2. Single-phase Full-wave Mid-point Rectifier.
3.8.3. Single-phase Full-wave Bridge Rectifier .
Three-phase Rectifiers . . . . . . . . . . . . . . . . : .
3.9.1. Three-phase Half-wave Diode Rectifier .. , .
3.9.2. Three-phase Mid-point 6-pulse Diode Rectifier ' .. 3.9.3. Multiphase Diode Rectifier . . . . . . '.. .
3.9.4. Evolution of Three-phase Bridge Rectifier
3.9.5; Three-phase Bridge Rectifier .. ' . .
3.9.6 .. Three-phase Tw,"lve-pulse Rectifier Filters 3.10.1. Capacitor Filter (c-filter) 3.10.2. Inductor Filter O-filter) 3.10.3. I-C Filter . . . . . . . . .
THYRI STORS . . . . . 4.1. Terminal Characteristics of Thyristnrs . . . . . 4.1.1. Static I-V Characteristics of a Thyristor . ", . . 4.2. Thyristor Turn-on Methods .. . . . . . . . . . . . 4. 3. Switching Characteristics of Thyristors . . . . . . . . 4.3 .1. ' Switching Characteristics During Turn-on . . 4.3.2. Switching Characteristics During Turn-off .. 4.4. Thyristor Gate Characteristics . . . 4.5. Two-transistor Model of a Thyristor .. 4.6. Thyristor Ratings . . . . . . . . 4.6 .1. Anode Voltage Ratings '. 4.6.2. Current Ratings . . .. 4. 7. Thyristor Protection . . . . . . 4.7.1. Design of Snubber Circuits . 4.7.2. Overvoltag Protection . . . 4.7.2.1. Suppression of Ovet"Voltages. 4.7.3. Overcurrent Protection .. . .. . , 4.7.4. Gate Protection . . . . . . . . . . 4.8, Improvement of Thyristor Characteristics 4.8.1. Improvemetlt in dil dt Rating . . 4. 8.1.1. Higher-gate Curr~nt .. . 4. 8.1.2. StructuTal 'Modification of The Device..
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4.8.2. Improvement in dvl dt Rating . 4.9. Heating, Cooling and Mounting of Thyristors · 4.9 .1. Thermal Resistance . . . . . . . . 4.9.2. Heat Sink Specifications . . . . . . 4.9.3. Thyristor Mounting Techniques .. 4.10. Series and Parallel Operation of Thyristors 4.10.1. Series Operation . . . . . . . . . 4.10.2. Parallel Operation . . . . . . . . ~ . ' " 4.11. Other Members of the Thyristor Family . . . 4.11.1. PUT'. (Programmable Unijunction Transistor) 4.11.2. SUS (Silicon Unilateral Switch) 4.11.3. SCS (Silicon Controlled Switch) . . . . . 4.11.4 . Light Activated Thyristors . . . . . . . . 4.11.5. . The Diac (Bidirectional Thyristor Diode) 4.11.6. The Triac .. . . . . . . . . . . . . . . 4.11.7. Asymmetrical Thyristor (ASCR) .. . 4.11.8. · Reverse Conducting Thyristor (RCT) . 4.11.9. Other Thyristor Devices 4.12. Gate Turn off Thyristor (GTO) .. .. . . 4.12.1. Basic Structure · . . . . . . 4.12.1.1. Turn-on Process . . 4.12.1.2. Turn-off Process . . . 4.12. 2. Static J- V.Characteristics . . 4.12 .3. Switching Performance ' . 4.12.3.1. Gate Turn-on. . . 4.12.3..2. Gate Turn-off.. . 4.12.4. Comparison Between GTO and Thyristor . 4.12.5. Application of GTOs .. .. 4; 13. Static Induction Thyristor (SITH) . .. . 4.13.1. Basic Structure . . . ; . . . . . 4.13.2. Turn-on and Turn-off Processes 4.13.3. Application of SITH and Comparison with GTO . 4.14. Firing Circuits for Thyristors . . . . . . . . . . .' . . . . . 4.14 .1. Main Features of Firing Circuits . . . . . . . . . . 4.14.2. Resistance and Resistance-capacitance Firing Circuits 4.14.3.. Unijunction Transistor (UJT) 4.15. PUlse Transformer in Firing Circuits. . . . . 4.16. Triac Firing Circuit . . . . . . . . . . . . . . . 4.17. Gating Circuits for Single-Phase Converters 4. 17. 1. Gate Pulse Amplifiers . 4.17.2. Pulse Train Ga~ing 4.18. Cosine Firing Scheme . . . . . ".
5.
THYRISTOR COMMUTATION TECHNlQUES . . . . . . .. 5. 1. Class A Commutation: Load Commutation ; . .. '.
. 5.2. Class B Commutation : Resonant-pulse Commutation .
5.3. Class C Commutati on : Complemen~ Commutation .
5.4. Class D Commu tation : Impulse COlIllllutatio 5.5. Class E Commutation : External Pulse Commutat;on 5.6. Cl838 F Commutation: Line Commutation . .. . . .
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6.
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PHASE CONTROLLED RECTIFIERS 249
6.1. Principle of Phase Control. . . . . . . . . . . . . . '. 251
6.1.1. Single-phase Half-wave Circuit with RL Load 253
6.1.2. Single-phase Half-wave Cir cuit with RL Load and Freewheeling Diode 256
6.1.3. Single-phase Half-wave Circuit with RLE Load 26~ 6.2. Full-wave Controlled Converters . . . . . . . . .' . . . . . . . . . . . . .
263
6.3. Single-phase Full-wave Converters . . . . . . . . . . . . . . . . . . . . .
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6.3.1. Single-phase Full-wave Mid-point Converter (M-2 Connection) 265
6.3.2. Single-phase Full-wave Bridge Converters . . . . . .. . 265
6.3.2.1. Single-phase Full Converter (B-2 Connection) . 268
6.3.2.2. Single-phase Semiconverter. . . . . . . . . . .
6.3.2.3. Analysis of Twc-pulse Bridge Converter with 270
Continuous Conduction. . . . . . . . . . . . . 275
6.4. Single-phase '!\vo-pulse Converters with Discontinuous Load . 275
6.4.1. Single-phase Full Converter with Disc'ontinuous Current . 276
6.4.2. Single-phase Semi converter with Discontinuous Current 278
6.5 . Performance Pa rameters of Two-pulse Converters 278
6.5.1. Single-phase Full Converter . . . . . . . . . . . . . . 280
6.5.2. Single-phase Semiconverter . . . . . . . . . . . . . .
283
6.6. Single-phase Symmet rical and Asymmetrical Semiconverters .' 283
6.6.1. Single-phase Symmetrical Semiconverter . 284
6.6.2. Single-phase Asymmetrical Semiconverter . . 285
6.7. Three-phase Thyristor Converter s . . .. . . . . . . . 286
6.7.1. Three-phase Half-w ave Cont rolled Converter 286
6.7.1. 1. Three-phase 1'1-3 Co verter with R Load. 288
6.7.1.2. Three-phase M-3 Converter with RL Load. 294
6.7.2. Three-phase Full Converter s 298
6.7. 3. Three-phase Semiconvarlers . . . . :: -. . . 308
6.7. 4. Multi-pulse Controlled Converters . . . . . . 309
6.8. Performance P arameters of 3-phase Full Converters 313
6.9. Effect of Source Impedance on the Performance of Converters 314
6.9.1. Single-phase Full Converter . . . . 317
6.9.2. Three-phase Full Converter Bridge 3'24
6.10. Dual Converters . . . . . . . . . . 325
6.10.1. Ideal Dual Converter .. . 326
6.1 0.2. Practical Dual Converter 331
6.11. Some Worked E amples . . . . . . CHOPPERS. 7.1. P ':ciple of Chopper Operation . 7.2. Control Stra tegies . . . . . . . . 7.2.1. Time Ra tio Control (trc). 7.3. Step-up Chopper s . . . . . . . . .
7.4. Typ~ of Chopper Circuits . . . . 7.4. 1. Firs t.quadrant, 01' TyJls-A, Chopper . 7.4.2. Second-qu adran t, or Type-B, Chopper . 7.4.3 . Two-quadrant Type-A Chopper, 01' 'li.rpe-C Ch oppe 1 .4.4 . Two-quadrant Type-B Chopper, or Type-D Chop er 7 .4.0. Four-quadrant Chopper, or Type-E Chopper . . . . I
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7.5. Steady State Time-domain Analysi of Type-A Chopper 7.5.1. Steady State Ripple . . . . . . . . . 7.5.2. Limit of Continu ous Conduction . . 7.5.3. Computation of Extinction Time tx 7.5.4. Fourier Analysis of Output Voltage 7.6. Thyristor Chopper Circuits . .. .. . 7.6.1. Voltage-commutated Chopper 7.6.2. Current-commutated Chopper 7.6 .3. Load-commutated Chopper. 7.7. Multiphase Choppers . . .. .. .. .
8.
INVERTERS . . . . • . . . . . . . . . • . .
8.1. Single-phase Voltage Source Inverters: Operating Principle 8.1.1. Single-phase Bridge Inverters . . . . . . . . . . . 8.1.2. Steady-state Analysis of Sin gle-phase Inverter '. . 8.2 . Fourier Analysis of Single-phase Inverter Output Voltage 8.3. Force-commutated Thyristor Inverters . . . . . . . 8.3.1. Modified Mcmurray Half-bri dge Inverter . 8.3.2 . Modified Mcmurray F ull-bridge Inverter '. 8.3.3. McMurray-Bedford Half-bridge Inverter 8.3.4. McMurray-Bedford Full-bridge Inverter 8.4. Three Phase Bridge Inverters . . . . . . . . 8.4.1. Three-phase 180 Degree Mode VSI 8.4.2. Three-phase 120 Degree Mode VSI 8.5, Voltage Control i Single-phase Inverter 8.5.1. Ext ernal Control of ac Output Voltage 8.5 .2. External Control of dc Input Voltage 8. 5.3. Internal Control of Inverter ! . 8.6. Pulse-widt h Modulated Inverters . 8.S.1. Single-pulse Modulation .. 8.6.2. Multiple-pulse Modulation . 8.S. Sinusoidal-pulse Modulation (SPWNl) . 8.6.4. Realization of PWM in Single-phase Bridge Inverters. 8.7. Reduction of Harmonics in the Inverter Output Voltage . . 8.7. 1. Harmonic Reduction by PWM . , .. . .. , . , .. 8.7}. Harmonic Reduction by Transformer Connections. &7 .3. Harmonic Reduction by Stepped-wave Inverters 8.8, Current Source Inverters . . . . . . . .. . , . . . . . . . . 8.8.1. Single-phase CSI with Ideal Switches ... . , .. , 8.8.2. Single-phase Capacitor-commutated CSI Wit h R Load 8.8.3. Single-phase Auto-sequential Commutated Inverter (I-phas e ASCl) . 8.9. Series Invert;1 . , . . .. .. . . . . . . 8.9. . Basic Series Inverter . . . . . .
8 ,9.2. Analysis of Basic Series lnverte
8. 9.3. Modified Series Inverter . , B.9 . . Half-Bridge Sa 'as In 1erter 8.10. Single~Phase Parallal Inverte 8.10.1. Analysi3 of Panillel Inverter. B.ll . Good Inyerter . . . . . . , . . .
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AC VOLTAGE C ONT~OLLERS . . . 9.1. Principle of Phase Con trol . . . . 9.2 . Principle of Integral Cycle Control . . . . 9.3. Single-phase Voltage Controllers . . . . . 9.3.1. Siilgle-Phase Voltage Controller with R Load 9.3.2. Single-Phase Voltage Controller with RL Load. 9.4. Sequence Control of ac Voltage Controllers . . . . .. . . . 9.4.1. Two-stage Sequence Control of Voltage Controllers . 9.4 .2. Multistage Sequence Control of Voltage Controllers . 9.4.3. Single-phase SinWloidal Vol~age Controller . . . . . .
504-531 504 508 511 512
CYCLOCONVERTERS • . . . . . . , .• ~ . ,; . • . . . . . . ; .. 10.1. Principle of Cycloconverter OperatioI:l' . .. . . . . . . . . . . 10:1.1. Single-phase to Single-phase. Cireuit-'step-up Cycloconverter 10.1.1.1. Mid-point Cycloeonverter.. . . . . . . . . . . . . . . .
10.1.1.2. Bridge-type Cycloconverter. . . . . . . . . . . . . . .
10.1.2. Single-phase to Single-phase Circuit-step-downCycloconvertel' 10.1.2.1. Mid-point Cycloconverter.. 10.1.2.2. Bridge-type Cycloconverter. . . .. . 10.2. Three-phase Half-wave Cycloconverters . . .; . . . . . .' 10.2.1. Three-phase to Single-phase Cycloconverters . 10.2.2. Three-phase to Three-phase cycloconverters 10.3. Output Voltage Equation for a Cycloconverter 10.4. Load-eommutated Cycloconverter. . . . . . . . . .' .
532-550 533 533 533 .
SOME APPLICATIONS . . . . .' . . . • . . . 11.1. Switched Mode Power Supply (SMP S) 11. 1. 1. Flyback Converter . . . 11.1.2. Push-pull Converter .. 11.1.3. Half-bridge Converter . 11.1.4. Full-bridge Converter. 11.2. Uninterruptible Power SUpplies 11.3. High Voltage dc Transmission. 11.3.1. Types of HYDC Link . . 11.3.2. Bipolar HYDC System . 11.3.3. Control of HVDC Converters. 11.4. Static Switches · . . . . . . . . . . 11.4.1. Si ngle~phase ac Switehes . 11.4.2. de Switches .. . . . . . . 11.4.3. Desigri of S ta tic Switchei! . 11.5. Static Circuit Breakers . .. . . . 11.5.1. Static ae Circuit Breaker s . 11.5.2. Static de Cir cuit Breakers 11 .6. Solid State Relays . . .. . . . 11.6.1. DC Solid Sta te Relays 11.6.2 . AC Solid State Relays 11.7. Reson n t Convertars . . . .. . 11.7. 1. Zero-cu · ent Switching Resonant Converters . 11.7.1.1. L-type 2 CS Resonan ~ Converters. .
551:-582 551 552 554 555 556 557 559 560 : 561 562 564 565 567 567 588 588 . 569. 570 .571
517 523 . 523 527 527
534 534 534 536 538 538 541 543 548
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575 578 581
11.7.1.2. M-type ZCS Resonant Converter. 11 .7.2. Zero-voltage-switching Resonant Converters .. 11.7.3 . Comparison Between ZCS and ZVS Converters
12.
.'.
ELECTRIC DRIVES . . . . . . 12.1. Concept of Electric Drive 12.2. DC Drives . . . . . . . . . 12.2. 1. Basic Perfonnance Equations of dc Motors 12.3. Single-phase dc Drives . . . . . . . . . . . . . . . . 12.3.1. Single-phase Half-wave Converter Drives 12.3.2. Single-phase Semiconverter Drives . 12.3.3. Single-phase Full Converter Drives . 12.3.4. Single-phase Dual Converter Drives 12.4. Three-phase dc Drives . . . . . . . . . . . . . 12.4.1. Three-ph~.se Ha1f-wave Converter Drives. 12.4.2. Three-phase Semiconverter Drives . 12.4.3. Three-phase Full-converter Drives . 12.4.4. Three-phase Dual Converter Drives. 12.5. Chopper Drives . . ' . . . . . . . . . . . . . . 12.5 .1. Power Control or Motoring Control 12.5.2. Regenerative-braking Control . 12.5.3. Two-quadrant Chopper Drives . 12.5.4. Four-quadrant Chopper Drives 12 .6. A.C. Drives . . . . . . . . . . . . . . . . 12.7. Inductio .-Motor Drives . . . . . . . . . 12.8. Speed Control of Three-phase Induction Motors 12.8.1. Stator Voltage Control . . . . . . . . . 12.8.2 . Stator Frequency Control .. . . . . . . 12.8.3 . Sta tor Voltage and Frequency Control 12.8.4. Stator Current Control . . . . . 12.8.5. Static Rotor·resistance Control 12.8.S. Sl'p-power Recovery Schemes . 12.8.S.1. Static Kramer Drive.. 12.8.6.2. Static Scherbius Drive. 12.9. Synchronous Motor Drives . . .. 12.9.1. Cylindrical Rotor Motors 12.9.2. Sall nt-pole Motors . . . 12.9.3. Reluctance Motors. . . . 12.9.4. Permanent-Magnet Motors . 12.10. Some Wor ed Examples . . . . .
13.
POWE
FACTOR IMPROVEMENT
13.1 . Effect of Poor Power-factor .. . . . . . . 13.2. Methods of Reactive Power Compensatio 13 .2.1. Capacitor Banks . . . . . . . . . . 13.2.2. Synchr6nous Condensers . . . . . 13.2.. Thyristor Controlled Reactors ('1'CRS) 13.3 . Static VAx CompensaUlr (SVc) 13.4. Som Worked Examples .. . . . . . . . . .. .
5 83~674
583 584
585 587
588 590 592 597 597
598 599 602 610 610 611 616 618 61 9 620
621 624 624 626 628 632 639 645 645 651 652 653 656 658 659 660
.....
675·'890 675 676 677 678 678 680 .'". 684
(xvi)
. .. , . .. 691·698
Appendix: A - FOURIER ANALYSIS . . .
. . . . . 699
Appendix: B - LAPLACE TRANSFORMS . App endix: C - OBJECTIVE TYPE QUESTIONS .
Power Semiconductor Diodes and Transistors .
Diode Circuits and Rectifiers . . . .
Thyristors . . . . . . . . . . . . . . .
Thyristor Commutation Techniques
Phase Controlled Rectifiers
Choppers . . . . . . . .
Inverters . . . . . . . . .
AC Voltage Controllers.
Cyc1oconverters ..
Some Applications .. .
E lectric Drives . . . . .
Power Factor Improvement
700-170
700
703
711
723
726 741
749
757
762
766
767
769
REFERENCES INDEX
771
••••••••••••
" ,
••••••••••••••••••••••••
1
773
Chapter 1
Introd ct·O ... -....... •.......... ~
~
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In this Chapter • concept of Power Electronics • Applications of Power Electronics • Advantages and Disadvantages of Power-electronic Con verters • Power Electronic Systems • Power Semiconductor Devices
Types of Power Electro nic Converters
• Power Electronic Modules .
"' . . . . . . . . .
_ _ .. .
.
...... .
.
.
.. .
. . . . . . ... .
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.
-. . . . . . .. .
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The object of this chapter is to discuss briefly the concept of power electronics, applications of power electronics and the types of power converter s desc:t ibed in this book. 1.1. CONCEPT OF POWER ELECTRONICS
I
I
Power electronics belon gs partly to power engineers and partly to electronics engineers [2] . Power engineering is mainly concerned wi th gen eration, trans mission, eli tribution and utilization of electric energy at high efficiency. Electronics engineering, on the oth er h and,. is guided by distortionless production, transmission and reception of data and signals of very low power level, of th e order of a few watts, or milliw atts, without m ch consideration to the efficiency.. ill· addition, apparatus associated with power engineering .s based mainly on electromagn etic principles wh ereas that in electronics en gineering is based upon physical phenomen a in yaCUUID , gases/vapours and semiconductors. . Power electronics is a subject th at con cerns the appl" cation of electronic-priIlciples into situations that are rated at power level rather than signal level. It may also be defined as a subject that deals with the apparatus and equipmen t working on the principle of electronics but r ated at powe r level r ather than signal level. F or e::Iample, semiconductor 'power swit ches such as thyri tors, GTOs etc. work on the principle of electronics (movement of holes and electrons), but h ave the na.m e power attached to them only as a description of their power ratings. Similarly, diodes, mercury-arc rectifiers and thyratrons (gas-fill ed triode), high-pow er .level devices, fann a part of the s"Llbject power electronics ; because their working is based on the physical phenomena in gases and vapours , an electronic process. As the inclusion of all such power-rated electronic equipments would be a voluminous task, the present book is devoted to the study of semi-condu ctor-based power-electronic compon ents and systems only. It shocld be understood that the techniques used in the design of high-efficiency and high-energy level pow electronic circuits are quite different from those employed in the design of low-effici ency electronic circuits at signal levels. .
.
Th e era of m odern power electron ics began with th e inv.en tion of silicon - controlled-r ectifier (S CR) by B ell Laboratories in 1956 . Its prototype was introduced by GEC in 1957 and subsequent ly, GEC i n trod Llced SCR-ba ed system s commercially in 1958. Since then , there have been eme:rgence of many new power semiconductor devices. Power el e ctron~:: sys .e . ~ today incorp ora te p ower semi :::onductor d 9vice ~ well as microelectronic inte.,.:3 ci ell cuit3 .
2
P owe Electronics
[Art. 1.2J
The term, 'converter system', in gener al, is used to denote a static device that converts ac to dc, dc to ac, dc to dc or ac to ac. Conventional power controllers based on thyratr ons, mercury-arc rectifiers, magnetic amplifiers, rheostatic controllers etc. have been replaced by power electronic controllers using semiconductor devices in almost all applications. The development of new power-semiconductor devices, new circuit topologies with their improved performance and their fall in prices have opened up wide field fOT the new applications ofpower electronic converters. Ajudicious use of power-semiconductor devices in conjunction with microprocessors or microcomputers has further enhanced the control strategies and synthesizing capabilities of the power electronic converters. It is said that power semiconductor devices can be regarded as the muscle and the microelectronics as the intelligent brain in the modern power electronic systems. For controlling the power flow to load, all power semiconductor devices, used in a power electronic converter, are either fully-on or fully-off. In other words, all semiconductor devices in power-electronic converter operate as switches. Wilen the switch is fully-on, power semiconductor device handles large current (divided by the load impedance) and negligible voltage drop across it. When the switch is off, the device handles negligible current vrith full- voltage across it. Therefore, a power semiconductor device, during on and off periods, has very low power loss in it as compared to the power delivered by the source to load. This results in higher energy efficiency of the power electronic converter system. At the same time, low energy loss in the semiconductor device can easily be removed by its efficient cooling. This all has contributed to the widespread use of power electronic converters in the power conversion and control systems. Table 1.1 lists various applications of power electronics. Tills list is however not exhaustive. No boundaries can be earmarked for the applications of power electronics, especially with the present trend of integrated design of power-semiconductor devices, microprocessors and the controlled equipment. The power ratings of power-electronic systems Tfu.J.ge from a few watts in lamps to several hundred megawatts in HVDC transmission systems. It is believed that in the early twenty-first cent ury, 60 t o 80% of the electric power consumed in utility systems .vill pass through power-electronics and this figure will event ually reach 100% in the future. Tab le 1 .1. Some Applica tion3 of Power El ec tronics 1. Ae rosp a ce:
Space shutt le power supplies, satellit e power supplies, aircraft power systems. 2. Commercial : Advertising, heating, air conditi oning, central refrigerlltion, comput er and office equipment, uninterruptible power supplies, elevators, light dimmers and fl ashers . 3. Industrial: Arc and industrial furn aces, blowers and fans, pumps and compr essors , industrial lasers, transformer-tap changers, r olling mills, textile mills, excavat ors, cement mills, welding. 4. Residential : Airconditioning, cooking, lighting, spa ce heating, refrigera tors , electric-dooT openers, dryers, fans, personal co mputer s , oth er entert ainment equipmen t, v cu um cleaners , was hing an d sewi n g machines, light dimmers, foo d mixers, electric blankets, food-vvarm ar trays. 5. Telecommuni ca t::i.on : Battery charger s, power supplies ( dc an d UPS ). 6. Tra nsp ortation; B ttery chargez:s, traction ont 01 of electri c vehicles , electri c locomoti ves) street cars, t rolley buses, subways, automotive electronics. 7. Utility systems : High voltage dc transmission (HVDC), exdtation systSID3, VAR compensation, stati c circuH bTeak ers , fans and boiler-fe d pump.s, Buppl a-mentaJ'l] ene gy sys elD.3 ( solar, -;vind ).
In traduction
l---\ rt.
.4]
3
1.3. ADVANTAGES AND DISADVANTAGES'OFFOWEB-ELECTRONlC CONVER The a dvantages possessed by power-electronic system.:; are as und er : (i) High efficiency du e t o low loss in power -sem icond ctor devices. (ii) High reliability of power- electronic conver ter systems. (iii) Long life and less maint enance du e to the abse e of any moving parts . (iv) Fast dyn am ic respon s e of the p ow er-e lectr oni c systems as compared to electromechanical converter systems. (v) Small size and less weight r esult in less flo or space and therefore lower installati on cost. (vi) Mass p oduction of power -s emiconductor devices has resulted in lower cost of the converter equipment. Systems based on p ower electronics, however, suffer from the following disadvantages: (a) Power-electronic convert er circuits have a tendency to generate harmonics in the supply system as well as in the load cir cuit. In the load circuit, the perform ance of the load is influenced, for exam.nle, a high harm on ic content in the load circuit cau ses commutation problem s in de m achines , in creased m otor heating and more accoustical noise in both dc and ac machines. So steps must be tak en to filter these out from the output side of a converter. In the supply system , the harmonics distort the voltage w avetorm and seriously influence the performance of oth er equipm ent connected to the same supply line. In addition, the harmonics in the su pply line can also caus e interference in a udio- and video-equipment (called raclio interference ). It is , th erefore , n ec'essary to insert filter s on the inpl t sid e of a con verter. (b) Ac to dc and ac to ac con verters operat e at a low in p ut power fa ctor under certain
operat ing conditions. In order to avoid a low pf, some special measures have t o be adopted. (c) Power-el ectronic con trollers h av e low ov erload capa city, Th ese conv erters must , ther efore, be r ated fO T taking m omentary over loads . As such, cost of power electr onic controller m ay mcrease.
(d) Regeneration of power is difficult in power electronic converter systems.
The advantages posses ed. by power electronic converters far outiiveigh their disadvantages mentioned above. As a consequence, semiconductor-based converters are being extensively employed in systems where power flow is to be regulated. As already stated, conventional p ower controllers used in many inst allations have already been replaced by semiconductor-based power electronic controllers.
The m ajor compon en ts of a powe eledronic' syst em are shown in the form of a block di gram in Fig. 1. 1. M ain power s ou ce may be an a c supply syst e _ or a de supply ·ys tem. The outp t fr om the power electronic circui t may be variabb~ dc, 0 ac voltage, or it may be a var iable voltage and fre quency. In gen er al, the ou tput of a pow er electronic conver tor ci cu 't depends u pon the r equirements ofth e lo ad. F or example) iftb e load is a de m ot or, the converter output must be adju stable dir ect voltage. In case the load is 3-vhase induction m otor, the converter may ha v ~ adju st able voltage and fr equ ency at its outp ut terminals. The feedb ack component in Fig. 1.1 me asures a param eter f he 10 ' d, say sp eed in case of a r ota ting machin e, and comp ares it with th command . The diff:: ence of the two, throug the digital circuit components , controls f e i stant oftu..rn-on of semicondu.:toT d evices forming the
Power Electronks
[Art. 1.5J Ma in Power
Sou re e
~
Command
..
Control Unit
-p
Digital Circuit
Power Electronic Circuit
-
.j
Load .j
Feedback Signal I
Fig. 1.1. Block diagram of a typical power electronic system.
solid-state power converter system. In this manner, behaviour of the load circuit can be controlled, as desired, over a wide range with the adjustment of the command. -
I.S. POWER SEMICONDUCTOR DEVICES Silicon controlled rectifier (SCR) was first introduced in 1957 as a power semiconductor device. Since then, several other power semiconductor devices have been developed. Must of these semiconductor devices are listed below in Table 1.2 along with their circuit, or device, symbol and present maximum ratings. Table 1.2. Maximum ratings o{power sem iconductor devices S .No.
Device
1
Circuit symbol
, 1.
Diod e
2.
Thyri.stors
AC
~
OK
Ao·
~G
oK
,
(a) SCR
(6) L'SCE
Ao I
I (e) AS CRIRCT
~o
(d)
GTO
A
~\
~G
~G K,A
Upper operating (req. (kHz)
5000 V/5000 A
1.0
7000 V/5000 A
1.0
6000 V/3000 A
1.0
2500 V/400 A
2.0
5000 V/3000 A
2.0
I
oK
0;-<
K
~ or~~~:
-
Voltage I current ratings
I
i
r
[Art. 1.5J
In trod uc tion
Device
S.No.
Circuit symbol
Voltage I curren t
ratings
5
Upper operating freq. (kHz)
(e) S1TH
2500 Vl500 A
100 .0
1200 V/40 A
20.0
1200 VIIOOO A
0.50
1400 V/400 A
10.0
1000 Vl50A
100.0
1200 Vl300 A
100 .0
1200 Vl500 A
50.0
(j) MeT
M~f-2---it§r ~1
(g) Triac
1G
3.
Transistors
(a)
8
8 npn
onp
E
(b)
c
c
BJT
!
E
MOSFET
(n-channel)
(c)
D
SIT G
s (d)
IGBT
In the above table, the various abbreviations are; SeR (silicon controlled rectifier), LASeR (light-activated SCR), ASCR (asymmetric al SeR), RCT (revers e conducting thyristor), GTO (gate-tUIIl off thyristo~) , SITH (static induction thyrist or), MeT (MOS controlle thyTis or), BJT (bipolar junction transisto ), MOSFET (metal-oxide semicon ductor fiel d effect transistor), 81 (static indudi on transistor) and IGBT (insulated gate bipolar tnms ·stor). (
6
Power Electronics
[Ar t. 1.6]
Based on (i ) t urn-o n and turn-off characteristics, (ii) gate signal requirements and (iii) degree of controllability, the power semiconductor devices can be classified as under: (a ) Di od es . Thes e ar e uncontrolled rectifying devices. Their on and off states are controlled by power supply.
(b) Thyri st ors. These have controlled turn-on by a gate signal. After thyristors are on, they remain latched-in on-state due to internal regenerative action and gate loses control. These can be turned-off by the power circuit.
Controllable swit ches. These devices are turned-on and turned-off by the application of control signals. The devices which behave as controllable switches are BJT, MOSFET, GTO, SITH, IGBT, SIT and MCT. (c)
Triac and RCT possess bi-directional current capability whereas all other remaining devices (diode, SCR, GTO, BJT, MOSFET, IGBT, SIT, SITH and MeT) are unidirectional current devices.
1.6. TYPES
~!. POWER
ELECTRONIC CONVERTERS
A power electronic system consists 0: one or more power electronic copverters. A power electronic convert er is made up of some power semiconductor devices controlled by integrated circuits. The switching characteristics of power semiconductor devices permit a power electronic converter to shape the input power of one form to output power of some other form . St atic pow er converters perform these functions of power conversion very efficiently. Broadly speaking, power electronic converters (or circuits) can be classified into six types as under: 1. Di ode Rectifi ers. A diode rect ifier circuit converts ac input voltage into a fix ed dc voltage. The input volt age may be single-phase or three phase . Diode rectifiers find wide use in el ectric t rac tion , batt ery charging, electroplating, electrochemical pr ocessing) power supplies, 'vveldin g an d uninterruptible power supply (UPS ) systems.
2. Ac to d c convert ers (Phase-controlled r ectifi ers). These convert constant ac voltage t o vari able dc outp ut voltage . These rectifiers use lin e voltage for their commutation, as such these are also called line-commutated or naturally-commutated ac to dc converters. Phase~controlled converters may be fed from I-phase or 3-phase source. These are used in dc drives, metallurgical and chemical industries, excitation systems for synchronous m achines etc. 3. D C to d-c converter s (DC Ch oppers). A dc chopper converts fixed dc input voltage to a controllable de outpu t voltage. The chopper circuits r equire forced, or load, commut ation to tu rn-off the thyristors . F or lower power cir cuits, th yristors are replaced by power transistors. Classifi at ion of chopper circuits is dependent upon the type of commutation and also on the direction of power t1ow. Choppers find ·wide applications in dc driv es, subway cars , troll ey t rucks, battery-dri ven vehicles etc. 4. DC to a ,c converters (in'Verte!'s). An inver t er converts fixed de voltage to a variable voltage. The outpu' may be a variable voltag_ and variable freq eney. These converters use lin v, load or forced commutation fo turning-off the thyristors. Inverters find wide use in induct ion-m otoT and synchronous-mot or dTives . indu ction heating, UPS, HVD C trans mission etc. At pr esent, conventional thyristors are als o being r eplaced by GTOs in high-power applicati ons and by power transistors in w-pO\vei applic ations. BC
[Axt. 1.7]
Intro uction
7
5. AC t o ac convert ers. These convert fixed ac input voltage into va_iable ac output voltage. These are of two types as under: (a) AC' voltage controllers (AC voltage regulators). These converter circuits convert fL ed ac volt age dir ectly to a variable ac voltage at the same fre quency. AC volt age controller employ two thyristors in antiparallel or a triac. Turn-off of both the devices is obtained by ine commutation . Output voltage is controlled by varying the firing angle delay. AC voltage controllers are widely us ed for lighting control, speed control of fans , pumps et c. (b ) Cyclocon verters. These cir cuits convert input power at one frequency to output power at a differen t frequ ency through one-stage conversion. Line commutation is more common in these converters, though forced and load commutated cycloconverters are also employed. These ar e prim arily used for slow-speed large ac drives like rotary kiln etc . 6. Static switch€ s. The power semiconductor devices can operate as static switches or contact ors . Static swit ches possess many advantages over mechanical and electromechanical circuit breaker s. Depending upon the input supply, the static switches are called ac static switches or dc static switches.
1.7. POWER ELECTRONIC MODULES
'
.
,':
A power electroni c converter m ay require two, four or more semicon ductor devices depending upon the circuit configuration. For example, a single-phase half-bridge inverter requires a power module consisting of two power semiconductor devices; a full-converter (or H -bridge converter) r equires a power module having four semiconductor devices; a three phase full converter n eeds a power module having six semiconductor devices. Thus, a power electronic converter can be ass embled from power modules instead of from individual semiconductor devices. A power module h as better performance characteristics as compared t o conventional devices so far as thei.r switching characteristics, operating speed and losses are concerned. Gate drive circuits for individual devices or power modules are also commer cially available. As a r esult of these developments, now inte ligent m odules h ave come in the mar 'et . Inte lligent m odule, also caned smart-power; is state-of-the-art power electronics and it consists of power module and a periph eral circuit. The peripheral circuit compris es of interfacing of power module with the inputJoutput through pr oper isolation from low-voltage signal and from high-voltage power circuit, a dri.ve circuit, protection and diagnostic circuitry against maloperation like excess current, over volt age etc, microcomputer control and controlled power supply. The user has merely to connect the existing supply and the load terminals to the smart-power. At pr esent, int elligent modules are being u sed extensively in power electr on ics. It is r eported that there are more than twenty manufacturers of int elligent modules.
Po wer semiconductOi d evices form the heart o f modern power electronics, A p o wer electronics engineer must understand the device thoroughly fo r e fficient, reHabl and cost-effective d esign of power c onverters. For this p urpose, ch a pter 2 Is devoted 'to the study of pO'-ll/e r semiconductor d iodes, 'transistors a nd MeT. C hapte r 3 deals with diode c ircuits a nd recti ners . In ch a pter 4, are discussed in detail t he thyristor characteristics and its control strat gles . Thyristor c ommut ation t echniques are describ ed In c hapter 5. O t' er power elec tro nic converte rs mentioned in this c hapte are d escribed in d etail in c hapter 6 o nward. L -_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ __ __ __ _ _ _ _ __ _ _ __ _ _ __ _ _ _ _ __ _ _ _ _ _ _ _
~
8
Power Ele ctronics
[Prob .5]
PROBLEMS: . . 1.1.
.
.
Wha t is power electronics? Discuss briefly the concept of power electronics. (6) What. is a converter? Illustrate yo ur a nswer with examples. (e) Giv e the reasons lE:ading to the wid esp read use of power electronic converters . (d) Enumerate at least ten applications of power electronics. 1. 2 . (a ) Give the advantages and disadvantages of power electronic converters . (6) Describe a power electronic system with its general block diagram. (e) List the following power semiconductor devices along with their circuit symbols and m aximum ratings: Diode, SCR, GTO, SITH, MCT, BJT, SIT, IGBT. 1. 3. (a ) Discuss the various types of power elect ronic converters . (0) Compare a diode with a thyristor (e) List the semiconductor devices which possess the capability of withstanding (i ) bidirectional current and (ii ) unidirectional current. (d ) Give the differences between a triac and a thyristor. 1.4 . (a) Giv e the differences between an ac voltage controller and a cycloconverter. (6) Wha t is the difference between thyristors and controllable switches? Make a list of unc ontrolled ar.d controllable switches. (e) How ma ny semiconductor devices are required for (i ) H-bridge converter and ( ii ) three phase full converter. (d ) vVhat is power electronic module ? De ~c rib e smart power. (a )
Chapt er 2
ower Se ic nductor
Diodes and ansistors
..... -.. -...... .. ... -.............. ..•. .. ... ... ..•.••••• . •••• •. .........•...•. •.• ••... .. .. ...• •.. - ~
In this Chapter • • • • • • • • • •
The p-n Junction Basic Structure of Power Diodes Characteristics of Power Diodes Types o f Power Diodes Powe Transistors Power Mosfets Insulated Gat e Bipolar Transistor Static Induction Transistor (SIT) Mos-controlled Thyristor (MCT) New Semiconducting Materials
.- ..... ....... -.. ...... .. .. .. ... -....... -..... ... ........• .•. . ... ....-... .. ... .... ... .. ~
-. ~
~
.. ...
A low-power diode, called signal diode, is a pn-junction device. A high-power diode, called power diode, is als o a pn-junction device but with constructional fe at ures somewhat different from a si gn al diode. Lifewise, power t r ansistors also differ in construction fro m signal transisto s . The voltag e, current and power ratings of power diodes and transistors ar e much high er than the corresponding ratings fo r signal devices. In addition, power devices operate at lower switchin g speeds whereas signal diodes and transist ors operate at higher switching speeds. Power s emi conductor devices are u s ed ext ensiv ely in power-electronic circuits . Some applications of power diodes in clude their use as freewheeling diodes, for ac to dc conversion, for rec overy of t rapp ed en e gy etc. Power t r ansi stors, used as a switch i ng device 'n power-electron ic cir cuit s, must operate in the saturation region in or der that their on-state voltage drop is low. Thei r applications as switching elements 'n elude dc choppers and inverters. Th e obj ect of t hi s chapte r is t o desc r ib e pow er diodes, pow er t r an sis t ors a nd MO S -con tr oll ed t hy be or (MC T). A th yris to r is mOTp. im port ant compo n ent of pow er sem" condu ctor devices;it is, therefore, discussed in detail in Chapter 4. 2:.1 _ ...
-
06
-n1N~~lION·!.'
~~~ _
~~,'
.. .: ;I '
- r.:~?
r I
.
,.~I:.
....
~~;' ..." ,,,,..,.'1jJ::'~'"'" . ., .,:.;".', t' ~'J '.
,
J
~. •
; ~r
A -n junction for ms t he basic buil ding block of all pow er semicondu ctor devLes. It is , th er efor e, worth while h er e to r eview this junction at an introductory l ev el. A p-n junc bon ' s form ed when p -type semi con ductor is br otlgh t in me tallur gi cal , or physical, comac wi th n -type sem icondu ctor. A p -r egion h as great "f can en tra tion of holes whereas n-region has more electron-concent r lltion . In p-region, free h oles are called m aj or ity
10
Power Electronics
[Art. 2.1]
carriers anci. fr ee electrons min ority carriers. In n-region, fre e electrons'are called maj ority carriers where as free holes are called minority carriers. Doping densities in p and n type semiconductors may be different. As such, p-type material m ay be design at ed p+,p or p-; similarly n-type mater ial as n\n- etc. Rough guidelines for labelling of p as p + , p - etc and n as n- , n + etc are a s under: (a ) If doping (or acceptor) density in p-type semiconductor = doping (or donor ) density in n-type semicondu ctor, then it is called pen junction. For example, if doping density in both p . a bou tl0 16 cm -3 t10 - 3 ·JunctIon . .IS terme d and n 1ayers IS ·0 17 cm, pen'Junc t'IOn. (b) If doping density in p-region is much greater than that in n-region, it is called p+n junction. For example, if doping densities are 10 19 cm- 3 in p layer and 1017 cm- 3 in n layer,. then it is termed p + n junction. (c) If doping density in n-type is less than that given in part (b ), the junction is called 13 p"'n- junction. For example, if doping densities are 10 19 cm- 3 and 10 cm- 3 for p and n types respectively, then p + n- junction is formed . (d) If both p and n-layers are heavily doped, it is called p'" n+ junction an d if very lightly doped, a p-n- junction is formed. For example, if drensity is 10 19 cm- 3 in both p and n layers, p'" n'" junction is formed. In general, p + indicates highly doped p r egion, ' n- lightly doped n region and so on. 2.1. 1. Depletion Layer 'When physical cont act between p and n regions is made, fr ee electrons in 11, material diffuse across th e junction into p m aterial, Fig. 2.1 (a ). Diffusion of each electron from n t o p , leaves a positive ch arge behind in the n-region near the junction . Similarly, diffusion of each hole from p t o n , leaves a negative charge behind in the p r egion near the jun ction. As a result of this diffusion , n region near the junction becom es positively charged ~dp region in the vicinity of junc tion becomes negatively ch arged, Fig. 2. 1 (b). These charges establish an electric field across the junction . When this field. ,grows str on g enough, it stops furth er diffu sion. Some electr ons , as these diffuse from n t o jI ie~ine '~±h h oles in p-region and disappear. Similar recombin ation occurs in n-region . . Immob ile ions
Holes ( diffuse
Electrons diffuse
,
I
p~
n
Ii I\!I
p
t
~
Junct:o n
Widt h of deplet ion layer
(a l
( b)
Oeple: ion- lcy er width increases
Deplet ion- layer wIdth d?':: "eCses
--t
p
-+
- + - +
n
r--
P
t "I - - ++ - - ++ - - - ++ ++
( l /
n
11/
I:
n
f--
+
d)
F ig. 2. 1. A p en junction showing (0) direction of holes and electrons diffusion den} tion region (c ) effect offorward biasin g an d (d). effect orr'everse bi asing.
. (b )
.
[. r t. 2.2J
Power Semiconduc to r Diodes and Transistors
. 11
When electric field s tops fur ther diffusion, charge carriers (h o es and electr ons) don't move . As a consequence , opposite charges on each side of the junction produce immobile ions , Fig. 2.1 (b). The region extending in to both p and n semiconductor layers called depletion region or space-charge region . The wid h of deple tion r egion, or depletion layer , is of th e orde r of 5 x 10- 4 mm . In equilibrium, there is a potential differenc of 0.7 V acros the depletion region in silicon and 0.3 V across the depletion r.egion in germa.!lium.. This potential differen ce across the depletion layer is called barrier potential.
is
When positive terminal of a battery is connected to p-type material and negative terminal to n-type material, Fig. 2.1 (c), th~ p-n junction is forward biased. Positive terminal of the battery sucks electrons from p material leaving holes there. These holes travel through p material towards the n egative charge at p-n junction al'.d thus neutralize partly this negative charge. Similarly, negative t erminal of the battery injects electrons into n layer. These electrons move through n material , reach the p -n junction thereby neutralizing partly the positive charge. As a res' It, width of depletion region gets r educed . . In case p material is connected to negative terminal of the battery and n material to positive t erminal of battery, then it can be deduced th at width of depl etion layer in.creases, Fig. 2. 1 (d) . A ris·e in jun c60n temperature also decre ases width of depletion layer. As the barri er potential depends on width of the depletion layer, the barrier potential decreases with rise in junction temperature. For p('lwer semiconductor devices, it should be kept in mind that (i) a junction with light ly doped layer on its on e side requires large breakdown voltage and (ii) a juncti0:l with highly doped layers on its both sides r equires low breakdown voltage. 2.2. BASIC S,TRUC'tURE: OF_POWER DIODES .
Power diodes differ in st ructure from signal diod es. A signal diode. constitutes a simple p-n .. junction as shown in- Fig. 2.1. The in tricacles in constructing powe diodes arise 'from the need to make them suit able for high-voltage and h igh-current applicat ions. The practical realizat' on and th resulting structur e of a power diode is shown in Fig. 2.2 (a). It consists of heav'ly dop8d n" substrate". On this substrate, a lightly doped n- layer is epi texially grown . Now a heavily doped p "'layer ... y Anode is diffused into n- layer to form th e anode of powe r diode , Fig. '2.2 (a ). This shows that n layer i the basic structural feature not found in signal di odes. Th e fu ch on of n - ·lay er is to :Dri1 ! absorb the depl etion layer of th e rev ers e biased I region ..l p"'n- j u nc tion J 1 . The br eak-d own volt age n '" Sub strut £' needed in a power d' ode governs the thi kness of n- layer; greater t he br eakdown voltaD'e, more the n- layer thickness. The drawback of -Co hod !? K. cothodi' n-layer is to add signi lcant ohmic resistance to (a) ( b) th e dio d e wh en it is cond u cti g a forward current. This leads to large p ower dissipation in Fig. 2.2. (a) Structura fea tures of power diode the diode ; so proper cooling arrangements in and (b) i 3 cir uit symbol. large diode ra tings are es ential.
12
Power Electroni s
[Art. 2.3]
The circuit symbol of a power diode, shown in Fig. 2.2 (b), is t he same as that for a signal diode. The modifications in the context of diode, presented above, makes them appropriate for high-power applications. As diode, or p-n junction, is the basic building block of all other power semiconductor devices; same basic modifications should be implemented in all low-power semiconductor devices in order to raise their power-handling capabilities. 2.3. CHARACTERISTICS OF POWER DIODES
\
As stated before, power diode is a two-terminal, p-n semiconductor device. The two terminals of diode are called anode and cathode, Fig. 2.2 (b) and Fig. 2.3 (a). Two important characteristics of power diodes are now described.
2.3.1. Diode i-v Characteristics· When anode is positive with respect to cathode, diode is said t o be forward biased. With increase of the source voltage Vs from zero value, initially diode current is zero. From Vs = 0 to cut-in voltage, the forward-diode current is very small. Cut-in voltage is also known as threshold voltage or tum-on voltage. Beyond cut-in voltage, the diod e current rises rapidly and the diode is said to conduct. For silicon diode, the cut-in voltage is around 0.7 V. When diode conducts, there is a forward voltage drop of the order ofO .S to 1 V. :.. For low-power diodes, current in the forward direction increases first exponentially with voltage and then becomes almost linear as shown in Fig. 2.3 (b). For power diodes, the forward current grows almost linearly with voltage, Fig. 2.3 (c). The high magnitude of current in a power diode leads to ohmic drops that hide the exponential part of i-v curve. The n- region, or drift region, forms a considerable drop in the ohmic resistance of power diodes. Forward voltage .drop
l)D
+- J
r
'I""" Cathode
Anode
i.
o
Vs
:'1 ,
(a)
F ig. 2.3.
(b)
(a)
(c)
(d)
A forwar d-biased power diode. i-v characteristics of (b) signal diode (c) power diode and (d) ideal diode.
When cathode is positive with respect to anode, the diode is said to be re verse biased. In the reverse biase d condition , a small reverse curren t called leakage current, of t he ord er of microam pel"es or m ill iamperes (for large diodes) flows. The leak age cur-rent is almost independent of the magnitude of reverse voltage l..m til this voltage reaches breakdown voltage. At th is r everse breakdown, voltage remains almost constant but rever e current becomes quite high-limited only by the ext ernal c'rcui t r esist ance, A l arge r everse br eak down voltage , as ociated with h igh r evers e current, leads to excessive power loss that may destroy the diode. This shows that reverse b eakdown of a power diode must be avoided by _ope a ting it below the specific peak reverse r epetitive voltage VRRM . Fig. 2.3 (c ) illustrates th.e i-u charac teristi cs of ;, Som e
au thors WT1te u-; characteristics
Power Semlcondueto Diodes and
r
[Art. 2.3]
ransistors
13
power diode and V RRm . For an ideal diode, the i-v characteristics are shown in Fig. 2.3 (d). Here, volt age drop across conducting diode, VD = 0, revers e leakage current = 0, cut-in voltage = 0 and r everse breakdown voltage V RRM is infinite. Diode manufacturers also indicate the value of neak inverse voltage (PlV) of a diode . This is the largest reverse voltage to which a diode m ay be subjected dur ing its working. PlV is the same as V RRM' The power diodes are now available wi th forwar d curren t ratings of 1A t o several thousand amperes and with reverse voltage ratings of 50 V to 5000 V or mo e.
2.3.2. D iode R everse Recovery Chara cteristics After the forward diode current decays to zero, the diode continues to conduct in the reverS E; direction because of the pres s re of stored charges in the dep letion region and the semiconductor layers. The reverse current flows for a time ca lled reverse recovery time t,.,.. The diode regains its blocking capability until reverse recovery current decays to zero. The reverse recovery time trr is defined as the time between the inst ant forward diode current becomes zero and the instant reverse recovery current decays to 25% of its reverse peak value I RM as shown in Fig. 2.4 (a). . The reverse recovery time is composed of two segments of time ta and t b, i.e. t r, = ta + tb' Time ta is the time between zero crossing of forward current and peak reverse current I RN/· During the time t a , charge stored in depletion layer is removed. Time tb is measured from the instant of reverse peak value I RM to the instant when 0.25ljV.;f is reached, ' .. Fig. 2.4 (a). During t b , charge from ~trr --i the semiconductor layers is removed. (a ) : ' t a - tb -...j The shade d area in Fig. 2 .4 (a) represents the stored ch arg e, or o~--~~__~--~'~~:~··~~~----------t reverse recovery charge, QR which
must be rem oved during the reverse
recovery time t rr . The r at' 0 tb1ta is (b)
v, called the softness factor or S-factor. o ~~~==~~~~-L----------: t T h is factor i s a measure of t h e , voltage transients that occur dt1 ring Power I the time diode r ecover . Ita usu al loss in : I i ~ ! value is unity an d this indicates low (c) dio de ssssssss
Fig, 2.4. Reverse recovery characteristics
lar ge os cill at ory ov er voltages. A (a ) varia tio n of forward current if diode with S -factor equal to one i (b) forward v oltage drop vf and (c) power loss in a dio de. called soft-reco very diode and a diode with S-fact or less than one i called snappy -re co very diode or fast· re covery diode. In Fig. 2.4 (6) is shown th e wav form of forward-v Itage drop vI' across the diode. The pro uct 0 vr and i f gives the powe loss in a clio e. Its vari.ation is shown in Fig. 2.4 (c ). The av erage value of vI' i/ gives th e total power 10:>':: in a diode. Fig. 2.4 (c) reve' Is that major power loss in a diode occurs during th period tb · It is no ticed fro Fi g. 2.4 (a) that pe ak inverse current IRM can be exp esse as I
:
'
:
I
: : :
.1 I
I ' I
I
I
:
'
1
I
I
I
.. (2 .1 )
l-l
Power Electro ks
[Art. 2.4]
where ddi is the rate of change of reverse current. The reverse recovery characteristics of Fig. t . . 2.4 (a) can be taken to be triangular. Under this assumption, storage charge QR. from Fig. 2 .4 (a ), is given by 1
· ."u·t rr: QR =-I :2 R
... (2.2)
or If tiT == t a , then fr om Eq. (2.1),
di
IRM
... (2.3)
= t rr · dt
From Eqs. (2 .2) and (2 .3), we get di::;: 2 Q R rr dt trr
t .
'. . [2 t
or From Eq. (2.1), with ta == trr>
'-
Q
R
]1/2
rr - (di / dt ) we get di
:[ . 2 QR
IR.Id = trr . dt = (dUdt)
.
I RM
. -_[2Q R .(diJ~1I2 dt
]112
di dt
...(2.5)
It is seen fr'om E qs. (2.4) arid (2.5) that reverse recovery time trr and peak inverse current di
are dependen t on storage charge and rate of change of current dt" The storage charge
depends upon the forward diode current IF. This shows that reverse recovery time and peak inverse current depend on forward field, or diode, current. A power-electronics engineer must know peak reverse current IRM stored charge QR' S-factor, PN etc in order to be able to design the circuitry employing power diodes . These parameters are u sually specified in the catalogue supplied by the diode manufacturers.
2.4. TYPES· OF~ POWER DIODES Diodes are cl assifi ed according to their reverse recovery characteristics. The three types of power diodes are as under: (0 General purpose diodes (ii ) Fast recovery diodes (iii ) Schottky diodes. These ar e n ow described briefly. 2.4.1. General purpose Diodes
Th es e diode have rel atively high r everse r ecovery time, of the order of about 25 ~lS . Their current r atings vary from 1 A t o several t hous and amperes and the r ange of voltage rating is fr om 50 V to ab ou t 5 kV. Applic ation3 0 power diodes of th is type inclu de battery ch arging, electr ic a action, electroplating, welding and unin terTUp tible power supplies (UPS).
Power Semjcon ductor Diodes and T ansistors
(Art. 2.5]
2.4.2. Fast.recovery Diodes T h e diodes with low rever s e recovery time, of about 5 ~s or le ss , are cl assifi ed as fast-recovery diodes. The£e are used in choppers, commutation circuits, switched mode power supplies, in ducti on heating etc. Th eir current ratings vary from about 1 A to several t housand a mperes and voltage r atings from 50 V to about 3 kYo For volt age ra tin gs below about 400 V, the epitaxial process is used for diode fabrication . These diodes have fast recovery time, as low as 50 ns. For voltage ratings above 400 V, diffusion technique is used for the fabrica tion of diodes. In order to shorten the reverse-recovery time, platinum or gold doping is carried out. But this doping may increase the forward voltage drop in a diode . 2.4.3. Sc hottky Diodes This class of diodes u se metal-to-semiconductor junction for rectification purposes instead of p-n junction. The metal is usually aluminium and semiconductor is silicon. Therefore, a Schottky diode has aluminium-silicon junction. The silicon is n-type. When Schottky diode is forward biased, free electrons in n material move towards the Al-n j unct ion and th en travel thr'ough the metal (aluminium) to constitute the flow of for war d current. Since metal does not have ~ ny holes, this forward current is due to the movement of electrons only. As the metal has no holes, there is no storage charge and no-reverse recovery time . It can , ther efore, be said that rectified current flow in a Schottky diode is by the movement of m aj ority car riers (electrons) only and the turn-off delay caused by recombination is avoided . As such , Schottky diode can switch off much faster thanp-n junction diode. As comp ared t o p-n junction diode, a Sch ottky diode has (i) lower cut-in volt age, (i i ) .~l gher everse leakage current and (iii) higher oper ating fr equency. Their revers e voltage r a tings ar? limite d to about 100 V and forward current ratings vary from 1 A to 300 A. Applications of Schottky diode include high-fr equency instrumentation an d switching power supplies .. Th e electrical and thermal ch aracteri.stics of power diodes are similar t o those of thyrist ors wh ich are descr ibed in chapter 4. 2.5. POWER TRANSISTORS
P ower d iodes ar e uncontrolled devi ces. In oth er words, their t u rn-on an d turn -off ch aracteristics ar e n ot under con trol. Power t ransistors, however, possess co n t rolle d charac teristics. These are turned on when a current signal is given to base, or control, terminal. The tran ist or reonains in the on-state so long as cont r ol signal is present . When this control si gnal is r em oved, a power transistor is turned off. Power transist ors are of our types as under : Bip olar junction transis ors (BJ Ts) Metal-oxide-semicondu ctoT fi e d-effect transistors (M OSFET s ) (i ii ) Insulated gate bipolar transistors (I GBTs ) an d (iv ) Static in duction t r ansistors (SITs ). (i )
(i i)
These four types are now described one after th e other .
2.5.1. B ip ol ur Jun ction Transist ors A bipolar transistor is a three-layer, t'i 70 j un ction npn or pnp semiconductor device. With on e p -r egion sand...vichecl by two n -regions, Fib ' 2. 5 (a ), np n transistor is obtained. With two p-regi ons sandwichin g one n-region , Fig. 2.5 (6), p np transist or is obtain ed . The ter m' bipolar
16
Power Electronics
[Art. 2.5]
denotes that the current flow in the device is due to the movement of both h oles and electrons . A BJT h as three terminals named collector (e), emitter (E) and base (B). An emitter is indicated
by an arrowhead indicating the direction of emitter current. No arrow is associated! with base .or collector . Power transistors of npn type are easy to manufacture and are cheaper also. Therefore, use of power npn transistors is very wide in high-voltage and high-current applications. Hereafter, npn transistors would only be considered. Collector
C
C
Collector
Ie P
IB 8ase
8 ase
iB
g
n
()
B
p
IE (
Emitt€.'1
Emitter
E
E
(a) (b) Fig. 2.5. Bipolar junction transistors (a) npn type and (b) pnp type.
2.5.1.1. Steady-state Characteristics. Out of the three possible circuit configurations for a transistor, common-emitter arrangement is more common in switching applications. So, a common emitter npn circuit for obtaining its characteristics is considered as shown in Fig. 2.6 (a).
Inp ut characteristics. A graph between 'base current IB and base~emitter voltage VBE gives input ch ar acteristics. As the base-emitter junction of a transistor is like a diode, IB vers us V BE graph resembles a diode curve. When collector-emitter voltage V CE~ is more th an V CEl' base curr en t, for the same VEE' decreases as shown in Fig. 2.6 (b).
18
Ic
VCE 2>'1CE !
;
VCEI
1
Ic
Ia5 >184> --- >181
las
I
I
Rc ~Ic
7!:,
- 1+
+ B
I/
f lVaE
t ) Fig. 2.6.
(a) non
I
I
VC
Re
1S4
I
,
ISJ
,I
W
Satu ration region
J
VCC I
I
I
I
I82
I
I al
I
IE
I
,
I
/
1
1
1e=0 VSE
Leako ge current
(b )
transistor circuit characteristics, (b) input charact eristi cs and
VCE
Cu t·ol1
r~gion
(c ) (c)
output characteristics.
Ou tp ut haracteristics . r graph between coll ector current Ie and coll ector-emitter vol ts.ge VCE gLe3 output character istics of a tr sis tor. F r zero base current, i.e. for IE = 0, as VeE is increased, a sm all le akage (collector ) curren t exist as shown in Fig. 2.6 (c). As the base current is increased from.lB = 0 t o 1Bl. I B2 etc, coll ector cun snt als rises as shm-'nl in Fig. 2.6 (c) .
(Art. 2.5]
Power Semiconductor Diodes and Transistors
17
Fig. 2.7 (a) shows two ofthe output characteristic curves, 1 for 1B = 0 and 2 for 1B ;f. O. The initial part of curv'e 2 , characterised by low VeE, is called the saturation region. In this region, the transistor acts like a switch. The flat part of curve 2, indicated by increasing VeE and almost constant Ie is the active region. In this region, transistor acts like an amplifier. Almost vertically rising curve is the breakdown region wh,ich must be avoided at all costs.
Ie
Breakdown
Collector
Saturation pOint
vee
A
Rc --
.(
Active
If.
2
n
--
Is
Brea!
Electron flow
p .
Base -.--'"
Load line Saturation
region
n
I
, ( 1,"0
I E~
B :=;) eEO B .--"",
V
. Vee
CE
Emitter
(b)
(a)
Fig. 2 .7. (a) Output characteristics and load line for npn transistoF and (b) electron flow in an npn transistor.
For load resistor R e , Fig. 2.6 (a), the collector current Ie is given by V ee - VeE Re
Ie=-~---=~ . ~
This is the equation of load line. It is shown as line AB in Fig. 2.7 (a). A load line is the locus of al1 possible oper ating points. Ideally, when tTansistor is on, VeE is zero and Ie =V eelRe· This collector current is shown by point A on the vertical axis . When th e transistor is off, or in the ' cut-off region, Vee appears across collector-em itter terminals and there is no collector current. This value is indicated by point B on the horizontal axis. For the resistive load, the line joining points A and B is the load line. Relation bet wee n a and ~. Most of the electrons, pr oportional to IE given out by emitter, reach t h e collector as shown in Fig. 2 .7 (b). In other words, collector current I e, though less than emi tter current IE, is almost equal to IE. A symbol a is used to indicate how close in value these t wo currents are. Her e a, called forward current gain, is defmed as _ Ie a= -[
.
...(2.6)
E
As I e < IE, value of a varies fr om 0.95 to 0.99. In a transistor, bas e current is effect ively th e input current and collector current is the outpu t current. The r atio of collector (output) C Trent Ie to base (input) curre nt IB is kno as the curre ne gain ~.
18
Powe r Electronics
[Art. 2.5J
As IB is much smaller, ~ is much more than unity; its value varies from 50 ~o 300. In an other system of analysis, called h parameters, hFE is used in place oJ~. ... (2.7)
Use of KCL in Fig. 2.6
(a )
gives ... (2.8)
IE=Ie+IB
Remember that emitter current is the largest of the three currents, collector current is almost equal to, but less than, emitter current. Base current has the least value. Dividing both sides of Eq. (2 .8) by Ie, we get
IE IB -=1+ Ie Ie
or and
1=1+1:. u ~ a
~=l-a
...(2.9)
u=--L
...(2.10)
~+1
Transistor S'\yitch. Transistor operation as a switch means that transistor operates either in the saturation region or in the cut-off region and nowhere else on the load line. As an ideal switch, the transistor operates at point A in the saturated state as closed switch with VCE = 0 and at point B in the cut-off state as an open s"vitch with Ic = 0, Fig. 2.7 (a). In practice, the large base current will cause the transistor to work in the saturation region at point A' with sm all saturation voltage VCES ' Here subscript S is used to denote saturated valu e. Voltage V eEs represents on-state voltage drop of the transistor which is of the order of abou t 1 V. When the control, or base, signal is reduced to zero, the transistor is turned off and its operation shifts to B' in the cut-off region, Fig. 2.7 (a) . A sman leakage current ICED flows in th e collector circuit when the transistor is off. For Fig. 2.6
(a ),
KVL for the circuit consisting of VB' RB and emitter gives
VB or Also , from Fig. 2.6 (a), or
VBE = 0 VB - VBE I B =--=- RB Vec = VCE +Ic Rc VCE =Vec - Ic.Rc =Vee - ~ IB Re
RBIB -
~ Rc == Vec - RB (VB -
Also or
VCE = V eB + V EE VCB = VCE - VBE
VBE )
... (2.11)
...(2:12)
.. (2. 13)
If V CES is the collector-emitter saturation vol tage, then collector current l cs is given by
Vcc - V eEs Rc
1cs == - ---::::,.---
... (2.14) .
[Art. 2.5]
Power Semiconductor Diodes and Transist ors
19
and the corresponding value of minimum base cur r ent, that produces saturation, is Ies IBs=T
... (2. 15-)
Ifbase current is less than I BS , the transistor operates in the active region, i.e. somewhere between the saturation and cut-off points. Ifbase current is more than I Bs , V eEs is almost zero and collector current from Eq. (2 .14) is given by Ies = Vee/Re . This shows that collector cunent at saturation remains substantially constant even if base current is increased. With base current more than I BS ' hard drive of transistor is obtained. With hard saturation, on-state losses of transistor increase. Normally, the practical circuit is designed for hard-drive of transistor and therefore, base current IB is greater than I BS given by Eq. (2.15 ). The ratio of IB and I Bs is defined as the overdriv~factor (ODF) . .~ I .. .(2.16)
=--.!!...
ODF
I BS
ODF may be as high as 4 or 5. The ratio of Ies to IBis called forced current gain .
~r where
Ies
~f = I B < natural current gain ~ or hFE
... (2.17)
The total power loss in the two junctions of a transistor is PT
... (2.18)
= VBE I B + VeE Ie
Under sat urated state, V BES is gre.ater than V CES , t his means BEJ is fo rward biased . Further Eq. (2.13 ) shows that V eB is ne~ative under sat u.rated conditions , therefore, CBJ is also forward biased. In oth er wor ds, under saturated conditio ns, both ju nct ions i.n a power transistor are forwar d biased. Example 2.1. A bipolar transistor show n in Fig. 2.6 (a) h as current gain ~ =40. The load resistance Rc = 10 .0, d c supply voltage Vec = 130 V an d i npu t voltage t o base circu it, VB = 10 V. For V CES = 1.0 Vand V BES = 1.5 V, calculate. (a) the value of RB for operation in the saturated state, (b) the value of RB for an over drive factor 5, (c ) forced- current gain and
(d) power loss in the transistor for both parts (a) a nd (b) .
Solut ion . H ere ~ = 40, Re;" 10 .0, V ce = laO'V, y"'B = 10 V, VeE ~ (a )
= 1.0 V and V BES = l.5 V
From Eq. (2 .14), for operation in the sa turated state, 1 r ~ ~ ,:,
= V cc ~ V CES'= Re
13 0 - L 0 10
= 1 2 qo A
From Eq. (2. 15), base current that produ ces saturaGion , I
BS
= 1cs =12.90 = 0 322 - A ~. 40 . ;)
Value of R B for I Bs = 0.3225 A is given by Eq. (2.11) as,
R = V B - VBES = 10 ~ 1.5 = 2 6 3r:.7 !J B I BS 0 .3...... . 0 1. 9 ')5
(A r t. 2.5]
2
Power Electr onics
(b ) Base current with overdrive, fr om Eq. (2.16), is '.
I B = ODF X I BS = 5 x 0.3225 = 1.6125 A
.. (c)
RB
=
10 - 1.5
1.6125 = 5.27
n
Forced current gain, from Eq. (2.17), is
~f= Ies = IB
12.90 1.6125
= 8, which is less
than the natural current gain
~ = 40.
(d ) Power loss in transistor, from Eq. (2.18), is
PT
= V BES lBE + V CES Ics
For normal base drive, P r = 1.5 x 0.3225 + 1.0 x 12.9 = 13.384 W . With overdrive, P r = 1.5 x 1.6125 + 1.0 x 12.9 = 15.32 W It is seen from above that power loss with hard drive of transistor is more. 2.5.1.2. BJT Switching Performance. When base current is applied, a transistor does not turn on instantly because of the presence of internal lC capacitances. Fig . 2 .9 shows the various switching waveforms of an npn power transistor with resistive load between collector and emitter, Fig. 2.8 .
When input voltage VB to bas.e circuit is made - V 2 at to> junction EB or EBJ is reverse biased, v BE = - V 2 , the transistor is off, ia = Ie = 0 and vCE = Vee , Fig. 2.9. At time t I , input voltage VB is made + VI and iB rises to IB l as sh own in Fig. 2.9. After t 1 , base-emitter voltage VBE begins to rise gradually fro m - V 2 and collector current ic begins
l -
1) B
t ___. .
t B
~
T
T
i" " " BE
-I
Fig. 2.8. npn transistor with
resistive load. to rise from zero (actually a small leakage current I CEO exists as shown in Fig. 2.7 (a)) and collector- emitter voltage VeE starts falling from its initial valu e Vee. After some time delay t d , called delay time, the collector current rises to 0.1 I es , veE falls from Vee to 0.9 Vee and vBe reaches VBES = 0.7 V. This delay time is required to charge the base-emitter capacitance to V BES = 0.7 V. Thus, dp.lay time td is defined as the time during which the collector current rises from zero to 0.1 Ies an d collector-emitter voltage falls from Vee to 0.9 Vec. . After delay time td , collector current rises from 0.1 Ies to 0.9 Ics and veE falls from 0.9 Vee to 0.1 Vee in tim e t r . This time t,. is known as r ise time which depends upo n transistor junction capa citances. Rise time t,. i defined as the time during which collector curr en t ris es from O.lIes t o 0.9 Vee and collector-emitter voltage falls fro m 0.9 Vee to 0.1 Vce. This shows that total turn-on time ton = t d + t r . Value of t on is of the order of 30 to 300 nano seconds. The transistor remains in the on, or saturated, state so long as input voltage stays at Vl' Fig. 2.9 (a).
In case transistor is to be turned off, then input voltage vB and inpu t base urrent iB are r eversed . At time t 2, inpu t voltage v B to bas e circuit is reversed from VI to -- V 2. t th e same tim e, base cu rren t changes from I Bl to -IB2 as shown in Fig. 2.9 (b). Negative base current 13 '2 rem oves excess carri ers from the base . The time ts r e quir ~ d to r emove these exc ess carri r is
Power S emicondu ctor Di.odes and
[Art. 2 ..5]
ransistors
2
\
o
Ivl
!
- 1/2
(a)
[-'12
--
t
I
T =1-
~
tS f (b)
IF ~
o ~------~--~r_IB~------~~~ I . -_I-B2~.~!--------~~ ~;-L_B~I--~t~ I [[57' ,
tl
t2
+3 : ' i
( c)
(d)
(e)
o
'
o I 'Ice
~ ~ tn ---l-~
t
ton
tl
t3
F ig. 2.9. Switching waveforms for npn power transistor of Fig, 2.8.
called storage time and only after t s , base curr en t IB2 begins to decrease towards zero . Transistor comes out of saturation only after ts' Storage time ts is usually defined as th e time during which collector current falls from Ies to 0.9 Ies and collector-emitter voltage VeE rises from V eEs to 0.1 Vee, Fig. 2.9 (d) and (e). Negative input voltage enhances the process of removal of excess carriers from base and hence reduces the storage time and therefore, the turn-off time.
.
After t S ) collector current b egins to fall and collector-emitter volt age st rts building c p. Time tr, called fall time, is defined as th e time during which collector current drops from 0.9 Ies to 0.1 Ies and collector-emitter voltage rises from 0.1 Vee to 0.9 Vee, Fig. 2.9 (d) and (e) , Sum of storage time and fall time gives the transistor turn-off time torr i.e . t ofr ;=' s + tr The vari ous waveforms during transistor swi tching a e shown in Fig. 2 .9. In t h is figu 8; tn = conduction period of tt-ans 'stor, to = off period, T = 11f is the periodic time andfis t he switc ing fr equency. 2 .Ll .1.3. Sale Oper ating Are a. The safe operating ar ea (SOA or SOAR) of a power tran sis tor specifies the sa e operating limits of coll ect or current l r. versus collector-emitte voltage V Cg For _eli able op er ation of the transistor, the collector current and v oltage m ust always lie within this a. ea. Actually, two types of safe operating areas are specifie d by the m anufacturers , F BS OA 3. d RBSOA.
22
Power El ectronics
[Art. 2.5]
The forward-base safe operating area (FBSOA) pertains to the transistor operation when base-emitter jun ction is forward biased to turn-on the transistor. For a power transistor, Fig. 2.10 shows typic-al FBSOA for its dc as well as single-pulse operation. The scale for Ic and VCE are log~thmic. Boundary AB is the maximum limit for dc and continuous current for Vc~ less than abouf S0 V.For VeE for more than 80 V, collector current has to be reduced to boundary BC so as to limit the junction temperature to safe values. For still higher V CE, current should further be reduced so as to avoid secondary breakdown limit. Boundary CD defines this secondary breakdown limit. Boundary DE gives the maximum voltage capability for this particular transistor. For pulsed operation, power transistor can dissipate more peak power so long as average power loss is within safe limits of junction temperature. In Fig. 2.10 ; 5 ms, 500 ~s etc. indicate pulse widths for which transistor is on. It is seen that FBSOA increases as pulse-width is decreased. It should be noted that FBSOA curves, as given by the manufacturers, are for a case temperature of 25°C and for dc and single-pulse operation. In order to take into consideration the actual working temperature and repetitive nature of the pulses, these curves must be modified with the help of thermal impedance of the device . . 120
100 I-:A----~B'
100 80 .
I B =-5A
Ic(A)
60
40 0
1~,0----------JO~0~--------~ E 1000
VCE('1)---
Fig. 2.10. Typical forward biased safe
operating area (FBSOA) for a power
transistor (logarithmic scale)
20 0 0
100
200
300
400
500
600
VCE(V)
Fig. 2.11. Typical reverse-block safe operating area (RBSOA) for a power transistor.
During turn-off, a tr ansistor is subjected to high current and high voltage with base-emitter junction reverse biased. Safe operating area for transistor during turn-offis specified as reverse blocking safe operating area (RBSOA). This RBSOA is a plot of collector current versus collector-emitter voltage as shown in Fig. 2.11. RBSOA specifies the limits of transistor operation at turn-off when the base current is zero or when the base-emitter junction is reverse biased (i.e. with base current negative). With increased reverse bias, area RBSOA decreases in siz e as shown in Fig. 2.11. Example 2.2. For a power trans istor, typical s witch ing waveforms are shown in Fig. 2. 12(a). The various parameters of the transistor circuit are as under: \lce =220 v~ V CES ::::: 2 V, lcs = 80 A , td::::: 0.4 1lS, tr::::: 1 ~s, t,.. = 50 ~s, ts '=.= 3 ~l S, t f ::::: 2 ,us, to = 40 ,us, f = 5 kHz. Collector to emitter le akage curre nt itlaic= 2 mA. De termine a verage p ower loss due to collector current during t on and t n . Find also the peak . instantaneo us po wer loss due to collector current du ring turn· on time.
[Art. 2.5]
Power Semiconductor Diodes and Transistors
23
Solution. During delay t ime , the tim e limits are 0 S t Std. Fig. 2.12 (a ) shows that in this time, ic (t) = ICEO and vCE (t) = VCC. :. Instantaneous power loss during delay time is P d (t) = ic VCE = IcEO VCC = 2 x 10- 3 x 220 = 0.44 W Average power loss during delay time with 0 ~ t S td is given by 1 Pd = T
ftd .
1 =T
ftd0 I
.
0 LcC t )·
VCE (t) dt
CEO · Vcc dt
=f · I CEO . Vcc . td
,,;, 5 x 103 X 2 X 10- 3 x 220 x 0.4 x 10- 6
= 0.88 mW
f = ~ =frequency ofiransistor switching
where
o ~ t S tn
During rise time,
ic
I
= tCS. .
(t)
t
r
and
VCE
=[ Vcc
-
(t)
Vcc - VCES ] tr .t
:. Average power loss during rise time is
p
=lft ICS.t[v _ VCC-VCES . t]dt
r T Ot cc t r
r
r
VCC . Vcc - VCES ] 3
= f . I cs . tr [-2- =5
X
10 3 X 80 x 1 X 10- 6 [2~0 -
22~ -.2J = 14.93~ W
ioBl.~------------~I]IBS - - T ='/1
t.. I
----l:f -----_
t
•
I
I
Ic s t
T
vee
lu-~~~~~~~~~ . 0 (... t on..l--- t n - --
Fig. 2.12.
(a)
-I-
t
Switching wave arms for Examples 2.2
an
2.3 .
24
Power Electronics
[Art. 2.5]
Instantaneous power loss during rise time is
l
PI' (t ) = Ics . t {VCC _ Vce - V CES . t tr tr 2 Ics . t I c s . t == - t - Vcc 2 [Vcc - VCESJ I' t,.
J
.. .(i)
d PI' . '.tm at whi ch mstantaneous . ' . dt (t) == O· gIves bme power 1oss d urmg tr wou Id b e maXlmum.
It is seen from Eq. (i) that
VCC·t r 220xlxlO- 6 tm = 2 [Vcc - VCES] = 2 [220 _ 2] = 0.5.046 Ils Peak instantaneous power loss P rm during rise time is obtained by substituting the value of t == tm in Eq. (i ). P
lcs Y1:c ' tr Ics (Vcc ' tr)2 [Vcc - VcEsl I'm tr 2 [Vcc - VcEsl - ~t; 4 [Vcc V cEs12 ==- .
=
2
Ics . Y1:c == 80 X 220 = 4440 4 W
4 [Vcc - VCES] 4 [220 -.2J .
Total average power loss during turIVon
Pon == Pd + PI' = 0.00088 + 14.933 =14.9339 W
During conduction time, 0 ~ t ~ tn
ic (t) ==lcs and VCE (t) == V CES
Instantaneous poy.,rer loss during tn is
P n (t) == ic . VCE == 1cs . VCES == 80 x 2 = 160 W
Average power loss during conduction period is
1
Pn=T
ft
n •
0
.
Lc,vcE · dt=flcs,VCES·tn
= 5 x 103 X
80 x 2 x 50 x 10- 6 == 40 W.
Example 2.3. Repeat Example 2.2 for obtaining average power loss during turn-off time and off-period, and also peak instantaneous power loss during fall tim e due to collector current.
Sketch the instantaneous power loss for period T as a fu nction of ime.
Solution. During storage time, 0 ic (t)
~ t ~ t s '
=Ics and VCE (t) = V CES
Instantaneous power loss during ts is
Ps (t)
= ic (t) VCE (t)
=Ics' VCES == 80 x 2 == 160 W
Average power loss during ts is
P s = ~ J~' Ics' V CES . d t = f · ICE ' V eEs ' ts == j
)<
10 3 x 80 x '2 x 3 )( 10- 6 == 2:4 VI!
1
[A rt. 2.5]
Power Semi co nduc tor Diod es an d Transistors
25
l
. f a 11 t'lIne, 0 ::; t ::; t , ~c . (t)::::: [I' cs - Ics -t I CEO 'J t D Urlng f
f
Du ring t f , ICEO is negligibly small in comparison with I cs ,
..
iclt)=Ies [l-il vCE(t)=
and
vCCCE· -V s t tf
Average power loss during fall time is
tr
=1.. f T
P
0
Ics [ l -
i J'[VCC - VCES . t] dt tf
tf
Ics Ics] :::::f(Vcc - V CES ) , tf [ 2-3 Ics :::::f' tf ' s
Wcc -
-' VcEsl 1
::::: 5
10 3 X 3 x 10- 6 X 80 x ~ x (220 - 2) ::::: 43.6 W
X
Instantaneous power loss during fall time is
PP) =Ies
dPN)
~::::: 0
[l-ifee ~fVeES
t]
'
gives time tm at which instantaneous power loss dur ing tr would be maximum ,
Here tm = tf l 2. :. Peak instantaneous power dissipation during tf is
P
=1
cs
fm
(1_11(VCC-VCESJ:::::Ics(VcC~VCES) 2)
2
4
,'
= 80 (220 - 2) ::::: 4360 W
4 Total average power loss during turn-off proces is
Pof{ = P s + P =2 .4 + 43.6 : : : 46 ·W
f n
Durin g off-period, 0
::;J ::; to,
ic(t)
=ICEO
and uCE(t)::::: Vce
Instantaneou s power loss during to is
Po(t) -= ic . VeE =I CED ' Vec ::::: 2 x 10- 3 x 220 : : : 0.44 W
Ave rage poweT loss during to is
1 flO
Po = T ::::: Q
0 X
Po(t)dt -= f I CEO ' Vee ' to
10 3 X 2,x 10- 3 x 220)( 4 0 x 10- 6::::: 0.0 88 W
26
P ower E.lectronics
[A r t. 2.6] P (t)
4440.4 W
160W
,----- --\ - - --
0.44 W
t.
td
' - - - - - " O.44W - - ------- ------ - --- --- ____- __--- __ ____- 4 - ____+ ___ t
- --
,L-~---+------
--~
--+----- tn + ts
_ - - - - _.....I •.-..-.tf
.....41 .... to
j
~--~--------T=~ ----------------------~ f
. Fig. 2.12.
(b)
Sketch of instantaneous power loss in a transistor for Examples 2.2 and 2.3 .
Total average power loss in power-transistor due to collector current over a period T· is PT = Pon + Pn + Poff + Po = 14.9339 + 40 + 46 + 0.088 = 10l.022 W. From the data obtained in Examples 2.2 and 2.3 , the power loss variation as a function of time, over a period T, is sketched in Fig. 2.12(b). Example 2.4 . Apower transistor has its switching waveforms as shown in; Fig. 2.13. If the a uerage power loss in the transistor is limited to 300 W, find the switching frequency at which this transistor can be operated. Solution.
Energy loss during turn-on
200 V
~~-.-.~
t Oil
=J
iC'
o
VeE
_
dt
I.
= J:t~~ x 10' t)(Vc~- :coc x 10' t Oil
Ics=100A
!' :
VCC=200V
I
t)dl
'
= J (2 x 10 6t ) (200 - 5 x 10 6t ) dt o
= 0.1067 watt-sec
Fig. 2.13. Switching waveform for Example 2.4
Energy loss during turn-off
=J;"ff (100 ~ 16°0° x 10' t)( 27°5° X 10' t)dt
= 0.1603 watt-sec
Total energy loss in one cycle
= 0.1067 + 0.1603 = 0.267 W-sec
Av erage power loss in transistor
= sWitching frequency x energy loss in one cycle
... Allowable switching frequency,
300
f= 0.267 = 1123.6 Hz 2.6.: PO~1'I~~Ts l
:
-
.
','. '.::' ~:
..-.
, .:
A metal-oxide-s emiconductor fi eld-effect tran ~ ist o ( IOSFET) is a recent device dev elop ed. by com bin ing t h e areas offield-effect con cept an d MOS technolo gy,
[Art. 2.61
Power Semicond ctor Diodes and Transistors
21
A power MOSFET has three terminals called drain (D), source (8) and gate (G) in place of the corresponding three terminals collector, emitter and base for BJT. The circuit symbol of power MOSFET is as shown in Fig. 2.14 (a). Here arrow indicates the direction of electron flow. A BJT is a current controlled device whereas a power MOSFET is a voltage-controlled devic e. As its operation depends up on the flow of majority carriers only, M08FET is a unipolar device. The control signal, or base current in BJT is much larger than the control sign al (or gate current) required in a M08FET. This is because of the fact that' gate circuit impedance in MOSFET is extremely high , of the order of 10 9 ohm. This large impedance permits the MOSFET gate to be driven directly from microelectronic circuits. BJT suffers from second breakdown voltage whereas MOSFET is free. from this problem. Power MOSFETs are now finding increasing applications in low-power high frequency converters. Voo
Load
!11---f\AA.----.. R
Jom;,
-U
Gate
J S
(a)
-
Current
Silicon di oxide
p-subs trate
So urc ? (b)
Fig. 2.14. N-channel enhancement power OSFET (a) circuit symbol and (b ) its basic str ucture.
Power MOSFETs are of two types; n-ch annel enhancement M08FET and p-channel enhancement MOSFET. Out of these two types, n-channel enhancement MOSFET is more common because of higher mobility of electrons. As such, only this type of 'lOSFET is studi ed in what follows. A simplified structure of n-channel planar MOSFET of low power rating is shown in Fig. 2.14 (b). On p-suhstr ate (or body), two h eaVIly doped n+ regions are diffused as shown. An insulating layer of silicon dioxide (Si0 2 ) is grown on the surface. Now this insulating layer is etched in or der to embed metallic sour ce and drain terminals . Note that n'" regions make contact with source and drain t erminals as shown. A layer of m etal is also deposited on Si0 2 layer so as t o form the gate ofMOSFET in between source and drai n terminals, Fig. 2.14 (b). When gate cir cuit is open, junction be t\ve en n'" region b;,!.ow drain and p-substrate i reverse biased by input voltage V DD' Therefore, no current flows from drain to source an d lo ad . 'When gate is made positive with r espect to source, an electri c fi eld is established as shown in Fig. 2.14 (b). Eventually, induced negative charges in the p-SLlbstrate below Si0 2 layer are formed thu causing the p layer below gate to become an induced n layer. These neg ative charges, called e ectrons, form n-ch ann el between t vo n '" egions and current can floY'; from drain to sour ce sh own by t he arrow. If V G L made lore pos 'tiv e, induced n-channel becom es m ore deep an therefore m ore current fl ows from D to 8. This shows that drain curr ent ID is enhan ced by the gradu al increase of gate voltage , h ence th e nam e enhancemen t NIOS ET.
as
28
P ower Electronics
[Ar t. 2.6]
The m ain disadv antage of n-channel planar MOSFET of Fig. 2.14 (b) is that conducting n-channel in between drain and source gives large on-state resistance. This leads to high power dissipation in n-channel. This shows that planar MOSFET construction of Fig. 2.14 (b) is feasible only for low-power MOSFETs. The constructional details of high power MOSFET are illustrated in Fig. 2.15. In this figure is shown a planar diffused metal-oxide-semiconductor (DMOS) structure for n-channel which is quite common for power MOSFETs. On n+ substrate, high resistivity n-layer is epitaxially" grown. The thickness of n- layer determines the voltage blocking capability of the device. On the other side of n+ substrate, a metal layer is deposited to form the drain terminal. Now p regions are diffused in the epitaxially grown n- layer. Further, n+ regions are diffused in p regions as shown. As before, Si0 2 layer is added, which is then etched so as to fit metallic source and gate terminals. A power MOSFET actually consists of a parallel connection of thousands of basic MOSFET cells on the same single chip of silicon.
Metal
Source
Source
Silicon dioxide ~:--~~§--~~~~~~~§l----f::
n
]
0 rift Region
~---------+~----------~
n+ substrate Curren t path
Fig. 2.15. Basic structure of an-channel DNIOS power MOSFET.
When gate circuit voltage is zero, and V DD is present, n- - p- junctions are reverse biased and no current flo'ws from drain to source. When gate terminal is made positive with respect to sou rce, an electric field is established and electrons form n-channel in the p- 'regions as shown. So a current from drain to source is established as indicated by arrows. With gate voltage increas ed , current I D also increases as expected. Length of n-channel can be controlled and therefore on-resistance can be made low if short length is used for the channel . f\n examination of the basic structure of n-channel DMOS power MOSFET (P MOSFET) reveals th at a parasitic npn bipolar junction transistor exists between the source and dr ain as :,hown in Fig. 2.16. The p body acts as the base, n+ layer as the emitter (or source) and n layer as th e co llec tor (or drain) of this BJT. Since source is connected to both base and emitt er of parasitic BJT) the saUTce short circuit both base and emitter. As a r esult, p ot enti al differ ence between base and emitter of the par asitic BJT is zero and therefor e) BJT is ahvays in the cut-off s Late. "' A mi. . \lie of s ilicon ato ms and pe nta vale nt atoms, deposited on wa fer, forms a layer of n -type se micon ductor on hea ted s urfa ce . This iayer is called epitaxial layer.
[Art. 2.6]
Power Semicon du ctor Diodes an d Transistors
Also, vertical travel from source to drain indicates the existence of a parasitic diode as shown on the right in Fig. 2 .16. The parasitic diode , with source acting as anode and drain as cathode may be used in half-bridge or full-bridge rectifiers. The parasitic diode also shows that reverse voltage blocking capability of PMOSFET is almost zero . This in-built diode is an advantage in inverter circuits. In Fig. 2.15, source is negative and drain is positive. Therefore, electrons flow from
29
Source
So urce
Load np n BJT ~1==='=-----~__.-/
Drain
O---~-----'
source to n+ layer, then through n-channel of Fig. 2.16. PMOSFET showing parasitic BJT and p layer and further through n- and n+ layers
parasitic diode .
to drain. The current must flow opposite to the flow of electrons as indicated in Fig. 2.15. Since the conduction of current is due to the movement of electrons only, PMOSFET is a majority carri er device. Hence, time delays caused by removal or recombination of minority carriers are eliminated during the turn-off process of this device. PMOSFET with a turn-off time of 100 ns are avaiiable. Owingto its low turn-off time, PMOSFET can be operated in a frequency r ange of 1 to 10 MHz. -., 2.6.1. PMOSFET Characteristics The static characteristics of power MOSFET are now described briefly. The basic circuit diagram for n-channel PMOSFET is shown in Fig. 2.17 where voltage and currents are as indicated. The source terminal S is t aken as common termin al, as usual, betw een the input and output of a MOSFET. ~
l.oed
Ro
--=- "D O
s
.t.
8 10 12
"G5
(6 )
(a)
Fi g. 2.17. N -chan n el po\ver l'vIOS FET
6
(a)
circ uit di agram and (6) its typical transfer char· cteristic.
Characteristics. This characteri tic shows the variation of drah'1 current I D as a function of gate- source voltage Ves. Fig. 2.17 (b) shows typical transfer characteristics for n·channel PMOSFET. Threshold voltage VeST is an important parameter of MOSFET. T'v'GST is the minilmun positive voltage between gate and source to induce n-channel. Thus, for threshold voltage below VCST' device is in the off-state . Mf\gnitude of VCST is of the order of 2 to 3 V (6 Ou tp u t Chara cterlstics. PMOSFET outpu t characteri: tics, sh own in Fig. 2.18 (a ), indicate the varia tion of drain current ID as a fu ncti on of ,h ain-sour ce vol tage V DS' wi b crat e-soUf.e voltage Ves as a p aram e ter. For low values of Vns . t h~ gr aph be tyveen [D - 1/D:· 15 a lm ost 'linear ; hi.:) in dica tes a constan t value of on -resi.'5t ance R DS := V DS' /]D For bven \- ,5, if (a. ) Transfer
30
P ower Elec tronics
[ r t . 2.6]
dram current is nearly constant. A load line intersects the output characteristics at A and B. Here A indicates fully- on condition and B fully-off state. PMOSFET operates as a switch either at A or at B just like a BJT. When power MOSFET is driven with large gate-source voltage, MOSFET is turned on, V DSON is small. Here, the MOSFET acting as a closed s\lvitch, is said to be driven into ohmic region (called saturation region in BJT). When device turns on, PMOSFET traverses i D - V DS characteristics from cut-off, to active region and then to the ohmic region, Fig. 2.18 (a) . When PMOSFET turns off, it takes backward journey from ohmic region to cut-off state.
V DS is increased, output characteristic is relatively flat, indicating that
Ohmic ~ Active ~
CJ
I
0'>
~
g
--~'----------~--~ ~ c
V GS
~
o
~~------------4~a
.0
....
~--~~--------~I ~
~ :::J
....o a
VGST
::;
V
.~
VI VGSP
~
c
o
tdn~ tr ~
.A-------.........;~---__1
C
,
5
a
iO
.....:j
II......----~-
w hen VGS
<
I .,i' ~----,------H... , , ,
t)os
Drai -source volta ge _ B . Cut- off
t
I
:
I
ID
I
t
VG ST
F ig. 2. 18. (a) Output characteristics of PMOSFET.
I
~ :
, :
Fig. 2.18. (b) Switching waveforms for
PMOSFET.
. . (c) S wit ching characteristics. The switching characteristics influ·~nce d to a large extent by the- internal capacitance of the
of a power MOSFET are device and the internal impedance of the gate drive circuit. At turn-on, there is an ini.tial delay tdn during which input capacitance charges to gate threshold voltage V GST . Here tdn is called turn-on delay time. There is further delay t r , called rise time, during which gate voltage rises to V GSP, a voltage sufficient to drive the MOSFET into on state. During tr,drain current rises from zero to full-on current I D . Thus , the total turn-on-time is t on = tdn + t r . The turn-on time can be reduced by using low-impedance gate-drive source . As MOSFET is a majority carrier device, t rn-off process is initiated soon after removal of gat e voltage at time t 1 . The turn-off delay time, t df, is the time during which input capacitance discharges fro m overdrive gate voltage VI to V osp. The fall time, tf' is the time during which input capa . tan c discharges from V osp to threshold voltage. During t f l drain current falls from ID to zero. So when V Gs ::; V GST , PMOSFET tum-off is complete. Switching waveforms for a power MOSFET are shown in Fig. 2.18 (6).
2.6.2. P 10 SFET Applica ti ons The on -resistance of MOSFET . cre ase wit h volt age rating; this makes the device very lossy at high-culTent applications. Since the 01 -resi tance has positive t emperature coefficient, pa aIlel operat' on of PMOSFETs i r latively easy. The positive t emperature co.efficient also reduces the second breakdo ;vn effect ' PMOSFETs.
P ower Semi on duct or D iodes at;lcl Transistors
[Ar t. 2.7J
31
PM OSFETs find applica tions in high-fr equency switching applications , varying from a fev{ watts to few kWs . The device is very popular in switched-mode power s upplies an d inverters . These are , at present available with 500 V, 140 A ratings.
2.6.3. Comparison of P MOSFET with BJ T
The three terminals in a PM OSFET are designated as gate, source and drain. In a BJT,
the conesponding three term inals are base, emitter and collector. A PMOSFET has sever al features different from those of BJT. These are outlined below : (i) BJT is a bipolar device whereas P MOSFET is a unipolar device. (ii) A PMOSFET has high input impedance (mega ohm) whereas input impedance of BJT is low (a few k ilo-ohm). (iii ) PMOSFET has low er switching losses but its on-resistance and conduction loss es are more. A BJT has higher switching losses but lower conduction loss. So, at high frequency applications, PMOSFET is the obvious choice. But at lower operating frequencies Cless than about 10 to 20 kHz ), BJT is superior. (iu) P MOSFET is voltage controlled device whereas BJT is current controlled device, (v) PMOSFET has positive temperature coefficient for resistance. Th is makes parallel operation of PMOSFETs easy. If a PMOSFET shares increased Cl1Trent initially, it heats up faster, it resistance rises and this increased resistance causes this current to shift to other devic es in parallel. A BJT has negative temperature coefficient, so current sharing resistor s are necessary during parallel operation of BJTs. (vi) In PMOSFETs , secon dary breakdown does not occu. , because it has positive t em perature coefficien t . As BJT has negative temperature coefficient, sec ond ar y breakdown does occm , I n BJT, with decrease in resistance with ris e in temperature , the curren t'increa~ s. This increased current over the s ame a rea results in hot sp ots and breakdown of the BJT. (v ii) PMOSFETs in high er voltage ratings have more conduction loss, (viii) The state of the art PM OSFETs are available with ratings upon 500 V, 140 A whereas BJTs are avai1 able wi th ratings upto 12 00 V, 800 A.
2'.7. INSULATED GATE BII-OLAR TRANSISTOR (lGST) IGBT has been developed by combining into it the best qualities of both BJT and PMOSFET. Thus an IGBT possesses high input impedance like a PMOSFET 3."'1d has low on-state power loss as in a BJT. Further, IGBT is fre e from second breakdowll problem present in BJT. All these m erits have made IGBT very popular amongst power-electronics engineers. IGBT is also known as me tal oxide insulated gate transistor (MOSI GT), conductively-modulated field effect transistor (COMFET) or gain-modulated FET (GEMFET). I " was also initially called insulated gate transistor (IGT ),
2.7.1. Basi Struct ure Tg, 2.19 illustrates the basic structure of an IGBT. It is constructed vir tu ally in the sam e manner as a pO 'vver MOSF ET. There is, however, a m ajor differ ence in the substr ate, The n+ 1ayer su bstrate at t he drain in a P MOSFET is now sub tituted in th e I GBT by a p+ layer substr ate called collector C. Like a pow er MOSFET, an IGBT has also thou sands of bas 'c SITucture cells con ected approp. 'a ely on a single chip of siheon. In IGBT, p + substrate is called injection layer because it injects holes into n- 1 yer. The n,- layer is called drift r egion. As in other semiconductor devices, thickness ,)f n- bYE:T clet 'rmine_ the voltage blocking capability ofIGBT. The p layer is called body ofIGBT The T' - lay~r in bet-.v·een p'" and p regions serves to accommodat e the dE'pletion layer of pn - jlU'lction i.e. j1.llction J 2 ,
32
Power Electronics
[Art.. 2.7]
E
Emitter
Load p
p ---------~j2
J21-------,.,.;--/
Drift layer
n
jlr----------4-+---------~jl
Injection laye r
p+
p + substrate
Metal
Current path C
Collector
Fig. 2.19. Basic structure of an insulated gate bipolar transistor (lGBT)
2.7.2. Equivalent Circ~i~ An examination of Fig. 2.19 reveals that if we move vertically up from collector to emitter, we come acrossp+, n;"',p layers. Thus, IGBT calf'b-e thought of as the combination of MOSFET and p+n-p transistor Q1 as shown in Fig. 2.20 (b). Here Rd is resistance offered by n- drift region, Fig, 2,20 (b) gives an approximate equivalent circuit of an 1GBT, E
G C Dr ift region resistance, RD ~
P"
P
~
G p+
p+ substrate
C
E
( b)
(a)
c
Mai n current th
,..--p_O_o----:!1---'----- -
-p~-- l
c
~Pa r osi\ic
: thyr ist r
, I-:-_.....
\lh l j-(p- _Od y
U
I resI5 .QI)Ce )
E i
,L _____________________ I ~
(e )
(d)
Fig. 2.20. IGBT (a) basic s tructure sho.. .ving par asiti c tr ansistors and tbyristor (b) a pproxima te equivaler t circuit (c) exact equivalent circui t a .d (d ) ci cui t symbol.
Power S emiconductor Diodes and Tra nsistors
[. rt. l:-7]
3
Fig. 2.20 (a) also shows the existence of another path from collector to emitter; thi.s path is collector, p+, n-,p (n-channel), n+ a..'1d emitter. There is, mus, another inherent transistor Q2 as n-pn+ in the structure ofIGBT as shown in Fig. 2.20(a ). The interconnection between two transi tors QI and Q2 is shown in Fig. 2.20 (c ). This figure gives the complet e equivalent circuit of all IGBT. H ere Rby is the resistance offered by p region to the flow of hole current I h · The two transistor equivalent circuit shown in Fig. 2.20 (c) illustrates that an IGBT st ructure has a parasitic thyristor in it. Parasitic thyristor is also shown dotted in Fig. 2.20 (a). Fig. 2.20 (d ) gives the circuit symbol of an IGBT. 2.7.3. Working When collector is made positive with respect to emitter, IGBT gets forward biased. With no voltage between gate and emitter, two junctions between n- region andp r egion (i.e. junction J 2 ) aTe reverse bias ed; so no current flows from colledor to emitter, Fig. 2.19. When gate is made positive with respect to emitter by voltage V0, with gate-emitter voltage more than the threshold voltage V OET of IGBT, an n-ch annel or inversion layer, is formed in the upper part of p region just beneath the gate, as in PMOSFET, Fig. 2.19. This n-channel short-circuits the n- region with n+ emitter regions . Electrons from the n+ emitter begin to flow to n- drift region through n-channel. As IGBT is forward biased with collector positive and emitter negative, p+' collector region injects holes i1).~o n - drift region. In short, n - drift region is flooded with electrons from p-body region and holes from p+ collector region . With this, the injection carrier density in n- drift region increases considerably and as a result, conductivity of n - region enhances significantly. Ther efore, IG BT gets t urned on and begins to conduct fo rward current Ie. Current Ie, or IE , consists of two cur rent compon ents: (i) h ole current Ih due to inj ected h oles flowin g fr om collect or, p+n-p transistor Q I' p-body region resistance R by and emitt er and (ii) electr onic current Ie due to injected electrons flowin g fr om collector, injection layer p +, drift region n-, n-ch annel r esistance R ch • n+ and emitter. This me n s that collector, or load, current Ie = emitter current IE =lh + Ie. Maj or compon ent of collector CUlTent is electronic current Ie' i. e. main cw'rent path fOT collector, or load, current is throu gh p+, n-, drift resistan ce Rd an d n-channel resistance R eh as shown in Fig. 2.20 (c ). Th erefore, t h e voltage drop in IGBT in its on-state is V eE .on = Ie· Rch + Ie · R d + Vj I =Volt age drop [in n- channel + across drift in n- r egion + across forward biased p +n - junction J 11 Here Vj l is usually 0.7 to 1 V as in a p-n diode. Th e voltage drop Ie . R eI! is du e to n- channel resistance, almos t the same as in a PMOSFET. The voltage dr op Vd{= Ie ,Rd in I GBT is much less than that in PMOSFET. It1.s du e t o substantial increase in t...~e conductivity caused by inj ec tion of electron ~ and h oles in n- drift r egion. Th e conductivi ty 'ncrease is th e main reason for low on-stat e voltage drop in I GBT than it is in PMOSFET 2.7.4. Latch-up in IGBT I t is seen from Fig. 2.20 (a) an d (c ) that I GBT structure has two inheren t transistors Q1 and Q2' which constitute a parasitic thyrist or. When IGBT i.~ on, th e hole-current flows through transistor p+n-p andp-body esiscance R by ' Ifload curren tic is large, h ole component of ~ urrcnt I" would also b large . This larg current WOL1ld increa e the voltage drop Ih . R by which IT ay
3'-1
[Art. 2.7J
forw ard bias the base p -emitter n + junction of transistor Q2' As a consequence, parasitic transistor Q 2 gets turned on which further facilitates in the turn-on of parasitic transistor p~n-p labelled
The parasitic thyristor, consisting of Q 1 and Q2' eventually latches on through regenerative action , wh en sum of their current gains C'J. 1 + C'J.2 reaches unity as in a conventional thyristor (discussed in chapter 4). With parasitic thyristor on, IGBT latches up and after this, collector emitter current is no longer under the control of gate terminal. The only way now to turn-off the latched up IGBT is by forced commutation of current as is done in a conventional thyristor. If this latch up is not aborted quickly, excessive power dissipation may destroy the IGBT. The latch up discussed here occurs when the collector current ICE exceeds a certain critical value. The device manufactures always specify the maximum permissible value of load current ICE that IGBT can handle without latch up. At present, several modifications in the fabrication techniques are listed in the literature which are used to avoid latch-up in IGBTs. As such, latch-up free IGBTs are available. Ql'
2.7.5. IGBT Characteristics The circuit of Fig. 2.21 (a) shows the various parameters pertaining to IGBT characteristics. Static I-V or output characteristics of an IGBT (n-channel type ) show the plot of collector current Ie versus conector~emitter voltage V CE for various values of gate-emitter voltages V CE1 , V CE2 etc. These characteristics are shown in Fig. 2.21 (b). In the forward direction, the shape of the output characteristics is similar to that ofBJT. But here the controlling parameter is gate-emitter voltage VCE because IGBT is a voltage-controlled device. When the device is off, junction J 2 blocks forward voltage and in case reverse voltage appears across collector and emitter, junction J 1 blocks it. In Fig. 2.21 (b), V RM is the m aximum reverse breakdown voltage. The transfer characteristic of an IGBT is a plot of collector current Ie versus gate-emitter voltage VCE as shown in Fig . 2.21 (c) . This characteristic is identical to that of power MOSFET. When VCE is less than the threshold voltage V OET ' IGBT is in the off-state. Ie ( A)
Ie
VCE I. > VCEJ et c VCEI.
R
c
+
E
~
VCEJ
Vec
V~E 1
(a )
(a)
oX
D
en
~.9
.~~ Cli
Vcu
(/
(
0
u
'1 CE I
.!!1 '0 u
V
VRM
F ig. 2.21. 1GBT
( A)
~ 0 -0
0
(
1+-=-
Ie
c
0
VCE
VC E
(c)
( b)
circuit diagram (b) static I-V ch aracteristics and
VCE r
(c)
transfer characteris tics.
2.7.6. Swit ching Ch a r acteristics . S witch ing char a cteristics of an I GBT during turn-on an d turn-off are sketched in Fig. 2.22. The turn-on tim e is defined as the time bet\oveen th e i 3 ~an s of forwa rd blocking to fo nv ard 011- 5 ate (7 ). Turn-o time is compos ed of delay time tdT'. and ris e t ime tr> i.e. tOT'. = tdn + t ,.. The delay tim e is d efined as the t ime for the collector- emitter voltage t o fall fr om VCE to 0.9 V CE ' Here V CE is the initial collector-emitter voltage. Time t d n may also be defined as t he t ime for th e
[Art. 2.7]
Power Semicon du tor Djodes and Transistors
I
~I tdl
veE. ic
I I I
1 1
I
I
I
t
I
~
-Y' ttl : 1 1 1
I
Ic
I I
I I
1 I
0.91e
1
-
I
35
VCE
'
I
I
1 I
1 1 1
VCE=V CC
I I
I I
1 I
I
I
O.lIc
--
ICE
VCES
t
F ig. 2.22. IGBT turn-on and turn-off characteristics.
collector current t o rise fro m its initial leakage current ICE to 0.1 I c , Here I c is the fin al value of collector current . The rise time tr is t he tim e during wh ich collector -emitter voltagefalls from 0.9 VCE t o 0.1 VCE oIt is a.ls o defined as th e t ime for the collector curren t to rise from 0.1 Ic t o its fin al value Ic. After time to"" the collector cur rent is Ic and the collector;emitter voltage falls to small value called conduction drop = V CES where subscript S denotes saturated value. The turn-offtim e is somewhat complex . It consists oft hr ee intervals: (i ) delay time, t d{ (i i ) initial fall tim e, tr1 an d (iii) fin al fall time, tj2 ; i. e. t off =td{+ tn + tj2 .The delay time is the time during which gate voltage falls from VGE to thresh old voltage V GET, As VGE falls to VGET dur ing t df, the collector current falls fro~ Ie to 0.9 Ie. At the end of tdf, collector-emitter voltage begins to rise. The first fall tim e tfl is defined as the time during which collect or cu rrent falls from 90 t o 20% of its initial value I c , or the time during which coll ector -emitter voltage rises fr om V eEs to 0. 1 V CE' The final fall time tf2 is the tim e during which collect or current falls from 20 t o 10% of c, or the tim e uring which collector-emitt er voltage rises fr om 0.1 VCE t o fin al value VCE. see Fig. 2.22.
2.7.7. Appli cation of IGBTs
-
IGBTs are widely used in medium power applicat ions su ch ,s dc and ac m otor drives, UPS system s , poweT supplies and drives for s olenoids, rel ays and contactors. Though IGBTs are somewhat mor e exp en siv e t han BJTs, yet t h ey aTe be com ing p op lar because of low er gate-drive requirem ents , lower swit ching losses and smal er ;:,nubber circuit r equirem ents. IGBT converters are m ore effi ci ent with less size as well as cost, as compar ed to converters based on BJTs . Recen t ly, I GBT inverter induction-motor drives u sing 15-20 kHz s\vitching frequ en cy are findin g favour whe.e audio-n oise is obj ectionable. In most applic ations, IGBTs
36
P ower Electronics
rArt. 2.8J
will eventually push out BJTs. At present, the state of the art I GBTs of 1200 V, 500 A ratings, 0.25 to 20 flS turn-off time with operating frequency upto 50 KHz are available.
2.7.8. Comparison of IGBT with MOSFET
The relative merits and demerits of IGBT over PMOSFET are enumerated below:
(i) In PMOSFET, the three terminals are called gate, source, drain whereas the corresponding terminals for IGBT are gate, emitter and collector. (ii) Both IGBT and PMOSFET possess high input impedance. (iii) Both are voltage-controlled devices . (iv) With rise in temperature, the increase in on-state resistance in PMOSFET is much pronounced than it is in IGBT. So, on-state voltage drop and losses rise rapidly in PM OSFET than in IGBT, with rise in temperature. (v) With rise in voltage rating, the increment in on-state voltage drop is more dominant in PMOSFET than it is in IGBT. This means IGBTs can be designed for higher voltage ratings than PMOSFETs. In view of the above comparison, (a) PMOSFETs are available upto about 500 V, 140 A ratings whereas state of the art IGBTs have 1200 V, 500 A ratings and eb) operating frequency in PMOSFETs is upto about 1 MHz whereas its value is upto about 50 kHz in IGBTs .
2.8. STATIC INDUCTION TRANSISTOR (SIT) SIT is a high-power high-frequency semicondu ctor device. It is the solid-state version of triode vacuum tube. SIT was commercially introdu ced by Tokin Corporation of Japan in 1987. Basic structure of SIT is shown in Fig. 2.23 (a) and its symbol in Fig. 2.23 (b) .. It is basically n+n-n device with a buried grid-like p+ gate structure. It h as short n-channel structure and p'" gate electrodes are buried in n-n epi-layers as shown. The buried gate structure gives lower gate-source chann el resistance, lower gate source capacit ance and lower thermal resistance. SIT is normally an on device, i.e. if V GS = 0 with V DS present, electrons (majority carriers) would flow from source s to n, pass through gate .0+ electrodes and would then continue their Electron f low ~
--- -----------1 i Source (5)
o
G
s c urren t ftow
( b)
Fig, 2.2 3. (a) B si c structure of SIT
(b )
device symbol
Power Semiconduc tor Diodes and Transistors -
tAr t. 2.91
-
37
journey t hrough n-, n+ and reach drain as shown in Fig. 2. 23. (a ). The drain cu rrent 1D would flow fr om D to S as shown. If Ves is negative, p+n j un ction s get reverse biased. As a result, depletion layer is form ed around p + electrodes and this r educes the current flow from its value when Ves = 0, Fig. 2. 24 (a). At some higher value of everse bias voltage V es , the deplet.ion layer would grow to such an extent as to cut-off th~ channel completely, F ig. 2.24 (b ) and load current iD would, therefore, be zero. ~
__....-----c 5 51
51 VGS-=-
Ip; T~)r®l\i;J;r~'
Lood
Vos
I D\p\etl oiIOyer \
-=
Vos ~------h-------~
'------------<"1 lO
( a)
D (b)
\
F ig. 2.24. (a) Lower reverse bias, load current iD redu ced due to depletion layer (b higher reverse bias, expande d depl etion layer stops current flow.
Although, the device conduction dr op is lower than that of equivalent s eri.es-parallel oper atio of P MOSFETs , the essentially large drop in SIT makes it unsuitable for gener al power -electronic applic ations . F or example, a 1500 V, 180 A SIT ha a channel resistance of 0.5 n giving 90 V conduction drop at 180 A. An equiv al ent thyristor or GTO drop may be . ar oun d 2 V [11J. Though conduction drop in SIT is abn or m ally high, the turn-on and turn-off tim es of the device are very low. For the SIT cited above, typicaf ton and tofr are 'rround 0.35 ).lS . High conduction drop associated with very low turn-on and turn-off times re~ult in low on-off energy losses. Thus, SIT is being used in high-power, high-frequency applications such as MlFM transmitters, induct ion heaters , high-voltage low-current power supplies, ultrasonic generators etc. SITs with 1200 V, 300 A rating with ton and toff around 0.25 to 0. 35 JlS and 100 kHz operating frequency are available. SIT is a m ajority carrier (electrons only) device, therefore SOA is limited by junction temperature. As channel r esistance rises with tem per at ure, parallel operation of SITs is easy. SIT is normally-on d evice, n ormally-off device is under development.
An M CT is a new device in the fi eld of semiconductor-controlled devices. Ie is basically a thyristor wi h two 10 SFETs built in t o the gate structure. One MOS 'ET is us ed for turning on the MGT and th e other for turning off the devic e. An MeT is a h igl -frequency, higb-power, low- conduction drop switching device. An MCT co bines into i t the fea tures of both con ventio. a1 fo ur-laye thyris tor a ving reg en rative action and MOS-gate stru cture . However, in MeT, an cde is the r eference \-vith r espect to which !?.II gate signals are applied. In a conventional SCR, cathod e is t h e re ~erence t erminal for o-a te signals . I
38
P owe r Electronics
[Art. 2.9] Ano de 0
Si02 /:// /////////////1
Gate n
tTrlY
Si02 I 1://,0//Wff//////';;
+ p
~~t--
J,
n
h
-~
Gate
"- {On- FET Chan nel or p-Ch annel)
n (Of1- FET Channel or n-Channel)
pp buffer
J3
n+ substrate
~Metal Cathode
Fig. 2.25. Basic structure of an MeT.
The basi c structure of an MeT cell is shown in Fig. 2.25. A practical MeT consist of thousands of these basic cells connected in parallel, just like a PMOSFET (7, 8). This is done in order to achieve a high-current carrying capacity of the device. The equiva1ent circuit of MeT is shown in Fig. 2.26 (a) . It consists of one on-FET, one off-FET and two transistors. One on-FET, a p-channel MOSFET and the other off-FET, an n channel 1-,'IOSFET, represent MOS-gate structure of MeT. The npnp structure of MeT is represented by two transist ors npn and pnp as shown in Fig. 2.26 (a). An arrow t owards the gate t etminal in dicates n-channel MOSFET and the arrow away from the gate terminal as " the p-chann p-l IOSFET. The two transistors in the equivalent circuit indicate that there is regenerative fe edback in the MeT just as it is in an ordinary thyristor. Fig. 2.26 (b) gives the circuit sym bol of an MeT. An MeT is t urned-on by a negative voltage pulse at the gate with respect to the anode and is turned-off by a positive voltage pulse. Working of MeT can be understood better by referring to t he equivalent circuit of Fig. 2.26 (a) .
Turn-on P oce ss. As stated above, MeT is turned on by applying a negative voltage pulse at the gate with r espect to anode. In other words, for turning-on MeT, gate is made negative with respect t o anode by th e voltage pulse between gate and anode. Obviously, MeT must be initially forward biased and then only a negative voltage be applied . With the application of this negative vol tage pulse, on-FET (p-ch ann el) gets turned on whereas off-FET is already off. With on-FET on , current begins t o flow fro m anode A, through on-PET and then as the base current and emitter current 0 npn transistor and then to cathode K. This turns on npn transistor. As a result, collector current begins to flow i n np n transistor. As off-FET is off, this collector cu rre nt of npn "ransisto r a cts as the bas e curr ent of p np t ransistor. Subsequently, pnp transistor is also t urne d on . On bo th t he tran s 'stors ar e on, regenera ti ve action of the connection sch eme takes place and the thyristor or Me T is turned on . Note that on-FET and p np transistor are in parallel when MeT is in conduction state. During the time MeT is on, base cur ·ent 0 np n transistor fl ows m ainly through p np t ransistor because of i ts better conducting property.
(Ar t. 2.101
Power Semic on duc tor Diodes and T r ansjstor
;---~_ _ _ __
__ A [nOde
I G~r-----~--~
Got,
+ 14 '/
ot
, fOIT1
39
I
pnp
Off - FET
(n-channel)
~ -7 V
On- FET (p- channel)
np n
(b)
(a) K
Fig. 2.26. MeT
(a)
Cathode
equivalent circuit and
(b)
circuit symbo~
Turn·off process. For turning-off the MCT, off-FET (or n-channel MOSFErl') is energized by positive voltage pulse at the gate. With the application of positive voltage pu 1 :e, off-FET is tu rned on an d on-FET is turned off. After off-FET is turneq on, emitter-base terminals of pnp transistor are shor t circuited by off-FET. So now anode current begins to flow through off-FET and therefore base current of pnp transistor begins to .decrease. Further, collector current of pnp transistor that forms the base current of npn transistor also begins to decrease. As a consequence, base currents of both pnp and npn transistors, now devoid of stored char ge in their n and p bases respectively, begin to decay. This regenerative action eventually t urns off the MCT. An MCT h as the fcllovving merits: (i) low fo rward conduction drop, (ii) fast turn-on and turn-off times, (iii) low switching losses and (iu) high gate input impedance, which allows simpler design of drive circ uits. Main disadvantage of MCT is its low r everse voltage blocking capability. MeT was commercially introduced in 1992. At that time, it was predicted that its use as a power -semiconductor device would be so vast that it might challenge the existence of most of the other devices like SCR, BJT, GTO, IGBT etc. This has, however, not :happened because an MeT has (i) limited reverse-biased SOA and (ii) its switching frequency is much inferior to IGBT. At present, MCTs are being promoted for' their use in soft switched converter topologies, where these inferiorities do not inhibit their use. 2.10~ NEW '
.
SEMICONDUCTING MATERIALS .. -. . . . At pr es ent, silicon enjoys - monopoly as a semicor.ductor material fo r the commercial pr oduc ti on of p ower-control dev ices. This is becau se silic on is che aply available an d semic ondu ctor devices of any size can be easily fabricated on a ingle silicon chip. There ar e, however, new types of materials lik e gallium assenic (C;'al\s), silicon carbide and diam ond which possess the desir able properties required for swit ching devices. At pTe en t, st ate-of-the-art techn ology for these materials is primitive ompar ed with silicon, and many more years of research invest ment are required befor e thes0 materials become commerc ' ally viable for t he production of power-controlled devices. Diode, power MOSFET and thyristor made from silicon carbide have be en established in the labora tory a d are expected to be commercially available very soon. Superconductive materiah may also b<:: used in the manufa cture of such devic es. but w .rk in this direc:tion has not yo.t been r eported.
40
Power Electronics
[pro b. 2]
G rm anium is not used li n the fabrication of thyristors because of the following reasons: (i) Germanium
has much lower thermal conductivity; its thermal resistance is, there fore, more . As a c~nsequence, germanium thyristors suffer from more losses, more temperature rise and therefore lower operating life. (ii) Its breakdown voltage is much less than that of silicon. It means that germanium thyristor can be built for small voltage ratings only. (ii i) Germanium is much costlier. than silicon.
PROBLEMS 2.1. (a) Why are semiconductor materials designated as p+, p-, n-, n+? Explain. (b) What is p-n junction? Discuss the formation of depletion layer in p-n junction. (e) What is barrier potential? How are depletion layer and barrier potential effected by temperature? 2.2. (al Explain the effect of forward bias and reverse bias on the depletion layer in a p-n junction. (b) How is the magnitude of breakdown voltage effected if a junction has highly doped (i) layers on its both sides and (ii) layer on its one side only. (e) Describe the structural features of power diodes. How do these differ from signal diodes? . 2.3. (a) What is a diode? Discuss i-v characteristics of power, signal and ideal diodes. (6) Describe reverse recovery characteristics of diodes. Show that reverse recovery time and peak in'!erse current are dependent upon storage charge j:md rate of change of current. ;'-2.4. (a) Describe the various types of power diodes indicating clearly the differences amongst
them. (b ) What is cut-in voltage in a diode? What are other terms used for cut in voltage? (e) Discuss the following tenus for diodes:
Softne ss fa cto r, PlY, reverse recovery ti me, reverse recovery current.
(d) For a power diode, the reverse recovery time is 3.9 ,us and the rate of diode-current decay is 50 A /~ s. For a softness factor of 0.3, calculate the peak inverse current and the storage charge. [Ans. (d) 150 A, 292.5 ~Cl 2.5. (al Discuss th e power loss in a diode during the reverse recovery transients. (b) The forward characteristic of a power diode can be represented by vf= 0.88 + 0.015 if' Determine the average power loss and nus current for a constant current of 50 A for 2/ 3 of a cycle. r 1 2 T/ 3 2 ] Hint. (6) With T as the time of a cycle, average power loss =T 0 vf' lf db '3' ufIf etc
l
f
[Ans. (6) 54.33 W, 40.825 A) 2.6 . (a) Enumerate the types of power transis tor s along with their circuit symbols . (b) Wh at
is a bipola r junction transistor? 'vVhy is it so called?
Describe t he types of BJTs with their circuit symbols.
(c) De me a an d ~ for BJT and develop a relati on between the two . Why is a less than 1 and ~ m OTe t han 1 ? (d ) Why is it preferrable to use hard drive for BJ T ?
2.7 . (c'z ) vVhat is th e difference bet we en ~ an orced ~f for BJTs ? (6) Whflt are th e conditio ns un der which a transistor operates as a switch? Discuss hard -drive and over drive fact or for BJT. (c ) Show L at colle ctor current at sat urati on remains s ubstan tially con3:ant even if base curren t is i n ~r ea sed.
(Prob. 2]
Powe Semiconduc tor Diodes and Transistors
41
2.8. A bipolar tr ansistor, with current gain P= 50, has load eBistance Re = 10 Q , de supply voltage Vee = 120 V and input voltage to base circuit, VB = 10 V. For VeEs = 1.2 V and VBES = 1.6 V, calculate (a ) the value of RB for operation in the saturated state ( b ) the value of RB for an or drive factor 6 (e) forced current gain and (d ) power loss in t he tr ansistor for both parts (a ) and (b) . [Ans. (a) 35.354 n ( b ) 5.892 n (c ) 8.33 (d ) 14.6362 W, 16.537 W]
2.9. (a) Explain the s.witching performance of BJT with relevant waveforms. Indicate clearly turn-on and turn-off times and their components. FBSOA and RBSOA for BJTs.
(b ) Describe
2.10. (a) Describe the input and output charac teristic for a BJT. Show the region of the '. t ransistor characteris tic where it acts like a switch. . ( b) Typical switching waveforms for a power transistor are shown in Fig. 2 .27. Show that switch-on energy loss is given by Vee · l es 6 ton
Also obtain an expression for the average value of switch-on loss. (c) Derive expr essions for the switch-off ener gy loss and also for its average value fo r the wa veforms shown in Fig. 2.27 .. [ Ans (b)
\C~::
~
~
I :, I i : I
I,
!I
iI
t!
Vee
i,
t
..
I
I
~.L-~~;_.______________~I~__~L-~
tton...! . "
T
I-tolfj
F ig. 2.27. Pertaining to Frob. 2.10
t
(b ).
Vee ' I es . Vee ' I es Vee ' I es f . ] 6 f to.•. (c) . ?".. ..:' t off, 6 tofr
2.11. In case les =80 A, Vee = 220 V; ton =1.5 ~s and t off = 4 ~s for the switching wavefor ms shown in Fig. 2.27, find the energy loss during switch-on and swi tch-off intervals. Find als o the aver age power loss in t he power tr ansistor for a switching frequency of 2 kHz. Derive the expression5 s-ed. [Ans 4 .4 mWs, 11 .73 mWs, 32 .267 Wl
2.12. ( a ) For the typical switching waveforms shown in Fig. 2.27 for a power transis tor, find expressions that give peak ins tantaneous power loss during ton and t off intervals respec tively. (b) In case I es = 80 A, Vee = 220 V, ton = 1.5 !-LS and tofr = 4 !lS, find the peak value of instan t ane ous power loss during t 0 1'. and torr intervals re~ pectively. (Ans. (a )
Ies , Vee Ies ' Vee 4 ' 4 (b ) 44 00 "V, 44 00 Wj
2.13. A power transistor is used as a switch and typical waveforms ar e shown in Fig . 2.1 2(a ) Th e par ameters for the tran"Tht or circuit are as under :
Vee = 2 00 V, VeEs =2.5 V, les"" 60 A, t d = 0.5 ).ls, tr = 1 ~s ,
tn = 40 !ls, ts = 4 )l s, tf "" 3 ).lS, to = 30 )lS, f= 10 kHz. Coll ector to emi tt er lea age current"" 1.5 JmA.
De termine ave ra ge pow er loss du e to collector current durin t on and t n . F in also t e p eeL
instant aneous power 1053 du e to collector curren t eluring turn-on tim e.
Sket ch t h e ins a tan eo u power loss duri g t on an d t n' (Ans. 20 .50 15 W, 60 Ii , 3037.97 Wl
42
Power Electronks
[Prob.2] 2.14. Repeat Prob. 2.13 for obtaining average power loss during turn-off time and offperiod, and also peak instantaneous power loss during fall time due to col lector current. Sketch the instantaneous power loss
during turn-off time and off-period.
2.15. Fig. 2.28 shows the switching charac teristics for a power semiconductor t device . Derive the expressions for ener gy loss during turn-on and turn-off periods, and also f.or the average Fig. 2.28. Pertaining to Prob. 2.15 . switching loss. Sk.etch the variation of power loss during turn-on and turn-off periods. For Vs = 220 V, Ia = 10 A, tl = 1 ~s, t2 = 2 ~s, t3 = 1.5 ~s and t4 = 3 ~s , find t.he average value of power-switching loss in the device for a switching frequency of 1 kHz.
. [AnS. ~ Vs . Ia
(tl
+ t2],
~ Vs Ia
(t3
+ t 4),
!
Vs Ia {(ton + torr),
7.5W ]
2.16. (a ) Explain the constructional details and working of low-power MO~FET and power MOS FET and bring out the differences between the two. (b) Discuss the transfer and output characteristics of power MOSFETs . 2.17. (a ) Describe the switching characteristics of power MOSFETs. (b) Compare power MOSFETs with BJTs . 2.18. (a ) Discuss how conduction takes place in PMOSFET of n-channel type. (b) Explai n the formatio~ of parasitic BJT and paras itic diode in a PlY~QSFET. Can parasi tic ~ diode be used in some power electronic applications ? 2.19. (a ) What is an IGBT ? What are its other nam es ? Give it" basic st ructural features. How does it differ in stnicture from PMOSFET ? (b ) Derive the approximate and exact equivalent circu its of an IGBT .from its structural details . Also describe its output and transfer characteristics. 2.20. (a) Describe the working of an IGBT. How does latch-up occur in an IGBT ? (b ) Give a comparison between an IGBT and a PlvIOSFET. 2.21. (a) Explain switching characteristics of an IGBT. (b) Discuss why PMOSFET has no reverse blocking voltage whereas an IGBT has. (c) Why a re IGBTs becoming popular in their applications to controlled converters ? Enum erat e some applications of IGBTs . . 2.22. (a) What is SIT? Give its basic structural details . E plain its working with relevant diagra ms . (b ) Tho ugh SIT is not suitable for general power -electronic applications , yet it is being used in som e specific applications. Explain . 2.23. (n ) Descri be t he basic structure of MOS controlled thyristor (M e T). Give its equiv a lent circu it and e plain t h e turn-on and turn-off processes. (b) Giv e the m er its and demerits of MCTs . In what type: o. _Appli cations are MCTs being promoted at present? (c) Discuss briefly about the new semicond ucting m ateri als . 2.24. Deduce to show t hat the energy loss during turn-on of a power t ransistor is given by (VI/6 ) T joules, wher e V :::: off-sta t e voltage, 1 = on-s tate current and T = turn-on time. Assume t he change of V and I to be li near over T. ' H ence, ca lculate t he tur n-on loss of a power transistor for which the voltage an d current, durin '" the process of t urn-on, chang e linearly from 300 V to zeTO V a nd zero A t o 200 A respectively in 2 j.l.S. [Ans. 10 m W -3 or 10 x 10- 3 J ]
Power Semiconductor Diodes and Transistors
[Prob. 2J
43
2.2 5. Read the following statements carefully and indicate the power semiconductor device' eac
statement represents . (a ) two-terminal three-layer device ( b ) majority carrier devices (c) bipolar devices (d) negative pulse turn-on device (e) on operation in ohmic region (f) normally on device (g) on-state in saturation region (h) two-terminal two-layer device (i) uncontrolled turn-on and turn-off device (j) controlled turn-on and turn-off devices. [Ans. (a) Power diode (b) PMOSFET, SIT (c) Diode, BJT, IGBT, MeT (d) Me T (e) PMOSFET (f) SIT (g) BJT (h) signal diode (i ) Diode (j) BJT, MOSFET, IGBT, SIT, MeT)
Chapt r 3
tifiers
Diode. Circuit s and ........ .. ~
~.~~
...•..•.... ...... .••••.•••.........•..•....•••. •..•... ........... ..•............. ~
~
~
In this Chapter • • • • • • • • • •
Diode Circuits with DC Source Freewheeling Diodes Diode and L Circuit Recovery of Trapped Energy Single-phase Diode Rectifiers Zener Diodes Performance Parameters Comparison of Single-phase Diode Rectifiers Three-phase Rectifi8rs Filters
................ .• ...•................................ ....
~
.. •.. . ...........•............. ........ ~
~
A rectifier is a circuit that converts ac input voltage to de output voltage. Semiconductor diodes are used extensively in power electronic circuit s for the conversion of power from ac t o dc. A rectifier employing diodes is called an uncontrolled rectifier, because its ·average output voltage is a fixed dc voltage.
In t his chap ter, first diode circuits involving different combinations of R, L and C are studied , and then diode rectifiers are describe ror simplicity, the diodes are considered as ideal swi tches. An ideal diode has no forwar a'volt a ge drop and reverse recovery time is negligible.
3.1. DIODE CIRCUITS WITH DC' SOURCE In this section, the effect of switching a de source to a circuit consisting of diode and different circuit parameters is exa mined. Th e conclusions arrived at can then be applied to similar situations encountered l ater in power-electronic circuits. 3 .1.1. Resistive Load
'1
In the circuit of Fig. 3.1 (a ), when switch S is closed, the current rises instantaneously to V / R as shown in Fig. 3.1 (b). Here Vs is the dc source ·v oltage andR is t he loa d esist ance . 'When switch S is opened at t 1, the current at once falls to zero, Fig. 3.1 (b) . Voltage Un across diode is zero during th e tilne diode conducts and is equal t o + Vs after diode stops conducting. 3.1 .2.
Re Load
A circui t with de SOUTce, diode and RC load is sho'wn in Fig. 3.2 (0). Vlhen switch S is closed a t t = 0, KV'L gives
L
45
[Art. 3.1]
Diode Circuits an d Rectifiers
J J="---4-' Iv,IR
(a)
Fig. 3.1. Diode circuit with
R
load
(a)
circuit diagram and
t
!VS
(b)
(b) waveforms.
t
.. ..
'.
Vs
R
is
1 !:------L_--=::::~_ _
Vb
o
+-
t
+
R
Ys
l~-~----JCTl~c (b)
(a)
Fig. 3.2. Diode circuit with RC load (a) circuit diagram and (b) waveforms.
. . . . 1 "[1(S) q(O)] Vs Its Laplace transform is R 1(s) + C + '= .
.
8
S
As the initial voltage acros s C is zero, q(o)
.. :(3.1)
S
= O. With this, Eq. (3.1) becomes
.[ 1]
1(8) R + - =Vs -
Cs
or .
i(t)
Its Laplace inver se is The voltage ac ross capacit or is
CVs
=
(" .
1 . RC s + RC
1(s)
=V
_S .
R
s
)=-RVs ..
S
1 =--1 + RC
e- tlRC
II .
1 uc(t) = C 0 .clt
... (3.2 )
Vs
= RC
=Vs (1- e- U RC )
II e 0
-tiRe
elt ... (3.3a )
46
Power Electronic
[Art. 3.1]
...(3.3b)
where "t = RC is the time constant for RC circuit. From Eq. (S.Sa), initial rate of change of capacitor voltage is given by
(dd~c]
=[V~ . e- tiRe. RIC] t=O
=:;
... (3.4)
t=O
source voltage, Vs RC =---;----:---- (d v/dt)t=o In Fig. 3.2 (b), current through the circuit and voltage variation across C are shown. 3.1 .3. RL Load When switch S is closed at t = a in the RL and diode circuit of Fig. S.3 (a), KVL gives
Time constant,
R i +L
~~ = Vs
... (3.5)
O~-----------------L--~t-
(a) Flg. 3.S. Diode circuit with RL load
(b) circuit diagram and (b) waveforms.
(a)
With initial current in the inductor as zero, the solution of Eq. (3.5) gives -!! t
V
i(t)
= ~ (1 -
e
... (S.6)
L )
Initial rate of ris~ of current is
_lit]
di
= (Vs -·e L
dt t = 0
The vol tage across L is . For RL circuit, ;
vL(t)
L
t
di
=L -dt = V
...(3.7)
3
.
=0
_li t (1
L
...(3 .8)
¥
='t is the time constant. The waveforms of current through the circuit and
voltage across indu ctance L are sketched in Fig. 3.3 (b ). It mllilt be noted that the behaviour of circuits in Figs. 3.1 to 3.3 is not affected whether a diode is used or not. It is because, in these circuits, the current i does not h ave a t endency to . reverse; i.e . current i remains unid'rectibnal,
[Ar~.
Diode Circuits and Rectifiers
47
3.1J
3.1.4. LC Load A diode circuit with dc source voltage V s' switch S and load LC is shown in Fig. 3.4 (a). When switch S is closed at t = 0, the voltage equation governing its performance is given by
f 'dt = V
1 L di dt + C
Its Laplace transform ~s
L
s
.s
C
As the circuit is initially relaxed, i(O)
[ sc1]
= 0 and Vc
1
V
=_s s S
(0) = 0 or q(O) = C . vc(O)
L [8 /(s) _ i(O)] + l.l/(S) + q(O)
=0
/(s) sL + - =Vs
or
S
1
..
Vs
/(s) =
L'
2
S
1 " Vs Let we = ~L C . ThIS gIves /(s) = -L--' . We
1 + LC
we 2 2 S + We
_
Ic
= Vs . "'V T.L
11
.
we 2 2 S + We
. :' ~ -
vcs;: . 1 I
o Vc
0 5
vD
+~-
•
•
t
1-4---to---~
I•
t
t~= 'T{I Wo
vL
. ~
I
•
f
+
+
t
tVL
+
Vs
c
\
T1\)C
V~l
lia
t-vs
0
t
.
(b)
(a )
F i g . 3.4. Diode circuit with L C load (a) circui t diagram and (b) wavefor~s .
Its La plac e inverse is Here
CDO
i(t) = Vs .
~ s'n wot
...(3 .9)
= ~lc is called reson ant fre quency of the circu·t. Capacitor voltage is given by 1
uc(t)
Jt
1 .t
_
/c
= C 0 i (t) . dt = C J0 Vs' "'V L =Vs (1 - cos CI), t )
sin
Wo t .
dt ... (3,10a)
48
Power Electronics
[A rt. .]
Voltage across inductance is given by
.
VL(t) =L
When Wo to = 1t or when to . vc(to ) = 2 Vs and vL(t O) = - Vs Here to
di(t) dt
... (3.10b)
=Vs cos Wo t
=1tlwo,
from Eq . (3.9), i(to)
=1t/wo = conduction t ime of diod'e
=0
and from Eq. (3. 1 0a), ·
= 1t ";LC
From Eq. (3.9), circuit or diode curr ent at to/2
=-1t2 W
o
attains a peak value of
Ip = Vs . ,jCI L as shown in Fig. 3.4(b). Voltage across diode, soon after diode stops conduction at to is given by VD
Waveforms of i(t),
=-
VL - Vc
+ Vs = 0 .- 2 Vs + V s =
- Vs· 1t
Vc,
.
vL and VD are sketched in Fig. 3.4 (b). It is seen that at to/2 = - 2 , . Wo
diode current reaches peak value, Vc = Vs and vL = O. Also at to =1t/wo =1t (";LC), diode current decays to zero and capacitor is charged t o voltage 2Vs ' Soon after to, voltage across L is zero and diode voltag,e VD = - VS'
'---' Example 3.1. For the circuit shown in Fig. 3.5 (aJ, the capacitor is initially charged to a voltage Vo with upper plate positive. Sw itch S is closed at t = O. Derive expressions for the current in the circuit and voltage across capacitor C. What is the peak value of diode current? Find also · the energy di ssip ated in the circuit.
S olution. When switch S is closed, KVL gives R i+
~ f idt= 0
Its Laplac e transform , including the initial v()ltage across capacitor, is
R 1(8) + 1..
C
[!.ill _ CV o] = S
0
8
1(s) [R + ~] = Vo Cs ·s . V
or
Its s olution, as per Art. 3. 1.2, is·
5
V
i(t)
ti RC =~ R e-.
.
ik
t
T
t
0
0
l)c
+ R
t
liR
-Vo
L
(a ) (b) F ig. 3.5, P ertaining to Example 3.1 (a) circuit diagram and (b ) waveforms.
•
Diod e Circuits and Rectifiers
[Art. 3.1]
49
_ Vo
:. Peak diode current
-R
ft.
1 v c(t) = C 0 ldt - Vo
Capacitor voltage,
=
~]t Vo . e-tiRC. dt _ Vo CoR
.
-IIRC
= - V0 e
Current i(t) and voltage vc(t) are sketched in Fig. 3.5 (6).
/ n e r gy dissipated in the circuit =
~ CV~ Joules
Example 3.2. In the diode and LC network shown in Fig. 3.6 (a), the capacitor is initially charged to voltage Vo with upper plate positive. Switch S is closed at t::; 0. Derive expressions for current through and voltage across C. Find the conduction time of diode, peak current through the diode and final steady-state voltage across C in case Vs = 400 V, Vo = 100 V, L = 100 W1.a'ni:1·~C = 30·~ Determine also the . ;. voltage across diode after it stops conduction. Souti on. When switch S is closed, KVL for Fig. 3.6 (-li}.gives ·- ,
l]'d t::; V
L di dt + C
l
S
Its Laplace transform gives L [s I(s) _ i(O)l + 1.. [I(S) + C · V o]::; Vs
C S S s
As 'nitially i(O) = 0, the above equation becomes
I
I(s) sL + L
s
Ie] = Vs -s Vo
I
I
I
' -7I /2w o-:
:
I
r 1
5 +
t
t
UL
+
vc 1C
~
Tlvc
(a) Fig. 3.S. Pertaining to Exa mple 3.2
t
(b) (a)
circuit diagram and (6) waveforms .
50
Power E lectronics
(Art . 3.1 ]
This equ ation in s-domain can be solved as in Art 3.l.4. Its solution is i(t)
= (Vs -
Va) .
~~in
1 (Vs - Va)
vc(t) = C
(Do
t
~ft . (Dot dt + Va ,T sm jl.J
0
(Vs - Va) (1- cos CDot) + Vo vc(t) = Vo vc(t) = Vs and at (Do t= 11:, vc(t) = 2 (Vs - V o) + Vo =
At
(Do
t = 0,
At
(Do
t
= 11: / 2,
Diode conduction time, Peak current through diode,
to
Ip
n
. .
=-(Do =11: ~LC =·~30 x 100 =(Vs = 300
Va)
= 2Vs -
Vo·
x 10- 6 = 54.77 ~s .
~
3 ~ 10°0
= 164.32 A
Steady state voltage across C occurs when (Dot a = n.
Voltage across diode, after it stops conducting, is given by vD
==-
v L - Vc
+ V~ = 0 ~ (2Vs - V o) + Vs
=- Vs + Vo =- 400 + 100
300 V.
k'
Example 3.3. In the circuit sho wn in Fig. 3.7 (aJ, the capacitor has i nitial voltage Va with upper plate positive. The circuit i.s switched ed t = O. Derive expressions for Cllrrent and voltage across capacitor. Find the conduction time for diode' and steady-state capCiCi tor voltage. Sqlution. The voltage equation for the circuit of Fig. 3.7 (a), after s\vitch S is closed at t = 0, is di+ 1 ~'d t=O Ldt C
f
:~~ ~ 71: /
Wo
t
.;
)0
I I
i
Vc
~.-;.. ' _--
s
i
t
I I
U D t-
O F
: I
I
to
=~o--~-_-V~~I-_
_ _
G
(a) Fig. 3.7. Pertaining to Exampl e 3.3
(6) (a)
circuit iagl'am and (6) waveform_ .
t
..
Diode CircuJts an d 'Re ctifiers
[Art.
.1 ]
51
Its Lapl ac e transform, including initial voltage across capacitC?r, is "
1(s) . sL + 1. [1(S) _ CVo] = 0
C
l (s)
s
[SL
s
+~] = Va sC s
Here minus sign is put before Va, because for the direction of positive current flow, polarity of Va is opposite. Solution of above s-domain equation, from Art. 3.l.4, is .
i(t) =
-
Va
~ sin Wo t
. -\I!cft L a sm wot dt -
1 vc(t) = C . Va
Voltage across C is Diode conduction time,
to =~ 000
Steady-state capacitor voltage Voltage across diode,
'-
Va' = - Va cos wot
= 1t ..JLC
=- Va cos 1t =+ Va vn = - Va .
'. Waveforms for i, Vc arid Vn are sketched in Fig. 3.7 (b ). . .. Example 3.4. In the circuit shown in Fig. 3.8 (a), capacitor is initially charged to voltage V o with upper plate pos itive . Sketch waveforms ofi, Vc, VL and in after the switch S is closed.
Solution . When switch S is closed, capacit or C begins to discharge through Land C. For :' obtaining i , Vc expressions, refer to Example 3.3.
i = Va
Therefore,
~ sin wot , vc(t ) = - Va cos ooot -\Irc t . dtd sm. wot = Va cos ooot = - vc'
di vL = L dt = L . Va .
and
Nli
.
~ vocoswo t
I + Vo
:
. 10
:
.
L
~neigy
{ o _..... tLlp J...C V~ .. , 0 2
'2
(a ) (b) Fig. 3.S. P er tainin g to Exampl e 3.4 (a) circui . diagram and (6) wave for ms
52
Power Elec tron ics
[Art. 3.1]
The waveforms for i, vc, uL and iD are shown in Fig. 3.8
i CV6
and i = O. At t =T/ 4, current i reaches peak value Vo
stored in C gets transferred to L as
( b) .
At t
=0, energy stored in Cis
~ = ]p'
Vc = vL
=
°
and energy
~ L];. Soon after T 14 (CDo T 14 =1t/ 2), as VL tends to reverse,
diode D gets forward biased. Current Ip now begins to flow as iD through D and as i through L. If there were no resistance in this closed path, current Ip would continue flowing unabated. In practice, inherent resistance in the closed path would cause this current to decay exponentially.
~ Example 3.5. In the circuit of Fig. 3.9 (a), current in inductor L is the waveforms of i, Vc and vL after switch S is opened at t = O.
Solution. When switch S is opened at
t
=0,
]0
before t = O. Sketch
current 10 begins to flow in the path
D, C and L. KVL for this path is
°
J.
1 ~ dt = di + C L dt
wt
o
10
L
wI
c
s
+
wi
(6)
(a)
F ig . 3.9. Pertaining to Example 3.5. (a) Circuit diagram and (6) waveforms .
. L [sf " (s) - i. (0)] + -1 [ -] (s) - -cVo Its Laplace transform IS . C s s
It is given that initially io
=10 and Vo = O. Therefore, we get
I(s) [ sL + or
I(s) Its La place inverse is i(t) Also
and
1= °
s~ 1 =. LIo
=]0
2
s
s +
2' CDo
1 where CDo = ~LC as before.
=]0 cos CDot
vc
= ~ Jidt = ~ ~ l o cos CDr! . dt = 10 ~ sin CDot
vL
=L
~~ = L
:t
(10 cos CDot)
= -10 ~ sin CDc t
[Art. 3.1J
Diode Circuits and Rectifiers
53
At wt =~, current i tends to reverse, but diode D blocks ~his current reversal. Also , at
wt =
~, capacitor is changed to 10 ~sin'~ =10 ~ = Vc and voltage across inductance is
V L =-Io
1f.
Thus, after wt
=~, capacitor voltage remains constant at 10 ~ whereas voltage across L
becomes zero because current is now zero. Energy stored in inductance as
i L I~ at wt ....
= 0
gets
2
transferred to C at wi
~ n/2 as ~ CV; ~ ~ C ( ~ J. [0
3.1.5. B.Le Load
A diode in series withRLC circuit is shown in Fig. 3.10 (a). KVL for this circuit, when switch S is closed at t = 0, is given by di + C 1 f'd V R l· + L dt l t = s
i
r
5
R
~
Critically damped
+- uR
lis
+
tv
i. C
i
•
Underdamped
'~.~ , .< overd cmped
,,
L
+
J
lvc
,,
.
,"
"
-.
...... .
-- t
(b)
(a)
Fig. 3.10. Diode circuit with RLC load (a ) circuit diagram and (b) waveforms.
With zero initial conditions, the Laplace transform of above equation is
[
1]
I(s) R +sL+ sC
Vs
:=
S
Vs
or
I(s)
=L'
Here s2 + ~ s + L~ = 0 is t he char acteristic
1: R 1 s + L s + LC 2
equ~ion in s-domain. The roots of this equation
are
s =- ~ ± -1(fLJ or wh er e
.s= -s± ~ R
- 2L
1 LC ... (3 .11 ) ... (3 .12)
54
Power Electronics
(A rt. 3.1]
...(3.13)
~ called the damping factor.
js called resonant frequencyjn fad/sec
= ~r-t--c-_
-(~-J-2
w,<
and
~2
=-}w6 Also
Wo
.
...(3.14)
= ringing frequency in rad Isee.
=-) w; + ~2
Depending upon the values of ~ andwo, the solution for the current can have three possible solutions. Case 1. In case ~ < wo, it is seen from Eq. (3.11) that the roots are complex and the circuit
is said to he underdamped. The two roots are Sl
=- ~ + j
wr and
j wr
s2 = - So -
and the current is given by
i(t)
=
V _$-.
wr L
e-~I
sin
W
t
... (3.15)
r
Case 2 . If S > wo, the two roots are real and the circuit is said to be overdamped. The two
roots are and the solution for current is .(t )
l
Case 3. In case ~
The roots are
81
v
S = L ol~ 'I ~ -
.
2 .
Wo
. h 01(,. 2 2) sm 'I <; ~ _Wo . t
-- --
= wo, the roots are equal and the circuit is said to be
... (3.16)
crit'(cally datnped.. .-__
=s2 =- ~ and the solution for the current is .(
l
t)
Vs -;t =r' t . e .,
... (3.17)
Waveforms of current for the three different levels of damping are sketched in Fig. 3.10 (b ).
E xample 3.S. For the circuit of Fig. 3.10 (a), the data is as under: R
= lOn,
L
= 1 mH, C = 5~, Vs =230 V
The circuit is initially relaxed. With switch closed at t = 0, determine (a) current i(t) (b) time of diode (c) rate of change of current at t = O.
c o n~ u c tio n
Sol ution . (a) From Eq. (3 .12),
S= 10 x
Fro m Eq. (3. 13),
1000 = 5000 2x1 1 1 105 Wo = 'LC = [ -3 - 6J1I ,) = ~ O = 14142 .136 rad/s " 1 x 10 x 5 x 10 - '1 ~U
]1/2
10
Fr om E q. (3 .14),
WI'
=[ 1~0
- (5000)2
= 13228. 76 r adls
[Art. 3.2J
Diode Circuits and Rectifiers
Here as
~ <
wo, the circuit is underdamped. The current is, therefore, given by Eq. (3.15 ).
x 1000 . -500Ot. in (13228 76)t 13228.76 x 1 e s .
'(t) = 230
~
= 17.3864· e- 5000t . sin (13228.76 t) (b)
Diode stops conducting when
t1 = 1t
.. Conduction time of diode,
1t
t1
di
.
" J::) e-f,t] -di = - Vs . - [-;./ e ~ .
(c) From Eq. (3.15),
dt
1t
=
I t=
a
= Vs
- 230 x 1000 1
L -
= 230000 A/s ,
_
.
3.2. FREEWHEELING DIODES In Fig. 3.3 (a), steady state current, after switch Sis clos.ed, is equal to V/R. Energy stored
in inductance L
is~. L (V/R)2 . lfthe switch S
is now opened, current V/R would eventually
decay to zero. As the current V/R tends to decay with the opening of switch S, a high reverse voltage appears across switch as well as the diode. High voltage acr oss switch leads to spark across the switch contacts, thus dissipating the stored energy: In the process , the diode, subjected to high reverse voltage, may get damaged. In order to avoid such an occurrence, a diode FD, called freewheeling, or flywheel, diode, is connected across load RL as shown in Fig. 3.11. (a). For understanding how FD comes into play, the working of circuit of Fig. 3.11 (a) is divided into two modes. .
Mode I : When switch S is closed in Fig. 3.11 (a) at t = 0, current flows through V s ' S, D, Rand L as shown in Fig. 3.11 (b). In this circuit, current i is given by
VS i=R
_!it
.. .(3.18)
(1 - e L)
Final value of current,
D
D
~I
+
:R! Vs
FD
Vs L
. i.
(a )
~t
R
R
FD L
L
(b) Fig. 3.11. Circuit of Fig. 3.3 with freewheeling diode
LI
(c)
~ L
+
56
t
P ower Electronics
[Art. 3.3J
Mode n : When switch S is opened at = 0, cu rr ent in the circuit tends to decay
and so a voltage L
Current
~~ is induced! in L which
forward biases freewheeling diode. The current is, therefore, transferred to the L _ _ _ _ _ _ _ _ _ _L __ ______ _______ circuit consisting of FD, Rand L as shown I I I time I---- Mode I -----.-I._0- Mod e IT--I in Fig. 3.11 (c). In this circuit, current is given by Fig. 3.11. (d) Current variation in the circuit of Fig. 3.11. V s --/ R . . ".(3 .19) II = R ' e L ~
~
The current i1 will eventually decay to zero exponentially in mode II of Fig. 3.11 (c). The current build up during mode I and current decay during mode II are shown in Fig. 3.11 (d).
3.3. DIODE AND L CIRCUIT Consider the circuit of Fig. 3.12 (a) where dc source feeds L through diode D. A freewheeling diode FD is connected across L. When switch S is closed at t = 0, KVL gives
V =L di S dt .
or
Vs
... (3.20)
l=yt
This shows that current i rises linearly with time t . In case switch S is opened at t 1 ,Joad VS current L tl begins to flow through FD. As there is no resistance in the circuit formed by
. . V S L and FD, current continues to flow at its constant value of L tl. Energy stored in the inductance is
~ [V:2 L tl . S
.L
= ~. ~ tl joules. Current waveforms
L
are shown in Fig. 3.12 (b).
t"
s
o
·1 FD
t
..
L
itd
t (a) Fi g. 3.12. Diode circuit with FD and L load
(6) (a)
circuit ciiagram and (6) waveforms .
I.
[Art. 3,5]
Diode Cir cuits and Rectifiers
57
3.4. RECOVERY OF TRAPPED ENERGY In the ideal circuit of Fig. 3.12 (a), the energy stored in the inductor is trapped. This trapped energy is not dissipated even when FD conducts because circuit does not contain resistance. The best way of utilization of this trapped energy is to return it to the source . In this manner, net energy taken from the source is reduced and the system efficiency improves. One way of returning this trapped energy back to the source is to add a second winding closely coupled with the inductor winding as shown in Fig. 3.13. A diode D is also placed in series with the second winding. The inductor now behaves like a transformer. The two windings are so arranged that their polarity markings are opposite to each other.
o i,
+
+
(a)
(6)
F i g. 3.13. Energy-recovery circuit (a) switch S closed and (6) switch S opened.
vVhen switch S is closed, current i begins to flow and energy is stored in the inductance of primary winding with Nl turns. The polarity of the secondary winding voltage V 2 is as shown. The diode D is reverse biased by voltage (V2 + Vs). When switch S is opened, polarities of voltages VI and V 2 get reversed, the diode is now forward biased by voltage (V2 - V s )' As a result, diode begins to conduct.a current i 1 into the posi tive terminal of source voltage Vs and so the trapped energy is fed back to the source. Energy fedback to de source = Vs x current i 1 dependent upon (V2 - Vs). The energy stored in L of NI turns is transferred to secondary winding of N2 turns from where it is fed back into th e de source.
3.5. SlNGLE·PHASE DIODE RECTIFIERS Rectification is the process of conversion of alternating input voltage to direct output voltage. As stated before, a rectifier converts ac power to de power. In diode-based rectifiers, the output voltage cannot be controlled. In this section, uncontrolled single-phase rectifiers are studied. The diode is assumed ideal as before. A rectifier m ay b h alf-wave type or full-wave type. A half-vllave rectifier is one in which cur rent in anyone lin e, connected to ac source, is un idirectional. However, a full-wave rectifier h as bidirectional current in any one line connected to ac surface. .
58
[Art. 3.5]
Power Electronics
A rectifier may be one-pulse, two-pulse, three-pulse or n-pulse type . The in any rectifier-configuration is obtained as under: pulse number
m~mber
of pulses
= number of load current
(or voltage) pulses during one cycle of ac source voltage. 3.5.1. Single-Phase Half-wave Rectifier This is the simplest type of uncontrolled rectifier. It is never used in industrial applications because of its poor performance. Its study is, however, useful in understanding the principle of ". rectifier operation. In a single-phase half-wave rectifier, for one cycle of supply voltage, there is one half-cycle of output, or load, voltage. As such, it is also called single-phase one-pulse rectifier. The load on the output side of rectifier may be R, RL or RL with a flywheel diode . These are now discussed briefly. R load: The circuit diagram of a single-phase half-wave rectifier is shown in Fig. 3.14 During the positive half cycle, diode is forward biased, it therefore conducts from wt = 00 to CDt = 11:. During the_positive half cycle, output voltage Vo = source voltage Vs and load current io = VoiR . At w t = 1t, . Vo = 0 and for R load, io is also zero.As soon as Vs tends to become negative after wt = 11:, diode D is reverse biased, it is therefore turned off and goes into blocking state . . Output voltage, as well as output current, are zero from w t =1t to (j) t = 21t. After w t = 21t, diode is again forward biased and conduction begins. (a)
( a ).
,471
, I
: ~.
~ '
-
"
I
27[371 I
VD
is
to
+
+
us :
R
1[Vmsinwt]
I
VD
tVa
I
I I
I
- .~. - I
4rc
wt
..
I
h
+.-- .
wt
I I
I
,,I
wi
..
o~----~-.~~----~-----~---
wi
ia
(a)
(b)
Fig. 3.14. S ingle-phas e half-wave d 'ode rectifier with R load (a) circuit diagram and (b) waveforms.
For a resistive load, output curren t io has the same waveform as that of the output voltage Va- Diode voltage VD is zero wh en diode conducts . Diod e is reverse biased fr1fm w t = 1t to CDt = 21t as shown. The waveforms of us' Va, ia and VD are sketched in Fig. 3.14 (b). Here source voltage is sinusoidal i.e. Us = Vm sin wt. KVL for the circuit of F ig. 3.14 (a) gives Vs = Vo + VD' Average valu e of output (or load) voltage,
Vo ;
~s: Vrn sin w t d(w III
[Art. 3.5]
Diode Circuits and Rectifiers
V ' Vm . m I cos w t I ~=. - 2n n Rms value of output voltage, Vor
=[
1 2n
TC
= Vm fTC -::r2n [ 0
2
,..(3.21 )
]1/2
\
fa ~ sin
59
wt . d (wt )
lV2
'
1- cos 2wt . d (wt ) 2
Vm
.. .. (3.22)
- 2 H ere the subscript 'r' is used to denote rms value . Average value of load current, Vo Vm 10 = - = -
... (3.23 )
. ' _ Vor _ Vm lor - R ~ 2R
.. .(3,24)
nR
R
Rms value of load current,
.Peak value of load, or diode, current _ Vm - R
... (3.25)
, Peak inverse voltage, PlY, is an important parameter in the design of rectifier circuits.
P1V is the maximum voltage that appears across the device (here diode ) during its blocking state. In Fig. 3.14, PIV = V m = -f2 . Vs = -v; (rms value of transformer secondary voltage ). It is seen from the waveform of source current is (or io) that the transformer has to handle de component of is' It leads to magnetic saturation of the transformer case, therefore more iron 10ss8s , more transformer heating and reduced efficiency. . .
..
Power delivered to resistive load ___
=V or
Input power factor
.,
= (rms load voltage) Vm Vm
lor
V~ V; = 4R :::: 2R
?
=1;,. R
... (3.26)
Power delivered to load Input VA Vor . lor Vor -f2vs . =v . ] =Y-= 2V =O.707lag.
=
s
~
= 2 . 2R
(rms load current ).
or
S
S
(b) L load: Single-phas e half-wave diode r ectifier with L load is shown in Fig. 3.15 (a ). When switch S is closed at wt = 0, diode starts con ducting. KVL for this circuit gives \ '
dio
io =;:
or
f sin w t . dt
Vm
= - - cos wL ,At wt
or
.
v s = Vo =L -dt - = V m sm w t
..".
= 0,
io:::: 0 , ..
' wt+A
.. (3. 27 a )
60
Power Electronics
[Art. 3.5J
Substitution of the value of kin Eq. (3.27 a) gives V io = wZ (1 - cos wt) dio
Output voltage, Source voltage
Vs
va = L (it
=L
Vm . wL [sm CDt]
... (3 .27 b)
.
CD
= V m sm CDt =Vs
and both output voltage Vo and output current io are plotted in Fig. 3.15
(b ).
Average value of output voltage, Va = 0
cut
wt
wt
(a) (b) Fig. 3.15. Single-phase one-pulse rectifier with L load (al circuit diagram and (b) waveforms.
The output current io consists of de component and fundamental frequency. component of frequency w. Peak value of current 1max occurs at wt = 7t Imax
Average value of current,
. Vm . 2Vm = wL (1 + 1) wL
=
.. .(3.28)
1 J27t V 10 = 27t 10 w'i(1- cos CDt) d(wt)
Vm
1
= wL =
.. .(3.29)
2 Imay:
Rms value of fu n dament al current, I Ir is given by
r1 [V \2 J2Tt w1:)
I lr 127t
10 = T2 . wL = wI. = ...[2 Vm
]1/2.
0 (coswt)2d(CDt)
Vs
... (3.30)
'\
Rms value of rectifie d curr nt = [ 102 + 121rJ1I2
+i~r +
l I
=1.225 10
.. .(3. 31)
61
[Art. 3.5J
Diod e Circuits and Rectifiers
Voltage across diode, . vD = 0. (e) C Load: In Fig. 3.16 (a), when switch S is closed at wt== 0, th e equation governing the behaviour of the .circuit is . lO
Output volta ge,
Vo
dv
d
.
= C dts == C dt (Vs sm w t) = wC Vm cos wt
...(3.32)
= ~ Jidt =V m sin wt = v s = VC
wt
7r
,
2rr
4rc
wt
,,,
wt
I I
wcvm ': I I
tJo + -:- ~
io
:1
0
rr,
tJot
+
n:
0
vcr _C
.'37[/ 2 I
~
I
3rr/ 2 2rr
3rr 7rr/2 :
wI.
v o=uc
io (0 )
(a)
Fig. 3.16. Single-phase half-wave di ode r ectifier with C load (a ) circuit diagram and (6) waveforms .
Capacitor is charged to voltage V m at wt
="21t
and subsequently this voltage remains
. constant at V m' This is shown as Vo = Vc in Fig. 3.16 (b ). Capacitor current or load current is maximum at wt ::: O. Its value at wt shown. 1t
The diode conducts for 2 w seconds only fr om wt = 0 t o wt
=1t/ 2 , diode voltage Vo + Vs = - V m + V m sin wi
voltage is , therefore, zero. After wt vD
= -
= V (sin wt -
vD
=
°is
w CVm as
= ~ . During this interval,
di ode
is given by ... (3.3 3)
1)
For Eq. (3.33), the time origin is r edefi ned at wt = 1t/ 2 . After wt
= 1t/ 2,
diode voltage is plotted as shown in Fig. 3. 16 (b ). At wt = 31t,
Average v alu e of voltage
2
a cro ~ s
diode,
Zit
VD
=~ J J..:-t
0
= V :::
V m (~in wt - 1) d (rot)
2Vs
G'D = -
2 V", .
.Power E ectronics
Rms value of fundamental component of voltage across diode, 21t
VIr
]112
.
=[. 2~ fo ~ sin 2 OJt d(OJt)·
. ==
V
~
" .(3.34 b)
,R ms value of voltage across diode
-= -Jr-"v""'-1-+-v-'i"-r = 1.~'25 V m Example 3.7. Find the time required to deliver a charge of 200 Ah through a single-phase half-wave diode rectifier with an output current of 100 A rms and with sinusoidal inpu.t voltage. Assume diode conduction bver a half-cycle. ;
Solution. For I-phase half-wave diode rectlfier, rms value of output current,
. V
R I or = ~ 2R = 100 A or V m. = 200 -
The charge is delivered by direct current 10 which is given by
= 200 R = 200 A 7tR 7tR 7t
Also .Io x time in hours = 200 Ah
\ :. Time required to deliver this charge
200 x 7t .
= 200 hrs = 7t = 3.1416 hrs I = Vm o
Example 3.8. A single-phase 230 V, 1 kW heater is connected across single-phase 230 V, 50 Hz supply through a diode. Calculate the power delivered to the heater element. Find also the peak diode current and input power factor.
H eater resIstance, . R · S o 1u t Ion.
2
230 n = 1000
Rms value of output voltage, from Eq. (3.22), is V or
=..J2 x 230 2
Power absorbed by heater element
- V;r
=Ii:=
2 X 230 4
2 X
1000 230 2
= 500 W
Peak value of diode current, from Eq. (3.25), is given by
..J2 x 230 2
Input power factor
230 _ Vor _ - V -
x 1000 = 6 .1 478 A
..J2 x
2
230
1 x 23 0 = 0.7 07 lag.
s
R E Load : Single-phase half-wave diode rectifi er with l oad resistance R and load counter emf E is shown in Fig. 3.17 (a). If the switch S is closed at OJt = 0° or when Vs = 0, then diode would not conduct at OJt = 0 because diode is reverse bi ased until SOUTce voltage Vs equals E. \¥hen V m sin 8 1 = E, diode D st arts conducting and the turn-on angle 8 1 is given by (d)
.-I (E] \V
e1 -- SIn
m
... (3.313)
63
[Art. 3.5J
Diode Circuits and Re ctifiers
wt
E
,
o '
+
eI
t' Uo I i
to (77 -8 1)
i' 11:
wt
i :
!i
-~r
i
i i
,
\@j
'
:
\-E
wt
(6)
(a)
Fig. 3.17. Single-phase half-wave diode rectifier with RE load (a) circuit diagram and (6) wave forms . .
.
'
<...
The diode now conducts from wt = 8 1 to wt = (&.= 8 1), i.e, conduction angle for diode is (n - 28 1) as shown in Fig. 3.1 i ( b). During the conduction period of diode, th~\oJtage equation. for the circuitjs V m sin wt = E +--i-o R
. V m sin wt - E
or ...(3.37) to = R Average value of this current is given by
1 Io = 2rtR
[f
8
it -
6
1
(Vm sin wt - E ) d(wt)
1
1
1
= 2rtR
[2
Vm cos
81 -
E
(n - 28 1)J
... (3. 38 )
Rms value of the load current of Eq. (3.37) is
[~ fT8t -8
I = or 2n
t
[Vm sin wt - E ]2 . ]112 R d(wt)
t
=[~ f:- (V~ sin 2nR 61
2
wt + E2 - 2 Vm E sin wt) d(wt )]112
1
.
~ [2~2 i(V; + E') (n- 28
112
'UOo
1)
1]
+ V; sin 28 1 - 4 Vm E cos 91
..(3.39:
Power delivered to load,
Supply pf
P = E 10 + 1;1' R "vatts Power deli vered t o 10ad =------------------------------------ (S ou rce volt age) (rms value of ;:;o urc e current ) 2
=
E 10 + l or R
... (3.4 0)
'
.. (3 .41)
64
P ow er Elec troni cs
[Ar t. 3.5]
It is seen from Fig. 3.17 (a) that at rot = 0°, UD = - E and at rot = 81 , uD = O. During the period diode conducts, U D = O. When rot = 3rr/2, Us =- V m and UD:;:: - (Vm + E). Thus PIV for diode is . (Vm + E ) .
Ii'
,
.., Example 3.9. A dc battery of constant emf E is being charged through a resistor as shown in Fig. 3.17 (a). For source voltage of 235 V, 50 Hz and for R = SQ., E = 150 V, (a) find the value of average charging current, (b) find the power supplied to battery and that dissipated in the resistor,
(c) calculate the supply pf, (d) find the charging time in case battery capacity is 1000 Wh and (e) find rectifier efficiency and PIV of the diode.
Solution:
(a)
The diode will start conducting at an angle 81 , where 8 1 -
. -1 sm
-274660 .J2150 x 230 .
Average value of charging current, from Eq. (3.38), is
10
~ 2n ~ 8 [2 . ,f2 x 230 cos 27.466
0 -
150 (1t - 2 x
= 4.9676 A (b)
'.
Power delivered to battery = E 10
2~.:g6 x n Jl
= 150 x 4.9676 = 745.14 W
Rms value of charging current, from Eq. (3.39), is
I"
~ [ 2n ~ 64{ (230' + 150') (n - 2 x 27 A66 x 1~0 J+ 230' sin 2 x 27.466 .
0
. - 4,f2 x 230 x 150 cos 27.466
}
r'
0
~ 9.2955 A
Power dissipated in resistor
=l~r R = (9.2955)2 X 8 = 691.25 W
(c)
From Eq. (3 .41), th e supply
pf
= 745.14 + 691.25 = a 672 1
CT
230 x 9.2955 . ao
(d) (Power delivered to battery) (charging time in hours) = Battery capacity in Wh. 1000 :. Charging tim e = 745.14 = 1.342 h
(e)
Rectifier efficiency
= Power deliver ed to battery
Total input power =
. (f) PIV of dio de (e)
745.~::·~~1.25 x 100 = 51.876%
:;:: Vm + E = -{2 x 230 + 150 = 475.22 V .
RL Load: A sin gle-phase one-pulse diode rectifier fe eding RL load is shown in Fig.
3.18 (a). Current i o continu es t o flow-even aft er source volta ge Vs has become negative; this i s be caus e of the pr es ence of inductance L in the load circuit . After positive half cycle of source
65
[Art. 3.5)
ectifiers
Ar~Q
A
ArroA,
wt
wt Va
to
l<;
+
+
0
tVR I\)
Vs=vmsinwt
+
wI
Vo 0
Va
wt
tVL (b)
(a)
Fig. 3.18. Single-phase half-wave diode rectifier with RL load (a) circuit diagram and (b) waveforms.
voltage, diode remains on, so the negative half cycle of source voltage appears across load until load current io decays to zero at wt = ~ . Voltage vR =io R has the same waveshape as that of io· Inductor voltage VL = Vs - vR is also shown. The current io flows till the two areas A and Bare equal. AreaA (where Us > vR) represents the energy stored by L and area B (where Us < uR) the energy released by L. It must be noted that average value of voltage VL across inductor L is zero. When io = 0 at wt = ~; uL = 0, l.iR = and voltage Us appears as reverse bias across diode D as shown. At ~, voltage vD across diode jumps from zero to Vm sin ~ where ~ > 11:. Here ~ = y is also the conduction angle of the diode. Average value of output voltage,
°
1 . Vo = 21t Vm
= 21t
f:
Vm sin wt· d(wt)
(1 - cos ~)
.. .(3.42)
Average value of load or output current
Vo Vm 10 = Ii = 21tR (1 - cos ~) A general expression for outpu t current io for 0 < wt <
... (3.43) ~
can be obtained as under:
When diode is conducting, KVL for the circuit of Fig. 3.18 (a) gives
dio
dt = Vm sin wt
Rio + L
The load, or output, curren t io consists of two components , one steady s tate component is and the other transient component it. Here is is given by
:X
V
is = ~R2
2
sin (wt -~)
66
Power Electronics
(Art. 3.5)
where¢ = tan- 1 ~ and X = wL. Here <» is the angle by which rms current Is lags source voltage
I
. Vs ' The transient component it can be obtained from force-free equation
di t Ri +L-=O t
dt
.
R
Its solution gives it = A e'" T, t
Tota.l solution for current io is, therefore,given by V ' R . io is + it sin (wt - ¢) +A e-T,t
=
=f
...(3.44)
where . . Z = ..JR2 +X2 . .
Constant A can be obtained from the boundary condition afwt ~ O.
Atwt = 0, or at {=O, io = O. Thus, from Eq. (3.44)
Vm . 0=:- -y .sin ~ +A
I
I.
I
.·· ·Vm
. . .•
·A=Zsm4>
I.
. Substitution of A in Eq. (3.44) gives
io'=~m [sin (wt -
I.
<») + si'n
~. e-¥tJ
wt ~ ~ It is also seen from the waveform of io in Fig. 3.18 (b) that when wt condition, Eq. (3.45) gives for 0
sin
o'f
.. .(3.45) . '.
~
(~- <») + sin ¢ , exp [
-1 ~] =
=~,
io == O. With this
0
The solution of this transcendental equation can give the value of extinction angle
~.
(f) RL load with freewheeling diode*: Performance of single-phase one-pulse diode rectifier with RL load can be improved by connecting a freewheeling diode across the load as shown in . Fig. 3.19 (a). Output voltage is Uo = Us for 0 ~ wt ~ n. At wt = re, source voltage Us is . zero, but output current io is not zero because of L in the load circuit. Just after wt = re, as Us tends to reverse, negative polarity of Us reaches cathode of FD through conducting diode D, whereas positive polarity of Us reaches anode of F D direct. Freewheeling (or flywheel) diode FD, therefore, gets forward-biased. As a result, load current io is immediately transferred from D to FD as Us -tends to reverse.' After wt = re, diode, or source, current is = 0 and diode D is
subjected .to reverse voltage with PlV equal to V m at CDt = 3re, 7; etc. 2 After wt = n, current freewheels through circuit R, Land FD. The energy stored in L is now dissipated in R. When energy stored in L = energy dissipated in R, current falls to zero at CDt = ~ < 2re. Depending upon the value of Rand L, the current m ay not fall to zero even when CDt = 2re, this is called conti uous conducti on. But in Fig. 3.19 (b) , l0 2.d CUTrent decays to zero before CDt = 2re; load current is ther efore discontinuous. * Freewheeling diode is aiso called bypass diode or commutating diode .
Diode Circuits and Re ctifiers
67
(A rt. 3.5]
wt
2rr
371':
wt
:~: '~I
I
1
2rr
.
I
31T:
wt
io is
ISkr=l
+
D
I
I
I
I
I
I
I
i~
I
rI
O~FO
I
to-
I
I
I
D --jFO
I
I
i
wI
I
3n
2/f
.I
..
I
wt
(b)
(a)
Fig. 3.19. Single-phase one-purse diode rectifier with RL load and fre!twnee1ing diode
O\v (a) circuit diagram and (b) waveforms.
The effects of using freewheeling diode are as under: It prevents the output (or load) voltage from becoming negative. (ii ) As the energy stored in L is transferred to load R through FD, the system efficiency is improved. (iii) The load current waveform is more smooth, the load performance, therefore, gets better. (i)
The w aveforms for
Us ,
Va, io, vD,
is
and ifd are drawn in Fig. 3.19 (6).
The expression for the load current io can be obtained from Art. 6.1.2 if required. It is seen from Fig. 3.19 (b) that 1 J~ = V sin wt d(wt) o 21t 0 m
average output voltage,
V
and average load current,
I a -....!!!:.- rtR
V
V
=....!!!:.-
1t
... (3.46) ... (3 .47)
(g) Single-phase fun -wave diode rectifier : Th ere are two types of full-wave diode
rectifiers, one i centre-tapped (or mid-point) full -wave diode rectifier and the other is full-wave diode bridge recti§,er. These are now described briefly. (i) Singl e-phase full-wave mid-point diode r ectifier: Fig. 3.20 (a) illustrates a single-phase full-wave mid-point rectifier using diodes. The turns ratio fTom each sec ondary to primary 's taken as l.ill·ty fOT simplicity. vVhen 'a' is positive with respect to ' b', or m id-point 0, diode Dl con ducts for 1t r adi ans. In the next half cycl e, 'b' is positive with respect t o 'a' , or . mid-poin t 0, and ther efore diode D2 condu cts. The output voltage is shown as Vo in F· g. 3.20 (b) . The waveform for output current io (not sh own in the figur e) 15 gimilar to va waveform.
68
Power Electronics
[Ar t. 3.5]
Vm sin wI
wt
.
wI
wI
vOl
is
+-
wI
0 Vs (\J
R
io
--+ Vo
02
wI
+-
v02
(a) (b) Fig. 3.20. Single-phase full-wave mid-point diode rectifier (a ) circuit diagram and (b) waveforms .
When 'a' is positive with respect to 'b', diode D2 is subjected to a reverse voltage of 2vs ' In the next half cycle, diode Dl experiences a reverse voltage of 2vs' This is shown in Fig. 3.20 (b) as VDl and VD2' Thus, for diodes Dl and D2 , peak inverse voltage is 2V~. Waveforms of Fig. 3.20 ( b ) show that for one cycle of source voltage, there are two pulses of output voltage. So single-phase fun-wave diode rectifier can also be called single-phase two-pulse diode rectifier. Source current waveform is is also shown in Fig. 3.20 (b). When Dl conducts, current in secondary flows upward from 0 to a and therefore, primary current is must flow downward to balance the secondary mmf from 0 to 7t rad . 'When D2 conducts, secondary current flows downwar d from 0 to b, therefore, prim ary current is must flow upward to balance the secondary mmf from It to 27t and so on. Average outpu t voltage,
...(3.48 a )
Average output current, Rms value of output voltage, .. .(3.48 b)
[A rt. 3.5]
Diode Circuits and Rectifiers
69
Va lor=Ji
Rms value of load current,
= Vor . lor =l~r . R
= Va . lor
Power delivered to load Input volt amperes :. I nput
pf=
~.~
Va' lor
= 1
(ii) Single-phase full-wave diode hridgerectifier : A single-phase full-wave bridge rectifier employing diodes is shown in Fig. 3.21 (a). When 'a' is 'positive with respect to 'b', diodes Dl, D2 conduct together so that output voltage is Vab' Each of the diodes D3 and D4 is subjected to a reverse voltage of Vs as shown in Fig. 3.21 (b). When 'b' is positive with respect to 'a', diodes D3, D4 conduct together and output voltage is vba' Each of the two diodes D1 and D2 e~perience a reverse voltage of va as shown.
A comparison of Figs. 3.20 (b) and 3.21 (b) reveals that a diode in mid-point full-wave rectifier is subjected to PIV of 2Vm whereas a diode in full-wave bridge rectifier has PIV of V m only. However, average and rms values of output voltage are the same for both rectifier configurations.
i 01
is
"
1: 1
to
01
03
a R "- ' .
04
b
02 '1. 0 2
l Vo
J
wt
(a) Fig. 3.21. Single-phase full-wave diode bridge r ectifier
(b) (a)
circuit diagram and
(b)
wavefonns.
For the waveforms of diode current i Dl or iD2 in Fit 3.21 (b) and also for iD3 , iD4 for the circuit of Fig. 3.20 (a) (not shown in Fig. 3.20 (b) ), the average and rms values for diode current are obtained as under:
. I J~ ~ Average va1ue of di ode curren t) ID = -2 1m sin rot . d (cut) =
.. (3.49 a)
n o n
Rms value of diode current,
ID • = [
Peak r epetitive diode current,
]m
2~ ( I~ sin' wt d (wt)
Vm =R
f'_ I;
... (3.49 b) ... (3.49 c)
70
Power Electronics
[Art. 3.5]
It can similarly be shown that average value of voltage across each diode in Fig. 3.20 (b) is
2Vm Vm. The corresponding rms values of voltage across eac hd'10 d e - and that in Fig. 3.21 (b) is TC
TC
.
V is V m = {2 Vs in Fig. 3.20 (b) and ~ = Vs in Fig. 3.21 (b). Three-phase rectifiers using diodes are discussed in APt. 3.9. Example 3.10 is formulated to illustrate the effect of reverse recovery time on the average output . voltage. . Example 3.10. In a single-phase full-wave diode bridge rectifier, the diodes have a reverse recovery time of 40 ~s. For an ac input voltage of230 V, determine the effect of reverse recovery time on the average output voltage for a supply frequency of (a) 50 Hz and (b) 2.5 kHz. Solution. Single-phase full-wave diode bridge rectifier is shown in Fig. 3.21 (a) and output voltage Vo is shown in Fig. 3.21 (b). If reverse recovery time is taken into consideration, the l..io diodes · D 1 and D2 will not be off atwt = TC in Fig. 3.21 (b), but will continue to conduct until
t=~+t'T as .depicted in Fig. 3.22. The w
t
reduction in output voltage is given by the cross-hatched area. Average value of this reduction in output voltage is given by Fig. 3.22. Effect of reverse recoverY time on output voltage.
t
V r
JV
= 1. TC
0
Vm
=-
sin wt d (cot)
rr
7t
m
.
...(3.50)
(1 - cos wtrr )
With zero reverse recovery ti m e, average output. voltage, fr om Eq. (3.48),
IS
VO = 2..f2 x 230 = 207.04 V 7t
For (:3.50), is . (a)
f= 50 Hz and trr = 40
the reduction in the average output voltage, from Eq.
).lS,
V
Vr
= ~ (1 - cos 27t f trr ) 7t
= -!2 x 230 (1 - cos 27t X 50 X .
.
40
X
10- 6 X 18 0J
1t
1t
.
=8.174mV Percen tage reduction in average outpuc voltage = (b) For
8.1~~;.: 0~0-3 x 100=3.948x 10- 3%
f = 2500 Hz, the reduction in t he average output voltage, from Eq. (3 .50), is
. (1 - cos 21t X 2500 X 40 V ,. = ~ x 230 1t
= 19 .77 V
X
10- 6 X -180 1t
I J
[Art. 3.5]
Diod e Circuits and Rectifiers
71
Percentage reduction in average outpu t voltage ~ 21;;, ~074 x 100 = 9.594%.
tI
It is seen from above that the effect of reverse recovery time is negligible for diode operation at 50 Hz, but for high-frequency operation of diodes, the effect is noticeable. .
t.r Example 3.11, A single-phase full bridge diode rectifier is supplied from 230 V, 50 Hz source. The load consists of R= 10 n and a large inductance so as to render the load current constant. Determine . ·(a) average values o{output voltage and output current, (b) average and rms values of diode currents, .
.
(c) rins values of output and input currents, and supply pf. ·Solution. The circuit diagram and relevant waveforms for this uncontrolled rectifier are shown in Fig. 3.23.
Vmsinwt
wt
I
. I
·i
I
I·
01 D2---L 0304 -.l--01 02+0304-( I I I .
\)0
.
I.
'I
I~
I
,I
"
iOl
. io
+ 01
03
is
w 't.
+
is 04
I
I '.
I
lIo
1 (b)
. (a)
·Fig. 3.23. Pertaining to Example 3 .11 (a )
i
-i1o
rIo .
02
i
.\
I
(a)
circuit diagram and (b ) waveforms .
Avetage value of output voltage,
Vo
2V
.
-~
.
= ~ = 2 '1 2 x 230 = 207.04 V . 7t . 7t . .
· Average value of output current,
10 = Vo == 207.04 = 20 704 A . R 10 . (b ) Average value of diode curren t , 1 . 7t I ')0 '7 04 I DA
= ~7t = ; =...
'2
= 10.352 A
'
wt ·
......
72
Power Electronics
[A r t. 3.5]
"
Rm s value of diode current, I Dr =
-"--j
a
{¥ 1 1t 21t 0
10
= 12=
20.704
T2"
= 14.642 A
As load, or output, current is ripple free, rms value of output current = average value of output current =10 = 20.704 A
-~ . ='J --;-= 10 = 20.704 A 1t
Load power = Vo 10 = 207.04 x 20.704 W
= Vs Is cos
Inp ut power .. 230 x 20.704 x cos
~' . Supply pf = cos
Rms value of source current, Is
"
o v Example 3.12. A diode whose internal resistance is 20 n is to supply power to a 1000 .0 load from a 230 V (rms) source of supply. Calculate (a) the peak load current (b) the dc load current (c) the dc diode voltage (d) the percentage regulation from no load to given load. (I.A.S., 1983 ) Solution . A voltage of 230 V supplying power to 1000.0, through a single diode, is shown .in Fi g. 3.24 (a ). Waveforms for the source voltage, load current ioaiid diode voltage vD are shown in Fig. 3 .24 ( b ).
wt
Va
+ - - io
•
wt
+ R=1000 .n.
•
wt
(a )
(b )
F ig. 3.24. Pertaining to Example 3.12 (a ) circuit diagram and (b) waveforms. (a )
It is seen from th e w aveform of i o that peak load current 10m is given by Vm -{2 x 230 10m = R + RD = 1020 = 0.3189 A
H ere R ~
= load resistan ce and
RD = internal resistance of diode.
.
(b ) f) C load curren t,
1
10
= 21t =
1 om
n
f lt 0
=
.
l orn sm wt d (wt)
0.101 51 A
[Art. 3.6J
Diode Circuits a n d Rectifie rs
(c)
1
V D =10 RD - 27t
DC diode voltage,
= (d )
V
At no load, load voltage,
on
V 01
At given load, load voltage,
10 RD - Vm 7t
230--12 sin wt d(wt)
= 0.10151 x 20. -
230.J2 = - 101.5 V 7t
= Vm = -f2 x7t 230 = 103.521 V . 7t . = 230-f2 7t
1000:: 101 49·1· V .
x 1020
=Vall -:- Val
:. Voltage regulation
f:
Von
x 100 = 103.521- 101.491 = 1.961%. 103.521
3.6. ZENER DIODES Zener diodes are specially constructed to have accurate and stable reverse breakdo'pn . voltage. Circuit symboi' for Zener diode is shown in Fig. 3.25 (a) . When it is forward biased, it behaves as a normal diode. When reverse biased, a small leakage current flows. If the reverse voltage across Zener diode is increased, a 'value of voltage is "reached at which re verse breakdown occurs. This is indicated by a sudden increase of Zener current, Fig. 3.25 (6). The voltage after reverse breakdown remains practically constant over a wide range of Zener current. This makes it suitable for use as a voltage regulator to furnish constant voltage from a source whose voltage may vary noticeably.
1+1 Cathode
Breakdown voltage
Rs
I
+
+v
-V Vz
Vs
Zener vo ltage Anode (a)
-I
Io
Is
/
+
1z Load
V"Z
Vo= Vz
~ (c )
(b) (a) circuit
Fig . 3.25. Zener diode symbol (b) I-V characteristics
(c ) use as a voltage re~lator.
For the operation of Zener diode as a voltage regulato~, (i) it must be reverse biased with a voltage great er than its breakdown, or Zener, voltage and (ii) a series res ist or Rs, Fig. 3.25 . (c) is necessary to limit the r everse cu rrent t hrough the diode below its rated value. If V z = voltage across Zener diode, th en it is seen ftom Fig. 3.25
is
(c)
that source current 15
74
[Art. 3.6]
Power
V Load, or output, current, 10 = Iiz where R
E lectronic~
= load resistance. Current through Zener diode,
1z = Is - 10 Power rating of a Zener diode is V z . 1z. These are available in a voltage range from few volts to about 280 V. Example 3.13. Design a Zener voltage regulator, shown in Fig. 3.26, to meet the following specifications: . Rs Load voltage = 6.8 V, Source voltage Vs is + 20 V ± 20% and load current is 30 rnA ± 50%. 1L
Is The Zener requires a minimum current of 1
+ z VL :6.BV RL mA to breakdown. The diode D has a forward Vs +lmA voltage drop of O. 6Y.
-
Solution. When source volbige is maximum ~d load current is minimUm, then source r·esistance should be maximum. .
Fig. 3.26. Pertaining to Example 3.13 .
. Vs _mlu =VL + (1L - mill + 1z ) Rs -max R _ = (20 X 1.2) - 6.8 = 1075 n . s - ml1l [3 0x O.5+1]xlO- 3 Similarly,
Vs -min = V L + (IL -max + 1z) Rs -mill
R _ = (20 X 1.2) - 6.8 = 200 n 5 - mm [30 X 1.5 + 1J X 10- 3
Maximum load resistance ,
= _ V_L_=
h
max
Minimum load resistance,
RL _mill .
=1
30
mill
V
L L - max
=
X
6 .8 =453.30
0.5 X 10- 3
6.8 30 X 1.5 X 10- 3
=151.5 n
The voltage rating of the Zener diode is 6.8 ·~
0.6
= 6.2 V .
. E x ample 3.14. The complete circuit shown in Fig. 3.27 (a) rep resents a 25 V de voltmeter
where G is a PM MC galvanometer having full-scale deflection current 1{sd = 200 micro-A and
resistance R o = 500 ohms, and D is a 20-V Zener diode. Find R l and R 2. W hat is the function of the diode D in this circuit? . (GATE, 1990) .
Solution. Current through galvanometer, 1
- I _ Zener voltage 2R .... R
(sd -
or
2 '
G
R 2 :°500 = 200 X 10- 6 6
or
X 10 R.2 -- 20200 - t)'- 00 --
09 oJ
.
5k
n
~~
,
[Art. 3.7]
Diode Circuits and Rectitiers R2
RI
+
+
II
75
lz 12 RG=5 00n
D
(b)
(a) Fig. 3.27. Pertaining to Example 3.14.
As Zener diode current is not specified, let it be assumed zero. Therefore, from Fig. 3.27 (b ),
II ~12 =lz == 0
cir
II =12 = 200)lA.
11 = 25 - 20 = 200 x 10- 6 Rl
Also
5 X 10 6 Rl = 200 = ~5 kn
or
Function of Zener diode is to provide a constant voltage to the ga1vanometer circuit. Whenever voltage across this diode exceeds 20 V, it conducts and the excess current is shunted away frtlm galvanomete r G. So here diode D prevents overloading of the PMMQ galvanometer.
3.7. PERFORMANCE PARAMETERS
'
.
The input voltage to rectifiers is usually sinusoidal. It is desired that the output voltage from a rectifier should be constant with no ripples in it. This, however, is not the case. This shows that the rectified output voltage is m ad up of constant de voltage plus harmoniC components. The waveform of input and output currents depend on the nature ofload and the rectifi er configuration. In order to evaluate the overall performan ce of rectifier·load combinations, certain performance parameters relatin g tp their input and output must be known. The object of this article is to defi n e the various pe-rform ance parameters (or indices) relating to input as well as output voltages and currents .
3.7.1. Input Performance Parameters The variou s parameters relating t o the source (or input ) side of t he converter-load combin ation are defined below : (i ) I nput power-fact or. Input voltage taken fr om power-supply undertaking is generally sinusoidal. However, ac input current is usually non-sinusoidal. Under such a condition, only the fund a mental component of inpu t current t akes pa t in extr acti g mean a c inpu t power fr om the source .
The inpu t power factor PF is defined as the r atio of mean input power (real power) to the total r ms input volt amperes (apparent power) given t o the conver t er (or rectifier) system .
If Vs = rm s value of supply phase volt age.
I, = r m s value of supply phase current in cluding fundamental and h armonics I s 1 = r ms value of fundam ent al component of supply cu rren t Is an d rp 1 =
phase an gle bet ween supply vol age V s an d fu ndam en tal componentIs1 of supply eu r ent Is ; see Fi g. 3.28 ;
Power Electronics
[Art. 3.7]
76
then , the input power factor, as per the definition, is given by PF = mean ~c input power = real power, Vs' lsI' cos (h total rms mput voltamperes apparent power, Vs . Is lsI
...(3.51)
=y .cos CPl s
For a given power demand, if input pf is poor, more input volt-amperes and hence more input current are taken froin the supply. . (ii) Input d i splacement factor (DF). As stated above, the phase angle between sinusoidal supply voltage Vs and fundamental component lsI of supply current Is is CPl' This angle CPl' shownin Fig. 3.28, is usually known as input displacement angle. Its cosine is called the input displacement factor DF. ..
DF= cos
...(3.52)
DF is also called fundamental power factor. (iii) Input current distortion factor (CDF). It is defined as the ratio of the rms value of fundamental component lsI of the input current to the rms value of input, or supply, current Is'
. I CDF=~ Is
... .(3.53)
It is seen from Eqs . (3.51) to (3.53) that PF = (CDF) x (DF)
or input power factor = (input current distortion factor) x (input displacement factor)
(iv) Input current harmonic factor (HF). Non-sinusoidal input, or supply, current is made up offundamental current plus current .components of higher frequencies . The harm onic fa ctor (HF) is equal to the rms value of all the harmonics divided by the rms value of fundamental component of the input current. If Ih = rms value of all the harmonic-components combined tTl
12
='i 1 s - sl
[L
Isn] HF= Ih = ~ -I~1 = n=2 ...(3.54) then , as per the defmition, lsI lsI lsI · where, Isn = rms value of nth harmonic content. H armonic factor is a measure of the harmonic content in the input supply current. HF is also known as total h armonic distortion (THD). Greater the value of HF (or THD), greater is the harmonic content and hence greater is the distortion of input supply current. .
Also,
HF =
~(~)2 - 1 = ~~-1 Is1 CD"
... (3.55)
Higher va1ue of input distortion factor CDF indicates lower magnitude ofharinonic content in the source current. Non-sinusoidal input current can be esolved in to Fourier series as under : a ...(3 .56) i =; + (an cos n wt + bit sin nwt) It = 1,2, 3, ... 00
L
=~ + where a o ::: T2
r
i(t).dt, an = T2
r 0
o
77
[Art. 3.7J
Diode Circuits and Rectifiers'
I
n
...(3.57)
en sin (nwt + <1>n)
=1,2,3, ...
i(t). cos nwt. dt and bn = ~
r
i(t). sin nwt. dt
0
Cn=[ a!; b! rand $n =tan- 1 ( ~: ) ( 3 5 8 1 Crest Factor (CF). Crest factor for input current is defined as the ratio of peak input currentlsp to its rms value Is. (v)
,, ,,'
wt
Fig. 3.28. Waveforms for source voltage US) source current is, fundamental component .isl of source current and ~l = phase angle between Us and is1 ' .. .(3.59)
CF is used for specifying the current ratings of power semiconductor devices and other cgmponents.
3.7.2. Ou tput performanc e Par ameters
..
, ,"
The load, or output, voltage and the load (or output) current at the output terminals of ae to de converters are unidirectional but pulsating in nature. Fouriers series is used to express these output quantities in terms of its two components, namely (i) average (or de) value and (ii) ae component superimposed on de value a s under:· t\
In general, average value of output quantity y is , Yo
+T
= Y de = ~ J y. dt tj
and its rms value is,
1 Y or = T [
[1 tj
+
TY 2 dt ]112
wh ere y = instantaneous v alue of the function in terms of t and T = tim e period for one cycle of y variation . . . Output de power, P dc
= (average outpu t
voltage, Vo) x (average output current, 10 )
::: Vjo where subscript '0' denotes output de valu es.
Output ae power Pac = Val"" 101' where subscript "or" denotes Ims value 01 output qU2.lltitiE':: .
... (3 60)
78
P ower Electronics
[Art. 3.7]
The various output parameters are now defined below. (i) R e ctification r atio 11. Rectification r atio, also called efficiency of a converter, is defined
as the ratio of de output power Pde to ac output power Pac'
Pdc
...(3.61)
11= Pac
Rectifier ratio is also known as rectifier efficiency or figure of merit. In case Rd = forward rectifier resistance, then 11
=
Pdc
... (3.62)
2
Pac + lor Rd
(ii) Effective, or ripple, value of the ac component of output voltage is given by
Vr=~~r- ~
...(3.63)
where VI' is called ripple voltage, or effective value of ae component of output voltage.
(iii) Form factor (FF). It is defined as the ratio of rms value Vo/' of output voltage to the dc value Va ~utput voltage. -~V FF=~
.. .(3.64)
Va
FF is a measure of the shape of the output voltage. The closer FF is to unity, the better is the dc output voltage wavefor m . For constant de output voltage, rms value of output voltage, Val' = average value of output voltage, Vo' Voltage ripple factor (VRF). It is defined as the ratio of ripple voltage VI' to the average output voltage Va. (i v)
V,.
VRF = Vo
.. .(3.65)
r
Substituting the value of VI' from Eq. (3.63) in Eq. (3.65) gives
VRF =[
or
(t J' -
FF ='/VRF2 + 1
1
=VFj" -
1
... (3.66 a) ... (3.66 b)
(v) P er-unit average ou tpu t voltage. It is defined as the r atio of the average output voltage Va for any value of triggering angle to the average output voltage Vom for zero-degree firing angle.
.. .(3.67)
(vi) Current ripple fa c tor {CRF). It is defined as the ratio of r m s value of all h armonic components of output current to the de component ] 0 of the output curren t.
... (3.68)
." .
[Art. 3.8]
Diode Circuits and Rectifiers
79
Here lor = rms value of output current including de and harmonics,
Ir = rms value of all harmonic components of output current
10 == de .component of output current.
Note that l~r == I; + I~. (vii) Transformer utilization factor (TUF). If V 2 (= V s) and 12 (= Is) are respectively the
rms voltage and rms current ratings of the secondary winding of a transformer, then TUF is defined as ... (3.69)
:. Transformer VA rating
... (3.70)
Lower the TUF, higher is the transformer VA rating required.. It is desirable that a rectifier produces a perfect de output voltage so that (i) rms value = de value (ii) FF = 1.0 (iii) ae component of output voltage = 0 (iv) HF = 0 (vi) PF = 1.0 and TUF = 1. 3.S. COMPARISON OF SINGLE·PHASE DIODE RECTIFIERS
.
In this article, the performance parameters of single-phase diode rectifiers feeding resistive loads are evaluated. The rectifier types discussed are I-phase half-wave rectifier and I-phase full-wave mid-point and bridge types . The performance parameters are then collated in tabular form.
3.8.1. Single-phase Half-wave Rectifier This rectifier, when feeding a resistive load, Fig. 3.14 (a), has waveforms for source voltage v s , output voltage Vo and output current i o in Fig. 3.14 (b). Its various performance parameters are obtained as under:
F rom Eq. (3.21), de output voltage, Vo =
Vm IT
. and! d e output current,
V
where 1m = ;
I = -~ =~
TtR
o
Tt
= maxi mum values of de current as shown in Fig. 3 .14 (b) .
Output de power, From Eq. (3.22), rms output voltage, and rms output current, Output power, Rectifier efficiency,
.. .(i)
v
m V. or=2
_ Vm _ [m 2.R - 2
I
or -
Pac
V mI ,.
= VOl' l or = -4P dr
TJ == p -=- = at;
FO :"lTI
factor,
V,,/m
- 21t
.. .(ii)
4.i
. ~l Y r'; l"l
== ~ Tt
V V FF= -E!.. =_2 y- -~ - .5708 Vo '). y'm, 2 ~
.
= 0.4053 or 40.53 %
80
Power Electronics
[Art. 3.8]
Ripple voltage, Voltage ripple factor, Also, Since source voltage
Vi
is a sine wave, its rms value, Vs
V
=~
Load current io waveform is the same as that of source current is' . . . I : . Rms value of source current, Is = rms value of output current, lor = ;. VA rating of transformer
:. Transformer utilization factor, TUF A TUF = 0.2865 means that VA rating of transformer is· i~F times the de power output. For a load of 100 watt, a transfo~er having a ratIng
ofO.;~~5 = 349.6 VA would be required .
PIV=-v2Vs =Vm Peak value of source current, Rms value of source current, Crest factor,
1m I s =1·or =_ 2 .
Isn 1m CF =-= =- x 2 = 2 Is 1m
3.8..2. Single-phase Full-w ave Mid-point Rectifier Its circuit diagram and various waveforms are shown in Fig. 3. 20. Its different performance parameters are obtained as under. From Eq. (3.48), de output voltage, and de output current,
Out put de power ,
R ms output voltage,
Rms output curren t,
Output ac power,
[Art. 3.8]
Diode "Circui s and Rectifiers 4
P
Rectifier efficiency,
T1 'I
2
81
8
= -Pdc =-2 V m I m ' --=-=08106 V I 2 . ac
Vor
m
1t
Vm
1t
m
1t
1t
FF=--Y=T2' 2V = iT2= 1.11
o m
Form factor,
2
2
1/ 2
V,=~v;,-v!=[(~ J _(2:," J 1 =0
Ripple voltage, Voltage ripple factor,
V 1t VRF=;- = 0.3077 Vm x 2 V o
3077
Vm
.
m
= 0.483
VRF = ~FF2 - 1 = ~1.112 -1 = 0.482
Also, TUF can be obtained as under:
V
Rms value of voltage for each secondary winding
=-#
. Note that current in
e~ch
secondary winding flows for half cycle only. . . I :. Rms value of current in each secondary winding
=; "
VA rating of secondary winding
= 2 [voltage rating of each secondary winding] x [current rating of each secondary winding] Vm 1m V m 1m . =2 X12x 2=~=0.707 V m 1m
Primary winding current is, how ever, made up of both positive and negative half cycles. :, Primary rms current Primary rms voltage Primary VA rating :. Average VA rating of transfo rmer = 0.5 +2°. TUF =
707
. V m1m = 0.6035 Vm 1m
. P dc 4 ' = --.2 . V m 1m average VA rating of transformer It-
PlV for each diode =2Vm Peak value of source current,
Isp
Rms value of source current, Is
X
1 - 0 672 0.6035 V m 1m - .
=1m 1m =T2
:. CF of input cur ent
3.8,3. Single-phas.e fuU-w Its circuit diagram is given in Fig. 3.21 (a). It is seen fr om Fig. 3.20 (6) and 3.21 (b) th at waveform of out ut (or load) type Vo and out current io are identic al in bo th ill! - 2 and B- 2 types of diode rectifiers. Therefor e, in aingle-phase B - 2 diode rectifi er also,
~1
Power Electronics
[Art. 3.8]
--~----~---------------------------------------------------------
2 Vm 21m Vm 1m Va = --11:-' I" = 11: ' Vor = 12 and lor = .J2 This shows that the rectifier efficiency, FF, ripple voltage V,., VRF are the 3ame for both types of diode rectifiers . However, PIVof diode in single-phase B- 2 rectifier is Vm whereas it is 2Vm in I -phase M - 2 rectifier. . Vm T UF: Rms value of source voltage Vs = -J2
Rms value of source current, VA rating of transformer
TUF=
4 2 8 . ' . ::-V 1 x--=-=08106 . VA rating of transformer 11: 2 m m V m 1m 11: 2 .
Pdt:
Source current waveforms for both types are identical, therefore CF = .f2. A comparison of three types of I-phase diode rectifiers discussed above is given in the table below where Vm = -Ii" V" H2re Vs = rms value of sinusoidal source voltage and f = source frequency in Hz. I
S.No.
1.
Half-wave (o r one-pulse)
Parameters
I
DC output voltage,
3.
I
Vcltage ripple factor, VRF
5.
Rectification efficiency, 11
I
i
Rms value of output
6,
Transformer utilization factor, TUF
7.
Peak inverse voltage, PIV
8.
Cr est factor, CF
9.
10 ,
Bridge (B- 2)
Centre-t,-zp (N!2) ~
'fm
"
Ivoltage, V o,. I Ripple voltage, V,.
4.
I
- Vm
V" 2.
Full-wav e (or Two pulse)
I
'2 V
- - in It
It
IT
Vm -
Vm
Vm
2
I
T2
"2-
0.3856 Vm
I
0.3077 Vm
0.3077 Vm
0.482
0.482
81.06%
81.06%
~I
1.211
40 .53% ,
I
0.2865
0.672
0.8106
Vm
2Vm
'11i !
2
~
-J2
Number of diodes
1
2
Ripple frequency
f
2f
I
-.
,
4
2f
It is seen from the above table that both u ll-wav e diode r ectifi ers (i)
are better th an the half-wave rectifier in so fa r as vol tage ripple factor, nctificaticn efficie ncy, T UF and cr est factor ar e concerned,
[Art. 3. ]
Di ode Circuits a nd Rectifiers
83
(ii) have average output voltage double of that of the half-wave rectifier (for the same
input voltage), (iii) have ripple frequency double of that of half-wave rectifier.
For both the full-wave rectifiers, the following is observed from the table . .
of B-2 rectifier is superior than the M - 2 type. Therefore, transformer required in M - 2 configuration is bulky and weighty. (ii ) PlV of diodes in B-2 rectifier is half of that of the diodes used in M ~ 2 rectifier. (iii) B-2 rectifier requires four diodes whereas M - 2 requires only two diodes (iv) Overall, a bridge rectifier using four diodes is more economicaL (i) TUF
.I
Example 3.15. A load of R = 60 n is fed from I-phase, 230 V, 50 Hz supply through a step-up transformer and then one diode . The transformer turns ratio is two. Find the VA rating of traTLsformer. Solution. The half-wave diode rectifier uses a step-up transformer therefore, ac voltage applied to rectifier = 230 x 2 = 460 V = Vs
Vm
Average value ofload voltage, Vo = p
Output dc power,
It
=
-J2 x 460 It
= 207.04 V
2
= V; = 207.04 = 714 .43 W de R 60
It is seen from the table that TUF for I-phase half-wave diode r ectifier is 0.2865.
= TUF Pde = 714.43 = 249365 VA 0.2865 .
:. VA rating of transformer
So choose a transformer with 2.5 kVA (next round figure) rating.
Example 3.16. A 230 V; 50 Hz supply is conne cted to a I-phase transformer which feeds a diode bridge as shown in F ig. 3.23 (a). Primary to secondary turns ratwIor transformer is 0.5 and load RL has a ripple free current 10 = lOA. Determine (i) average value of output voltage (ii) input current distortion fa ctor (iii) inp ut displacement factor DF (iv) input power factor (v) input current harmonic factor HF (o r T HD) and (vi) crest factor. Solution. Waveforms for supply voltage vs' constant load current io = 10 = 10 A and source current is are shown in Fig. 3.23 (b). Rms value of input voltage to bridge rectifi er,
~l = 2~0 = 0.5 2
s
vs = 230 = 460 V 0.5 The sourc e current, or input current, is can be expressed in Fourier series as under :
-
is::::: Ide
+
2. (ancos n oot + bn sin noot)
n'"
1,3,5
Here Ide = de v alue of source current
.. .(3.56)
Power Elec tronics
[Art. 3.8]
84
It can also be stated from the waveform of is that as the area of positive and negative half
cycle are equal,~average value of is i.e. Ide = O.
1 .2 TC an = - J io(t). cos nrot. cosnrot. d(rot) 1t
0
= -2
fTC
1t
0
21 10 cos nwt. d(wt) = - 0 nIt
tt
bn = 1. 1t
io
=~
sin nrot d(rot)
(t)
ISIn ' nwt Ilt0 =:: 0 It
0
r 0
lor a 11 n.
J:
10 , sin nwt. d(wt)
210 21 [- cos nrot]oIt = --- 0 [1 - cos n1t] nIt n.1t 410 for n = 1, 3, 5 ....... (for odd values of n)
n1t
=-
bn =:: 0
and
for
= 2, 4, 6 ....... (for even values of n)
Substituting the values of Ide' an and bn in Eq. (3.56), we get
L=:: 41oSinnrotand
=::tan-1[jL]=::0 , nIt n b n
.
~s
.
(l)
410
=1t
I
1. 1 .. 1. sm rot + 3" sm 3wt + 5 8m 5rot + "7 sm
[.
.
Average value of outp ut voltage, Vo ..~
2{2 x 460
2V
m =-I= t.1t
twt + ... ]i
= 414.08 V 41
(ii) Since fundame n t al component of input source current --..R sin rot is in phase with source 1t
voltage V m sin wt, the displacement angle
= O. Also, from
:. Input displacement fact or, DF = cos $1
= cos
0°
above, <1>1 = O.
= 1.
(iii) Rms value of fund am ental component of source current, lsI
~
of source current, I, [ I;: R
1'" ~ ~ 10 A. 10
410 . . lsI 1 Input current dIst ortIOnal factor, CDF = -I = --;;::-2 x -I s
(iu) In pu t
(v)
HF
(u i)
~
= ~o X :J2. A. Rms value
It."~
r~ [(019 1r~
0
= 12 x 2 = 0.9 1t
pf= CDF x DF = 0.9 xl = 0.90 (lagging)
THD
~
[ (
£" J' - 1
J' -
0.4843 or 48.43%
Crest fa ctor. H ere Jsp = 10 = 10 A and I s = 10 A.
..
10 CF= 10
= 1.00
;,., Examp le 3.17. A single-phase B ·2 diode rectifier is required to supply a de output voltage J of 230 V to a load of R = 10 n. De termine the diode ra ti ngs and transfo rmer rating required
for this configuration.
Solu ti on.
Average, or d e output voltage,
2V.7l 2{2. Vs
V0 = - = =230V 11:
'It
\
[Art. 3.9]
Diode Circuits and Rectifiers
:. Rms value of input voltage to rectifier
= transformer secondary voltage,
E
=232 Average load current,
I
2
R
r
1m sin rot. d(rot)
and rms value of diode current,
2555 V .
10
3613 A = -f2 x10255.5 =.
It is seen from the waveform of diode current diode current is 1t o
Vs
= Vo = 230 = 23 A o
· fd'od MaXlmum l e current, 1m = Vm R vaIue O
IDAv = 21
=
1t
85
from Fig. 3.23 (b) that average value of
=1m = 36.13 = 11.50 A 1t
IDr
iDl
1t
J:
=[ 2~ I~l sin
l/e
2
wt.
d(rot)
I
.
=; =36213 = 18.07 A
]
PN =-f2vs =-f2 x 255.5 = 361.3 V I
Transformer secondary current Transformer rating [Check:
.
= ~ = 3~3 = 25.55 A = Is =Vs Is = 255.5 X 25.55 = 6528 VA = 6.528 (kVA) Pdc = Vo 10 = 230 x 23 = 5290 W
:. Transformer rating
P de TUF
5290
= - - =- - = 6.530 VA = 6.53 kVA 1
0.81.
.
Thus, diode ratings are : I DAv = 11.50 A, I Dr = 18.07 A
Peak diode current, 1m = 36.13 A and PIV = 361~3 V and transformer rating = 6.528 kVA.
3.9. THREE-PHASE RECTIFIERS The highest possible value of average output voltage from a single-phase full wave (ectifier is 2V ml1t = 0.63662 V m' At the same time, single-phase rectifiers are suitable up to power loads of about 15 YW. For higher power demands. three-phase rectifiers are preferred due to the following reasons : (i) Higher dc voltage (ii) Better TUF (iii) Better input pf (iu) Less ripple contBnt in output current; therefore better load performance and (v) lower size of filter circuit parameters because of higher ripple frequency. Three-phase rectifier are classifIed as under: (a) Three-phase half-wave rectifier (b) Three-phase mid-point 6-pulse rectifier (c) Three-phase bridge rectifier and (d) Three-phase 12-pulse rectifier.
These are now described one after the other.
. 3.9.1. Thre ,e-PhaBe Half-wave Diode Rectifier C'r cuit diagram of a three-phase h alf-wave rectifier using three diodes is sho'wn in Fig. 3.29. It uses a 3-phase transfonn er with primary in delta and se condary in star. Th e primary in delta provides a path for the triplen harm onic currents. This st abilizes the voltages on the
86
Power Electronics
[Art. 3.9] vD I
+-
1.5 cc-
La 01
A
• p
to
A
+
•
C
Lb
02 R
b
1.e
1 J Vo
03
B
n
Fig. 3.29. Three-phase half-wave diode rectifier with common cathode arrangement.
secondary star. The three diodes 01, D2 and D3, one in each phase, have their cathodes connected together to common load R. Neutral is used to complete the path for the return of load current. As the cathodes of three diodes are connected together, circuit of Fig. 3.29 is also known as common-cathode circuit for a 3-phase half-wave rectifier. The three-phase supply voltage is shown as va (= Van' voltage between a and n), vb' Vc in Fig. 3.30 (a). "Us
Ve"
(a)
rt /2
0
"
---- ( b)
10 3
I
-'
...
"
5rr /2
. ,;
~;,"
. . . . --._.-
",,,,'<. .......
'.
/'"
",'",
Dl
. . . . . ----*"---~"" 02
03
-- Vb
(e )
Vmp
........ ,
.'*'...."
wt
",
. . . . . . _.
01
Uc .-'-'-'"
Va ~
......... ~
/
~ ...
/
0.5 Vmp
Vo ltage of termina l P -~
"
wt
Voltage of neutral n
tC~ ie La 1.b ic ta ( d)
o
I
Vmp
mp=1f' 17/ 2
"
Tl
0
wt
. .151LD1
(e)· ~ . ~_~__ la=i5=___ iO1 __________~______~~L-~ _________ _
_____ wt
I
~---------------2rr=36 0 '--------------~'
150 '
( I)
270 '
O r--.----------~~--~------_r~---------,------------~---w--t
Fig. 3. 3'0 (a ) Line t o eutral so r ee voltages (6) dio de conduction (c) loa d volt age sourc e cur r ent and (f) volta ge across diode D l.
(d )
load current (e )
81
[Art. 3. 1)]
Diode Circuits an d Rec tifiers
The rectifier element connected to the line at the highest positive instantaneous voltage can only conduct. In Fig. 3.29, a diode with the highest positive voltage will begin to conduct at the cross-over points of the three-phase supply. It is seen fro m Fig. 3.30 (a) that diode Dl will conduct for CDt = 30° to CDt = 1500 as this diode senses the most positive voltage :Jet' as compared to the other two diodes, during the interval. Diode D2 will conduct from wt = 150° to 270 0 and diode D3 from CDt = 270° to 390°. The conduction of diodes in proper sequence is shown in Fig. 3.30 (b). When a diode is conducting, the common cathode terminal P rises to the highest positive voltage of that phase and the other two blocking diodes are reverse biased. The voltage Vo across the load follows the positive supply voltage envelope and ha.s the waveform as shown in Fig. 3.30 (c). It should be noted that voltage of the neutral point 'n' is taken as zero and is given by the reference line wt. The voltage of point P in Fig. 3.29 is shown by va' Vb' Vc etc above the reference line in Fig. 3.30 (c). The de load voltage v(J varies between V mp ( = maximum phase voltage) and 0.5 V mp' It is observed that for one cycle of supply voltage, output voltage has three pulses, the circuit of Fig. 3.29 can therefore be called a 3-phase 3-pulse diode rectifier or 3-phase half-wave diode rectifier. Voltage variation across diode D1 can be obtained by applying KVL to the loop consis ting of D1, phase 'a' winding and load R. So VDl - Va + Vo = 0 or VDl = Va - Vo When diode D1 conducts, CDt = 30° to CDt = 150° in Fig. 3.30 (j).
Vo
= v o'
..
vDl
= Va -
Va
= O.
This is
shown from
wpe.n diode D2 conducts, Vo = Vb. .. vnl = Va - Vb wt = 180°, vb= 0.866 V mp and va = 0,:. vDl = 0 - 0.866 V mp = - 0.866 V mp CDt = 210°, Vb = V mp and Va = - 0.5 V mp ' :. vDl = (- 0.5 -1.0) V mp = - l.5 V mp CDt = 240°, Vb = 0.866 V mp and Va = - 0.866 V mp ,:' vDl = (0 - 0.866 - 0.866) V mp = - \f3Vmp CDt = 270°, Vb = 0.5 V mp and Va = - V mp • :. vDl = (- 1 - 0.5) = - 1.5 V mp ' When D3 conducts, V Dl = va - Vc and variation of voltage unl from wt = 270 to wt = 390 is
At At At At
0
0
obtained as outlined before. It is seen from Fig. 3.30 (f) that peak inverse voltage across diode D1 is ,[3' V mp ; in general prv = '1/3 V mp for each of the three diodes D1, D2 and D3. AB in single-phase rectifiers, the average output voltage in a 3-phase diode rectifier can be obtained by considering the output voltage over one periodic cycle.
For a 3-phase diode rectifier of Fig. 3.29, the peri.odicity is 120° or 27t/3 radians as per Fig. 3.30 (c). Here the output volt age comprises of phase voltages Va' Vb, Vc and its average (or m ean) value V o is given by V = Are a over one periodicity (shown cross-hatched in F ig. 3.30 (c) o P eriodicity .
= p ena.~.lCl't y
r at
Va •
d (CDt)
In th e above expression, va is zero at CDt = 0; t herefore, V mp sin wt is written for v~ . Further Va appears from OJt = 30° to 150° in the ou tput voltage waveform; these are, therefore, the limits of integration and periodicity is 120 0 = 21t/3 r adians. 3 f5it/6 . 3\f3 3-16 . 3 .. (3 71) Vo = -2 Vmp sm CDt. d (CDt) = -2- V mp = - 2- Vph = -2 . V ml n rr~ n 1t 7t where V mp = maximum val e of phase voltage, ~ = -f2 Vph and V mi = m aximum value of line voltage, VI = € . V mp = '1/6. Vph
88
Power Electronics
[Art. 3.9]
=[ 2 1 3 j/t/6 (Vmp sin wtf d(wt) ]112
Rms value of out put volt age, Vor
1t/
l!/6
=[ 21t 3 V~p I t _ x 2' w
5/t/6
sin 2wt 2
]112 = 0.84068 V mp I
I
... (3.72)
/t/6
Ripple voltage,
VI'
= -lv~1' - V~ = V mp -lO.84068 Vr VRF = Va
2
-
0.827
2
= 0.151
v.nip
0.151
= 0.827 = 0.1826 or 18.26%
FF = VOl' Va
= 0.84068 = 1.0165 0.827
Rms value of output current, lor = ~r
= 0.8~068
V mp
= 0.84068 Imp
V mp where Imp = R = peak value of load, or output, current.
313
313
Pdc = Va 10 = 21t V mp' 21t Imp Pac
=Vor . lor = (0.84068)2 V mp Imp =Pde = [ 313 )2x
Rectifier effid en cy
Pac
21t
1 = 0.96765 0.84068 2
or 96.765%
During the interval wt = 30 0 to wt = 150°, source current is = ia and the periodicity of source current is 21t radian s as shown in Fig. 3.30 (e).
=[21
:. RMS value of source current, Is Rms value of source voltage, Vs
1t
[/t/6
/t/6
(Imp siD: wt)2 d (wt)
]1!2 = 0.4854 Imp
V
mp =T2 = 0.707 V mp
Transformer h as three phases, therefore,
= 3 Vs Is = 3 x 0.707 V mp x 0.4854 Imp = 1.0295 V mp I mp
VA rating of t r ansform er
DC output power, ..
Pd ,
~
TUF =
[
3;;
J
x V mp Imp
~ 0.684 Vmp Imp
. Pdc Transformer VA ratmg
As stated before , Pl V for each diode
= 0.684 Vmp . Imp = 0.6 644 1.0295 V mp . Imp
=-{3 v mp
Following observations can be made fr om th e above analysis. E ach diod e conducts for 1200 only. (ii) Th ese are three pulses of output voltage, or output current, dur ing one cycle of input voltage. It is, therefore, calle 3-ph ase t hree-pul e diode r ectifier. (i i i) Current in th e transfor mer secondary is unidir ectional, t herefore, de exists in the transform er second ary current. As a result , transfor mer cor e get s saturat ed leading to more iron losses and redu ced effi ciency. (i)
[Art. 3.9]
Diode Circuits and Rectifiers
89
3.9.2. Three-phase Mid-p oint 6-pulse Diode Rectifier This rectifier is also called six-phase half-wave diode rectifier or three-phase M-6 diode rectifier. It is seen from the previous section that the performance of three-phase three-pulse diode rectifier is better than the single-phase two-pulse diode rectifier so far as magnitudes of average output voltage and ripple content are concerned. From this, it appears to be logical to think that a rectifier with more number of pulses per cycle would give an overall improved performance. 1.b2
A 0---..,----.......
06
b2
A
c u - - - -__
Cl
R B~----------~
Vo
1
02
ic 1
1
05
Fig. 3.31. Three-phase mid-point 6-pulse diode rectifier . .
Fig. 3.31 shows a three-phase mid-point 6-pulse rectifier using six diodes. A three-phase transformer with primary in delta and secondary in double-star. is used. One diode in each phase is connected as shown. Note that secondary of each phase winding is in two halves. The mid-points of the three secondary windings are connected to form the neutral n. Six-phase supply is available from six terminals aI' C2, b l , a2, CI and b2 · Phase voltages Val, Vbl> Vel are phase-displaced by 120°, similarly V aZ , V bZ , Vcz are displaced by 1200 • But voltages Val and VaZ are out of phase by 180 0 • However; Val, V cZ , are out of phase by 60 as shown in Fig. 3.32. . th - V ' 60 0 Th erefore, if v a l -- V mp sID rot , env c2- mpsm(rot - )
'1b2
vel
0
vbI
= V mp sin
(rot -120°) , va2
= V mp sin (rot vb2 = V mp sin (cDt Vcl
240°) = -
= Vmp sin (rot V mp
F i g. 3.32. Six-phase ' CI 331 vo It ages fior F lb' . .
180°) = -
sin (rot - 60°) = -
Val
Vc2
300°) =- vbl Here V mp = maximum value of per phase voltage The waveform of six-phase voltages , Val' VcZ, Vbl .... are sketched in Fig. 3.33 (a). As before, a diode sensing the highest positive anode potential gets forward biased and conducts. Therefore, from wt = 0° to wt = 60 0 , voltage Vb2 is the highest positive , therefore, diode D6 conducts; from wt = 60° to wt = 120, diode. Dl conducts and so on, Fig. 3.33 (a) and (b). Each diode conducts fo r 6 ". It is seen fr om Fig. 3.33 (c) that load voltage Vo is m ade up of Vb2 from wt = 0° t o 60° ; Val from rot = 6 0 to 120 0 and so on. Also, Va varies between Vmp and 0.86 6 V m p ' Periodicity of output voltage 1,'0 is 60°. and
90
Power Electronics
[Art. 3.9]
wI
( b)
'--'0
06
01
vb 2
VOl
I
02 vn
03
D4
05
06
01
vbl
v a2
vel
ub2
val
02 Ve 2
(c)
0 wI Lo
1.b2
i.(2
tal
i 02
t bl
lb2
te I
Lal
te2
(d)
rn "1'" 0
60
0
=Lal ( e)
42 0 0
1200
ri"'
,",.i,.iol
0
VD I~
!.
211' =360 180
0
wI
0
0
270 0
'"'
wI
277
(0 -0.866 Vmp
Fig. 3.33. (a) Line to neutral (or phase) source voltages (b) diode conduction (c) load voltage (d ) load current (e) source current and (f> voltage across diode Dl for 3-phase M - 6 diode rectifier of Fig. 3.31.
Average output voltage, __ 1_ /3
frt /3
Vo -
1t Rms value of output voltage,
rc/3
.
.
_
V mp sm rot. d(rot) -
3 V mp
...(3.73)
1t 112
V or =[
1
1t
/3
=[ 3V21t
frt / 3
mp
rt/ 3
2
.
(Vmpsmrot) d(rot)
{~_ sin 240 3
•\ 112
0 -
2
sin 120
0
1 ]. J 2
Ripple voltage,
1t
o
FF = ~r o
= 0. 95~8 x 1t =1. 009 .
095'"8 V . U mp
1/2
V, =~~, - ~ =V O.95582_( ~ I ] V ) VRF = V = 0 . 04~8 x =0.043 or 4 .3% mp [
=
= 0.0408 Vmp
... (3.74)
(Art. 3.9]
ectiflers Val"
Rms value of output current, lor = If::::
Pd,
0.9558
R
V mp
91
=0.9558 Imp
J
~ V, I, ~( ~ Vm~lmp
Pac = Vorior =(0.9558) V mp Imp
Rectifier efficiency
= ( -3 J2 X 1t
1 2 = 0.9982 or 99.82% 0.9558 .
The source current is in phase a l has the same waveform as for the current ial , Fig. 3.33 (d) and (e). It is seen that periodicity of is is 21t radians.
:. Rms value of source current, I, VA rating of transformer DC output power,
~[2~ ( : ' (Imp sin wt)'. d (wi) Vm
f' ~
0.39 Imp
.
= 6 Vs Is = 6 ~ x 0.39 Imp = 1.655 V mp Imp 3V
P dc
TUF
3
mp =Vola =-1tX -Imp = 0.912 V mp Imp
1t
P
0.912
'.
·5""
= vn ratmgdc0 f trans.f = -655 = 0.551 or 1. lTA
•
::> .1%.
Voltage variation across any diode, say Dl, can be obtained as done in 3-phase 3-pulse diode \ rectifier. Therefore, VDl = Val - Vo 0 When D1 conducts, va = Va l ' :. VDl = Val - Val = 0 from wt = 60 to 120°. When D2 conducts from wt = 120 0 to 180°, va = vc2 ' :. vDl = Val - vc2 At wt = 150°, vc2 = ¥mp an d Lla l = 0.5 V mp ; :. vDl =(0.5 - 1.0) V mp = - 0.5 V mp At wt =It, vc2 = 0.866 V mp and Val = 0, ; . VDl =- 0.866 V mp 0 When D3 conducts from wt = 180 to 240°, Vo =Vb I, :. VDl = Val - Vbl At wt = 240°, Vbl = 0.866 V mp and Val =- 0.866 V mp ' :. vDl =(- 0.866 - 0.866) V mp = - -{3 V mp When D4 conducts from wt = 2400 to 300 0 Vo = va2 :. vDl = Val - Va2 At wt = 270°, Val = - V mp and va2 = V mp' :. vDl = (-1- 1) V mp =- 2 V mp
Similarly, vDl waveform can be obtained when diodes D5, D6 conduct, this is shown in Fig. 3.33 {f;. It is seen from vDl waveform that PIV for each diode = 2 V mp ' The above analysis reveals the following. Quality of output is sup erior as compared to 3 pulse rectifier, because RF is 4.3% and FF is close to unity. (ii) TUF is poor as compared to 3 pulse rectifier ; it is because of lower value of conduction angle (= 60°) for each phase and diode of this rectifier. (iii) Output frequency is 6f; siz e of filter, if required, is therefore reduced. It may be observed from Fig. 3.33 that each phas e winding carries un directional current and there are six pulses during one cycle of sourc e volt age. That is why it is called six-phase half-wave diod e rectifier. At any time, only one secondary phase winding, say phase a 1, carries current, this gets reflected downward in the primary delta for m mf balance. After furth er c.ut = 1800 , secondary-phase ,
(i)
92
Power Electronics
[Art. 3.9]
win ding a2 carries current, this gets reflected upward in the primary delta. This shows that both secondary and primary windings handle alternating current during one cycle of source voltage; there is therefore no magnetic saturation of the transfcrmer core as it is in the transformer used in 3-phase half-wave rectifier. ~ 3.9.3. Multiphase Diode Rectifier For three-phase three-pulse rectifier, or three-phase half-wave rectifier, each phase conducts for 27t/ 3 radians of a cycle of 21t radians. Three-phase M-6 rectifier may be considered as 6-phase half-wave rectifier and it is seen that each phase of this type conducts for 27t / 6 rad. In general, in m-phase half-wave diode rectifier, each phase and diode would conduct for 2 1tl m rad and number of output voltage pulses p would be equal to number of phases m. For m more than three, an m-phase diode rectifier would have delta-connected primary and the secondary would have mid-tapped m/2 windings . The number of diodes is equal to m. Fig. 3.34 shows a few pulses of output voltage waveform for m-pulse half-wave diode rectifier. Each phase conducts for 21tl m or 21tl p radians, because number of pulses p = number of phases m for half-wave rectifiers. With time origin AA' taken at the peak value of output voltage in Fig. 3.34, the instantaneous phase voltage is v == Vmp cos rot = "-f2 Vp h ' cos rot where Vph = rms valu e of per-phase supply voltage . Waveform of output voltage Vo in Fig. 3.34 shows that in m-phase half-wave diode rectifier, conduction occurs from - ~ to ~, or from - ~ to 7tlp with time origin at AA' and periodicity m m p . is 27tl m, or 27tl p, radians. f-_K
-!P ,I
kR--" A/
P
.
i
~ I,
! /::: ! v ~ :
A
:
c..!t
~271/P--1 '
.
Fig. 3.34. Output voltage waveform for m-phase half-wave diode rectifier.
1
:. Average value of output voltage, Yo = -2-
[ IP
1tl p _ re l p
Y mp
cos rot d (rot)
= Ymp . E.. sin ~ 1t p Rms value of output voltage,
Y or = [ L2
[
1t
IP
.. .(3.75 )
(Ymp cos rot) 2. d (rot)
]112
- 'relp
y2 [IP = P' mp (1 + cos 2rot ) d (rot) [ 41t _ 'itlp
]
1 2)J
=V mP1rL21t (~p + -2 sin ~ p
...(3 .76)
= V mp : 1 R mp
Maximum value of load current Aver age value of diode current,
, 112
I
1
D
=~
fTC IP - rtlp
1~ cos cot. d
(rot )
=
Imp 1t
1t
sin
P
... (3 .77 )
[Art. 3.9]
Diode Circuits and Rectifiers
Rmsvalue of diode current, IDr =[2. 11t r-iP
(Imp
cos wt)2
d (wt)
93
]1/2
ltl p
./1\ 'Y'
~c..
(J
=1
\L.
mp
[ =1 ( !!. + -1 sin .J!. 2 21t p 2 P
J]1/2
... (3.78 )
~
Example 3.18.A step-down delta-star transformer, wiih per-phase turns ratio 5, is fed from 3-p hase, 1100 V, 50 Hz source. The secondary of this transformer, th rough a rectifier, feeds a load R = 10 n. Calculate the average value ofoutput voltage, average and rms values of diode current and power delivered to load in cas'e the rectifier is (a) 3-phase, 3-pulse type and (b) 3-phase M-6 type. Solution. (a) 3-phase three-pulse type.' Per-phase secondary voltage
1100 _'"
Vph == - ·-5- = 220 V and V mp = '12 X 220 V. From Eq. (3.71), or from Eq. (3.75) withp == 3, average value of output voltage, 3-{3 3,(3 . Vo;::;: 21t V mp == 21t X ~ x 22 0 = 257.3 V Maximum value of load current, V"' · 1 = .:..!!!i!. == '12 x 220 mp R 10
. =..J2 x 22 A
From Eq. (3.77), average value of diode current, ID = 22: -f2' sin From Eq. (3.78), rms value of diode current, lIn
~ 22 xv'2[ ;n(~+ sin;20°
Jr
~ = 8.575 A
= 15.10 A
···
From Eq. (3 .76), rms value of output voltage,
0
sin 2120 3 (1t Vor == 220 x..J2 [ 21t - "3 +
Power delivered to load
==
;r
==
J]112 = 2.61.52 V
26~g22 == 6839. 3 watts
(b) 3-phase M- 6 type.' P er-phase secondary voltage,
Vph
220
="'"2 = lIOV and V mp =..J2 x llO V
From Eq. (3.73), or fr om Eq. (3."';) withp == 6, average outpu t voltage is
Vo
=~1t x -Y2 x 110 =148.53 V
Maximum value of load current,
I
mp
= VRmp = ..J2 10 x 110 = 'if2 X
From Eq. (3 .77), average value of diode current, ID
11 A
= -{2 x 11 ::;in 11:/ 6 = 2.4755 A 11:
94
Power Electronics
[Art. 3.9J
I l6·
. 1 ( 7t sin 60 Fnm Eq. (3.78), rrns value of diode current, In,. == 11 x --f2 L 27t + 2 From Eq. (3.76), rms value of output voltage, V
Power delivered to load
~ = 110 x '1'2[ ;n (~ + sin;oo ) =
;r
=
)]112
= 6.069 A
r;
148.66 V
14~~62 = 2209.98 = 2210 W
3.9.4. Evolution of Three-phase Bridge R ectifier Before studying the 3-phase bridge rectifier, let us first examine the evolution of this rectifier type. . Three-phase half-wave diode rectifier, with common cathode configuration, has already been discussed in Art. 3.9.1. For the circuit of Fig. 3.29, the conduction of diodes Dl, D2, D3 indicated in Fig. 3.30 (b), is again shown in Fig. 3.36 (b), just for the sake of convenience in understanding the evolution of 3-phase bridge rectifier. . ', • . Consider now a three-ph&se h alf-wave diode rectifier with common-anode arrangement as shown in Fig. 3.35. In this circuit, diode will conduct only during the most negative part ofthe supply voltage cycle. This rrieans that a diode will conduct when the neutral i,s positive with respect to terminal a, b or c. Therefore, for the supply waveform of Fig. 3.36 (a), qiode D5 would conduct from CDt = 0° to CDt == 90° as the voltage Vb is the most negative for this m terval. From CDt = 90° to CDt = 210°, voltage Vc is the most negative, therefore diode D6 must cO,n duct during this interval as shown in Fig. 3.36 (c). Similarly, diod~ D4 would conduct from CDt = 210° to CDt = 330 0 and so on. E ach diode conducts for 1200 (as in common-cathode configu'ration). The load voltage waveform v o' der ived from Fig. 3.36 (a), follows the negative supply voltage envelope as shown in Fig. 3.36 (d) for the diode configuration of Fit;>. 3.35. Voltage of neutral n is fixed at zero by th e reference r ne rot in Fig. 3.36 (d). Th e voltage orterminal Q of Fig. 3.35 is shown by vb, V c ' Va et c below th e reference line in Fig. 3.36 (d) . \ . .
a
DL. A
0------..... 0
A
05
c o--
,..."
R
b
06 I
8 O-------------~
n
1 Vo
1
Fig. 3.S5. Three-phase half-wave diode rectifie with common anode arrangement.
The three-phase half-wave rectifier circuits of Figs. 3.29 and 3.35 can be connected in series ' as shown in F:ig. 3.37 (a). An examination of this series connected circuit r eveals that load current can exist even without neutral wire n. Fo e ampl e, when diode Dl is conducting from wt = 30° to W: = 150° in Fig. 3.36 (b ), the return path for the curren t is through diod e D5 fr om wt 30 C to 90° and through dio e D6 from cot = 90 C to 1500 I s ee F ig. 3.36 (b) and (c) . Supply point
=
95
[A r t. 3.9]
Diode Circuits a nd Rectifiers
(a)
( b)
1 03
(c)
I
01
05
D5
va
03
D2
IT
12
06
05
OL.
Vo ltage oi neut ral n
0
01
Vol tage of terminal Q
2rr
1T
(d)
ub
Dc
Fig. 3 .36. (a) Line to neutral source voltages (b) diode conduction for
Fig. 3.29, (e) diode conduction for Fig. 3.35, (d) load voltage.
'a' connected to the anode of Dl is the same as that connected to the cathode of diode D4. The
neutral wire can thus be eliminated and cathode terminal of D4 can· be connected to anode of Dl. Thus, the circuit of Fig. 3.37 (a) can be redrawn as shown in Fig. 3.37 (b). This circuit can further be rearranged to that shown in Fig. 3.37 (c). The only difference between Fig. 3.37 (a) and Fig. 3.37 (b) and (c) is that load voltage is equal to line to neutral voltage in Fig. 3.37 (a) and it is line to line volt age in Fig. 3.37 (b) and (c). The circuit configuration shown in Fig. 3.37 (c) is called 3-phase full waue bridge rectifier, or 3-phase six-puLse bridge rectifier. Note that diodes Dl, D2, D3 of the bridge would conduct when supply voltage is the most positive, whereas diodes D4, D5, D6 would conduct when supply voltage is the most negative. Diodes Dl, D2, D3 may therefore be called a positive diode group and D4, D5, D6 a negative diode group. The voltage across load would always be the direct emf with the polarity of P positive and that of Q negative as indic a ed. 01 00-
02 b
03 c
C
01
b
02 p
p
c
n
D3 0
L 0
b
A
D
A
0
c·
L 0
b
A
c'
D
D2
D6
06 c
Q (a )
(b)
(c 1
+
05
-t
L
(d)
F ig. 3.3'7. Evol ution of 3-phase six-pulse rectifi er (a ) circuits of F igs. 3.29 and 3.35
connected in series, (b) circuit obtaine d fro m (a ), (e) circuit of
(b) rearranged and (d) diode numbering scheme alt ered.
I~
96
Power Electronics
IArt. 3.9]
It may be seen from Fig. 3.36 (b) and (c) that conduction sequence of diodes is D1 (from positive group), D6 (from negative group), D2, D4, D3, D5 etc. In order that sequence of conducting diodes is D1, D2, D3, D4, D5, D6, D1. .... (easy to remember), circuit of Fig. 3.37 (c) is redrawn in Fig. 3.37 (d) with diodes numbered D1, D3, D5 for positive group and D4 (1 + 3), D6 (3 + 3), D2 (5 + 3 - 6) for the negative group. Line to neutral, or phase, voltages of Fig. 3.30 (a) or 3.36 (a) are redrawn in Fig. 3.38 (a) as va, Vb' vC ' Fig. 3.36 (b) and Fig. 3.36 (c) are combined and drawn in Fig. 3.38 (b) but with diode- numbering scheme of Fig. 3.37 (d) . It is seen that for wt = 0 to 30°, diodes D5, D6 conduct together, for wt = 30° to 90°, diodes D1, D6 conduct together and so on. Each diode conducts for 120°. At the instant marked 1 (when wt = 30°), diode D6 is already on, conduction of diode D5 stops and that of diode D 1 begins. The magnitude ofload voltage VI at instant 1 is, therefore, given by
= V rnp sin 30 (from .va) + V mp sin 90° (from Vb) = 1.5 Vm~~''"' At the instant marked 2, the load voltage has a magnitude ofV2 = V mp sin 60 + V rnp sin 60 VI
=f3vmp ' At the instant marked 3, Here
V mp
V3
= 1.5 V rnp
= maximum value of phase
(or line to neutral) voltage.
The voltage of the load terminals P and Q of Fig. 3.37 (c) or (d) is shown in Fig. 3.38 (a). This figure also reveals that at the instants marked 2, 4, 6, 8, 10 etc, the load voltage has a magnitude off3Vmp ' At the instants marked 1, 3,5,7,9, 11 etc, the magnitude ofload voltage is 1.5 V mp' The load voltage, or the rectified output volt age, Va can therefore, be plotted as shown in Fig. 3.38 (c). In this figure, voltage ofterminal Q is shown at zero potential by straight reference line wt, whereas the potential of terminal P is shown by line voltages v cb' vab' Vac etc. In fact, if voltage waveform of terminal Q in Fig. 3.38 (a) is made a straight li~~, Fig. 3.38 (c) Vo ltage of terminal P
I wt (a ) Voltage of ter minal Q
D1
I (c) /
"
;' /w, i
30'
/
I
04
02
06
+ve group
I 02
-ve grou p
./
/ '
/
01
05
03
,":2:3 3C' '4
1.5 Vmp
.JJV m p
(.
I
6
1 cycle =277 =360
a 9 0
I 10
11
1'2
wt
.1
Fig. 3.38. (a ) 3-phase input voltage waveforms (b) conduction sequence of positive an negative group of diodes (c ) output vo ta e wav eform of 3-phase six-pulse diode bridge of Fig. 3.37 (d) .
Diod
[Art. 3.9]
Circuits and Rectifiers
97
is obtained. It should be remembered that in Fig. 3.38 (c); vab, v ae ' v bc et c are line voltages, whereas in Fig. 3.38 (a) ; Va' Vb, Vc are phase voltages. The dual subscript ab in vab should be taken to denote that as per the first subscript 'a', diode connected to phase termin al 'a' from positive group, i.e . D1 conducts. Asper the second subscript 'b', diode connected to phase terminal 'b' from the negative group, i.e. D6 conducts. For example, for output voltage Vcb, diode D5 from positive group and diode D6 from negative group conduct. Note that each diode conducts for 120°. Fig. 3.38 (c) reveals that there are six pulses for one cycle of supply voltage. Thus , three~phase bridge rectifier of Fig. 3.37 (d) can be called 3~phase six~phase diode rectifier or 3-phase B-6 diode rectifier. Here B denotes bridge and 6 denotes the n umber of output~voltage pulses per cycle of source voltage.
c.) -1-
It is thus seen from above that when two 3~phase 3~pulse rectifiers are connected in antiparallel, a 3~phase 6 pulse rectifier (or a 3~phase bridge rectifier) is evolved.
;;f.v
3.9.5. Three-phase Bridge Rectifier
Power circuit diagram for a 3~phase bridge rectifier using six diodes in shown in Fig. 3.39. The diodes are arranged in three legs. Each leg has two series-connected diodes. Upper diodes Dl, D3, D5 constitute the positive group of diodes. The lower diodes D2, D4, D6 form the negative group of ·diodes. The three~phase transformer feeding,.the bridge is connected in delta-star. This rectifier is also caned 3~phase 6~pulse diode rectifier, 3-phase full-wave diode rectifier, or three-phase B-6 diode rectifier. . . Positive group of diodes conduct when these have the most positiv e anode . Similarly, . n egativE) group of diodes would conduct if these have the most n egative anode. In other words, diodes D1, D3, D5, forming positive group, would conduct when these exper ience the highest posi tive voltage. Likewise, diodes D2, D4, D6 would conduct when these are subjected to the m ost negative voltage.
+ 03
A 0 > - - - -....
05
lo
R
A
Co--~....
06
02
Bo-------------~
Fig. 3.39. Three-phase bridge rectifier using diodes.
It is seen from the source voltage waveform V s of Fig. 3.40 (a ) that from rot = 30° to 1500 , voltage va is more positiv e than the voltages v b' vc' Therefore, diode D1 connected to line 'a' (as p.er subscript 'a' in va) counducts during the int erval rot = 30° to 150°. Likewise, from rot = 150° to 27 0°, voltage vb is m or e positive as compared to va) V c; therefo e, diode D3 connect ed to line 'b ' (as per the subscript 'b' in Vb) conducts during this interval. Similarly, diode D5 from th e positive group co .ducts fr om cut = 27 00 t o 390 0 and so on. N ote als o that from (ct ,. 0 to 30° , Vc is the most positive, therefor e, diode D5 from t he positive group conducts for this
98
Power Electronics
[Art. 3.9]
interval. Conduction of positive group diodes is shown in Fig. 3.40 (b) as DS, Dl, D3, DS,Dl etc. Voltage ve is the most negative from rot =90 0 to 210 0 • Therefore, negative group diode D2 connected to line 'e' (as per subscript 'e' in ve ) conducts during this interval.Similarly, diode D4 conducts from 210 0 to 330 0 and diode D6 from 330 0 to 450 0 and so on. Note also that from rot = 00 to 90 0 , Vb is the most negative, therefore diode D6 conducts during this interval. Conduction of negative group diodes is shown as D6, D2, D4, D6, etc in Fig. 3.40 (b). During the interval rot = 00 to 300 , it is seen from Fig. 3.40 (b) that diode D5 and D6 conduct. Fig. 3.39 shows that conduction of D5 connects load terminal P to line terminal e; similarly, conduction of D6 connects load ter minal Q to line terminal b. As a ·result, load voltage is Vpq = Vo = line voltage veb (first positiv~ subscript corresponding to e and second negative subscript corresponding to b) from rot = 00 to rot = 300 • Likewise, during rot =30~ to. ,90 0 , diodes Dl and D6 conduct. Conduction of diode D1 connects P to a and D6 to b. Therefore, load voltage
(al .
01
(b)
03
02 Vab
Vae
.
01
05 OL.
06
+ 'Ie group 102 -'Ie group
liae
Vbc
(c)
,
,, I
o
217
ia or is .
wt
(0)
--130· ~30~
i..Do'l
1 T - - - -,
~I
. L I~
~06 01 ~ 01
oz--l
21T
wI
30'
O r-~-----------'--------~~----------~~--------~-w~t (t)
.,
(t
. Fig. 3.40. Three-phase diode-bridge rectifier 3-phase inpu t voltage waveform (b) conduction sequence of diodes (c) output voltage waveform . (d) input current waveform (e) diode current wave onn through Dl and (j) vo tage variation across cHode Dl .
[Art. 3.9]
Diode Circuits and Rectifiers
99
during this interval ·is Vo = line voltage vab' Similarly, for interval 90° to 1500 , diodes D1 and D2 conduct and Vo = line voltage Vac; for interval 150° to 210°,.diodes D3 and D2 conduct and Vo = line voltagevbc and so on . Output, or load, voltage waveform is drawn by a thick curve in Fig. 3.40 (c). Average value of load voltage, Vo
= peno . 1d'lClty . J~ vab . d a
(cut)
1
It is seen from Fig. 3.40 (c) that the value is 60 0 or 1t/3 rad.
OfVab
at rot = 0 is V ml . sin30° and its periodicity
.. .(3.79 ) 1t
1t
1t
where V ml = maximum valu e ofline voltage V, = rms value of line voltage and Vp
= rms value of phase voltage.
Average value of output voltage, VOl can also be obtained as und~r.: .
to 120°. It is because the voltage7 pulse area required extends from rot = 60 0 to rot = 120 0 for the sin rot function. 3V 3 Vo = V ml . sin cut. d (rot) = ~ ... (3.79) 1t ~3 1t
(i) Take any sinusoidal wave and integrate it from 60°
j7t/3
(ii) For a cosine function cos cut, voltage pulse of 60° duration extends 30° to the left of its peak and 30° to t he right of its peak. . .. 3 _ -Vo=-
[16 Vmlcos(j)t . 3V . .d(cut)=~
1t -~6
...(3 .79)
1t
[~ j7t/3~l sin2cut. d (rot) ]112
Rmsvalue of output voltage, V or=
1t lt/3
~ = [
.Bipple voltage, Voltage r ipple factor,
V,
27t/3]1/2
;1 I rot - s:n: I .
3 2
cut
.
lt/ 3
~ "V;, - V; ~[ 0.9558 ~ )' 2
- (
r
r - 0.0408 - 0 042 7 - -V 497C1.y O VRF -: V - 3 /1t - . or.... o
FF = Vor = 0. 9558 V mt = 1.0009 Vo ~V 1t
rol
Rm value of output current, l or = ;r = 0_ 9~58 V m l = 0.9558 1m l
= 0.9558 Vml Vml
...(3.80)
~ 0.0408 V
ml
Power Elec.tronics
[Art. 3.9]
100
2
=(~ 1V
P d<= V,I.
Pac = Yo,.
'
ml : Iml
(6.9558) V ml .Iml
Vor =
", =, (3)21 ' "1t x .O.9558 =0.9982 or 99.82%
Rectifier efficiency
V
' it = {:3v 'R
mp. It is seen from
For a resistive load, peak current through each diode is Im[=
the waveform of line current iu (or transformer secondary current is) that (0 periodicity of this current is 1t rad. and (ii) this current has two pulses, each of 60° durati~n, for each periodicity of 1t rad . .. Rms
valu~
of line current
= rms Is = ,
Transformer VA rating,
value of transformer secondary current
r2rr / 3
2
I - JI
I[ 1t
2
. 2
Iml sm rot . d (rot)
rr/ 3
.
]li2 = 0.7804 ,, Imt , ,
Vml ' =3Vs I s = 3 x ~ x 0.7804 I ml
'
2
..
TUF =
de
P ' VAra~ingoftransformer
... (3.81) ",
' "
=(,2.) x 3 x fa , =0.9541 0.7804 1t
It is seen from diode-current waveform of Fig. 3.40 (d ) that average value ofdiode current . . . ~
is
2 ]27t13 ' . I D =-2 Im1.sm 1t
Rms value of diode current, I Dr =[ 22
1t
rr/ 3
tJ(/ 3 rr/ 3
rotd(rot)
"
I~ll sin
2
lml =
rot . d (rot) '
.~ .(3~. 82)
7t
]112= 0.582 lml
...(3.83)
"
A conducting diode has zero voltage drop across it. Let it be required to sketch in Fig. 3.40 If), the variation of voltage across diode D1 belonging to positive group in Fig. 3.39. For de output voltage Vab' vae ; diode Dl conducts, therefore, ' voltage across diode D1 is " zero, i.e. vDl = 0 from rot = 30° to 150° and this is shown in Fig. 3.40 (f). After rot :: 150°,diode D3 conducts for 120 0 with output voltage vbc' Vba ' Now cathode of Dl gets conriected to supply terminal b through conducting diode D3 for a period of 120 0 , whereas its anode is already connected to supply terminal 'a'. Therefore, voltage across D1 from wt = 150° to 270.0 is , 9 v Dl = va - V b as per KVL CVm - Va + Vb = 0). This voltage Vab during interval of rot ,= 150 to 270° reverse bia es diode D1 and therefore, voltage Vab is shown bel'ow the reference line in Fig. 3,40 (f> . Examin ation of Fig. 3 40 (c) reveals that waveform Vba from rot = 150 6 to '270° , when reversed, becom es v ab in Fig. 3.40 (f>. /
After wt = 270°, diode D5 conducts for 120 0 an d now voltage VDl = Vac from wt = 270 to 390 • This voltage is sh own as v ac and as a reverse bias across diode D'1, so vae is sk etched· below the re f~.ren ce line in Fig, 3.40 (f>. As before, voltage wa veform 'vcu from rot = 27 0'0 to 390 0 in Fig, 3.40 (c), when reversed, becomes Vae in Fig. 3.40 (f; , 0
Fig. 3.40 (() revesals th a t PIV fo di ode D1, or any other diod e, is {:3 Vmp = value of lin e vol tage .
Vml
= m aximum
10J .
[Art. 3.9]
Diode Cir cuits and Rectifiers
An examination of waveforms in Fig. 3.40 (a) and reveals that line voltage Vab leads phase voltage va by
300 ~ Similarly, vbc leads vb by 30 0 .and VQl leadsv c by
. ..30 0 • A phasordiagramshowing 3-phase phase voltages
. .. V~, Vii, Vc and the cqrresponding 3-phase line voltages
Vah (=Va - Vb);Vbc (= V~ - V c), Vea (= Ve -:- Va) is drawn
· in Fig. 3.41. Line vol~ge Vba is 180 0 away from Vab as
· shown. Similarly, line voltages Vcb and VClC are shown in
Fig. 3.41, thus resulting in a six-phase system of line
voltages Vcd" Vue' Vbc ' Vba,Vca ' V cb ' Phasor diagram of
. Fig, 3.41 reveals thatline voltage Vab leads phase voltage
Vet by 30 0, similarly V&: leads Vb by 30 0 and Ven leads
0 V~ by 30 ; this matches with the waveforms in Fig. 3.40
Fig. 3.41. Six-phase line voltage as . expected. Vab. Vae:. Vbe: etc for secondary winding of delta-star transformer of Fig. 3.39. Note that Va = Vb= Vc ~ phase voltage, Vph and
Vab =V be == V ca = Vba =Vcb = Vae = line voltage, VI'
(c)
· 3.9.6. rhree-phase TWelve-pulse Rectifier · It has been stated before that as the number of pulses per cycle are increased, the output de waveform gets improved. So, with twelve pulses per cycle, the quality of output voltage waveform would definitely be improved with low ripp!e content.
Fig. 3.42 show~ th e circuit diagram for a 3-phase twelve-pulse rectifier using a total of twelve diodes. A 3-phase transformer with two secondaries and on e delta-connected primary .feeds the diode rectifier circuit. One secondary winding is connected in star and the other is
in delta. Star-connected secondary feeds the upper 3-phase diode bridge rectifier 1, whereas
the delta·connected secondary is connected to lower 3-phase diode bridge rectifier 2. Each
bridge rectifier uses six diodes as shown. The two bridges are series connected so that net
· output, or load, voltage va = output voltage of upper rectifier, VOl + output voltage of lower
r~ctifier V02'
~,.
+
110 I
Ao---e.... L
0
A
C
+
A
110
0
0----. \)02
80-- -
----
2
Fig. 3.42. Three-phase twelve-pulse Tectifier.
J
I
lQ2
Power Electronic
[Art. 3.9]
If Val, V bl , Vel are phase for the upper star, then upper line voltage Vabl (= Val - V bl ) would lead Val by 30 0 as i~ a three-phase bridge rectifier Of Fig 3.39. Similarly, line voltages V bel' Veal would lead by 30 0 their corresponding phase voltages V bl and Vel respectively. In Fig. 3.43 (b) are shown all the six line voltages Vab l ' V acl ' V bcl ' Vbal, Veal , Vebl for the upper star-connected secondary. _Vertical
(cY'
(b)
(a)
:--Vertica I
Fig. 3.43. Voltage phasor diagram for (a) primary line voltages (b) line voltages
for secondary star and (c) line voltages secondary delta.
Phase voltage V a l of upper star must be in phase with primary line, or phase, voltage as per the tran sfor mer principle. Likewise, line, or phase, voltage Vab2 of the delta-connected secondary must be in phase with VAl' All the secondary line voltages V ab2 , V ~2' V bc2 ' V baZ , V ea2 , Veb2 for the delta-connected secondary are shown in Fig. 3.43 (e). The line voltages V abl for upper rectifier and V ab2 for lower rectifier are phase~displaced by 30 with Vab1leading Vab2 by 30 0 • In case input line voltages to upper rectifier 1 and lower rectifier 2 are superimposed ; line voltages Vabl' Vab2' V acl ' Vac2 etc would be obtained; these line voltages would be phase-displaced from each other by 30 0 •
VAl
0
In Fig. 3.44 (a) are sketched waveforms for voltages available across upper-star-connected seconda~y.
Six-pulse de output voltage
UOl
obtained from upper rectifier 1 is shown in Fig. 3.44
(b); this is identical to the output voltage waveform of3-phase B-6 rectifier of Fig. 3.39. Lower
r ectifier 2 also gives six-pulse de output voltage V02 as shown in Fig. 3.44 (e). As V ab2 lags 0 . V abl by 30 , peak of vab2 of v 02 is also shown lagging p eak of vabl of VOl' Since the two rectifiers are serie-s-connected, net output, or load, voltage Vo = UOI + v 02 is obtained by adding the corresponding ordin ates of VOl .a nd v02' Note th at waveforms of VOl and V02 are phase-shifted from each other by 300 , therefo e, waveform of output voltage Vo consists of twelve pulses per cycle of supply voltage. Peak value of output voltage, Uo =V ml cos where
V ml
50
+ Vml cos 15° = 1.932 V ml
= m aximum
value of line voltage availab e from star-conn ected, or delta-con nected, s econdary.
For the sake 0 convenience, let the p eak val ue of t welve-puls e output voltage Vo be den oted by Vp (= 1.932 Vmi)' P eriodicity of Vo is 300 = 11:/6 radians .
103
[Art. 3.9]
Diode Circuits ·and Rectifiers
(a)
wI
(c) O~
__
~
____
~
________L -_ _ _ _ _ _
~
_ _ _ _ _ _ _ _~ - ~ -_ __ _ _ _ _ _~
wI
2 ,/ , "
N
(d)
,
-J +
.J / I~
~
" ,'
. "
3
4
5
6
8
10
12
I'~
, I
I"
:
I
'
!
: I
'\,
\
I
\
yp = l ,~32Vmi\ .: :
1.866 Vmi \
'\
,f
\\
[:~-75-.-~-T~3~~·~~--~ : ·~.~I~n~--~--------~2~rr------------~-1 90·
'1
90·
Fig. 3.4.,1, Waveforms for (a ) voltages across star-eonnected secondary (b) six-pulss output
voltage vO l from upper bridge 1 (c ) six-pulse output voltage v02 from lower bridge 2 and
, (d) resultant twelve-pulse output voltage Va obtained from ~01 + V02'
.
6 :. Average value of output voltage, Vo =-
fIOS.
"
,
' Vp sin rot , c!. (rot) = 0.98861'6 Vp
1t ,75·
'
\
= 0,,988616 x 1.932 Vml = 1.91 Vml
Rms value of outp
volt age,
VOl' =
... (3. ,8 3)
[~1': J75·os· ~ sm2ox d (c.ct) ]112 =0.988668 Vp ,
R ipple voltage, Voltage ripple factor,
I
=0.988668 x 1.932 Vm = 1.9101 Vml .. . (3. ~4) 2 2 Vr = ~V~ - V! = [1.9101 - .91 ]1/2 Vml = 0.019545 ml , Vr 0. 019545 . V ml ' VRF =--V = 191 V =0.01023 or,l.023% o
,x
m!
'
Power Electronics
[Art. 3.9]
104
FF =
Form' factor,
~: = \~~~1 = 1.00005
As voltage ripple factor is sufficiently small, the output voltage from 12-pulse rectifier is almost pure de voltage. '
. A comparison of various 3-phase diode rectifiers discussed above is given in the table below:
[VI I = maximum value of line voltage = ..J3 V mp where V mp = maximum value of line to neutral,
n , . or phase, voltage].
!
I S.No.
1.
Parameters
I
DC output voltage, Vo
2.
6-pulse rectifier B-6 type
12-pulse rectifier
3 ..f3vntp 3 Vml or- · re re .
1.91 V ml
6-pulse rectifier M-6 type
3-pulse rectifier 3Vml
3 V mp
2re
re
or
..J3vml re
'1
1.9101 V ntl
Rms output voltage, VOl'
0.4854 V ml
0.55185 V ml
0.9558 V ml
Ripple voltage, VI'
0.0872 Vml
0.02356 Vml
0.0408 Vml
4.
Voltage ripple factor, VRF
0.1826 or 18.26%
0.043 or 4.3%
0.0427 or 4.27%
5.
Rectifier efficiency, 11
96.765%
99.82%
99.82%
-
6.
TUF
0.6644
0.551
0.9541
7.
PIV
V ml
1.155 V ml
V ntl
-
8.
Form factor, FF
1.0165
1.0009
1.0009
~. 1--
0.019545 V ntl . 0.01023 or 1.023%
-
1.00005
It is seen fr om this table th at quality of dc output voltage is improved significantly with twelve-pulse rectifier. Out of 6-M and 6-B configuration s , 3-phase 6 -B rectifier is f>. su~r lOr so far as average output voltage, TUF and PlV are connected . .Q., 0 Exam ple 3.19. A 3-phase bridge rectifier, using diodes , delivers power to a load of R = 10 n at a de voltage of 400 V Determ ine the ratings of the diodes and of the three-phase . delta·star transformer. .
.r:
3~pha~e
Solution. It is given that de output voltage, 3V V o = -1t-ml= .400 V : . Maximum value ofline voltage,
Vml
=. 4003x 7t = 418.88 V
:. Rms value of phase voltage for share-secondary, V = s
ml
'l2Vx Ts = 418.88 T6 = 171 .OV
M aXlm ' um va 1u e 0 fl oa d current, 1m = R V ml = 418.88 10 = 41.89 A
F rom Eq. (3.81), rms value of phase current, Transformer r atin g
Is = 0.78041m = 0. 7804 x 41.89 =32.691 A = 3 Vph . lph = 3 Vs ' Is = 3 x 171 )< 32.691 = 16770. 5 VA 2rr/3 .
Rms val ue of dio ecurrent,
I Dr
= [ 22
7t
f
r:/ 3
l~! sin 2 wt . d (wt )
] 1/2
[Art. 3.9]
Diode Circuits and Rectifiers
= [ I!
21t
(~+ -v3) ]112 = 0 55181 3 2 .
105
m
= 0.5518 x 41.89 == 23.115 A 1 f21t / 3 . 1m 41.89 ID =1m sm wt d (rot) = - = --~-~ = 13.334 A
Average value of diode current,
1t 1t/ 3
1t
1t
= 1m = 41.89 A, PIV = V ml = 418.88 V
Peak diode current
. . 400 ·
Pd = Vo 10= 400 x 10 = 16000 W
Check .'
Pd,
:. Transformer rating
16000
=TUF = 0.9541 = 16769.73 VA
.
..r Example 3.20. A
3-phase bridge rectifier charges a 240- V battery. Input voltage to rectifier is 3-phase, 230 V, 50 Hz . Current limiting resistance in series with battery is an and an inductor makes the load current almost ripple free. Determine (a) power delivered to battery and the load (b) input displacement factor (c) current distortion factor (d) input p ower fa ctor (e) input H F or THD (f) transformer rating.
Solution. (a)
Here
Vml
V
Arrange output voltage,
o
=..f2 X Vl =..f2 x 230 V = 3 V ml = 3..f2 x 230 = 310.56 V 1t
7t
.
But
Vo =E +laR
. Vo - E :. Average value of battery chargIng current, 10 = R
= 310.568 -
240
= 8.82 A
=El~ =240 x 8.82 =2116.8 W ' .
P d = Elo + l!r. R
Power delivered .to battery
Power delivered to load, S ·n ce.load current is ripple free , 10r= 10 = 8.82
I
P d = 240 x 8.82 + 8.822 x 8 = 2739.16 W
Fig. 3:40 (d) shows that phase-a current ia , or transformer secondary current is, wou ld be constant at 10 = 8.82 A from fiJi = 30 0 to · 1500 and -: 10 from 210 0 t o 330 0 and so on . As positive and nega ive half cycles are identical, average value of is;" 0, -i.e. Ide = '0. (b) For ripple fr ee load current,
all
or
. a l
=it2 J·S1t/6/t/6 10 cos n rot . d (rot ) =
210
5 11:/6 1t2 f 1t/6 10 cos wt. d (rot) = 1t [sin 150
2 n
hI = -
f5/t/6 /6
It
.
210
10 sm rot. d (rot) =-
7t
sin 30°] = 0
0 -
[- cos 150
0
+ cos 30°]
From Eq. (3.56), fun amentsl component of source current is givE::n by .l til
2..[3 I0 ' = --;sm rot an d !PI =tan-
nput disp cem ent factor, DF =cos !PI = 1
1[
0 ~
1= 0
0
2-v3 =10 1t
106
Power Electronics
[Art. 3.10]
(e)
. Rms value of fundamental component of source current, lsI
12 x21t Rms value of source current, Is = .[ . ~ x 3 -.
1/ 2
.
Current distortion factor, CDF
lsI
=1; =
]
2{3 X 10 1t.
r
"12
2{3 · 10
=--;t x T2
1f. 3" .-
'.
'.
· 2
=
10
.
{3
x {2. 10
3
= n= 0.955
(d) Input pi= CDF x DF = 0.955 xl = 0.955 (lagging) (el HF
=THD =[(::1]'
-
1
=[ ( 0.0~55 )2
-1
r
=0.3106
(n Transformer rating ={3Vs .Is :::;: {3 x 230 x ~x 8.82 =2868.4 VA Also, transformer rating =
:;;F
=
20~:;4114 = 2876.92 VA
3.10. FILTERS
A rectifier should provide an output voltage that should be as smooth as possible. In practice, however, output voltage fr om rectifiers consists of de component plus ae component, or ae ripples. The ae component is made up of several dominant harmonics. It is more so in single-phase rectifiers with R load . The ae component does rio useful work. For example, in a de motor, it is the de current that produces the required torque; in a battery, energy is stored · due to de current only. AC ripples in rectifier output current do not contribute to motor torque, . or to the energy stored in the battery. AC component merely causes more ohmic losses in the circuit leading to reduced effi ciency of the system. This shows . that it is of paramount importance to filter out the unwanted ae component present in the rectifier output. For this purpose, filters are used. When used on the rectifier output side, these are called de filters; these tend to make the de output voltage and current as level as possible. The more common de filters are of L, C and LC type as shown in Fig. 3.45 (a), (b) and (e). L
1+
Rectifi er
Vo
1
,, I'
1+ Vo
Re ctifier
, I
Filt er
Filter
(a)
( b)
L
L
+
1+
Rec tifier
Re ctifier
1)0
JFi lter (c)
J
.,
Fi lter
I; (d )
Fig. 3.45. (a ), (b ) and (e) de filters, (d) ac n Iter .
VO
Diode
Cj r~u its
[Art. 3.10]
and Rectifiers
107
The non~sinusoidal output current in rectifier circuit causes the supply line current to contain harmonics. For reducing these harmonics in the supply current, ae filters are used at the output terminals of rectifier circuits. Fig. 3.45 (d) shows an ac filter of LC type. An inductor L in series with load R, Fig. 3.45 (a), reduces the ae component, or ae ripples, considerably. It is because L in series with R offers high impedance to ae component but very low resistance to de. Thus ac component gets attenuated considerably. A capacitor C across load R, Fig. 3.45 (b), offers direct short circuit toae component, these are therefore not allowed to reach the load. However, de gets stored in the form of energy in C and this ·allows the maintenance of almost constant de output voltage across the load. In this article, a simple design of L, C and LC type de filters is presented.
3.10.1. Capacitor Filter (C-Filter) Fill-wave, or two-pulse, rectifier is more often used than a half-wave, or one-pulse, rectifier. In the present discussion, therefore, single-phase I Filter full-wave diode rectifier is only examined. Its ripple frequency is 2 f, where f is the supply frequency. A capacitor C directly connected D3 across the load, as shown in Fig. 3.46, serves to smoothen out the de output wave. v 5
Fig. 3.47 shows the steady state waveforms
pertaining to Fig. 3.46. Fig. 3.48 (a) gives the
circuit model of Fig. 3.46. S ource voltage
Vs = Vm sin wt is sketched in Fig. 3.47 (a). Load
voltage Vo is shown in Fig, 3,47 (6). In this figure,
D2
.'.
Fig. 3.46. Single-phase full-wave diode
rectifier with capacitor filter.
from wt = 0 to wt = e, source voltage Vs is le3s than capacitor voltage Vc = Vo; therefore diodes Dl, D2 are reverse biased and cannot conduct, During this interval, i.e, from rot = 0° to cot = 8, capacitor discharges through load resistance R. At (Dt = 9, Vo = Vc = V2 as snown in Fig. 3.47 (b), Soon after wt = 9, source voltage Us exceeds Vo (= v c ), diodes Dl, D2 get forward biased and begin to conduct. a result, source voltage charges capacitor from V 2 to Vm atwt=7t/2, Fig. 3.47 (b) and Fig. 3.48 (b). Soon after wt = 7t/2, source voltage begins to decrease faster than the capacitor voltage; it is because capacitor discharges gradually through R. Therefore, after wt = 7t/2, diodes Dl, D2 are reverse bi ased and capacitor dis ch ar ges through R as shown in Fig. 3.48 (e). The capacitor voltage falls exponen tially, Fig. 3.47 (b). In the next half cycle, Vc = Vo = V2 at wt = (7t + 8)~ Just after wt = (7t + 8), Vs > vc ' diodes D3, D4 get forw ar d biased and begin to conduct. The capacitor voltage rises from V2 t o Vm at wt = 37t/2, Fig. 3.47 (b ). It is seen from this figure that voltage dro - from m aximum to minimum is V m - Y2' or peak to peak ripple voltage, V rpp = V m - V 2 ·
As
In Fig, 3.47 (e) is drawn the profile of ripple voltage with the help of Fig. 3.47 (b) , A horizon tal line at a h eight ~ (Vm + V 2 ), from r eference lin e wt in Fig. 3.47 (6) is now taken as the r eference lln·e in F ig. 3 .47 (c) fo plotting voltage pr ofil e U r • AB st ated before, peak to peak r ipple vo tage is V rpp
= Vm -
V2
and peak ripple voltage
V rp ::::
Ripple voltage i se en to be almost triangular in shape.
%(Vm -
V 2)
as shoV'rn in Fig. 3.47 (c).
108
, Power Electr onics
[Art. 3.10]
Vs
=Vm sin wI
0
wI. '
,5u/2
(a)
Vo
Uo=uc A
wt
0' 9.-+-t,
wCV~cos
e
0~~--~oL~-L~--~---1------~2~rr---r---T-L---------w~t
,
(d)
p
Capacitor disch,arging
,
't
I
'
:! : I "0 I (e)~: : :l! j •
iI I
I I
.
r
! v'2 O R.,.. L N
o
ILP
i
'
71/2
. '
•
I .
iI I : :!
f
-, .,
• ;
I
~
311/2
'
:
'
:
I
! I
571/2
277
;
JIl
wI
~ig. 3.47. Waveforms for (a) source voltage (b) load voltage with arid without filter
(c) ripple voltage (d) capacitor current and (e) load current for the circuit of Fig. 3.46. ...
~' .
Charging of Capacitor. From rot ,= e to 1t/2, capacitor charges from V 2 to V m . The equivalent circuit for capacitor charging of Fig. 3.48 (b) gives the charging cunent ic as under :
.
, ": , ' du, tc =. C dt
The charging current stored in C at wt = n/ 2 is ~
ic
. C dt (Vm sm oot)
=
d
=w C Vm cos IDt
at wt =1t/2 is roCVm cos 90 0
CV!..
(3
=0,
but
.. . .85 Dc
= V m' Therefor e, energy
)'
Discbarging of capacitor. KVL for the circuit model of Fig. 3.48 dis~harging
)
gives
~ fi
dt+Ri= 0
(c )
for capacitor
,
0
vsbcL
to .
tc C
R
wt
0
T
01
02
~I
~
us~
1
Lo
'fc
'f'
wI 0
fa)
.
109
[Art. 3.10]
Diode Circu its ancf Rectifiers
1+
+
c
Vm
Va
J(c)
(b)
Fig. 3.48. Single-phase full-bridge diode rectifier (a) circuit model (b) charging and (c) discharging.
In this equation, time origin is taken at rot = 1t/2. Laplace transform..of this equation is l.[I(Sl_ C'Vm]+R.I(S)::O
C
S
S
.
V
or
1(s) = ;
1
1 s+ RC
V
V
R
R
i(t) = -..E!. e- tiRe :: ~ e-
where't' =RC
tit
v o = R . i(t) = V 111e- tit =V m e- 1I RC
Load voltage.,
.....
~"
.. . . - '. . : .,
....
.
~
Pe ak to peak value of ripple voltage, V''PP:: Vm - V 2 = Vm - [voat t == t2j'=-V~ - Vm e- 12/t . It is known that e- x = 1 - x.
.
. V,pp
Charging time T
=
tl
=Vm -
Vm ( 1 -
~~ ) = VR~'
is ·usually small, it can therefore be
negl~cted.
7' therefore t2 if and peak to peak ripple voltage, V,pp = =
. V,pp Peak value of rIpple voltage, V rp = 2
2
As a result, t2
=; . But
~c
Vm
= 4 (RC
Variation of ripple voltage, Vr • shown in Fig. 3.47 (c), is not a sine wave, therefore rms va ue 0 V,. of r ipple voltage can be approximat ely found from the relation ... (3.86) .
.
It is seen from the waveform ~ output voltage vo ' Fig. 3.47 (b), that variation of ripple .voltage VI' shown in Fig. 3.47 (c) is almost triangular and average value Vo of output voltage is us ually taken as . Vo= maximum valu e of V m-pe ak value of triangu~ ar ripple voltage
. = Vm -
V,.p = Vm - 4
¥inim um value of load voltage, V 2 = Vm - Vl'pp
~C = V
=V m [
1 2
m [
1-
4kc 1
kc 1
... (3 .87)
110
[Art. 3. 10]
Ripple factor RF is given by
ripple voltage, Vr
RF'= - - - - - - - - - average output voltage , Va '=
1
Vml 4~.fRCx V ,
m
[1 __ 1_]=:T2l4' fR 4fRC
...(3.88)
C-l]
Its simplification gives
1]
C '= 4' ifR [ 1 + F. RF
...(3.89)
Waveform of capacitor current is as under:
At cot
VC
'=
OA
0 , le '= OP = -R=-R ,
, At ' wt = 8, ic = 0' K
' O'BV
=R = R2;
"
,
both OP and 0' K are shown negative because capacitor
discharges from wt = 0 to wt = 8. During charging time t 1, ic (= w CVm cos rot) follows the cosine wave with peak value wCV:m cos 8 at rot = 8 and ic '= 0 at wt = n/2. Current ( is positive as the capacitor is getting charged. Soon after the maximum value of V m at wt
=n/2, capac.'tor begins to discharge and i c
" '= -
V
I Lp where I Lp
= ;. At wt
'=
" (n + 8),
Vc
decays to
V 2 and lc .=- O'K as before.
Waveform of load current io in Fig. 3.47 (e) is identical with the waveform of va in Fig. 3.47 (b). At wt = 0, io
= ~A, at wt =- 8, io =
i,
at .wt
= n/ 2, io =- ;
and so on.
Eq. (3.86) shows that if C is increased, ripple voltage gets reduced,. But high value of C increases the amplitude of charging current as per Eq. (3.85). A high charging current would entail hig~er current rating of diodes. This leads to increased cost of the rectifier-filter circuit. Thus, a compromise between the value ofC and the magnitude of ripple voltage must be made . E xample 3.21. A 's ingle·pha se diode B -2 rectifier is fed from 2 30 V, 50 H z source ahd is connected to a load of R '= 400 n.
(a) Design a
capacitor~filter
so that the ripple factor of the output voltage is less than
5%.
(b ) W ith the va lue of C obtained in part (a ) , determine t/.. ~ average value of output voltage. (c ) Dete rmine the average value of output voltage without C- filter.
S olution . (a ) From E q. (3 .89), the value of capa citor C to limit the ripple factor RF to 5% 1S
l[
C = 4 fR
11 1 [
1 + '12. RF
1] ='
=- 4 x 50 x 400 1 + -.1"2 x 0.05
189.3 )IF.
111
[Art. 3.1 0J
Diode Circuits an d R ectifiers
(b) From Eq. (3 .87), average value Vo of 'the output voltage, with filterC, is 6
V o =Vm [1---.L]=2 4 {Re ' 3 0'-'2[ '1L. 1 _ 4 x 50 x 10 400 x 189.3 ]=303745V . (c)
Average value of output voltage without C-filter is . 2V _r;;; V = __ m = 2 '12 x 230 = 207.04 V o
1t
.
1t
·It is seen from this example that use of C filter has reduced the RF from 48.2%·to 5% and at the same time, average output voltage Vo has increased from 207.04 V to 303.754V.
3.10.2. Indu ctor Filter (L-filter) · An iriductor filter connected in series with the resistive load serves to provide the requisite filtering. In a resistive load, current waveform is identical with voltage waveform. An inductance in series with R load does not allow sudden changes in the load current. As a consequence, load current profiles becomes noticeably smooth. This has the effect of reducing . the current ripple factor. . '. _.
· Fig. 3.49 (a ) shows an in ductor filter L connected in series with R-Ioad in a I-phase two-pulse diode rectifier. Fig. 3.49 (b) shows the rectified voltage Vo and the load current i o ' It . is seEm that inductor has a smoothing effect on the load current profile. iOl
:-Filter ...., :,,
.,'
10
+ D1
is
, to
D3
Vo
R
Vs= Vmsinwt
D4
D2
(a )
F ig. 3.49. Single-phase diode bridge rectifi er (a ) circuit diagram and ( b ) waveforms for load voltage, load current , diode Dl current and sour :-" current.
From wt = 0 to wt = 1t; diodes Dl, D2 conduct and fr om wt = 1t t o 21t; diodes D3 , D4 conduct. The load current is continuous. ,The output v oltage appendix A), Vo
Vo
of the 2-puls e r ectifier can be analysed in to Fourier series as (see
2Vm . 4 Vm , 4 Vm . 4 Vm = - - - 3-' cos 2 wt -15 cos 4wt -3'"' cos 6 1t
1t
whereVm = maximum v alue of source voltage.
. 1t
0 1t
wt
.. .(3.90)
112
Power Electronics
[Art. 3.10]
Average load voltage,
v = 2Vm
Average load current,
10=]i =nR"
','t
o
V0
2Vm
.Zn = ~R2 + (n
Load impedance for nth harmonic,
roL)2
Magnitude of second harmonic load current] Similarly, fourth harmonic current, Load current io can, therefore, be written as . Lo
2Vm
= --;if -
4Vm
4 Vm
.
3n ~R2 + (2roL)2 cos (2wt - 92) - 15n ~R2 + (4 roL)2 cos (4 wt - 94)
where first, second and third terms of above current expression are dc, second~harmonic and fourth-harmonic components of i o' Rms value of harmonic-current components, or ripple current,Ir is
J2 '
4 Vm
Ir= [ ( 3n'l2
.
[
1
(
R2+(2roL)2+ 2
4V~J = ' ( -:;c;J2
4 Vm
15nT2
[1
J2
.
.' -
J1/2
. ]]
.
1
R2+(4WL)2+'~~ '
1 · 1.
J' 3 2 . R2 + (2 WL)2 + 15 2 ' R2 + (4 roL)2 + .. '
,.. (3 ,91)
Second-harmonic component seems to be the most dominant component of I,,, Therefore, ' . neglecting higher-order even harmonics, we get rms value of ripple current,
1 -1 _ 4 Vm r - 2 - 3 n -fi '-iR2 + (2 roL)2 4 Vm re R . 1r : , 'Current-ripple factor, CRF - - x- - 10 - 3'Jt '\[2"JR2 + (2 roL)2 2 Vm
= 3 x 42 {2' 'IR 2 +R(2 roL)2 = 0.4715.. ~R2 +R(2wL)2
... (3.91 a)
...
(3 92) ,
For good filtering, roL »R. Thus, neglecting R in the denominator ofEq. (3.92), we get .
.
R
CRF =0.4715. 2wL Fo 50 Hz supply,
. R
=0. 236 wL
R CRF = 0.236 (2re X 50) L = 7.51
X
10-
4
~
It is seen from above that (i ) reduced current ripple r equires large value of L, (i i) as the load is reduced (or R increased), ripple current inc eases, (i ii) indu ctive filter is preferred where load resistance R is consistently low or load ' current is invariably high , (iv ) induc tor in the load circuit introduces time delay of the load CUl'l'ent with respect to the load voltage, (v) as the current rofile becomes more smooth r transformel' utilization factor is im proved.
[A rt . 3.10]
Diode Circuits an d Rectifie rs
113
Comparison between C and L filters. It is worthwhile at t his stage to compar e a C-filter with an L-filter. . (a) If load
resistance is low, ripple factor for L-filter is low and pigh for C-filter. (b) In both C-filter and L-filter, time constant should be large for better waveform, i.e. for low ripple factor, 't should be high. .' . ' . . . . .. ' ' . . (c ) For C-filter, if R is increased, 't (= RC) increases, and therefore, ripple factor gets reduced . · . . . Cd) For L-filter, if R is lowered, 't (= LIR) increases, therefore ripple factor becomes low. This shows. that C-filteris suitable for loads having low curr ent (highload resis,: tance) consistently. D·filter is suited for loads requiring high load current (low R) consistently. . .' . ". '. ' . (e) A high value of C reduces ripple factor but increases the chargingcurrEmt and the diode-current rating. . ' . . .. . . .' (f) Ahigh value of L reduces the ripple 'factor, but .a delay is introduced in the response . .
,
.
'
.
Inductor filter is bulky, weighty, expensive and causes extra ohmic loss as compared to C-filter. Besides, L-filt er is noisy in nature. . . Example 3.22.A single-phase two-p1..J,lse diode rectifier has input supply of230 V, 50 Hz and the load resistance R = 300 n.Calculate the value of inductance to bf!.connected in series with R so as to limit the current ripple factor to 5%. Find the value of L in case R = 30 O. Determine also without L . the value of CRF . . . Solution. From Eq. (3.92), current ripple factor (CRF) with L is .
.
R
CRF= 0.4715. ~R2+ (2rol.J
0.05 =0.4715
· [3002 + (2 L = 4.4755 H .
For R = 3000,
For R
= 300,
.
~
0.05 = 0.471<>
.1
~
' .
.
300
X
21t x 50
X
L)2] 112
30
2
'130 + (2 x 21t x 50 x L ) .' L = 0.4477 H
CRF without filter L can be obtainedby putting L = 0 in Eq. (3.92). This gives CRF = 0.4715 without L . ' . ' . 3.10.3. L- C Filter An L~C filter consists of inductor L in series with the load and capacitor C across the load. This filter possess es the advantages of both L-filter and C-filter. In addition, ripple factor in L-C filter has lower value than that obt ained by either L-filter or C-filter for the same values ofL and C. ' Fig. 3.50 (a) show the use of L -C filt er for reducing th e ripple from the output voltage of a single-phase fu ll-wave di ode rectifier. Its equivalent circuitis g1venin Fig. 3.50 (b). The inductor L blocks th e dominant harmonics. Capacitor C provides ,aneasy p ath to t h e nth harmonic ripple currents . In order t h a t capacit or yields an easy path for harmonics, load impedfu'1ce R must be much ' greater than nth h armonic capacitive reactance; i.e. R that capacit or provides effe ctive filtering if
~ n~~ ' It has been fOlllld in practice
Power Electronics
[Art. 3.10]
114
L
L
01
1+
+
03
von
C
Vn
R
Vnl
rY'j . wt
04
J l
R
1
02
(a)
Fig. 3.50. (a)
1)on
( b)
Single-phas~ diode bridge rectifier with L - C filter circuit diagram and (b) its equivalent circuit.
R=~
...(3.93 a)
n roC
In case load consists of R and LL in series, then Eq. (3.93 a) becomes
2 .fR v + ( nro L L )2 = .. 10r'l
n
I ••
...(3 .93 b) .
.
UJ\.., . .
Under the condition ofEq. (3.93), effect of-loadR, or Rand LL' can be ignored during further analysis. Therefore, nth harmonic current in Fig. 3.50 is
I" =
v-n
1
nroL--
nroC where V" = rmsvalue of nth harmonic of rectifier output voltage. Thus, nth harmonic component ofload voltage VOrl across filter C in Fig. 3.50 (b) is
Vo" =
[n~I~T 1v, = [ (nro}2-;C - 1 nroC
1
v,
...(3.94)
Total ripple voltage due to all harmonics is Vr
=[
£ ,~" ]112
... (3.95)
n = 2,4,6,
The Fourier series analysis of output voltage of I-phase full-wav e rectifier is given in Eq. (3.90). The average value of output voltage Vo = 2Vm/1t. It is seen fromEq. (3.90) that second harm onic is the most dominant component. Therefore, other harmonic components can be neglected.
-
F rom Eq. (3.90), the rms v alue V 2 of secon d harmonic voltage is
4 Vm . V 2 = 31t . T2 From Eq. (3 .94) for n =2, ripple voltage is
V - V02 -
r-
rL (2ro)2-LC1 - 1 ]1/2
...(3 .96)
115
[Prob . 3]
Diode Circuits and Rectifiers
The value of filter capacitor C can be obtained from Eq. (3.93) as
C=~
...(3.91a)
2CJJR
C-
or
- 2m ~R2
10
... (3 .97b)
+ (2m LL)2
The VRF is defined as
Vr
V2
1
VRF = Vo = Vo . (200)2 L C _ 1 =
= -f2 [ 3
2
1
(200) LC -1
[
4 Vm ) 3 1t . "l2
.
[
1t)
2Vm
1
x (200)2 LC - 1
1
. . (3.98)
I
Once filter C is obtained from Eq. (3.97), the value of filter inductor L can be calculated from Eq. (3.98) for a specified value of VRF. Example 3.23. A single-phase two-pulse diode rectifier has input supply of 230 V, 50 Hz. and a load resistance R =50 n and load inductance LL = 10 mHo An L·C filter is to 'be used on the output side so as to reduce the output voltage ripple to 10%. Design the LC filter. Sol u t ion. From Eq. (3.97 b), the value of filter capacitor Cis
10
C = 2 X 21t 50 ~502 + (200 1t x 10 x 10- 3)2 = 315.83)lF
From Eq. (3.98), VRF =
or
(2001t)2 x 315.83 xL
or
L
PROBLEMS
;"
O~. = {2 [ 1 1 3' (2001t)2.Lx315.83x 10- 6 -1
=~ x 0\ + 1
.
= 0.045822 H or 45.822 mHo I
'. . Vo volts and (b) - Vo volts.
-
',"
•
_
3.1. Capacitor in the circuit of Fig. 3.2 (a) is initially charged with (a) For both these parts, determine the expressions for current in the circuit and voltage across
capacitor. Sketch the waveforms for current as well as capacitor voltage .
What is the final value of voltage across capacih>r~in each case?
[ Ans.
(a)
Vs-Vo -tiRe · -tiRe V R e . ' Vo+(Vs-Vo)(l-e ), s
(b ) Va ; Vo
e- tiRe, _ Vo + (Vs + V o) (1 - e- tiRe), V s ]
3.2. A diod e is connected in series with LC circuit. If this circuit is switched on to de source of voltage Vs at t = 0, derive expressions for current thr ough and voltage across capacitor. The capa citor is initially charged to a voltage of - Vo. Sketch waveforms for i, Ve, vL and uD' In case this circuit has Vs = 230 V, Vo =50 V, L =.0.2 mR and C = 10 IlF, detennine the diode
~:~:ction time, diode peak [::~:t::::'::o~i:::t~V:H~::t~:';::::o)C::~cic::~~: - '\10, 140.496 11 3, 62.61 A 510
V, - 280 V]
116
Power Electronics
[Prob.3J
3.3.
the circuit shown in Fig. 3.4 (a), the circuit is initi ally rel axed. If switch S is closed at t = 0, sketch the variations of i, uL, Ue and uD as a function of time. Derive the expres sions describing these functions . (b ) For part (a), Vs = 220 V, L = 4 mH , C = 5 IlF. Find the diode conduction time and peak diode c urrent. Determine also ue, uLand uD after diode stops conducting. fAns. (b) 0.444 ms, 7.778 A, 440 V, 0, - 220 V1
(a) For
3.4. For the circuit shown in Fig. 3.5 (a) ; Vo = 230 V, R = 25 nand C = 10 IlF. If switch S is closed at t = 0, determine expressions for the current in the circuit and voltage across capacitor C.
Find the peak value of diode current and energy lost in the circuit.
Derive the expressions used. [Ans. 9.2 e-4000t, - 230 e-4000t, 9.2 A, 0.2645 watt- sec]
3.5. In the circuit shown in Fig. 3.51, switch S is open and a eurrent of 20 A is flowing through the freewheeling
diode, Rand L. If switch S is closed at t = 0, deter
m(fie the expression for the current through the
switch . [Ans i(t) = 22 - 2 e- lOOOtJ
r
5
tOll.
VS =220V FD 3.S. (a) Describe how the energy trapped in an inductor tOmH can be recovered and returned to the source . (b ) A 230 V, 1 kW heater, fed through single-phase
half-wave diode rectifier, has rated voltage at its Fig. 3.51. Pertaining to Prob. 3.5. terminals . Find the ac input voltage. Find also [Ans. (b ) 325.32 V, 460 V, 8.696 AJ PlV of diode and peak-diode current.
3.7. (a ) In the circuit shown in Fig. 3.52, a PMMC ammeter is placed in series with diode and a PMMC volt meter across the diode . Take PMMC instruments ideal. Find the readings on t hese instrument s . Derive the expressions used for obtaining thes e readings. (b) If PMMC ammeter is replaced by MI ammeter, find its reading. PM iVlC [Ans. (a ) 10.352 A, 0 V (b ) 12 .6812 A] ,------( V } - - - - - ,
D
D O.lH
230V, 50 Hz
l,UF
C
PMMC
Fig. 3.52. Pertainin g to Prob. 3.7.
r7 'f/
V
F i g. 3.53. P ertaining to Prob. 3.8.
3.8. (a ) In the circuit of Fig. 3.53, ideal PMMC voltmeters are placed, one across capacitor and an other across diode as shown. Find the voltmeter readings . Obtain the expresoions used for det ermining these readings. (0) In case PMM C voltmeter 2 is repla ced by MI voltmeter, find its reading. [Ans. (a) 325.22 V, 325 .22 V (0 ) 398.3 94 V]
3.9. A ba ttery is ch arged by a single-phase half-wave diode rectifier. The supply is 30 V, 50 Hz and th e battery emf is const ant at 6 V. Find the resistance t o be inserted in series with t he battery to limit the chargin g current to 4 A. Take a voltage drop of 1 V across diod e. Derive th e expression used . I .
Dra\v waveform of vo ltage acr oss diode and find its P lV. t e . - 1 ,pi x 30 et c. I n . 1 = sm
[H'
[Ans.
2~
7
.
[2Vm cos 8 1 (E + 1) (It - 2 61 )], 2.5467 n, 49.42 V]
Diode Circuits and Rec tifiers
[Prob. 3]
117
3 .10. (a) A single-phase half-wave uncontrolled rectifier is connected to RL load. Derive an expres sion for the load current in terms of Vm' Z, ro etc. (b) For part (a), Vs == 230 V at 50 Hz, R = 10 n, L = 5 mH, extinction angle = 210°. Find
average values of output 'Voltage and output current.
v"3,11.
(a) A
single-phase half-wave diode rectifier feeds power to (i) RL load and (ii) RL load with freewheeling diode across it. Describe the working of this rectifier for both these parts with relevant waveforms and bring out the differences if any. Hence point out the effect of using a freewheeling diode. (b) For part (a), Vs = 230 V at 50 Hz, R = 20 n, L = 1 H. Find the average values of the output voltage and output current with and without the use of a flywheeling diode. [Ans. (b) With freewheeli.n g diode : Vo = 103.52V and 10 = 5.176 A
Q o'f"" .. 3.12.
[Ans. (b) 193.172 V, 19.3172 AJ
Without freewheeling diode: Extinction angle
p not known, so
Va' 10 cannot be calculated 1
circuit shown in Fig. 3.54, the output current io is considered constant at 10 because of large L. Sketch the waveforms of vs , io, va, iD' ifd , and is' For the above circuit, find (i) average values of output voltage and output current, (ii) average and rms values of freewheeling diode current, (iii) supply pf. [Ans. (b) (i) 103.52 V, 26.76 A (ii) 13.38 A, 18.925 A (iii) 0.6364Iag]
(a) For the
(b)
to
1s
io +
o
RI
2n 230 V. 50 Hz
+
FD
Vo
i fd
L
Vs
Z
V:0
10
R
SOV
F ig. 3.54. Pertaining to Prob . 3.12.
Fig. 3.55. Pertaining to Prob. 3.13.
3.13. For the circuit shown in Fig. 3.55, Vs = 160 V, V z = 40 V and zener dio.de current varies from 4 to 40 rnA. Find the minimum and maximum values of Rl so as to all~w voltage regulation for output current 10 = zero to its maximum value 10m , Also calculate 10m , [Ans. 3k n, 30k a, 36 mAl -
/3.14 . (a) Enumerate the input perfo rmance par ameters of a rectifier. Discuss how the performance of a rectifier circuit is influenced in case these parameters have low, or high, value. (b) Define input power factor, displacement factor DF and current dist ortion factor CDF for /.....a rectifier system an d show that input power factor = CDF x DF. 3.15. (a ) Defin e input current har monic factor (HF) and crest factor . Express (HF) in terms of _ /"current distortion factor.~f HF is more, what does it i rucate in a rectifier system . (b) Define t he following terms : Rectification ratio, T' pple volta ge, form factor, voltage ripple factor, current ripple fa ctor ,/ and tran sformer utilization factor. 3. 6. F or a single-phase half-wave diode rectifier feeding a res 'stive load R, fmd the values of recti 1e efficiency, form factor, voltage ripple factor, transformer utilization factor and crest fa ctor. / 3. 7. A sin gle-phase half-wave d'ode r ·ectifier is de igned to supply de outp ut voltage of 23 0 V to a load of R = 10 n. Calcula te the ratings of di ode and transfonner for this circuit arra gem ent . [ADs. I DAV =23 A, ID; =36.1 3 A, PN = 722. 6 V, Trans. rating = 18 .462 kV,\ ]
118
Power Electr onics
[Prob.3]
3.18. A single-phase full-wave mid-point diode rectifier feeds resistive load R. For this circuit, . determine rectifier efficiency, form factor, voltage ripple factor, transformer utilization fac tor and crest factor. D\,f;,4 , How does this 'rectifier circuit differ from single-phase full-wave bridge rectifier?
V t..3. 1 ~ (a) Why are three-phase rectifiers preferred over single-phase rectifiers?
(b) For a 3-phase half-wave diode rectifier feeding load R, obtain the following: , Average output voltage, rms output voltage, VRF, FF, TUF andPIV 3.20. Describe the evolution of three-phase six-pulse diode rectifier from 3-phase three-pulse diode rectifiers with appropriate circuits and waveforms. Hence, derive an expression for the average output voltage of 3-phase six-pulse diode rectifier. 3.21. Describe a 3-phase M-6 diode rectifier with a circuit diagram and relevant waveforms for resistive load R. Hence, derive expressions for ave·rage and rms values of output voltage and obtain therefrom . V o f" VRF, FF, rectifier efficiency and TUF. , ¥ 3.22. A3-phase mid-point 6-pulse di.ode rectifier feeds a load of 10 n at a de voltage of 400 V. Find the ratings of diodes and the three-phase transformer. [Ans. IDAV=: 6.667 A, IDr =: 16.337 A, PIV =: 837.76 V, Trans. rating·=29.038 kVA] /
.-.-
~
.
..........
Describe a 3-phase full-wave diode-bridge rectifier with a circuit diagram and relevant
waveforms for load R .
Hence, derive expressions for average and rms values of output voltage and obtain there from
VRF, FF, rectifier efficiency and TUF.
3.24. A 3-phase full-wave diode rectifier feeds a load requiring constant current 10 and is supplied from a 3-ph ase delta-star transformer. (a ) Sketch input volta ge waveforms for Vab, v~, vbc etc., taking vab zero and becoming positive at rot =: O. (b ) Sketch waveforms for currents for the three diodes of positive group a'nd phase current of the t ransformer secondary. (c ) l<"'r om th e wave orm of secondary phase current , determine current distor~ion factor CDF and THD. [Hint. Here al
=: -
..f3 10 , hI =.9. 10 , etc.] 1t
[Ans. (e) 0.955,0.3106]
1t
3.25. A 3-phase full-wave diode rectifier delivers power to an inductive load which takes ripple-free . current of 120 A. The source voltage is 3-phase, 400 V, 50 V, 50 Hz. Determine the ratings of diodes, power delivered to load and the rms value of source current. [Ans. 1DAV = 40 A, 1Dr = 69.284 A, PIV = 565.6 V, 64813.2 W, 97.98 AJ 3.26. Describe a three-phase 12"pulse diode rectifier with circuit diagram and appropri ate waveforms. Hence derive expressions for average and rms values of output voltage. Fro m 0\ these, obtain VRF and FF. 3.27. (a ) 'Wh at ar e the advantages of 3-ph ase bri dge rect ifier over 3-phase M-6 rectifier. (b ) For a 3-ph ase p-pulse diode rectifier, prove the following: Av era ge output voltage,
Vo
= V mp ' E. sin ~
1t
and rms outpu t voltage, V = V or
mp
P
[.E.. (~+ 1. sin 21t )11/2 21t p 2 . P I
where Vmp = m aximum value of per-pha e su pply vo1t~ge.
3.28. (a) ·W ha t ar the fu nctions offilt ers in rectifier circuits ? Distinguish between de and ac filters . ( q ) Explain how the in ductance Land capaci tan e C pI y th eir rol e in r educing the harmonic . contents in r ectifi er circuits.
Diode Circuits and Rectifins
[Pr ob. 3]
119
3.29. A single-phase full-wa ve diode rectifier feeds R with 3. .capacitor C directly connected across load. Describe the opera tion of C as a filter with relev'~mt voltage and current waveforms . Show that peak ripple voltage is
~ (Vm - V 2).
3.30. A single-phase two-pulse diode rectifier feeds Rand C in parallel. Explain charging and discharging of capacitor C and derive expressions for ripple factor and the value of filter capacitor C. 3.31. A single-phase diode bridge rectifier is fed at 230 V, 50 Hz. The load is R = 200 n shunted by a capacitance of 300 IlF . Neglecting all losses, detennine the average value of load voltage, VRF, maximum and minimum value ofload current, peak capacitor current and average load [Ans. 298.12 V, 0.0589, Imax = 1.6261A, Imin = 1.3551 A, 30.651 A 1.4911 AJ current. 3.32. A single-phase two-pulse diode rectifier feeds load R with an. inductor L in series with it. Describe the working of L as filter with relevant voltage and current wavefonns. Derive expression for current ripple factor and show that for 50 Hz supply, CRF = 7.51 x 10- 4.
~.
3.33. A single-phase full-wave diode rect ifier has mean output vo.ltage of 200 V ·and the load resistance is 400 n. Determine the inductance required to limit the amplitude of second-har [Ans. L = 3.48 Hl monic current in the load to 0.06 A. 3.34. A single-phase full-wave diode rectifier with L-C filter feeds load R. Describe its working and derive expressions from which the parameters of L-C filter can be obtained. 3.35. (a) Compare C-filter with L-filter. In what tyPe of applications are the two types usually preferred? (b) A single-phase two-pulse diode rectifier is fed from 230 V. The load is R = 200 n. Design an LC filter so as to get voltage ripple factor of5.89%. Find the rms value of ripple voltage. [Ans. C = 79.58 IlF, L = 0.28654 H, 12.195 V]
FOUR
Thyristors ----.. _--.- ... ... •. .. .........• ... •............ -_ .................. .. ... ....... ....... ... .... ..... -,
In this, ::hapte r • Term ina l C h aracteristics at Thyristors • Thyristor Turn-on Methods
• • • • • • • • • • • • • •
Switc hing Charac teristics of Thyristors Thyristor Gate CharacteristiCs Two-TranSistor Model of a Thyristor Thyristor Ratings Thyristor Protection Improvement of Thyristor Characteristics Heating, Cooling and Mounting of Thyrts,tors Series and Para ll e l Operation of Thyristors Other Members of the Thyristor Family Gote Turn oN (G .T.O.) Thyristor Static Induction Thyristor Firing Circuits tor Thyristors Pulse Transformer in Firing Circuits Triac Firing C ircuit
...... -.......................... __ ................................................ ................ As stated before, Bell Laboratories were the first to fabricate a silicon-based semiconductor device called thyristor. Its first prototype was introduced by GEe (USA) in 1957. This company did a great deal of pioneering work about the utility of thyristors in industrial applications. Later on, many other devices having characteristics similar to that of a thyristor were developed .. These semic~mductor devices, with their characteristics identical with that of a thyristor, are triac, diac, silicon-controlled switch, programmable unijunction transistor (PUT), GTO , RCT etc. This whole family of semiconductor devices is given the name thyristor. Thus the term thyristor denotes a family of semiconductor devices used for power control in dc and ac systems. One oldest member of this thyristor family, called silicon· controlled r ectifier (SCR), is the most widely used device. At present, the use of SCR is so vast that over the years , the word thyristor h a.s become synonymous with SCR. It appears that th e term thyristor is now becoming more common than the actual term SCR. In this book, the term SCR and thyristor have been used at r andom for the same device SCR. Other members of thyristor fam ily are also discussed in this chapter. A thyristor bas characteris tics similar to a thyratron tube. But fro m the construction vi:w poin t, a thyris to r (a pnpn device) belongs to transistor (p np or n.pn device) fam ily. The name 'thyristor ', is der ived by a combination of the capi tal letters fr om THYRa.tr on and trans ISTOR This mea.TJ.S that thyristor is a solid state device like a transist or and has charFlcteristics s im ilar to that of a thjTatron tube. The pr esen t-day r eader may not be fa miliar with thyratr on tube as this is not being taught these days. Actually, the n:lme 'thyri stor ' ca me in to exis~ence afte r a
Thyristors
[Ar t.
~. l l
121
. fo rmal' (i"ecision taken at a conference held by lEe (International Electrotechnica l Comml.:ision) in 1963. Prior to that, it was called s ilic on con trolled r ectifie r. or SCR. It appears that commission must have evolved the name 'thyristor' as discussed above. At this conference, the definition of thyristor was decided as under : (0 It constitutes three or more p-n junctions. It has two stable states. an ON-state and an OFF-state and can change its state
(ii)
from one to another. As per this definition, thyristor now includes a lar ge variety of sem iconductor devices ha ving similar basic characteristics. The object of this chapter is to discuss the thyristor ch aracteristics and other related topics useful for their industrial applications.
4.1.
TER~llNAL
CIWL\CTERISTICS OF THYRISTORS
Thyristor is a four layer, three-junction, p -n-p -n semiconductor switchin g device. It h as three terminals ; anode, cathode and gate. Fig. 4.1 (a) gives constructional det~ils of a typical thyristor. Basically, a thyristor consists of four layers of alternate p-type and n-type silicon semiconductors forming three junctions J l , J 2 and J 3 as shown in Fi g. 4.1 (a). The threaded portion is for the purpose of tightening the thyristor to the frame or heat s ink with the help of a nut. Gate terminal is usually kept near the cathode terminal, Fig. 4.1 (a). Schematic diagram and circui t sym bol for a thyristor are shown respectively in Figs. 4.1 (b ) and (c ). The terminal connected to outer p region is called anode (A), the terminal connected to outer n region is called cathode and th at connected to inner p region is called th e gate (G)' For large current applications, thyristors need better cooling; this is achieved to a gTeat extent by mounting th em onto heat sinks. SCR rating h as improved considerably since its introduction in 1957. Now SeRs of v oltage r ating 10 kV and an rm s current rating of 3000 A w.i th cor responding power-handling capacity of30 l'Irw are available . Such a high power thyristo r can be switched on by a low voltage supply of about 1 A and 10 W and this gives us an idea of the immense power amplification capability (= 3 x 105) of this device. As SeRs ar e solid state devices, they a re compact, possess high reliability and hav e low loss. Because of these useful featu res, SCR is almost universally employed these days for all high power-controlled devices.
----
Thrl!odli!
stud
Anode (Aluminium )
Anode
P
p
n
n
1-- ,I----1 J, 1---II--jJ, Gai, Galt terminal weJdtd to p r tgion Ca ll'lodo:
Fig. 4. 1.
A
G
p
G
n
Col ho(J~
W ~I (0) Con3tn.:ctional d ·':ili!.s (b, Sch~matic diagrum
A
K
K
~
and
(e)
circut t symbol of a
thy :-iSt8r.
122
Power EJectronics
[Art. 4.1J
An SCR is so called because silicon is used for its construction and its operation as a rectifier (very low resistance in the forward conduction and very high resistan ce in the r everse direction) can be controlled. Like the diode, an SCR is an unidirectional device that blocks the current flow from cathode to anode. Unlike the diode, a thyristor also blocks the current fl ow from anode to cathode until it is triggered into conduction by a proper gate signal between gate and cathode term inals.
For engineering applications of thyristors, their terminal characteristics must be known . In this article, their static [. V characteristics, dynamic characteristics during tum·on and turn-off processes and their gate characteristics are discussed. 4.1.1. Static I-V Characteristics of a Thyristor An elementary circuit diagram for obtaining static I- V characteristics of a thyristor is shown in Fi g. 4.2 (a ). The anode and cathode are connected to main sour ce th rough the load. The gate and cathode are fed from a source E.I which provides positive gate current from gate to cathode,
Fig. 4.2 (b) shows static I -V characteristics of a thyristor. Here,vo is the anode voltage across thyris tor terminals A , K and 10 is the anode current. Typical SCR J-V characteristic shown in Fig. 4.2 (b) reveals that a thyristor has three basic modes of operation; namely, reverse blocking mode, forward blocking (off-state ) mode and forward conduction (on-state) mode. These three modes of operation are now discussed below : Reverse Blocking Mode. When cathode is made positive with r espect to anode with switch S open, Fig. 4.2 (a ), thyristor is reverse biased as shown in Fig. 4.3 (a). J unctions J 1, J 3 are seen to be reverse biased whereas junction J 2 is forward biased. The device behaves as if two diodes are connected in series with reverse voltage applied across them. A s mall leakage current of the order of a few milliampe res (or a few microamperes depending upon the SCR r ating) flows. This is reverse blocking mode, called the off·state, of the thyristor. In Fig. 4.2 (b), reverse blocking mode is shown by OP. If the r everse voltage is increased, then at a critical +10
K Forward conduction mode (on ·stote )
1--ElO~A~DUr-
+
A Reverse !E.'C~oc;e \:urr ent
E
•
-v,
. '> ,.......
RevE'r;i.' blocking modE'
>
a
V30: forwar d bri'okover voltage - [0
~)
Forward Iec.lIcge currml
For .....ord bloc~in9 mode
" 3i1' Reversi' ':lr ... oltc!own voltc!ii.' C,p GatE' curren t
~)
Fig. 4.2. (e) Elementary circuit for obtrtining thyristor I-Y characteristics (b ) S~atic I-Y charac:eristics of!l thyristor.
[Art. 4.2]
Thyristors
123
breakdown level, ~alled r ever se breakdown voltage A Forward V BR • an avalanche oc cur s at J 1 and J 3 and th e \rokog« rever se current increases rapidly. A large current currEnt associated with V BR gives rise to more losses in the p SCR. This may lead t o thyr istor dam age as the J, junction temperature may exceed its permiss ible n n temperature ris e. It should , therefore, be ensured J, 0---<>'< P p that maximum. working r everse voltage across a o--c"'C G J, thyristor does not exceed VBB.. In Fig. 4.2 (b ), reverse G n n ava lanche region is shown by PQ . When reverse ___ Rqversil voltage applied across a thyristor is less than VBRI leakage current the device offers a~ high impedance in the reverse ,. + K direction . The SCR in the reverse blocking mode may K therefore be tre ated as an open switch. (a ) (b) Note that [ .V characteristic after avalanch e Fig. 4.3. (a ) J 2 rorward biased and J 1.J3 · bl k · d ' revers e biased (b ) J 2 reverse biased and · b rea k d own d uring r eve r se oc 109 mo e I S fi db' d applicable only when load resistance is zero, Fig. 4.2 Jl' .~~ orwar lase. (b) . In case load resistance is present, a large anode eui+ent associated with avalanche breakdown at V SR would cause substantial voltage drop across load and as a result, ]. V characteristic in third quadrant would bend to the right of vertical line drawn at V BR ·
"-!.
Forward B locking Mod e: When anode is positive with r espect to the cath ode, with gate circuit open, thyristor is said t o be forward biased as shown in Fig. 4.3 (b). It is seen from this figure that junctions J h J 3 are forward biased but junction J 2 is reverse biased. In this mode, a s mall current, called forward leakage current, flows as shown in Figs. 4.2 (b) and 4.3 (b ). In Fig. 4.2 (b) , OM represents the fonvard blocking mode ofSe R. As the fonvard leakage curren! is small, SCR otTers a high impedance. Therefore, a thyristor can be. treated as an open switch even in the forward blocking mode. " F orward Conduction mode. 'When anode to cathode iorward voltage is increased with gate circuit opep, reverse biased junction J 2 will h ave an avalanche breakdown at a voltage called fOl'ward breakover voltage VBO ' After this breakdown, thyristor gets turned on with point }.If at once shifting to N and then to a point anywhere between N and K. Her e NK represents the forward conduction mode. A thyristor can be brought from forward blocking mode to fo rw'ard conduction mode by turning it on by applying (i) a positive gate pulse between gate and cathode or (ii) a forward breakover voltage across anode and cathode. Forward conduction mode NK shows that voltage drop across thyristor is of the order of 1 to 2 V depending upon the rating of SCR. It may also be seen from N K that voltage drop across SCR increases slightly with an increase in anode current. In conducti on mode, anode current is limited by load im pedance alone as voltage drop across SCR is quite small. This small voltage drop VT across the device is due to ohmic drop in the four layers, In forward conduction mode, thyristor is treated as a cl osed switch.
4.2.,TIlYRISTOR TtJRN.ON lUETHODS
., , . • .. .
'"
With anode positi',e with r espect to cat.hode, a thyris tor can be turned on by anyone of the follow ing techniques: (a) Forward voltage trigging (b) gate triggering (c) du / dt triggering (d) t emper ature triggering and (eo ) light trigg=ring,
[Art. 4.2J
Power Elt!ctronics
These methods of turning·on a thyristor are now discussed one after the other.
Forward voltage triggering. When forward voltage is applied between anode and cathode with gate circuit open, junction J 2 is reverse bianed. As a result, depletion layer is form ed across junction J 2 . The width ofthis layer decreases with an increase in anode-cathode volta.ge. If forward voltage across anode·cathod e is gradually increased, a stage comes when the depletion layer acr oss J 2 vanishes. At this moment, reverse biased junction J 2 is said to have avalanch e breakdown and the voltage at which it occurs is called forward breakolJer voltage VaG- The name forward breakover voltage is given because at this voltage VBo• i-u characteristic breaks over and shifts to its on-state position wi th breakover current IBO' At this voltage, thyristor changes from off-state (high voltage with low leakage current) to on-state characterised by low voltage across thyristor with large forward current. As other junctions J t • J 3 ar e already fOT'Hard biased, breakdown of junction J 2 allows free movement of carriers across three junctions and as a result, large forward anode-current flows. As stated before, thi s forward current is limited by the load impedance. In practice, the transition from off-state to on-state obtained by exceeding V BO is never employed as it may destroy the device. (a )
The magnitudes offorward breakover and reverse breakdown voltages are nearly the same and both are temperatur e dependent. In practice, it is found~ that V BR is slightly more than V BO' Therefore, forward breakover voltage is taken as the fin al voltage rating of the device durin g the design of SCR applications. After the avalanche breakdown, junction J 2 loses its reverse blocking capability. Therefore, if the anode voltage is r educed below V BO SCR will continue conduction of the current. The SCR can now be turned off only by r educing the anode current below a certain value called holding current (defined later). Gate Triggering. Turning on of thyristors by gate triggering is simple, reliable and efficient, it is therefore the most usual method of firing the forward biased SCRs. A thyristor with forward breakover voltage (say 800 V) higher than the normal working voltage (say 400 V) is chosen. This means that thyristor will remain in forw ard blocking state with normal working voltage across anode and cathode and with gate open. However, when turn-on of a thyri stor is r equired , a positive gate voltage between gate and cathode is applied. With gate current thus established, charges are injected into the inner p layer and voltage at which forward breakover occur s is r educed. The forward voltage nt which the device switches to on-state depends upon the magnitude of gate current. Higher the gate current, lower is the forward breakover voltage. (b)
When positive gate current is applied, gate p layer is flooded with electrons from the cathode. This is because cathode n layer is heavily doped as compared to gate p layer. As the thyristor is forward biased, some of these electrons reach junction J 2• As a result, width of depletion layer near junction J'l, is reduced. This causes the junction J 2 to breakdown at an applied voltage 10v-/eT than the forward bre akover voltage VBo.lfmagnitude of gate current is increased, more ele ctrons would reachjunctionJ'l" as a con sequence thyristor would get turned on at a much lower forwar d applied voltage . Fig. 4.4 (a ) snow3 that for gate current 1, = 0, fo rward breakover voltage is 1,/30' For gate current lll' forward br eakover, or turn-on voltage is VI which is less than VBO ' For 1,2 > 1,1' forward breakdover voltage i3 further reduced to V2 < VI ' For 1,3 > 1z'2 1 the forward breakuver voltage is VJ < V2 , Fig. 4.4 (0). Th e effect of gate current on the forward br eakover voltage of a
,
[M •. 4.2J
Thyristors
Forward voltcgi/' drop
N
------
p
ReV E'r$i/' leakage curren t
\ 0' "
t
I g]
-
VJ <
Forward leokcge current V2
o - I,
~)
a
101 102 10 3
Gat¢ Curr¢n t ~)
Fig. 4.4. Effect of gate current on forward breakover voltage.
thyristor can also be illustrated by means of a curve as shown in Fig. 4.4 (b) . For I, < oa, forward breakover voltage remains almost constant at VBO ' For gate currents 1,1,18'1 and 1,3' the magnitudes of forward breakover voltages are ox = Vi' OJ = V2 and oz =V3 respectively as sh own in Fig. 4.4 (a ) and (b). In Fig. 4.4 (a), th e curve marked I, =0 is actually for gate ::urrent less than 00. In practice, the magnitude of gate current is more than the minimum gate current required to turn on the SCR. Typical gate current magnitudes are of the order of 20 to 200 rnA. Once the SCR is conducting a fofward current, reverse biased junction J 2 no longer exists. As such, no gate current is required for the device to remain in on-state. Therefore, if the gate current is r emoved, the conducti on of current from anode to cathode remains unaffected. However, if gate current is reduced to zero before the rising anode current attains a value, called the latching current, the thyristor will turn-off again. The gate pulse width should therefore be judiciously chosen to ensure that anode current rises above the latching current. Thus [atckin current may be deemed as the minimum value of anode current which it must attmn uring turn-on process to mam am con uc Ion w en gate signal is remove .
-
Once the thyristor is conducting, gate loses control. The thyristor can be turned-off(or the thyristor can be return ed to forward blocking state) only if the forward current falls below a low-level current call ed the h olding current. ,!,hus h olding current may be defined as the minimum value of anode current below which it must fall for turning-off the thyristor. The Iitching current IS hIgher than the holding current. Note _that latching current is associated . \V1th turn-on prct:ess and holdin[..c urrent wi ~h turn-off process. It is usual t o take latching current as two to three tImes the h oiding current tIl. In mdUstria::t applicatioos, holding current (typically 10 rnA) is alm ost taken as zero. (c )
~~ Triggering. Wi th forward voltage across th~ anode
and cathode of a thyristor, the
two oute r junction J 1• JJ are fo rw ard biased, but inner junction J 2 is revers e biased. Thi5 revers e bias ed junction J,!, Fig. 4.3 (b), h as th e characteristics of a capacitor due to charges exis tin g across the jLlOction. In other words, space-charges exist in the depletion region nea r junction J 2 and th~:-efore junction J 1 ~e ha.ves like a capacitance. If for-vard voltage is su ddenly
126
Power Electronics
[Art. 4.21
applied, a charging current th rough junction capacitance CJ • may turn on the SCR. .AJrnost the entir e suddenly applied forward voltage Vo appears acrcssjunction J 2 • the char ging current ir: is , ther efore, given by
. _
'e-
_ dV. dCj dt - dt (Cj . Va) - CJ dt + V(l o dt
As the junction capacitance is a lmost constant.
.
... (4.1 a )
dC · dt is zero and current ie as
dVe
... (4.1 b)
'e= CjTt
Therefor e, if rate of rise of forward voltage dVa / dt is high, the ch arging current ic would be more. This charging current plays the role of gate current and turns on the SCR even th ough gate signal is zero. Note th at even if Va is small, it is the rate of change of Va that plays the role of turning-on the device. (d ) Temperature Triggering (Thermal Trigge ring) . During forward blocking, most of the applied voltage appears across revers e biased junction J 2. Thi s voltage across , J 2• associated with leakage current, would raise the temperature of this junction. With increase in temperature, width of depletion layer decreases. This further leads to more leakage curr ent and t herefore, more junction temperat ure . With the cumulative process, at. some high temperature (within the safe limits), depl etion layer of r everse biased junction vanishes and the device gets turned on . (e) Light Triggering. For light.triggered SeRs, a recess (or niche) is made in the inner p·layer as shown in Fig. 4 .5 (a ). When this recess is irradiated , free charge carriers (pairs of h oles and electrons ) are generated just lik e when gate signal is applied between gate and cathode. The pulse of light of appropriate wavelength is guided by optical fibres for irradiation .
If the intensity of this light thrown on the t ecess exceeds a certain value, forward-bi ased SCR is turned on . Such a thyristor is known as light-activated SCR (LASCR). LASCR may be triggered with a light sou r ce OT with a gate signal. Sometimes a combination of both light source and gate signal is used to trigger an SCR. For this, the gate is biased with voltage or current slightly less than that required to turn it on, now a beam oflight directed at the inner p·layer junction turns on the SCR. The ligh~ intensity required to turn·on the SCR depends upon the voltage bias given to the gate. Higher the voltage (or curren t) bias, lower the li ght intensity required. Light· trigger ed thyristors have now been used in high-voltage dir ect current (HYDC) transmission systems. In these several SCRs are connected in series· parallel combination and their light-triggering h as the advantage of electrical isolation between power and control circuits. Example 4.1 . Discuss what would happer. if gate is made positiue with respect to cathode during the reverse blocking of a thyristor. S olution . Before answering t his question, it is worthwhile to know in terna l details of a thyristor.
8
little more about the
Fig. 4.5 (b) shows cross- section of a conventional cen tr e-gate t hyri stor. In this figure , a pproxi mate doping densities in num ber per cubic centimeter are also indicated for all t he four layers. For example, for PI layer, the doping den sity is 10 19 per cm 3 .
[Art. 4.3]
Thyristors
127
A node
.
A, •
J
p J,
n Light -
, ,
10"cm-1
P, (pot)
pp " "
n
K
J
n,(n-) 10" to 5x
l0"cm-l
JJ
r,10"cmn,
n,(n+)1
, J, J
10"cm- 1
P1(p ot )
l
J Cathode
(a)
Gote (b)
Fig. 4.5. (0) Elementary LASeR (b) Structural details of conventional centre-gate thyristor. For semiconductor devices, it should be kept in mind that (i) junction with lightly doped layers (at least on one side of the junction) requires large breakdown voltage, (ii) junction with highly doped layers on both sides requires low breakdown voltage. When thyristor is in forward blocking state, junctions J 1, J a are forward biased whereas j unction J 2 is r everse biased. As layer n l is lightly doped around j unction )2' depletion r::egion of junction J 2 extends mainly into n} layer. Therefore, n 1 layer is made to h ave larger width to withstand the high voltage during forward blocking state. For reverse voltage on the device; junctions J 1, J 3 are r~verse biased and J 2 is forward biased. As layers P2, n2 across junction J 3 are heavily doped, J 3 h as low breakdown voltage. Layer nl being lightly doped as compared to layer Pl,juDction J 1 has large break'down voltage. As a consequence, during the reverse blocking of a thyristor, junction J 1 suppor ts most of the r everse voltage. Even during blocking, the depletion r egion extends into tne nl layer. This shows that width of layer nl absorbs most of the voltage during forward blocking mode and also during the reverse blocking mode of a thyristor. If positive gate voltage is applied between gate and cathode during the rever se blocking of a thyristor, blocking property of junction JJ disappears as J 3 has low breakdown voltage. As a result, reverse voltage appears across junction J l ' Positive charge carriers are now inj ected into the n 1 layer of reverse biased junction J 1. This causes an increase in the reverse leakage current . The flow of large leakage current associated with high reverse voltage results in increased power loss across junction J 1 and heat thus generated may raise the junction temperature above the allowable maximum and this may destroy th.~CR. Such an happening can be avoided if no positive gate voltage is applied between gate and cathode during the reverse blocking of SeR. Some manufacturers do specify the maximum positive voltage (usually less than 0.25 V) that can exist between gate and cathode during the reverse blocking of a thyristor.
4.3. SWITCHING CIIARACTERlSTlCS OF TH'KRIS'fORS Static and switching characteristics of thyristors are ahvays taken into consideration for ec onomical and r eliable design of converter equipm ent. Static characteristics of a thyristor
128
Power Elec lronics
fArt. -l·JI
have already be en examined. In this part of the section; switching , dynamic or transient , characteristics of thyristors are discussed. During turn-on and turn-off processes, a thyristor is subjected to different voltages across it and different currents through it. The time variations of the volta ge across a thyristor and the current through it during turn-on and turn-off processes give the dynamic or switching characteristics of a thyristor. Here, first switching characteristics during tum-on are described and then the switching characteri stics during tum-off.
4.3.1. Switching Charact eristi cs during Turn-on A forwa rd-biased thyristor is usually turned on by applying a positive gate voltage between gate and cathode. There is, however, a transition t ime from forward off-state to forward on state. This transition time, called thyristor tum-on time, is defmed as the time during which it changes from forward blocking state to final on-state. Total tum-on time can be divided into three intervals; (i) delay time tel, (ii ) rise time tr and (iii) spread time tp ' Fig. 4 .8. (i) Delay time td : The delay time t d is measured fr om the instant at which gate current r eaches 0.9 Ig to the instant at which anode current r eaches 0.110' Here 16 and 10 are r espectively the final values of gate and anode currents. The delay tim e may also be defined as the time during which anode voltage falls from Vo to 0.9Vo whe re Vo =initial value of anode voltage. _-'\n other way of defining delay time is the time during which anode current rises from forward leakage current to 0.110 wher e 10 = final value of anode current. With the thyristor initially in the forward blocking state, the anode voltage is OA and anode current is small leakage current as shown in Fig. 4 .8. Initiation of turn·on process is indicated by a rise in anode current from small forward leakage current and a fall in anode-cathode voltage from forward blocking voltage OA. As gate current begins to flow fr om gate to cathode with the application of gate signal. the gate current has non-uniform distribution of current density over the cathode surface due to the p layer. Its value is much higher near the gate but decreases r apidly as the distance from the gate increases, see Fig. 4.6 (a). This shows that during delay time t d , anode current flows in a narrow region near the gate where g-ate current density is the highest.
The delay time can be decreased by applying high gate curren t and more fo rward voltage . between anode and cathode. The delay time is fraction of a microsecond. A
P J, n
'%
l
.~
JJ
n
G
K
la)
(b )
Fig. -1.6. la ) Distribution of gate and anode currents durin g delay time (b) Conducting area of cathode (i) during td (ii ) afte r tr (iii) after tp'
Thyristors (ii)
[Art. Uj
129
Rise time t,.: The rise time t,. is the tim e taken by the anode cur rent to rise fr om 0.1
Ia to 0'.9 I a' The rise time is also defined as the time required for the forward bloc kin g off-state
voltage to fa ll from 0 .9 to 0.1 of its initial value OA. The ris e tim e is in versely pr oportional to the magnitude of gate current and its build up rate. 'Thus t,. can be reduced if high and steep current pulses are applied to the gate. However, the main factor determin ing t,. is the nature of anode circuit. For example, for series RL circuit, the rate of rise of a node current is slow, ther efore, t,. is more. For RC series circuit, dU dt is high, t,. is therefore, less. From the beginning of ris e time t,., anode current starts spr eading fr om the narrow condu cting r egi on near the gate. The an ode current spreads at a rate of about 0.1 mm per \ g microsecond [2]. As the rise time is small, the anode current is not able to spread over the entire cross-section of cathode. Fig. 4.6 (b ) illustrates how anode current expands over cathode surface ~rea during tum-on process of a thyristor. Here \the thyri stor is taken to have single gate electr~e a way fr om the ce ntre of p -layer. It is seen that anode current conducts over a small conducting l chann el even after t,. - this conducting channel area is however, greater than that during t d-. Fig. 4.7. Typical waveform for gate current.
During ris e time, turn-on losses in the thyristor are the highest du e to high anode voltage (Va> and large an ode current (I a) occurring together in the thyristor as shown in Fig. 4 .8 . As these losses occur only over a small conducting r egion , local hot spots may be formed and the device may be dam aged. (iii) Spread time tp : The spread time is the time taken by the anode current to rise from 0.9Ia to I f1" It is also defin ed as the time for the forward blocking voltage to fall fr om 0. 1 of its ini tial value to the on-state voltage drop (1 to 1.5 V ). During this time, condu ction spreads over the entire cross-section of the cathode of SCR The spr eading interva:l depends on the ar ea of cathode and on gate structure of the SCR After the spread time, anode current a ttains steady state value and the voltage drop across SCR is equal to the on-state voltage drop of the order of 1 to 1.5 V, Fig. 4 .8. . Total turn-on tim e of an SCR is equal to the sum of delay time, rise time and spread tim e. Thyristor manufacturers usually specify the rise time which is typically of the order of 1 to 4 )l sec. Tota l tum~on time depends upon the anode circuit par a meters an d the gate signal wavesh ap es. Durin g turn-on, SCR may be consider ed to be a charge controlled device. A certain amount of charge must be injected in to the gate r egion for th e thyri stor conduction to begin. This charge is dir ectly pr oportion al to the value of gate cu rren t. Therefor e, higher the magnitude of gate curren t, the lesser time it takes to inject this charge. The turn·on time can therefor e be reduced by usin g higher values of gate currents. The magnitude of gate current is usually 3 to 5 times the minimum gate current required to tri gger an SCR ·When ga te current is sever al times high er than the minimum gate current required, a thyristor is said to be hard-fired or overdriven. Ha rd -firing or ouerdriuing of a thyri3 tor reduces its tum-on tim e and enhances i ts dildt capability. A typical waveform for gate current, that is wide ly used, is shown in Fig. 4.7. This wavefor m has high er initial -nIue of gate current wi th a very fast rise time. The initial high value of gate curren t is then r educed to a lower value where it stays fa r sever al micrasecan d3 in order to avaid unwanted turn-off of the device.
130
[Art. 4,3J
Pow e r Electronics
4.3.2. Switching Characteristics during Turn-off Thyristor tu m-off means that it has changed from on to off state and is capable of blocking the forwa rd voltage. This dynam ic process of the SCR from conduction state to forward blocking state is called commutation process or turn -off process. Once th e thyristor is on, gate loses control. The SCR can be turned off by reducing the anode current below h olding curren t· , If forward voltage is applied to the SCR at the moment its anode current falls to zero, the device will not be able to block this forward voltage as the carriers (holes and e lectrons) in the four layers are still favou rable for conduction . The device wi ll therefore go into conduction imm ediately even though gate signal is not applied. In orde r to obviate such an occurrence, it is essential th at the thyristor is reverse biased for a flnite period after the anode current has reached zero.
I
VOltog., Vg
Gate pulse
, Anode voltoge LIe and gote curr ent i g
A
On state voltagE' drop across SCR
~nodp
--O.IVe I I I , I current ]
+r--I[
Anode curre-nl begins 10 decrease
,
I
1
I ,
---r---,
!
I)
i
-i't't.Iorf-T-tOI\+i
. Reverse voltage due to powt>r circuit
Ii; .
POWt>f Ioss 'I ( Ucla I
I
I
I'
[ di
C:lmmutation I I
I
" I
I
Recovery
j'
I
! 'I
!
,R ecombination
,I ' i
tJ
,!
I·
t l,
,
I I
ts
:
I
"
I i..- Steady stole ! --I. I·I - operation t
t
I
I
0 .1 10
Forward leakoge currpn!
I t, ---i '/ t-- tq -----'1 I , ,
" [I i 'I"~,~_I_ I' Q_~_"Q_O_C_~_"_._n~'
Ie
. ....
A~--l g
0.9 I9
o, ~
-. . . ". ~ .... ~ " :.i '~•
--··0.9Ve OA =Ve= lnitial anode voltage
I
r-
I
t tr - l -tg l
t
I
j
; - - tq I ,
t , - - --'
' 1
I .
I: Time In Microsec
fig LS Thyristor voltage and cu rrent wavl! fo r rru du:ing turn-on and turn-off proces.s-es . 'Th )" C;1:l be achie\"erl thNl ugh nat ural comm u ta tion or forced com:":"lutation.
Thyristors
[Art. 4.3)
131
The turn-off time tq of a thyristor is defined as the time between the instant anode current becomes z.ero and the instant SCR regains forward blocking capability. During time tq , all the excess carriers from the four layers of SCR must be removed. This removal of excess carriers consists of sweeping out of holes from outer p-Iayer and electrons from outer n-layer. The carrier s around junctionJ2 can be r emoved only by recombina tion . The tum-off time is di vided into two intervals; r everse r ecovery tim e tTT and the gate recovery time t,r; i.e. tq = tTT + t,r The thyristor characteristics during turn-on and turn-off processes are sh own in one Fig. 4.8 so as to gain insight into these processes. At instant t 1• anode current becomes z.ero. After t 1, anode current builds up in the reverse direction with the same dildt slope as before t 1. Th e reason for the rever sal of anode current after tl is due to the presence of catTiers stored in the four layer s. The r everse recovery current r emoves excess carriers fr om the end junctions J 1 and J 3 between the instants tl and t3' In other words, r everse recovery current flows due to the sweeping out of holes from top p-Iayer and electrons from bottom n-Iayer. At instant t 2• when about 60% of the stored charges are r emoved from the outer two layers , carrier density across J 1 and #3 ..begins to decr ease and with this reV,e rse recovery current also starts decaying. Th~ reverse current decay is fast in the beginning but gradual thereafter. The fast decay of r ecovery current causes a 'reverse voltage across the device due to the circuit inductance. This reve r s~ voltage surge appear s across the thyristor terminals and may therefor e damage it. In practice, this is avoided by using protective RC elements across SCR. At instant t 3. when reverse r ecovery current has fallen to ne arly zero value, endjunctionsJ1 and J 3 recover and SCR is able to block the reverse voltage . For a thyristor, r everse r ecovery phen omenon between tl and t3 is similar to that of a r ectifier diode. At the end of r everse recovery period (t3 - t 1), the middle junction J 2 still has trapped
'3.
charges, the refore, the thyristor is not able to· block the forward voltage at The trapped charges around J 2• i.e. in the inner two layers, cannot flow to the external .circuit, therefore, these trapped charges must decay only by recombination. This recombination is possible if a reverse voltage is maintained across SCR, though the magnitude of this voltage is not important. The rate of recombination of charges is independent of the external circuit pa.ram eters. The time for the recombination of charges between t3 and t4 is called gate recouery time t,r' At instant t", junction J 2 recovers and the forw ard voltage can be reapplied between anode and cathode. The thyristor turn-off time tq is in the range of 3 to 100 )lsec. TI:e turn-off time is influenced by the magnitude of forward current, dildt at the time of commutation and junction temperature. An increase in the magnitude of these factors increases the thyristor turn-off time. If the value of forward current before commutation is high, trapped charges around junction J 2 are more. The time required for their recombin ation is mor e and ther efore tu::-c..-off time is increased. But turn-off time decreases with an increase in the magnitude of reverse voltage. particularly in the range of 0 to - 50 V. This is because high reverse voltage sucks out the carriers out of the junctions J I , J 3 and the adjacent transition region::; at a fa3tcr r ate. It is evident from above that turn-off time tq is not a constant parameter of a thyristor. Th e th)..-ristor turn-off time tq is applicable to a.T). indi,,;dual SCR. In actual practice. thyristor (or thyristors ) fonn a part of the power circuit. The rum-off time provided to the thyristor by th~ practical circuit is called circllit tum ·offtime t l! ' It is defined a.5 the time bet\",een the instant anode current becomes zero and LI-J.e iIlstant reverse voltage due to practical circuit reaches zero, sec Fig. 4.8. Time tc must be greater than tq for reliable turn-off, otherwise the device may turn-on at a n UndE'3U-ed instant, a proc ess called commutation failure .
Power Electronics
[Art. 4.4]
132
Thyristors with slow turn·off time (50 - 100 ~sec ) are called converter grade SeRs and those with fast tum-off time (3 - 50 lJ.Sec) are called inverter.grade SeRs. Converter-grade SCRs are cheaper and are used where slow turn-off is possible as in phase-controlled rectifiers, ae voltage controllers, cycloconverter s etc. Inverter-grade SeRa are costlier and are used in inverters, choppers and force-commutated converters.
4.4. THYRISTOR GATE CHARACTERISTICS The forward gate characteristics of a thyristor are shown in Fig. 4.9 in th e form of a graph between gate voltage and gate current. Here positive gate to cathode voltage V, and positive gate to cathode current I, represent de values. As gate-cathode circuit of a thyristor is a p-n junction , gate characteristics of the device are similar to that of a diode. For a particular type of SCRs, Vg-Ig characteristic h as a spread between two curves 1 and 2 as shown in Fig. 4.9. This spread, or scatte r, of gate ' characteristics is due to inadvertent difference in the doping levels of p and n layers . '!'he gate trigger circuitry must be suitably des igned to take care of this unavoidable scatter of characteristics. In Fig. ·4;9, curve 1 represents the lowest voltage values that must be applied to turn~on the SCR. Curve Z gives the highest possible voltage ." . values that can be safely applied to gate circuit. Each thyristor has maximwn limits as Vgm for gate voltage and Ip for gate current. There is also rated (average) gate power dissipation PIfI1IJ specified for each SCR These limits should not be exceeded in order to R.void permanent damage of junction J 3 , Fig. 4.3. There are also minimum limits for V, and I , for reliable tum.-on, these are represented by ~ and ox respectively in Fig. 4.9. As stated before. if Vp . I,m and Ptau are exceeded, the thyristor can be destroyed. This shows that preferred gate drive area for an SCR is bcdefghb as shown in Fig. 4.9.
Vo
Minimum ga te voltage a nd current to trigger an SeR. Vgm , I,m - Maximum permissible gate voltage and current. 00 Non- triggering gate voltage. Oy. ox -
-.. y
C~~'~----------~I~~ ---~ gm
Ig
Fig. 4.9. Forward gate characteristics of thyristor.
A non-triggering gate voltage is also prescribed by the manufactur ers of SeRs. This is indicated by oa in Fig. 4.9. If fl.!·""ing circuit generates positive gate signal prior to the desired i. n.stant of triggering the SCR, it should be ensured that this un wanted sipal is less than the non-triggering gate voltage 00 . At the same time, all spur ious a T noise s ign als sh ould be less than the voltage oa. The design of the firing circuit can be carried ou t with the help of Figs . . t I 0 and 4.11. In Fig. 4.10 (0 ) is shown a trigger circuit feed ing power to gate-cathode circuit. For this circuit,
Thyri5lors
[Act.
A
Trt gg~r circuit , -__________ .x,
+
-
+ G
R,
+ E,
I
E,
R,
,, ,,
, L-______L-~-J K 1l ______________ ---J,
;7
10
I
I1
133
A
,r -1
~.~l
l _____ __ __________
(a )
Vo
~ -
, J,
K
(b )
Fig. 4.10. Trigger circuit connected to gate-cathode circuit of an SCR.
E,
=Vg +lg Rs
... (4 .2a )
where
Es= gate source voltage Vg = gate-cathode voltage I N = gate current and R, =gate-source resistance The internal resistance R, of trigger source should be such that current (E/R~ ) is not harmful to the source as well as to the gate circuit when SCR is turned on. In case R, is low, an external resistance in series with R s must be connected. A resistance Rl is also connected across gate-ca thode terminals, Fig. 4.10 (b ) , so as to provide an easy path to the flow of leakage current between SCI{ terminals. If I ClHn and V Nmn are the minimum ga te current and gate voltage to turn- on.SCR, then it is seen fr om Fig. 4. 10 ( b ) that current through Rl is Vgmn l R 1 and the trigger source voltage E$ is given by
E,+,m" + VIC JR,+ V,m" For low-power circuit3, it is customary to obtain the operating point by utilizing the V-I characteristics of both source and the device. In view of this , for selecting the operating poin t for the circuit of Fig. 4. 10, a load line of th e gate sou rce volt age Es = OA is drawn as AD in Fi g. 4.11. Here OD = trigger circui t short circuit current = E,I R J . Let us consider a thyristor who5e V g-Ig char acteristic is given by curve 3. In tersecti on of load line AD and Vg -Ig curve 3 gives the operating point S. Thus, for this SCR, gate voltage =PS and gate current = OP. In ord er to mini mis e turn -on time and j itte r (u:1reliable tu m-on ), the load lin e and hence the operating point 5, whi ch may ch an ge fr om S! to 5 2. mus t be as cl ose to the P gCl(' curv e as poss ible . .-\t th e same tim e, the oper ating !Joint
.. .(4.2b )
V~
V
gm
3
1 c "" E
"
" 0
g
_.!-_~ Sl~·" - ·'
1:01
;g 0
p
Igm 0
Gate Curnmt-
Fig. 4. 11. Choice of ga te circu it po:-a:neters
Ig
P ower Electronics
[Art. 4.4]
134
S must li e within the limit curves 1 and 2. The gradie nt of the load line AD (= OAI OD) will give the required gate source resistance Ro!. The minimum value of gate source series resistance is obtained by drawing a line AC tangent to Pgau curve. Gate drive requirements in terms of continuous de signal can be obtained from Fig. 4.11 . However, it is common to use a pulse to trigger a thyristor. For pulse widths beyond 100 I-lsec, the de data apply [1]. F or pulse widths less than 100 ~sec , magnitudes of gate voltage and gate current can be increased, see Example 4.2. As stated before, thyristor is considered to be a charge controlled device. Thus, higher the magnitude of gate current pulse, lesser is the time to inject the required charge for turning-on the thyristor. Therefore, SCR turn-on time can be reduced by using gate current of higher magnitude. It sh ould be ensured that pulse width is sufficient to allow the anode current to exceed the latching current. In practice, gate pulse width is usually taken as equal to , or gt:eater than , SCR turn-on time. If T is the pulse width as shown in Fig. 2.12 (a), then T ~ ton With pulse triggering, grea~er amount of gate power dissipation can be allowed ; this should, however, be less than the peak instantaneous gate power dissipation Pgm as specified by the manufacturers. Frequency of firing (or pulse width) for trigger pulses can be obtained by taking pulse of (i) amplitude Pgm (ii) pulse width'T and (iii) periodicity T 1. Therefore,
P T
~>P Tl grlli
Pgm " T f>P - gau
or
Pgau
. or
... (4 .30 )
fT $Pgm
wher e
f = ~ =fre quency of firing, or pulse repetition rate, in Hz,
and
T = pulse width in sec.
,
or f= ~;-P.!;"~,T· Pgm
In the limiting case,
A duty cycle is defined as the ratio of pulse-on period to periodic time of pulse. In Fig. 4.12 (a), pulse-on period is T and periodic time is T 1. Ther efore, duty cycle 5 is given by T o=-=(T
T,
I
'E ig
-,• u
>l
SLLLLLULL____-11111LLL-___ (0)
Ie)
Fig. 4.12. (a) Pulse gating a nd (b ) high-frequency carrier gating of SCRs , (c) Thyristor protection against reverse overvolta ges .
Thyristors
[Art. U J
p From Eq. (4.30 ),
135
.
-.E 'S
Ii
Pgm
.. .( 4.3 b)
Sometimes the pulses of Fig. 4.12 (a) are modulated to generate a trai n of pulses as shown in Fig. 4.12 (b). This technique of firing the thyristor is called high·frequency carrier gating. The ad vantages offered by this method offiring the scas are lower rating, reduced dimensi ons and therefore an overall economica1 design of the pulse transformer needed for isolating the low power circuit from the main power circuit. For an SCR, Vim and I,m are specified separately. If both of these are used for pulse firing, then Pim may be exceeded and the thyristor would be damaged. For example, GE·C35 thyri stor has Vim = 10 V and I,m = 2 A. If both these limits are pl aced on C35 , the power dissi pation is 20 W. But this is far excess of the specified Pgnt = 5 W. It should be ensu red that (pulse voltage amplitude) (pulse current. amplitude) < Pgnt . There is also prescribed a peak reverse voltage (gate negative with respect to cathode ) that can be applied across gate-cathode terminals. Any voltage signal, given by the trigger circuit (or by any interference), exceeding this prescribed limit of about 5 to 20 V may damage the gate circuit. For preventing the occurrence of such hazard s, a diode is connected either in series with the gate circuit or across the gate-cathode terminals as shown in Fig. 4.12 (c ). Di ode across the gate-cathode terminals, called clamping diode, prevents the gate-cathode voltage from becQming more than about 1 V. Diode in series wi th gate circuit prevents the fl ow of negative gate source current from becoming more than small reverse leakage current. The magnitude of gate voltage and gate current for triggering an SCR is inversely proportional to junction temperature. Thus, at very low temperatur es, gate voltage and gate current must have high values in order to ensur e turn-on. But P,m should not be exceeded in any csse;The res istor R 1. connected across gate-cathode terminals, Fig. 4.10 (b ), also serves to bypass a part of the thermally-generated leakage current across j unction J 2 when SCR is in the forward blocking mode ; this improves the thermal stability of S'CR. Example 4.2. (a) The average gate power dissipation for an SCR is 0.5 W. The allowable gate voltage variation is from a minimum of 2 V to a maximu m of 10 V. Taking ar;e rage gate power dissipation constant, plot allowable gate voltage as a fun ction of gate current. (b) If SCR of part (a) is triggered with gate pulses of du ty cycle 0.5, find the new valLle or average gate power dissipation. Solution . (a) Here F or For
v"Ig = 0.5 W
Vi = 2 Y, Ig = 0.5/2 = 0. 25 A to Vg = 10 Y, Ig = 0.5/ 10 = 0.05 A For other values of ga ts voltaot? V, in between 2 and 10 V, gate current 19 is obtained and plotted in Fig. >c:: - 4 4.13 showing the variation of V, 83 a fun cti on of I, for CJI > constant p,cu. 2 (b) For this example, T I =2T in Fig. 4.12 (a)
. __ ..
......,;.
I: -
I f = 0 .5 W.
so th at B = 0 .5. For dc values, V
F or puls e firing, Fi g. 4.12 (a ), the average gate power dissip ation can be ob tain ed fr om the r elation
a
,, : :
p
:; 1, ..-- ..' I..,
P9 C ~ :0' 5 W
..• • -l..J ___~_ ' :
,
•
:" : . ::.-:+t.::-:1::~::.-::::_. , ... , I .' , 1 :
t: :
"
0.05
:
t
0.1 0.15 0.1 Ig in A -
0.25
Ig
Fig. ·U3 . Perta ining to E:, :'.mpl i:! 4.2.
Power Electronics
136
where UN' iJl are th e ins tantaneous val ues of gate voltage and gate current. Therefore, for this e xampl e, av erage gate power dissipation is given by
V, · I, · ;;. = (o.o)!
=0.25 W.
As thi s is less th a n the all owable Piau, higher values of ti" i, can be used fo r the pu lse firin g of SeRs. Example 4.3. For an SeN. the gate-cathode characteristic has a straight-line slope of 130. For trigger so urce voltage of 15 V and allowable gate power dissipation of 0.5 watts, compute the gate-source resistance. Sol uti on . Here V, I,=0.5W V and -:.L = 130 Ig
1301;=0.5
.
Thi s gives
I, = [0.51130) ln~ Q.052 = 62 rnA
:. Ga te voltage. F or th e gate circuit,
V, = 130 x 62 x 10- ' =8.06 V E, =I, R, + V, = 0.062 R , + B.06 = 15 15 - 8.06 R, = 0 .062 = 1l1.94l1.
Exam p le 4.4. The trigger circuit of a thyristor has a source voltage of 15 V and the load l ine has a slope of - 120 V per ampere. The minim um gate current to tum- on the SCR is 25 mA Compute (a) source resistance required in the gate circu it, . (b) the trigger voltage and trigger current for an average gate power dissipation of 0.4 watts. Solution. (a ) The slope of load line gives the required gate source resis tance. From the load line, se ri es resis tance required in the gate circuit is 1200. (b) Here V, I, = 0.4 W EJ =R)c + V, F or th e gate circuit,
..
or
15 = 120 I + 0.4 g Ig
120 Ii - 15 Ig + 0.4 = 0 Its so luti on gives I, =38.56 mAo or 86.44 rnA ..
V I
= 0.4 X
10' 38.56
= 10 37 V .
V - 0.4 X 10' _ 4 6? - V ,- 86.44 - . .. 1 •
or Choose Vg = 4.627 V and I,
=86.44 rnA for minimum gate current of
25 rnA.
Example 4.5. Foran SCR, gate-cathode characteristic is gi uer. by Vi = 1 + IO I~. Gate sou.rce ()oltage is a recta ng ula r pu lse of 15 V with 20 )J sec du. retiol!.. For an average g ate power d issipation 0/ 0.3 \V and a pec k g ate -dri ve pOlL"er ol5 ~v, compute (a) the resistance to be connected in series with the S CR gate, (b) the triggering fre quency and
[A rt. 4.4]
Thyristors
137
(c) the duty cycle of the triggering pulse. Solution. (0) Here V, = 1 + 10 I,. For pulse-triggering of SCRs, (P eak gate voltage) (peak gate curr ent) during pulse-on period = peak gate drive power, P,m' As the gate pulse width is 2.0 ~ sec (less than 100 ~ sec), the dc data does not apply. Had the gate pulse width been more than 100 j.LSec, the relation (1 + 10 I.) I, = 0.3 W will hold good . But as the dc data does not apply, we have here
(1 + 10 I,) I, = 5 W
1Of,+I,-5=0
or
I,
Its solution gives, :. Amplitude of current pulse During the pulse-on period,
or
E,=R, Ig+V, 15 = R, I, + 1 + 10 I,
<.
15 - 1
R, 0.659 -10= 11.2440
(b )
Pgm
:. Triggering frequency, (c)
= 0.659 A. =0.659 A
Pr fJ II
= fl"
f=
H ere T= 20 ~sec
0;:ig' =3kHz
B=fT= 3 x 10' x 20
Duty cycle,
X
10"= 0.06, 0 '
Example 4 .6. Latching current for an SCR, inserted in between a dc uoltage source of200
V and the load, is 100 mA. Compute the minimum width of gate·pulse current required to turn-on this SCR in case the load consists of (a) L =0.2 H, (b) R = 20 n in series with L =0.2 H and (c) R = 20 n in series with L = 2.0 H. Solution. (a) When load con sists of pure inductance L, the voltage equation is
E= L
..
0100 =
di dt
~og t
or
d'L = E Lt
or
t=
0.1 x 0.2 200
Thus, minimum gate-pulse is 100 ~sec (b) The voltage equation for R·L load is
· L -di E =R L+
dt
. E( '!!.')
or
t=R 1 -
or
t = 100.503
eL
or
or
~s e c
: . Minimu m ga te-pulse width is 100.503 ).l5eC
. -E t L
L=
= 100 ~sec
138
Power Electronics
[Art. 4A]
. -E t=
(e)
R
(1 -!!,) -e
L
0.1 = 22000 (1 _ .-1o ,)
or
or
t =1005.03 ~see .
This example shows that if load resistance is increased from zero to 20 n, the gate-pulse width remains almost unaffected. But with an increase in indu ctance from 0.2 H to 2 H, the gate-pulse width becomes 10 times its previous value. Example 4.7. Th e gate current of a forward biased SCR is gradually increased from zero until the device is turned on. It is obserlJed that gate CUTTent, just prior to the instant of ttlrn -on, is 1 mA and soon after SCR goes in.to conduction, gatt. current decays to about 0.3 rnA. Discuss how it happens. Solution . When anode of an SCR is made positive with respect to cathode, a small voltage E'g genera ted internally, appears across the gate- cathode terminals, Fig. 4.14 (a ). The magnitude of E'g depends upon applied anode voltage and the device geometry. In the gate-c athode equivalent circuit of Fig. 4.14 (0), R is the static non-linear gate resistance. If the SCR is turned on by applying a positive gate signal, then the equivalent circuit for the trigger circuit is as shown in Fig. 4.14 (bl. Here E, is the gate voltage ~e nerated internally due to the flow of anode current. The magnitude of E'g is mu ch smaller as compared to Eg. For a typical SCR, E'g = 0.05 V and Eg '" 0.7 V.
~~;~~i~ '{ ----r-----J'-v--\-"-r---~'_.:G9-,_,;,lo!..., +
Go--~
R
+
R
+
Es
+
Eg
E'o K (b)
(a )
Fig. 4. 14. Pertaining to Example 4.7.
Before the SCR starts conducting, gate current I;:: E,I(R + R J ). Alt er SCR goes into conduction, I, =
I,' = ER-E' + R: . As E;
is very small,
E - E + R:' Voltage Eg is quite large as
R
compar ed to E'g, therefore , gate current is reduced fr om a higher value of 1,' to a lower value oi I,. In ca3e E, is reduced to zer o, gate current becomes n egative with its value equal to
-E
Iz" = R +
A Under this condition of E •
J
= 0, the voltage appearing across the gate-cathode
J
term in als i5 Eg - Ig" R. E xa m pl e 4.9. Gate-cathode characteristics of a thyristor ha ue a spread g iuen. by the following two relations:
Ig = 2.1 x 10- 3 V; and Ig = 2.1 X 10- 3 V;·S
"
[Act.
Thy ristors
~.5J
139
T he gate source voltage is 16 Vand load line has a slope of - 128 V I A. Calculate the trigger voltage and trigger current for an average gate power dissipation of 0.5 W Are the val ues of V" I, obtained here justified? Discuss. Solut ion. Slope of load line gives gate-source resistance, R. and E, =I, R.+ VI
= 128 n. Here Vg Ig = 0.5 W
16= I ,x 128+ 0/ g
or
128
I; -16 I, + 0.5 =0
= 62.5 rnA and V, =a v So point 8 in Fig. 4.11 has V, = 8 V and Ig = 62.5 rnA. For the same V,; I, = 2.1 X 10- 3 gives mor e I" therefore it represents curve 1 6.11 . Point 8 1 on this curve can be obtained from V, I , = 0.5 Wand I, =2.1 x 10- 3 ~ It solution gives I ,
V;
..
I, = 2.1 x 10- 3 (
Point 8 2 can be obtained from V, l j .
I, = (2.1)
X
10- 3
~,5
.
of Fig.
.
J
orI, = 80.67 rnA and V, .= ~:198 V
=0.5 W and I, = 2.1 x 10- 3 V:·5
_),.5
(
~gO
or I, = 56.01 rnA and Vg = 8.93.V.
Since point 5 (8 V, 62.5 rnA) lies in between 5, (6.198 V, 80 ..67 rnA) and 5 , (8.93 V, 56.01 rnA) as desired, the calculated values of V,; a V and I,l =6'2'.5 rnA are j ustified.
.
~
-
.....
4.5. TWO·TRANSISTOR MODEL OF A THYRISTOR The principle of thyristor operation can be explained with the 'use of its two-tr ansistor model (or two-transistor analogy). Fig. 4.15 (a) shows schematic diagrum"o-f a thyristor. From this figure, two-transistor model is obtained by bisecting the two middle layers , along the dotted line, in two separate halves as shown in F ig. 4.15 (b ). In this figure , junctions J 1 - J 2 and Jz - J 3 can be considered to constitute pnp and Itpn transistors separately. The circuit repres entati on of the two-transistor model of a thyristor is shown in Fig. 4.15 (c lIn the off-state of a transistor, collector current Ie is rel a ted to emitter current IE as
Ie = CJ. IE + l cBD wher e a is the com mon· base cu. rrent g ain and ICBO is the comm on· base leak age current of collector- base j un ction of a transis tor. For transis tor QJ in Fig. 4. 15 (c), emitter current Ie = anode current Ia a nd Ic = collector current le i' Th er efor e, for Q t, wher e an d
ICI = a lIa + I eBol al = common- base current gain of Ql Ieaol = co mmon- base leak age cu rr en t oi Q!.
...(4 .4 )
Power Electronics
[Art. 4.5]
140
A
A
A
r,
1,
1,
p n G 1,
I
P J
,,, , ,,
n J
P ,, P
J,
,
P n
n
n 1"
J, G [,
Jl
P
J
P
1"
I
n
161=lc2
a,./3,
J
n
p,-0,
, ,
1"
J,
G 1,
1"
"-
I "~ n
P a2'PZ 0, n
r.
K
K
(a)
(e)
(b )
Fig. 4.15. Thyristor (a) its schematic diagram, (b) and (c) its two·transistor model.
Similarly. for transistor Q2, the collector current IC2 is given by I C2 =
... (4.5 )
ct:z lie + I CB02
ct:z =common-base current gain of Q2 I CB02 =common-base leakage current of Q 2 and I. = emitter current of Q2The sum of two collector currents given by Eqs. (4.4) and (4.5) is equal to the external circuit where
current f a entering at anode terminal A.
:. I", =IC1 +IC2 or
=Ci! 1(1. + I CBOl + ~ I. + I CB02 .. ,( 4 .6 ) When gate current is applied. then lit =10 + I , _Substituting this value afl ll in Eq. (4.6) gives 10 =Ci l It:! + I CBOI + ct:z (10. + 1,) + 1eB CY.2 1(1
Ig+ ZCBOl +iCBCYl ... (4 .7 ) '" 1- (a l +
or
I
=' (l2
I
•,
[Art. 4.5J
Thyristors
141
mechanisms for turning-on a thyristor are now discussed below : (i) GATE Triggering : With anode positive with respect to cathode and with gate current I, = 0, Eq. (4.6) shows that anode current, equal to the forward leakage current, is somewhat more than I CBOl + [CB02' Under these conditions, the device is in the forward blocking state. Now a sufficien t gate-drive current between gate and cathode of thyristor, or the transistor of Fig. 4.15 (e) is applied. This gate-drive current is equal to base current 182 = I , and emitter current III of tr ansistor Q2' With the establishment of emitter current l it of Q2' current gain ~ ofQ2 in creases and b ase current Is2 causes the existence of collector current I C2 = ~2 182 = ~2 1" This amplified current I C2 serves as the base current 181 of transistor Ql ' With the flow of l Bl, collector current lCI = ~1 IBI = ~I ~2 I , of Q1 comes into existence. Currents lSI and l CI lead to the est ablishm ent of emitter current I II ofQ 1 and this causes cu rrent gain Ct.} to rise as desired. Now current I, + I CI '= (1 + ~t~2) I , acts as the base current of Q2 and ther efor e its emitter current IJr. =ICt +1, ri se. With the rise in emitter curr ent 111. , ~ ofQ 2 increases and this further causes IC2 = ~2 (1 + ~l~) I, to rise, As amplified collector current I e'! is equal to the base current of Qt, current gain a l eventually rises further. There is thus established a regenerative action internal to the device. This regen erative or positive feedback effect causes al + ~ to grow towards unity. As a consequence, anode current begins to grow towards a larger va lu e limited only by load impedance exte rn al to th e dev ice. When regeneration has grown sufficiently, gate current can he withdrawn. Even after II is removed, regeneration continues . This characteristic of the thyristor makes it suitable for pulse triggering. Note that thyristor is a latching device. After thyristor is turned on, all the four layers are filled with carriers and all junctions are forward bi ased. Under these conditions, thyristor has ver.y low impedance and is in the forward on -state. (ii) Forward-uoltage triggering : If the forward anode to cathode voltage is ' increased, the collector to emitter voltages of both the transistors are also increased. As a result, the leakage current at the middle junction J 2 of thyristor increases, which is also the collector cu rrent of Q2 as well as QI ' With increase in coUector currents ICI and lc2 due to avalanche effect, the emitter currents of the two transistors also increase causing a l + ~ to approach unity. This leads to switching action of the device due to r egene rative action. The forward-voltage triggering for turning-on a thyristor may be destructive and should therefore be avoided. (iii ) d u/dt triggering: The reversed biased junction J 2 behaves like a capacitor because of the space.charge present there. Let the capacitance of this junction be Cj . For any capacitor,
i
= C ~~. In
--
case it is assumed that entire forward voltage ull appears across rev erse biased
junction J 2• then charging current across the j unction is given by duo i =C·-
) dt This ch argi ng or displ ac ement current across junction J'). 13 collector currents of Q'l and Qt· Cu rrents I C'l , I Ct will induce emi tter curren t in Q2' Qt. In case rate of ri se of anode voltage is large, the emitter currents will be large and as a res ult, 0.1 + ~ will approach un ity !e ~.d ing to eventu al switching action 01 th e thyri stor. (iu) Te mperatltN! triggerin.g : At high temperature , the forward leakage current aCr0.55 j~lDction J 2 rises. This leakage current se rves as the collector junction current of th e compon ent
.u
j.
[Act.
-I.6J
Power Electronics
transistors Q 1 an d Q2' Therefore, an increase in leakage current l eI' I~ leads to an increase in th e emitter currents of QI' Q2' As a result, (exl +~) approaches unity. Consequentiy, switching acti on of thyristor takes place. ( 1I ) Light triggering: When light is thrown on silicon, the electron-hole pairs incr ease. In the forward-biased thyristor, leakage current across Jz increases which eventually incr eases a, + ~ to unity as explained before and switching action of t.hyristor occurs. As stated before , gate-triggering is the most common meth, d for turning-on a thyristor. Light-trigger ed thyristors are used in HYDC applications.
4.6. THYRISTOR RATINGS Thyristor ratings indicate voltage, current, power and temperature limits within whi ch a thyristor can be used without damage or malfunction. Ratings and specifications serve as a link betwee n the designer and the user of SCR systems. For reli able operation of a thyristor, it should be ensured that its current and voltage ratings are not exceeded during its working. One of the major di sadvantages of thyristors is that they have low thermal time constant. If a thyristor handles voltage, current and power gr eater than its specified ratings, the junction temperature may rise above the safe limit and as a resu lt, thyristor may get damaged. Therefore, when SCRs are selected, some safety margin must be kept in the form of choosing device ratings somewhat higher than their normal working values. The manufacturers of thyristors make a comprehensive list of the voltage, current, power and temperature r atin gs after carefully testing the device. If SeRs are operated under these specified conditions, no damage will be done to SCRs. The object of this section is to discuss the various SCR ratings. A thyristor has several ratings such as voltage, current, power, dul dt , dil dt, turn-on time, turn-off time etc. For correct application of the device in thyristor circuits, a kn owledge of these . r ati ngs is desirable. Some subscripts are associated with voltage and current ratings for convenietlce in identifying the m. First subscript letter indicates the direction or the sta~e : D -+fo rwa rd-b lockin g r egio n with gate circuit open; T --ton-state; R ~re verse F ---tforward. Except for the gate G, second subscript letter denotes the operating values. W -}workin g valu e; R -+repetitive value ; S ---tsurge or non-repetitive value; T -+trigger Third s ubscript letter I\;/ indicates the maximum or peak valu e. Ratings with less than three subscripts may not follow these rules. Gate ratings involve the s ubscript G. Subscript A usually stands for anode and subscript AV for average. 4.6.1. Anode Voltage Ratings A thyri3tor is made up of fou r layers and three junctions as shown in Fig. 4.1 (b). Th e middle junction J'2 blocks the forwa r.l voltage whereas the two end junctions J ,. JJ block the reverse voltage. The anod e voltage r atings indicate the values of maximum voltages tha t a thyristor can withs tand without a breakdown of the junction area with gate circu it open. For ac systems, the supply voltage may n ot be a smooth sine wave. The voltage transients may occur regularly or at random as shown in Fig. 4.17 (a). The diffe r ent anode voltage ratings are as un de r : (i) VDIr .\{ -Peck woding fo rward-blocking uoltage. It spec ifi es t he maximum fonvard-blocking volt age that a thyristor can withstand during its working. Fig. 4.1 i (a ) shows th at '-li·,~·'\1 i3 equal to the ma.:timum value of the sine voltage wave.
[Mt. 4.6]
Thyristors
143
(ii ) V DUM -Peak repetitiue forward-blocking uoltage. It refers to the peak transient voltage that a thyristor can withstand repeatedly or periodically in its forward-blocking mode. The rating is specified at a maximum allowable junction temperatUre with gate circuit open or with a specifi ed biasing resistance between gate and cathode.
Voltage V DRM is encountered when a thyristor is com mutated or turned-off. It may be recalled t h at during turn-off process, an abrupt change in reverse recovery curr ent is accompanied by a spike voltage L
:~
; this is responsible for the appearance of
VDR.\I
acr oss
thyristor terminals. (iii) V DSM -Peak surge (or non-repetitiue) forw ard-blocking voltage. It refers to the peak value of the forvtard surge voltage that does not repeat. Its value is about 130% of V DR.I/, but V DSM is less than forward breakover volta'ge V BO as shown in Fig. 4.17 (b).
",VOS M
(a )
(b )
Fig. 4.17. Anode voltage ratings during the blocking state of a thyristor. (iv ) VRWM -Peak working reuerse voltage. It is the ma.'Ximum r everse voltage that a thyristor can withstand repeatedly. Actually, it is equal to the peak negative value of a sine voltage wave, Fig. 4.17 (a). (v) V RR .\! - Peak repetitive reverse uoltage. It specifies the peak reverse transient voltage that may occur repeatedly in the reverse direction at the allowable maximum junction temp erature. The transient lasts for a fraction of the time of one cycle, Fig. 4.17 (a). The reason for the periodic appearance of V RJUf is the same as for V DR.\!' (vi) VRSM - Peak su rge (or non-repetitive) re ve rse voltage . It represe nts the peak value of the reve rse surge voltage that does not repeat. Its value is about 130% of VRRJf . But VRS.\I is less than r everse breakover voltagc-"BR as shown in Fig. 4.17 (b ). Both VDS.\I and VRS.\f ratings can be increased by conn ecting a diode in series with a thyristor. Th e anod e vo ltage ratings listed above from (0 to (iii) pertain to forw ard ' blocking voltages whereas from (iu) to (vi) belong to reverse blocking voltages; a thyristor must be J.ble to suppOrt these voltages safely with gate circuit open . (vii) V T - On-state voltage d rop. It is the voltage drop between anode and cathode .....ith specified forward on-state current an d junction tempe ratu re. Its value is of the ord ~r of 1 to 1.5 V. (v iii ) Foru'ard du /d t rating. If rate of ri se of fo n~ard anode-to-cilthode vol tage !.:i high, thyr : ~t o r mny turn on eve n when
"
Power Electronics
[A r t. 4,6J
144
there is no gate signal and voltage is less than fonvard breakover voltage. When a thyristor is in the forward blocking mode, the applied voltage appears across junction J 2 as junctions J 1 and J 3 are forwa rd-biased. The reverse biased junction behaves like a capacitor. When forward voltage is sudd~nl y appli ed to the device , a charging current C;. du l dt begins to fl ow which may turn on SCR as explained in Art. 4.2. A high value of du/dt , at which a thyristor just gets turned on is called critical rate of rise of anode voltage or forward du / dt rating of the device. If applied du / dt exceeds this critical value, thyristor gets turned on . For applied du / dt lower than forward du / dt r ating, thyristor remains in forward blocking mode. (0 )
(b) anode.to.cath~de
The forward du / dt ratin g depends on the junction temperature; hi gher the junction temperature, lower the forward du / dt rating of the device . In pr actice, du /dt triggering is never employed as it gives random turn-on of a thyristor. This type of triggering also leads to destruction of the device through high junction temperature. (ix) Voltage safety factor (V SF ) ' It is defined as the ratio of peak r epetitive reverse voltage (VRRM ) to the maximum value of input voltage. V ..
= Peakrepetitive r everse voltage (VRRM) SF..f2
x rms value of input voltage
Voltage safety factor is usually taken between 2 to 3. (x) Fin ger voltage. It is the minimum value of forward bias voltage between anode and cathod e for turning- on t he device by gate triggering. The magnitude of finger voltage is somewhat more than the normal on-state voltage drop in the thyristor. 4.6.2. Current Ratin gs A thyristor is made up of semiconductor material, its the rmal capacity is therefore quite small. Even for short overcurrents, the junction temperature may exceed the rated value and the device may be damaged. As the j u nction tempe r ature is dependent on the c,-!rrent handled by a thyristor, a correct (a) cho ice of curren t r atings is essential for a long workirrg life of the device. In this part of ,,".,/Temp.curve ,", " \ '. the article , current r atings of SCRs are , ;. discussed fo r both repetitive and c ", , " I , ,, non-repetitive type of current waveforms. ~!' ---Tj=125 ·C B --f- -- - - -- :~"! ~--- - - 1-Average on-stote current (lTAV)' The o ,:-.-I • forward voltage drop across conducting SCR Anodll: ,, ~urnnt ,, is low, th ere fo r e power loss in a thyristor depends pri mar ily on forward average \ I on- state cu r rent ITAv. For th e purpose of Conduction °f- T --...! I ang lll: illust ra ting the si gnificance of average ~ 2T = 3aO e' ---I ( b) on -s tate cur rent, consider a continuous dc current 0.4. fl owing through the SCR, Fig. Fig. 4. 18. Variation of junction temperature with 4.18 (a ). Me: th e application of this current constant anode current ic and wit h a: t = 0, junction temperatu re begins to rise rectnngulo.r wnve of io .
' .,
.
r- '--
,
----..
[Art. 4.6]
Thyristors
145
until finally it reaches its rated value Tj = 125°C. As the SCR has low thermal time constant, final ·temperature of 125°C is reached in a relatively short time. Suppose now that anode current is of rectangular waveshape with conduction angle 180' (
=
ir
x 360' ) as shown in Fig.
4.18 (b) . If the rectangular wave bas. average value equal to the constant current OA in Fig. 4.18 (a ), then current amplitude of rectangular wave in Fig. 4.18 (b) is DC = 2 times OA . As the SCR has short time constant, junction temperature in Fig. 4.18 (b) is likely to exceed the allowable
temperature of 125°C and this is not desirable. In order to limit the temperature to 125°C for rectangular wavefonn of anode current; there are two techniques, (i) provide better cooling to the thyristor or (ii) reduce the pulse amplitude from OC. As per the second method, pulse amplitude of anode current is r educed from OC to some lower value on (say), so that junction temperature remains within limits, Fig, 4.18 (b). But a reduction in the amplitude of rectangular wave would result in a lower value of average anode current. This means that for the temperature rise to remain within limits, SCR must be rated at a lower value of average forward current I TAv when it is conducting a pulsed anode current than when it is carrying a constant dc. This shows that thyristor is derated when it handles rec tangular or square wave of anode cumnt. The effect of conduction angle on anode current ITAv is depicted in Fig. 4.19 (a) for rectangul ar waves. The avera.ge on·state power loss Pa ~ in this figure is apprOximately given by Pall
= (forward on-state voltage across a thyristor) x I TA V
It can be obtained more accurately fr om the relation Pall
=~
J(instantaneous voltage across SCR) (instantaneous current through SCR) dt
where T = periodic time of the anode current waveform. The rms current for an SCR is constant whatever the conduction angle may be. But average current is given by (Irrru I FF') where FF is the form factor of the current waveform. The conduction angle for sine wave is defined in Fig. 4.19 (b). For the same conduction angle , the
180'
t
1,
3:
3:
.E
•
> 0.0
> 0.0
ITAyil A _
(a )
li..,W in A _ (b)
Fi g. 4.19. Avera ge on·state power dissipa tion Pa ll as e. fiJn ction of IT~\ ·..,for (0 ) rectangular wa've and (b) hal i·w:.w e s inuJoid .
146
Power Electronics
[Art. 4.6]
form factor for sine wave is higher than , for the rectangular wave (see Examples 4.8 and 4.9), This means that average current for sine wave will be lower than it is for the r ectangular wave for the same de (or rms ) current. The derating of the SCR is therefore more for sine waves than for the square or r ectangular waves . The effect of conduction angle on average current is depicted in Fig. 4.19 (b) for sine wave. For 180" conduction angle, the anode current in Fig. 4.19 ( b ) is less than that in Fig. 4.19 (a ). This diagram is applicable for I-phase half-wave circuit (o r I-phase one-way, one pulse circuit). Curves shown in Fig. 4.19 are supplied by the manufacturers ofthyristors and are valid for supply frequency range 0[50 to 400 Hz. The curve marked dc in Fig. 4.19 (0 ) is applicable when anode current is continuous dc. The current for different conduction angles are terminated at different valu es of average current in Fig. 4.19. For example, for 30" conduction angle, I TAV terminates at [Idcl(form factor)] = (I dcI3.46 4) in Fig. 4.19 (0) for rectangular wave !lnd at (ldc / 3 .979) in Fig. 4.19 (b) for sine wave. At these terminal points, maximum rms current rati ngs of the device is reached . Table 4.1 gives different values of form-factor for different conduction angles of the half-wave sine waveforms.
Table 4.1. Form Factor tor Sine Waves Condu.ction angle
15'
3(¥'
45'
60'
90'
1200
1800
FCJrm factor
5.650
3.9812
3.233
2.7781
2.2214
1.878
1.5708
Curves of Fig. 4.19 are applicable when load is purely resistive. In case load i5 inductive in nature, these curves should be modified. With an improvement in the waveform , i.e. with waveform becoming more smooth, the form factor decreases and as a consequence, higher average on-state current I TAy can be handled by the device. RMS on·state current (IRMs)' By definition, for direct current, rms value I RMS or I rmJ = average or de value, I dc ' Heating of the resistive elements of a thyristor, such as metallic joints, leads and interfaces depends on the forward rms current l rnu. The rms current rating is used as an upper limit for constant as well as pulsed anode current ratings of the thyristor. Its value is equal to Ide of Fig. 4.19 (a). The value of the rms forward current for an SCR remains the same for different conduction angles . Average current, however, is dependent on conduction angle as shown in Fig. 4.19. For example, for 180" conduction angle, the form factor for half-sine wave is n: / 2, therefore average current is 2 I de/ X or 2 I''M /x. This means that for 180" conduction angle, thyristor circuit should be designed to carry an average current of 2 I dJrt instead of I de (or Ir'M)' The derating of the SCR below the dc value depends upon the current waveshape and it is defined as under : SCR derating below de value
=ld' - :; =li
1- F~ )
...(4.3 )
where FF is the form factor of the waveform. Its value is always more than one. For rectangular wave, FF is less as compared to its value for sine wave for the same conduction angle. Eq. (4.3 ) reveals that SCR derating below dc is less for rectangular wave than for the sine wave. The average current IT_4v fo r other conduction angles can be comput ed as discussed above. The significance of ITA'" and l'nl3 can be highlighted with an example. Suppose ma.'<..imum
35
r m3 curren t for a thyristor is 35 A. F or 120" conduction angle for sine wave, IT.4v = 1.875
Thyris tors
1M!. 4.6J
147
= 18.637 A.
This means that thyristor can h andle an average curr ent oi 18.637 A for 12 0 ~ conduction angle and its temperature will rem ain within limits. Suppose an ammeter is placed in s eries with the SCR for measuring the average current. Now decrease the conduction angle t o 30" but with average current as measured by the ammeter remaining unchanged at 18.637 A. But a n average current of 18.637 A at 30° conduction angle would require an rms current oi I rml = 18.637 x 3.9812 = 74.1976 A. But such a large value ofrms current would caus e large ohmic losses and is, therefore, certainly going to destroy the SCR. This shows that as conduction angle is reduced, I TAV must be lowered acco rdingly so that rms current is not exc eeded beyond its rated value and the SCR is not damaged . The current ratings I TAV and I,rru are of repetitive type. They are dependent on maximum junction temperature. If better cooling is provided to a thyristor body, th es e ratings can be upgraded . As stated above, power loss in a thyristor and its heating is dependent upon the rms current. Manufacturers also provide curves showing the variation of case temperature Te with
t
, '"
oU
t
120
u
•
~
v
~I ~
~
E
E
!
'••"
•• 0 v
0
v
E
E
0
~
•0
•
~
ITAV in A _
I TAV in A ----" (b)
(0 )
Fig. 4.20. Maximum allowable case temperature T ern as a function of frAY for (a ) rectangular wave and ( b) for balf-wave sinusoid .
average on-state currentITAv Fig. 4.20. These curves can be obtained :from Fig. 4.19 provided 9.k (thermal resistance betw~en junction and thyristor case)" in °C/ W is known. If Tj is the junction temperature, then
TJ.- Te = 9·"jt: . P0 11 For SCRs, Tj is usually 125°C. Taking 9.k = 0.15°CI W for dc current of200 A ; PfJ IJ = 300 W, fr om Fig. 4.19 (a), is obtained for I TAy = 200 A.
125-Te =0.15 x 300 or Te= 80°C This poin t is pl otted in Fig. 4.2 0 (a ) as A. For 80 A dc, Pa u = 100 W. 125 - T, = 0.15 X 100 or T, = 1l 0' C Thi s point is plotted as B in Fi g. 4. 20 (a ). For 180° con duction angle , for Ir..w = 140 A.
P
fJIJ
= 225 W fr om Fig. 4.19 (a ). T(
= 125 -
0.15 X 225
• Far und !rs tandi ng the term th C!rmal rC!sistance read Ar t . 4.8.
=91. 25°C
Power Electronics
(Art. 4.6]
148
This point is plotted as C in Fig. 4.20 (0 ), Other points can be plotted accordingly for rectangular as well as half-wave sinusoids to obtain the curves of Fig. 4.20. These curves indica te that for junction temperature Tj = 125"C, lower the average on-state current I TAv • greater is the case temperature that can be allowed for the same conduction angle. For example, fOT sine wave with 180" conduction angle, for I TAv = 120 A the case temperature Tem =91 DC ; for !TAV = 80 A the case temperature Tem = 104°C and so on. Surge Current Rating. When a thyristor is wolking under its repetitive voltage and current ratings, its permissible junction temperature is never exceeded. However, a thyristor may be subjected to abnormal operating conditions due to faults or short cirtuits. In order to accommodate these unusual working conditions, surge current rating, I TSM (pea,k non-repetitive on -state current), of thyristors is also specified. A surge current rating indicates the maximum possible non-repetitive, or surge, current which the device can withstand. Higher currents caused by non-repetitive faults or short circuits should occur once in a while during the life span of a thyristor to prevent its degradation. Surge currents are assumed to be sine waves with frequency of 50, or 60, Hz depending upon the supply frequency. This rating is specified in terms of the number of surge cycles with corresponding surge current peak. Surge current rating is inversely proportional to the duration of the surge. It is usual to measure the surge duration in terms of the number of cycles of normal power frequency of 50 or 60 Hz. For example, a three-cycle surge current rating for a period of 60 msec (3 x 20 msec) for 50 Hz supply consists of three conducting half-cycles, each followed by a n off-period. Thre e different surge current ratings are provided by the 1 manufacturers ; as for example, ITSM =3000 A for "2 cycle, ITSM =.2109 A for 3 cycles and
IrsM =1800 A for 5 cycles. One cycle surge current rating is the peak value of allowable non-recurrent half-sine wave of 10 ms~c duration for 50 Hz. For duration less than half-cycle i.e. 10 msec, a subcycle surge current rating is also specified. This rating for 50 or 60 Hz supply is the peak value for a part of the half-sine wave. The subcycle surge current rating I.b can be determined by equating the energies involved in one cycle surge and one subcycle surge as follows :
l;b ' t =1' . T l,b =I
or
~
... (4.8 a)
where
T = time for one half--<:ycle ofsupply frequency, sec I =one-cycle surge current rating, A l ,b = subcycle surge current rating, A t = duration of subcycle surge, sec f.:j r 50 Hz supply, T = 10 msec I 1 l,b = 10 . Tt
... (4.8 b)
,
I-t rating. This rating is employed in the choice of a fuse or other protective equipment for thyristors. The r ating in terms of amp2-sec specifies the energy that the device can absorb for a short time before the fault is cleared. It is usually specified for overloads lasting for less than , or equal to, one-half cycle of 50 or 60 Hz supply. The ft r ating is given by the relation . (rms value of one-cycl e surge current )2 x time for one cycle
...(4.9)
Thyristors
[Art. 4.6]
149
~ an example, Pt rating for 4 A(rms) SCRis 10 amp2.sec and for 35 ASCRis 100 amp2.sec. In order that a fuse (or other protective equipment) protects a thyristor reliably, the Pt rating of fuse must be less than the ft rating of the series-connected thyristor. dil dt rating. This rating of a thyristor indicates the maximu~ rate of rise of current from anode to cathode without any harm to the device. When a thyristor is turned on, conduction starts at a place near the gate. This small area of conduction spreads to the whole area of junction. If the rate of rise of anode current (dUdt) is large as compared to the spreading velocity of carriers across the cathode junction, local hot spots will be fonned near the gate connection on account of high current density. This causes the junction temperature to rise above the safe limit and as a consequence, SCR may be damaged pennanently. Therefore, a limit on the value of dUdt at turn·on is specified in amperes per microsecond for all SCRs. Typical values of dUdt are 20 to 500 A / J.l sec. Other ratings . In addition to the voltage and current ratings ofthyristors discussed above, there are some other ratings as under : (a) Latching and holding currents,
(bl Turn-on an~ ;urn-off times, . (c) Gate circuit voltage, current and power ra~gl': These ratings have already been discussed in Art. 4.1 to Art. 4.5. Detail ed ratings of any SCR during on-state and off· state can be obtained from the manufacturers by quoting the specification sheet number. . Example 4.9. The specification sheet for an SCR gives maximum nns on-state current as 35 A. If this SeR is used in a resistive circuit, compute average .on-stqte current rating faT half-sine walle current for conduction angles of (a) 180° (b) 90° and (c) 30°. Solution. For half-sine wave of current as s'h own in Fig. 4.21 (a ), .
150
[Ar t. 4.6]
Power Electronics
I
- I rm $
_
FF -
TAV-
35 x 2 _ 22 282 A It
-
.
(b ) For 90° conduction angle, 8 1 = 90°
and
Form factor
(e) For 30° conduction angle, 8 1 = 1500
la,
I,m.
I
= 2;
[1 + (- 0.866)]
= 0.02 13227 1m
[~: { 1"2 + ~ (- 0.866)}f2 =0.0849035100
=
(b)
(0)
Fig. 4.21. Pertaining to (a) Example 4 .9 and ( b) Example 4.10.
- 0.0849035 1m - 3 9818363 - 0.02 13227 1m - .
Form factor
35 I TAv =3.98 184
=8.7899A.
Example 4 .10. Repeat Example 4.9 in case the current has rectangular waveshape.
Solution. For the rectangular waveform of current shown in Fig. 4.21 (b),
~Lx 360
Conduction angle
nT
360°
or
n
=~C;-o-n-;d-u~ct;Ci~on-a-n-gl;-e
I =I x T =l
Here
nT
a ('
I
n
_l-I 2 x T]U2 _ I
rms -
nT
-Tn
[Art. 4.6)
Thyristors (a) F or
18 0~
151
360 conduction angle, n=-=2 180 I Iau =2 and
I 2 .m =T2 ' 1=,2
Form factor
35 I TAV = T2 = 24.7487 A 0
(b ) For 90 conduction angle,
n
= 360 90 =4
I I ou ="4 and
4 ="2I ' 1=2
Form factor
35 I TA V =2"=17 .5A ' (c) F or 30 ° can d uctlon angI e,
n _- 360 12 -_ 12 I I",u = 12
and
I '
:
I,.17U = ill
... .
I 12 = Form factor = ill ' T = , 12
35
ITAv ~ill
= 10.1036 A.
Example 4.11. An 8CR has half-cycle surge current rating of 3000 A for 50 Hz supply. Calculate its one·cycle surge current rating and I 2t rating. Solution. Let I and I,b be the one-cycle and sub-cycle su rge current ratings of the S9 R r~spe ct ively. Then equating the energies involved in them, w e get
PT=I;b ' t .. 1 2 1 r x 100 = (3000) x 200
or
or
1=
3~0 = 2121.32 A
2
From Eq. (4.9) ,
2 It
. ratmg
=
I,
1 (--:J2 3000 ) x 100 1 = 45000 Amp 2 . sec. x 2f=
Thus the SCR has one·cycle surge curr ent r ating of 2121.32 A and I 2 t rating of 45000 amp2.sec. Example 4.12. In the circuit of Fig. 4.22, the thyristor is gated with a pulse width of 40 microsec. The latching current of thyristor is 36 rnA.. For a load of 60 nand 2 H, will the thyristor get turned on ? Check. If the answer is negative, how this difficulty can be overcome for the given load. Find the m axim um va lue of tht~ remedial parameter shown dotted. S olution. The current through load and thyris tor is
. V' (l-e_!!.,)
LT ='R
For the circuit sh own,
-: ~
T JOO V -.l.-.
60n
2H ~
,, ,,
,,, l,,, ,, ,'
,,
"~
,,
L _ _ _ _.._____ -',
Fig. 4.22. P:rtaining to Example 4.12.
L
. -_ 300(1 tr 60 - e. 60 2 ' '' '''' ' ) -- o. 996 x 10· ' -- o. 996mA .
Power Electronics
(Art. 4.61
152
This shows that for a pulse width of 40 ~s, the anode current rises to 5.996 rnA which is far less than the latching current of 36 rnA. So thyristor will not get turned on. The remedial parameter, shown dotted in Fig. 4.22, should be resistance, say R I • because current can rise in resistance without any time delay. The value ofRl can be obtained as under: . 36 10· J 300 300 (1 tr= x =T+ 60 -e·0.00'2,)
,
300
k
3
R, = 30.004 x 10 = 9998 n = 9.998 n.
or
Example 4.13. During forward conduction, a thyristor has static J- V characteristic as shown by a straight line in Fig. 4.23. Find the average power loss in the thyristor and its rms current rating for the following load conditions: (a) A constant current of 80 A for one-half cycle.
100 A
(b) A constant current of 30 A for one-third cycle. (c) A half-sine wave vf peak value 80 A. Solution. It is seen from Fig . .4 .23 that for any current i a • the voltage drop across thyristor is . - liT
- 0.8 . 08 0012' = O.8 + 2.0100 x La = . +. La
Constant current of80 A for one-half cycle is shown O.SY 2.0Y Vo ia =80 A, the voltage drop across Fig. 4.23. PertainiIlg to Example 4.13. thyristor is liT = 0.8 + 0.012 x 80 = 1.76 V. From the waveforms of ia • vT shown in Fig. 4.24 (a), the average on-state power loss in thyristor is (a)
m Fig. 4.24 (a ). For
1
Pau = T
JTI2 . 0 VT' ta . dt
=~ J0
~2
1.76 X80dt =1.76~:OX
T
=70.4W
Waveform of ia gives the rms current rating of thyristor as
-J 80~;
T = 56.577 A io
:I
I
~,,,
I
BOA
t:-TI1~
.I
,
"
v,
VT
TI<' 0
-----I T
Til (aj
/'
, ,,, ,,
.
t
,,
w'
•
,,
,,
,
C.
O.96Y
- ---6 ~8\T --
•
1. (oj
F ig. -\,.24. Curr ent a nd voltage wave forms pertai ning to Example 4.13.
wt
(Art. 4.7)
Thy ristors (b )
Here
UT
153
= 0.8 + 0.012 x 30 = 1.16 V
P = 1.16x30 x T = 115W au 3T . Rm s current rating
=
30 x
fa
=
17.321 A
Half·sine wave of peak: value of 80 A, Fig. 4.24 (b), can be expressed as ta = 80 sin wt . UT = 0.8 + 0.012 x 80 sin wt = 0.8 + 0.96 sin wt From the waveforms for ia and UT shown in Fig. 4.24 (b ), the average on·state power loss is given by (c)
P"
J: = 2~ J :
= 21.
(0.8 + 0.96 sin wi) (80 sin WI) d (wl) 64 sin WI . d(wI) + 21.
J:
76.8 sin' WI . d(wI)
76.81 w1 - siri2W1 1" . 1 =-x 54 1-coswII' +-2n: 0 . 4n: 2, 0
Rms current rating
=20.372 + 19.2 =39.572 W = Imu = 80 = 40 A. 2
'.
2
4.7. THYRISTOR PROTECTION Reliable operation of a thyristor demands that its specified ratings are not exceeded. In practice, a thyristor may be subjected to overvoltages or overcurrents. During SCR turn-on, dUdl may be prohibitively large. There may be false triggering of SCR b.y high value of duldt . A spurious signal across gate-cathode terminal s may lead to unwanted turn-on . A thyristor must be .protected against all such abnormal conditions for satisfactory and reliable operation of SCR circuit and the equipment. SCRs are very delicate devices, their protection against abnormal operating conditions is, therefore, essential. The object of this section is to discuss various techniques adopted for the protection of SeRs. (a) dUdt protection. When a thyristor is fOIVIard biased and is turned on by a gate pulse, conduction of anode current begins in the immediate neighbourhood of the gate-cathode junction, Fig. 4.6 (a ), Thereafter, th e current spreads acro ss the whole area of j Ullc ti on. The thyristo r design permits the spread of conduction to the whole junction area as rapidly as possible. However, if the rate of rise of anode current, i.e. dil dt, is large as compared to the spread velocity of carners, local hot spots will be formed near the gate connection on account of high current dens ity. This localised heating may destroy the thyristor. The r ~re, the rate of rise of anode current at the time of turn-on must be k ept below the specified limiting value. The value of dUdt can be maintained below acceptable limit by using a small inductor, called dUdt inductor, in series with the anode circuit. Typical dildt lim it values of SeRs are 20-500 A/'tl sec. The method of determining inductance of dildt inductor is illustrated in Example 4.14. Local spot heating can also be avoided by ensuring that the conduction spreads to the whole are a as rapidly as possible. This can be achi eved by applying a gate curren t nearer to (but never greater than ) the m ~:dmum specified gate current. (b ) duldt protection. It has a!:-eady been discussed in .- \rt.-t 2 that ir'ra: : of rise of sudden ly applied voltage across thyristor is hi gh, th e devic e may get turn ed on. S_lch phe no mena of
154
Power Electronics
[Art. 4.7J
turning-on a thyristor, called du l dt turn-on must be avoided as it leads to false operation of the ' thyristor circuit. For controllable operation of the thyristor, the rate ofrise offol"'Nard anode to I cathode voltage dVa/ dt must he kept below the specified rated limit. Typical values ofduldt are' 20 - 500 V/ ~sec . False turn-on of a thyristor by large du / dt .;:an be prevented by using a sn ubber circuit in parallel with the device,
4.7.1. Design of Snubber Circuits A snubber circuit consists of a series combination of resistance R. and capacitance C, in parallel with the thyristor as shown in Fig. 4.25 . Strictly Rs Cs speaking, a capacitor C, in parallel with the device is Discharge sufficient to prevent unwanted dul dt triggering of the turnnl SCR When switch S is closed, a sudden voltage appears across the circuit. Capacitor C, behaves like a short circuit, therefore voltage across SCR is zero. With the passage oftime, voltage across C, builds up at a slow rate LOAD such that duldt across C, and therefore across SCR is les~ than the specified maximum du l dt rating of the device. Here the question arises that if C, is enough to prevent Fig. 4,25, Snubber circuit across SCR. accidental turn-on of the device bydul dt, what is the need of putting R, in series with C, ? The answer to this is as under. Before SCR is fired by gate pulse, C. charges to full voltage V,. When the SCR is turned on, capa ci tor d ischarges through the SCR and sends a current equal to V, I (resistance ofl ocal path formed by C, and SCR). As this resistance is quite low, the turn-on dil dt will tend to be excessive and as a result, SCR may be destroyed. In order to limit the magnitude of discharge current, a resistance R, is inserted in series with C, as shown in Fig, 4.25, Now when SCR is turned on, initial discharge current V,IR, is relatively small and turn-on dUdt is reduced. In actual practice; R" C, and the load circuit parameters should be such thatduldt across C, during its charging is less than the specified du l dt rating of the SCR and discharge current at the turn -on of SCR is within re~son~ble limits. Normally, R C, and load circuit parameters " values. form an underdrunped circuit so that duldt is limited to acceptable The design of snubber circuit parameters is quite complex, Here only an approximate method. of their calculation is presented in Example 4.14. In practice, designed snubber parameters are adjusted up or down in the final assembled power circuit so as to obtain a satisfactory performance of the power electronics system. Example 4.14. Fig. 4,26 (a) shows a thyristor controlling the power in a load resistance R L . The supply uoltage is 240 V dc and the specified limits for dUdt and duldt for the SCR are 50 A I J.LSec 'and 300 V 1 )JS~c respectiuely. Determine the ualues of the dil dt inductance and the snubber circuit p arameters R, and C,. Solu tion. Snubber circuit parameters R, and C, are connected across SCR and dildt inductor L in series with anode circuit as shown in Fig. 4.26 (b). When switch S is closed, the rapacitor behaves like a short circuit and SCR in the forward blocking state offers a very high r esistance. Th erefor e, the equivalent circuit soon after the instant of closing the switch S is as shown in Fig. 4.26 (c). For this circuit, the voltage equation is V,
=(R, + RLJ i + L ~:
... (4.1 0
oj
[Art. 4.7]
Thyristors
155
Snubber Cs ,1-'" cir(.lJit
r-- --- ---------- -.
i,
Rs
!
S
L
:
L
-~
- - -- - - - - - - - -
L
5
~<>------L~LroW ,
la)
(b)
(e)
Fig. 4.26. (a) Thyristor in series with RL (b) Thyristor protection with L and R~> C! (c) Equivalent circuit of Fig. 4.26 (b) at the instant switch S is closed.
Its solution gives, where In Eq. (4.10 a ), t is the time in seconds measured from th e instant of closing the switch. From this equation, di dt
=1 0e- II-r:: . .!.= or
R, + RL-._tl -r: V, R, +RL L. '
V, _II~
=r e
o.
The value of dil dt is maximum when t =
. ..(4.]0 b )
or
L=
The voltage across SCR is given by.
or or
= 240 x 10- ' = 4 8 H
Y,
(dildt)m~ Vo
=
50
. ~
Rs i 0
di dv o _ dt - R, . dt
(d:;L R,.(~:L
... (4.]] )
=
From Eq. (4. 10 b) and (4.11),
(dUoL dt or
=
R,' Y, L
"-
(dUol
R = -L - , Y, dt
... (4.12)
= -48 '- x300= 6a ~
240
The circuit of Fig. 4.26, consisting of R, L, C, should be fully analysed to determine the op timum values of sn ubber circuit par ameters R,> C,o The analysis of this circuit shows that resistance R, can be ob tained from the relation {9J
R,=2~#,
Power Electronics
[Art. 4.71
156
where ~ is the damping factor (or damping ratio). In order to limit the peak voltage overshoot across thyristor to a safe value, damping factor in the range of 0.5 to 1 is usually used. For optimum solution of the problem, ~ is taken to be about 0.65. .
,
c'= (~:JL=(2 X~.65J x 4.8 x 10-·=0.2253~F
It is seen from Fig. 4.26 (b) that when switchS is closed, capacitor C, is charged to dc supply voltage before the SCR is triggered.. Now when the SCR is turned on, capacitor C, will discharge a maximum current of V,IR, and total current through thyristor will be (V,IR, + V.r I RLl . It should be ensured that this current spike is less than the peak repetitive current rating (1TRM) of the SCR. Thus if R. is small, the current spike contributed by the discharge of C, will be large. In order to r educe this spike, R. is normally taken greater than what is required to limit d uldt . At the same time, value of C. is also reduced so that energy stored in C, is small and the snubber discharge does not harm SCR when it is turned on. Thus, in the present case, R, may be chosen somewhat higher than SO, say I on and C, somewhat less than 0.2253 ).1F, say 0.15 ).1F. The adoption of the new value of R, demands a new value of L . From Eq. (4,12),
L=
R, . V, = 103xO~40 = 8 ~H (d val dt}rrwu
This value of inductance is more than that required to limit di l dt to 50 A/j.lsec. For ac circuits, maximum value of input voltage (V m ) can be used in place of V, in Eq. (4.12) for computing R,. Example 4.15. A thyristor operating from a peak supply voltage of 400 V has the following specifications : . Repetitiue peak current. I, = 200 A,
(dildt)~ = 50 A/~. (~~ l~ = 200 V/~.
Choosing a factor of safety of 2 for 1p ' T~
(~: Land (~~
L'
design a suitable snubber circuit.
m inimum value of load resistance is 10 n. 200 Solution. For a factor of safety of 2, the permitted values are I p =-=100A. 2
(
diL 50 = 20- AI~, (duldt)m~ = 2 200 = 100 V I IlS · dt ="2
In order t o r estrict the rate of rise of current beyond specified valu e, (di/dt) inductor must be i:Jerted in series with thyristor. From Example 4.14, V,
L = (dil dt)=u =
400 x 10-' 25 = 16 ~H
R =L(dul =1 6 X 10-' x 100 =40 , VJ dt u 400 10- 6
Before thyristor is turned on, C, is charged to 400 V. 'When thyristor is turned on, the peak current through the thyristor is 400 400 = 140 ~ 10 + 4 ..
..
[Art. 4.7]
Thyristors
157
As this peak current through SCR is more than the pennissible peak. current of 100 A, the magnitude of R, must be increased. Taking R, as 8 n, the peak current through the SCR 400 = 400 10 + 8" = 90 A, less than the allowable peak current. So choose R, = 8 o.
C.=(¥,J'L=(\j3)' x 16 x lO-· =0.4225~
Also
The value of C, may be lowered as discussed in the previous example, so C, may be taken as 0.30 ~F. At the instant switch S is closed, Fig. 4.26, thyristor is open circuited and current through C, is given by du V, C, dt == R, +RL
o3
or
. x
or
10- 6 du = 400 dt 10 + 8 du 400 -d =-18 x t
1 6 = 74.07 V/ ~s 0.3 x 10-
Since designed value of(du / dt) is less than the specified maximum value oflOD of C, chosen is correct. So choose L = 10).lH, R, =8 n and C, = 0.31J,F.
V/ ~s ,
value
Example 4.16. Thy ristor shown in Fig. 4.27 has ilt rating of20A 2s. 1ftenninal A gets short-ci rcuited to ground, calculate the fault clearance time so that SeR is not damaged. f\) 230 /2 sin 3141 Ion Solution. The worst possible fault current I should he considered for calculating the fault clearance time. Maximum fault current occurs when source voltage is at its peak = 230..J2V. Fig. 4.27. Pertaining to Example 4.16. When terminal A gets short-circuited to ground, the resistance offered to s our ce x l . . 230-12 x 11 . =1 + 10 10 + 1 =21/ 11 O. Assummg m8Xlmum fault current = 21 A to remam constant during the short clearance time
tr:.
we get
f;',' f;' (230~ · dt=
..
x 11)' . dt =20 A' s
t, =20[ 230N x
III
x 1000 ms = 0.6892 ms.
4.7.2. Overvoltage Protection Thyristors are very sensitive to overvoltages j ust as other semi-conductor devices are. Overvoltage transients are perhaps the main cause of thyristor failure. Transient over voltages caus e either maloperation of the circuit by u nwante d turn-on of a thyristor or permanent damage to the device du e to reverse breakdown. A thyristor m ay be subjected to inter nal or extern al overv oltages ; the former is caused by the thyris tor oper ation whereas the latter comes fro m the supply lines or the load circuit. (£) lnte rna! oue ruoltag es. Large volt ages may be gen er a ted intern ally dur in g the commutation of a thyris tor. After thyristor an ode curren t reduces to zero, anode current
158
[Art.
·1.71
Power Eleclronics
rever.ses due to stor ed charges. This reverse recovery curren t rises to a peak value at which time t he SCR begins to block. After this peak, reverse recovery current decays abruptly with large dildt . Bec au se of the s eries inductance L of th e SCR circuit, large transient voltage
L
~~
is produced. As this internal overvoltage may be several tim es the breakover voltage of
the devi ce, the thyristor may be destroyed permanently. (ii) External oueruoltages. External overvoltages are caused due to the interruption of current fl ow in an inductive circuit and also due to lightning strokes an the lines feeding the thyristor systems. Wben.a thyristor converteT is fed through a transformer, voltage transients ar e likely to occur when the transformer primary is energised or de-energised. Such overvoltages may cause random turn on of a thyristor. As a result, the overvoltages may appear across the load causing the flow of large fault currents. Overvoltages may also damage the thyristor by an inverse breakdown. F or reliable oper ation, the overvoltages must be suppressed by adopting suitable techniques . 4.7.2.1. Suppression of overvoltages. In order to keep the protective components to a minimum, thyristors are chosen with their peak voltage ratings of 2.5 to 3 times their normal peak working voltage. The effect of overvoltages is usually minimised by using RC circuits and non-linear resistors called uoltage clamping devices. .
The RC circuit, called snubber circuit, is connected across the device to be protected, see Fig. 4.30. It provid es a local path for internal overvoltages caused by reverse recovery current. Snubber circuit is also helpful in damping overvoltage transient spikes and for limiting d u/dt across the thyristor. The capacitor charges at a slow rate and thus the rate of rise of forward voltage (d u/dt) across SCR is also reduced . The resistance R, damps out the ringing oscill ations between th e snubber circui t and the stray circuit inductance. Snubber circuits are also connected acr os s trans fonner secondary terminals to suppress overvoltage transients caused by switch ing on or switching off of the primary winding. As snubber circuits provide only parti al protection to SCR against transient overvoltages, thyristor protection against such over-lOltages must be upgraded. This is done with the help of voltage·c1amping devices .
• ~
Current
"
,, ,
o
.~
a: ~
6
.
,
~
c
-•
'E Rrsistonet.,
~ !; u
"i
~
,,
'
"
. .,
\ ,.
,
"
". Vo llage
• •,, ,, ••
.
,,
u
\
\ '",
Prok loull current -...... without any 1 luse ,."'0 ",
, ,,,
'.
.'
\
•\ I
c
o L~let :inqL !,rein~ I IIml , t,;J"'" tme, t o ~C [ £or l nc:;J
In)
lime
lim!!:
Ib) an.! yolt-r:!3istance char acteri s tics of voltage-damping de'lice
Fig. L2J. (e ) Vol ~ ·~t:l~e re (b ) Action of currant· limiting fu se i n en !I.e circuit.
Thyristors
[A ct. 4.7]
159
A voltage-clamping (V. C.) device is a non-line ar resistor connected across SCR as shown in Fig. 4.30. The v.e. device has falling resistance characteristic with incr easing voltage, F ig. 4.28 (a). Under normal working conditions of voltage below the clamping level, the device has a high resistance and draws only a small leakage current. When a voltage surge appears , the v.c. device operates in the low resistance region and produces a virtual short circuit across the S CR. The incr eased current associated with virtual short circuit produces an increased voltage drop in the source a nd lin e impedances and as a r esult, voltage across SCR is clamped to a safe value. After the su rge en ergy is dissipated in the non-linear resistor, the operation of the V.C. device returns t o its high resistance region . Selenium thyrector diodes, metal oxid e varistors or avalan ch e diode s uppre ssors are commonly employed for protecting the thyristor circuit against overvoltages. As the voltage clamping ability of a thyrector is inferior to those of metal oxide varistor and avalanche-diode suppressor, use of thyrector is on the decline. It has already been stated that RC snubber is not enough for overvoltage protection ofSCR. In practic, therefore , a combined protection co nsisting of RC snubber and V.C . device is provided to thyristors as shown in Fig. 4.30.
4.7.3. Overcurrent Protection Thyristors h ave small thermal time constants. Therefore, if a thyristor is subjected to overcurrent due to faults, short circuits or surge currents ; it~ junctt'm}\emperature may exceed the rated value and the device may be damaged. There is thus a need for the over current protection of SCRs. As in other electrical systems, overcurrent protection in thyristor circuits is achieved through the use of circuit breakers and fast- acting fuses as shown in Fig. 4.30. The type of protection used against overcurrent depends upon whether the supply system is wea.1( or stiff. In a weak supply network, fa ult current is limited by the source impedance below the multi-cycle surge current rating of the thyristor. In machine tool and excavator drives, if th e motor stalls due to ov erloads, the curren t is limited by the source and motor impedances. The filter inductance commonly empl oyed in dc and ac drives may limit the rate of rise of fault curr ent below the multi cycle su rge current rating of the thyristor. For all such systems, overcurrent can be interrupted by conventional fuses and circuit breakers. However, proper co-ordination is essential to guarantee that (i) fau lt current is interrupted befor e the thyristor is damaged and (ii) only faulty branches of the n etwork are isolated. Conventional protective methods are, however, inadequate in electrical stiff supply networks. In such systems, magnitude and rate ofri se of current is not limited because source ' has negligible impedanc e. As such, fault current and therefore junction tempera ture rise wit hin a few milliseconds. Special fast-acting current-limiting fuses are, therefore , required for the protection of thyristors in these stiff supply ne tworks . The operation of fa5 t-acting current-limiting fuse is"illustr ated in Fig. 4.28 (b ). These fuses and thyristors are found to have similar th ermal properti es, their co-ordination is therefor e sim pler. Th e current-limiting fuse consists of one or more fine silver ribbons having very short fusing tim e. In Fig. 4.28 (b ), fadt is shown to occur at zero crossin g of the ac sine wave, i.f . at t = O. 'W ithout fuse, the fault current would rise upto A and then would follow dotted curve, reach pea.1( valu e D and then decrease as shown. A prope rly selected current limiting fuse melts at A. An arc is then struck. F or a brief interval after A, the cu rTent continues to ris e depending upon th e circuit parameters nnd the fus e design . This current reaches a pea.l.;, value , called peck Ie : :n-ough current, which i3 indicated by point B in Fig. 4.28 (0). _ ote that pe ak let through current is considerably less than th e pe":lk fault current without the fuse , the latter is indicated by point D . ..\fter the point B, arc resisnnce L'1Cre a3es and faul t current decr eas es . At point C, arcin g s tops and the fault current is clear ed. Tp e t otal clearing ti me t~ i:i the su m of melting ti me t", and a rc ing time t;:, i .~. t: = t", - t:: .
160
[Art. 4.7]
P ower Electronics
Proper co·ordination between fast-acting current-limiting fuse and thyristor is essential. A fuse carries the thyristor current as both are placed in series. Therefore, the fuse must be rated to carry full-load current plus a marginal overload current for an indefinite period. But the peak let through current of fuse must be less than the subcycle surge current rating of the SeR. The voltage acr oss the fuse during arcing period is known as arcing, or recovery, voltage. This voltage is equal to the sum of source voltage and the emf induced in the circuit inductance during arcing time tao If the fus e current is interrupted abruptly, induced e.m.f. L
~~
may be
high; as a result arcing voltage would be excessive. It should therefore be ensured during fuse design a~d co·ordination that arcing voltage is limited to less than twice the peak supply voltage. In case voltage rating of the fuse is far in excess of circuit voltage, an abrupt current interruption would lead to dangerous overvoltages. When both circuit break er and fast·acting current-limiting fuse are used for overcurrent protection of SCR, Fig. 4.30, the faulty circuit must be cleared before any damage is don e to the device. A circuit breaker has long tripping time, it is therefore generally used for protecting the semiconductor devi ce against the continuous overloads or against sur ge currents of long duration. A fast·acting C.L. fus e is used for protecting thyristor s against large surge currents of very short duration. The tripping time of the circuit breaker, the fu sing·time of the fast-acting fuse must be pr o ~e rly co·ordinated with the rating of a thyristor. In order that fuse protects the thyristor reliab ly, the [2t rating of the fuse must· be less than that"Of the SCR. Electronic crowbar protection. As thyristor possesses high surge current capability, it can be used in an electronic crowbar circuit for overcurrent protection of power converters using SCRs. An electronic crowbar protection provides rapid isolation of the power converter before any damage occurs. Main fusr
-
Fig. 4.29 illustra tes the basic principle of electronic crowbar protection. A crowbar
+ Crowbar thyristor'"
thyristor is connected across the input de
Convtr· t .. Powrr
l jb- Got!Ci rTri~f!r cuit
l
0
A terminals. A current sensing resist or detects the value of converter current. If it r- 0 exceeds preset value, gate circuit provides Currrnt the signal to crowbar SCR and turns it on in '-srnSlng rrglSt rr a few microseconds . The input terminals Fig. 4.29. Elementary electronic crowbar circuit. are then short-circuited by crowbar SCR
I J
Over cur rem Protection ,.. . ______ _. .. . ___ \1 . ., \
, :
C. 8 .
i '
;I
Snubber cirC'Jit , ~-.- .. :\.. _..., , ,
~ inductor
, UUV
F.A.C.L.Fi
L __._ •• ____ • __ ••••• .i
Gol e Protection
,.----------- -- .---'> .- - .--,
I
H 5
J
I
SeR
L~ : ~
c >
\
' ,:
,
, ,,, , ... . _. _ _ _ _ __ __ - - - • _ _ _ - - - ...I
~ - .-
-_. --- -
Fig. 4.30. Circuit component:.; showing the thyristor protection. C.B.-Circuit breaku ; F.A. C.L.F.-Fa3t acting C'.l."'Tent limiting ftlse ; H.S.-Heat sin.\. ; ZD-Zener diode.
Thyristors
(Art. 4.7J
161
and it shunts away the converter overcurrent. The crowbar thyristor current depends upon the source voltage and its impedance. After some time, main fuse interrupts the fault current. The fus e may be replaced by a circuit breaker if SCR has adequate surge current rating. 4.7.4. Gate Protection Gate circuit should also be protected against overvoltages and overcurrents. Overvoltages across th e gate circuit can cause false triggering of the SCR. Overcurrent may raise junction temperature beyond specified limit leading to its damage. Protection against over-voltages is achieved by connecting a zener diode ZD across the gate circuit. A resistor R2 connected in series with the gate circuit provides protection against overcurrents. A comm on problem in thyristor circuits is that they suffer from spurious, or noise, fi ring. Turning-on or turning-off of an SCR may induce trigger pulses in a nearby SCR. Sometimes transients in a power circuit may also cause unwanted signal to appear across the gate of a neighbouring SCR. These undesirable trigger pulses may turn on the SCR leading to false operation of the main SCR. Gate protection against such spurious firing is obtained by using shielderl cables or twisted ~ate leads. A varying flux caused by nearby transients cannot pass through twisted gate leads or shielded cables. As such n o e.m.f. is induced in these cables and spurious firing of thyristors is thus minimised. A capacitor and a resistor are also connected across gate to cathode to bypass the noise signals, Fig. 4.30. The capacitor should be less than 0.1 ~F and must not deteriorate the waveshape of the gate pulse. Example 4.17. For the circuit shown in Fig. 4.31, 10 A o.15 )JF (a) calculate the maximum ualues of dil dt and d ol dt for the SCR, (b) find the rms and atJerage current ratings of the SCR for firing angle delays of90"and 150" and
SCR
/2. 230 s in 314\
(c) suggest a suitable voltage rating of tit. SCR. Solution. (a) From Eq. (4.10),
(~;L =[~)
2"
Fig. 4.31. Pertaining to Example 4.17 .
= '1'2 · 230 = 21.685 A/ ee. 15 x 10- 8 ~ From Eq. (4. 11),
(~~L =R· (~;L = lOx 21.685=21~;85V/~ec
(b ) For 1 5~, X L = 314 x 15 x 10- 8 = 0.00471
R
n. As
this value of X L is much lower than
= 2 n, the current is primarily limited by 20. ..
I
-
= '1'2 . 230 = 115 . '1'2
2
For firing angle delays of 90° and 150:», the conduction angles are 90° and 30C respectively and from Example 4.9, the respective values of iorm factors are rt/{2 and 3.98184. '1'2 . 115 · '1'2 :. For firing angle delay of 90°, iT,w = =73.211 A
•
an d fo r firin g angle del ay of 150°,
'1'2 . ·115 ITAv= 3.98184 =40.8l!4 A
162
[Art. 4.7]
Power Electronics
RMS current rating of the thyristor is 115...[2 = 162.634 A for any conduction angle, but average currents aTe 73.211 A for conduction angle of 90° and 40.844 A for conduction angle of 30°. (e) Voltage rating of the SCR = (2.5 to 3) times the peak working voltage = (2.5 to 3) x 12·230 =813.173 V to 975.807 V. So a voltage rating of about 900 V may be chosen for the SeR. Example 4.18. For the circuit shown in Fig. 4.32 (a), the initial voltage across capacitor is Uc (0) = - 100 V. Sketch the time variations of i, ULI Vel iD and iL after the thyristor is turned onatt=O. Solution. When the thyristor is turned on at t =0, the voltage equation for the circuit is 1 f 'd L di dt + C t t
=V
6
Its Laplace transform is sL
or
1 [1(5) C· ",(0)] V, 5 =-;·1(5) + C • + 1(. ) =
V, L
"N .
1
1 s+LC 2
300
=T '
1
lIvLC 2
1
~LCs+LC
.-
Vl'Yc
3001/
.,~ L....".L---'~_-'-_+-~
T(1~______L-~~~~ o
iLk:: . "2 (b)
(0)
Fig. 4.32. Pertaining to Example 4.18.
300 . Its Laplace inverse is ,'(t) = "',;sm "'ol
h were
1
"'0 = ~LC
diet)
=L dt =300 cos "'ol Vc = V, - UL = 200 - 300 cos toot
"L
The current and voltage waveforms are as shown in Fig. 4.32 (b). At 1i: / 2, uL tends to reverse and as a result, diode D gets forward biased and current i L starts flowing through D as iD ' uL is therefore zer o from 1t/2 to 'It. Voltage u~ remain,:; 200 V and current i zero from 1t/2 to 7t as shown in Fig. 4.32 (b ).
Thyristors
[Art. 4.8J
163
4.8.. IMPROVEIIIENT OF THYRISTOR CHARACTERISTICS It has be en explained in Art. 4.3.1 that rate of growth of anode current during rise time of t or: is hi gh, but cathode-conduction area is small. High rise of anode current in a thyristor, associated with high anode voltage, causes more losses. These high losses, occurring over a sma ll cathode-conduction area during rise time, may result in hot spots leading to the destru ction of the device. A high value of du / dt may turn on the thyristor at an unwanted instant which is undesirable. This all prompts us for an improvement in di/dt as well as d u/dt ratings of thyristors. A boost in these ratings can be made by doing some structural modificati ons in thyristo.rs; this is explained below. 4,8.1. Improvemen t in dildt Rating The rate of rise of anode current (di/dt) in a thyristor depends primarily on the initial area of cathode conduction during rise time. This implies that if initial cathode conduction area is increased, the di/dt rating also gets improved. There are two methods of doing this, (i) by us ing a higher-gate current (ii) by intermixing the A gate-cathode regions.
'.
. " 4.8.1.1. Higher-gate current. At the start of turn on, if P1lot Man !hyris~or thynstcr higher-gate current is applied, turned-on area of cathode surface / .' . ......... has to be more for handling this higher-gate current. As a consequence, initial cathode-conduction area for allowing anode curr e n~ to pass through it, increases, and this is what is desired. The widely used gate current profile is shown in Fig. 4.7. However, FJom gc!e..eny ~ elfe!.: .! K big.her gate current should not be obtained from a gate drive circuit . The usual way of accomplishing this goal of higher-gate Fig. 4.33. l'o'fain thyristor turn current is through the use of a pilot thyristor shown in Fig. 4.33. 2n by pilot thyristor. When pilot thyristor is turned on, a high value of gate current flows from anode A, pilot thyristor and gate-cathode terminals of main thyristor for switching Hoo. . ' 4 .8.1.2. Structural m odification of the device . As stated above, the di / dt rating of a thyristor can be improved by having more cathode-conduction area during delay and rise time of totl ' This can be achieved by higher.gate current (already discussed ) and by modifying the gate-cathode geometry. This alteration consists of iritermixing, or interdigitating gate and cathode regions . The effect of this structural change can be r.ealized by examining the initial conduction process firs t (i) in side-gate thyristor and then (ii) in centre-gate thyristor.
...
A side-gate thyristor is shown in Fig. 4.34 (a). When gate current is applied , the gate·cu rrent density is higher near the gate terminal. As a r esult, cathode-conduction area is small during delay and rise time . This :=.llows that initial conduction occurs over a nar row channel n ear the gate terminal as shown in Fig. 4.34 (a). Refer ence to Fi g. 4.6 and its rele'l;mt write·up is also helpfuL A cen tre-gate thyristor is shown in Fig. 4.34 (b ). \-Vhen positive gate pulse is appli ed, the gate current flows fr om gate to cathode in all possible direc tions covering a ring·shaped area on the cathod e surface as indic a.ted in Fig. 4.34 (b). Exam ination of Fig. 4.34 (0 ) and (b ) shows that initial ar ea of cathode conduction is very large in centre-gate thyristor as compared to that in sid e·gate thyristor. Thi3 illustrates that initial area of cath ode conduction can be enhanced significantly by intermixing th: gate-cathode reg:ons approp riately.
164
Pow er Elec tronics
[Art. 4.8] Anode
Anode
r-~P'~~~::::=1-An~e current f----tt-----j J,
p' Al"IOde
current
r - - t t ----1 J, !.!---:--j"
~;;:=\I J' -t==== f--+--+---j J, J,
Cothode
Initiol cathode conduction
area
(aJ
( bJ
Fig. 4.34 . Initial cathode-conduction area in (a) side-gate thyristor and (b) centre-gate thyristor.
"
Anode
Another configuration indicating the intermixing, or interdigitating of the gate-cathode regions is illustrated in Fig. 4.35.
p.
4.8.2. -Improvement in dv / dt rating
It has already been discussed in Art. 4.2 (du/d/
p
triggering) that when du/dt is large, high charging currents flow through the reversed biased junction J 2 which may turn ,,""".,,-L--.,-'-..-'......-'-..-'..-' on the thyristor: The effect of capacitor-charging current, or Gele du /d t, can be minimised by using cathode-short structure show n in Fig. 4.36 (a). Cathode-shorts are realized by Colnode overlapping metal on cathode n+ regions with a narrow Fig. 4.35. Interdigitating of p-region in between. Fig. 4.36 (a) shows metallization M and gate-cathode regions in a thyristor. N which form the cathode. Most of discho current dvldt
r
,g,
p'
-
p'
J, o·
o· J,
p
"
o·
Cathode short K
K (a)
., 0
RA
I
FIR I
I/
r rJ
0-
G
~ ICcthode shor·t ..l..l (b )
Fig. 4.36. Thyristor cathoda-3horts (a) 2!ementary intermi:ring
(0)
advanc::!d intermixin g.
Thyristors
(Art. 4.9J
165
In normal structure, discharge current d uldt (acting as gate curren t) flows through J 3 junction, Fig. 4.3 (a), and leads to spurious tum of of SCR. In cathode-short structure, most of the discharge current (or displacement current) passes through narrow p channels in between cathode n· regions as shown in Fig. 4.36 (a ). Junction JJ shares only a negligible amount of dvl dt current. A little discharge current flowing through J 3 junction (and acting as gate current) is too small to turn on the device. Thus, higher valuos of dul dt are now pennissible with cathode-short structure. The thermally generated leakage current across junction J 2 also does not pass through gate-cathode junction J 3 . Therefore, current injection across gate-cathode, or J 3 , junction is drastically reduced, hence the total discharge current, or dul dt , can be larger without' turning-on the device. It can, therefore, be inferred that cathode-short structure improves du / dt rating of the thyristor. 1.9. IIEATING, (,OOLING ANI!
~\OUNTING
(W THYHlSTOnS
Some power loss occurs in a thyristor during its working. The various components of this power loss in the junction region of a thyristor are as under : (i) Forward conduction loss (ii) Loss due to leakage current during forward an~ reverse blocking (iii) Switching losses at turn-on and turn-off (iu) Gate triggering loss At industrial power frequencies between zero and 400 Hz, the forward conduction loss, or on-state conduction loss, is usually the major component. But switching losses become dominant at high operating frequencies. These electrical loss es produce thermal heat which must be removed from the junction region. The thermal losses and hence the temperature rise of the device increase with the thyristor rating. The cooling of thyristors, therefore, becomes
more difficult as the SCR rating increases.
.
The heat produced in a thyristor by electrical loss is dissipated to ambient fluid (air or water) by mounting the device on a heat sink. When heat due to losses is equal to that dissipated by the heat sink, steady junction temperature is reached. Thyristor heating and hence its junction temperature rise is dependent primarily on current handled by the device during its working. As Buch, current rating of thyristors is often based on thermal considerations. 4.9.1. Thermal Resistance Thermal energy, or heat, flows from a region of higher temperature to a region of lower temperature. This is similar to the flow of current from higher to lower potential in an electric circuit. There is thus an analogy between thermal-power flow and current flow as given in tabular form below : Eledrnal quantit~s
The rmal quantilus
1.
Heat, J or Ws
1.
2.
Temperature difference, "C
3. 4.
Tbennal power, or rate of heat transfer, W Thermal resistance, "CIW
2. 3. 4.
Charge, C or As Potential diffeT~Dce , V Current. or rate of charge transfer, A Electrical ~sis tancc, VIA or ohms.
It is seen fr om above that thennal resis tan ce, an alogou3 to electrical resis tanc.e, is the resistance offer~d to thennal. power flow. Thermal r esis tance is denoted by 8. If p o'~er loss,
166
Powe r Electron ics
[A r t. 4.91
Pall in watts, causes the temperature of two points to be at T l °C and T 2 °C wher e Tl > T2• then t hermal resistance is given by 9 12 = Tl - T , ' CIW
... (4. 13 )
p~
The heat gener a ted in a thyristor due to internal loss es is t aken to be developed at a j unction within the semiconductor material A simple arrangement of thyristor, its case and h ea t si nk is shown in Fig. 4.37 (a) . Various temperatures and thermal r esistanc e are also indicated in this figure. The heat flow from thyristor junction to ambient fluid is as under:
from th e junction to thyristor case, thermal resistance 9jc (ii ) from t he thyristor case to h eat sink, thermal resistance 9" an d (iii) fr om the h eat sink to the surrounding ambient fluid (air or water), thermal resistance 9SA ' (i)
Tj a jc
9 0s
, (b)
(a)
Fig. 4.37. (0 ) Thyristor with its case and heat sink (b ) thermnl equivalent circuit for a thyristor.
There is thu s therm al resistance 9jc between junction temperature T, and case temperature Te' Simil arly, ther e is th erm al r esistance eel between Tc and sink te mp erature T! and 901A between T, and ambient temperature Tk Using the electrical analogy, a thermal equivalent circuit depicting the flo'.'V or h eat from junction to ambi ent fluid can be drawn as shown in Fig, 4.37 (b). Her e P au is the average r ate of heat generated at a thyristor junction and is analogous to constant-cu rrent source. Here p au
= Tj -Tc ~ Tc -T, =T,- TA Bjc e BrA Cl
_ TJ.-TA 9jA
... (4.14)
wh er e e', A= 9·JC + BCol + B_.. is the total thermal r es istance between junction and ambient. ;on .
The junction-to-case thermal resistan.:e 9jc is specified in th e thyristor data sheet. The cas:-to-sink therma l r esistance BI'S depends on the size of the device case, flatness of the case surface, the clam ping pressure and the use of conducting greas e beh. .·een the in terfaces. Usual value orge, vari es between O.0 5 ~C /W and O. 5D C/W. In addition to Bjc , thyristor data sheet also
.
Thyris tors
[Art. 4.91
167
specify 9u assuming correct installation procedure and use of the interface thermal lubrication. The sink·to-am bient thermal resistance 8sA. is independent of the thyristor configuration. The parameters on which e. . . depends are heat sink material, surface area and fmish of the heat sink, volume occupied by heat sink and the type of cooling (air cooling or water cooling). For naturally cooled heat sink, aaA may be equal to O.5°CIW and this valu e would be lower for better cooled heat sinks. The difference in temperature between junction and ambient can be written from Fig. 4.37 as Tr T.=P.,(9.. +9,,+ 9..)
...(4.15)
Eq. (4.15) shows that for maximum valu e ofTj (= 125°C), POll can be increased by reducing BaA. This means that by providing efficient cooling system to the SeR, the power dissipation capability of the device can be increased. 4.9.2. Heat Sink Specifications The thyristor data sheet specifies maximum junction temperature Tj t thennal resistances Bjc arid eell. The manufacturers of heat sinks provide catalogs in which sufficient data on heat sink is available. Fig. 4.38 gives typical data in the forI\l. of curves for standard heat sinks of aluminium extrusions. These curves relate temperature difference (T lI - T A) in °C between heat sink and ambient uersus average power dissipation POll in watts. 100
h
i
t
,v
Cur vt
Sink
dl"""';;'on«'""1
a ..... . . ... 3.2 xIOx7.5 b _·._. __ .... 3. 2xl0xI2 . 5 ~, ·------· .. 10xIOxlO
h----.......
IOxlOxl0 1·------ .. • 15.5xI5. 5_22.5
160
120
Pay in watts ___
Fig. 4.38. Standard heat sink ratings of aluminium extrusions.
In order to illustrate the use of these curves, choose a particular h eat sink and read POll and (T, - TA ). Then the thermal resistance of th e h eat sink to ambient is' calculated as T lI - TA
~=p
••
( ~l m
For maximum specified temperature Tj (usually 125°C) and a known ambient temperature TA , the permi3sible value of POll (with e)C and eu already known) is calculated fro m Eq. (4.15) with e.sA computed from Eq. (4.16). Ii this Pov is different from tha t chosen earher from Fig. 4.38, another heat sink with other values of Pall and (TJ - T,,) is tried until Eqs. (4.15) and (4. 16) are satisfied. After deciding the value of POll' use this value oi Pall in Fig. 4.19 to obtain permissibl e value of average current rating for a given conduction angle and current waveform.
168
[Art. 4.9)
Power Elec tronics
In the seconu m ethod of heat sink design, first average armature current is determined from a known current waveform and conduction angle. Corresponding to this average current, P cz,; is read off from Fig. 4.19. For this P fW thermal resistance aaA is determined from Eq. (4.15) as temperatures Tj (= 125°C), TA and 9ft. eu are already known . This computed value of 8sA is used to obtain temperature difference (T.. - T A ) between heat sink and the ambient from Eq. (4.16). Using these values of Pau and (T, - TA ), an appropriate heat sink is selected from Fig. 4.38, the details of which are usually supplied by the manufacturers. In the third method of h eat sink selection, first compute average armature current as done in the second method. For this value of average cUrrent, obtain Pou from Fig. 4.19 and case temperature Tc from Fig. 4.20 for the known current waveform and conduction angle. An
examination of Fig. 4.38 r eveals that the sink temperature T, in terms of case temperature T( is given by ...(4 .17)
As ambient temperature is known. (T, - T),) can be calculated. Now, with the knowledge
of Fau and (T. - T A ) . a cl:.oice of suitable heat sink can be made from Fig. 4.38. Heat sinks are made from metal with high thermal conductivity. Aluminium is the most commonly used metal. Copper, being a costly metal. is seldom used as a heat sink. material. Heat dissipation from heat sink takes place primarily by convection. As such, thyristor cooling by convection can be made more effective by enlarging the cooling surface area by providing the heat sink with peripheral fins. Heat dissipation also takes place by radiation. Heat sinks are usually provided with black anodized. finish to enhance the heat dissipation by radiation. Sometimes the size of naturally-cooled finned heat sink may become large. In such a case, size of the heat sink can be reduced by using forced air cooling which involves a fan ~l owin g air over the fins . With forced air cooling, heat-removing capability of the finned heat sink incr eases by a fact or of two t o three. For dissipating large loss es 10 high-power thyristors , water-cooling is usually employed·to get a compact size of the heat sink. 4.9.3. Thyristor Mounting Techniques Internal pow er losses in a thyristor cause high thermal stresses which further give rise to mechanical for ces. A thyristor must be braced to withstand such mechanical forces. In addition, SCR mountin g must be so designed as to facilitate heat flow from junction to the case. Depending upon the low or high power ratings of thyristors, there are five major mounting techniques for SCRs as described below : (0 ) Lead-mounting, For load-c~ent rating of about one ampere', lead-mounted SeRs are used, Fig. 4.39 (a ). Such SCRa do not require any additional cooling or heat sink. Their h ousings
dissipate sufficient heat by radiation and convection. (b) Stud-mounting. This type of construction shown in Fig. 4.39 (b) is very widely us ed due
to its fl exibility and ruggedoess . The threaded stud forms the anode. The SCR is attached to a h eat sink by means of threaded stud and nut. Thus anod e gets electrically connected to the heat sink. If electrical connection between anode and h eat sink is undesirable then mica or PTFE washers are used in betwee"n the joini."'lg surfaces. Both .mica and PTFE conduct heat easily put act as insulators to electricity.
(c ) Bolt·down mountin.g. This is als o called flat-pack m ounting. This type of device mounting has tabs \vith one or more holes. Sometimes the hole is provided in the middle a3
169
[Art. 4.9]
Th:.. ristors
c
Teb
H. S.l.
Holt
(lI HOlE
er
© ~
c·
C AG (0 )
CA G (0)
(b)
C 0
Cool ing fins
G a
,
HE
A
SCR •
~ H .S.
~
G
C HE: T . SINK
~)
Fig. 4.39. Different SCR mountings and heat sink.
shown in Fig. 4.39 (c). Bolts are pu shed through these holes so as to mount the device on to heat sink with nuts etc. In case the device is to be insulated from the heat sink, a thin insulating mica or PTFE washer is used between the device and heat sink and the bolt is made up of nylon. This type of mounting is used for small and medium ratings.
(d) Press-fit mounting. Press-fit (or pressure-fit) package is designed for insertion into an appropriate sized hole in the heat sink. The insertion may be done by using a vice and pressing the device into the hole using wooden block etc. For large sizes, the insertion is carried out by means of a hydraulic r am. This type of mounting is used for large rated thyristors. Fig. 4.39 (d) illustrates press-fit mounting of SCRs. (e) Press-pak mounting. This type of mounting is also called "disc" or "hockey-puck" mounting because of its shape. The SCR is clamped between t\l'O heat sinks, Fig. 4.39 (e) .and external pressure is applied evenly so that there is no deformation of any part. The heat sinks may be air, wateT or oil coolfi'd. Such type of mounting is used for thyristo rs of very high current ratin gs.
170
[Ac!. 4.9J
Power Electronics
Example 4.19. The data sheet for a thyristor gives the following values: T· Jm = 12S'C 9j<=O. l S·C/ W Be.. =O.075°C/W (a) For average power dissipation of 120 W: check whether the selection of heat sink g from Fig. 4.38 is satisfactory. Use first method of heat sink selection with ambient temperature of 40· C. (b) A sinusoidal voltage source 0(230 V. 50 Hz feeds power to a resistiue load of R = 2 n. For a firing angle delay of zero degree, choose a suitable heat sink and find the circuit efficumcy. (e) For the heat sink chosen in part (a), compute case and junction temperatures in case the firing angle delay is 60°. Solution. (0 ) For the' heat sink g , Fig, 4.38 gives a value of T .. -TA=54°C for Prlu =120W. From Eq. (4.16),
9. . = :2~ =
O.4,· CIW.
125 - 40 p o. = (0.15 + 0.075 + 0.45) - 125.93 W.
From Eq. (4.15),
As this computed value of Pall is different from the previous value of 120 W, another heat sink, say f, for which T, - TAo = 58°e for Pav = 120 W should be tried .
From Eq. (4.16),
58 9... = 120 = 0.483
From Eq. (4.15),
Po,
125 - 40
= (0.225 + 0.483) = 120.06 W .
This shows that selection of heat sink f is satisfactory. For Pall = 120 Wand for sinusoidal current, Fig. 4.19 (b) gives average current rating for the thyristor as 80 A for 1800 conduction angle or ex = 00 , 74 A for a = 60° and 68 A for cr. = 900 • (b) F or a = 0 0 , conduction angle is 180 0 • Here second method of heat·sink. selection is used .
I TAo V=
1 f ' Vm . Vm 'l'2x 230 2it 0 Ii: sm rot · dlcot) = rtR = 7t X 2
.. A
=51.77 =. . 2
For this current, P ou from Fig. 4.19 (b) is 90 W . From Eq. (4.15), From Eq. (4.16),
9...
= 1259~ 40 -
(0.225)
= 0.7194'
CIW
T. -TA =90xO.7194=64 .75· C
For T, - TA = 64.75°e and Pr;J.u= 90 W, Fig. 4.38 shows that heat sink c should be selected. The use of third method of heat. sink selection is also demonstrated. First I~A~ is calculated as in the second method. F or this current, Pall = 90 W from Fig. 4.19 (b) and T( =112°e from Fi g. 4.20 (b). Now sm.< temperature from Eq. (4.17) is T. = 112- 90 x 0.075 =105.25'C and For T, - TA = 65.25°C and Pcw = 90 W, Fig. 4.3B shows that heat sink c should be chosen. As expected, this agrees with the choice made by the use of second m ethod.
Thyrist nrs
[Art. 4.9J
Power delivered to load,
V;
P~I,R~R
Vr =rms value of load voltage
wher e
v,~[;
,
13225
= 13225 + 90 ~ 0.993 pu or 99.3%.
:. Circuit efficiency (e)
,
171
For ex = 60°, conduction angle is 180° - 60° = 120° 1 ITAV =21t ~
J' ViSinc.ot . d(oot)=21tR V (1+coscx) a
230>12 , 2< . 2 (1 + cos 60 ) =38.82 A.
For I TAv =38.82 Aand conduction angle of 120°, 4.38 for h eat sink e. T, - TA = 46'C T, ~ 40 + 46
~
POll
from Fig. 4.19 (b) is 52 W and from Fig.
86' C.
From Eq. (4. 17), case temperature,
T" = T8+ Pall' Su = 86 + 90 X 0.075 = 92.75°C and junction temperatur e,
T) =T" + Pall'
S)C
= 92.75 + 90 X 0.15 = 106.25°C.
This example demonstrates that selection of heat sink by second and third methods is more simpler than by the first method.
Example 4.20. For a thyristor, maximum junction temperature is 125°C. The thermal resistances for the thyristor-sink combination are Sjc = 0.16 and SCI = O.OBoe I W. For a heat·sink temperature of 70°C, compute the total average power loss in the thyristor-sink combination. In case the hea t sinh temperature is brought down to 60°C by forced cooling, find the perce ntage increase in the device rating. Solution . From the equivalent circuit of Fig. 4.37 (b ) T j = T~ + POll (Sjc + 9c~)
125 - 70 p. ol ~ 0.16 + 0.08 = 229.17 W Thus total average power los5 in the thyristor-sink combination is 229.17 W. With improv ed cooling,
P" , ~
125 - 60 0.24
=270.83 W.
Thyristor r atin g is proportional to the square root of aver age power loss. :. P ercentage increase in thyristor rating = "270.83 - " 229.1 7 100 = 8 71a '1229.17 x . ".
172
[Art. 4.10J
Power Electronics
4.10. SERIES AND PAHALLEL OPEHATION OF THYRISTOnS seR ratings have improved considerably since its introduction in 1957. Presently. SCRs with voltage and current ratings ·o f 10 kV and 3 kA are available. However, for some industrial
applications, the demand for voltage and current ratings is so high that a single SCR cannot fulfll such requirements . In sucn cases, SeRs are connected in series in order to meet the h.v. demand and in parallel for fulfilling the high current demand. For series or parallel connected SeRs, it should be ensured that each SCR rating is fully utilized and the system operation is satisfactory. String efficiency is a tenn that is used for measuring the degree of utilization of SeRs in a string. String efficiency of SeRs connected in series/parallel is defined as string efficiency _ Actual voltage/current rating of the whole string - [Individual voltage/current rating of one SCR J [Number of SCRs in the stringJ In practice, this ratio is less than one. For obtaining highest possible string efficiency, the SCRs connected in s eries/parallel string must have identical J- V characteristics. As SCRa of the same ratings and specifications do not have identical characteristics, unequal voltage/current sh aring is bound to occur for all SeRs in a string. As a consequence, string efficiency can never be equal to one. However, unequal voltage/current sharing by the SCRs in a string can be minimised to a great extent by using external equalizing circuits. Even in a string provided with external equalizing circuits, the string efficiency is less than unity. For a given system, if one extra unit is added to the series7paraUel string, the voltage/current shared by each device would become lower than its normal rating. The use of this extra unit will certainly improve the reliability ofthe string though at an increased cost. A measure of the reliability of string is given by a factor called derating factor DRF defined as under : DRF = 1 - string efficiency
For example, for a string voltage of 3300 V, let there be six series-connected SeRa, each of 600 V rating. :. String efficiency
and
3300 , / = 600 x 6 = 0.917 or 91.7 % DRF = 1 - 0.917 = 0.083 or 8.3%
If one extra unit is connected in series with the same system voltage, then string efficiency 3300 = 600 x 7 = 0.786 or 78.6% and DRF = 1 - 0.786 = 0.214 or 21.4%. With the addition of extra SCR, DRF has increased from 8.3% to 21.4%, indicating higher reliability of the string, though at an extra cost. The object of this section is to study the problems concerning the serieslparallel operation ., of SCRs and to discuss the measures adopted to overcome these proble m ~. 4.10.1. Series Operation When system voltage is more than the voltage rating of a single thyristor, SCRs are conne cte d in seri es in a string. As stated b e fore , thes e SCRs should ha ve thei r J - V characteristics as close as possible. On account of inherent variations in their characteristics, the voltage shared by each SCR may not be equal. For ins tance, consider two SeRs with their static 1- V characteristics as shown in Fig. 4.40. For SCRl, leakage resistance (= VI I lo) is high whereas for SCR2, it is low (V 2/ Jc). F or the same lea.\cage current 10 in the series connected
Thyristors
[Art. 4.10]
173
----l (b)
(a)
Fig. 4.40. Series connected SeRa. SCRa, SCR1 supports rated voltage VI whereas SCR2 supports voltage V 2 < VI' Each SCR in Fig. 4.40 is rated for a forward blocking voltage of VI volts which is always less than its forward breakover voltage. Here VBOI and VB02 are the forward breakover voltages for thyristors 1 and 2 respectively. It is seen from Fig. 4.40 that two SCRs can support a maximum voltage of VI + V 2 and not the rated blocking voltage 2V1. The string efficiency for two series connected SCRs of Fig. 4.40 is, Therefore,
VI + 2V
V, _!( V,)
I
-2 I +V
I
This shows that even though SCRs have identical ratings, voltage shared by each is not the . same and string efficiency is therefore less than one.
A uniform voltage distribution in steady state can be achieved by connecting a suitable resistance across each SCR such that each parallel combination has the same resistance. This will require different value of resistance for each SCR which is a difficult propositi'on. A more practical way of obtaining a reasonably uniform voltage distribution during steady state working of series-connected SCRs is to connect the same value of shunt resistance R across each SCR as shown in Fig. 4.41. This shunt resistance R is called the static equalizing circuit. Magnitude of parallel resistan ce R can be obtained as follows U] . . Consider n thyristors connected in series as shown in Fig. 4.41. Let SCR1 has minimum leakage current 1bmn and each of the remaining (n -1 ) SCRs have the same leakage current 1bmx > hmn. An examination of Fig. 4.40 (b) reveals that an SCR with lower leakage current blocks more voltage. As SCRI has lower leakage current, it will block voltage Vom (say) which..Js more than that shared by each of the other (11. - 1) SCRs. Here Vbm is the maximum permissible blocking voltage of SCR I. It is seen from Fig. 4.41 that
!t =1-1bmll and 12=1-101T'':: where I =total string current Voltage across SCR1 is Vbm=IIR
174 [Art. 4.10] Power Electr onics --~--~----------------~.. ,~ : ------------- -. ~
Onl: SCR wit h leokOl}« current
, ,,I
I
( Ibmn
,, ,,
,i__ (n-1) SC R, with ',akag' cumnt Ib~
!":~
..J"' :
St ri~
I
I
I,
R
'
I,
,
R
~Vbm-J
,.-c'
I .r'" ,
I
I,
cu r r. l:nt
"I, ,
I
,,
,
I,
R
R
,
... ;.~---------- St r in 9 voltagl: Vs -------------.~,
Fig. 4.41. Static-voltage equalization for series-connected string.
Voltage across (n -1) SCRs
=(n -
1) 1.JI
0
For a string voltage V.. the voltage equation for the series circuit of Fig. 4.41 is
where As
V, =llR+ (n -1) RI, =V'm + (n - 1) R(l -1,=) =Vbm + (n -l)R [1 1 - (Ibnu:-IbmnH = V'm + (n - 1) RI 1 - (n - 1) R . t> I, 11 Ib =I bmx - I bmn RI 1 = V bm , V.. = n V bm - (n - 1) R . 11 Ib
R
or
= n Vbm -
V..
.. .(4 .18)
(n - 1) . t> I,
The SCR data sheet contains only maximum blocking current I bmx and rarely 11 lb' In such a case, it is usual to assume 11 Ib =Ibrru: witll Ibmn = O. With this, the value of R calculated from Eq. (4.18 ) is low . than what is actually required. The value of minimum leakage, or blocking, current Ibm may be acquired from manufacturers if required, but data sheet does not give its value. Once the value of R is €alculated, its power rating is given by
V;
PR =1f where
V,
=rms voltage across
R.
It is likely that SCRs do not have identical dynamic characteristics. In such a case, series -connected SCRs will have unequal voltage distribution during the transient conditions of turn-on, turn-off and high frequency operation. The dynamic characteristics of two SCRs during turn-on are shown in Fig. 4.42 (a ) where it is assumed that turn-on time ofSCR2 is more than that of SCR1 by 11 td . Before these two SCRs are gated, string voltage V, is shared as V, 12 by each thyristor as shown. Now both SCRs are gated at the same time . F,:, SCR1 has less turn-on tim e, it gets turned-on at instant tlo whereas SCR2 is yet off. Voltage across SCR1 drops from V,/ 2 to alm ost zero. At the same inst~t t I , voltage across off SCR2 win boost from V, / 2 to almost full V s' Thus , the voltage shared by two SCRs are unequal. Mer instant t I , voltage V, a cross SCR2 m ay turn it on in .c ase V, is greater than its breakover voltage. S CR2 will , however, get turn ed on a t time (t t + 11 td ) as assumed> Fig. 4.42 (a ). During t urn-off, thyristor characteristics are shown in Fig. 4.42 (b ). SCR1 is as sumed to h ave less turn-off time t q l than that ofSCR2, i. e. tql < t q2 . At instant t 2 , SCR1 h as recovere d and is pa.5sin g through zer o vol tage whereas SCR2 is developing revers e recovery volt age xy. At
0;
[Art. 4. 10J
Thyris:ors
175
ing loge ... tt.ld-et
J
i
Anode voltage V,
,,I !
, I ,
I , I f, I
",
,,: ,
:: ~l1 ,., .,
,
,, I
,
An ode 'ojloge
,
I':
, ,--
~- 2 . ,,Y , I tl: '
, :C
Anode tunen\
: i
:" I
:
,
:
,I :: : :, ,,t q2-----! , tq 1"""",, ':
v,
,
T
F,
: I
t2
, :
,. 1/ : , .,...2 .. ~
(a)
,I
(b)
,
"
Fig. 4.42. Unequal voltage distribution for two series connected SCRs during (a)
turn-on and (b) turn-off.
instant t l in Fig. 4.42 (b), both SCRs are developing different reverse recovery voltages given byab for SCRl and ac for SCR2 as shown, so the two SCRs have unequal voltages across them at tl ' It is thus seen that .SCRs with different characteristics during tum-ofT time suffer from unequal voltage distribution during their turn-off processes. It may thus be concluded from above that series-connected SCRs do suffer from unequal voltage distribution acros s them
during their turn-on and turn-off processes and also during their high-frequency operation which means more frequent turning-on and turning-off of the devices. A simple resistor as shown in Fig. 4.41 for static voltage equalization cannot maintain equal voltage distribution under transient condition . Duritig turn-on and turn-off, the capacitance of the reverse biased junctions determines the voltage distribution across SCRs in a series connected string. As reverse biased junctions are likely to have different capacitances, called self-capacitances, the voltage distribution during tum-on and turn-oft: periods would be unequal. Voltage equalization under these conditions can, however, be achieved by employing shunt capacitors as shown in Fig. 4.43. This capacitance has the effect of removing the inequalities in thyristor self-capacitances. In other words , during tum-on and turn-off periods, the resultant of shunt capacitance and self-capacitance of each SCR tend to be equal for each of the series connected SeRs. Thus the shunt capacitors playa dominant role in equalizing the voltage distribution across the series-connected thyris tors during their turn-on and turn-off processes. When any SCR is in the forward blocking state, the capacitor connected across it gets charged to a voltage existing across that SCR. When this SCR is turned on, capacitor discharges heavy current through this SCR. For limiting this discharge current spike, a damping r es istor Rc is used in series with capacitor C as shown in Fig. 4.43. Resi stor Rc also damps out the high frequency oscillations that mily arise du e to the series combination of R e , shunt capacitor and circuit inductance. Combination of Rc and C is called the dynamic equaliz ing circuit and is shown in Fig. 4.43 (a ). Note L1.at the function of Rc and C us ed in Fig. 4.43 is to equalize the voltage during dynamic (or transient) cond itions and to protect the thyris tors against high du ldt.
176
Power Electronics
[Act. 4.10} Dynamic equall zing circuit \. r- -- . ---,
, j
,,
' D,
R
,,
c>
,,
;, C =, c
R
,• : ,,, , : ,
==i,
~
·,•,
'D! 2) ,,,
(4)
,! ,,
,
,, ,, ,
j
:
r- ' - - - ~- ' ~- - .
,, , Rc
R C
, ,
I
\
·,,,,
, :
,
,
r ----
, ,
,:•
',{ , ,
,
Stot ic equoliz ing circu it
I
D~
L. ____ -C __
~_-
,
R
,•, ,
__ J,
,, Rc
I
D 2
>R C
,
,\ Revers e recovery curr en'
R
I
' It.. Re:ve:rst
reCO,Yery
current (b)
Fig. 4.43. Dynamic and static equalizing circuits for series-connected SCRs.
A diode D is also placed across Re- When forward voltage appears.,diode byPasses Rc during charging time of the capacitor C. This makes the capacitor more effective' ui'Yoltage equalization and for limiting du/dt across SCR. However, during capacitor discharge,Rc comes into play for limiting the current spike and rate of cbange of current di/dt. During turn-off period, when all SCRs are developing r everse voltage, the reverse recovery current ir flows through all series connected SeRs as shown in Fig. 4.43 (a). However, if one SOR recovers early, it will not allow the passage of ir from ~he other SORs. If SORI is assumed to recover fully and earlier than other SORs, then reverse recovery current due to other SCRs can pass through R connected across SCRI as shown in Fig. 4.43 (b ). In this figure, i, may flow through C, Rc also in case the conditions are favourable. For simplicity, only two SORs are shown in Fig. 4.43. The existence of reverse recovery current is desirable as it facilitates the turning-off process of the series-connected SOR string.
•)'
Value of capacitance C shown in Fig. 4.43 can be obtained as under : In series connected SORs, voltage unbalance during turn-off time js more predominant than it is during turn-on time, Fig. 4.42. Therefore, choice of capacitor C is based on the reverse recovery characteristics of SORs . In Fig. 4.44 (b) are shown reverse recovery characteristics for two SORs of Fig. 4.44 (a ). SORI is assumed to have short reverse recovery time as compared to SCR 2. Shaded area 11 Q, proportional to the product of current and time, is the difference in the reverse recovery charges of two thyristors 1 and 2. Under this assumption, SORI recovers firs t:; it, therefore, goes into blocking state and does n ot allow the passage of excess charge 11 Q left on SCR2. This charge 11 Q can, however, pass through C as shown in Fig. 4.44 (a).
The voltage induced by 11 Q in the capacitor C, connected across SCRI, is 11 QI C; whereas no voltage is in duced by 11 Q (= Q 2 - QI) in C connected across SOR2. There is thus a difference in voltages, equ al to Q2 ~ Q1 = I1CQ, t o which the two shunt cap acit ors are charged. The thyristor with the least reverse recovery time will share the hi ghest transient voltage, say V bm . As stated above, the voltage difference to which the two shunt capacitors are charged
, .. (
[Art. 4.10)
Thyristors
17i
Anode current
r-' . ~
'---c. -,
V,
.r
•i, =c +
~I
~ __ _
~
_ .J
t,
1
.r +
.
~2 •• ,•
=c
LRcvcrsc recoonry curr~nt
=
(a)
(b )
Fig. 4.44. (0 ) Flow of reverse recovery current if SCRl recovers 6rst (b ) Variation of reverse recovery characteristics fOr"two SCRs of.Fig. 4.44 (0 ). during reverse recovery time is 6 QI C, the transient voltage shared .by slow thyristor 2 must be
V'm - '"cQ (less than V'm shared by fast thyristor 1). Thus, in Fig. 4.44 (a),
voltage across fast top thyristor 1 , VI = Vbm
Q and voltage across slow bottom thyristor 2, V2 = Vb~ - 6 C : . String vol tage,
or
an d
Vs
=VI + V 2 =Vb," + Vbm -
6
Q
C
V'm=~ (v. + "'cQ) V, = V'm - "'! = ~ (V. - '"cQ)
In order to aid the reverse recovery process of the thyristor s in a string, the string voltage reverses in polarity as shown in Fig. 4.44 (a ). Now consider that there are n series-connected SeRa in a string as shown in Fig. 4.45 . If top S CR h as characteristics similar to SCRI of Fig. 4.44 (b) and the r emaining (n - 1) SCRs h ave ch aracteristics similar to SCR2 of Fig. 4.44 (b), then SCRI would r ecover first and support a voltage Vbm. The charge (n - 1) 6 Q fr om the remaining (n - 1) thyristors would pass through C connected across top fast SCRI and as a result, a voltage (n - 1) 11 QI C would be induced in C. As befon.::·excess charge contributed by each on e of the (n -1 ) thyristors is 60 Q, therefore, the voltage aCrOSS each one of the slow thyristors is
(V.
m -
"'cQ ) as shown in
Fig. 4.45. Thus, fo r
a string of n series-connected thyristors, voltage across fast top thyristor I,V } =Vbm voltage across each one of the slow thyristors . V2 is
v.-, = Vbm - ~ C and voltag e across (n - 1) slow thyris tors
=(n -
1) Vo;!
178
Power Elec tronics
[MI. 4.10]
=(n-l) (Vbm -~!) String voltage,
V,
=V, + (n -
1)
V,
= Vbm + (n -1 )(Vbm Its simplification gives Vbm
and
~CQ)
=~[V (n -l)~ Ql n ,+ C
C = (n -1) ~ Q. n V bm Vs
... (4.19)
Voltage across each one of the slow thyristors, in terms of VJ • is given by
V,
=
(Vbm -
-
-, ....... .
J
~cQ) 2
=V, + (n n
or
V,
1) ~
C
V-~ , C =
v,
<
-
or or
V bm
J
+
= 1'1
C= ,-
" t =, " · ,i C = ....... . · "· C ~
•
' -c
Fig. 4.45. String having n-senes connected "thyristors .
(n - 1). ~Q
C
=l[v n
Ci= , ,-
__ • •• .l
~. ~Q and this must
be supported by all SCRs together, which is equal to n.Vbm - V
----r
;:: (n_~ t. Q
,
Durin g the turn- off process, the source voltage V, must reverse to aid the r everse recovery current. The transient voltage whi ch each SCR must be able to withstand is Vbm - However, total voltage acting across th e circuit consisting of VJ • thyristors n, 4,
n. V bm -
3
.. .(4.20)
n
3, 2 and top C, and per KV1, is V, + (n
!
)
Q_~
nC
~
]
,+
(n -l)·~Ql C
C = (n -1). ~Q n,Vbm V,
... (4. 19)
4.10.2. Parallel Operation 'When current required by the load is mor e than th e rated current of a single thyristor, SCRs are connected in parallel in a string. For equal sharing of currents, I-V characteristics of SCRs during forward conduction must be identical as far as possible. In Fig. 4.46 (a ) are shown two SCRs in parallel and their characteristics during forwa rd conduction are shown in Fig. 4.46 (b ). For paralle\.connected SeRs, voltage drop Y r across them must be equal. Fig. 4.46 (b ) shows that for the same voltage drop Vr SCRI shares n rated current I ! wh ereas SCR2 carries current 12 much less th an th e rat ed curren t I t The total current ca rried by th e unit is I , ,.. 12 and not the rated current 21, as required. Therefore, s tring effi cien cy is given by
[Art. 4.10]
Thyristors
,
]79
· ~:·-· · i.
I ••..•..••• b t V,
!
2
.• _ ••..9
, ' .. ., ,
,•
LL..J....-;'vc-,- - - V (bl
(0)
dJQ
.-ia-1 , I
.
.
(el ,
Fig. 4.46. (a ) and (b) Parallel operation of two thyristors (c)
Dynamic resistance decreases as junction temperatul'O:: risesr
11+1'=!(1+ I,) 211 2 II
I
Now consider n parallel connected SCRs. For satisfactory operation of these SCRs, they should get turned on at the same moment. The importance of their simultaneous turn on can be explained with an example. Consider that SCRI has large turn·on time whereas the r emaining (n -1) SCRs have low turn·on time. Under this assumption, (n - 1) SeRs will turn on first but one SCR1 with longer t~·on time is likely to remain off. The voltage drop across · (n - 1) SCRs falls to a low value and SCR1 is therefore subjected to this low voltage. For a given gate drive power, anode to cathode must have Some minimum forward voltage, called finger voltage, for a thyristor to turn·on. If voltage across SCRl drops to a value less than its finger voltage, then this thyristor will not tum on. As a consequence, the remaining (n - 1) SeRs, w~ich are already on, will have to share the entire load current. A$ such, the~e SCRs may be overloaded and damaged because of heating caused by overcurrents. If one SCRI in a parallel unit carries more current than. other SCRs, then this SCRI will have greater junction temperature rise. As a result, its dynamic resi~.tance (= dVrldIa ) during forward conduction, Fig. 4.46 (c) decreases and this further increases the current shared by this SCR. In Fig. 4.46 (c), dynamic resistance is oalab wd current shirred is 1'. Because of junction temperature rise, its dynamic resistance decreases to oalac and current shared by SCRl increases to 1". This process of anode current rise becomes cumulative and subsequently the junction temperature of SCRI exceeds its rated value; as a r esult SCRl is damaged. This sequence of events may engulf another SCR and in this manner all SCRs in the string may be destroyed permanently. Therefore, wh.3 SeRs are to be operated in parallel, it should be ensured that they operate at the same temperature. This CfUl be achieved by mounting the parallel unit on one common heat sink.. Unequal current distribution in a parallel unit is .also caused by the inductive effect of current carrying conductors. Vlhen SeRs are arranged unsymmetrically as shown in Fig. 4.47 (a), the middle conductor will have more inductance because of more flu.'C linkages fro m two nearby conductors. A3 a consequence, less current flows through the middle SCR as compar::d to ou ter two SCRs . This unequal current distribution can be avoided by mounting the SC R.s symmetrically on the heat sin.., ., shown in Fig. 4.47 (b).
180
Power Elec tronics
(Art. 4.10)
+ (Heat sink
2
oJ
o 1
(a) (b)
(b)
(c)
Fig. 4.47. Parallel operation ofSCRs (al unsymmetrical arrangement and symmetrical arrangement on heat sinks (c) current equalization by the use of reactor .
In ae circuits, current distribution can be made more uniform by the magnetic coupling of the parallel paths as shown in Fig. 4.47 (c). The tapped point A is the mid point of the reactor. H anode currents are such that II = 12 , then flux produced by two halves of the reactor oppose each other. As A is the mid point, opposing flux linkages cancel and there is therefore no voltage drop in the reactor. If currents II and 12 Bre unequal, say 11 > 12• then resultant flux linkages are not zero. These flux linkages induce emfs in Ll and L2 as shown. Emf across reactor L l opposes the flow of 11 whereas that across L2 aids the flow of 1a. There is thus a tendency to buck 1! and boost 12 so as to minimise the unbalance of currents in the parallel unit . When three or more SeRs are connected in parallel, reactors can be arz:anged accordingly so as to minimise the current unbalance. Example 4.21. A string of four series·connected thyristors is provided with static and dynamic equalizing circuits. This string has to withstand an off-state uoltage of 10 k V. The static equalizing resistance is 25000 n and the dynamic equalizing circuit has Rc = 40 nand C =0.08~. The leakage currents for four thyristors are 21 mA, 25 mA, 18 rnA and 16 rnA respectiuely. Determine voltage across each SCR in the off-state and the discharge current of each capacitor at the time of turn-on. Solution. Let 1 be the string cur r ent in the off·state. Then c'urrent through static' equalizing resistanc e R of25000 n is (I-leakage current), current through each SCR is its own leakage current and no current flows through series r')mbination of Rc and C. :. Voltage across Voltage across SCR1 Voltage .across SCR2 Voltage across SCR3 Voltage across SCR4
R
=voltage across each SCR =(1- 0.021) x 25000 =VI =(I - 0.025) x 25000 =V, =(I - 0.018) x 25000 =V, = (I - 0.016) x 25000 = V,
The sum ofV l , V 2 , Va and V" gives 25000 (41 - 0.08) = V1 + V 2 + Va + V" =string voltage, 1000 0 V or 1= 0.12 A From above, voltage across SCRI
= (0.12 -
0.02 1) x 25000 = 2475 V Similarly V2 = 2375 V, Va = 2550 V an d Vol = 26 00 V.
..' '.
Thyristors
[Art. 4.10J
181
Discharge current through SeRt at the time of turn on . = ;~ =
2:~5 = 61875 A
Similarly, discharge currents through thyristors 2, 3 and 4 are respectively 59.375 A, 63.75 A and 65 A.
Example 4.22. SCRs with a rating of 1000 Vand 200 A are available to be used in a string to handle 6 k V and 1 kA. Calculate the number of series and parallel units required in case derating factor is (a) 0.1 and (b) 0.2.
Solution. (a) Derating factor, DRF = 1 - string efficiency ..
0.1=1-
6000 =1 - 1000 n, x 1000 np x 20.0
: . Number of series-connected SCRs, 6000 n, = 1000 x 0.9 = 6.6" 7
Number of parallel-connected SCRs, 1000 x 0.9 = 5.5" 6
np = 200 (b)
As above, number of series -connected SCRs, 6000 n, = 1000 x 0.8 = 7.5" 8
and number of parallel-connected SeRs. 1000 x 0.8 = 6.25 " 7
np = 200
With higher value of DRF. more SCRs are required and therefore voltage and current shared by each device are lower than their normal rating. This increases the string reliability though at an increased investment. Example 4.23. It is required to operate 2S0-A SCR in parallel with 350-A SCR with their respective on-state voltage drops of 1.6 Vand 1.2 V. Calculate the valu.e of resistance to be inserted.in series with each SCR so that they share the total load of 600 A in proportion to their current ratings. Solution. Dynamic resistance of 250-A SCRt Dynamic resistance of 350-A SCR2 Let R be the resistance inserted in series with each SCR. With this, current shared by 12 +R SCRI = 600 350 . ~ 250 Tota l res13tance 1. 6
and current shared by
SCR2 = 600
? ~O
+
_0.
R
Total resls tance
~ 350
182
Power Electronics
[Act. 4.11]
1.2
From above, Its simplification gives
350+
R
1.6 R 250 +
250 5 =-=350 7'
R = 0.004 n.
Thus the resistance to be inserted in series with each SCR is 0.004 n. Example 4.24. Discuss the conditions which must be satisfied for turning-on an SCR with a gate signal. Solution. Conditions which must be satisfied for turning-on SCR with a gate signal are . as under: (a) An SCR must be forward-biased. It means that anode must be positive with respect
to cathode. (b) Gate pul'Se width must be more than the turn-on time of an SCR. This will ensure that anode current exceeds the latching current before gate signal is removed.
Cc) Anode to cathode voltage must be more than finger voltage. A finger voltage is that voltage below which an SCR cannot be .turned on with 'a gate signaL (d) Magnitude of gate current must be more than the minimum gate current required to turn-on a thyristor, otherwise the thyristor tum-on will not be reli able. (e) Magnitude of gate current must be less than the maximum gate current allowed, otherwise gate circuit may. be damaged.
(f) The gate triggering must synchronize with the ac supply.
4.11. OTHER MEMBERS OF TilE THYRISTOR FMIILY The term thyristor includes all four-layer p-n-p-n devices used for the control of power in ac and dc systems. The silicon controlled rectifier is the most popular member of thyristor family. There are several other members of thyristor fQ.ll1i1y like PUT, SUS, SCS, triac, diac etc. All these devices, except triac, are low power devices. Several new devices have been developed and added to the thyristor family. These recently developed thyristor devices are asymmetric thyristor CASCR), reverse conducting thyristor (RCT), static induction thyristor (81TH), gate-assisted tum-off thyristor and gate turn-off (GTO) thyris~or. MOS-controlled thyristor (MCT) has already been described in Chapter 2. The object of this section is to discuss other members of the thyristor family. 4.11.1. PUT (Programmable Unijunction Transistor) It is a pnpn device like an SCR. But the major difference is that gate is connected to n-type material near the ar..ode as shown in Fig. 4.48 (a). PUT is used mainly in time-delay, logic and 8CR trigger circuits. Its largest rating is about 200 V and 1 A. Circuit symbol and I - V characteristics of a PUT are shown in Fig. 4.48 (b) and (c) respectively.
In a P UT, G is always biased positive with respect t o cathode. "'Vhen anode voltage exceeds the gate volt age by a bout 0.7 V,junction J 1 gets forward biased and PUT turns on. When anode . voltage beco mes less than gate vo ltage, PUT is turned off.
.
.
[Art. 4,IIJ
Thyristors
18J
A
P G
A
J,
10
n p
n
J2
10
---'iT
G
Va
Va
J3
L_ _ _ _.
..~J K
~
K
,
~)
~
Fig. 4.48. (a) Schematic diagram (b) circuit symbol and (e) J- V characteristics of a PUT.
4,11.2, SUS (Silicon Unilater a l Swit ch ) A SUS is similar to a PUT but with an inbuilt low-voltage avalanche diode between gate and cathode as shown in Fig. 4.49 (a). Because of the presence of diode. SUS turns on for a fixed anode-to-cathode voltage unlike an SCR whose trigger voltage and/or current vary widely with changes in ambient temperature. SUS is used mainly in timing, logic and trigger circuits. Its r atings are about 20 V and 0.5 A. Circuit symbol. equivalent circuit' and I - V characteristic of an SUS are shown in Fig, 4.49 (b), (c) and (d) respectively, . _' A A
A 10
P
---+1
n
G
G p
D
n
10 G
va
__ J
- Va
L. Va
K K K ~
W
00
~
Fig. 4.49. (a) Schematic diagram (b) circuit symbol (c) equivalent circuit and Cd) 1-V characteristics of an SUS.
.!
,I ,
4,11,3, SCg (Silicon Con trolled Switch) SCS is a tetrode, i.e. four electrode thyristor. It has two [~tes, one anode gate (A G) like a PUT and an other cathode gate (KG) like an SCR. In other words, ses is a four layer, four terminalpnpn device; with anode A, cathodeK, anode gateAG and cathode gate KG , Fig. 4.50 (a ). SCS can be turned on by either gate. Circuit symbol and J - V characteristic of an 8eS are shown in Fig. 4.50 (b) and (c ) respectively. When a negative pulse is applied to gate AG, junction J 1 is forward biased and 8 eS is turned on. A positive pulse atAG will reverse bias junction J 1 and turns off the SCS . A positive pulse at gate KG turns on the devi ce (just like an S CR) and a negati ;> e pulse at KG turns it off Gust like a G,T,Q,),
184
Power Electronics
[Art. 4.IIJ A
.
P
J,
ON ~.
~
n AG
J,
KG
AG - Vo Vo
puls.e
h
0"
K&
n
L
-10
K
(0)
Fig. 4.50 .
~L
A
ON
P
pulse
r
OFF
pulse
(b)
(~)
(c)
Schematic diagram (b) circuit symbol and
(e) J. V
characteristic of an SCS.
Its ratings are about 100 V and 200 rnA. This can be operated like an OR gate. Its applications include : (i) timing, logic and triggering circuits
(iO pulse generators
(iii) v:oltage sensors
(iu)
oscillators etc.
4.11.4. Light Activated Thyristors The circuit symbol and I - V characteristics of light. activated thyristor, also called LASeR, are shown in Fig. 4.51. LA SeRs are turned on by throwing a pulse of light on the silicon wafer of thyristor. The pulse of appr opriate wavelength is guided by optical fibres to the special sensitive ar ea of the wafer. If the intensity of light exceeds a certain value, excess electron·hole pairs are generated due to radiation and forward-biased thyristor gets .t urned on. 10
A
-Vo
K
-I.
(0 ) (b) Fig. 4.51. (a) Circuit symbol and (b) I-V characteristic of LASCR
The primary use of light-fIred thyristors i5 in high-voltage hi gh-current applications, static r eactive-power compensation etc. A light-fired thyristor has complete electri cal isolation be tween the light-triggering source and the high-voltage an ode-ca thode circuit. Light-activated thyris to rs are available up to 6 kV and 3.5 kA, wi th on-state voltage drop of about 2 V and with light-triggering requirements of5 m\V.
Thyristors
[A n . 4.11]
185
4.11.5. The Diae (Bidirection al Thyristor Diode ) A cross-sectional view of a diac showing all its layers and junctions is depicted in Fig. 4.52 (a ). If voltage V 12• with tenninal 1 positive with respect to terminal 2, exceeds break-over voltage Vaal then structure pn pn conducts. In case term in al 2 is positive with respect to terminal 1 ~d when V 21 exceeds breakover voltage V B02 • structure pn pn' conducts. The term 'diac' is obtained from capital letters, DIode that can work on AC. Fig. 4.52 (b) gives the circuit symbol and Fig. 4.52 (c) the I-V characteristics of a diac. It is seen that diac h as symmetr ical breakdown characteristics. Its leads are interchangeable. Its turn-on voltage is about 30 V. When conducting, it acts like a low resistance with about 3 V drop across it. When flat conducting, it acts like an open switch. A diac is sometimes called a gateless triac.
10
,
p
,I
, , ,I
n
,
!,
.
p
,
:~
n
I
n'
pnpn
-------------
p
~~V':O~2;=======_+~~~==~~~ -Va Vaal Va
n
------po-p"';
p
-10
2
00
00
00
Fig. 4.52. (0) Cross-sectional view (b ) circuit symbol and (c) ,!- V characteristics of a di ac.
4.11.6. The Triac An SCR is a unidirectional device as it can conduct from anode to cathode only and not from cathode to anode. A triac can, h owever, conduct in both the directions. A triac is thus a bidirectional thyristor with three terminals. It is used extensively for the control of power in ac circuits . Triac is the word derived by combining the capital letters from the words TRIode and AC. Wh en in operation, a triac is equivalent to two SCRs connected in antiparallel. The circuit symbol and its characteristics are shown in Fig. 4.53 (a) and ( b ) respectively. As the Ie MT2 positive I g'l>!q l > l,j O
19'1
191
Igo =O
MT2
- lgO
-19 1
-1 92
.\\T 2 n egc:i 'l l!
G MTI
(0 )
-1 0 .
(b)
F'ig. 4.53. (a ) Circuit symbol and ( b) static 1- V ch aracteris tics of a t:ri ~c.
186
Power Ele ctronics
(Art. 4.111
MTI triac can conduct in bo th the directions, the terms anode and ca thode are not applicable to triac. Its three terminals are usually designated as MTl (main terminal 1), MT2 and the gate by G as in a thyristor. For understanding the operation of the triac, its cross-sectional view sh owing all the layers and junctions is sketched in Fig. 4.54. The gate G is near terminal MTl. .The cfoss-hatched strip shows that G is connected to N3 as well as P2. P. Similarly. terminal MTI is connected to P2 and N z i terminal MT2 to Pi and N 4 . N, With no signal to gate, the triac will block both half cycles of the ae appli ed voltage in case peak value of this voltage is less than the breakov cr voltage orvBD l or V BD2 of the triac, Fig. 4.53 (b). The triac can, however, be turned on in each half cycle of the applied Fig. 4.54. Cross-sectional voltage by ap plying a pos itive or negative voltage to the gate with view of a triac. respect to terminal MTl. For convenience, terminal MTI is taken as the poi nt for measuring the voltage and current at the gate and MT2 terminals. The tum-o n process of a triac can be explained as under : (i) MT2 is positi ve and gate current is also positive. When MT2 is positive with respect to MTI, junction PI NI, P2 N2 are forward biased but junction N I P2 is reverse biased. When gate terminal is positive with respect to MTl , gate current flows mainly through P2 N2 junction like an ordinary SCR, Fig. 4.55 (a ). When gate current has injected sufficient charge into P 2 layer, r everse biased junction NI P z breaks down just as in a normal SCR. As a result, triac starts conducting through P I N I P z N 2 layers. This sh ows that wh~n MT2 ' and gate terminals are positive with respect to MTl, triac turns on like a conventional thyristor. Under this condition, triac operates in the first quadrant of Fig. 4.53 (b). The device is·m ore sensitive in this mode. It is recommended m ethod of triggering if the conduction is desired in the first quadrant. . (ii) MT2 is positive but gate current is negative. When gate tenninal is negative with r espect to MTl , gate current flows through P 2 N 3 junction , Fig, 4.55 (b) and reverse biased junction Nt Pz is forward biased as in a normal thyristor. As a result, triac starts conducting through P , NI P 2 N3 layers initially. With the conduction of P l Nl P z N 3, the voltage drop acr oss this
'. 'I 1,
1, 1,1 Ig MTI ( - )
G
1,
., I "
I
P,
MT2(r )
MTt i - )
G
, N,
N,
11
1,
N,
1, P,
- _.Pz--J IMiol
conduction (P,N, P2 NJ)
Finol
N,
conduction
(P, N, P, N,) P,
MT 1(+)
G
N,. ..
"
'ilt
",
conduction (PzN,P, N,)
", 10112(+)
Io4TI ( ... )
G
.,
1, P,
Finol
P,
1,
", P, N,
MT 2{-)
1 MTz( - )
(e) more sen3itive (b) (e) (d ) more sensitive Fig. 4.55. Turning-on process in a triac. Final conduction is through PI N ) P,!N,! in (0) and (b) and through P,! I'll PI N .. in (c) and (d).
[Art. 4.11 )
Thyristor s
187
path falls but potential of layer between P2 Ns rises towards the anode potential of MT2. As the right hand portion of P 2 is clamped at the cathode potential of MTl, a potential gradient exists across laye r P2 , its left hand region being at higher potential than its right hand region. A current shown dotted is thus established in layer P2 from left to right. This current is similar to conventional gate current of an SCR. As a consequence, right-hand part of triac consisting of main structure PI NI P2 N2 begins to conduct. The device structure PI Nl P 2 N J may be regarded as pilot SCR and the structure P INI P2N2 as the main SCR. It can then be stated that anode current of pilot SCR serves as the gate current for the main SCR. As compared with tum-on process discussed in (i) above, the device with MT2 positive but gate current negative is less sensitive and therefore, more gate current is required. (iii) j}IT2 is negative but gate current is positiue. The gate current Ig forward biases P2 N2 junction Fig. 4.55 (e). Layer N 2 injects electrons into P 2 1ayer as shown by dotted arrows. As a r esult, reverse biased junction N I P I breaks down as in a conventional thyristor. Eventually the structure P 2 Nt PI N4 is completely turned on. As usual, the current after turn-on is limited by the extemall oad. As the triac is' turned on by remote gate N 2 , the device is less sensitiue in the third quadrant with positive gate current. (iv ) Both MT2 and gate current are negatiue. In thi.s mode, Na acts as a remote gate, Fig. 4.55 (d). The gate current 16 flows from P2 to Na as in a normal thyristor. Reverse-biased junction N t PI is broken and finally, the structure P2 N} P l N4 is turned on completely. Though the triac is turned on by rem ote gate N3 in third quadrant, yet the device is more sensitive under this condition compared with turn·on action with positive gate current discussed in (iii) above. It can, therefore, be concluded from above that :
.
.
sens itivity of the triac is greates t in the first quadrant when turned on with positive gate current and also in the third quadrant when turned on with negative gate current, . (ii) sensitiv ity of the triac is low in the first quadrant when turned on with negative gate current and also in the third quadrant when turned-on with positive gate current. (i)
Thus the triac is rarely operated in first quadrant with negative gate current and in the third qu adrant with positive gate current. As the two conducting paths fr om MTI to MT2 or from MT2 to MTI interact with each other in the structure of the triac; their voltage, current and frequency ratings are much lower as compared with conventional thyristors. At present, triacs with voltage and current ratings of 1200 V and 300 A (r ms) ar e available. Triacs are used extensively in residential lamp dimmers, heat control and for the speed control of small single-phase series and induction motors. . A tri ac may sometimes oper ate in the rectifier mode rather than in the bidirectional mode. This may happen due to the fan owing r easons: For a given value of'positive gate current, a triac may turn on with MT2 positi ve in first quadrant bu t may fail to turn on with MT2 negative. (b) 'Wi th constant negati ve gate current, the triac m ay t urn on with MT2 negative in third quadrant but may not turn on with MT2 positive. The re ctifi er-mode can be overcome by increasing the value of gate current. (a)
188
[Art. 4.12J
Power Electronics
4.11 .7 . Asymmetrical Thyristor (ASCR) A conventional thyristor is able to block a large reverse voltage, but this blocking capability is not required in several industrial applications. For example, in voltage source inverters converting de to ae and in some chopper circuits, a freewheeling diode is usually connected in antiparallel across each thyristor. This fr eewheeling diode damps the thyristor voltage to 1 to 2 V under steady state conditions. An asymmetrical thyristor, or ASCR, is specially fabricated to have limited reverse yoltage capability ; thi s permits a reduction in tum-on time, turn-off time and on-state voltage drop in ASCR A typical ASCR may bave r everse blocking capability of 20 to 30 V and forward blocking voltage of 400 to 2000 V. ASeRs with turn-off time half of that of a similar rated conventional SCRs have been developed. Fast turn-off ASCRs minimize the size, weight and cost of commutating components and permit high frequency operation (20 KHz or mor e) with improved efficiency. 4.11.8. Reverse Conducting Thyristor (RCT) A A reverse conducting thyristor is a special case asymmetrical thyristor with a monolithically integrated antiparallel diode on the same silicon chip. This construction reduces to zero the reverse blocking capability of RCT. A current pulse through the diode part of the chip turns off ReT. The arrangement of ASCR and diode in a single device reduces the heat sink size and leads to compactness of the converter, The undesirable stray loop inductance between ASCR and diode is also eliminated and unwanted reverse voltage transients across ASCR are avoided ; this leads to better tum off behaviour of RCT. RCTs with 2000 V and 500 A ratings are available. For high-performance inverter and chopper circuits, RCTs can now be tailor-made. 4.11 .9. Other Thyristor Devices
G
, Fig. 4.56. Reverse conducting
thyris tor
Gate-assisted turn-off thyristor (GAT) is. a normal four-layer thyristor, but its turn-off is achieved by applying a negative gate drive across gate-cathode terminals. In order to reduce the turn -off time appreciably, the gate-cathode junction is highly interdigitated so that stored charges can be removed more effectively from the base region . GAT thyristors are extensively employed in TV deflection circuits at fr equencies around 20 kHz with turn -off times as low as 2.5j.l sec for 200-V devices. Gate-turn-off thyristor (GTO) and static induction thyristor are described in the next two articles. The latest semiconductor device to enter the family of thyristors is integrated-gate commutated thyristor (IGCT). IG CT is basically a hard-switched GTO. IGCT with 4500 V, 3000 A ratings are available. Its adv:lntages over GTO are (i ) lower conduction drop . (ii) faster switching speed, (iii) monolithic by-pass diode, (iu) snubberless operation and (u) ease of series operation \121.
, 4.12. ~TE:I'URN OFF TIIYRISTOR (GTO) Conv entional thyristors (CTs) are nearly ideal switches for their use in power-electronic applica tions . These can easily be turned on by positive gate current_Once in the on-state, gate 105:5 control. CTs can now be turned off by expensive and buLlty commut ation circuitry. This shor:coming of thyristors limit their use up to about 1 kHz applic ation:;. These drawbacks in thyris tors has led to the development of GTOs. A GTO is a more versatile power-semicon ductor device. It is like a. CT but with added features in it. A GTO can easily be turned off by a negative gate pulse of appropri ate amplitude. Thus, a G'T O is a pn pn device th3.t can be turned-on by a positive ga.te cu rrent and turned-off by n nega tive gat e current at iw gate cathode terminals.
Thyristors
[Art. 4.12]
189
Self-turn off capability of GTO makes it the .most suitable device for inverter and chopper applications. . 4.12.1. Basic Structure A GTO is pn pn, there terminal device with anode (A): cathode (K) and gage (G), Fig. 4.57 (a ). The four layers are p ·np+ n· as shown. In CT, anode consists of p. layer, but in a GTO, anode is made up of n· type fingers diffused into p . layer. Fig. 4.57 (e) gives two alternate circuit symbols for GTO. Since GTO is a four layer pn pn device just like CT, it can also be modelled by two--transistor analogy as shown in Fig. 4 .57 (b ). The four layer s have different doping levels indicated by p +np+n.+ . Transistor Q 1 is p +np· type and transistor Q2 is np+n+ type, with p. emitter of Q1 as anode A and n+ emitter of Q2 as cathode K.
4.12.1.1. 1'urn·on Process. A GTO is turned on by applying a positive gate current Ig in the reference direction shown in Fig. 4.57 (b). As GT O is forward biased, regeneration process starts as in a CT. Current gains ai'
A
"
n' l " 1n'1 " n " ~ rn'l 1
I. J J1 JJ
A
n
"
G
A
1" IS2
v
n
P'r 0, n'
I,
G
•G
r.,
Gate (6)
~
.
I"
I"
I Ca thode (K)
p+ l.....J Q,,.-I
,
, ,. ~)
(b)
Fig. 4.57. Gate turn-off thyristor (a) basic structur e (b) two-transistor analogy and (e) circuit symbol.
A
4.12.1.2. Turn·off process The turn-off process in GTO is quite different from th at in a CT. The two-transistor model is analysed for understanding the tu rn- off process in a GTO. For co nvenien ce , the two-transistor E'odel of GTO is redrawn in Fig. 4.58. From Eq. (2.7 ), I c, = p,. I.,
and
I CI
From Eq. [2.6).
and
= P, . Ial
let = Cll l EI
I C2 = ~. IE2
As stated before, for initiating the turn-off process in a GTO, a nega tive gate current Ig' is applied across gate-cathcde terminals as shown in Fig. 4.58.
K
Fig. 4.53. Two-transistor model fo r GTO with negoative gate current 1/
190
Power Electronics
[Act. 4.12J
Now KCL at anode M in Fig. 4.58 gives l ei-I; -182= 0 IB2=Icl-Ig' =al I,, -I;/
or
... (4.20 )
Fig. 4.58 also r eve als that f a = l eI + IC2 ... (4.21 )
oc
I When saturation in Q2 has occurred, 182 = ~, For initiating the turn-ofT process, Qz must
be brought out of saturation. This can be accomplished only if 182 is made less than IC2 /~2' So, when [82 < (Ic2/~2)' Q2 would shift to active region and regener ative action would eventually turn-olfthe GTO.
. :. For turnmg off of Q 2 (or GTO),
IB2
I~
<~
Substituting the value of 182 from Eq. (4.20) and IC2 from E q. (4.21) we get
al Ia - I,' < ; 2 (1 - (Xl) ,
-i,<
(1 - 0. 1)
~2
fa
fa
.
-alIa
Substitution of
oc
I g' > I , [ a, +: : - 1
or
1
... (4 .22)
_ In order that gate current I ,' for turning-off GTO is low, CX:! should be made as near to unity as possible whereas 0. 1 sh ould be made small. Th e turn -off gain is defIned as the turn- off the GTO_
I, ~Qii = I;
:. Turn -off gain. The turn -off action in GTO
rati c~f
Ca.!1
anode current I" to gate current Ig' need ed to
ex,
= 01 + O:z -
1
... (4.23)
now be explained as under:
F ig. 4.58 shows that Ia =IJt and Eq. (4.22) gives Ii' more than I!. So when negative gate current la' flows be t.,... een gate-cathode terminals, net base current (182 - 19') is reversed, exc ess (i)
carriers are drawn from base p ~ region of Q2 and collector current l e1 of Q1 is diverted into :he
rhyristors
[Art. 4.11J
191
external gate circuit. This r emoves base drive of transistor Q2' This further rem oves ba3e current I Bl of transistor Q1 and the GTO is eventually turned off. (ii) As stated above, a low value of negative gate current requires low value of u i and high
value_of U 2 ' Low value of current gain a l of Q 1 can be achieved (a ) by diffusing gold or other heavy metal n base of Q 1 transistor (6) or by introducing short-circuiting n+ fingers in the anode p. layer as shown in Fig. 4.57 (a), (c) or by a combination of both the techniques li sted in (a) and (6) here. Techniques (a) and (6) are described below. (1) Gold -doped GTO. A go ld-doped GTO retains its reverse blocking capability. A Gold-doping also reduces turn-off time, therefore, these GTOs are suitable for high-frequency operation. However, gold-doped [, p' 1 GTOs suffer from more on-state voltage drop for a given current than a similar CT. ,,AnOC;? s ho rt (ID Anod e-sh orted GT O. The short-circuiting fingers, also called anode-shorts, leads to short-circuit of the emitter P+ (anode A) with base n of Ql transistor as shown in Fig. 4.59: For anode or emitter current la' effective emitt!3r current 1£1 is reduced I, because of anode short. This further decreases collector current Y' 11 I C I ' Therefore, effective current gain a l of QI' now given by c.o---J._--:,c:_-{~ ~_1 l Cllla gets r educed. So by anode-short structure, Ct l is reduced .. ' but ~ remains unchanged as desired. Anode-short, howeve r, reduces r everse voltage blocking Fig. 4.59. Tw o-trrm.~i s tor capability. With reverse biased GTO,junction that blocks' reverse model of GTO with voltage is J 3 only. Junction J 1 blocks no voltage because of n+ nnode-s ho n _
'I
fingers in between p + anode. As J 3 junction has large doped layers p., n+ on its two sides, J 3 has lower breakdown voltage, of the order of about 20 to 30 V. The above is summarised below: Gold-doped GTO
Anode-shorted GTO
1.
More on-state voltage drop
1.
Low on-state voltage drop
2.
High re .... erse-voltage blocking capability
2.
Low reverse-voltage blocking capability.
3.
Suitabl e for high-frequency operation
3.
Suitable for low-frequ ency operation
4.12.2. Static I·Y Characteri stics The static I-V characteristics of a OTO is identical with that of a conventional thyristor. Latching current for GTO is, h owever, several amperes , S'-';": 2A, as compared to 100 -500 rnA for a conventional thyristor of the same rating. If gate current is not able to turn on the GTO , it behaves like a high-voltage, low gain transistor with considerable anode current. This le ads to a n oticeable power loss under such conditions. In the reverse mode, revers e-voltage blocking capabil ity of GTO is low, typically 20 to 30 V, because of (i) anod e shorts and (ii ) large doping densities on both sides of rev erse blockingjunction J 3 , Fig. 4 .57 (a). 4.12 .3. Switching Performance A basic gate drive circui t for a GTO is shmv n in Fig. 4.6 0 (0) . For turn ing· on a GT O, fir s~ transistor TRI is turned on, this in turn s . ."i tches on TR2 to apply a positive gat e-curr ent pul.:5e
192
Power Electronics
lArt. 4.12]
to tum on GTO. For turning off the GTO, the tum-off circuit should be capable of outputting a high peak current. Usually, a thyristor is used for this purpose. In Fig. 4.60 (a), turn-off process is initiated by gating thyristor Tl. When Tl is turned on, a large negative gate current pulse tUrns off the GTO. 4.12.3.1. Gate turn-on. The turn-on process in n GTO is similar to that of a conventional thyristor. Gate turn-on time for GTO is made up of delay time, rise time and spread time like a CT. Further, tum-on time in a GTO can be decreased by increasing its forward gate current as in a thyristor. In Fig. 4.60 (b), a steep-fronted gate pulse is applied to tum-on GTO. Gate drive can be removed once anode current exceeds latching current. However, some manufacturers advise that even after GTO is on, a continuous gate current, called back porch current 19b as shown , should be applied during the entire on-period of GTO. The aim of this recommendation is to avoid any possibility of unwanted turn-off of the GTO.
,--,TR'
Spl~~ vo!tag~
", Tail current
R,
c, K L
i
TI
i
(a)
Fig. 4.60. Gate turn-off thyristor
... t .. P
Gale current
(b) (a)
basic gate drive circuit and
(b)
switching characteristics.
4.12.3.2. Gate turn-off. The tum-off characteristics of a GTO are different from those of an SCR. Before the initiation of turn-off process, a GTO carries a steady current la. Fig. 4.60 (b) . This figure shows .a typical dynamic turn-off characteristic for a GTO. The total turn-off 'time tq is subdivided into three different periods; namely the storage period ,(t,), the fall period .(t;) and the tail period (t f ). In other words,
tq =t, + t;+ t/ Initiation of turn-off process starts as soon as negative gate current begins to flow after t = 0 at instant A. The rate of rise of this gate current depends upon the gate circuit inductance L and th e gate voltage applied. During the storage period, anode current 1a and anode voltar.e (equal to on-state voltage drop) r emain constant. Termination of the storage period is indicated by a fall in If! and rise in Va' During t" excess charges, i.e. holes, in p '" base are removed by negative gate current and the centre junction comes out of saturation. In other words, during storage time t., the negative gate current rises to a particular value and prepares the GTO for turning-off (or commutation) by flushing out the stored carriers. After ts' anode current begins to fall r apidly and anode voltage starts rising. As shown in Fig. 4.60 (b), the an od e current falls to a certain value and then abruptly changes its rate of fall. This interval during which anode current fall s rapidly is the fall time t'l Fig. 4.60 (b ) and is of the order of 1 ~sec {4J. The fall
[Ml. 4.13J
Thyristors
193
period l; is measured from the instant gate current is maximum negative ta the instant anode current fall s ta its tail current. At the time t = l .. + l ,. there is a spike in voltage due ta abrupt change in anode current. After t,-. anode current ig and anode volta.ge Ug keep moving towards their turn-off values for a time t/ called tail lime. After t,. anode current reaches zer o value and Ug undergoes a transient overshoot due to the prese nce af RJ , C.. and then stabilizes to its off-state value equal to the source va ltage applied to the anode circuit. Here R J and C.. are the snubber circuit parameters. The turn -ofT process is complete when tail current reaches zero. The over shoot voltage and tail current can be decreased by increasing the size of C but a compromise with snubber loss must be made . The duration oft, depends upon the device" chara cteristics [4] .
4.12.4. Comparison b etween GTO and thyristo r A GTO has the following disadvantages as comp ar ed to a conventional thyristor : (i ) Magnitude oflatching and holding currents is mor e in a GTO. (ii ) On state voltage drop and the associated loss is more in a GTO. (iii) Due to th e multicathode structure of GTO, triggering gate current is higher than that required for a conventional SCR. (iu ) Gate drive circuit losses are mor e. (v ) Its r everse-voltage blocking capability is les s than its forward· voltage blocking capability. But this is no disadvantage s o far as in~erier and chopper circuits ar e concerned . In spite of all these demerits, GTO has the fon owing advantages over an SCR : (i)
(ii ) (iii) (i v ) ( v)
(ui)
GTO has faster switching speed. Its surge current capability is comparable with an S CR. It has more dil dt rating at turn-on . GTO circuit configuration has lower size and weight as campared to thyristor circuit unit. GTO unit has higher efficiency because an increase in gate·drive power loss and on-state loss is more than compensated by the elimination of forced -commutation losses. GTO unit has reduced acoustical and electromagnetic n oise due to elimination of commutation ch okes.
4 .1 2 .5 . Application of GTOs
In view of the above facts, GTO device are now being used in (a) high·perfonn anc e drive systems, s uch as the field-oriented control sc heme used in rolling mills , robotics and machine tooh [41. ( b) tracti on purposes because oi their lighter weight and (e ) adju 5 ~3.bl e-fr2:que ::-lo)· inverter dr ives . At present, GTOs with r atings ':1P to 5000 V and 3000 A are available .
-1.13. SFATIC INDUCTION TIlYRIsroR ISITB)
.
'
A static in ducti on thyristor, or 81TH, is a three terminal sel f· controlled devjce just like a GT O. It wa s co mmercially introdu ced in Japan in 1988. A si mil a r de-.-i: e, kn own as field ·controlled thyristor (FCT), or fi eld- controlled diode (FCD ), was de'/ elopea ea rEe r by Gene ral Elec tric but could not be comm ercially launched . H owever, commercial use 0[5I1 H is being promoted by Japanese unive rsities and industries .
194
Power Electronics
[.-\ rt • •. 13)
4. 13.1. Basic Structure The basic structure of 81TH is shown in Fig. 4.61 (a) and the device in Fig. 4.61 (b ). It is primari ly ap Pnn- diode, withp+ grid-like gate electrodes buried in n layer. Comparison of Fig. 2.23 (a ) and Fig. 4.61 (a) reveals that device structure of SITH is ana1ogous to SIT except that a p. layer is added on the anode side. In addition, n" type ftngers are diffused in p'" anode layer just as in a GTO.
\,--<>001('
A
, (b )
(a)
Fig. 4.6 1. Static induction thyristor
(a )
ba!ic structure and (b ) device symbol.
4.13.2. Turn-on and Turn-off processes A sim plified structure shown in Fig. 4.62 is employed for explaining the turn on and the turn-off processes in a 81TH. When anode is forward biased with gate·cathode voltage Vg equal to zero, the device behaves like a diode. Load current i(J flows from anode to cathode as p"n junction is forward biased, Fig. 4.62 (a). This shows that 81TH is a normally-on device just like SIT. When gate is reverse biased with respect to cathode, i,e. when Vg is negative, a depletion layer ..."ould be formed as shown in Fig. 4.62 (b), This dElpletion layer blocks the flow of anode current. With varying negative gate bias, the magnitude of anode current can be controlled. Cothode
Vg
5
. ~
v, L
o
•o
~
L-==---iJ Anoae (a )
(b)
Fig. 4.62. 81TH (a ) on-conditio n when gate voltage V, is zero and (b)
off-condition when
Vg
is nega tive.
. "
"
Thyristors
195
If 51TH is reverse biased, with cathode positive and anode n egative, electrons can flow from anod e, intermixed n'" layer, n , through p + grid, n '" and finally to cathod, Fi g. 4.6 1 (a). Th us, reverse current from cathode to anode can exist when 51TH is r everse biased. Thi s shows that 81TH does not have any reverse blocking capabi lity due to emitter-shorting (p+ lay er interdigitated with n+ layers at the anode ).
4.13.3. Application of 81TH and comparison with GTO At present, SITHs with 2500 V/500 A ratings and 100 kHz operating frequency are available. 8ITHs with higher power ratings and with normally-off characteristics are likely to be developed in the near future. Their use in induction heating, high fr equency link dc-ac converters and HYDe converters is being prom oted by Japanese organisations. When compared with a GTO, a 81TH (i) is normally-on device unlike GTO (ii) has higher conduction drop (iii ) has lower turn-off current gain, typically 1 to 3 instead of 4 to 5 for GTO (iu) has higher switching frequency (u) the du l dt and dtldt ratings are higher and (ui) has more SO A, (s afe operating area). 4.J.f. FIRING CIRCUITS FOR THYRISTORS An SCR can be switched from off-state to on-state in seve ral ways; these are forward-v oltage triggering, du l dt triggering, temperature triggering, light triggering and gate triggering, see Art. 4.2. The instant ofturning on the SCR cannot be controlle~ by the first three meth ods listed above. Light triggering is used in some applications , particularly in a s er ies-connected string. Gate triggering is, however, the most common method ofturn ing on the SCRs, because this method lends itself accurately for turning on the 5CR at the desired instant of time. In addition, gate triggering is an effic ient and reliable method. In th is secti on, firing ci rcuits for thyristors are studied in detaiL
-.
4.14.1. Main Featu res of Firing Circui ts As stated above, the most common method for controlling the onset of conduc tion in an SCR is by means of gate voltage controL The gate control circuit is also called firing, or triggering, circuit. These gating circuits are usually low-power electronic circuits. A firing circuit should . fulfil th e follOwing two functi ons . Shi eldll:d ~ cobles AC
"'0
Pulse g~n~rQlor
nput
Jlf1.
--+
,
PuiS! amplif i er
+-
-L..
J1Jl
-
Pu ! s~ trons to rm~ r
ll)
Pulse transtorme r
f
Pu isil tr ansfor mer
J
SCR
SCR
DC power supply
v Contro l CirC 1Ji!
,
,
v
,
Dr iV'lf ci rcuit
Fi g. 4.63. A ge::e: al lnyou t of t~ e firi ng circuit scheme fo r SCR.,.
SCR
'----y----' PO'Nllr (: jrcuit
Power Electronics
[An. 4.141
196
(i) If power
circuit has more than· one SCR, the flring circuit should produce gating pulses for each SCR at the desired instant for proper operation of the power circuit. These pulses must be periodic in nature and the sequence of firing must correspond with the type of thyristorised power controller. For example, in a single-phase semiconverter using two SeRs, the triggering circuit must produce one firing pulse in each balf cycle; in a 3-phase full converter using SLX SeRs, gating circuit must produce one trigger pulse after every 60° interval. (ii) The control signal generated by a firing circuit may not be able to turn-on an SCR It is therefore comm on to feed the voltage pulses to a driver circuit and then to gate-cathode circuit. A driver circuit consists of a pulse amplifier and a pulse transfonner.
A firing cir cuit scheme, in general, consists of the components shown in Fig. 4.63. A r egulated dc power supply is obtained from an alternating voltage source. Pulse generator, supplied from both ac and dc sources, gives out voltage pulses which are then fed to pulse amplifier for their amplification. Shielded cables transmit the amplified pulses to pulse transform ers . The function of pulse transformer is to isolate the low-voltage gate-cathode circuit from the' high-voltage anode-cathode circuit. Some firing circuit schemes are described in this section .
4.14.2. Resistance and Resistance-Capacitance Firing Circuits Rand RC firing circuits are not in commercial use these days. These are presented here for th e sake of highlighting the basic principles of triggering the SeRs. They otTer simple and economical firing circuits (3J. (0)
Resistance firing circuits. As statad above, resistance bigger circuits are the simplest
and most economical. They however, suffer from a limited range of firing angle control (O~ to 90°), great de p end en ce on temperature and difference in performance between individual SeRs. Fig. 4.64 shows the most basic resistance biggering circuit. R2 is th e variable r esistance, R is the stabilizing resistance. In case R2 is zero, gate current may flow from source, through load, R 1, D and gate to cathode. This current should not exceed maximum pe"rmissible gate current 19m. R 1 can therefore, be found from the relation, Vm
where
It is thus seen that func tion of Rl is to limit the t c. ~
0
LOAO
il rv "'s= Vmsinwt
b
R,
I,
~R'
,
.,I
D
Vm
R , S Igm or R , ~ Igm ... (4.24 0 ) V", = maximum val ue of sourc e volta ge
gate curr ent
,,'--'0---"';.
>R
J
!
j
t
Fig. 4.64. Resistance firing circuit.
safe val ue as R2 is varied .
Resis ta nce R should have such a value that maximum voltage drop a cross it does not exceed maxi mum possible gate voltage V,Ir.. This can happen only when R2 is zero. Under this condi tion, Vm R S V,flll 0:
R < Vi m' R l
- V It!
-
VJ m
..'< 4.24 0)
197
(Art. .U4]
Thyristors
As resistances R t , R2 are large, gate trigger circuit draws a small current. Diode D all ows
the flow of current during positive half cycle only, i.e. gate voltage amplitude of this dc pulse can be controlled by varying R 2.•
u, is half-wave dc pulse. The
The potentiometer setting R2 determines the gate voltage amplitude, When R2 is large, current,i is small and the voltage across R, i.e. tlg = iR is also small as shown in Fig. 4.65 (0). As V,p (peak of gate voltage u,) is less than Vgl (gate trigger voltage), SCR will not turn on. Ther efore, load voltage tlo = 0, io =0 and supply voltage U~ appears as UT across· SCR as shown in Fig. 4.65 (a>" . Note that trigger circuit consists of resistances on ly, II, is therefore in phase with source voltage u~ . In Fig. 4.65 (b ), R2 is adjusted such that V,p =V,I. This gives the value of
v,
V;
I
',I i,(
, •
Vgt ;
, ,
'
'
, ,, ,
:
:
:
,
. - . .;-.--: . - . ......:... j. - ~
V~ i
"
"
VlI'Isinwl
v" ., v,. w' ,WI•
,, : wI'
"
:
,U
,
,
Vgp=Vgt
;
·
:, wt
"bN i,t N
.-.-,
N,
, ,
: I
: wi
: I
T\ ,
,
,
:wl
:I
:i
,,
r:::::s
210' ;
:6 .,
• ;wt
"t
: WI
: : wi wi
Cl.
=90'
(1<
90'
00 ~ W Fig. 4.65. Resistance firing of an SeR in a half-wave circuit with dc load (a) No triggering of SCR (b) a = 90° (c) a. < 90~.
firing angle as 90 0 • The various current and voltage waveforms are shown in Fig. 4,65 (b ). In Fig. 4.65 (c ), Vgp > V~. As soon as va becomes equal t o V,! for the first time SCR is turned on, The resistance triggerin g cannct give firing angle beyond 90 0 • Increasing v, above Vg l turns on the SCR at firing angl es less than 900 • \"Vhen Vg r eaches Vgl for the first ti me, SCR fires, gate loses control and v, is reduced to almos t zero (about 1 V) value as shown. It may also be seen that firing a1,;{le can never be equal to zero degree however large VzP may be ; it can, of course, be brought nearer (2 0 _4°) to zero degree firing angle. A relationsh.ip between pea1t gate voltage VzP and gate trigger vol tage VGt may be expressed as follows: V gp sin
a
= V,I
or
a
=sin- 1 (Vj IIVgp)
• Some s~ud'!n !.J argue th3t in every posi:l're cycle of source, died e ci rcui~ will :e act ive l1:ld will therefore dro'l! curren: [roo. s:lurce. The current will ca u.se '/olt:!g! d!1)p liD acreS! load iL-"~ .::erefore, v" ..nd i .. should be shown in. F i~. ".5~ (a l. Ac~ually, load resilta."lce (3 few I,lh::!.s) in ccr::::Jar:so:1. w:~h R\ ... R:l " R (km is quite ,:nail. Ther'! ;",,!e, current d uring the posi tive cycle of source is negligibly S;';"I:lll ;:!.::d li!t'!wise ')~ across the load.
Power Elect ron ics
[Art. 4.14]
198
Since
vgp = 'R;-,-+-" Vrn R ROc,-+- R-;<
. _, [VI' .IR,VR + R, + R l] '
a=sm
m
As V,I,R 1, R and Vm are fixed, a"'" sin- 1 (R 2) or a"", R 2.
This shows that firing angle is pr oportional to R 2. As R 2 is increased from small value (i.e. small a), firing angle increases. In any case, a can never be more than 90°. As the firing angle control is from 00 (approximately) to 90 0 , the half·wave power out put can be controlled from 100% (for a = 00 ) down to 50% (for a = 900 ).
Exampl e 4.23 . Discuss what would happen to the circuit of Fig. 4.64 in case load is shifted between terminals a and b. Solu tion. In the circuit of Fig. 4.64 , when SCR is on, voltage vT across it is almost zero (actually about 1 to 1.15 V) and therefor e voltage across R I , R 2, D, R is a lso nearly zero. As a result, trigger supply voltage v, is reduced to zero after SCR turn-on. There is thus hardly any gate current and the associated gate power loss is zero during the time SCR is cond ucting in Fig. 4.64 . In case load is shifted between terminals a and b, the circuit ·may still operate. But after SCR turn-on, the circuit comprising of R 1, R 2, D and gate to cathode would be subjected to source voltage. This would cause an increased gate current and the associated gate power loss would be mo re during SCR turn on . Such an happening would certainly burn out the gate circui t and destroy the SCR. This shows that load should never be connected between t erminals a and b in Fig. 4.64. (b) R C firi ng circuits. The limited range of firing angle contr ol by resistance ·firing circuit can be overcome by RC firing circuit. There a re several variations ofRC trigger circuits. Here only two of them are presented.
RC half-wave trigger circuit. Fig. 4.66 illustr ates RC half-wave trigger circuit. By varying the value of R, firing angle can be controlled fr om 00 to 1800 • In the negative half cycle, (i)
capac itor C ch arges through D2 with lower plate positive to the p eak supply voltage V m at wt = - 90°. After wt = - 90 0 , source voltage v, decreases from - Vm at wt = - 90 0 to zero at wt = 0°. During this period, capacitor voltage Uc may fall from - Vm at wt = - 90° to some lower value - 0 a at wt = 0" as shown in Fig. 4.67. Now, as the SCR anode voltag e passes through zero and becomes positive, C begins to charge through LO A 0 variable resistance R from the initial volt age - oa. \-Vhen capacitor charges to positive voltage ~ual to 02 . R~ gate trigger voltage VS/I SCR is fired and after this, '1 capacitor holds to a small positive voltage, F ig. 4.67. rv vs =Vmslnwt Diode D1 is used to prevent the breakdown of 01 ca thode to gate junction through D2 during the negative half cycle . .'\n examination of Fig. 4.67 re veals that firing a ngle can never be zero and Fig. 4.66. RC half-wave trigger circui t.
Y
VC f C
180'.
Thyristors
199
In the range of power freque ncies , it may be empirically shown [3J that RC for zero output voltage is given by
=±
1.3 T RC > - 2 -w
where
T=
... (4.25)
7
= period orae line frequency in seconds.
The SCR will tri gger when u~ =Vgt + ud' where ud is the voltage drop across diode D1. At the instant of triggering, if Uc is assumed constant, the current I, t must be supplied by voltage source through R, D1 and gate to cathode circuit. Hence the maximum value of R is given by u,:::?: RIgt+u c u~ ~RI,t + Vgt +ud u.- V,t - ud R5 I
or or
... (4.26)
II
where Us is the source voltage at which thyristor turns on. Approximate values of Rand C can be obtained from Eqs. (4.25 ) and (4.26).
.l',0104-1.1 , ,
wI
,, ,,
".
.-' 0:
~Q~
I
". wI
2rr
wi
0
: 37f
I
i
wi
~:-
I
", 1 0
wi
w
v,,
I I,
,I,
I
wI
~f..-
11
3<
~)
Fig. 4.67. Waveiorms for RC bali·wave trigger circuit of Fig. 4.66 (a) high value of R (b) low value of R. ' When SCR triggers, voltage drop across it falls to 1 to 1.5 V. This, in turn, lowers the voltage across R and C to this low value of 1 to 1.5 V. Low voltage ;:,.cross S CR during conduction period keeps C discharged in positive half cycle until negative voltage cycle across C appears. Tl-!is charges C to maxi mum negative voltage - Vn: as shown in Fig. 4.67 by dotted line. In Fig. 4.6 7 (a ), R is more, the time taken for C to charge from - 0 0 to (V,t + ud ) == V.ft is more, firing angle is more and therefore average output voltage is 10....·. In Fig. 4.67 (b). R is less, firing an.gi e is low and therefore ave rage output voltage is more . (iO RC full · wa ve trigger cirCtlit. A simple RC trigger circuit giving fuI1 wave ou:rut voltage is shown in Fig. 4.68. Diodes DI- D4 fOlm a fu1l "N£!.v e diode bridge. In this cireui:, t he initial 4
4
200
[Mt. 4.14J
Power Electron ics
voltage from which the capacitor C charges is almost zero. The ca pacitor C is set to this low positive voltage (upper plate positive) by the clamping action of SCR gate. When capac itor char ges to a voltage equal to Vgt, SCR triggers and rectified voltage ud appears across load as uo. Th e value of RC is ca l culated by the empirical r elation [31,
RC
~ 50 T2 "
157
...(4.27)
· W
01
As per Eq. (4.26), the value of R is given by U -
R«
s
V
[,.
8
T
_oj
t
where UJ is the source voltage at which thyristor turns on. In Fig. 4.69 (a), firing angle a is more than 90 0 and in Fig. 4.69 (b) , a < 90 0 •
+fv
vd vs=Vm sin wt
N
•
:..t
Fig. 4.68. RC full·wave trigger circuit.
4.14.3. Unijunction Tran sistor (UJT) Resistance and RC triggering cir cuits described above giv e prolonged pulses. As a result , power dissipation in the gate circuit is large. At the same time , R and RC triggering circuits cannot be used for automatic or feedback control systems. These difficulties can be overcome by the use of UJT triggering circuits. Pulse triggering is preferred as it offers several merits over R and RC triggering. Gate characteristics have a wide spread, Fig. 4.9. Pulses can be adjusted easily to suit such a wide spectrum of gate characteristics. The power level in pulse triggering is low as the gate drive is discontinuous, pulse triggering is therefor e more effici ent. As pulses with higher gate current are pennissible, pulse firing is more reliable and faster. In this section, first U lT is described along .....ith its charact eristics and then its use as a r elaxation oscillato r for t riggering SCRs is presented. v,
"
Vm sinwt
wt
!
_
.
....l.
.'*- - ~~ '.. _.*.. - . ...•. ; , vo
" '' I "e ', I. v.gt "e :! , I '
Cl
.
,
' La: : k..cc I
:
I
: .
Vjvj l (a )
wt
I
'I
.
i
wT
j " I I ' i Vd~
Vd~dl I Ii I I '-
.I
WT
wi'
"1N N N J etr"'
vT w t~
I' lj
I'
i'
j
I
!,
1 /1
!
i/1
(b )
Fig. 4.69. Waveforms for RC half·wave trigger ci r C~lit of Fig. 4.68 (0 ) high valu e of R (b ) low value of R.
I wi
Thyristors
[Art. 4.U ]
201
An UJT is made up of an n-type silicon bas e to which p-type emitter is embedded , Fig. 4 .70 (a). The n-type base is lightly doped whereas p-type is heavily doped. The two ohmi c con tacts provided at each end are called base-on e B1 and base-two B 2 • So, an UJT has three terminals, n amely the emitter E, base-one Bl and base-two B 2 . Between bases B1 ano. 8 2 , the unijunction behaves like an ordinary r esistance. RBI and R B2 are the internal resistances respectively fr om bases Bl and B2 to eta point A, Fig. 4.70 (a). Its symbolic representation is given in Fig. 4.70 ( b) and its equivalent circuit in Fig. 4.70 (c). When a voltage V BS is applied across the two base tenninals B I and B 2, the potzntial of point A with respect to 8 1 is given by V AB l
V BB
RBI
=R 81 + R B2 . RB i = R 81 + R B2 . V 8S =n VBB
R
where n = R
81
B~ is called the intrinsic stand-off ratio, Typical values ofn are 0.51 to 0.82. + B2
Interbase resistance RBB = RBI +- RB2 is of the or der of 5-10 kn . Thi s resistance Ras can easily be measured by a multimeter with emitter open, As stated before, RBB is brok en up into two r esistances, RBI between emitter and base B 1• and RB2 in between emitter and base B 2 . Since emitter is nearer to 8 2 , resistance RB2 is less than the resistance RBI '
.,
8,
Eto-pcint
€"to -~ in~
p-type E
E
E
1+
.
QS 1.
n-type
v
1.,
8,
w
~)
•
1-
",
\
:
"B2
...,
A~
,
~
.../
.S.
i VB B
"~ 'IB8 QEl I
, ~
-
-
~
F ig. 4.70 . (a) Basic structure ofUJT (b) symboli c representation and (c) its equi"alent circu it.
The operation ofUJT can be und erstood with its equivalent circuit of Fi g. 4.70 (c). A:, UJT is usually operated with both B2 and E biased positive with respect to reference base terminal B!. DC voltage source VB8 between B2 and Bl is constant . DC source v-EE in ser ies with resistance RE is considered as l?put to tl~e UJT. Both V BB and VEE are shown in Fig. ~71 (a ) where UJT equivalent circuit is sh own inside the dotted rectangle. As before VAB I
V ••
=VA = R Bl + R B2 . RBl =11 · V SB
Th e magnitude of voltage V, can be varied by regulating external r esis tancz R E . _!.5 long a.5 emitte r voltage Vr < ll. V BB , the E - Bl Wlijunction (or p - n junctior!) is reverse biased and emit ter current 1, is negative as shown by curve PS in Fig. 4.il (b) , The region PS of very low current is tr eat ed as 'off' state of UJT. The resistance behveen E - B I j unction is therefore \·er.y high. At ;lOint S, 1, = 0, drop across R z is zero, th erefore 'llf = sour ce volta;;e, i. e. OS = V~ = VEE '
202
[A ,t. 4.14]
Power Electronics
Actually, off·sta te ofUJT extends to a point where emitter voltage V, exceeds VA' or n .V!JB' by diode voltage VD in E - BI junction. So when V, =fl . VBB + VD• point B is reached and E - B I junction gets forward biased to allow forward current through the diode. Here V D is the forward voltage drop across E - B I junction (usually 0.5 V). Point B is called the peak point . Voltage Vp and current Ip pertaining to point B are called peak·point lIoltage and peak·point current respectively. By varying R E • Vr is increased till V, approaches Vp . At this peak point, V, =Vp = fl .VBB + YD. thep ·emitter begins to inject h oles from the heavily doped emitter E into the lower base region B t . As n type base is lightly doped , the holes rar ely get any chance to recombine. The lower base region B I is, therefore, filled up v.;th additional current carriers (holes). As a r esult, resistance RB l of E - B1 junction decreases. The fall in R BI causes potential of eta point A to drop.
v,
UJT
i·········-···········
region
Saturation rf9!On ~
,
~-<';.'---'-~
eqlJlvclent ---.: CirCUlI '
Negative re sistance
Cut· olf region vse
j.R lood Une ; '~B(peakpoin l )
( nV
.~ S8
+v ) D
1 ::
VOl' , ·.:::.·· · ··· ····T · ·· ··-········ ········-~'I 5
: :
'~ ~
RI load line ''':'' J
.. •.. .::. ~
v,
a
'.:
•
:
~~~,
:
v
:.~ ~
:L__ •• _ . _• ••• ____ BI• _.~:
P 0 101
b
·····i•x
···· .. ·.········. xl
volley point
Iv ~)
00 Fig. 4.71. UJT
'0
~~,
~~
L. V ~::::"••C, __ :
:
.: ..... __ _ ....1 VI . : : i
.'...
I,
equivalent circuit with Vss and VEE , and (b) typical static V - / characteristics. (a)
This drop in VA' in turn, causes Vf (= VA + V D) to fan . As VEE is constant, fall in Ve gives rise to more emitter current Ie (= (VEE - Ve) / RE). This increased Ie inj ects more holes into region B 1• the reby further reducing the resistance RBI and so on. This regeneratiue or snow· balling effect continues till RBi has dropped to a small value (from about 4 kn to around 2 to 25 !l). The emitter current, limited by external r esistance R E • is then giv en by
=~V;.E=-E_-~V!!-D
I r
R BI +RE
'Nhen RBI has dropped to a very small value, indicated by point C in Fig. 4.71 (b ), the UJT has r eached 'on' state. At point C, entire base region Bl is saturated and resistance RBI cannot decr ease any more. This point C is called the valley point ; Vu and Iu are the corresponding emitter potential and curr ent. After UJT is on, or after valley point is r eached, an increase in Ve is accompanied by an increas e in I f; this is indicated by curve CQ. At point Q, Ve is a little more than its valley poin t voltage Vu. Between points B and C, em itter voltage Vr falls as It incre ases; UJT, therefore. exhibits negative resistance between these two points. The negative
Thyristors
203
[Art. 4. 141
resistance region between peak and valley points in Fig. 4.71 (b) gives UJT the switch ing characte ristics for use in SCR triggering circuits. At the valley point, the current is given by VuIRB l' Valley-point current , aLso called holding current, keeps UJT on. When emitter current It falls below l u, UJT turns off. UJT oscillator triggering. The unijunction transistor is a highly efficient switch ; it5 sw itching tim e is in the range of nanosecond s. Since UJT exhibits negati v~ resistance characteri stics, it can be used as a relaxation oscillator. Fig. 4.72 (a ) shows a cir cuit diagram with UJT work1ng in the oscillator mode. The external resistances R 1. R 2 ar e smaH in comparison with the internal resistances RBI' RB2 of UJT bases. The charging resistance R should be such that its load line intersects the device characteristics only in th e negati ve res istance region. In Fig. 4.72 (a), when source voltage VBB is applied. capacitor C begins to charge through R exponentially towards '\lBB ' During this charging, emitter circuit of UJT is an open circuit. Th e capacitor voltage Uel equal to emitter voltage u t • is given by u~=Ut=
V BS
The tim e constan t of the charge circuit is
(1-e-
t l =
Re.
When this emitter voltage u t (or IJ e) r eaches the peak-point voltage Vp (= 11 V BS + VD)' the unijunction between E - B I breaks down. As a result, UJT turns on and capacitor C rapidly discharges through low resistance RI with a time constant t 2 = R IC. Here t a is much smaller than tl' When the emitter voltage decays to the valley-point voltage Vu, emitte r current (V1/(R B1 + R 1 falls below I I) and UJT turns off. The time T required for capacitor C to charge from initial voltage VI) to peak-point voltage Vp • through large r esistance R , can be obtained as under :
»
Assuming or
... (4.28 )
v, VE3
Capocitor discnorging
v,
R,
R S·,
E
C
V,
Bl
I
R'
.,
,.
00
Fig. 4.72. UJT oscillator
(a )
Connection diagram and
W
(b)
Vol tage wave rorms.
204
[Art. 4.14)
Power Electronics
In cas e T is taken as the time period of output pulse duration (neglecting small discharge time ), then the value of firing angle a 1 is given by
a, = roT = wRC In
1 I -~
... (4.29 )
where w is the angular frequency of UJT oscillator. The amplitude of pulse voltage is obtained by drawing a load line B b for Rl 'a s shown in Fig. 4.71 (b). The vertical projection of Bb, equal to xy, gives the voltage pulse amplitude. With the discharge of capacitor, the operating points Band b move towards C. For pointsp and q, the pulse amplitude is XlYl' Eventually, point C is reached at which puls e voltage is zero, then the operating point shifts to a , Fig. 4.71 (b ). The potential of eta point A is 11Vss , but that of the emitter is Vu which is less than 'lVBS ' As a result, E-B} unijunction is reverse biased and ceases to conduct, the UJT turn s off and goes into blocking mode. Capacitor C now again charges from V t = Vu to voltage 11 V Ba + VD, E - B} unijunctioll breaks down and the above cycle repeats. If the output voltage pulses are used for triggering an SCR, resistance R} should be su ffi ciently s mall so that normal leakage current drop across R IO when UJT is off, is not able to tri gger the S CR. In other words,
R
Vas ' R l R R < SCR trigger voltage V,t
BB+
1+
2
where
R Bs = R st + R'B2 The emitter-diode forward characteristics vary with temperature in such a manner that VD decre ases and Ras increases with temperature. In order to provide compensation against this thermal effect, the value of R2 used in Fig. 4.72 should be calculated from the relation
10'
R, =-TjV
... (4.30)
SB
- ..
The width of triggering pulse is sometimes taken equal to R }C. In case load line forR intersects the UJT characteristics in the region CQ, Fig. 4.71 (b), the intersecting point will result in stable operating point and the circuit then cannot work as an oscillator. This fact fixes the maximum and mini~um values of charging resistor R and the osc illator output frequency. Th e maximum value of R is determined by the peak-point values Vp and Ip- When voltage across C r each es V p • the voltage acrossR is V BS - Vr _ V BB
R n fU -
-
Vp _
I
p
V-SB -
-
(TlVSB + VD)
I
... (4 .3Ia)
p
The mimmum value of R, governed by valley-point values VII and I I,; is given by
.
R o~
=
V ss - Vu I
... (4.31 b )
"
E xamp l e -1.2 5. A rela.xa tion oscillator using an UJT, Fig. 4. 72 (a), is to be designed fo r triggering an S CR. The UJT has th e following deta :
=0.72, Ip = 0.6 rnA , Vp = 18.0 C'2rrent !dc.i. emitter open = 4.2 mA. II
V, VI: = 1.0 V , I v = 2.5 TT'. A, Ra3 = 5 k n , No r m a l leak age
•I
[ML 4.14)
Thyristors
The firing frequency is 2 kHz. For C = 0. 0-1
~}
205
compute the ualues of R, Rl and R'!..
Solution. The valu e of charging resistor R, from Eq. (4.28), is T
R= A s V D is not given,
.
C In -1=-~
=
1
(C In 1 = ~
10'
=
= 9.82 kn
2000 x 0.04 In 0.; 8
Vp =T)Vss
V
_.'::e_18.00_2-V 11 ~ 0 .72 -
ss -
;:)
10'
From Eq. (4.30),
R, = 0.72 x 25 V BB
With emitt er open.
R,
= 555.55 n
=
Leakage curre nt (R 1 + R 2 +R BS )
=
25 3 4.2 x 10-
-
5000 - 555.55
n = 396.83" 397 n
Example 4.26. If the firing frequency of the SCR in Example 4.25 is changed by uarying charging resistor R, obtain the maximum and minimum values of R and the corresponding frequencies . Solution. From Eq. (4.31),
R
= max
R .
= 25.0 -
1.0 = 9 6 kn 2.5 X 10- 3 .
mm
From Eq. (4.28),
r. mm
and
,
25 (1-0.72) = 11.67kn 0.6 X 10- 3
=_1_ =_~1,---:- R e i _1_ m= n 1-~ .. 10' = 1 =1682 .8 Hz" 1.683I;Hz 11.67 x 0.04ln 0.28
fm= =
T m:u:
10' 1 = 2045.7 Hz" 2.05 kHz 9.6 x 0.04 1n 0.28
Synchronhed UJT triggering (or Ramp triggering). A synchronized UJT trigger circuit using an UJT is shown in Fig. 4.73. Diodes Dl - D4 rectify ac to dc. Resistor Rl lowers Vde to a suit able value for the zener diode and UJT. Zener diode Z functions to clip the rectified voltage to <-'itandard level Vl" which remains constant except near the Vde zero, Fig. 4.74. This volt age V: is ap pli ed to the charging circuit RC. Current it charges c a p a~i to r C at a rate deter mined by R. Voltage across capacitor is marked by ue in Figs. 4.73 and 4.74. "Then voltage VI; reaches th e unijunction threshold voltage TlVr • the E - Bdunction ofUJT breaks down and the capacitor C discharges through primary of pulse tran sfor mer sending a current i',! as shown in Fig. 4.73. As the curren t i2 is in the form of pulse, windings of the pulse transfor mer have pulse voltages :g thei r secondary terminal s. Pulses at the two second J. :-Y wi:-_:iings feed the same in-pha se pu13e to two SCRs of a full- wav e circuit. SCR with positi ve anode vol tage would turn
206
Power Electronics
[Act. U4]
01
~
+
+
03
<
i, ~
~R /. E
e
Vd,
04 ~
zr; ~
\:
02
-
+
V,
-
;: C
'~
Puln Troosf.
le j)8 " I,
GI
c,
TO SCR
~G2
-
GAT ES
~c,
Fig. 4.73. Synchronised UJT trigger circuit.
on . As soo n as the capacitor discharges, it starts to recharge as shown. Rate of rise of capacitor voltage can be controlled by varying R. The firing angle can be controlled up to about 150:1. This method of co ntrolling the output power by varying charging resistor R is called ramp control, open·loop control or manual con trol. As the zener diode voltage V z goes to zero at the end of each half cycle, the synchronization of the trigger cir cuit with the supply voltage across SCRs is achieved. Thus the time t, equal to a / w, when the pulse is applied to SCR for the first time, will remain constant for the same value of R. Small variations in the supply voltage and frequen cy are not going to effect the circuit operation.
1111
Pulst voItac;j!
•
Puis! voltage
",",,
, :,, ,
,,-
iii
~)
, " "
I
,,
:: "
I'
1111 " " "
:2
,, i ,,
" ':"
I'
""" ",, I I 2 ,' ,
1111
•
~
Fig. 4.74 . GenerBtion of output pulses for the circuit of Fig. 4.73. Here, t = ct/w.
In case R is r educed so tha t Uc reaches UJT threshold voltage twice in each half cycle as shown in Fig. 4.74 (b), then there will be two pulses in each half cycle. As the first pulse will be able to turn-on the SCR, second pulse in each cycle is redundant. Ramp-and-pedestal triggering. Ra mp and pedestal triggering is an improved version of syn chr onized-UJT-oscillator triggering . Fig. 4.75 shows the circuit for r amp-an d-pedesta! triggering of two SCRs conn ected in antiparalleJ for controlling power in an ac load. This trigger ci rcuit can als o be us ed for triggeri ng the thyri3tor s in a single-phase semiconverter or a 5ingle-phase iull converter. The various voltage waveforms are shown in Fig. 4.76. Zener diode voltage V": is constant a t ita thr es~-h old voltage. R2 acts as a potentia! divider. Wiper of R2 controls the value of pedestal voltage Vpd. Diode D allows C to be quickly charg ed to Vpc! through the low r es is tan ce of the upper portion of R 2. The setting of wiper on R,! is such
,
Thy r istors
[Acl. 4.14J
207
LOAD
+ Y,
, 01
+
+
R,
b
<
' 03
R
D
~VmSinwt
V"
Z'2
~ V R,
,
04
\
VC:l
02
-
-
>,
!=
i-
,
C=
Y,
/
/
~
R,
,. (
g.
Y,
nr
~.
\
T2
0
c........"O
Fig. 4.75. Ramp and pedestal trigger circuit for ac load. that this value of Vpd is always less than the UJT firing point voltage 11 V:. When wiper setting is such that V pd is small, Fig. 4.76 (al, voltage Vz charges C through R. When this ramp voltage VI; reaches 11Vz, UJT fires and voltage u ' through the pulse transformer, is transmitted to the i gate circuits of both SCRs Tl and T2. The forward biased SCR Tl is turned on. After this, Vc reduces to Vpd and then to zero at wt = n. As til; is more than Vpd. during 'the charging of capacitor C through charging resistor R , diode D is r everse biased and turned off. Thus Vpd does not effect in any way the discharge of C through UJT emitter and primary of pulse transformer. From o to n, Tl is forward biased and is turned on. From n to 21t, T2 is forward biased and is turned on. In this manner, load is subjected to alternating voltage tlo as shown in Fig. 4. is.
Y'~m'inw. ~ .I w'
i'~. V,.
Ib,
,
V"
V,
1
v,
I'
'
I . Vp '
jJI
k I .
I
vo~~JJ
I . Wi I 1
1
1
U (0 )
U_
,I
I
Vdc
"
- ~~iz I I W.
1
~IW'
v. •
WI
Y ,
Y'F~2
•
I
I
I
Y'~~~ .
I N_ w!
I
'
1
V,
I'
.
'I' r -..... 1
I
_ ,_
v,~ ,_ ... ~, ___I[ Y, V~'._.L + ~' !
. !
1 Vg Q2
I
i
V
I
·
'
I v ' J lIJl
pd
'
I
'
rtVz . ,.
1 I
w
ltUt'"'
r
I
Yo
-- I---wr(b )
Fig. 4.76. Waveforms for ramp-and·pedes tal circuit of Fig. 4.75. Wit h the setting of wiper on R.2, pedes tal vol tage Vpd on C can be adjusted. With low ped estal voltage across C, r amp charging ofC to Tl "llz takes longe r tim e, Fig. 4.76 (0 ) and firi ng angle de by is therefore more and output voltage is low . With high pedestal on C, voltage- ramp
208
Power Electronics
(Ar!. 4.15]
charging of C through R r eaches nVz faster, firing a.Y1g1e delay is smaller, Fig. 4.76 (b) and ', output voltage is high. This shows that output voltage is proportional to the pedestal voltage, The time T required for the capacitor to charge from pedestal voltage Vpd to nV: can be obtained from the relation
,= Vpd+( V, - Vpd)(l-.-TI Re )
~V
Note th at (V:
~
Vpd ) is the effective voltage that chnrges C from Vpd to nVz' From above Vl-VDd
... (4.32 )
T =RC In V, (1 _~) and the firing angle delay
U2
is given by Vz - V Dd
...(4.33 )
ex, = w RC In V, (1 _ ~) 4.15. PULSE TRANSFORMER IN FIRING CIRCUITS
Pulse transformers are used quite often in firing circuits for SCRs and GTOs , This transformer has usually two secondaries. The turn ratio from primary to two s~condaries is 2 : 1 : 1 or 1 : 1 : L These transformers are designed to have low winding resistance, low leakage reactance and iaw inter-winding capacitance. The advantages of using pulse transformers in triggering semiconductor devices are : the isolation oflow-voltage gate circuit from h igh-voltage anode circuit and (ii) the triggering of two or more devices from the same trigger source. (i )
A square pulse at the primary terminals ofa pulse t ransformer may be transmitted at its
secondary terminals faithfully as a square wave or it may be transmitted as a derivative of the input waveform. The conditions governing the operation of a pulse tranSformer in these two functional modes are now examined, A general layout of the trigger circuit using a pulse transformer is shown in Fig. 4 .77 (0). Here the function of the diode is to allow the flow of current after the pulse period (i .e: when the transistor is off) so that energy stored in the primary of pulse transformer is dissipated. I In Fig. 4.77 (a). the trans,istor is acting simply as a switch, turning on when the pulse applied to its base is at its high level, thereby connecting the de bias VB to the transformer primary. The advantage of this arrangement are two fold: (0 ) There need not be a variable strength pulse generator since the pulses may be of the same amplitude and the strength of the generated pulses may be incr eased simply by varying the de bias voltage. (b) The operation of the circuit becomes independent of the pulse characteristics sinc e..!he only rol e the pulse plays is to turn-on or turn·offthe transistor. Therefore, there is no effect of pulse dis tortion (e.g. pulse edges or any spike superimposed on the puls e) on the working of this circuit,
In Fig, 4.77 (a), RL limits t h e current in the prima ry circuit of puhe tra.'1sformer. Its equivalent circuit is drawn in Fig, 4.77 (b ), where L is the magnetizing inducta:.lce of the pulse tr ansformer and Rg is the resistanc e of gate-cathode ci~cuit of an SCR. Fig. 4.77 (c ) 5hoW3 the transfer of R, to pulse transformer prim ary as R, =
[~:
J
Rg. This circuit can be analysed by
Thyristors
[MI. US] Rl
209
Puls e tra nsf
L
R,
V,
l
N, : N2
+
(b )
•
R,
T2
I
•
I,
t Pulse
i
V,
' • R,,(-N,N, )R
(e).
trans former
IUl G
Vo
RO)
L'
..
I.
..
.' . , .... .
..
(
applying Thevenin's theorem at the terminals a circuit, where
R,
Vo= VB R , + R L The voltage equation for Fig. 4.77
(d )
b.
Fig. 4.77
(d )
is the Thevenin's equi .... alent
and
is
di V0= R O l· + L dt Rl R 1RL . di VSR 1+RL = Rl + RL,+Ldt
or
) ~;
L VB= RL i + L (R ';,R
or
Its solution is given by
i=
~: [ 1 _ e- L (:I :tRt ) t]
The voltage across L appears as the outpu t voltage. The magnitud e of thi s voltage from pulse trans former is
or
where
~
o W- ~
.,
(a )
I
I
L
...(4.34 )
.
21 0
[Act. 4.1;]
Power Elec tronics
Depending upon the values of Ro and L , t her e are two functio nal modes of pulse transformer.
Jio > 10 T, where T is th e pulse width (Fig.
(a ) If L is so large as compared with R o that
4.78) of the input signal at G, then from Eq. (4.34),
-V Rl e e - B Rl + RL
Fort=O, and fo r t = T,
(t IlO T)
R, "0= V. R ,+ R L "T = V. R .
Rl 1+
R "
- 0.1
L
= 0.904 V.
R
Rl
R
1+ L
= 0.904"0
Thus the fall in the pulse level during the transmission through the pulse tran sform er at
t = T is very small. This shows that when ;
o
> 10 T , the input pulse is faithfully transmitted
as square pulse a t the output terminals of pulse transformer as shown in Fig. 4.78 (a).
B
Fort=O, and· for t = T, This shows that for
Rt Rl+RL
~o < ~, then from
Eq. (4.34).
'e -( l O/7')t
R, "o=V' R ,+ R L Rl
-10
eT = V. R, +RL e
;0 ro, <
R1 = 0.0000453 · VB R, +RL = 0.0000453'0·
the input pulse is transmitted in the fo rm of exponentially
decaying pulses as sh own in Fig. 4.78 (b). It is seen that for a step rise in input voltage, the pulse transformer output is a positive puls e. In other words , the input signal is trapsmitted as a derivative of the input waveform for a step rise. Likewise, for a step fal l in input voltage, a negative pulse appears at the pulse transformer output. Fig. 4.78 (b). The operation of th~ pulse transform er in this mode can be achieved by uS.mg a small value of L , i.e. by us ing an air core for the pulse transformer. L --!-_ __ + --!___--'-_ +.t
.,
II
v91
It can thus be inferred from above that the deciding v factor in the waveshape of the output pulses from a pulse transfo r mer is its inductance . If the pulse tr ansformer has large inductance, the pulses are fai thfully r eproduced and ifthe inductance is small, the pulses are exponentially decaying pulses. The negative going puls es : an be easily removed by using a clipper.
9
f
(O)!
l
1
1----.J1\l.l. . - - - .r-J.:I\>---.,yrr- . ,
h--r=:y
to)
Fig. 4.78. Output voltage wa veform
of a puls e transform er for L
(0 ) Ro > l OT
L
T
and (b) Ro < 10
Thy r is tors
[Act. •. 16[
211
The amplitude of t he trigger voltage at the secondary te rminals of pu lse t ransformer is
N2
Vg=N- · V· R 1
R} 1+
R
l.
The magnitude of VB should be large enough to pr oduce t rigger voltage Vgr at the gate circuit of SCR fo r its r eliable turn on, i.e.
N, V. R , N 1 ' R I + R l. ~Vg,
V. ~V~~: (l + ~~)
or
R, =
But
(~: JR,
V. ~V" ~: [1 +(~:J' ~:1
(435J
In practice, exponentially decaying t rigge r pu lses of Fig. 4.78 ( b ) are preferred due to the . following reasons: {O This pulse waveform is suitable for inj ecting a large charge in the gate circu it fo r reliable turn on . (i i )
The duration of this pulse is small, therefore no signi fi cent heating of the gate circuit
is observed. (iii) For the same gate·cathode power, it is permissible to raise VB t o a suitable high value so that a hard·drive of SCR is obtained. A device with a ha rd-drive can withstand high dU d t at the anode circuit which is desirable. (iu) The size of th e pulse t ran sfor me r is r educed . For an ext ended pulse, large L (with iron-cor e) is r equired which increases size and cost of the pulse tr ansformer . 4.16. TRIAC FIRING CIRCUIT
.
A tr igger ing circuit for a triac using a diac is discu ssed in this. sectio n. Fig. 4.79 shows a t ri ac fi ring circuit employing a diac. In this circuit, resistor R is variable whereas resistor Rl has constant resistance. When R is zero, R } protects the diac and triac gate fr om getting exposed to a lmost full supply voltage. :1 Resistor R2 limits the current in t he diac and r--'i:i!co;;."' }.!.-r-- - - ----; triac gate when diac turns on. The value of C and R, potentiometer R are so selected as to give a firing angle range of n early 0:' and 180:'. In practice, however, a triggering angle range of 10:' to 170= TriOC" is only possible by the firing circuit of Fig. 4.79.
,-vo
Va ri a ble resistor R controls th~ charging time of the capacitor C and therefor e the firing D!ce angle of the triac . When R is small, the charging t ime con st a nt, e qu al to (R 1 + R ) C. is small. The refor e, sou rce voltage charges capacitor C to Fig. 4.79. Firing circuit ror a triac u5ing n due.
212
Power Electronics
[M t. 4.16J
diac trigger voltage earlier and firin g angle for triac is small. Likewise, when R is high , firing angle of triac is large. When capacitor C (with upper plate positive) charges to breakdown voltage Vd: of diac, diac turns on. As a consequence, capacitor discharges rapidly thereby applying capacitor voltage Vc in the form of pulse across the triac gate to turn it on. After triac turn-on at firing angle cr., source voltage v, appears acr oss the load during the positive half cycle for (n - cr.) radians. When Vs becomes zerq at rot = n, triac turns off. After OJt = It, uJ becomes negative, the capacitor C now charges with lower plate positive. When Vc r eaches Vell of disc, diac and triac turn on and v , appears across the load during the negative half cycle for (n - cr.) radians. At Wi = 2n, triac turns off again and the above process repeats. The waveforms for v" vc. vT and Vo are shown in Fig. 4.80 (a ) for minimum R and in Fig. 4.80 (b) for maximum R . Here VJ is the source voltage, Vc is the voltage across capacitor, vT is the vo ltage across triac and Vo is the output or load voltage. After triac turn-on, capacitor C holds to a small positive voltage.
v, , I
I, I Vc
:
----~~ .__ -
:
.I I·
. ,,
I
,
wt
••
l - -,,- /1~dr ~- -- -~:-~I _____1: ;/1
•
.. __
--J
'
!
----;1----- :l Vdl1 --:- ,' --::r-:1 Vo
-.
--' ra---i, ''( 2Ifter), '" i.
i VT
I
( 1f+0) I
i
.'2J't
:I 'K
:
I
,
. W
I
•
:, II ! wt I
I wi
~ (a )
~
Fig. 4.80. Waveforms for triact firing circu it using Q diac with pot . R adjusted to minimum and (b) pot. R adju ~t(' d to maximum.
The wave forms shown in Fig. 4.80 ar e for ~o... . ideal circuit components in Fig. 4.79. In fact, this circuit produces unsymmetrical waveform f Ol" the positive and negative half cycles of load vo ltage. This asymmetry is, to some extent, R due to tria:.-characte ristics but it is mainly due ru us::VmSlnt.lt t o hystere sis present in the capacitor. This means that when lI, is zero, Vc is not zero. In c c, other words, capacitor retains some charge of the in iti al voltage applied across its p lates wh en 30u r ce voltage falls to zero. The Fig. 4.81. Commercia! tri ne firi ng ci rcu it wav eforms fo r posi tive and negative half cycles U!~ing a diac. ca n, h owev er, b e made symmetrical if e.d dit ion al r es istance R3 and capacitor C1 are employed as shown in Fig. 4.81. This circuit is
Thyristors
[Ac'. 4.16]
213
commercially used for controlling the power in lamp dimmers, heat convertors, speed control of fans etc . For inductive loads, snubber circuit must be used across the triac. Example 4.27 . The firin g circuit for a triac using a diae, Fig. 4.79, ha.s the following data : R I = 1000 n, R = zero to 25000 n. C = O.SIJF. V, = 230 Vat 50 Hz, Diae breakdown voltage = 30 V.
Find the magnitude of maximum and minimum firing·angle delays for the triac. The effect of load impedance may be neglected. Solution. When the diac is not conducting, the current through R I • Rand C is given by V,
I,=Z
Z=[(R +R)'+(~)'
where
1
, ]'12
r
10002+(2'X;~·XO.5)
Z= [
When R =0 :
= [1000' + 6366.2'1"' = 6444.3 n I I leads V, by an angle
R, +R
"=\ -, lIwC =t _16366.2 =81 07' an an . 1000 . • 230 LO' I, = 6444.3 L _ 81.07'
:. Voltage across capacitor
V, =J, · X, = 230 L81.07' x 6366.2 L _ 90' = 227.2 L - 8.93' 6444.3 . u, =,[2 . (227.2) sin (Oli - 8.93')
or When capacitor voltage a l is given by
Uc
reaches the breakdown voltage of the dine, the triac firing angle
u, = ,[2. (227.2) sin (a, - 8.93') = 30 V or
al When R = 25000 n
:
_ -
. -1 sm
'12
893' -143' x30227.2 +. .
Z = [26000' + 6366.2'1"' = 26768 n
$.' tan-'
[;~~~;] = 13.76'
230 LO'
I, = 267 68 L- 13 .76' or When ..
or
ii;a
76 230 ' x 63 66.2 L- 90' = 54.7 L - 76 .24' 2 Uc =..J2 x 54 .7 sin (wt - 76 .24°)
V,
Uc
equal 30 V, let the firing angle be ~ . . 2 x 54 .7 sin (~-76 . 2 . P ) = 30 V
ex, = sin-
1
(12 ;~4.7 ) + 76.24' = 99.06'
Thus the ma'
214
[Art. 4.17[
P ower Ele ct ron ics
4.17. GATING CIRCUITS FOR SINGLE·PHASE CONVERTERS A gate trigge r circuit fo r thyristors in phase-controll ed r ectifi ers sh ould possess the following : A circuit for the detection of zero crossing of th e input voltage. Gener ati on of trigger pulses of required waveshape. (iii ) DC power supply for pulse amplifier. (iu) Gate tri gger circuit isolation from the line potential by means of pulse transformers or optocouplers. A general block diagram for gate trigger circuit for single-phase converter is shown in Fig. 4.82 . The gating circuit consists of synchronizing transformer, diode rectifi er, zero crossing detector, firing-angle delay block. pulse amplifier, gate-pulse isolation transformer and pow er circuit for the con verter. (i)
(ii)
Synchronizing mid-tapped transformer steps down the supply voltage suitab le for zero crossing detector and for .delivering de supply Vee to gate trigger circuit. The zero crossing Power circuit
1
Sourc.e vollage
~
3
~
Ov,
..
, It
2
~
•
~ ;;
=C vcc
-
~ Sync;;tz ing Transf .
-'ICC
I Zero crossing Detector
Input (ontrol sigl"lol
I Firing OI"Igle
I
Vi
D~ l ay
Pulse amplifie r
Vj
Gale pulse isolation transforme r
=:~G' ~~ V" =:G ~
V9 1
hr
~ C'
-vc., Fig. 4.82. Blc c:~ dia -rram of':! th}Tisto, g";!.ting circuit.
•
Thyristors
1M!. 4.17]
2t5
Synchronizing Trans!. vol tage
r o
I
1
Romp voi,tage from zero C. O.
o
,i.'
).
I
i
I
I
I
I .
•
! I
I
wI
. I
f-.=f1
2.
'
b~ --n I
wI
Control voltage Ec
,
Uik·-b o
2_ I
(211'..-u ) 311'
I
1:--:0
'-If
WI
I
i
I
wI
Fig. 4.83. Waveforms for the ci rcuit of Fig. 4.82.
detec tor conve rts ac synchronizing input voltage into ramp voltage and synchronizes th is ram p voltage with the zero crossing of the ac supply voltage 8S shown in Fig. 4.83. In the;firing- angle delay block, the constant amplitude r amp voltage is compared with con tr ol voltage Ee. When rising ramp voltage equals contr ol voltage E e , a pulse signal of con trolled duration is generated as shown in Fig. 4.83. Th ese signals are indicated as u, for thyri stors 1 and 2 and uJ fo r thyristors 3 and 4 for the power circuit of Fig. 4.82. If Ec is lower ed, firin g angle dec reases and in case Ec is rai se d, firing angle increases. This shows that firing-delay an gle is dir ectly proportional to the control signal voltage. The pulse output from the firin g-delay angle block are n ext fed to a pulse amplifier circuit. The amplified puls es ar e th en used for triggering thyristors 1, 2, 3 and 4 through gate-pulse isolation trans formers as shown. 4.17.1. Gate Pulse Amplifiers Pulse output from integrated circuits (lCs) may be directly fed to gate-cathode circuit of a low-power thyristor to turn it on. But in high-power thyristors, trigger-current requirement is high. The refore, pulses derived from rcs must be am plified and then fed to thyristor fo r its reliabl e turn on. In a thyristor, anode circuit is subjec ted to high voltage whereas gate circuit works at a low voltage. Th erefore, an isola tion is essenti al betwe en a thyristor and the gate-pulse gen er ator. As stated befor e, this isolation is provided by an optocoupl er or a pulse transform er. A pulse-amplifi er circuit for amplifying the input pulses is sh own in F ig. 4.84. It consists of a MOSFET (or a tran sistor ), a pulse transfor mer fo r is olation and diodes 01, D2. Wh en a voltage of appropriate level is applied to the gate of MOSFET, it gets turned on. :\3 a resu lt, most of the de voltage Vee appeC'.rs acr oss transformer pr imary a.'1d corresponding pulse voltage is induced in the transformer secondary. This amplified pulse or;. the secon dary side is applied to gate and cathode of a thyristo:" to turn it on. When pu lse signal appl ied to the gate of MOSFET goes to zero, MOSF ET turns off. Th e prim nry current due t o Vee ten ds to fall and li kewise flux in core also tene!.s to decrease. Due to th is tendency, a vo ltage of opp os ite polarity is indu ced in both prima.ry and seco ndary windings of pu lse transformer. Diode DI an the
Zl6
Power Electronics
[Art. 4.1 7]
secondary side of pulse transformer prevents the flow of negative gate current due to the reverse secondary voltage when MOSFET is off. Reverse voltage in primary, however, forward biases diode D2 when MOSFET is off. Current flow is thus established in the circuit consisting of primary, Rand D2. As a consequence, energy in the transformer magnetic core gets dissipated in R and the core flux gets reset. In case pulse width at the secondary terminals is to be increased. then a capacitor C is connected across R as shown in Fig. 4.84 (b). vco
A Pulse Trans
V~t=15V 0
0
~: .,
)
0'
G
o
~ N, iN,
0
N,
02
.J tG
N,
~
02
K
v g, R
~
P ~15 e ,i gnal
II-
(al
MOSFET
K
~ ~~JMOSFET
rr,
(
=!=
R
..
R,
t
Pulse si gnal
(bl
Fig. 4.84. Pulse amplifier circuit using a MOSFET for a thyristor trigger circuit (a) short-pulse output (b ) long-pulse output.
4.17.2. Pulse Train Gating Pulse gating is not suitable for inductive, i.e, RL loads, because initiation of thyristor conduction is not well defined for these types ofloads, This difficulty for such situations can be overcome by triggering the thyristor continuously. Continuous gating, however, suffers from some disadvantages like increased thyristor losses and distortion of output pu lse due to saturation of pulse transformer by continuous pulse. In order to overcome these shortcomings of continuous gate signal, a train offuing pulses is used to tum on a thyristor. A pulse train of gating signal is also called high-frequency carrier gating. A pulse train can be generated by modulating tl~e pulse width at a high frequency (10 to 30 kHz) as shown in Fig. 4.85. A circuit for generating a pulse train is shown in Fig. 4.85 (a) . This circuit consi .;;!;:! of an AL'ID-Iogic gate, 555 timer, MOSFET, isolation pulse transformer and diodes Dl, D2. The pulse signal Vi, obtained from the thyristor trigger circuit and shown in the top of Fig. 4.85 (b ), is fed to AND gate. The output ut of the timer 555 as shown is also fed to the A.!'\fD gate . T he duty cycle of the timer should be less than 50% in ord er to allow the transform er fl u x to reset . The pulse signal Vi and timer output V t are processed in the AND gate to get the waveform output Ux as shown. The output from A-'N'D gate is ~hen appli ed to pulse amplifier circuit to augment the amplitude of Ux to Ugif.' The amplified output waveform ug .\: is then app lied across gate-cathode terminals of a thyristor to turn it on.
[Ad. 4.1 81
Thyristors
217
A
Pu lse Trans
Vee
01
•
•
N,
N,
G
!
\
02
R
v,
I
",k
K
I
V'hnn nhnnnhnnnhn V'h nnn pnno ; ,
I I
I
•
::IMDSFET
R, Timer
(a) Fig. 4.85. Pulse train gating
(a)
(b) circuit (b) waveforms.
4.18. COSINE FIRING SCHEME Cosine firing scheme for thyristors in single· phase converters is shown in Fig -t86. The synchronizing transformer steps down the supply voltage to an appropriate level. The input to this transformer is taken from the same source from which converter circuit is energized. The output voltage Ul of synchronizing transformer is integrated to get cosine-wave U2. The dc control voltage Ec varies from maximum positive Ecm to maximum negative Ecm so that firing angle can be varied from zero to 180°. The cosine wave U2 is compared in comparators 1 and 2 with Ec a nd - Ec. When Ec is high as compared to v 2 • output voltage u3 is available fr om comparator 1. Same is true for comparator 2. So the comparators 1 and 2 give output pulses 1J 3 and V.I respectively as shown in Fig. 4.87. It is seen from this figure that firing angle is
JK FLIP FLOP
Inverter
tor 2
Fig . 4.86. Cos ine firing sch em~ for trigge r:ng thyristo rs.
I----v
21S
I"'t.
Power Electronics
~.181
governed by the intersection of u2 and E,. When E, is maximum, firing angle is zero. Thus, firing angle a in terms of V2m and E, can be expressed as V2m cos a =E, or
a
where V 2m
=cos- 1 (~)
...14.36 )
V'm
= maximum val ue of cosine signal
1)2'
' I
Vm sinwt
4n wI
"!
I
I
o
~
wI
wI
!'< 3lT-+ot) 'r-----;--+---jl
wI
:.... C211'-tQ)
wI
•
",I
2"
Fig. 4.87. Waveforms for cosine firing scheme of Fig. 4.86 .
The signals 1)3' u 4 obtained from comparators are fed to c1ock·pulse generators 1, 2 to get clock pulses us, Us as shown in Fig. 4.87. These signals 1)5, U6 energise a JK flip flop to generate output signals I)j and ur The signal ui is amplified through the circuit of Fig. 4.85 (a) and is then empl oyed to tum on the SCRs in the positive half cycle. Signal Uj' after amplification, is used to trigger SCRs in the negative half cycle. For a single.phase full converter, average output voltage is given by '
2Vm
,
... (4.37)
Vo = - - cosa
Substituting the value of a from Eq. (4.36) in Eq. (4.3 7), we get
m
1]
Vo =2V - - cos [ cos- IE, - -] = [2Vm - - . - - . E, 7t
V2m
1t
V 2m
,
Th} ristors
[Prob.4]
219 ... (4.38)
This shows that cosine firing scheme provides a linear transfer characteristic between the average output voltage Vo and the control voltage Ec. This scheme, on account of its linear tr ansfer characteristic, improves the closed-loop response of the converter syst em. This feature has mad e the cosine firing scheme quite popular in industrial applications.
PROBLEMS 4.1. (a ) What is a thyristor? How has this term been coined? Name the most popular thyristor.
v
/
(:'
I
I
~
4.2.
(b) What is the definition of thyristor as per IEC ? (c) Give constructional details of side-gate thyristor. Sketch its schematic di af;:Tam and the circuit symbol. (d) Sketch static I-V characteristics of a thyristor. Label the various voltage3. currents and the operating modes on this sketch. ./"
Des cribe the diffe rent modes of operation of a thyristor with the help of its stati c I-V characteristics. <--fb ) Enumerate the various mechanisms by which thyristors can be triggered into co nduction. Discuss the techniques which result in random turn-on of a thyristor. v' 4.3. (a ) Describe gate-triggering of a thyristor. Does the gate-current has any effect on the forward-breakover voltage? Discuss. lb) How does light triggering of a thyristor differ from gate triggering ? Whe re are LASeRs used ? lC-./ :? 4.4 . (a) Define latching and holding currents as applicable to an SCR. Show these currents on its static I-V characteristics. (b) What are the necessary conditions for turning-on of an SCR ? Discu~s. (c) Define turn-on and turn-off times for an SCR. 4.5. Sketch switching (or dynamic) characteristics of a thyristor during its turn-on and turn-off processes . Show the va riation of voltage across the thyristor nnd current through it during these two dynamic processes . Indicate clearly the various interva ls into which turn-on and turn-off times can be subdivided. Discuss briefly the nature of these curves. &>~4 :6 . (al Can a forward voltage be appl ied to an SCR soon after its anode current has fallen to zero? Explain. Ib) A forward voltage is applied to an SCR soon after reverse recovery current drops nearly to zero value. Discuss what would happen to the SCR. (ci Discuss the importance of dildt rating during the turn-on process of a thyristor. 4.7. (al Discuss the conditions which must be satisfied for turning on an SCR with a gate signal. (b) A thyris tor is conducting a forw ard current. Discuss the basic requirements for commutating (turning-off) this thyris tor. (0) Bring out clearly how the: anode curre nt expands over the cathode surface area during the turn-on process of a thyristor. 4.8. (0) Are the tum-on and turn-off times of a thyristor constant? On what factors do thes e de pend? (0) In an SCR, the anode curren.t rises li nearly fro m zero to II =- 100 A whereas anode voltage across S CR falls linearly from VI =- 600 V to zero during its turn-on time of i } = 5 !JS. Derive an e!'Cpression for the average power los3 in SCR for t} = 5 !-IS. Derive an expression for the average power loss in SCR for a t rigge!"ing frequency f . In case f""- 100 Hz, find the aVer:lge pow!:!r 1053 in SCR. IAns. 1/ 6 VI. /I .l I .!; 5 watts] (a )
220
Power Electronics l/
4.9. (a) Justify the statement, "Higher the gate current, lower is the forward breBkover voltage." (b ) Wh at is hard-driving for a thyristor? What are its advantages? Sketch a typical waveform for gate current for hard-driving the thyristor. (c) For an SCR, the gate-cathode characteristic is given by a straight line with a gradient of 20 volts per ampere passing through origin . The maximum turn-on time is 4 ~s and the minimum gate current required to quick turn-on is 400 rnA. If the gate source voltage is 15 V, calculate the resistance to be connected in series and the gate-power dissipation. Given that pulse width is equal to the turn-on time and the average power dissipation is 0.2 W, compute the maximum triggering frequency that will be possible when pulse firing is used. (Ans: (e ) 17.5 n, 15.625 kHzl 4.10. Ca) Draw thyristor gate characteristics showing the six gate ratings as specified by the manufacturers. Discuss these ratings. Indicate clearly the preferred gate drive area . Are there any other gate ratings in addition to the six mentioned above? If yes, describe this/these briefly. (b) The gate-cathode characteristic of an SCR is given by Vg:: 0.5 + 8Ig . For a triggering frequency of 400 Hz and duty cycle of 0.1, compute the val ue of resistance to be connected in series with the gate cireuit. The rectangular trigger pulse applied to the gate circuit has an amplitude of 12 V. The thyristor has average gate-power loss of 0.5 watts. [ Hint : (b) 0 =
~
or pulse width, T::
~O~:: 250 IJ.S. A1; T is more than 100 !ls, dc data apPlY]
IAns: Ca) Peak gate-power dissipation and peak reverse gate voltage (b) 44.23 OJ 4.11. (a) Draw the gate input characteristics ofa batch of thyristors indicating the upper and lower limi t loci and explain why this variation exists. (bl Draw the circuit model of a triggering circuit connected to the gate·cathode terminals of a thyristor. Explain the purpose of connecting a resistor across the gate circuit of an SCR. :./' Ccl A thyristor data sheet gives 1.5 V and 100 mA as the minimum value of gate-trigger voltage and gate-trigger current respectively. A resistor of 20 n is connected across gate-cathode tenninals. For a trigger supply voltage of8 V, compute the value of resistance that should be connected in series with gate circuit in order to ensure tum-on of the device. [Ans, (b ) 37.143 OJ 4.12. (a l Draw thyristor gate V-I characteristics indicating clearly the gate· drive limits. Explain, with the help of these char'acteristics, the selection of an 'operating point and the choice of gate circuit parameters. Discuss also how tum-on time and jitter can be minimised. ( b ) In case gating signal for an SCR consists of a train of pulses instead of continuous dc signal, explain how the frequency of triggering and other factors are decided . {cl A thyristor is triggered by a train of pulses of frequency 4 kHz and of duty cycle 0.2. Calculate the pulse width. In case average gate power dissipation is 1 W, find the maximum allowable gate power drive. I Ans: (c ) 50 !lS, 5 WI 4.13 . (a ) Di3cus:; the function of connecting a
diode across gate-cathode terminals, (ii) diode in series with gate circuit, (iii) a resistor acros.> gate·cathode terminals. (0 ) How e re the magnit udes o[gate-voJtage and gaLe-current influenced by temperat uTO rise in a thyristor? (c) During turn·off of a th),!,istor , idealized voltage and current waveforms are shown in Fig. 4.88. For a triggering frequen cy 0(50 Hz. find the mea n power loss due to turn·ofT loss . Aho obtain th e rev e!'3 ed recovery charge. (i)
[P,ob .4]
Thyris tors .
t,
221
I
2jJ~' I '
--!,).J~_I......;
O '-----~~'C------C;--_C~·_C,--300 A
,
-+'__ ___
OL-________ - : 2.1.15;"" ~ '
-100 '1
Fig. 4.88. Pertaining to Prob. 4.13 .
[Hint:
(e)
Pf)=~fV1Il.tl]
[Ans : (c) 1W, 900 IlCl
4.14. The spread in the gate--cathode characteristics of a thyristor is given by the foll owing two relations: I, =2.0 x 10- 3 V; and Ig =2.0 x 10- 3 V:·s For an average gate· power dissipation of 0.5 W, design the trigger circuit voltage and current fo r hard-drive. (Hint: Here I, =79.37 rnA and Vg = 6.3 V. Also I, =47.82 rnA and Vg = 10.456 V. For harddrive, choose I, say 75 m...o\ etc.1 [Ans. Vg = 5.56 V, Ig = 75 mAl
4.15 . (a l Discuss the two-transistor model of a thyristor. Derive an expression for the anode current and discuss therefrom the turn-on mechanisms of a thyristor . '. 4.16. (a ) Which cU.rTent rating of an SCR is the most important? ( b) What is the difference between repetitive-current and surge-current ratings of a thyristor? (c) What are V DR.lI and VRR.\f? Are these ratings different from each other for a thyristor? (d ) Is it possible to exceed rms current rating of an SCR? (e ) What are VDWM and VDR.\l? Which ruting is low? (f) An SeR has maximum rms current rating of 78.5 A. Find its maximum average current rating. (Ans: (a) RmscuITent (cl VDR.\/ = VRRJf (d) No (e) VDlVMisl'ow if) 50A1 4.17. 1o ) Describe the various anode voltage ratings as opplicBble to an SCR. Indicate these voltage ratings on a relevant voltage waveform . Ib) Discuss the signi ficance of duldt in case of thyris tors. (d Explain why an SCR is derated when it handles pulsed anode current 83 compared to its rating for constant dc current. The ave rage current rating of an SCR decreases as its conduction angle is reduced . / ' Id) Explain . >'
4.1 S~ Define th e following terms relating to SCR and discuss their significance : ~ 1i) Forward breakover vo ltage (ii) Peak inverse voltage (iii) critical rate of rise of voltage (iu) voltage safety factor (u) on-state voltage drop ( ui ) finger voltage. 4.19. (01 The derating of an SCR is more for sine waves than for the square (or rectangular) waves . Explain. Sketch the curves showing average power dissipation as a function of average forward current for different cond oldion angles for both sine and square waves. ( b) \Vho t is the effect on average curr~nt rating oi an SCR in C"52 inductance is in:;;erted in the anode circuit? Discuss.
222
[P.ob. (c)
~J
Power El ec tronics
The specification sheet for an SCR gives maximum rms on-s;:nte cuin nt us 50 A. If this SCR is used in a resistive circuit, compute its average on ·state current ra ting for conduction angles of30~ and 60 a in case current waveform is (i ) half-::;ine wave nnd (ii) rectangular wilve. [Ans: (cl 30 G (i) 12.56 A (ii ) 14.434 A; 60 (i) 18.00 A (ii ) 20.412 AJ Q
/
4.20 .
If a forward voltage is applied to an SCR which is below its breukover voltage , it may well switch on , particularly if the voltage is applied rapidly. Explain why this is so . Discuss how the effect mentioned above can be min imi zed . ( b ) A thyristor is placed between a constant dc voltage source of 240 V nnd resistive load R . The specified limits for dildt and du l dt for the SCR are 60 Nmicro second and 300 V/micro second respectively . Determine the values of the di l dt inductor and the snubber circuit parameters. Take damping ratio as 0.5. Discuss how these parameters may be modified to suit the working conditions in the circuit. Derive t he various expressions used. [Ans: (b) Computed values: 4 J..lH, 5 n, 0.16 J..lF modified values : 6.4 !lH, 8 n, 0.12 !lFI 4 .2 1. (a ) Snubber circuit for an SCR should primarily consist of capacitor only. But, in actual practice. a resistor is used in series with the capacitor. Discuss. (b) R, Land C in an SCR circuit meant for protecting aga inst du l dt and di l dt arc 4 n, 6 J..lH and 6lJ.F respectively. If the supply voltage to the circuit is 300 V, calculate permissible maximum value:;; of duldt and dildl. [Hint: (b) Rate of change of voltage across the thyristor at t = 0 when the supply is switched on is given by dUe di . l~e V, 300 _ = R - + - where I = - = = 7";) !\ etcl [ Ans: (b) 50 A/J..ls. 212 .5 Vl }lsJ dl S dt C ' C Rl 4 (n )
4.22. Following are the specifications of a thyristor operating from a peak supply of 500V: Repetitive peak current, Ip = 250 A
(~:
L
= 60
AI~s,
(d;,' L 200 YI~s =
Take a factor of safety of 2 for the three specifications mentioned above. Design a suitable snubbe r circuit if the minimum load resi3 tance is 20 n. Take; = 0.65. I An" 17 ~H, 6 n, 0.5 ~F I 4.23. (a) Discuss how a thyristor may be subjected to internal snd external overvoltages. Describe the methods adopted for suppressing such overvoltages in thyristor systems . ( b ) During the turn-off process in a thyristor, the revers e recovery current of 10 A is interrupted in a time interval of 4 !ls. The thyristor is connected in series · with an inductance of 6 mH with no resistance in the circuit. If the source voltage during turn-off process is - 300 V, calculate (il peak voltage across the thyristor when reo:erse current is interrupted and (i j ) the value of snubber ci rcuit resistance in case snubber capacitance C, '" 0.3).J.F and damping ratio is 0.65:[ Ans : (b) - 15.3 kV, 183.85 OJ 4.24.
Explain the methods adopted for the protection of SCRs against overcurrents. (b ) A thyris tor. having maximum rms on-state current of 45 A, is used in a resistive circuit. Compu:e its average on·s tate current rati ng fo r half·sine wave for co nduction angl es of ,";13 and ',"'( 12 . [Ans: (b ) 16.198 A, 20 .26 Al
{a )
4,25, (a ) Describe electronic crowbar prot
T hyr istor s
[Prob.4)
223
4.26. (a ) Enumer ate the various abnormal conditions against which tbyristors must be protected. (b ) Describe tbe significance of di l dt and duldt in (c) Describe, with the help of a circuit diagram, the functi on of various co mponents uded for the protection of gate circuit of a thyristor. 4.27. (a) Describe the methods employed for improving dildt rating in a thyristor. (b) Large dIJ l dt may turn on a thyristor at random . Describe how cathode·shorts in thyristors improve their dIJl dt ratings . 4.28. (a) Discuss briefly the different components of power loss that occur in a thyristor during i~s working . Which of the power loss component/components islare dominant at power frequencies a nd which at high freque ncies? ( b ) Give the concept of thermal resistance. Describe the analogy between thermal and electri· cal quantities. (e) Draw the thermal equivalent circuit for an SCR and discuss the various paramete rs involved in it. Cd) Describe anyone method of designing the heat sinks for thyristors. 4.29. Ca) For thyristors, various mounting techniques are based on their therm al considerations. Discuss these mounting techniques with releva nt diagrams. (b) A t hyristor is rated to carry full·load current wi th an allowable case temperature of 100Ge, for maximum allowable junction temperature of 125°C and thermal resistance between case and ambient as 0.5°C/W. Find the sink temper.aty re for an ambient tempe rature of 40°C . Take thermal resistance between sink and ambient as 0. 4 ~C/W. [Ans, (b) 88' C[
scas.
4.30. A thyristor is rated to carry an rms current of 100 A Its maximum allowable j unction t emperature is 125°C. (0) If this thyristor is made to carry direct current continuously, find the maximum allowable current rating of the SCR. (6) If th is SCR is used in a single-phase half-wave circuit with resistive load, find the maximum allowable average current for firing a ngl es of 0.1 = 30" and ~ = 120°. (c) For part (b ), determine the sink remperatures if average powers dissipated are 200 W for 0. 1 and 150 W fo r 0.2' The value of thermal impedances a re: a;e = 0.15° C/ W, Be", = O.07°C/ W for a1 and 9je=0 .16°C/ W,9c, =0.OaoC/ Wfora2' [An., (a) 100 A (b) 60 .273 A, 25 .118 A (0) 81 ' C and 89' CI 4.31. A thyristor string is made up of a number ofSCRs connected in series and parallel. The string has voltage and current rati ngs of 11 kV and 4 kA respectively. Th e voltage and cur~ n t ratings of avail able SCRs are 1800 V a nd 1000 A respectively. For a string efficiency of 90%, calculate the number of series and parallel connected SCRs. For these SCRs, maximum off·state blocking current is 12 mA . Determine the value of static equalizing resistance for the string. Derive the fonnula used for this resistance . [Ans: Series- 7. Para llel-5, R = 22 .22 kOJ 4.32. For the th}Tistors of ?rob. 4.31, maximum difference in their reverse recovery charge is 25 mi crocoulombs. Compute the value of dynamic equalizing capaci ta nce of this stri ng. Derive the formula used for the computation of this capacitance. [Ans: C == 0.094 ~FJ 4.33. Three s eries·connected thyTistors, provided with static and dynamic equalizing circuitd. have to withstand an off·st ate volta ge of 8 kV. The static equali zing r esistance is 20 kn and the dynamic equali. zing ci rcuit has Rc = 40 n a nd C == 0.06 ~I F . These th ree th}'1"istors have ieak,lg:!: currents of25 rnA, 23 rnA and 22 IT'_A. respectively. De:ermi ne volta ge across each SCR in the off· state and the discharge curren t of each ca pacitor at the time of t urn o ~ . iAns: 2500 V, 254 0 V, 2560 V; 62. 5 A, 63. 5 A. B.IA]
224
[Prob.41
Power Electron ics
4.34. In a power circuit, foul' SCRs are to be connected in series. Permissible "difference in blocking voltage is 20 V for a maximum difference in their blocking currents of 1 rnA. Difference in recovery charge is 10 ~C . Design suitable equalizing ci rcuit. IAns: Static equalizing resistance = 20 k 0; shunt capacitance = 0.5 ~Fl 4.35. (a ) Discuss how 8CRs suffe r from unequal voltage distribution across them during their turn·o n and turn-off processes. (bl A number of8CRs , each with a rating of2000 V and 50 A, are to be used in series-paraUel combination in a circuit to handle 11 kV and 400 A. For a derating factor of 0.15, cnlculate the number of SCRs in series and parallel units . The maximum difference in their reverse recovery charge is 20 microcoulombs. Calculate (il the value of dynamic equalizing capacitance and (iil the voltage across each of the slow thyristors in case one series-connected SCR is fast. [Ans: (b) n,:7, np=10, C=0.04).lF, 1500V)
4.36. Define string efficiency for series I parallel connected SCRs. Show that stri ng efficiency of two series connected SCRs is usuallv les s than one. Derive an expression for the resistance used for sta tic voltage equalization for a series connected strin g. 4.37. Describe how two series connected SCRs are subjected to unequal voltage distribution during their dynamic conditions. Derive an expression for capacitance C used in the dynamic equalizing circuit for n series connected SCRs. 4.38. Show that string efficiency for two parallel connected 8CRs is usually less than one. Discuss the problems associated with the parallel operation of 8CRs and how these are overcome. 4.39. (a) Describe briefly the following members of thyristor family. PUT, SUS, SCS Illustrate your answer with suitable diagrams. (b)
(c)
4.40. (a ) (b )
(c) 4.41. (a) (b )
4.42 . (a) (b )
4.43. (a ) ( b)
4.44. (a ) ( b)
(cl
Draw the cross-sectional view of the diac and explain how it can conduct in both the directions. Give the cross-sectional view of a triac and explain its turn·on process with relevant diagrams. Hence show that a triac is rarely operated in first quadrant with negative gate current and in third quadrant with positive gate current. Describe LASCR. Give its industrial applications. Discuss how a triac may sometimes operate in the rectifier mode. Enumerate the advantages of ASCR and RCT over conventional thyristors. What is a GTO ? Describe its basic structure. The turn-off process in a GTO can be described with its two·transistor model. Explain this in detail. Bring out clearly the difference between gold-doped GTOs and anode-shorted GTOs . Describe switching performance in a GTO with relevant voltage and current waveforms. Give the merits and demerits of GTOs as comp ared t o conventional thyristors . Define the following terms as applicable to GTOs and discuss their significance. Turn-oft' gain, backporch current. Give the application of GTOs. Describe the bas ic structure of a static induction thyris tor (81TH ). Explain the turn·on and turn-off processes in a SITH. Show that 5 1TH is a normally-on device. Compare 81TH with a GTO.
,
Thyristors
[P
.,
2' •
/' 4.45. (a) Discuss the features that the firing circuits for thyristors should possess.
Give the general layout of a firing circuit scheme and explain the function of various components used in it. (0 ) Describe the ,.,istance firing circuit used for triggering SeRs. Is it possible to get a firing angle greater than 90° with resistance firing? Illustrate your answer with appropriate wa.veforms. <'.46. (0) For resistance firing circuits show that firing-delay angle is proportional to the variable resistance. (b ) Resistance firing circuit is used for triggering an SCR in a laboratory. This SCR is destroyed by a batch of students inadvertently. A new SCR with the same specification number is installed. But it is found that maximum firing angle attained is 75 a only. Explain how the desired maximum firing angle of 90 ~ can be obtained. {Ans: (b) Increase Rl or R2 or else decrease R in Fig. 4.64J 4.47. (a) Draw RC half-wave trigger circuit for one SCR and discuss the function of the various
components used. Describe, with the help of wavefonns, how the output voltage is controlled by varying the resistance. Draw the voltage waveform across SeR also. (b) Describe RC full-wave trigger circuit for one SCR when the load is (i) ac type (ii ) dc type. Relevant diagrams and waveforms should be drawn to illustrate your answer. 4.48. (a) Compare an UJT ·ti.ring circuit with Rand RC firing circuits. (b ) A unijunction transistor, used in relaxation oscillator, has the following data: T1 =0.67, Iv = 10 rnA, Vu = 2.5 V, Ip;: 15 JJ,A An oscillator, with an oscillation frequency of 1 kHz, is to be designed by using this UJT. Compu~ the values of charging resistor and external resistors needed in the ba3e circuits. Take O:::O.4)J.F and forward-voltage drop of E - Bdunction as 0.5 V. Scu..-ce voltage is 24 V dc and triggering pulse width is 50 J,ls. [Ans: (b) R = 2.772 k 0 , RmCl:r = 495 k n, Rmin::: 2.15 k 0 , R, =621:9 n, Rl =125 nJ 4 .49. (a) Explain the working of an oscillator employing an UJT. Derive expressions for the
frequency of triggering and firing angle delay in terms of eta, chargir.g resistance etc. A relaxation oscillator, using an UJT, is to be designed for triggering nn SCR. The UJT has the following data : 11=0.7, Ip =0.5mA, Vp =15 .0V, Vu=O .BV, Iu=2mA, R BB =6k!l. Normal leakage current with emitter open = 3 mAo The firing frequency is 1.5 kHz. For C = 0.05 )J.F, compute the values of charging resistor and the external r esistors connected in the base circuits. Take forward-voltage drop of E - B 1 junction as zero. (e) If the frequency of firing the SCR in part (b) is changed by varying charging resistor R , obtain the maximum and minimum values of R and the corresponding frequencies. IAn" (b) 11.074 kn, 47 6.66 n , 666.67 n (c) 12.858 ill, 10.315 kn, 1.292 ffiz, 1.611 kHzl
(b)
4.50. (a) The intrinsic stand-off ratio for an UJT is 0.65. Its intel'bl.'!.se resistance is 10 k n. Calculate the values of the interbase resistances . (b ) Estimate the minimum and maximum values of chargi ng resistor in the UJT oscillator circuit for manu al trigger-angle control of a between 2tj" and 160~ fo r 50 Hz supply. IAns: (a ) 6.5 k 0, 3.5 k 0, (b ) 2.307 k 0, 1B.457 k OJ Assume C = 0.4 ).IF and 11 = 0.7. 4.51. (a) Draw and explain the working of an UJT oscillator. Discuss how th e ampli tude of output voltage pulse can be estimated in this oscillatnr.
226
Powe r El ectronics
[Prob.4J (b)
Using a 15-V supply to an UJT, design the oscillator circuit for a frequency of5 kHz . Da ta fo r UJT is as und er : ., = 0 .65 to 0.75, R B8 = 4 .7 .to 9.1 len 'fake C = 0 .04 J..LF. Missing data may be assumed. [Hjnt! (6) Ass ume leakage current = 1.88 mAl (An.: (b) R=4.153kO, R ,= 952.40, R I = 126.30, R_ = 6 kO, Rmin =3 k {min = 3460.8 Hz, (max r 6921 .6 Hzl
n.
4.52. A r elaxation oscillator using an UJT is' fabricated to generate pulses for triggering SCRs. When th e circuit is energised, the circuit fails to oscillate. What could be the plausibe causes of this failure? How can the circuit be made functi onal? [Ans. More V BS or less VB! than required ; R < R mill or R > R ma..:.:1 4.53. An UJ T of Fig. 4.71 (a ) has the following parameters :
., =0 .67, VD == 0.7 V,lu =3 rnA, Vu =1 V,lp = 12~. VBS =20 V (a) Find the value of VEE so as to turn-on UJT if ~E =1 kn. (b)
Find the value of
[Hint. (a ) V EE
to which it must be r educed to tum-off the UJT . V BB + VD+ /p . RE e~c. 1 IAns. (a ) 14.112 V (b) 4 VI
=.,·
VEE
4.54. Dr aw synchror..ized UJT trigger circuit using a zener diode . Describe it briefly with relevant voltage and' CUlTent waveforms. Explain how synchronization of the trigger circuit with the supply voliage across SCR is ' achieved . In case charging resistor is small so ·that the capacitor voltage reaches UJT thr eshold v ?lta ~e twice in each half cycle, explain how the circuit operation is influenced. 4.55. (a. ) Draw a circuit diagram for the ramp-and-pedestal trigger circuit used for a single-phase semiconver ter. Describe its operation with appropriat e waveforms . For this trigger circuit, derive expressions for the frequency oftrigg'!:ring and firing-angl e delay in tenus of eta, charging resistor etc. (b) A fi r ing circuit, u~in g ramp-and-pedestal triggering scheme, has the following dat a : Char ging resistor = 4 k 0, ch arging capacitor = 0.2 ~F. supply 0.75, zener-diodevoltage 15 V. fre quency =50 Hz, Com pute the magnitude of firing angle in case pedestal voltage is (i) zero and (ii ) 4 V. (Ans: (b) (i) 19.963' (ii) 15.496' 1
,, =
=
4.5 6. (a ) Describe the use of pulse transformer in the triggering ofSCR! and GTOs. With a suitable circui t, discuss the conditions under which the input pulse is faithfully transmitted or is tr ansmitted in the form of exponentially decaying pulse. Which of these two functional modes is preferred and why? (b ) The primary of a pulse transformer is connected in series with a t ransistor and a current limiting r esis tor R L . The data for the triggering circuit is .as under : RL = 500 0 , gat.e to cathode r esis tance = 200 n P rimary to secondary turns ratio
= ~, voltage requir ed to trigger the SCR = 3 V.
Comput.e the voltage a pplied to the circuit consisting of transformer primary, RL etc. Derive the expression used. [Ans: (b) 16.5 V] 4.57. (a) Describe the trigger circuit for a triac using a diac. ( b) A diac with a breakdown voltage of 35 V, Fig. 4.79, is used for t riggering a tri ac. This circuit h as RI = 1000 n, R = zero to 280 k n and C = 0.1 J,lF. For a supply voltage of 230 V, 50 Hz ; calculate the maximum and minimum values of firing-delay angles for the triac. The effect of load imped ance may be neglect ed. lAns: (b ) 156.5", 7.98 j Q
Thyristors
(Prob.4]
227
4.58. (0 ) Describe 9. gate trigger circuit for a single-phase full converter. Discuss how the adjustment 'of control voltage varies the firing-delay angle. (b) Describe a gate-pulse amplifier using' a MOSFET. 4.59. (a) Why is pulse-train gating preferred over pulse gating? ·Explain, with relevant circuit and waveforms, the pulse-train gating of SeRs. (b ) Why is the cosine-firing scheme so popular? Describe 8 cosine-firing scheme for the triggering of thyristors.
Chapter 5
Thyristor Commutation Techniques •
............ .. ... .... ...... .. ..... ....•. ............... ............•..• .........••.. ..... •• •....... In Ihis Chapter • • • • • •
Closs Class Closs Class Class Class
A Commutation: load Commutation 8 Commutation : Resonant-pulse Commutation C Commutation: Complementary Commutation D Commutation: Impulse Commutation E Commutation : External Pulse Commutation F Commutation: Une Commutation
.. ........ ..... __ ... ... ... ..... .. ...... .... ..... .. ................... . .. -- .. , ..... ..... __ ..•... .... . .
...
A thyristor is turned on by applying a signal to its gate-cathode circuit. For the purpose of power control or power cO:lditioning, a conducting thyristor must 'be turned-off as desired. As stated before, the tum-off of a thyristor means bringing the device from forward-conduction state to forward-bl ocking state. The thyristor turn-off requires that (i) its anode current falls below the holding current and (ii ) a reverse voltage is applied to thyristor for a sufficient time to enable it to recover to blocking state. Commutation is defined as the process of turning-off a thyristor. Once thyristor starts conducting, gate loses control over the device, therefore, external means may have to be adopted to commutate the thyristor. Several commutation te ch niqu es have been devel oped with the sale objective of reducing their turn-off (or commutation) time,
The use ofthyristor circuits in low-power converters has declined relatively. This is because of recent advances in semiconductor power devices leading to the availability of power transistors , GTOs and IGBTs. However, for high-voltage and high-current applications above about 1 kV and 0.5 kA,' thyristor circuits offer popular circuit configurations. The classification of thyristor commutation techniques, as reported by various authors, is not the same, Here, an attempt is made to refer to all these classification techniques. Primarily, the classification of commutation techniques is based on the manner in which anode current is reduced to zero and on the configuration of the commutating circuits. J'hyristor commutation techniques use resonantLC, or underdamped RLC circuits, to force the current and / or voltagf! oi a thyristor to zero to turn off the device. Several power-electronic converters employ the circuit co nfigurations used for describing the thyristor commutation techniques. TherefClre, a study of the various commutation techniques serves as an introduction and leat.ls to '" better und ers tand ing of the trans!ent ph enomena occuring in power-electronic converters under switching con ditions . The various commutati on techniques are now described in this chapter.
For achieving load commutation of a thyristor, th e commutating com ponents L and C are connected as shown in Fig. 5.1. Here R is th e load resis tance. For low v alue of R, Land C are
Thyristo r Commutation Techniqu es +
T
[Art. 5.1J
229
,
v. O~--~~~,----,~,f'~'-----~'~'~'~_---'-t
r-- --, o o
0
0
Io R
11--: , __ ;
ILood 0
L. _..:
i
T
l
v,
c
r0 oR
-,
ih O~
A\.
0
:lood Ic_ 0 ...J
,7
,,-t
'-"
connected in series with R , Fig. 5. 1 (a). For high value of R,lond R is connected across C, Fig. 5.1 (b). The essential requirement for both the circuits of Fig. 5.1 is that the overall circuit must
be underdamped. When these circui~ are energized from dc, current waveforms as shown on the right hand side of Fig. 5.1 are obtained. Itis seen that current i first rises to ma.umum value and then begins to fall . When current decays to zero and tencfs' to reverse, thyristor Tin Fig. 5.1 is turned-off on its own at instant A. ~ Load, or class-A, commutation is prevalent in thyristor circui ts supplied from a dc source. The nature of the circuit should be such that when energized from a dc source, curre nt must have a natural tendency to decay to zero for the load commutation to occur in a thyristor circuit. Load comm utation is possible in dc circuits and not in ac circuits. ClassA , or load, commutation is also called resonant commutation qr self-commutation. A practical circuit employing load commutation is a series inverter which is described in Chapter 8. Asimple example illustrating the basic principle of load commutation is given below: ' ..
I
Example 5.1. The circuit shown in Fig. 5.2 (a) is initially relaxed. The thy ristor T is turned on at t =O. Determine (a) conduction time of thyristor and (b) voltage across thyris tor and capacitor after SCR is turned off Ca lculate these values for L = 5 mH, C = 20 p.F and
V. =200 V. Solution. \vh;n thyristor is turned on , it behaves like a diode. Th erefo re, with S CR on, the devi ce acts like a closed switch , Fig. 5.2 (b). KVL for this circui t gives If 'd V L di dt + C l t = J lt3 solution, fro m Art. 3.1.4, is given by Eq. (3.9) which is r epeated here. i(l) = V.
~ sin ""t
... (5.1 )
230
P ower Electronics
[A r t. 5.2]
l(t)~
+
v. ~,
.
T L
v.
1
T
O~--------x~/2~~-,-------"~/~w-,----~t'" ••
C
I
-~----l _ (a )
i
+
2V,
T
O~I::::::=tt~,.~~~::::::::.!~1---'tl
v,
·-l_ _cTl~'
i
".O.T jf-_ _ _ _w_ ' _ _-f__.-.
-v,-,I_-
(b)"
t
Ie) Fig. 5.2. Ca ) ond (b ) Load commutation circuit (e) waveforms.
Here
Wo =
",Ie
is
~alled
the resonant frequency of the circui t.
Capacitor voltage, from Eq. (3.IOa) is
ii, (I) = V. (1- cos "'0 I)
... (5.2)
It is seen from above equations that at time t = to=n/Wo. i(t ) = O· anl tic: {t } = + 2 V;. TruS'''' shows that 1t/000 sec or 1t ..JLC sec after thyristor is clos ed at t = 0, the charging current becomes zero, Fig. 5.2 (c) and thyristor is, therefore, turned off on its own. Here 10 =conduction time of the thyristor
=. ~LC
....
~
... (5.3)
Voltage tiT across thyristor during its conduction time to is zero. When it stops conducting, Ur =- 2V~ + V, =- V" It implies that SCR is subjected to a revers e voltage afV, which helps in its recovery. For the circuit parameters given, the calculations are as under : R' t f . f th . ·t 1 esonan requency 0 e ClrcUl , COo = {S x 10- ~ x 20 x 10- 61 1/ 2 =
10' =Fa'
3162.27 rad/ , .
=~= 7t " 10 = 9.9346x 10- 4 s 10' o "'0 = 99.346-.1s. Volt age across thyristor after it is turned off =- V. =- 200 V. Conduction time of thyri stor,
5.2. CLASS B
CO~I1IIUTATION
t
: RESONANT-PUI.sE CO~I1IIUTATION
For explaining class- B, or resonant-pulse, commutation, refer to Fig. 5.3 (0 ). In th is figur e, sou rce voltage VJ char ges capacitor C to voltage V, with left hand plate positive as shown . Main
I
(
Thyristor Commu tation Techniques
[A rt. 5.2 ]
it 'gAt io 1
231
t
'"
t
110 t
,~wot=l(_ I' , , ,
TI
+
iT!
V,
+ ~(_.
t
--------+-----.---='-J'''''''''-, t( ,
I
V'
"
,
.,.i... . ~D
C
•
"PHI
ic
,
,
,I ,I
Vall
~
:
.
load
-" TA
' T1 tc lor ~ I
" ie1
+
"0
10
-v
T1
ON
-
:
,, _1..._. __ :
t, TA DN
w
I
,
, ,
I
t
II
t
t2 t) t, ts TA T1
OFF OFF ~)
Fig. 5.3. Resonant-pulse commutation (a ) circuit diagram
(b )
waveforms .
thyristor Tl as well as auxiliary thyristor TA are off. Positive direction of capacitor voltage Vc and capacitor current ic are marked. \Vhen Tl is turned on at t = 0, a constant current 10 is established in the load circuit. Here, for simplicity,load current is assumed constant. yptill time tv Uc = V r, ic = 0, i o=10 and iT! =10, Fig. 5.3 (b). F or in itiating th e commutation of main thyristor Tl, auxiliary thyristor TA is gated at t = tl . With TA on, a resonant current ic begins to flow from C through TA. L and back to C. This resonant current, with time measured from instant t l , is given by
ic =- V,
~ sin Wo t = -lp sin Wo t
Minus sign before 1p .sin 000 t is due to the fact that this current fl ows opposite to the r eference positive direction chosen for ic in Fig. 5.3 (a) . Capacitor voltage = V, cos Wo t
... (5.4 )
After half a cycle of ic from instant t 1; ic = 0, Uc = - V, and iTl =10 , •.<\fier Jt radians from instant t l , i.e. just after instant t2 • as ic tends to reverse, TA is turned off at t 2 . With Vc = - V" right-hand plate has pos iti ve p olarity. R es onant curr ent ic n ow b u il ds up thr ough C, L , D and T l. As this curren t ic grows opposite to forward thyristor current of Tl, net for ward curr ent in = 10 - ic begin s to decre ase. Finally, wh en ic in the reve rsed direction attain s the value 10 , fo rw ard current in T l (i n =10 - 10 = 0) is r educed to zero and the device Tl is t urn ed off at t J . For r eliable commutation, peak r esonan t current Ip must be greater th an load current 10 . As thyri stor is commutated by the gradual build up of r 2 ~o na.T}t current in th e reve rs ed
232
Power Electronics
[Art. 5.2J
direction , thi:i method of commutation is caUed current commutation, class-B commutation or resonant-pulse commutation. After Tl is turned off at t3' constant current 10 flows from V. to load through C, Land D . Capacitor begins charging linearly from - Va b to zero at t;l and then to V, at t5' As a result, at instant Is. when Uc = V,. load current io =ic =10 reduces to zero as shown. It is seen fr om the waveform of i( that main thyristor Tl is turned off when V,
or
~ sin 000 (t 3 -
"'0 (I, - I,)
where
-
Ip
t 2) = 1 0
=sin- 1 (~; J
... (5. 5)
=V, ~ =peak resonant current.
Main thyristor Tl is commutated at t3 ' As constant load current 10 charges C linearly from Vab at t 3 to zer o at t 4 , SCR Tl is reverse biased by voltage Uc for a period (t4 - t 3 ) = te' :. Circuit turn-off time for main thyristor, V••
... (5.6)
tl; =t~-t3 =C 10
Eq. (5.6) shows that tc is dependent on the load current. Waveform of capacitor voltage u~ reveals that the magnitude of reverse voltage Vab across main thyristor TI , when it gets commu tated, is given by ... (5.7)
Example 5.2. Circuit of Fig. 5.3 (a) employing resonant-pulse commutation (or class -B com m utation) has C =20 ~ and L =5 ~. Initial uoltage across capacitor is V, = 230 V. For a constant load current of 300 A, ~alculate (a) conduction time for the auxiliary thyristor,
(b) uoltage across t~ mam thyristor when it gets commutated and (c) the circuit turn-off time for the main thyristor. Solution. Peak value of resonant current, Ip = V ,-\f[ - /c = 230 -'I - [20= s 460 A 1
"'0 = ~LC
Resonant freq uency, (0 )
10'
= ~ 100 = 0.1 x 10' radl s
Conduction time for auxiliary thyristor n = -',L = --u
(0 ) From £q. (5.5),
"'0 (I, -
t,i=sin-
n
, = 31.416 ~s .
0.1 x 10 1
(~~~) = 40.706 or 0.71045 ra d.
Voltage across main thyristor, when it gets turned- ofT. is given by Eq. (5.7).
..
V•• = V, cos "'0 (I, - I,) = 230 cos (40. 706') = 174 .355 V
[Art. 5.3]
Thyristor Commutation Techniques (c)
Circuit tum-oiTtime for main thyristor, from Eq. (5.6), is Vo'
I, = I. - I, = C 1, = 20 x 10
_, 174.355 2 300 = 11 .6 4 ~s .
CO~lPLEMENTARY
5.3. CL\SS C COMMUTATION:
COMMUTATION
l,
------ ---_":\....._- -----+
v,
RI
u;t ITI
T1
-
~, i, ,r - V,+ T2
./
~
i -- - -~T- ·
(a )
iT I =ll"f" l'
(b)
,Ei:l
P,/RII II
,•
II'd! I
v,[~,+mb i !
1"
I, ,,
t-
f/"RI ,•
,
,
-v,
v, -v,
Vs (1- 2p -t/RrC)
v,
i,
T1
ON
, ,
-2V. e _ '/ R1C R, ,
I
I,
n OFF T2 ON
(c)
Fig. 5.4. Cla33'C commutation (a) and (b ) circui t diagrams (c) wa veforms .
233
234
Power Electronics
[Mt. 5.3J
-
In this type of com mutation , a thyristor carrying load current is commutated by transferring its load current to another in coming thyristor. Fig. 5.4 (a ) illustrates an :. arrangement empl oying complementary commutation. In this figure, firin g of SCR Tl commutates T2 and subsequently, firing of SCR T2 would turn offTl. Positive and negative directions of voltages and currents are marked in Fig. 5.4 (a ). In this figure, capacitor is supposed to be initially virgin i.e. uncharged. When Tl is turned on at V V t = 0, current through Rl is il = and through R 2 is ie = so that thyristor Tl current
R:
iT' =i, + i, =V,
(i, i;I +
through R2 fr om ue = O.
R:'
begins to flow, Figs. 5.4 (b) and (c). Capacitor C l?egins charging
~e charging current through the circuit V,
I
C and R2 is given by
V ic (t ) = R' . e- t/R~ c
,
and voltage across capacitor C is given by !Je(t)= V. (l_e-tlR,c)
Voltage across thyristor T2 is un = Uc (t) After some time , when transients are over, Ue = un = V, and ie d ecalys to zero . Also iTI = V, I R I • The waveforms for these currents and voltages are shown in Fig. 5.4 (e). Uc
When Tl is to be turned off, T2 is triggered. lfT2 is turned on at t l , then capacitor voltage ap plies a r everse potential V. across SCR Tl and tUrn s it off. In other words, at
0
t 1> Un = ,uTl =- V I '
'c· =- 2V, R l an d'.T.!. =V
I
(2R I + 1) In t h. e CirCUit .. R
2
'
.
conSIS t'log a f
V" R I , C and T2, the capacitor voltage changes from V, to - V, as shown in Fig. 5.4 (c).
Jic dt =V, 1 [I, CV,] V, C s s s
For this circuit, KVL gives
R t ic + ~
Its Laplace transform is
R ·1 (5)+ -
Its solution gives,
ie (t)=
1
c
(s)
- - - =-
2V R' · e-IIRLc
,
As. this current ic (t) flows opposite to the positive direction indicated in Fig. 5.4
(a),
... (5.8)
-1
Voltage across capacitor is =V.[2e-tlR,c_ll
... (5.9a )
Volt age across SCR Tl is
un = - Uc = V, [1- 2 e- tlRIC) ... (5.9b) Note that in Eqs. (5.8) and (5.9), time t is measured from the instant t,. The plots of capacitor current ic (t) from Eq. (5.8), and capacitor voltage u~ (t) and uTI from Eq. (5.9) are shown in Fig. 5.4 (c). Current in falls from its value V,
R ,C.
(i,
+ ~2) to V, I R, with time constant
Thyristor Commutation Techniques
[Art. 5.4J
'When transients are over after t l ,
uTI
= V"~ Ue = - V"~ ic = 0,
Un
= 0, i1'2 = V, I R 2 and i Ti = O.
When TI is turned on to commutate T2 at instant 1"i,.,=O, iT1 = Un
= - V"~
UTI
= 0 and
. Lc
=
235
V,(;, ;j} +
2VJ
R,
With the turn on ofT2 at t I , capncit~r voltage V, suddenly appears as reverse bias across Tl to turn it off. Similarly, at t 3 , capacitor voltage V, applies a sudden reverse bias across T2 to turn it off. On account of this, class·C' commutation is also called complementary impulse . commu tation. Waveforms for voltages and currents are drawn in Fig. 5.4 (c) . Waveform for UTl indicates that a reverse voltage - V, to zero appears across thyristor Tl for a certain period, This period, called circuit turn-off time tel for Tl is given by UTI
or
= 0 =V.
(1_2e-'J RtCj ...(5.100)
I" =R j Cln (2) Similarly, circuit turn-off time for T2 is I" = R, C In (2)
... (5.10 b)
Example 5.3. Circuit of Fig. 5.4 (a ), employing class·C commutation, has V, =200 V, R J = 10nandR2 = lOOn Determine (a) peak value of current through thyristors Tl and T2
(b) ualue of capacitor C if each thyristor has turn-off time of 40 )ls. Take a factor of safety 2. Solution. (a) An examination of Fig. 5.4 (c) reveals that
peak value of current through TI
= V,
and. peak value of current through T2
=V J
[~j + ;,] = 200 [110 + I~O]= 24 A
[.l. +.1.] RI
R'l
2 = 200 [ 10 + (b) From Eq. (5. 10 0),
c= =
Fro m Eq. (5.10 b),
,
I~O ] = 42 A
tel
Rjln (2) 2 X 40 X 10- 6 10 In (2) =11.542~F
_ 2x40xlO- 6 C = 100 In (2) 1.1042 ~F
So choose a capacitor oflarge size of 11.542 )IF.
5.4. CLASS D CQMl\IU.TATION , IMPULSE COMIUUTATION For explaining class D. or impulse, commutation, r efer to the circuit of Fig. 5.5 (a). In this figure, Tl and TA are called m.nin and auxiliary thyri3tors respectively. Initially, main thyristor T l and auxiliary thyristor TA are off and capacitor is assum ed charged to voltage YJ with up per pl a te positive. When T l is turned on at t = 0, source voltage
236
Power Electronics
[Art. 5.4]
i"b i9At~_
_ __
_
--!'I!!I "--_ _ _---,_
i,r----r:---+---, . Ir,
o
t
. v" In
10=10
i,'
i,
v,·+==c
t
+ VTA-
,V-
Y,
fA
L
0
t
A
0
.vLv
-
0
'-0 T1 ON
~)
•
tl T1 OFF TAON
t2 TA OFF
~
Fig. 5.5. Class-D commutation (a ) circuit diagram (b) waveforms .
V, is applied across load and load current 10 begins to fl ow which is assumed to remain constant. With Tl on at t = 0, another oscillatory circuit consisting of C, TI, L and D is formed where the capacitor current is given by .
Ic SLnWot . = IpSLnCllot ·
- t 'e· V = .. 'I
When (00 t = Tt, ie = 0. Between 0 < t < (It!wo), iTl = 1 0 + Ip sin Wo t , Capacitor voltage change s from + V, t o - VI co-sinusoidally and the lower plate becomes positive. At Wo t = 1t, ic = 0, iTl = 10 and tic =- V". Fig. 5.5 (b). At t l • auxiliary thyristor TA is turned on. Imm ediately after TA is on, capacitor voltage V, applies a reverse voltage across main thyristor Tl so that uTI = - V. at tl SCR Tl is rurned off and in = O. The load cun:ent is now carried by C and TA. Capacitor gets charged from - V, to V. with constant load current 10 , The change is, ther efore, linear from + V, to - V, as shOwn. 'W hen Ut =V" it = 0 a t t2 , thyristor TA is turned off. During the time TA is on from tl to t 2, Uc = uTl,i c = -10 and to =10 - For main thyristor T1, circuit turn-off time is tc as shown in Fig. 5.5 , (b).
ana
With the firing of thyristor TA , a reverse voltage V, is suddenly applied across Tl ; this method of commutati on is therefore, also called uoltage commutation. . With sudden appearance of reverse voltage across Tl, its current is qu ench ed; in fact the current momentarily reverses to recover the stored charge ofTI. As an auxiliary thyristor TA is used for turning-off the main thyristor Tl, this type of commutation is also known as auxiliary commutation.
[A rt. 5.5]
Thyristor Commutation Techniques
237
When thyristor TA is turned on, capacitor gets connected across Tl to tUrn it off, this type of commutation is, therefore, also called parallel-capacitor commutatLon . Example 5.4. Circuit of Fig. 5.5 (a) illustrates class-D commutation. For this circuit, V, =230 V, L = 20)J.H and C =40 I.J,F. For a constant load current of 120 A , calculate. (a) peak value of current through capacitance and also through main and auxiliary thyristors, (b) circuit turn-off times for main and auxiliary thyristors.
Solution_
(0)
When main thyristor Tl is turned on, an oscillatory current in the circuit
C, '1'1, L and D is set up and it is given by
tc (t) = V, . ~ sin 000 t :. Peak value of current through capacitor Ip=V,
~ =230~=325.22A
Peak value of current through main thyristor T1 = Ip + 10 = 325.22 + 120 = 445.22 A
Peak value of current through auxiliary thyristor TA
=10 = 120 A
Waveforms for vTl or ve in Fig. 5.5 (b) indicate ~hat circuit turn-off time for mam thyristor Tl is the time required for un or ve to change linearly from - V J to zero. (b)
.. I ,
=
V, Ct,
.. Circuit turn-off time for main thjTistor V, _ 6 2 30 t, = C 10 = 40 x 10 120 = 76.67 ~s An examination of Fig. 5.5 reveals that when Tl conducts and during the time upper plate of C is positive, um =- LIe i .e. auxiliary thyristor TA is reverse biased by LIe' This gives circuit turn-off time tel for TA Here
-~
= 2
n OJo
1
10'
"'0 = 'JLC =
10'
ho x 40 = ~800
Circujt turn -off time for auxiliary thyristor, It
tel
It
--f8()'O
= 2000 = 2 X 106 = 44.43 J..!..5.
In this typ e of comm utati on, a p1l1se of current is ob tained from OJ sepr.rate voltage sou rce tL :urn off the conducting SCE . The pea.~ value of this current puls e must be more than the load
238
Power Electronics
[Art. 5.6]
current. Fig. 5.6 shows a circuit using external-pulse TJ Tt T2 l commutation. Here V. is the voltage of the main source and VI is the voltage of the auxiliary supply. Thyristor + + Tl is conducting and load is connected to source V" VS v, Load C 2V, When thyristo r T3 is turned on at t = 0; V l • Tal Land C form an oscillatory circuit. Therefore, C is charged to a voltage + 2Vt with upper plate Fig. 5.6. External-pulse commutation cireui. positive at t = n ..JLC as shown and as oscillatory current falls to zero, see Art. 3.1.4, thyristor T3 gets commutated. For turning off the main thyristor TI, thyristor T2 is turned on. With T2 on, Tl is subjected to a reverse voltage equal to V. - 2V1 and Tl is therefore turned off. After Tl is off, capacitor disch arges through the load. 5.6. CLASS F COMMUTATION: LINE COMMUTATION
Tbis type of commutation is also 'known as natural commuta~ion. This c;an occur only when the source is ac, When an SCR circuit is energised from ac source, current has to pass through its natural zero at the end of every ·positive half cycle. Then ac source applies a reverse bias across SCR automatically. As a result, SC~ is turned ofT. This is called natural commutation because no external circuit is employed to turn-off the thyristor. This method of commutation is applied to phase-controlled converters, iine-commutated inverters, ae voltage controllers and step-down cyclocon verters. U,
Ur
+-T Vs =Vmsin wt
(a)
I
I
.
t
'
wt
~,~ . to t U ~ i io:
t,
to
+ U,
I , ......
~ ..... • ....
0
I
'b
~
R
"raj
.
2'1('
1"-1 ~
J1r
\.Ji
i
•
t.1I'wt
wt'
(b)
Fig. 5.7. Class F commutation (a) circuit diagram (b) waveforms.
A single-phase half-wave (or one-pulse) controlled converter employing line commutation is shown in Fig. 5.7 (a). In this figure, thyristor T is fired at firing DJlgle equal to zero, i.e. when oot = 0, u, = O. Load is resistive in nature. With zero degree firing-delay angle, the thyristor behaves like a diode. During the positive half-cycle, Uo = uJ and waveshape of load current io is identical with th e waveshape of Uo for a re3istive load. At wt =tt, v~ = 0, Uo = 0 and = 0; therefore T gets turned off at this in3tant. From rot =-1! to Wi = 21t , T is reverse biased for a period t, = tt/ ro sec, longer than the thyristor turn-off time t q . Here t, is called the circuit turn-off time. Another met·bod of classification of thyristor commutation technique is as under: (1) Line commutation: cl ass F (2) Lo a.d commutation : class A
'0
Thyristor Commutation Techniques
[Ml5.6J
239
(3) Forced commutation : class B, C and D
External-pulse commutation : class E. In line, or natural, commutation, natural reversal of ac supply voltage commutates the conducting thyristor. As stated before,line commutation is widely us ed in ac voltage controllers, phase-controlled rectifiers and step-down cycloconverters. In load commutation, L and C are connected in series with the load or C in parallel with the load such that overall load circuit is under damped . Load commutation is commonly employed in series inverters. . .. .' In forced commutation, the commutating components Land C do not carry load current continuously. So class B, C and D commutation constitute forced commutation technique!:!. As stated before, in forced commutation, forward current of the thyristor is forced to zero by external circuitry called commutation circuit. Forced commutation is usually employed in dc choppers and inverters. Example 5.5. In the circuit shown in Fig. 5.8, SCR is forced commutated by circuitry not shown in the fig ure. Compute the minimum ualue orC so that SCR does not get turned on due to re-applied dv / dt . The SCR has miniR= 50 !l mum chargi ng current of 5 mA to turn "it on and its junctio n capacitance is 25 pF. . Solution. Under steady state, SCR conducts a current V, 200 . . =R = 50 = 4 A and voltage across ldeal SCR = voltage v, across (4)
C=O.
\'lhen SCR is force commutateq. , capacitor C begins charging from source V, through R so that capacitor voltage u, (= uTl is .given by V, = V& [l_
Fig. 5.8. Pertaining to Example 5.5.
e- tI RCl
..
dU, ] = V . -
[du'l
or
dt
' 0=
V, RC
The rate or rise of capacitor voltage
current Cj
.(
~'
).0
Vc
...( i )
across SCR may be large. In case SCR charging
happens to be equ al to 5 mA, SeR will get turned on. Here Cj is the junction
capacitance ofSeR.
Cj (~':t O =5mA "
j
Substituting the value
of(~cl _
fro m Eq. (i) abo..·e, we get
, -0
V, • 10-3 CJ RC =oX
or
25 x 10- t2 _ 200 = 5 x 10- 3 bO X C
Power Electronics
[Art. 5.6J
240
C = 25
or
12
10. x 200 250 x 10" X
=0.02
~
F
In order to obviate turning on of SCR, the value of capacitance C should be less than 0.02 ~F . T1
...
Example 5.S. For a voltage or impulse commutated thyristor circuit shown in Fig. 5.9, capacitor is initially charged to V, with polarity as shown . Find the circuit tu rn -off time for the main thyristor in case C = 10~, R =Snand V.=200 Vdc.
. v, =i';c ...
vs=200V
.:Y"
TA.,..o R
.
Solution . When auxiliary thyristor TA is turned on, . . . ~ main thyristor Tl is turned off by means of capacitor Fig. 5.9. Pertammg to Example :l.6. voltage V. appearing as reverse bias. After Tl is off, KVL for the circuit consisting of V" C, TA and R in series is given by
Ji
R . i (I) + ~
Its Laplace transforme is R· I (s) + -1
C
(I) dl = V.
[IM v.] v. [ 1] s
- -C ·s
=s
I(s ) R+- = -2V.
or
sC
2V.s
I (s)
sC
2V,
=-s- . (RsC + 1) = "'j/"
s
1 1 s+ RC
Its Laplace inverse is The voltage across capacitor C is !.Ie
(t) =
~
fi
(t) dt
+ initial voltage across capacitor
"
JI
1 2V' e-tlRC_V =V [1 _ 2e- tlRCj =_ CoR " During the time auxiliary SCR TA is on, Ve = !.In =V, [1- e- 1IRC ) . The circuit turn-off time for Tl is the time taken by lie =Un to change from its value - V, to zero. .. or
O=Vs (l -2e- VRC j I, =RC In (2) = 5 x 10 x 10" In (2) = 34.6574 ~s.
Example 5.7. For the circuit shown in Fig. 5.1 (a), commutatillg elements L = 20 J,J1{ a.nd C = 40 ~ a.re connected in series with load resistance R =1 n Check whether self-commutation, or load commutation, would occur or not. Find the conduc tion time of the thyristor. Solution. It is seen from Art 3.1.5 that ringing frequ ency given by
w, =
1/ L~ ti )' = '" (
w,. in radh.ec , from Eq (3. 14), is
damped frequency of oscillat ion, w.
[Art. 5.6]
Thyristor Commutation Techniques
241
The condition for underdampir..g is that OOd > 0
L1C-( ~ J
or
>0
R<"'f
or
f4L
.
- f4L
4L 4 x 20 x 10- ' Here -C = • = 2. Therefore '''I ~C = ..f2 = 1.41 4 and R = 1 n . As R < 'I ~C ,' the 40 x 10 circuit is underdamped. Fig. 5.1 (a ) shows that thyristor stops conducting when Here
OJ,
, ]'12
10 12 1 X 106 = 2 x 40 - [ 2 x 20 )
:. Conduction time of thyristor, I, =
Example 5.B.
00, t 1 = n .
[
1t
OJ,
= 25000 radl sec
1t x 106 = 25000 ~ = 125.664 ~s .
For the circuit shown in Fig. 5.10 (a), d ul dt rating of thyristor T is 400 o. Calculate the ualue of
(a )
V I llS and its junction capacitance is 25 pF. Switch S is closed at t = C, so that thyristor T is not turned on due to du I dt.
(b ) In case maximum current through thyristor of Fig. 5.10 (a) is limited to 40 A, determine the ualue of R,.
I
Sol u tian . (a ) When switch S is closed, the equivalent circuit for Fig. 5.10 (a) is as shown in Fig. 5.10 (b) wh ere C = Cj + C, . The voltage rise across C is given by
r
_ v,
-
:200'1
Fh20n.
R ~
R,
I--
0,=1=
T
C;
--- v,
C,
=F 1=F
T
I (eJ
(a J
Fig. 5. 10 (0) Pertaining to Example 5:8. Equivalent circuit (b) for (0) at t = 0 and (e) for (a) when SCR turns on.
The voltag e variation across C is the same as that across C, or Cj of thyristor T . ..
UC= UT=V,[l - e-
tl t
]
T} = V, 400 = 200 (dO dl =, RC or 10-' 20 x C ..
5
. C= 10x lO- = 002-- F '" 400 . o~
C, = C - Cj = 0.025 x 10-' - 0.025x 10-".". 0.025 ~
Vi""hen switch S is closed, C, would be charged to voltage VJ 'w ith upper plate positive as shown in Fig. 5.10 (c). Now when thyristor is turn ed on , current iT a t that moment through T would be give n by (b)
VJ V, I T =R+ 1f
,
R, =10n.
or
Power Electronics
[Arl. 3.6]
242
Example 5.9. For the circuit shown in Fig. 5.11 (a), V,
=200V, L =0.2 mH, C =20!1F,
constant load current 10 = 10 A and capacitor C is initially charged to source voltage V~ with lower plate positiue. The auxiliary thyristor TA is turned all, at t =0 to commutate the main thyristor TI . Calculate (a) the time at which the commutation ofmoin thyristor Tl gets initiated (b) the circuit turn-off time for T 1. Comment on the conduction timl! 1)( auxiliary thyristor.
.,
I 01
,
--'-
. ~ F- c
T1
1
L
"
+
"
[,
-=- 'I, LOad
Load
[0 =lOA
(a)
(b)
Fig. 5.11. (a) Pertaining to Example 5.9 (b ) Circuit model when Tl is turned on.
Solution. (a ) When auxiliary thyristor TA is turned on at t =0, C begins to discharge through L. T1. TA and C. Eventually. upper plate of capacitor would become positive with Uc = V,. With TA on, current ic begins to rise as
:b4<.,.
•
I_
il l
Note that i" flows opposite to in so that iTl ~ 10 - ie for 0< t < t l , Fig. 5.11 (b) . When it; attains a value of 10 A, iTl = 0 and therefore. Tl
is
turned·off at
t}
is obtained from the relation,
. . rr f sin
H ere ..
or
OJ,
000 t}
o-o:..oo
i
:
o'-'''i,,,,----7":-'~'.--;as shown in Fig. iDo'l" : [Vs}t-Io] : i
5.11 (c). Therefore. time tl at which commutation ofT1 begins
V, -\J
:~
, .: , lo!---., t~ - - l.. i,
~
~
.
'-">c'~-'-..J-~-ccFig. 5.11 (c). Waveforms
= 10 A
pertaining to Example 5.9.
rc .
1 .1 10' 10' dV G.OO xlO- · = ~LC = \I 0.2 x 20 = ~40 an ,'I f = 200 \I 0.2 x 10-' = 40 A 40 sin 00 tl = 10 A 0
-I(
1 . t '=-sm 1
Wo
{.IQ . - ' (02-) -10 ) =--sm . <> = 1-<>. 9S1 5
40
10
(b) After t 1• as i" exceeds 10, diode D1 begins to conduct till time t 2 where time to of h alf cycle of i, is given by
t
~s
iDl
falls to 10 , Here
n .,.,. {.IQ = = n , LC =n = 198.693 ~s o 00 105 0
Circuit turn·off time faT Tl = conduction time of diode D1 = t2 - t} = to- 2 tl = te = 198.692 - 2 x 15.981 = 166 .73
~s
Thyristor Commutation
Techn~ques
[Art. 5.6]
243
Th e auxili ary thyristor TA will continue conducting .till + + capacitor charg~s to V, with upper plate positive and ie falls to ",=i= c ,,", zero. Example 5.10. The circuit of Fig. 5.12 can be used to explain -;; !;- v~ i.e ......,...,.... J ~ ici"lo '" class-B commutation. With p ositiue directions indicated for tie> R i e• La> iT and u']\ describe, with appropriate wa ueforrns of Lo' ie• tic. i T and uT.how current·commutation is achieved in this circuit. Derive expressions for i~ reuerse uoUage across SCR when it is Fig. 5.12. Pertaining to Example 5. 10. turned off and the circuit turn ·off time. State the assumptions made. Solution. The assumptions are as under: (i) Thyristor is initially off. (ii) Capacitor C is charged to source voltage V, with upper plate positive. (iii) Load current is constant and (iu ) SCR is an ideal devic e. Now. when T is turned on at t = 0, capacitor begins to discharge through C, T , L , Fig. 5 .14
.
(a ).
Ther efor e, an oscillatory current
i.e =
V,
~ sin Ulo t is e~tablished
Also, at 't = 0, load current 10 is set up so that
and
Vol cos Ulo t.
ue =
th~5tor current iT = 10 + V, ~ sin Wo t as
shown in Fig. 5.13 and Fig. 5.14 (a) .
i'O~
.
--
_
_
--'-_
_
It,
---' t
v,
-.-
c' ~,.3
iT :lO.,.i c
T
, o~o R ~ (a) O
L
R
·
~ \T:1 0-1<
" l.i . 1
t
" "
··, • • ---t,• l , ,.. •
o ~----------~~~~~
, T
"
'.,
I T''
0
,
v, "
vI
1
t
t,
I,
I
R
,b
,
T oil
Fig. 5.1 3. WaVefOTI!lS for the circ uit of Fig. 5. 12
Fig. 5. 14. (n) When T is tutn;d on at tDO . i... =l.+i. (b) t. < t < t., a nd ( r.) t,,< t< t".
244
[Art. 5.6J
Power Electronics
In Fig. 5.13, at 000 t =lt/ 2, ie = COo t l
-
Vs
~
Vc =
0, iT =1 0 + V,
~ (peak value),
liT =
0 and at
=Tt, ie =0, lJe =- V•• i T =10 and vT = O.
After time t l • ie r everses, therefore i T = 10 - it: begins to decrease, Fig. 5.13 and Fig. 5.14 ( b ).
At t 2• ic ri se s to 10 and in
= 0,
therefore T is turned off. Als o at tim t 2• it = 10 , i T = 0, lie = - ' Vab = - V, cos 000 (t 2 - tl)' UT = - Vob =- V, cos 000 (t 2 - t 1). Note that SCR Tis subje.::ted to reverse voltage V ab - Between (t3 - t 2), current ie =10 charges C linearly fr om Vob at t2 to zero at I, so that uT = 0 at Fig. 5.13 and Fig. 5.14 (c ). This gives
t,.
I =C Vob t3 - t2
o
:. Circuit turn-off time for T
=(t3 -
V.. t 2) = te =C T
,
Eventually C charges linearly from zero at t3 to V, at t, as shown in Fig. 5.13. Then ie reduces to zero. io = 0, ue = V.. (as before),(i T = 0, vT =V, after t,,_ .
..
-,
Example 5.11.1n. the circuit shown. in Fig. 5.15 (a), switch S closes at t ~ 0and opens after 10 ms. What will thi currerits in R and L, 8 ms after the switch S opens. A ssume 0.7 V drop across diode whenever i.t conducts. Solution. When switch S is closed, Ri + L
It solution Here
~: =V•
,=. V ' [l -e-I/'J R
is t
=R1 = 501 =.002 sec an d 1' = 200 50 [ 1 -
At t = 0.01 sec, i = 4 [1 - e- l12 ] io = 1.54 A inR andL .
= 1.574 A.
' ] e- -0.02
Therefore, when switch S ope~s after 10 ms ,
O.7V
00
W
Fig. 5. 15. (a ) and (b ). Per taining to Example 5.11
Counting tim e from the instant switch S opens, voltage across L forward biases diode, Fig. 5.11 (b) and current i1 begins to flow.
di 1 L dt + 0. 7 = 0
or
or i 1 - i 1 (0)= - 0.7 t
or
i 1 = i 1 (0) -0.7 t = 1.574-0.71
Thyristor Commutation T echniques
[Prob. 3]
245
Wh en I = 0. 008 s, current in L would be il = 1.574 - 0.7 x 0.008 = 1.5684 A Current in R at t = 0.008 s would iR
=O.
Example 5.12 .. In the circuit shown in Fig. 5.16, switch S closes at t = 0 and opens afte r 10 ms. What will be current in R, L and voltage across C, 9 ms after switch S opens. Assume diode to be ideal. Solution. 'When switch S is closed, Ri +L
~; =V" i=4[1_e- tI O,02]
and At t = 0.0 1 s, i
= 1.574 A as in the previous example. R~
50n
i,
'"F == C
;;~
L=IH
·1
t
(aJ
Fig. 5.16. Pertaining to Example 5.12
When switch S opens, iR = a and voltage
~croBs L , equal to L ~;: forw~d biases the ideal
diode which begins to conduct at once. An oscillatory circuit consisting of L, C an d diode is form ed in which current i 1 will decay to zero, Fig. 5,16 (b ). at time tl given by
11 =-"-=< ..JLC =< ,11 x 1 x 10-' =3 .141 ms
w,
_
Since tl is less than 9 ms, energy in L gets transferred to C as the current il in L decays to zero. l
v ' l L ·, c ='2
'2 C
'0
Vc = =
PROBLElUS
.
-1f _I
x Current when S opens, io 1
'l/1 x l0-
6
x 1.574 = 1.574 kV.
,
5.1.
Expl ain t he need of commutation in thyristor circuits. Wha t are the differen t methods of commu tation schemes? Dis cuss one of them, involving tw o thyristors . with a neat schematic and wa veforms. (b ) A circuit employing parallel·resonance turn-off (or c1ass-B commutation) circuit ha!l C =50 IJ.F, L = 20 IJ.H, VJ = 200 V and inicial voltage across capacitor is 200 V. Determine th~ circuit turn-off tim e for main thyristor for load R = 1.5 n . rAns : (b) 68 ~I3J
5.2.
(0. )
(0.)
Distinguish clearly be twee" voltage commutation and current co mmuta tio n in thyristor circui ts.
Power Eleci:ron ics
246
Discuss how the voltage across the commutating capacitor is r eversed in a commutating circuit. (c) For the circuit in Fig. S.3 (a), supply voltage V, = 230 V dc, load current 10' = 200 A. circuit turn-off lime for main thyristor::: 25 J.I.S and reversal current is limited to 150% of 10 , Determine the values of commutating components C and L . [An.: (e) C =29.166 ~F, L =17.143 ~Hl
(b)
5.3. In the circuit shown in Fig. 5.9, capacitor C is initially charged to V. = 200 V with polarity as indicated. Find the circuit turn-off time for mai n thyristor Tl after it is voltage commutated by thyris tor TA Load current is constant at 40 A and C = 10 ).IF . [Ans. 50).ls] 5.4.
(a ) (b)
5.5.
Explain the merits and demerits of self-commutation of SeR and its other methods of commutation. For the circuit shown in Fig. 5.10, given that the load current 10 to be com mutated is 10 A, circuit turn-ofTtime required is 40)Js and the supply voltage is 100 V, obtain the proper values of commutating components. Take peak resonant current equal to twice the foad current. lAns : (b) C:::: 4.619 )JF, L = 115.475 )JH]
Di sc uss , with relevant waveforms, class A and class D types of commutations employed for thyristors . (b) For the circuit shown in Fig. 5.10, peak thyristor current =2.5 times the constant load current, L = 18 )JH and C = 4 )JF. Find the time elapsed from the instant thyristor is turned on to the instant it gets turned off. rAns: (b) 32.852 }lsl (a.)
5.6. (a.) Enumerate the various commutation techniques used for thyristors. (b) Describe line-commutation and c!ass-E commutation for thyristors. Name the circuit configuration where line-commutation is employed. 5.7. (a. ) Discuss, with relevant waveforms, class B and class E types of commutations employed for thyristor circuits. (b ) A circuit employing resonant-pulse commutation has C:::: 20 )JF and L = 3 flH. The initial capacitor voltage = source voltage, Vl =230 V dc. Determine conduction time for alL'Ciliary thyristor and circuit turn-off time for main thyristor in case constant load current is (i) 300 .-\ nnd (ii) 60 A. rAns: (b) (i) 24.335 )JS, tc = 13.23)Js (ii) 24 .335)iS, tc = 76.273 IlS]
5.8. (a) Describe class-C type of commutation used for thyristors with appropriate current and voltage waveforms. (b) An impulse·commutated circuit is shown in Fig. 5.5 (a). In this circuit, capacitor is initially charged to source voltage VJ =200 V with upper plate negative . When auxil iary thyristor is turned on main thyristor gets commutated in 50 )lS. Find the value of C in case load r esistance i5 20 Q. Ir peak value of current through main thyristor is limited to twice the full-load current, calc ulate the value of commutating inductance. fAns: (b) 3.607 )IF, 1.4428 mHl 5.9. What is com plementary impulse commutatio n? Desc ribe this type of commutatio n with a circuit di agram and appropriate wavefonns . Derive expressions for current through and voltage across commutating capacitor. Find also the circuit tunloff times for the complementary th)T1stors. 5.10. (0 ) A ca pacitor C, initially charged to dc voltage VJ • is connected to inductance L through a thyristor. Determine (0 the peak value of current through thyris tor Bnd (ij) the maximum value of dildt through SCR. (b ) F' OT illustrati ng closs C commutation , ci rcuit of F ig. 5.4 (a ) is empl oyed wher e VJ :: 200 V and R 1 :: 10 n. Find the value of C 50 that thyris tor Tl i3 commu tat.i!d in 50 )l s .
[Prob. 5]
Thyristor Commutation Techniques
247
It is required that SCR T2 is turned off naturally when current through it falls below the
holding current of 4 mAo Find the value of R 2. [Hint: (b) When C is fully charged, current through T2
[ Ans: (a) V,
V
= holding current = R~ etc.J
.""£. L
_icV,
. ] Amp/sec (b) 7.2135 ~F. 50 kn
5.11. In the circuit of Fig. 5.4 (a) employing complementary commutation; V., = 200 V, RI = 20 n and R2 = 100 Q. Detennine the minimum value of C so that thyristors do not get turned on due to re-applied duldt. Each SCR has a minimum charging current of 4 rnA to turn it on and its junction capacitance is 20 pF. [Ans: 0.1 ~FJ
5.12. For current-commutated circuit of Fig. 5.3 (a ); VI = 230 V, L = 16 ~H and C = 5 ~F. Capacitor is initially charged to voltage VI with left hand plate positive. Auxiliary thyristor TA is turned on at t = O. Find the total time for which capacitor current i( exists. The peak resonant current is 1.5 times the full-load current.
. . [ Hint: In Fig. 5.3
(b), t5 - t J =
C
1
V, +V.b 10 etc.
[Ans , 58.047 ~I
Chapter 6
Phase Controlled Rectifiers ... . _- .-_ .. _-_ .... .............. ............. .. ... ..... -.... .. ... ... ... .. ...... --_. __ ......... ..... In this Chapter • • • • • • • • • II! •
Princ iple of Phase Control Full-wave Controlled Converters Single-phase Full-wove Converters Single-phase Two-pulse Converters with Discontinuous load Current Performance Parameters of Two-pulse Converte~ Single-phose Symmetrical and Asymmetrica l Semiconverters Three-phase Thyristor Converters Performance Parameters of 3-phose full converters Effect of Source Impedance on the Pe rformance of Converters
Dual Converters Some Worked Examples
Many industrial applications make use of controllable de power. Examples of such applications are as follows: (a) Steel-rolling mills, paper mills, printing presses and textile mills employing de motor drives. (b) Traction systems working on dc. (c) Electrochemical and electrometallurgical processes. (d ) Magnet power supplies. (e) Portable hand tool drives. (f) High-voltage dc transmissi on. Earlier, dc power was obtained from motor-generator (M G) sets or ac power was converted to de power by means of mercury-arc rectifiers or thyratrons. The advent of thyristors has changed the art of ac' to dc c·onversion. Presently, phase-controlled ac to dc converters employing thyristors are extensively used for changing constant ac input voltage to controlled dc output voltage. In an industry where there is a provision for modernization, mercury-arc rectifiers and thyratrons are being replaced by thyristors. In phase-controlled rectifiers, a thyristor is turned off as ac supply voltage reverse biases it, provided anode current has fallen to a level below the ""hCJ~ding current. The turning-off, or commutation, of a thyristor by supply voltage itself is called natural, or line commutation. In industrial applications, rectifier circuits make use of more than one SCR. In such circuits, when an incoming SCR is: tm-ned on by triggering, it immediately reverse biases the outgoing SCR and turns it off. As pha.5e-<:ontrolled rectifiers need no commutation circuitry, these are simple, less expensive and are therefo re widely used in industries where controlled dc power is required. In the study of thyris tor systems, SCRs and diodes are assumed ideal switches which means that (i) th ere is no voltage drop across them, (i i) no r everse current exists under reverse voltage conditions and (i ii) holding current is zero.
[.,,1. 6.J]
Phase Controlled Rec tifiers
2-'1)
Trigger circuits are not shown in SCR circuit for convenience. In this chapter, single~phase and three~phase controlled converters are described and the effect of source inductance on their performance is examined. Basic oper ating features of dual converters are also presented.
V/ 6.1. PRINCIPLE OF PHASE CONTROL
The simplest form of controlled rectifier circuits consist of a single thyristor feeding dc power to a resistive load R as shown in Fig. 6.1 (a). The source voltage is u, = Vm sin (llt, Fig. 6.1 (b). An
SCR can conduct only when anode voltage is positive and a gating signal is applied. As such, a thyristor blocks the flow of load current io until it is triggered. At some delay angle a, a positive gate signal applied between gate and cathode turns on the SCR. Immediately, full supply voltage is applied to the load as vo, Fig. 6.1 (b). At the instant of delay angle a, Va rises from zero to Vm sin a as shown. For resistive load, current io is in phase with va' Firing angle of a thyristor is measured from the instant it would start conducting if it were replaced by a diode. In Fig. 6.1, if thyristor is replaced by diode, it would begin conduction at wi = 0, 21t, 41t etc. ; firing angle is therefore measured from these instants . A firing angle may thus be defined as the angle between the instant thyristor would conduct if it were a diode and the instant it is triggered.
Firing pulses I
Vm sin Cl
/
}
wi
I
I I,
I Ii
V Vm
ii
wi
(io11 TO')
l,,(" , ,
wi
..
~~
+
R
~
1
,, WI
'0
1 ~
Fig. 6.1. Single-phase h:lli-wave thyristor ci rcuit with R load (0 ) circuit diagram and (0) vol tage and current waveforms
250
Power Electronics
[Art. 6.1]
A firing angle may also be defined as follows: A firing angle is measured from the angle that gives the largest average output voltage, or the highest load voltage. If thyristor in Fig. 6.1 is frred at wt = 0, 2n, 4n etc., the average load voltage is the highest ; the firing angle should thus be measured from these instants. A firing angle may thus be defined as the angle measured from the instan t that gives the largest average outpu t voltage to the instant it is triggered. A critical observation of Fig. 6.1 leads to the emergence of another definition of firing angle. Thus, a ring angle rna be defined as the angle measured fr om the instant SCR gets forward ~l..- ~ biased to the Instant it is triggere . Once the SCR is on, load current flows, until it is turned-off by r eversal of voltage at wt = n, 3n etc. At these angles of1[, 3n, 5n etc. load current frula to zero and soon after the supply voltage reverse biases the SCR, the device is therefore turned off. It is see n fr om Fig. 6.1 (b) that by varying the firin g angle ex, the phase relationship between the start of the load current and the supply voltage can be controlled ; hence the term phase control is used for such a method of controlling the load currents [3J . A single-ph ase half-wave circuit is one which produces only one pulse of land current durin g one cycle of source voltage. As the circuit shown in Fig. 6.1 (a) produces only one load current pulse for one cycle of sinusoidal source voltage, this circuit represents a single-phase half-wave · thyristor circuit . In Fig. 6.1 (b ), thyristor conducts from wt =ex to n, (2n + ex) to an and so on. Over the firing angle delay ex, load voltage tlo = 0 but during conduction angle (1t - ex), tlo =tis. As firing angle is increased fr om zero to n, the average load voltage decreases from the largest value to zero. Th e variati on of voltage acr oss thyristor is also shown as tiT in Fig. 6.1 (b). Thyristor remains on from Wi = a to n, (2n + a) to 3n etc., during these intervals ti T =0 (strictly speaking 1 to 1.5 V ). It is off fyom n to (2n + a), 3n to (4/t + ex) etc., during th ese off intervals vT has the waveshape of s upply voltage tI, . It may be obse rved that tI, =Vo + tiT. As the thyristor is r everse biased for 1t r adians , the circuit turn-off time is given by
t, =-w " sec where w = 2ft' and, is the supply frequen cy in Hz. The ci rcuit tum-off time tc must be more than the SCR turn-off tim e tq as specified by the manufacturers. Average voltage Vo across load R in Fig. 6.1 for the single-phase half·wave circuit in terms of firing angle ex is given by 1 Vo= 2n
f"Vm sin wt · d CI
(wt)=
V 2; (1 +cos ex)
... (6. 1) \
The maximum value of aver age output voltage Vo occurs at ex = 0". Vm Vm Vo·m = -·2= - 1t 21[
.--\lso, Average load current,
Vom VO = T ( l -rcos ex)
Vo
Vm
IO =Ji = 2nR (1 + cos a )
...( 6.2 )
lA,'.6.IJ
Phase Controlled Rectifiers
251
In some types ofloads, one may be interested in rms value of load voltage Vew Exam ples of such loads are electric heating and incandescent lamps. Rms voltage Vur in such cases is given by
v" = [1... J' V~ sin' w/ . d (WI)]" 2it tl
= ~~
[(n-a)+~ sin 2a
The value of r ms current lor is
r
... (6.3)
V"
l or = Jf
Power deliver ed to resistive loa d = (rms load voltage) (rms load current)
V;r
'J
= Vo'!or = If = 1~,R
Input volta mperes
... (6.4 )
= (r ms source voltage) (tota l rms line cur;eiit1 =V
J .
Input power fact or
F;m Eq. (6.3), input
..J2 ...In V; lor = 2R
[ (:t -
1 sin 2 a ] " a) + '2
to load V or ' l or Vor = P ower delivered -- V l or =VsInputVA p{ =
'I;n [ (n - a)
V
+
~ sin 2 a
r
J
•
... (6.5)
6.1.1. Single.phase Half.wave Circuit with RL Load Asingle-phase half-wave thyristor circuit with RL load shown in Fig. 6.2 (a ). Line voltage v, is sketched in the top of Fig. 6.2 (b). At wt =n , thyristor is turned on by gating signa l (riot-
,;j.
~~~
is
shown). T he load voltage Vo at once becomes equal to source voltage v, as show n . But the inductance L forces th e load, or output, current io to rise gradua lly. After some time, io reaches maximum value and then begins to decr ease. At wt = Tt , Uo is zero but io is not zero because of the load inductance L . After rot = Tt, SCR is subjected to reverse anode voltage but it will not be turned off as load current io is not less th an the holding current. At some angle P> It, io reduces to zero and SCR is turned off as it is already reverse biased. Afte r Wi = p, Vu = 0 and io = O. At wl = 2n + (,(, SCR is triggered again, Vo is applied to the load and load current develops as before. Angle ~ is called the extinction angle and (P - a) = y is called the conduction angle. The waveform of voltage LIT across thyristor T in Fig. ~2 ( b) reveals that when wt = a, vr = Vm sin CJ. ; from wi = a to P, vr = 0 and at Wi = ~, ur =V m sin ~ . As P> n, ur is negative at wt = p. T hyristor is therefore reverse biased fr om Wi = ~ t o 2n . Thu s, circuit turn- off time
tc = 2n; ~ sec. For satisfactory commutation, tc should be mor e than time. Th e voltage equation fo r the circuit of Fig. 6.2 (0 ), whe n T is on, is
dio V . . .. sin C!.'l :: R io+L dt
tq
the thyristor turn-off
,-, -,-
Power Electronics
[Art. 6. 1J
of+--'--;\--
wi
,, , ai---y..J.J
:,
. I.
+1
T
R
+ rv
wi
"5:
WI
I
"0
L_Vm_Si_nw_t_ _L::.J
U-:.
wi
(a)
(6)
Fig. 6.2. Single-pbase half-wave circuit with RL load (a) circuit diagram and (b) voltage and current waveforms .
The load current io consists of two components, one steady-state transient component it. Here i, is given by
compo~ent i.
and the other
i, == ~R2:)(l sin (Cilt - 41) where
q. = tan- 1 ~ and X = 00£. Here $ is the angle by which rms current I. lags V"
The transient component it can be obtained from force-free equation
. di t R I,+L Tt= 0 Its solution gives,
it =A e-(R IL)t
..
io = i.+t/=
i
V
sin(wt_¢l) +A-(RI L)t
wher e Z=,JR'+X" Constant.4 can be obtained from the boundary condition at oot = a.
At this time t =.!!. OJ io = O. Thus, . from Eq. (6.6),
0= or
A
V
i
= _
sin (a- 9)+ Ae- RalLw
V m. sin (a _ 41) eRa/tal. Z
... (6 .6)
[Art. 6.1 1
Phase Controlled Rectifiers
253
Substitution of A in Eq. (6.6) gives .
io =
~m sin (00/ -
sin (11 - 9) expo\ -
$) - ;
!
(00/ -(1) }
... ( 6.7 )
for
a
sin
(~ - $) =sin (11- $). exp \ -
!
(~ - Il) }
This transcendental equation can be solved to obtain the valu e of extinction angle case ~ is known, average load voltage Vo is given by
1 J~ Vm Vo = 21t 0. Vm sin rot d (rot) = ~ (cos ex - cos ~) Vm • Average Ioad current, 10 = 2TtR (cos ex - cos "') Rms load vol tage,
/
r
2~ J:V~ sin' d(w/ ) V[ (~ - (1) -"21 (sin 2~ - sin 2(1)]'12 = 2'1:
V., = [
00/ .
~.
In
.. (6.8)
. ..(69)
... (6. 10)
~s load current can be obtained from Eq. (6 .7) if r equir ed.
: .. /6.1.2. Single-phase Half-wave Circuit with RL Load and Freewheeling I;>iod e
The waveform ofl oad current io in Fig. 6.2 (b) can be improved by conneCting'a freewheeling (or flywheeling) diode across load as shown in Fig. 6.3 (a). A freewheeling diode is also called by-pass or commutating diode . At rot = 0, source vo ltage is becomi ng positive. At some delay angle a , forward biased SCR is triggered and source voltage vJ appears across load as vo' At wt = It, source voltage vJ is zero and just after this instant, as v, tends to rever se, freewhe eling diode FD is fonvard biased through the conducting SCR. As a r esult, load current io is immedi ately tran sferred from SCR to FD as vJ tends to reverse . At the same time, SCR is subj ected to r everse voltage and zero current, it is therefore turned off at rot = 1t. It is assumed that l!uring freewheeling period , load current does not decay to zero until the SCR is triggered again at (21t + a). Voltage dr op across FD is taken as almost zero , the load voltage Vo is, therefore, ze ro during the freewheeling period. The voltage variation ac ross SCR is shown as l)T in F ig. 6.3 (b). It is seen fr om this wave-form that SCR is reverse biased from wi = 1t to wi =21t. Therefore, circuit turn-off time is n
tc =-sec 00
The source current i, and thyristor current iT have the same w ave form as shown. Operation of the circui t of Fig. 6.3 (n) can be explained in two modes. In the first mode, called con duction m ode, SCR conducts from ex to Tt, 21t + a. to 3rt and so on and FD is reverse biased. Th e duration of this mode is for ((Tt - CX)/ wJ sec. Let the load current at the beginning of mode 1 be 10 , The exp ression for current io in mode I can be obtained as follows:
j'vl ode I : For conduction mode. the voltagt> equation is
.
.
dio
V m sm wt= R~o +L dt
Power Electronics
[Art. 6.1]
25-1
WI •
'!
1
i, ' [0
I"~
_ T .;.._ FD'
'I, '
ICl I: I
is
l<
T R
+
!!
T
i
• !
101
1
~ .' T i
1-:-
in !:
wt
T
I:
. :
I
!
wi
N4,,+a)
,
F_~f_d~_L--1~J
.
Wi
I .
I
,
Wi
;"Mod~~
I
•
:
, I
rvL.. _V___ '
•
101
I
i,
wi
.
(a)
(b)
Fig. 6. 3. Single-phase half-wave circuit with RL load and 8 freewheeling diode, (a ) circuit diagram and (b) voltage and current waveforms .
Its solution, already obtained in the previous section, is repeated here from Eq. (6.6 ) as
.
Vm
.
_
RIL
' o=ysm(wt-41)+Ae (
)t
At Wi =0., i o =10' i.e. att = a ,io=Io
..
A=[ 1
..
io=
'"
0-
~m sin(a-¢) }eRa/1ill.
~mSin {"'t-$)+ [Io - ;Sin {a-$)} exp {- f (t - ~}
.
(S. l!)
Note that for mode I, a s: Cilt s: n Mode II : This mode, called freewheeling mock, extends from n to 2n + a, 31t to 41t + a and so on. In this mode, SCR is reverse biased from it to 21t, 3n to 41t ... as shown by vol tage waveform Ur in Fig. 6.3 (b ). As the load curr ent is assumed continuous, FD conducts from 1t to (21t + a), 31i to (411: + a ) and so on. Let the current at the beginning of mode 11 be 101 as shown . As load current is passing through FD, the voltage equation for mode II is .
d io
O=RIO+ LCi/
[Art. 6.1]
Phase Controlled Rectifiers
Its s olution is
. II - (R/ L J t '0 =. e
At wi = n,
io =101 ,
l t giv es
A = 101 eR::/ wl.
..
io=IOlexp [
Note that for mode
n,
- ~I-: )l
,--
-0>
...(6.12)
Tt < wi S (2n + a )
Average load voltage Vo from Fig. 6.3 (b ) is given by 1 Vo = 21t
J' Vm sin ex
wt d (wt ) = V 2; (1 + C05 (X)
Vo Vm Average load current, I = - = - - (1 + cos (X) o R 21tR
.. .(6.13 ) ... (6. 14)
Note that load current io is contributed by SCR fr om a to Tt, (2n + 0 ) to 3n and so on and by FD from 0 to (X, 1t to (2Tt + a) and so on. Thus the waveshape of thyristor current iT is identical with the waveshape of io for wt = a to Tt, (2n + a) to 3ft and so on . Similarly, th e wave shape of FD current iid is identical with the waveform of io for wt = 0° to ex, n to (2ft + a) and so on. In Fig. 6.2, load consumes power Pl from source for a t o 1t (both Vo and io arc positive) whereas energy stored in inductance L is return ed to the source as power P2 for n to Jl (va is n egative a nd iois positive). As a result, net power consumed by the load is th e difference of these two powers PI and Pa. In Fig. 6.3, load absorbs power for ex to n, but for 1t to (21t + a ), en ergy stored in L is delivered to load resistance R through the FD. As a consequence, power consum ed by load is more in Fig. 6.3 . It can, therefore, be concluded that power delivered to load, for the same firing a ngl e, is more when FD is used. As volt-ampere input is almost the sam e in both Figs. 6.2 and 6.3, the inputpl(= power delivered to load/input volt-am pere) with the use ofFD is improv ed . .
v.;.-
It is also seen from Figs. 6.2 (b) and 6.3 (b ) that load current waveform is improved with FD in Fig. 6.3 (b ). Thus the advantages of using fr eewh eeling diode ar e input pI is improved (ii) load cu rren t waveform is improved (iii) as n r esu lt of (ii) , load performance is better and (i v) as energy stored in L is trans ferred to R during the freewheeling pe ri od, overall converter efficiency improves. (i)
It may be seen from Fig. 6.3 (b ) that fr eewheeling diod e prevents the load voltage Uo from becoming negative. \Vhenever load voltage tends to go negative, FD comes into pl ay. As a result, load current is t r ansferred fr om main thyristor to FD, allowing the thyristor to regain its forward blocki ng capabil ity. It is seen from F igs. 6.2 ( b ) and 6.3 ( b ) th nt sup pl y current i , taken fr om the source i~ uni directiona l an d is in the fo rm of de puls e.5 . Si n~ le phase h alf-wave converter thu s intr odu ces a dc co:nponent into th e supply line. TPi s is u nd ~ :;i r a bl= as it leads to .5atura tion of the supply transformer and other diffi culti es (harmonic ::; etc.). Th e3e shortc oming5 can be overcom e to som e ext e nt by the use of sin gle-ph as e iull wave circu its discu3s ed in Art. 6.2.
256
!'
V
[Art. 6.1J
Power Electronics
6.1.3. Single·phase Half-wave Circuit with RLE Load
A sin gle-phase half-wave controlled converler with RLE load is shown in Fig. 6.4 (a). The counter emf E in the load may be due to a battery or a de motar. The minimum value of flring angle is obtain ed from the relation Vm sin wt = E. This is shown to occur at an angle 9 1 in Fig. 6.4 (b).
where ... (6.15)
In case th}Tistor T is fired at an angle a. < 91• then E > V., SCR is reverse biased and therefore it will not turn on. Similarly, maximum value of firing angle is 9 2 =1t - 9 1• Fig. 6.4 (b ). During the interval load current io is zero, load voltage uo=E and during the time io ' is not zero , Uo follows v, curve. For the circuit of Fig. 6.4 (n) and with SCR T on, KVL gives the voltage differential equation as .. .(6.16)
The so lution of this equation is made up of two compon ents; namely steady-state current component i J and the transient current component it. For convenience, i. may be thought of as Y.
·
001
9,
I
.,
"i
i,
Firng pulses
a
I Wi
Yo
I,
FYT?]
io
T
Y.
'V
•
R
•
•
•
·i
~ a :""' y ...........:
~~
vT
" -----, (VmSina-E ) ~
J' ,
: --r
'.
!I---r. :
!I i.
wt
. I
I
! wt
Yo
Vm s in wt
L
E ~
a
W
Fig. 6. 4. Single-phase half-wave ci rcuit with RLE load (0) circui t dingram and (b) voltage and current wave forms.
P h:.ist: Cuntr OIl t:d K.:..:(ii:.:ts
[An. 6.11
257
the sum of i"l and i $'~' where is1 i3 the 3teaily 3'::<1te current due to liC source voltage acting alone and i s'1. is that dut: to de counter emf E acting alone. A3 in the p'ccsentation leading to Eq. (6.6), i~l due to source voltage V«. sin Wi is given by
If only E were present, t hen steady state current
i ,;2
would be given by
ij;l = - (EI R)
The t ransien t current i: is given by
i, _A - e-tlVL}l
Thus t he total current LO is given by
. t il
=
.
,~:
.,.
.
.
t.~ or '/
= V'II Z
.... = a, to . = O,L' .e. at t = a . = 0 . Thi s glVes . EAt u..-. W' LO .It = R
[
; f - - R. VJ.cX-p
V",r SlO . ((Ill_I"',• . .. ',o = --! -:'In (Ct -
z ·L
'
:iL
. ( S lO
,
)
Vc'
Zm.Sln (a -
,11 -
.
~ Wl-v.J ~
we -.;>
E A - iRI L )(
- R
~
T
e
) ] eRwLw
Er l -exf"" f,. -
-
' R. _:..
C1l( - ( l
~
)1] • '
,
... {ii. !. 7i ~.; .
\o.l.7J ~:s ~~~ 1 ~':::':::' tc I~:'- .:.c.:; u.t ~ ::.. r..~(. ::: c:-..1..:. .:'(. ... :. tiring ~fl g!.:: .:r. ;ina the 10;;.,J i(...:;.e.:J.~·:I:i.:= ::.r.:.gle y.
::.:-. gh; ~ .:lc~ .:r..:ls U$iVO
iV-lid emf E.
Aver 3.ge voltage across inductance is zero. Thus, a.ver&ge value of luad current can be obtained by integrating \Vm sin wt - E) / R oetween a ar..d ~ . T he average load curren t 10 is therefore given by
10 =
2~[ J: (V sin WI - E) d(WI) ] m
1 = 2JtR
Her e conduction angle'y =
~
[Vm (co. a - cos~) - E (~- a)l
- a . Putting
~
=y + a
... (6. 18)
in Eq. (6.18 ) gives
1
10 = 2JtR [Vm Icos a - cos (y + a») -E , yl Using the trigonometric relation, . ~ . l..::..! cosx - cosy = 2 Sill 2 SIn 2
the above e:cpression lor 10 can be written
a3
_
IO = 2~[ 2v sin ( a+~ JSin~-E ' Y] m
u( 619)
Average load voltage Vo is given by
Vo =E+loR=E +~ 2v sm ( a+n 3in ~ - y ' E ] m
= ~ l - J.. ) + Vr.mSinru +l2 )'inl2 2n ,
u.( 6. 20)
25:s
POWH E)cClr()nk~
[A n . 6.1 )
The above expression for the average lo&.ct voltage Vo can also be obtained as under : For periodicity 21t , extending from a to (2:t 1 Vo = :in
T
a ), we have
[J" VI{, sin u.
UJl . d(wt)
,
=
In case
-!:- i Va (cos a ~ T.:"
'
T
E (2rr T a .-
co.;~) T E (211
T
CI. -
13)1j
~)l
...:
is is made i::qual co (YT c.) in the a bvve expression , Eq. (6.20 ) can be lihrained.
If lead inductance L is :lero in Fig. 6.4 (ul, then extin..:;tion anglo:: 13 would be: .:'l.ual to tl! = :'t - 1, i.e. now 13 woula be leSS thz..r. n. Average V:iILle oilo~d current can still be oL~:i.ined from Eq. (6.1S' by substituting ~ =:! - al . Therefore, a V i;r~gE: load current 1", with L = 0, is
°
... (6. 18)
Its amplification gives J"
=[
2~21 (V; + E' ) (~ - Cl) - '1 (sin ~
Powe r delive red to load,
2
- 2 sin 2Cl) - 2 Vm E (cos Cl-
c~s ~) }
P = I;,. R + 19 E.
Supply power factor
J" . .
(6.2 1)
...(6.22 ) ... (6.23)
=
Tht! time variation of voltage across thyristor is ;;hown as uT in Fig. 6.4 (b). At WI = 0, u$ = 0, and therefore, uT = - E . At wt =ft, u, = E, therefore u = O. At Wi = a, Us = Vm sin a, therefore uT = Vm sin a-E. During the conduction angle y =(P - a), uT = O. At wt = p, ui h as reverse polarity. Therefore , just after thyristor is turned off at Wi = p, voltage UT =-IVm sin (p - TC ) + E I. It is also possible to write uT = V", sin P- Eat wf = Il, because Vm sin Pis negati ve fo r P> It. The magnitude ofma.ximum r everse voltage is (Vm + E ) as shown j
in Fig. 6.4 (b ). Fig. 6.4 (b) also reveals that circuit
v:
turn~off tim e is
21t + 91 OJ
~ sec.
Example S.l. A s ingle·phase 230 J kW heater is connecte d across J·phase. 230 V. 50 Hz supply through an SeR. For fir ing angie delay s of 45° and 90°, calculate th e power absorbed in the heater element Solut ion. Heater resi3tance
-I·
[Art. 6.1 ~
PhaSe Cuntrulll.! c1 Hc..:t ifi..::r:.
V:.' R
=
--'
;/ =
, ;!
"30
-
\
V'
~
' 155 .071
;.( :000 =
~ 5 .; .5 -:
\. ;,:::t::.
( 11- \' ) x 1000 = 250 \v::.tt ::..
~ :23~
Example 6.2. A de: I,c.!tery is charged rhro~gh a resisror R a.s .shOI4"n. in Fig. (;.5 fa). Derive C!xpre.s.sioH /'cJr the average uaLue of chargirtg curr.!!!! in rerms of VIII" E.,.!? etc. on. the usswflpticJft thut SeR is {ired cCmtiILuDusly (J.fI
.'~ I
R :::. ~
F", ,. .:1. ~:.: :,:~!.J.rI:.: l.~l!(J,':;·': ui ~:;{J V ,
~1 ~/'.!
j il:J. r/: .; L·~b(! (jia ;J':f~i;c. d:/.J.I~.; ./~d ~'~lfrt:H' rur
E :::. 150 V.
, ; " J.'! 1.•1 !f,,' ~"'''<1 r ( I.')
[;(j fi'.!.
::,·up.0 i,d !i.J tX1.!!":ry wId [ f.a[ d i:,;3ip'-l (.:ci
Cl1.h:u!,;:.!.! chi! .suppLy
if:
tit.! resi.stor.
pl
Solution. Fur the circuit of Fig. 6.5 \a ), t ho:
vol t:1~e
cq us.tion is
Vm sin we =E T ioR V sinwt-E . C..!.mo....:=;,;::-'--=-
I, =-
or
R
'o',; t
vm sinwt
I
R
T '0'$
E
'"'v Vmsin wl
1 J
(a)
Fig. 6.5 .
(c )
, ,,
e,
I!
'I:N-e,) to :
i
I
j
!!
WI
, :
•
\'~S'tnw t - E
i r (" ,
1/
'\ I
R
,
(b)
Power circuit diagram (0 1 various waveforms fo r Example 6.2 .
It is seen from Fig. 6.5 that SCR is tur ned on when Vm sin 8 1 = E and is turned off when Ym sin 8:! = E, where 8:! = It - 8 1, T he battery charging r equires only the average cu rrent 10 given by
260
Pow~r
(A n. 6.IJ
Electronks
1 = 21tH (2Vrn cos 8, - E(x - 28,)1 (a) Here
8
27 660 ,= sm. - , \12150 . 230 = .•
1 [ r;;_ ( 2x27.496xx ',1 _ 10 = 2 • . 8 2 · ,2 ·230 cos 27.466° - 100 • 180 ] = ·,.9076 A.
(b) Power supplied to battery
=
E10 = 150 x 4 .9676 = 745.14 W.
For finding the power dissipated in R, rms value of charging current must by obtained. From Eq. (3.39),
1"
=[ 2.164 {(ISO'''' 230') ( •
- 2 x 27.466 x
1~0
) T (230)' sin 2 x 27.466
.
- 4 · -12 · 230 · 150 cos 27.466t)1l1/2 = 9.2955 A.
.. Power dissipated in resistor = (9.2955)2 x 8 = 691.25 Watts . (CJ ~
_ 691.25 + 745 .14 _ 0 -70 I • . • 0 0 0 ,. ')~ __ - .0 ... :i1 o gTIl", .
From Eq. (6.23/, supply pf
-
_ .....
X
::;t. _~ oo
.
£~:imp~..:: -(; . ~ .
Ri!puJ."l Example 6.2 in case ihyri.sm ... is triggered al a {iring eui!.ry p ositivi! hUll cy:le.. Solution. (a ) Her.:: ex = 35".
P= 02 = r. - e1 = 180 -
.
a (,gL~
oj 3f," if!
.
27.466" = 152.534°
From Eq. (6.18), average charging current is given by 1,
= 2.: 8 [ {2 x 230 (cos 35'
- cos
152.53~O) -
150 (152.534 - 35°) x
1~0 1
= 4.9192A (b ) Power delivered to battery = E1, = 150 x 4.9192 = 737.88 W
From Eq. (6.21), rms value ofload current is
1o, = [ 2. ~ 64 { (230' + 150') (152.534 - 35) x
,
-
1~0
2~0 (sin 2 x 152.534 - sin 2 x 35) - 2. {2 .230 x 150 (cos 35 - cos 152.534°) ]
= 9.2874 A
.
Power dissipated in resistor = 9.28742 x B.= 690.05 W (c) Supply power ractor
= 6;~.g~ ~.~~~.:8 = 0.6685 lagging
Examples 6.2 and 6.3 demonstrate that an increase in the firing angle reduces the value of average charging current. rms current and the supply power factor. Example 6.4. A 230 V, 50 Hz, one-pu.lse SCR controlled converter is triggered at a firing angle of 40 and the load current e."Ctinguishes at an angle of 210t). Find the circui t tllm off tim e, average output uoltage and th~ average load current for Q
(0 )
R = 5 n end L
(b ! R
= 5
=2mH.
n, L =2 mE and E = 11 0 V.
r.
Phase Controlled Rectifiers
26 1
[Ac'. 6.1J
Soluti o n. (0) For this part, refer to Fig. 6.2. It is seen from this figure that circuit tUrn off tim e tc
_ 2"-~ _ (360-210) " -8333 00 - 180 X 2l't x 50 - . m-sec
-
From Eq. (6.8), average output voltage ,[2 · 230 _ Vo = 2l't [cos 40° - cos 210°) = 84.47 1 V Average load current (b)
10 =
~o = 84;77 = 16.8954 A.
Fig. 6 .4(b) shows that circuit tum-off time is
=_2_"_+_6-'.,_-.'-.~
t c
Here
(oJ
E . - 1 110 19 77. . - 1 V", e1 = sm = sm "'12 x 230 = . ·19.77 - 210) " t.: = (360180 X 2l't x 50
= 9 432 .
ms .
From Eq. (6.18), the average charging current is 1 r . 10 = 2" . sl ,[2 · 230 (cos 40· - cos 210·) -
no (210 -
•
40) 180
1
=6.5064 A. /.::-- :. Average load voltage, Vo =E • loR = n o+ 6.5064 x 5 = 142.532 V. Example 6.5. A single-phase transformer. with secondary uoltage of 230 V, 50 Hz, deliuers power to load R = 10 n through a half- waue controlled rectifier drcuit. For a fi ring.angle dela.y of 60°, d etermine (a) the rectification efficiency (b) fo rm factor (e) uoltage ripple factor (d) transformer utilization factor and (e) PIli of thy ristor. Solution. Here V, = 230 V,r =50 Hz, R = 10 a,a =60· From Eq. (6.1),
V '2 x 230 · Va = 2; (1 + cos a ) = 2Tt (1 + 005'60°) = 77 .64 V
I.
= 7~;4 = 7.764 A
From Eq. (6.3),
Output dc power,
P rk = V., I I)
Output ac power.
Pac = Vori.:r = 145.873 x 14.587 = 2127 .85 W
=
; 7.64 x 7.764 = 602.8 W
~: = 2~~;:~5 = 0.2333 or 28 .3370
(a ) Rectific ation efficiency
=
(b) Form facto r,
= V" = 145.873 =1 879
(c)
FF
Voltage r ippl e factor, VRF
\0'0
77 .6·1
.
= -IFF' - 1 =,1.t.8 79' - 1 =1.5908
262
Pow er Elec tronics
[A rt. 6.2J
(d )
TUF = V,l,
(e )
PIV =Vm
V .. I.
= V, l , = Vr I"r
602.8 230 x 14.587
=0.1797
="2. V, = "2 X no = 325 .22V
6.2. FULL-WAVE CONTROLLED CONVERTERS There is a large variety of SCR controlled converters (or rectifiers). One way of class,ifying these ac to dc converters is according to the number e! supply phases on the input side. As per this classification, the ac to de converters discu..c:sed in Figs. 6.1 to 6.4 are single-phase half-wave converters. Three-phase controlled rectifiers, as the name suggests, have three-phru:.e supply on their input side. these are discussed later in this chapter. The other way of classification is according to the number ofload current pulses per c::--c!e of source voltage. It is seen from Art. 6.1 that single-phase half-\vave controlled rectifie.r s proouce oJ'ly 011~ p'.llse of lead current dt.:.ri.ng Doe cycle of source voltage, these can L~er~fe"e be termed a" ~in~:.~~hase one-pulse converters. Thus. the controlied rectifiers d.isc uss~ ~'! :~;;s. 5.1 !o '3 ..4 c'\.r:,,: a!.l ~i_'1gJ'?-pha~e ~ne F'..tlo::o;: C':;';lverters.
. .'
- I LeAl) ~.----
(-7'
{a )
Fi£. 6.'3 .
(a ~ Sin:o::t~ - ~""';::-: !"j:-:-? .• :<:~ ..... ;,.~ -~':'ir..~ c':'"''-",rt!'":" .
U»
: 0 ·1'1
The di~ad \· ::'.:1t ~-::c~ ~f s\ngle-ph.<..~~ half-w-3.\-O::. ~r. sin~:l ~·t"ha ~ ~ ':'n'.:'- t"Jl,,? ,:::, n\- ~Tto:! ... are minimi sed by t.':1~ U~!? of s!ngk-t'_~a~'? f 'JJl \\·a.... ~. (lr s1nzl ~- pha ~e tW0 cu l :;:~. (';o,,\"e"~'."rs. In practice. th erE" arp t W I) bal':i...: :("\nfi~urations f')r f1JJ1-\\'a \-~ cC'ntrr,i:~J C':'('l\·~ rt <;: r s. One configuration uses an 1nput transformer with tv,;o \1.·indin!:.~ for ?ach ino:_lt oh ;\$~ "'1.0:1;:11:. Thi,. is cailed mid-point. ':()n. I.;ert.er. A sL'lgk-phase tW'J-pulse mid-poi ... t SCR cp';,\-<.:r',-:- r i~ ~ ~ ~wn in Fig. 6.6 (a' and a three-phase 6 PlJl~e mid-point converter in Fu::. t3 .G ~ 1: ' . ,- C "~st
f !>uoc-' y
Zr
-
•
zf
.,-
~
--,,'
\('0
'." .)
:,
tf tf
z:f
.: ,. . : - C 0. -
~
.
"
?S •
~n
z:f
i • (17)
' '2)
Fie:. 6. 7.
(e )
SingJe·pha5'e two-pulse
' h) Tlv(>€-phas",
-" ix- pul~!'
bridg~
converter and
bridJ?"€ con\"~rter .
---z:f •
z{ •
r-,
W
Phase Controlled Rectifiers
(Ad.6.3J
263
The second configuration uses SCRs in the fo rm of a bridge circuit. Single-phase full -wave, or two-pulse, bridge converter using four SCRs is shown in Fig. 6.7 (a ) and a t hree-phase six -p ulse bridge converter using si x S CRs in Fig. 6 .7 (b ). A bridge conver ter h as som e advantages over mid-point converter, these will be discussed after both these configurations a re studied in the n ext article. 6.:1. SINGLE-PHASE FULL-WAVE CONVERTERS
In single-phase two-pulse (or full-wave ) converters, voltage at the output terminals can be controlled by adjusting the firing angle delay of the thyristors. Mid-point or bridge-type circ uits may be used for ac to dc conversion. In this section, first mid-point and then bridge-type configurations are discussed with input from single-phase source.
/
. 6.3.1. Single-phase Full-wave Mid-point Converter (M-2 Connection) Th e circuft diagram of a single-phase full-wave converter using a centre-tapped transformer is shown in Fig. 6.8 (a ). When terminal a is positive with respect to n , terminal n is positive ..\ith respect to b. Therefore , VII" = V" b or Villi = - Ubll as n is the mid- poin t of seconda ry '.vinding. Equ ival en t circuit of this arrangement is shown in Fig. 6.B(b ). It is assumed here that load, or out put, current is continuous and turns ratio from primary to ea ch secondary is unity. VC"
W!
Q
+
T1
10 1
w!
vo • I
T1
•
--1[>(
.11
b:'= '
(0' v tlro
. -
'7 v - , -
g~ : -. h , I
-
n
.
-.
~I ~ -
'"\..
+
vl::on
-
: ... V"-.
e
12
R
l
(b )
+
2IJmsi no
i
(c)
Fi g. 6.8. S ingle-phase f'.ll1 - wav~ m id·point convert er (t.l) ci rcuit diag-rarn. (b) equ ival en t ci rcui t and (c ) va riO\lS volta ge and cu,pcn~ w:! ',· i!iorr.5'
264
P ower Electronics
[Art. 6.3] Thyri 5 tcr~
T l and T2 are forwa:d biased during positive and negative half cycles r espectively ; these are therefore t:riggered accordingly. Suppose 12 is already conducting. After Wi = 0, va." is positi\'e, I I is therefore forward biased and when trigger ed at delay an gle a . TI gets turned on. At this firing angle (x, supply voltage 2V"m sin ex reverse biases T2. thi s SCR is therefore turned off. Here Tl is called the incoming thyristor and T2 the outgoing thyristor . .~ the in:: ":':n i:-;g SCR i~ triggered. ac supply voltage applies reverse bias across the outgoing thyristor and turos it off. Load CUTTed is also transferred fr om outgoing SCR to incoming SCR. This process of SCR turn off by natu.ral reversal of ac supply voltage is called n a. tural or lin p. commu.tation. From the equivalent circuit of Fig'. 6.8 (b ), it i ~ s-een that if flo" = V m sin (Ot, then ub" =- u ,,~=- Vm ~i n Wi and r' '1'::. = I.: c" . "',.!, = 2V;"I. s;n (·,t .-~
When rot = ex, T1 is triggered . SCR T2 13 subj ected to a reverse voltage uob = 2Vm sin a as stated before ; current is transferred f!"om T2 t o T1 and as a result T2 is turned off. The magnit ude of reverse voltage across T2 can also be 0bt.ain~ ~ b;.-- applyin&, KVL t C' the loop /!fgb.p. of the equi-.:alent circuit of Fig. 6.8 (b) at ~:,,~ l'1st.ant Tl is trigge!"ed. Thus. I)T2 - '1 /:" . ...·011. - !..·Tl = IJ ~
~=~-~~~
With Tl conducting, un = O. Therefc!'e. the voltage across T2. at the instant
(t)t
=
a is gi...-en
by 1) T2
=- v.'" sin 0: -
VIn sin ex = -
2v-... sin a
This sh ows that SCR T2 is r ever.<;e biased by vo!t.age 2V.., sin a and it is therefore turned off at rot = a . Thyristor Tl conducts from a. to ;-:: • a. Mer ,.:)/. = 1':. Tl is reverse piased but it \...ill continue conducting as the f'Jrward biased SCR T2 is n ot. get gated. At WI. = j t . a. T2 is
triggered, Tl is reverse biased by voltage '.)f magnitude 2\lm sin 0 , current is transferred from Tl to T2. Tl is t herefore turned C'ff. At CJ)t = a. T2 is turned off anel it r e main~ r ever~e bias-<;;:d from (ot = a. to r.. thj ~ c ~.n be seen f!'om Fig. 6.8 (c). The turn-off time prC'. .-;.ded ?,y this circuit to 8CB. T2 1<:: t~'O'r.efo:-'£ gh-c.:n by r.-(l t ,, = - .
'"
...{'3.24'
~ec
Thyristor Tl is turned off at wt = 1't. a md Fig: . 6.8 {c) re .... eals ~h :;>.': T-;, is ~ '.l'Jjec ted to a reverse voltage from OJ!: = l't -I- a to Wi = 2l't. Therefore. this circuit provides a tu:-n-'Jff time to thyristor Tl as
te =
2;;; - (r,: • a ) ' :-.: - CI. (.)
=-CJ)-
which is the s ame as provid ed t l) thyrist or T2 : Eq. (6.24,). It is seen from voltage waveform I.'c . Fig . 6. &l.c) . that average value of -:)1.ltpu t \·ol~ a ge i!'=
gi ven by 1 'I/o =-Tt
. (1 --:
I
. (1
2\. Vn-. sin ox . d (oot) =~ cos a ;t
.. .(6.2.5 '
The circuit turn -off tim e t el E q. (6. 24), as provi ded by this circu it of F ig. 6.8 {<::.' mu st be gr ea te r t ha n SCR t urn-o ff ti me t'l a s gi ven in th e sp eci fi cation sh e!;' t. b case t e "-": t 7 . co mm uta tion fai lu re w ill occur and the whole s econdary win ding \\ill be sh ort circuited. During com m uta tion fai lur e. if th e .r a te of rl 3e of fa ult curren t is high. t he inc omin g SCR may be
Phase Controlled Rectifiers
[Art. 6.3J
265
damaged in case protective elements do n ot clear the fault. Fig. 6.8 (c) reveals that each SCR is subjected to a peak voltage of 2V m' , The follo..ving observations can be made from the above study. (i ) \Then commutation of an SCR is desired, it must be r everse biased and the incoming SCR must be forward biased. (i i) Wh~n incoming SCR is gated on, current is transferred fr om outgoing SCR t o incoming SCR. (iii ) The circuit turn-off time must be greater than SCR turn-off time. It is seen from above that thyristor commutation achieved by means of natural r ever~al of line voltage l called line or natural commutation, is simple ; it '\ s therefore employed in all phase-controlled rectifi er s. ac \'oltage controllers and cycloconverters. V 6.3.2. Single.phase Full-wave Bridge Converters Phase-controlled slr.gle-phase, or three-phase, full-wave converters are primarily of three types ; namely unc o:':".troll ed converters, half- controlled converters and fully- con trolled converters. An uncontrolled con uerter or rectifier u~es only diodes and the le.vel cf dc output voltage cannot be ccntrolled. A half-controlled conUl!rter or semiconverter uses a mixture of diodes and thyristors and ther e is a limited • v, control over the level of dc ou tput voltage. A f'J.llY-':l)ntrolled conli<:::rtl!r or r.J.ll cOnl.:erter uses thyristors only and the~e is a ""ider control over the level of dc output voltag~. • A semiconverter is one·quadrant converter. A one-quadrant converter ha::: one polarity of dc ou~p ut voltage and curren t at its output tenninals, Fi:?". 5.9 (a). AtwfJ.qu.adrrmt conv~er iF O>lO? i'1 wbich v('Itage polarity can reverse but current direction eanD.')t r everse because of the (. ) (b ) unidirectionO"j nature of thyristors. Fig. 6.9 Fig. 5.9. (a ) One-quadrant cenverter and (I,}. In this part of the se<:~on, single-phase (h: two-Quadrant . co~,'ert.er. bridge type full converte~ and semiconve:rt.ers &-r:'e studiei in detail. 6.3.2.1. Single-phase full converter (B·2 Co nnect ion) As.ingle-ph;:.~e full converter bridg2 using-fcur SCR~ is shown in Fig. 6.10 (0). The load is ass'.1med to be of RLE type, where E is the load circuit emf. Vo : ta~e E may be due to a battery in th e load circu it or may be g-enerated emf of a de motor. Thyristor p air Tl. T 2 is simultaneously triggered and n: r adians later. pair T3, T4 is gated t egeth er. When a is positive with respect to b. supply voltage waveform is. sho\. .'l1 as L''ll) in Fig. 6.10 (b). \Vhen b is positive with respect to 12, supply voltage waveform is shown dotteci ~ L·~g . Obvio1,lsly, L'tV! = - t: ,-,~. The CU':Tent dire c ~ion!' and voltage polarities sho ....'l1 i.n Fig. 6.10 (0) are treated as positive. Load c'Jrren t or output current it' is assumed continu'Jus over the working range ; thi~ means that l'n d is alw ays connected to the ac voltage source through the thyrist'J!"s. Between rot = 0 and C!.'l = CL ; T1, T2 are fo:ward biased through alre ady co nducting SCRs T3 and T4 and bl ock the iorward voltage. For continu ous current. thyristors T3, T4 conduct after oot = 0 ever. though these are r everse biased. ""nen iOT\l,; ard biased SCR.5 Tl. 1 2 are trig-gered at (·x = 0:, th-:-y get t·..trned on. A$ a result. supply "'oltage Vm sin n immediately appears across thyri~ ~o rs T3. T4 'a3- a "'€'vers~ bi as . thes~ are ther efore tur!'led off by naturaL or line. commutation. At the same ti:ne. load current io flowin.;- tnr()llgc. T3. T4 i ~ tran.~ fe r!'E'd to Tl , T2 at rot = CL . :\ote . ~.
I
266
[Art 6.3J
Power Electronics
that when TI, T2 are gated at wt =a , thes e SeRa will get turned on only if Vm sin ex > E . Thyristors T1, T2 conduct from rot = ex to 1t + ex. In other words, TI . T2 conduct for 7t radians. Likewise, waveform of current iT} through Tl (or in through T'2 ) is shown to flow 1t radians in Fig. 6.10 (b). At wt = n + IX, forward biased SeRs T3, T4 are triggered . The supply voltage turns off Tl, T2 by natural commutation and the load current is transferred from Tl, T2 to T3, T4. Voltage across thyristors T1, T2 is shown as un = uT2 and that across T3, T4 as un = uT4' Maximum revers~ voltage across Tl, T2, T3 or T4 is V m and at the instant of triggering with firing angle a, each 8-CR is subjected to a r everse voltage of V m sin a. Source current i, is treated as positive in the arrow direction . Under this assumption, source current is shown positive when Tl. T2 are conducting and n egative when T3, T4 ar e c0:t:lducting. Fig. B. 10 (b).
,,
E
,,.!
.
~ , ....'
,
,
,' I
WI
/ I
T1
I 13
I T2
T4
,
Output vollog~
Volo-!
I !
!
I r'
•
I .
WI
I:
wI
~.".~
.1.:.-,,+ .
i! .
!
I~
jj
lIoltog~ ocross TI
0'
T2
o'~2~~__~ I T:-'~~__~~__~____ WI
wI
(a )
(0 )
Fig. 6.10. (0) Single-phase full converter bridge with RLE load (b) voltage and current wavefonn.;: for conti n ~ous load current.
Phase Controlled Rectifiers
[Art. 6.3]
267
During a to ft, both t:, and i, are positive, power therefore flows fmm ac source to load. During the interval 7t to ( 7t .. a), L'~ is negative but i~ is positive, thE> load therefore returns some of its. energy to the supply system. But the net power flow is from ac sou rce to dc load becau~e ( " - (X) > (X in Fig. 6.10 (b) . The load terminal voltage, full-converter output voltage. Vo is shown in Fig. 6.10 (b). The average value of out put voltage Vo is given by
or
V I)
1
=;:
f!t . . ", V". .sin rot . d (wt ) = 7 2F (l
Rm s v:ll1..le of output voltage for be obtained as und er. ., 1 vo~=-
sin£le-phas~
J't.(t V",sm
"
2
V' [
O "
... (6.26)
/\/ -2. or B-2. controlled converter can also
Oo'l/.d(wt)
a
= 2ft- wt - "21 V
. 2
co~ a
, 1- · .. 1' v'... V' 1 5102(.)/ 'l J=T = ~ ... (6.27)
~ V,
Eq. ('3.25'1 show~ that if a> 9(F . v~ is v j rtegati\"'? T.,,;!' i~ illust r ated in Fi~ . ~ . H' (c ' . • where a I$: ~hown gT''? ater than ~(\~ . It" thi!' figu~ ,:, . ave rage te rmina l volta,£c \"-n is: o {,-- -X -~ )'negativc. If the l oad ci rcuit emf E is re-vcrseo. t his source E \\-ill feed power back T) T' 'J "''2 T4 OY1o u! '1"2 t" ac supply. This t'peration offu ll c('n"-erter is knl)wn <''$ irl\'ert~T ('perati-:-n ';'if the conv,:,rtc ... The f'J1I Cl)nv~rter with fir-ine angle delay gr~ater thar.. 90-: i~ called li ll'! -t:.omm.lJf. o t '!d int; uter. Such an 'Jpe r atiC' o I~ lJ $ed in t h e r l?g"'n~rative Ovtout tu ..... ,,~! i;I{':;lk.if'ls:' "'Nfe of a de mt:'tor in which ca.c:e ~hE.'n E i ... t:."'J n~ E' r '?mf of the dc m"tf'lr. _ 'C' __ ' \\ '. Duri"s: 0 to 0:). . ac ~('lJrce ,","ll t rt(' '' of FiJ;:. 6.1 0 r.:) inr Q.::- 30i n v .~!':~ r rr :e.c:. : h.\"ri~t o rs must be forwa rci
.,
,-
268
[Art. 6.3]
Power Electronics
biased and current through SCRs must flow in the same direction as these are unidirectional devices. This is the reason output current io is shown positive in Fig. 6.10 (c). As before, source current is is positive when T1, T2 are conducting.
The variation of voltage across thyristors T1, T2, T3 or T4 reveals that circuit turn-off time for both converter and inverter operations is given by
x-a t, =--5ec ro As both the types of phase-controlled converters have been studied, the advantages of single-phase bridge converter over single-phase mid-point converter can now be statecl. : (i) SCRs are subjected to a peak inverse voltage of 2 VI'll in mid-point converter and V 1'1\ in full conVE-rter. Thus for the same voltage and current ratings of SeRs, power handled by mid-point configuration is about half oft..~at handled by bridge configuration, see Example 6.6. (ii) In
mid-point converter, each secondary should be able to supply the load power. As such, the transformer rating in mid-point converter is double the load rating. This, however. is not the case in single-phase bridge converter. It may thus be inferred from above that bridge configuration is preferred over mid-point configurati on. Howe.... er, the choice between these two types depends primarily on cost of the various components, available source voltage and the load voltage required. Mid-point configuration is used in case the terminals on dc side hav~ to be grounded. ExamplE' 6.6. SCRs with peak forwcrd uoltage rating of 1000 V and auerage on·state current rating of 40 A are used in single-phase mid-point conuerler and single-phase bridge conuerter. Find the power that these two conlJerters can handle. Use a factor of safety of2.5. Solution. Maximum voltage across SCR in single-phase mid-point converter is 2Vm , Fig. 6.8. Th erefore, this converter can be designed for a maximum voltage of 21~~~5
=200 V.
:. Maximum average power that mid-point. converter can handle (2Vm =l-x-cos a
I
1 _ yTAV= 2x200 • x40x 1000=,·093kW ·0
SCR in a single-phase bridge converter is subjected to a ma.ximum voltage of"'... , Fig. 15.10. Therefore, maximum voltage for which this converter can be designed is
1000 =400 V 2.5 :. Max;mum average power rating of bridge converter 2 x 40C
= l OCO x x x 40 = 10.186 kW 6.3.2.2. Single.phase 'semiconverter, A single·phase semicon\' ~ rt e r ,ridge wi L~ t;.."·o thyristors and three diodes is shown in Fi g. 6.11 (a). The two thyristors are Tl, T2; the ~'O d.i odes are Dl, D2 ; the third diode connected across load is freewheeling d.iode FD. The load is of RLE type as for the full converter bridge. Various voltage and currant waveforms for this converter are sho\\>-n in Fig. 6.11 (b), where load current is assume d continuous over the working ra."1ge . After ~t = 0, thyristor T I is ion vard biased only when source voltage Vm sin IJ.)l exceeds E . Thus . T l is tri gger ed at a fir ing ang!e delay a such that V III sin a > E . With TI on. load ge ts
[.-\rt. 6. 3J
PhiJ.j ( Controlled Rl!ctiti ers
wt
wt•
wt ,
,W
~1T-cr-l
• wt
I !
D ,
'..
"
~TI .,.! T2 i . "f tt la
"""1r '
~"
I
1
T ,, ..;;,
Ib FO +
I
I
'---+- - 4 02
01
~)
, , ..."
:\ ~;r""~J
,,-----·~--Tr_=_."";;"1-:;:--i .1 -
is.
N r -,
:'\
wt
::1"11'
,
:(2,..,.« ) I
R
, :'
'T~[
wt
I.
lJ
i
"
I
'CJ
wt•
~
Fig. 6.11. Single-phase s emiconverter bridge (a) Power-eircuit diagram with RLE load and , (b) voltage nnd current wavefonns for continuow load current.
connected to source through Tl and Dl. For the period wt = ex to It, load current io flows through RLE, Dl, source and TI and the load termina.l voltage 1)0 is of the same waveshap e as the ac source voltage US' Soon after rot = It, load vol tage Uo tends to reverse as the ac source voltage changes polarity. Jus t as Uo tends to r everse (a t wt = It T ), FD gets forward biased and starts conducting. The load , or output, current ill is transferred from TI, DI to FD. As SCR Tl is reverse biased at wt = 7t T through FD, T l is turned off at rot =It +. The waveform of current iTt through thyristor TI is sh own in Fig. 6.11 (b). It flows fr om ex to It, 21t + ex to 31t and so on for an interval ofCn - a) radian s. The load term l11.als are short circuited through FD, ther efore load, or output, voltage tlo is zero during it < wt < (it ~ a). After WI. = It, during the n egative h alf cycle, T2 will be for.vard biased only when source voltage is more than E . At CL'i =It + a, source voltage exceeds E , 1'2 is therefore triggered. Scan after (It + a), FD is revers e biased and is therefore turned off ; load cu rrent now shifts from FD to T2, D2. At wi =21t, FD is again forward biased and ou tput current io is transferre d from 1'2, D2 to FD as explained before. The sourc e current !J is pC3itive fr om a to it whe :1 T I , DI conduct :lnd is negative fr om (It + a ) to 27t when T'2, D2 cond uct, see Fig. 6.11 (b ).
27u
[An. 6.-'1
During tn .:: Interv:.:al c£ to It, Tl an d. Dl conC1u..:t and ::.~. ".)urce ddiv.::r:: . ..:f.cr~y to the luau '::lfCuit. T hl::i cn.:r:;y is partl iiUy "tofo::d In indu .::: t:in .:: ~ L, r..:...rtl:iUy ::,t vfi:C il:::. i::":":[L" l ~' 2r1c: rgy in iUilCi-l.:lfCUlt I:mf E and parri'-llly dissipated ;AS ~..::::J.t .... ",
'r•• I
R . Dunn::: the tTt." ewho.:dmg: pl:nod It t li tit ..... oJ. " ..:n.:!rg"Y st u r ~.J :n ind •.u..:tan..:o.: b f .;:..:v .... <:f t::d. :in..:!. IS p art ially dl:isip;).t..:d 'itl R Hnd partially added tv ttl":
In
" ",
", .....T. i, ,, -, , , r ... -----., ,
cnl!rg)' stored In IV:id t:m t' E . No enerb'Y h Jcd. t.a..:: k t o the .sou r..:e during freewh eeling period .
For
V,~,
,,
i:i~micu n vl?-rt c r ,
the
vo! . .
" -',
'
"
sin
:!.wt
L;'", ~ ;t
.
V.! .r
1
=-.' I
-,
:. IJ.. _.; .• ......
-
CJ.)
T
---
Fig-. 6.11 . \cj Convt:ctcr uutput \' Cil~gc as a (unction ot' nnr,!::" ::in~l.:: t~r s.::mi' and
W ( - - - - , :': ! " :!
, 1. , .. ! Y,
,
--'r
n "
V ,,," =
,,
I
2vrr
," V:.· = -l J"'. V~, sm- we d (wt )
"
,,
i
and rms valu.: of Output voltage is
:1:1: '
'.0'::
,,
,.
1 "' , V VI/ = -n L VII.::'lnwc · d(oot)= ;" ~l .,.. t.:v"u)
= _ !.!.: '
..
,", ' .,
... _ -;; ;:.'T
-~-
•
..
~ r.
-
[uU,.;vnvo.:rti:r:il. ::.,,'n ')'" _..... -j Ct.)
T
-
'-.
: - -
::
"
Thl! vdriatiun of voltag!: acro::.s Tl and '1"2 i.::i abo dtlpn:to::d, in F ig'. 6 .11 ,bi. It i::. ",..:.:.:n irum thl::::.e w;lvli:forms that circuit·turn olf tim~ for the !)cmiconverter is
n-(.( t, = w-
s~c
The variation of average value of converter output voltage as a function of firing angle a is shown in Fig. 6.11 (c) for semiconverter and full converter, 6.3.2.3. Analy~is of two-pulse bridge converter with continuous conduction. In this part of th~ section, steady state analysis of single·phase two-pulse converter of both the types is presented. S ~ miconv e rter . During c{Jnduction period, the voltage equation for the circuit of Fig, 6.11 (a ), b
. dio uo = u~ = R La TLdt +E fo r
",(6,30)
Ct.
,
During fre ewheeling period, the voltage equation is
dio O=R io+LdtT E for
It
".(6.3 1)
< wt S (1t + a)
Eqs. (6.30 ) and (6.31) can be solved in ti me domain if requir ed. Und er ste3dy-state co nditions, only average valu e of output voltage and current are requ ired, Therefore, fr om these two 2quation ~, we get steady·state solution as
op ~ r ating
Va = Rio + E
",(6.32)
Pha..s~
Controlh:d
Vo = V m (1 T cos
where
I" =
0.)
"
3v~ragl! luad
=a\'~rage \'oltage appl il!d to the: load
\;uTI"t' nt ; E = L.>;,J..:1 cir.: uit . .·mt
In ca .:5 ~ loo.d is a dc motor, then E =Km w,,,; R a rm a ture current ; T" = K", I". T~
wh~ r e
211
[AI"(. 6.3}
R~ctili~l"s
=r (..
arm&tun:·~ir~uit rl;:~ is tan'::l! ;
f;. =1."
= electr omagnetic tor que in Nm.
Km = torque const ant in NmlA, or em! constant in V-sedrad Vo=r,)g TKmwm
V. -rJ
= "
or
Will
or
w /J;-
"
_ (V,,/:-c)( l
T
cos
K
QJ
_!..:.... T
m
K~
'"
Full·converter. Th~ voltage equati.>n, for the circuit of Fi g. (6 .10, (..
J
= t·
~
:.:: R~ \J
...\6. 33,
.: lCL I,
i.i
di J , L ",1:.' t.!t
2 V'" V\I =- - ClIS Q
where
"
In case load is a dc motor, Vo = r)g ,Km{J)", ... \6.341
or
A comparison of the waveforms for iTt and ioJ in Fig. 6.10 (b ), fo r n single-phase full· converter,
n.:v~al::l
tha t in =
~ io tor continuous
:. Average value of thyristor current
I., .....
or
load current .
= ~ ( aver3g~ v alu ~ of load current )
=-21 I "
... \6 .351
Th is. h owever. is n ot the case in single-phase semiconverter. Exa mp le 6.7 . A s ingle-phase fu ll conuerter bridge is connected to RLE load. The source volcage is 230 V, 50 H2 . The auerage load current of 10 A is conscant ouer the working range. For R = 0..1 nand L = 2m H, comp ute (a ) fi ring ang/e del"y for E = 120 V,
(0) fi ring angle dela:1 for E = - 120 V. Indicate which source is delivering pou:er to load in parts (a) and (b ). S ketch the time oi Ofl~p u t vo ltage and load cu rrer.! fo r both the parts.
~·C:l'ia!iofl..s
Ie)
h {'cse Otltp :a current is cssu fr'.:! d cO.'1stcn t, fi nd the inpflt pljor bo th parts (a) and (bJ.
[An, 6.3)
Solution. (0) For E = 120 V, tn !: fu ll conv erter is opera ting as a contl'olled rcctiilcr. " V ~co:sa=ETI;ft
or
"
.) . {2 . ->30
or
-
"
-
COS " = 120 T 10 x 0 .4 = 124 V
= 53.208 =- 53.2:1': For a = 53.:2 1", power tlows from ac 50urce to de lllsci . (b ) F or E = - 120 V, the full converter is operating as a l ine comm uta t t!C1 mvertcr. 0.
:2 . v'2 . 230
"
cos a. = - 120 T 10 x 0.4 =- 116 V
or a = 124.075=: U4.1;;'
For a = 1 24 . 1 ~ , the power flows from de source to ae load. Output volt::...ge and load current waveform:; for ex = 53.21 CI can be drawn by rererring t.o Fig. 6.10 ib) and for a = 124.P from Fig. 6.10 (c ). (c ) For constant load current, rms value ofload current I"r is 1,,=1, = lOA ..
V.. · /orcos¢l=El o+I!,R
. _ 120 x 10 T lO~ x 0.4 _ ~ ~.) · ll ~ ..~ "0 I " -u,., ... ':t a ., . :..;:S X u
F o'r ,.....' , = ~-'3 ,'_; 1' , ~'\..{"
Cc.!)V-
., . . .. _ :..20 X Iii - 4 0 _ 0 _,..
U = 12:4 .1;,
.... v:>
V -
2::s0 X 1u
-
?
" .:;
.:lv4 ... b o o
Example 6.8. (a; A :iingle-phase full conuerter d~li vus p&wer co a resistive load R. Por ac ;source voltage V$' show that average output voltage Vo is given by ..f2V$ Vo =- - (1 Tcas a )
"
Sketch the time variations of source voltage, output voltage, output current and voltage across one pair of SCRs. Hence find therefrom the circuit tum-off time. (b) For the converter of part (a), show that rms value of output current is giuen by
H
1,,=~ ("-a)+~Sin2a} Solution. Time variations of source voltage, load voltage and load current are shown in Fig. 6 .12. At wt = 1t , Vo = VI = 0 and for resistive load R, iJ
v.
=R
r
vo
w.
= 0 and io = R = 0,
soon after wt = r.:, supply voltage reverse biases TI , T2; this pair is therefore turned off. When T3 , T4 i3 triggered at wt =it + a , output voltage /,10 = uJ up to Wi : 21t. Note th a t n o SCR conducts during 0 t o a , 1t to (1t + a ) and so on, Fig. 6.12. F or th e ou tput volt a ge wa veform " 0' a ver age ou tpu t volta ge Va is V,
1
•
! I
' :'1:
=-1t J
Q:
V m. , in 0Jt ' d (col )
€v,
,
= - - (1..,.
C0 3
a)
Fig. 6.12. Per taining to Example 6.8.
Phase Controlled Rectifiers
273
[Art. 6.3 ]
The other waveforms can be drawn by referring to Fig. 6.10 (b ). (b ) Rms value of the output current can be obtained from the waveform io shown in Fig. 6. 12.
~ 1: (; R ~ 1:
I" = [
=V [
sin oX
J
d(oX )
r
(1- cos 2oX) d (oX)
]'12
=i[ ~ { (x-a)+~ sm 2a}]'" Example 6.9. (a) A single-phase controlled rectifier bridge consists of one SCR and three diodes as shown in Fig. 6.13 (a). Sketch output voltage waveform for a firing angle a for the SCR and hence obtain an expression for the average output voltage under the assumption of constant current. Show the conduction of various components as well. (b) Draw waveforms of current through Tl, Dl, D2 an.d D3 osewrW:tK conatq..n t load current. (c) For an ac s"Ource voltage of230 V, 50 Hz and firing angle of 45°, find the average output current and power delivered to battery in case load consists of R = 5 n, L = 8 mH and E =lOOV.
Solution. (a) F or the circuit of Fig. 6. 13 (a). output voltage waveform Vo is shown in Fig.
V '! ~~~----~---\----,-", ~
g~~T1D' ""'0203-!~~T1D '+ 0203-1[ ··,, 0 :I :' i I i[ I I , I rio i 0 , ,, i 1
·
ii I!
I
I,
io TI
OI
~1+----i
o
t,
i
03
, ,
i ,
~~+---~----~:--~,----~-fa:
W
(b)
wI
[
i01H: _ Fig . 6.13. (c) Ci rcui t di agrn::: for Ex nmple 6.9
I
,, ,
wI
Various
. H::)r" - - - - - I~' <;
vo1t~ge
211'
W
J,
and curre m 'l.'o·:::onn3.
274
Power Electronics
(Art, 6,) J
6.13 (b ). The co nduction of various. comPQnents is also indicated. It is seen that awrage value
of
Vo
is given by
"
Vo = 21. [ ( Vm sin wt , d(wt ) -
L"
Vm sin wt ' d(wt )
1 ~~ =
(3 + cos al
in
(b) The conducti on of various elements shown h~lps ~rawing the waveforms for currents through Tl, Dl. D2 and D3. For example, D3 conducts from wt = 0 to a, from 1t to?1t + a , from 31t to 47t + a and so.on ; this is shown· as iUJ in Fig. 6.13 (b).
Vo
(e)
=
'.
-12 230 2 (3 + cos 45°)
10 = i91.8~ - 100 o Power delivered to battery
'
=191.88 V
= i8376 A
'='tno = 100 x 18,376;" 1.8376 kW , ,, 1000 '"
Example S.lU. A single·phase full converter feeds power to RLE load with R = 6 n, L = 6 mH and E = 60 V . The ac source voltage is 230 V, 50 Hz . For continuous condtiction, find the average value of load 'current for a {iring angle delay 'of 50°_
In case one of the four SCRs gets ~pen circuited due to a fault, find the new value of avei age load Cllrrent taking the outpllt current as continuous. ." Sketch waueform for the ' new ' output 'volta'ge 'and indicate the conduction of various SCRs.
vsl :....
Solution.
,'-+-'---+----,I;-;---t.;--~ wt
2 Vm 2-12 , 230 Vo= - -cosa= cos 50"
• •
I
= 133,084 V
(
'0
o 10 = V "- E = 133,084 - 60 ,R 6 = 12,181A
I I I
.
('71'+0)
I I-
. ( 311 +0: )
S uppose SCR T3 in Fig. 6.10 (a ) is damaged and t a) fT wt is open circuited. With this, output v'oltage waveform T1T2 -I-- TIT 4 TIT, ~ Uo is as shown in Fig, ~ . 14. Initially, suppose Tl, T2 tT1T4 are conducting from 0. to 7t + 0.. At wi = 7t + Ct., when T3 T4 are gated, only T4 is turned on and as a result, load Fig. 6.14. Pertaining to Example 6.10. current fre ewh eeling through TI, T4 is zero till T l , 1"2 are triggered again at wt = 27t + a. For this waveform, average output voltage is given by 1 Vo =- 27t
Vo =
f' " V U
-12·23 0
•
In
sin wt . d(wt) = V; ' cos a .
_ cos ;)0 0
10 = 66,54: - 60
27 [' (2:J
-
= 66.542 V
= 1.0903 A
It is scen t hat load current is reduced radically wit h one SCR getting open circui ted . It is also observed t hat th:y-ri stor Tl remai ns on.
Phase
Controlled R ectifiers
[Act. 6.4 1
275
6.4. SINGLE-PHASE TWO·PULSE CONVERTERS WITH DISCONTINUOUS LOAD CURRENT So far, sin gle-phase two-pulse converters have been studied on the assumption of continuous load current. In practice, the output current may become discontinuous at high values of firing angle or at low values of load current. The term discontinuous is applied to the condition when load current reaches zero during each half cycle before the next SCR in sequence is fi red. The term continuous means that load current never ceases but contin ues to flow through SCRJdiode or their combination. The load performance deteriorates if load current becomes discontinuous. It is therefore preferable to operate dc load in continuous current mode. This is prom oted by having freewheeling action and using an external inductor in series with the load. In this article, working of both single-phase full converler and semi converter is studied with their load current discontinuous. 6.4.1. Single.phase Full Converter with Discontinuous Current Power circuit diagram for a single-phase full converter is shown in Fig. 6.10 (a ). For this converter, when SCR pair T1 T2 is triggered at rot = n, load current begins to build up from zero as shown. At some angle ~, known as extinction angle, load current decays to zero. Here P> It. As T1 and T2 are reverse biased after wt =Tt, this pair is commutated at rot = P when io = O. From a: to ~, output voltage Vo follows source voltage V!, From ~ to (rt + a ), no SCR conducts, the load voltage therefore jumps from Vm sin ~ to E as shown. At wt = 1t "T" Ct , as pair T3 T4 is triggered , load current starts to build up again as before and load voltage 1J0 follows u! waveform as shown. At It + p, io falls to zero, Vo changes from Vm sin (7t +~ ) to E as no SCR conducts. The source cu rrent i, is also shown in Fig. 6.15 (a). v,
lIab Ym
.
#_,f'''bO ", ' I E
wt
Ii
i,
, ,: I: TJt7I \;--TlT2 --:Y,__ T3T4 ~ !-T ITZ I No SCR J1 : II : 0
:
II ' II ;
,
I',
-t :-~
II :
:
I conduc ts (a ) rr < (3 < (rr t o ) II ; i oWe... /.l --oJ " I I, ': t: :, : y /? I I, ..... ( 1' + 1:: ): : ..... (2'1'-t OJ I : 1.
a'..
I :-' "0 ,
o
-----: ! : ir! .
,''
',:
'
"1 1 T3T-4
--+-1
I1
I' I I, t
'I
· ~ rr
1[ 11'+/3 )
11
I
'!
2"
(b)
ti l:
1 I •
P< r: a nd V m sin P< E
wt
I•
~
3JT
wt
4'11'
•
'"'
Fig. 6. r5. Voltage and CU ITe:1t wO\vi:::orm.s for di.scon:'inuo'.ls load currer.t for a s i..!'l.gle-pha~e fl!!1 co nverter.
175
[A r .. 6A]
Power Electro nics
Und er some conditions, load current may become zero at wt =~ , \vhere ~ is less than It . It is assumed here that V m sin P< E. At P, Vo jumps from Vm s in ~ to E . The waveforms for load current io and load voltage Vo are shown in Fig. 6.15 (b). No SCR conducts from ~ to (n + Ct. ) and during this interval, therefore Uo =E. From above , the following observations can be ma de : (i ) Conduction period, a < oot < ~, Tl , T2 conduct and ~' o = u~ . Also (Tt + 1t. From a to Tt, Tl Dl conduct and Vo = vS' At Wi = 1t, as v, tends to become negative, FD is iorward biased and starts conducting the load curren t. "Vhen FD condud~ fr:2ll\.1t to ~, Vo = O. Y,
:
,:
Vm
"
-",
"
.,
"h=
10
I"' i
: ( a)
/J--;,jI::
./""'-"....i .
1:
< /3«11" -to)
• ...-,. , ,.....-:\"
,{:r+/lJ :
.,
I Ii :
.~ I
I"
I' ~ I
"
"J!
:ig. 6.16. Voitage (l ~d C1.::-rent waverorms fer diseon:in :':Jus condu ctio:! for a sb gle·ph ::'!.se 5t!miconv er ter.
[An_ 6.J]
Phruc Coni roll ed Rectifie rs
277
From p to it + a, no circuit componen t condu cts, therefore l'O = F as shown in Fig. 6.16 (a ). Du:-ing ~ to it + a, as load current is zero, this makes the load cu:-rent discontinuous. When T2 is tr iggered at 1[ + a, io builds as shown. At 2it, FD is forwaru biased and starts conduc[.ing till it +~. During the time FD conducts, lIo = O. From 1t + P to (21t + Ct) , no circuit component conducts, t herefore LID= E. At l2n + a), T1 is triggered again and the above pmcess rcpeat.i . Source current i J is also shown in Fig. 6.16 (a). In case load current becomes zero before it , i.e. for P less than It, then the current and voltage waveforrrs are shown in Fig. 6.16 (b). Here Vm sin P is assumed less than E. Duri ng p to 7t + a, no cir:uit component conducts, therefore Uo =E . From the waveforms for single-phase semiconverler, the following observations are made: (0) When
It
<
P< It + a :
(i) Conduction period, ( ii ) Freewheeling peri od ,
(iii) Idle period, (b ) When
a < rot < It, TIDl conduct and lI a = lI J . Also for it + a
p < rot < 7t + a, no circuit component conducts, io = 0 and
P< 'It and V rn sin P< E
lIo
=E
:
(i) Conduction period, ex < cd < ~, T1D1 conduct and lIo = Vol' Also
for
7t
+ ex < rot < 7t + ~, T2 D2 conduct and lIo =v, and so on
(ii) Freewheeling period, absent and ifi! = O. (iii) Idle period , ~
io =0 and lIo =E.
< wt < It + ex and 1t + ~ < Wi < 21t + a, no circui t elem e nt condu cts,
The output voltage during discontinuous conduction is not given by Eq. (6.28) . Th~ load performance with discontinuous load current deteriorates as stated before. Average output volt age and current. For single~phase full COTlverter. for P> 7t or ~ < It , the average load voltage VD, obtained from VDwaveform in Fi g. 6.15 (a) or (b ), is given by V, =
[f
1 "
=Vm It where y =
Vm sin oll d (oll) + E (n + a -
p)J
a
[cosa-COSP) +E ( 1- 1 ) It
P- a = conduction angle
nR
V, , --. (cos a - cos ~) + R E ( 1 - it' 1} Average load current,lD=Ii =
;;i' . sin gl~ph3Se semi corti.lerler
......i th ~ > 7t , the average Quqmt voltage VD, obtained from vD\Va ... ;. ::.;.1 :: Fig. 6.16 (a), is given by
", =.!it
[r
• Cl.
11_ si n wi . d (CJ::) .;- E ( : -'- .:.: ".
= ': ,1 + 'os c:.) +E (1_11) It I.
A·;e rage load cu;--:~:'.t,
VO V E( -, , lJ = R = r_R (l -)- cos u:) + R 1 -+- ~ )
;)}
Power Electronics
[Act. 6.5J
278
6.5. PERFORMANCE PARAMETERS OF TWO·PUI.SE CONVERTERS
• parameters of single-phase full converter and single-phase Here the performance semi converter are derived from their input and outpu t waveforms, already f.~ ~ained . The load , or output, current is assumed continuous during these derivations. . In general, the instantaneous input current to a converter can be expressed in Fourier series as under : a
I
i(t ) = a" + /I '"
where
(an cos nffit
+ bn sin nwt )
1, 2, 3
a.
1 r" =-2 J
a,
=1.lto
• • i (I) d(",,)
t
itt)
cos nwt d
(wi)
and b, =1.
t
itt) .
sin nOlI. d
(wi)
fto The performance parameters are now obtained first for single-phase full-converter and then for single-phase semiconductor. 6.5.1. Single-phase full converter In Fig. 6.10 (b ), the variation of input current, or source current i~, from a to (rt + ex). from (rt + ex) to (2rt + ex) is continuous but not constant. Here i l is assumed to be ripple free with ." amplitude 10 during each half cycle. where 10 = constant load current. a
i l (t ) =10 +
I
CTI sin (nwt + 8T1 )
" '" I , 2, 3
where Now
C, =
~a; +b; and e, =tan- 1 ( : : ) 1
I . = -2 [ 11
a"
FHa I •. d (OlI) - j,aI , . d (0lI)1= 0 0:
11 .. 0:
=-111 [J"a 1 0
0,
I
=_ 0 nIt
cos nwt. d (wt ) -
J211: +CI !t+ CI
10 cos nwt. d (lOt)]
II sin nwt 11(+ Q- I sin nwt 12:t+ Q] 0: lI:~a
41
=- - -0
n,
'
SInn
. ex ......... forn
=
1. 3, 0- •...
= 0 ............ for'l = 2, 4, 6,.... . b" =-1 1 sin ncut. d (wt ) - f' ,-a 10 sin milt. d (cot ) it
[f'-" a
0,
1
n;*CI
=.!.. [ 1-cosnOll 1,a ·a n1l:
l -cDsn WI 1:'"1.... :1
41. _ = - cos n ex .................. forn= 1, 3, 0 , ...
n.
r -~-,:-"
=O ................. ............. .. for r. =2 , -1 6, .. .
C, = [ [ - : :. sin na )' + ( : ; cos ncr )' BTl
= tan- 1 [- tan n o l = -
n(.(
=
Phase Controlled Rectifiers
[Ar'- 6.51
displacement angle of nth harmonic current . • Cl '41 ,'. i. (t ) = ~ _ 0 sin (a wl - no)
279
=
£.
na l ,J, S
n7t
... (6.36)
.
The rms value of nth harmonic input current, from Eq. (6.36), is . 410 2 >/2Jo '. I - ..",...-cc - In -
"2 .rite - .
n1t
...(6.300 )
.
I - - 2>/2 . I. Rms va ue of fund~ e~tal curren~, 1' 1== _ ' rr = 0.,900,32 Ip
·[I;.n]1I2 = 1
I . Rms va ue of total input current, I. = . ~
0
,
Also 9 1 =: - a ,
p
Negative sign for 9 1 indicates· that,.fundamental, current lags the sourc e voltage . The various parameters are now obtained. -j'
_. "
From Eq. (3.52), displacement' factor ifF = cos 81 = cos (- ~) = cos a ... (6.37) . , , ! . 1. , • 2 10 1· 2.J2 From Eq. (3.53 ), current distortion factor, CDF = -I1 = Y-I = = 0.90032 I
;&':
,
- . 2..f2:· , · Power factor, PF = CDF x DF = - - cos' a n
Harmonic factor HF
•
it
0
1t
... (6.38)
.'. ' i
o~ THD =[· CDr 1~;' 1 .]'12 =[ (
From 8q ( 366 a) , voltage
r2 ]"2
25 )2-1 =0.48343-0; 48.34p%
riP,~le ~~ct~~_= [(~: J -) . , .
SubstItutIng the values of V~-fr6m' l!Iq , (6.27) apd V. fr0111 J;:q. (6. 26), we get
. [,-.Y.:;n-c.'"
-
1. '
- it
2
]'12
[2 1t
]"2
.. .(P9) VRF= . -. -X , - 1 ~ -1 2 ~ ~ . cos 2 a ' 8 cos2 a Active power input. Only ti:l,e.nns fundam ental component of input current contributes to the active input power to the ci;nverter. -
:. Active power input,
Pi = 'rms
~alue of ~ource voltage x rms fund amental component
of s ource current x displacement lactor =
Vs .161 . cos 9 1
=V•
Reactive pow er input,
(
2{2I. } cosa = -2Vm -]:) cOSet = V0 I0 1t 1t
...(6.4 0)
Qi = V,. 1st . sin et =V
•
(2{2I' )Sin e< =2V.. I , in a
V. =-I cos a
1t
0
u
1t
sin a = V I tan 00
C!
... (6.4 1 a ) .. .(6.41 0)
280
[Art. 6.5]
Power Electronics
6.5.2. Single-phase semi converter For this converter, th~ variation of input current i8 • in Fig. 6.11 (b), is shown continuous from a to n, (rt + a.) to 2 1t and so on, but i, is not constant. A!J before, i, is now assumed ripple free also with its amplitude 10 , Here
I,
l =-2 [f' 1t
a. =.! 1t
I, . d (wt) -
Cl
If
f"
I, d (WI)]
11:+0.
I, co. nWl d (WI) -
I
I, co. nWl. d (WI)J
II:+Q
Cl
= n~
r
=0
{I
sinnwt I~-I sinnCllt 1~+Cl]
2I, .
=- ~ sm n a ....... .. .. .. for n = I, 3, 5, .. . =0 ............ ..................... for n = 2,4,6, .. . b.
=.! 1t
If
I, 8in nOlt . d (Olt) -
a
r
n+a
2/,
= mt (1 + cos n a) ........ for n
I, • 8in nWl. d (WI)J
= I, 3, 5 •...
= 0 ............. .. ........ .......... for n = 2, 4, 6 •.. .
, ']'12 [ :~ ~in n a) + ( : ; (1 + cos n a) )
C. = (-
2'fi
= - - (1 + cos na)l/2
n"
It is known that 1 + cos n
e =2 cos2 ~e
2 "2 [ ,n a Cn = - - 2C08 n1t
2
]V2 =-cos410 no. 2
nn
J
a.= at n- 1[ -:;----;--",sin",:,!n,",::" ] B =an t -1 bn
/I.
= tan- 1
_
1 + cos no.
2 sin¥ cos 2
. (/) = ;. L
2
na sm . ( nwt-2" n"
4 I,
2
no
2 co. l,
T ]=_ net
~cos2'
J
... (6.42)
n _ l . 3.5
Rm s value of nth harmonic input current, from Eq. (6.42), is 41" no 2..J21" nQ l ,n = 'VI?_ • nit cos - 2 = n7t . cos -2
Rms fundamental circuit, I,} =
2"2.1, n
a . cos -
2
. .. (6043 )
.J
[Art. 6 ..5]
Phase Controlled Rectifiers
Rms value of total input current, Is =
l' (rc -
[
0
112
a)
]
TC
=
10 [
n: - a
l SI
11'-
- n- ]
9 1 = -a2
Also,
DF= cos 9 =cos ( - ~ )=cos % 1
CDF =
HF or THD
I
_,_I =
2 -12 I
n
I,
=[--.!,
r 2 -12 cos !!2 cos .!!VTt = 2 I , vn - a C1v~n"'(n-_-a;o )
. 0
-1 ]112
CDF
.
[
a)
- 1 ]112
8 cos 2 ex 2
1]"2
n (n - a)
... (6.4 5}
4 (1 +cos a) . 'a
=
Powedactor
=[ n(n-
2 v'2. cos 2" a =CDF x DF = ~'1 7t (7t - a ) x cos- 2 =
2
2[
TC ( TC _
VRF=[ ( ~:)' =
=
... (6.44 )
[
~m[( 27t
[[
a)
]112 cos22a
-1
r
)
Jt -
a +
(n - a) +
Sin2a] 2
1
2 sin 2 a
]
[
=
2
TC ( TC _
a)
]"2
Jt2
. (1 + cos a)
x ~ (1 + cos a )2 -
1 ]~.
112
n _ 1] .. .(6.47}
2(1+cosa)2 Power input,
Pi = V,. Id· cos 9 1 = V,.
2 -12. I , It
a
cosZ·cos
a Z
=
V, . I , . -12 1t
Vm
=Tt (1 + cos a) . 10 = Vo 10 Reactive power,
( 1 + cos a.)
.. .(6AS}
Qi = V, . l' l· sin 9 1 10 ex. a Vm • = V, . 2 "2 COS- sln "? =-10.sma TC 2 1! ~
=
AJ.s o,
... (6.46)
~
.. .(5.49
e)
[Reactive power required in I-phase full converter for the same l ol
Vm Vo = - ( 1 + cos a )
or -
I( I ;· Q: = 1 + ttcosC a· S!n
VI
It
0. =
Vm
0
it
Vo
= .,----"-l+ cosa
. a o· tan "2
282
Power Electronics
[Art. 6';J
Example 6.11. A single -phase full converter; connected to 230 V, 50 Hz source, is feeding a load'R = 10 0 in series with a large inductance that mahes the load current r?pple free. For a firing angle of 45 ~, calculate the input and output performance parameters of this converter. Solution. From Eq. (6.26), V, =
.
2V
---.!!!.
cos a =
.
2 x .[2 x 230
_ cos 40° = 146.423 V
I = V, = 146 .423 = 14 6423 A .
R
0
10
.
V"
From Eq. (6. 27 ),
V,,=;n= V, = 230V
As load current is ripple free, rms value of load current,IO,. = 10 = 14.6423 A.
From Eq. (3.60),
P"" = .V, I, = 146.423 x 14.6423 = 2143.97 W
Rms value of source current , l, = 10 = l or = 14.6423 A
P~ = V" x I" = 230 x 14.6423 = 3367.73 W
Output ac power,
From Eq. (3.6 1l, rectification efficiency = Vo,.
'r"!
From Eq. (3.64),
~: = ~i:;:;~ = 0.6366 ?r 63.66~
230
Fr = V, = 146.423 = 1.5708
VRF = -iFF' - 1 = A5708' -1 = 1.2114 As load current is ripple free, CRF = 0
From Eq. 13.660 ),
2.[2
I" = -
From Eq. (6.36 0 ),
•
x 14.6423 = 13.183 A
8, =-a=-·45' From Eq. (,.37),
DF = cos a. = cos (+ 45) = 0.707 Rms value of total input current, I, = 10 = 14.6423 A CDF = 0.90032, PF= CDF x DF = 0.90032 x 0.707 = 0.63653 lag THD or HF = [
1." - 1]112 = [ 1 t CD, 0.9003
- 1] = 0.48342
From Eq. (6 .40 ), active power = V, I, = 146.423 x 14.6423 = 2143.97 W From Eq. (6.41 a), reactive power
2Vm
.
= - - 10 SIn a
•
=
2.[2x230
•
. x 14.6423 x sm 45 0 = 2143.963 VAr
Al so, from Eq. (6.41b), Q I = VJ, tan ex = 146.423 x 14.6423 x 1 = 2143.97 VAr Example 6.12. Repeat E:co.mple 6. 11 for a single-phase semiconductor. V "ro x 230 Solution_ From Eq. (6.29), Vo = ---.!!!. (1 + cos a ) = (1 + cos 45°) = 176.72 V
•
•
I = V, = 176.72 = 17672 A o R 10 . F rom Eq. (6. 30).
V"
=V, [
H(.-
ex) + sin 2ex } ] 2
'.
".
[MI. 6.6]
Phase Controlled Rectifiers
~ 230 [ l or
H( ~ )+ n-
sin 90 } ] 2
283
~ 2193V
=10 (" .. load curr ent is ripple free ) =17.672 A
From Eq. (6.30),
Pdo ~ V, I ,
~
176.72 x 17.672 ~ 3122.996 W
Output ac power,
P ~ ~ V" I" ~ 219.3 x 17 .672 ~ 3875.47 W
. P Rectification efficiency
~ --"'- ~ 3122.996 ~ 0 8058 80.-8% Poe 3875.47 . or :;:,
FF~ V,,~ 219.3
'From Eq. (6.43),
1.241 V, '176.72 VRFdFF' - 1 _ -)"1.""247:1""---;-1 ~ 0.735, CRF ~ 0 45' 2..[2 l sI =--;- x 17 .672 x cos 2= 14.697 A
-~
-
I. ~ 17.672 'JT':4 · ~ 1'.304 A
.n
DF = cos 9 1 =cos ( _ !! ) =cos a 2 2
CDF ~ 2..[2 x cos 22;:'
=cos 45 =0.9239 2
~ 0.9603
[ +-~)] PF ~ CDF x DF ~ 0.9603 x 0.9239 ~ 0.88721ag
[1
THD = CDIi' -1 Power input Reactive power input,
~
]'12 =
[1
0.9603' - 1
]11' = 0 .2905
V, I, ~ 176.72 x 17 .672 =3122.996 W Vm
Q, = -
n
.
10 sm a
=..[2 xn 230 x 17 .672 x cos 45° =1293.79 Vp...r
6.6. SINGLE-PHASE SYMMETRICAL AND ASYMMEmlCAL SEMICONVERTERS A single-phase sem i-converter topology employing two SCRs and three diodes is commonly used in industrial application s and therefore, th e term "single-phase semiconverter" implies thi s conve rter only. Here single-phase semiconverters , or s ingl e-phase half-controlled converters, employing two SeRs and two diodes are also studied. Both of these semi converters hn.ve two legs. A single-phase semiconverter, employing one SCR and one ·diode in . each leg, is called 5ingle -p hase symmetrica.l semiconuerter. The other configuration, using two SeRs in one leg and two diodes in the other leg, is call ed single-phase asymmetrical semiconuerter. These are now studied briefly.
6.6.1. Single-phase symmetrical semiconv er ter It is also called sin gle- phase half-controlled symmelrical conv erter, or single-phase two-pulse symmetrical converter. As stated before , it has two legs, each leg is made up of one SCR and one diode in .:ieries as shown in Fig. 6. 17 (al. The \·oltage and current waveforms are drawn as shown in Fig. 6.17 (b ).
2g~
Powe r El ectronics
[Act. 6.6]
\
I i,
$",
T1
J
+
T2
a
". ~L
b
DI
D2
-
(a l Fig. 6.1 7. Single-phase symmetrical semiconverter
(a )
(bl circuit diagram and
(b )
waveforms.
From wi = 0 to rot = 0., let constant load current 10 free wh~el T2 Dl. Soon after wi = 0, T1 gets forward biased through T2. At a firing angle n, T1 is turned on. Load current shifts from T2 to T1; thyristor T2 is therefore turned off. At Wi = 0, T2 is subjected to a reverse voltage of Vm sin a which aids in the commutation of T2. Load current n oW flows through Tl, RL, Dl , b, scurce u!' a and Tl and load voltage Uo = uab' At. rot =Tt, U, = O. Soon after wi = It, b becomes somewhat positive with respect to a, D2 gets forward biased through D1; as a consequence, load current now freewheels through Tl D2 fr om rot = It to (It + (X). At rot = (It + a), forward biased T2 is tur n ed on, Tl is therefore, turned off. Current n ow fl ows through T2 D2 from rot = (n + a ) to 2Tt and Vo = Vb«' Soon after rot = 2 Tt, D1 is forward biased, therefore 10 now begins to fre ewhe el through 12 D1 and so on. Various waveforms for voltages and currents are drawn in Fig. 6. 17 (b ). During fr eewheeling periods, Vo = 0 be cause devices are considered ideal. Wave form for un shows that circuit turn-off time tc is given by
n- a t, = OJ - sec 6.6.2. Single-phase asymmetrical semiconverter For th is semiconverter, power ci rcuit diagram is drawn in Fig. 6.18 (a ) and th e relevant wav eforms in Fi g. 6.18 (6). An exa minat ion of Figs. 6. 17 (0 ) and 6.1S (0 ) r ev eals th at in 5 in gl~ - pha .5 e symm etrical semiconver ter of Fig. 6.17 (a) , the two ca thodes of S CRs T l and T2 are cO:l.:.ected togeth er, th es e ca thodes are, therefor e, at the same potential. As s uch. only one trigge ring ci rcu it is suffirien t for this s emiconverter. Wh en si ngle gate pu l.;e is appli ed co both the t:-tyri3to:'s. the S CR \vh ich is for ward-biase d at th at instan t will get ~u rn e d on . I n
".
P h a.~e
Co n troll ed Rectifiers
285
[Art. 6.7J
asymmetrical semiconver ter of Fig. 6.18 (a ), however, two separate triggering circuits must be employed. The thought-process, leading to the understanding of previous single-phase symmetrical semicon verter, can now be extended to this converter configuration also. For both symmetrical and unsymmetrical topologies, output voltage V" is given by Eq. (6.28) and its rms value Vor by Eq. (6.29). The other perform ance parameters can be evaluated as desi red. 0 ~r-1-~~~-----j~,~,-----,7'~-C",~, lrr
~~
: "~ - '~
01 :-T 'OI ~ 01 i--T202~ 01
:1.
.,
01'
'.
~v.
£"
0
,
c,
:.....11-0: ••
.01
II,
f.-. -101 """':' .
""
"0'1. I fI, i.
c
·02·
/ 317
2.
""
"I
.,
""
"
> 12
C,
-
(a )
Fig. 6.18. Single·phase asymmetrical semiconverter
( bl (a)
circuit diagram and (b) waveforms.
A comparison between single- phase two-pulse type semiconverters now be made as under :
ana·full converters can
Single-phase semic on verter requires two SCRs and three (or tw o) diodes but a single-phase full conver ter needs fr om SeRs. Single-phase semic onve rter circuits are. therefore, cheaper. (ii) Single-phase semiconverter offers one-quadrant operation, whereas si ngle-phase fu ll convertor can furnish two-quadrant operation . (iii) Freewheeling action in semiconverter circu its render power factor better than its value in fu ll conve rter ';ucuits. (i)
Performance parameters of semiconverter circuits a re superior than their corresponding val ues in full conve rter circuits.
Th e advantage3 of using thre e·ph ri3e cont rolled converters over singl e.pr.... se cont rolled converters are the san e (13 pn5se'::;5ed by 3-ph a3c diode rec tifiers over I-ph ~1.5e diode r -::c tiE~rs ?!lumer.J.ted 1:1 f'..rt 3.9. It i:i ?olso discu.:;sed there why thr ee·phase deltu·st::1: tr anAu rm~r i.5
186
Power Elcctronks
[Act. 6.7]
empl oyed for delivering power to three-phase converters . .~1 three-phase controll ed converters use line-commutation for the turning-off of thyristors. Three-phase thyristor conv·e rters may be classified as under : Three-pulse conver ters Six-pulse converters (c) Twelve-pulse converters. Six-pulse converters include 3-phase full converters, 3-phase semiconverters and six-pulse mid-point converters . These are now described one after the other. 6.7.1 . Three-phase Half-wave Cont rolled Converter (a)
(b)
This converter is also called 3-phase 3-pulse converter or 3-phase M-3 converter. This is now discussed with different types of loads. 6.7.1.1. Three-phase M-3 Converter with R load. Power circuit diagram of this converter is shown in Fig. 6.19 with resistive load R. A reference to the circuit of Fig. 3.29 and waveforms of Fig. 3.30 is of considerable help. If firing angle is zero degree, SCR T 1 would begin conducting from wt =30 0 to 1500 , T2 fr om wt =150 0 to 270 0 and T3 from rot =270 0 to 3900 and so 00. In other words, firing angle for this cont rolled converter would be measured from rot = 30 0 for TI, from = 150 0 for T2 and from wt = 270 0 for T3 as indicated in Fig. 6.20 (a). For zero degree firing angle delay, thyristor behaves as a diode and the voltage output waveform Uo is as shown in Fig. 3.30 (e). The operation of this cqnverter is now described for a < 30 0 a nd for a > 30 0 •
wt
u"
+--
"
TJ
lu,
A
<0
0
A
t,
11
C
«
TJ
B
R
T J
Fig. 6.19. Three-phase half-wave thyristor converter feeding R load.
Firing angle < 30°. The output voltage waveform uo' for firing angle less than 30° (say around 150) is sketched in Fig. 6.20 (b ), where T1 conducts from wt =30o +Ct to 5~ . . wt = 150 0 + a =""""6 + Ct., T2 from 1500 + a to 270 0 + Ct. an d so on. Each SCR conducts for 1200 • The waveform of load current
io
(not shown) woul d be identical with voltage waveform
r--,
Va.
5:1
3 Average value of outpu t voltage, Va = 21t
a-; Vmp sin wt d (wt) .. .(6.50 )
where
ma.ximum value of phase (line to neu tral) voltage V ml = maximum value of line voltage =..[3. Vmp Ct. =flring-angle delay
Vn:p =
[Art. 6. 71
Phase Co ntr olled Rectifiers .
"
O:" ~
IX: 0 for T3
0::0 lor T2
lor TI
I
lor il
0:0
.-.~ ........ t
1
1
2S7
:
'Imp
lal
Ibl wI
Ie}
o
I I
"0" 1 I
TJ--I~T l ~/I--- T 2
~---No
SCR
,cndlJ~IS
",'~ "1-
I
JOO'/I
,I---TJ-! 4
wI
Tl
I
I
Fig. 6.20 . Three· phlise 3·pulse converter with R load (a) line to neutral source voltages. Load. voltage waveforms for (b) 0 < a < 30 e and (c) a:> 30 e Vu 3 Vml Average load current, I" = Ii = 21t. R cos a
... 16.5001
Rms value of output, or load voltage is V"
=[
3 2.
r
+ 5 ""
a-+l't/6
v" = 3 V;., [ 1 wt
or
or
41t
3V~
=~
4.
[2.
v;,p sin' wt. d (wt) -j'" ••
0
1
+"" _I sin 22 wt
0.+,; / 6
-.f3 ] -+-cos2a 3
2
1 3 -.f3 cas2a V or=Vmp [ 2+~
or
=.J3vmp Rms Ioa d curre nt,
[1 -.f3
]'"
S+81tcas2a
]t" =V [1 -.f3
V"' [1S+8rtcos2a -.f3 ]'" [or = VO R , =[f'
ml
'6+8ltcas2a ]'"
.. .(6.51) ... (6.5 1 a)
Firin g angle :> 30". When firing angle is more than 30", Tl would conduct from 30 ~ + a to 180",1'2 from 150" + a to 300" and so on a3 show n in Fig. 6.20 (b ). For R load, when phase voltage v" reaches zero at rot = 180", current io = 0, '1'1 is th: refore turn ed off. Thus, Tl woul d conduct from 30" + a to 180". Same i3 true for oth er SCR3. This show,; that each SC R, for firin g angle:> 30", conducts for (150" - a) only. This also implies that for R load, m aximum pos.5ible value of[iring angle is 150". Waveform ofia agrees with!l(J wav eform, Fig. 6.20 (e ). Average value of load voltage,
288
Power Electronics
(Art. 6.7J
' . f • Vmp slnwt. d (wt) n a .. • 3V
3 V' = -2
mD
=~
3 Rms value of output voltage, Vor = [ 2
n
V
=
r
r :1. [(
~ = ';.
...(6.52)
( 1 +cos(a + 30') I
V!p sin2 wt. d (wt)
a +1tI6
mp
[( 5R 6
a )+
-
]112
~ sin (2 a n
/ 3) ]'"
... (6.530)
)+~
5 R -a sin (2a+R / 3)]," ... (6.53 b) 6 6 .7.1.2. Three·phase M·3 con vert er with RL load. In Fig. 6.19, load R is now replaced by load RL . The load inductance L is large so that load current is continuous and constant at 10 as shown in Fig. 6.21. For firing angle < 30 0 , Vo and Vor are given respectively by Eqs. (6.50) and (6.51). For the firing angle range of30° < a < 900 and 90 0 < a < 1800 , this converter behaves differently as described here. 30 0 < a < 90 0 • For firing angle in this range, let the firing angle be, say, 45 0 for which the various waveforms in Fig. 6.21 are drawn. Note that T1 conducts from 30 + a to 150 + a, 12 from 150 + a to 270 0 + a, T3 from 270 0 + a to 390 0 T a and so on. Thus, each SCR conducts for 120 0 • vsl 0 &0 for T1
I
v,
v.,
w,
T
T
1. 225 V.., 195"
l.22 SVm
1..-
21,0·
fJvm=
315·
,
' _O.t.t.8 Vml)
-1.67 V,.. ..
Fig. 6.2 t.
Th r c~- pha 3e
M·3 converter
wa~'efor:n3
fo r 30° < a < 90° for ripple-free load
C ~lrrer: ~
Phase Cont rolled Re(tifi crs
289
[-"I. 6.71
iTl (o r La> is not zero because of RL load. Therefore. T1 would continue cond ucting beyond cot = it . As such , va = V~ goes negative beyond Wi = It. When T z is t u rn e d on at cot = 150 0 + 0, load current shifts fr om Tl to T2 and a voltage ua - Vb = [Vm si n (150 + 0) - Vm sin (30+ 0.)1 appe a rs as rever se bias ac r oss T1 t o aid its commutation. SCR 1'2 co nducts from (150 0 + aJ to (270" + a) and 50 on . The waveform for in or io. i n or ib and i1"3 or ie a re as shown in Fig. 6.2l. ill Wi = it, ph ase voltage va is zero, but
The waveform Un fo r voltage ac ross Tl, on the assumption of firing angle 45 0 can be d rawn as und er ; When
T1
is on, un = va
= 0 from cot = 75 0 to 195 0 , Fig. 6.2 1.
- u'a
\Vhen T z is on , vTI = Vo - vb from rot = 195 0 to 315 0 and When T 3 is on, uTI = va - Vc from wt = 315" t o 435 0 and so on. When T2 is turned on at wt = 195 0 , un = va - Ub = - V mp sin 15 - V mp sin 75" = -1.225 V mp ; at oot = 210", vTI= -1.5 V mp ; at wt =240 0 , uTI =.J3vmp ; at wt=270",L'n=-1.5Vmp ; at wt = 300", uTi =- V mp sin 60- 0 = -0 .866Vmp ' At wt = 3 150 , un ~_: V I11P s in45" + V "'P sin 15"
=- 0.448 Vmp '
AJso, at wt
.
= 315
0
,
T2 gets turned-off whereas T3 is turned on. un = va - ue ~ - V mp sin' 45" + V mp sin 75 0 = -1.673 V mp '
This shows that at oot in Fig. 6.21.
=315
uTI at once changes from - 0.448
0 ,
~IIIP
to - 1.673 Vmfl a;;
.~ hown
At cd = 330 0 , un = - V mp sin 30 - V mp = - 1.5 V mp At Wi = 360e , un = 0 - 0.866 V mp At rot = 390", Un
= 0.5 V mp -
0.5 V mp = 0
At Wi = 420 ~ , un = 0.866 V mp At wi
=435
and also
0
,
un
=V mp sin 75
uTI = Ua - VII =
=- 0.866 V mp
0
a = 0.866 Vmp + V mp sin 150 ~ 1.225 V mp
0 a nd so on.
Average and rms values of output voltage are the same as give n by Eqs. (6.50) and (6.51) re5pecti vely. 90 ~ < a < 180". For firing angle in this r ange, let a be, say, 165°. Under the assumption of ripple free load current l a, the var ious waveforms are shown in Fig. 6.22. As the output voltage wav eform u a is be low the reference line, average value of Uo must be negative . It is also eviden t from Vo =(3 V m!/ 21t ) cos a that \yhen firing angle a is more than 900 , V o is nega tive . F o r. ~ a > 90 ~, three-phase 3-pulse converter operates as a line-commutated inverter which i:i possible only if the load ci rcuit has a de voltage source of reve rse polarity. as in a single-phase full converte, a lready discussed in Art. 6.3 .2. 1.
It is also seen from Figs. 6.21 and 6.22 that ave rage value of thyristor current. In. \'201ue of so un: e curr ent, 1.. .4. = (1" x 120)/360 = 10/ 3.
o ~_ .~!.,~
,:".' . ,.; ..
¥
~. , .. ,,\ _. . __ ... ' . n.l ""' ~ :)'
¥
_
t /
o. :,ourcecurran ,
_/
T.. -
_ [ J; x 120
~r -
360
11/2 fo J ="\3
= average
290
Power Electronics
1M!. 6,7 1
V't
_o::lorT 2 - --
r'
-0
!or
1.,
i1: \ \5'j~------"! ..,
...
c:
!!
o
r --r-Cl rOr i J - j
U~
.
...-:_~_
1.(
:
.
,
..
v,
,~
':t"
13
"0
_ _
I
v,
v,
•I
T2
T1
13-
T2 - - - " f - - l J -
11
I',
'' ''L
w,
~_
w,
+-2"/3--1
l: ~ 1------,
,.
w,
\~:o', IL---!L-_l''----..L-------------'--'''--_;_;;_ I', I" I---- 2/rI J ----I
wt
Fig. 6.22. Three-phase 3-pulse converter waverorms ror 90° < Cl < 180 0 for constant load current. Waveforms of i u , Lb, ic in Figs. 6.21 and 6.22 show that transformer windings have to carry de current which is harmful to the transformer. Th e problem can, however, be sorved by using delta-zigzag connection instead of delta-star connection as shown in Fig. 6.23 .
In this figur e, delta-zigzag transformer.feeds RL load through a 3-phase 3-pulse converter. Load curr e nt 10 enters the neutral n of secondary zigzag, divides equally in the three ha lf·windings, i.e . each half winding a, b, c shares a load current 10/ 3. Thi s current. fl ows through other half windings bl , e l , aI, through SCRs TI , T2 T3 and load RL as shown in Fig. 6 .23 . Note th at each secondary winding is separated into two equal ha\y es which are appropriately connected to result in zigzag seco ndary. lo {3 11
A
a
• I,
•
I~
~b ~J lQ~
e
,
,
I,p
I ,~
"
, I,
+
r1
v,
Ie '0-e
I, R
~
I QI 3 l
J,
F ig. 6.23. A rl elta-zigz.3g tn!:'!i>iorme:-
f~~ di r: g .1
lJ
3-pha3e 3-pulse thyristor CO!1','erte r.
I
[A.t. 6.7J
Phast: ConI roll ed R '! ctificrs
291
A care fu l observation of zigzag winding in Fig. 6.23 reveals that same phase winding, dividing into t wo halves as a , a l ; carries current 10/ 3 in both these J:uil.ves , but in opposite directi on s. S a me is tru e for phase b, c windings . Since each h alf, "of the three secondary windin gs , carri es direct current in opposite direction , their magnetic effects cancel each other. As a resul t, t he core flux and therefore core loss and temperat ure rise remain unaffected . This show s tha t a 3-phase 3-pulse converter can be used for energizing a dc load provid ed a delta-zigzag trans former is employed on its input side.
V .,.
Examp le 6. 13. A 3-phase M-3 conuerter is operated from 3-phase, 230 Vi 50 Hz supply with load resistance R =10 n. An auerage output uoltage of 50% of the maximum pos§.ible output uoltage is req uired. Determine (a) the firing angle (b) auerage and rms "Glues of load current and (c) rectification efficiency. So lution. (a) For R load, voltage is continuous when as 30 0 and average output voltage is given by 3 Villi V = - - cosCt , 2"
. um Jts maXlm
·bl e va I ue IS, .. Vo m = ~ 3 VmI = 3-1221C x 230 = 15-o.3 V
POSS l
= V;m = 15;.3 =77.65 V
Requ ired average output voltage
3 Vml V o =~ cos Ct = Y om cos 1
V o ="2 Vom
No\v
Thi:; sh'ows that actual value of Vo
= V"m cos Ct.,
=~ Y om = 77.65 V
a :. 0
= 60
cannot be obtained for a < 30 0 , but
for a > 30· . So by using Eq (6.52), we get
.ra
3 Vm 3V Vo = ~ [1 + cos (0 + 300 )J = 21C
..
0
0
[1 + cos (a + 30 )1
1 T3 [1 + cos (cu 30· )} = V
27t Vo 1 x-- = =" 3 Vml Yom 2 a=67.7.J
3
! = V, = 77.65 = 7 765 A o R 10 .
(b)
Rms voltage from Eq. (6.53 bJ is V. =
[.
:J~ W6" -C< )+ ~ sin (2 a +n/ 3) J
= V2X 230 r (5"_ 67. 7 x n ) 1. 510 . (2 x 677 60. J ] =10476-V 2 :.r;; 6 180 +2 . + . ;)
!. =
1°i·~65 :
10.477 A
· V, !. 77.65 x 7.765 0 -493 -493" (c J R eC '~l·Iiler e ffi llcle ncy = VorIlJr = 10. t 765 x 10.477 = .0 or o. 10 Exa mpl e 6.14. Deriue expressions for the auerage and rms outp ut uoltages for a 3-phase 3-plllse con trolled con uerter by using cosine /imctior. for the supply uoltage . A ssume continu ous conduction.
Powe r Electronics
l.-\ n. 6.7 J
S ol u tio n . The variation of output voltage fo r 3·phase 3·pulse converter is shown in Fig. 6.24 . For using the cosine function for the input voltage, t he origin must be taken when instantaneous voltage is maximum . So here origin is taken at ~O' as shown in Fig. 6.24. With ~O ' as the 'o rigin, VC! = Vrn cos rot and integration must be made from instant 1 where where WI
=( ~ + a
rot =- ( ~- a ) to
~c
,',
""'--j'-,~'/:)+I :- -''''/3:-i.-'!p--...L''''':-~-;'a::f~-
instant 2
--Ia!- .
J
} Thus, t he average output voltage
Vu for a 3-pRase M·3 converter is
Fig. 6.24 . Pertaining to Example 6.14
3 f,.. 3V V.=2" J ~ [ i_"(mpCOSWld(WI)= 2;P
=
-fa;"'"
3..J3vmp
I
- sinWi
l'-j[··~-. l
3Vml ' cos a = - - cos a
2 7t
.. .(6.50)
27t
Now, rms value of output vo ltage V"r is
.,
3 rl ~ o. V"r = 27t J
-(1-0. )v;,p cos- rot. d (rot)
v'2 _ 3 V~p
IIr - ~
sin 2rot wt +
Vor = ..f3 VIIIP [
or
2
I~
H,
- ( ~-o. )
,11'
1,f3
'6 + 81t cos 2(.( J
1,f3 =Vmt [ S+S1tcos2a
~
"./'"
I
]11'l
, .. (6.51 )
E x amp le 6.15. A 3-phase half-wave controlled converter is fed from 3·phase, 400 V, 50 Hz sou rce a nd is connected to load taking a constant cu rrent of 36 A. Thyristors hatle a voLtage drop of 1,4 V. (a) Calculate average vbalue of load lJoltage for a firing angle 0(30 0 and 60=, (b) Determine average and rms current ratings as well as PN of thyristors. (c) Find the auerage p ower dissipated in each thy ristor. Sol ution . Here, average output voltage, Vo = 3?Vm1 cos a _ !.I T; Vml = \'2 X 400 V and
For a fir ing ar:¥le of 30°, For (b )
Ct
= 60 0 ,
." VII = 3 ~ x 400 cos 30° - 1.4 = 232.474 V ." 3 \ 400 VII
=
X
2
;<
Aver age current r ating of SCR, ITA =
R • 1':15 curr ent
· r:l t mg
a rSCR .
cos 60 0
-
vT
= 1.4 V
1.4 = 133.63 V
I II 36 '3 = 3 = 12 A
I Tr -!!.-~-?0·8 - \'3 - -./3 - - . 1 ;)-
·'
_"'\
PIV of SCR = ,f3 V m, = Vml = V2 x 400 = 565.6 V tc .--\n~rag'e powe r dissi p
[Art. 6.iJ
Phase Controlled Rectifiers
~ Example
293
6.16. A 3-phase 3-pulse con verter; fed from de lta-star transform er, is con nected to a load requiring ripple free current. A freewheeling diode is connected across the load. Shetch waveforms for sou rce voltage, output voltage, load current, line current and free wheeling diode current. Obtain expressions for average and rms value of output voltage. thyristor cu rrent and freewheeling · diode current. Solution. A 3-phase 3-pulse converter feeding RL load and with freewheeling di ode across RL is shown in Fig. 6.25 (a ). For firing angle < 30°, free wheeling di ode does not come into play. So here, fir ing angle is taken , say 60°, just to illustrate how freewhe eling diode comes into playana to examine its effect on the performance of the converte r. At cot = Jt, as phase voltage Va tends to go negative, freew heeling diode gets forward biased through Tl. Therefor e, freewheeling diode starts conducting from rot = Jt till 1'2 is turned on at rot = 150 + cx. Similarly, when Vb and ve tend to go negative, freewheeling diode comes in to play, as shown in Fig. 6.25 (b) . Note that each SCR conducts for (150° - ex) and .freewheeling diode for (a - 30°). i. 11
Iv. ~
i,
~Tv.
T2
( 0--'-'
FO
L--_--"'--~ J-
Bo-- -- - - - J
I.
TJ
l'd
(a)
v. ~
O ~ ~v~~ ~ LTJ--l FD L.-
i i ----l FO L--.T 2
"I
-
ll:~ !
r- 15O - et
.
01
.
i:!
- ..! FO ~TJ _~ FO I--n
WI
II,
"' -
1
II I', I
n I---
L
---1CE-:O~
n
I
,,, 1\ 1
2 - )----...J
n
wI
..
(,
Fii · 6.25. Tr...:-ee· pr.:u<:l M·'; con . . erte: Ca ) circuit (b ) wevefo ..m E:ulmple ~ .0.
2 9~
Power El ectro nics
[Art. 6.7) 3 Average value of output voltage, V" = -2 It
(!.
ru
Rms thyristor current,
[(5n 6 - a )+"21
V ml
Vor = 2"1Tt
I TA
.
V mp sin wt . d (wt) ... (6.52 )
2n
=
[
.
S in
. (20 + 1t/ 3)
]';2
... (6.53 b)
n
6-a.
1" Average thyristor current,
:t/ 6
3 V mp [1 +OOS (CL + nI 6))
=
Similarly.
r
2Tt
2
I"
I" = 2n
1" 51t =21t[ 6 - CL]
]"2
{s--CL l 5n:
1
=
I, [ 2n
{s--CL lJ 51t
112
Average va lu e of FO current,
Rms value of FO current ,
6.7.2. Three-phase Full Converters If all the diodes of Fig. 3.39 are r eplaced by thyristors , a three-phase full ·converter bridge 3S shown in Fig. 6.26 is obt ained. The three-phase input supply is connected to term in als .4., B, C and the load RLE is connected across the output terminals of conv erter as shown. As in a singlephase full -converter , thyristor power circuit of Fig. 6.26 ' works as a three-phase ac to dc converter for firing angle delay 0° < a. :5 90:: and as three-phase line-commutated inverte r for 90° < a < 180°, A three-phase full T3 T5 R converter is, th erefore. preferred where regenerati on of A a 1/, power is required. The numbering of SCRs in F ig. 6.26 B 0-- --1---4 v
l
C"---lL-_-_-_-_-~+-_-_-_-_-~:__ L! "
4is ( = I1, + 3), + 3), 2( 3 - 6) for group the negatand ive 3, 6 (= 5 3for th =e 5 +positive .J group. This numbe ring scheme is adopted here as it T4 T6 T2 agrees with the seq uence of gating of the six thyrbtors in a 3-phase full conv erter. Fig. 6.26. Power ci rcu it for a 3-phase For a =0°; Tl, T 2, ......T 6 behave like diodes. This is shown in Fig. 6. 27 (a) . The sequ ence of con duction of full- converter feeding RLE lon d. SCRs T 1 to T6 is also ind icate d in thi s figu re. Note that for a = O ~ , T1 is tr igge r ed at wt = lt / 6, T2 at 90°, T3 at 150° and so on. The load voltage has, therefore, the waveform as shown in Fi g. 3.40 (c). For a =60", the conduc tion sequence of bYristors T1 to T 6 is sh own in Fig. 6.27 ( b ). H ere T1 is triggered at wt =30" + 6 0~ = 90e , T2 at 90 + 60 = 15 0~ and so on . If the conduction interval of va riolls thyri stors T1, T2, ..... T6 is shown firs t, then it becomes eas ier t q draw the voltage and current waveforms . N ote that each SCR conducts for 120", when T l is t rigge r ed , reve rse bias e d thyri stor T 5 is tu rned off a:"l.d Tl is tu rned on. T6 is al ready condu cting . .-\s Tl is connected to A and T 6 to B, voltage Vall appears across load . It vane.:: from 1. 5 Vm to zero as shown . H ere V~.D is the ma..'"imum value of phase \·oltage. When T2 is turned on, T6 is co m mu tated fr om th e negative grou p. T1 is already conducting. As Tl and T2 are conn ec ted to .-\ and C respective ly, voltage u:;~ 8}::pea r5 :lcross lO:ld. Its vahl e vaTi e.:: fr J ;r! 1.5 Vmp to ze ro as show n. This sequence of triggering is conti nu ed for other SeRs.
Phase Controll ed Rect ifiers
V[s
[Arl.6 .7J
295
t--- 120° ------i
"t T1
"0
I
vb
13
t
Vmp
v,
T5
I
" TJI
Tl
I E
0
wI
0=0'
j T4
T2
T2
T6
f---120o~
(al
ITS
v,
I T6
TI
I
I
T2
T3
I
I
T4
T3
T'
I
I
T6
I
TI
I
+~e
T2
_~e
GrOllp Gro up
TI
T'
O~4U~~~~~~~~~~~~~~~~~+-~~~~W~1 l·SVmp
1.5'v'mp
T3 T2
•
1 T6
T4
T5
T4
T2
I T6
T1
.,.V(Z GrOllp
I -~e Group
Fig. 6.27. Voltage waveforms and conduction of thyristors for a 3-phase full converter.
Note that positive group ofSCRs are fired at an interval of 120°. Similarly, negative gl:uup of SCRs are fired with an interval of 120° amongst them. But SCRs from both the groups are fired at an interval of 60°. This means that commutation occurs every 60°, alternatively in upper and lower group of SeRs. Each SCR from both groups conducts for 1200 • At any time, two SCRs, one from the positive group and the other from negative group, must conduct together for the source to en er gise the load. For ABC phase sequence of the three ~phase supply, thyristors conduct in pairs; TI and T2, T2 and T3, T3 and T4 and so on . The sequence of events in Fig. 6.27 can also be shown more conveniently ifline voltages, instead of phase voltages, are considered. In Fig. 6.28 (a) are shown line voltages Uab, uoe • Ubc , Uba etc. For a = 0°, SCRs TI, T2, ....T6 behave as diodes and the output voltage waveform is as shown in Fig. 6.28 (a ) by tlab Ucc , Ubc etc. In this figure, for ex = a, Tl is t1..1l1lJa. on at o..'l = 60 0 , T2 at wt = 120°, T3 at wt = 180° and so on. In Fig. 6.28 (a), therefore, firing angle is measured from wt = 60° for T1, from wt = 120 0 for 1'2, from wt =180 0 for T3 and so on. The question may arise in the minds oi the readers as to why TI, for a = 0, conduct.s from wt = 60" and not from wt = 00 • H ere the use of subscripts 06, ac , DC, ba etc corne to the rescu e of readers. As observed, the subscripts in sequence appe(,'r twice. When first subscrip t ap pears twice, the SCR in the positive group pertainin g to that line conducts for 1 20 ~ . Likevris e, when second subscript comes twice, the SCR in th e negud ve group pert aining to th at line conduct.s for 120°. For example, first subscript 'a' appe ars hvice in tl ab Uae ; th erefor e SCR from positive
296
Powe r
[Art. 6.7J
El~ctronics
group T1 will begin conduction when Uab appears i.e. at cot = 60". In uue ' uk second subscript 'C' appears twice, the refore SCR fr om negative group T2 will begin conductioQ..w hen va,' appears i.e. from cot = 120" in Fig. 6.28 (a). Similarly, first subscript 'b' appears twice in uk ub", so SCR from positive group T3 will begin conduction when Vbc appears i.e. from cot = 180 0 in Fig. 6.28 (0 ).
For a = 60°, T1 is turned on at cot = 60 + 60 = 120°, T2 at cot = 180'\ T3 at cot =,240° and so on. When T1 is turned on at rot 120°, T5 is turned off. T6 is al ready conducting. As 'P and T6 are connected toA and B respectively, lo ad voltage must be u"b as shown in Fig. 6.28 (b). ';Vhen T2 is turned on, T6 is commutated. As T1 and T2 are now conducting, the load voltage is u"c' Fig. 6.28 (b). In this manner, load voltage waveform can be drawn with the turning on or. off of uther SC Rs in seque nce. For a = 90°, the load voltage is symmetrical about the reference line wt, therefore its average value is zero. In Fi g. 6.28 (el, load current waveform i" is drawn on the
"<
I a ~O
.r
10
%
cr2~Oft
I
,
T\
T2
t ",b I
n
TI
T2
"'cb ~
"'oe ~
1
JI7
l
la)
TI
I
a,90'
Ie)
wt
~""J""'l ""'l """l
llokbO~CO ~Cb
° l ~~v
r '~L,
+ve grouo _ve grour
•
wi
""J""J '
db doc ;.{b:· ~bO ~ta ~(b
•
~V~\
V
)---0"]
wi
= 150C~
>--0",50'"
,
>--0,_ ' 50·-----<
v,
•
wi
~
o. :1 5
0"iI
~Q: , = 15 0° - - 1
d. ~l t b r.
t!)O
lee
itO
I---
I--
Cl:
6 =15 0 0-----'l-j
loa
ico
c --------____________________________________ i} Id )
l
lob
I"
~ ,
WI
c -,~------,-~~-~~-~~ ~ c:.;::::::::J wi
loa lcc Fig. 6.23 . Volt3 ge and current waveforms for a 3-phase rull -conwrter for different fir ing i!ngles . i boJ
ita
Ph os!: Contr olle d Rec tifiers
[Art. 6. 71
297
assumption that load is pure L . When v cu is peak positive, slope dio /d t is maximum positive so t ha t L. (dio/dt) equals peak positive ucn ' Similarly, dioldt is maximum negative wh en U ea iii peak negative. When current io has peak value, dio(dt is zero and likewise Ucu is zero as shown in Fig. 6.28 (c) . For CJ. = 150°, T1 is triggered at wi = 210 T2 at 270° and so on . T he output voltage waveform is shown in Fig. 6.28 (d ). It is seen from this figur e that average voltage is reversed , in polarity. This means that de source is delivering power to ae so urce; this is called line-comm utated inverter operation of the 3-phase full converter bridge . It may be seen from above t h at for a = o ~ to 90°, power circuit of Fig. 6.26 works as a 3-phase full converte r delivering power from ac source to dc load a nd for a = 90° to 180°, it works as a lin e-commu tated inverter delivering powe r from dc source to ac load. It can work in th e inverter mode only if the load has a direct emf E due to a battery or a dc motor. It should be noted that direction of current for both converter and inverter ope rati ons r ema in s fixed but the polarity of output voltage reve rse!';. Q
,
Source current iA in phase A is als o draw n in Fig. 6.28 (d) for 0.= 150 e . For the a rrow directi on indi cated in Fig . 6.26, iA is t reated as positive. Therefor e iA is positiv e wh en T1 is conducting , i.e . when first subscript for voltages or currents is 'a'. Likewise iA is negative when T4 is conducting, i.e. when the second subscri pt for voltages or currents is 'a' . Source current waveforms for other two phases can al so be drawn accordingly. For other fir ing angles , so urce currents can be drawn similarly. Expression for the a~e ra ge output voltage Vo v, ca n be obtained by referrin g to Fig. 6.29 where uab' v a,: etc. are sketched from Fi g. 6.28 ta) fo r ii ring anglo: de lay Ct. < 30°. Note that peri odicity of output vo ltage is it / 3 r ad ian s. Average va lue of output voltage is ob tained by finding the dashed area wI I • I 11 i abed ov e r a pe r iodic cycle , Fi g. 6.29. and then l--- -1 ----.- -, ....,0 ct +ttl dividin g it by the pe riodic tim e. With 00' as the Fig. 6.29. Output voltage waveform for a ori gin a t the maximum value of Ul.lb. Vo is given by 3-ph ase full co nverter.
(~ ." J
f
Vo =~ Vm 1coswt·d(wt ) -(~ -. ~
)
n) . ( +'6.] =-3Vn-ml cos a
ml
. ( 0.+'6 - sm -3Vn- [ sm
... (6.54)
Ct.
H ere V m: is the maximum value of line voltage.
If s ine function is used for the source voltage, then
wt = o.
uub
=V rnl sin wi
2:'1
'J+a
Vo = ~
f
l,lml
si n (,:)/;. dt wt)
!.'. - a
3
3\lm l
r (l 3"2. - (( ,)-
~ - - , - _ C,j S
C0 5
3Vml '1 - a l~~ =----;:CO.5 a
( n
because
Vub
=
a at
298
[ Ac <. ~ . 7J
P ower Electronics
Rm s value of out put voltage VIN" is
VM=[.:!.n: J Y"' v;.,si n' wid (wi ) ]"2
,
--(1
a v_ ~ .2
V!. : - 2 ml n
or
2.'1:
~ - o
J;- .. u
,
J
(1 -
cos 2wt ) d (wt )
3 [3"+2cos2a n ,f3 ]112
...(6 .54a )
Vor=Vml -"' fn
It is observed from Fig. 6.28 th at source cu rrent fo r phase A, i.e. iA (or fo r any other phase) flo ws for 120 0 fo r every 180 0 • Th erefore, in case outpu t current is assumed constant at / 0' the r ms va lue of source current is I = - II , 2n x l = 1 - ~ J \J o a it 0·" "3
Each SC R co ndu cts for 120 0 for every 3600 • Therefore, the rms value of thyristor curren.t is
ITI, :
_I
2 2.
\J I O T
1 X
2n:
_ It =Io-\J a
6.7.3 . Threc· Phase S emi converters In Fig. 3.39, if diodes Dl, D3. D5 are repl aced by thyris tors Tl , T2 , T3 respectively. a 3-phase semi co nverter bri dge of Fig. 6.30 is obtain ed. A fr eewhee ling diode FD, in parallel with RLE load , is connected acr oss the out put t e r min a ls of th e semiconverte r as shown . + Three-phase balanced supply is given to the three R " "t2 T3 inpu t termi na ls A, B , C of Fig. 6.30. The outpu t voltage Vo acr oss the load terminals A FO ~ L is con trolled by varying the firing angles of SCRs B T 1, T2 and T3. Th e diodes D1, D2 and D3 provide C merely a return path for the current to the most ,. 0 1 03 02 E~ negativ e line termina l. The se mico nve rter bridge operation for diffe ren t firin g (ln glcs is sh own in Fi g. 6.31 in t he form of Fig. 6.30. Power circuit for a 3-phase volt age a nd cu rrent wavefo rms. The conduction semi·converter feed ing RLE load. angles for the SeRs, diod es or FD are also shown . For a firin g nngle delay of a: O ~, thyristors T1, T2, T3 would behave as diodes and the output voltage I) f semicon verter would be sy mmetr ical six-pu lse per cycl e as shown in Fig. 6.3 1 (a). The outj:o-iJt voltage consisting of pulses uel.. Ua b' v ac ' v b.: etc. shown in thi.:i figure is similar to that shown in Fig. 3.40 (c). The output voltage consists of pulses I..'ab' L'.w t;b: ' ueo etc. as in Fig. 3.40 (c). \\:,e n the fi ring angle is delayed to a = 15° (say) as s hown in F ig. 6.3 1 (b), the triggering of S CRs T1, T2, T3 is delayed but return diodes Dl, D2, D3 remain un affected so that only alternate pulses are altered. The load current is continuous and h as little ripple . The FD doe5 no t c:>me into play fo r a : 15 ~. Each S CR an d diode conduct f:>r 1 20~. In Fig. 6.31 (0), L'cb is the load vol~3.ge from wt : 00 to 60 0 As the first subscrip t indic at es conduc ting eleme nt in th e posi tive grou p, u ~!> shows that T3 is al ready cO:1Ciucting through di od~ D2 o~· ne gative !oTfO Up . Voltages U:tO ' Vat" indic ate that , nc:ordl:1.g to the first :subsc ript, T 1 conducts fur 1:200 nnd i~ begin.:i to conduct at wt = 6,Y for Ct = J :\3 .ihown in Fig. 6.3 1 (a ).
'0
'.
YTI
'I;
f
[Art. 6.7 1
Phase Controlled Rec tifiers
Vo
Q:
e::O
0. : 0
T1
T2
0:0 TJ
2 ~9
0" : 0 TI
oo{ .:~':)." v:: 61 .. :~: .~b: I "~_: .~" __ t:~~_
(a)
0;,'
...
Vo
....,,/
Tl
1
Q
' .... '
..../ ' 1'0
12
r
E
r
r
1Q
"v
'- v '
•..•
'.
".
TJ
WI
r
•
1Q
•
Q: IS• I b)
WI
I
tJ
i'e
"
OI
1
1
OJ
1
"
01
1 1
~ 120·--+I
---.....
T1
-----.
" 1 1" ---...
TJ
T2
. 'H I group
-"(l I;jroup
01
02
~
Tl
V~ Q -W ~ ~ ~ ~'
WI
.L~
Ie)
Id )
WI
Ie)
" I g I:I "
01
Q2
rr
, 0
wI
F
! I b'.
0
21'l'
I! I
fl
F
02
0
• •
I:, I 3n
WI •
rig . f~31. Vol t3ge and current wa\'eforms for a 3·phase semiconverte r for different firing .mg!e:;.
Similarly, t:bc. (..'bu indicate that T2 conducts for 1 20 ~ and it begins to conduct at wt = 18 0~ for Ct. = O~ . An SCR with zero degree firing angle behaves like a simple diode. Thus, as per the definit ion of firing angle, it should be measured from wt = 50 ~ for T1 . from wi = I BO for T2. fro m Wi = 300 ~ for T3 and so on. For Ct = 60\ Fig. 6.3 1 (c ), the thyristors are fired so that current retu:ns through on e diode d'J.!'ing each 121.1 conduction period. For voltage l-'" " Tl and D3 c on dll c~ si multaneou.:;iy for I:.W; as show n. Si milarly, other elements conduct. P'O does not COr:1 ? into play even fo r n = 60' . Fu:[her note that .... oltage pu i:5es (Ja il. (..'be l",:o clo not appenr in the output voltage w;1 :eform for Ct. = 60°. It will be seen tha t for ex 2: 60;, \'oltag::! pu13es ~'ab ' t':", L'e:: are eli minated. Tr.;, l o ~ ..1 C1.1!'rent, assumed co ntinuous for a = 50'> , i.:i not :5nown in Fi g ".31 <('1. Q
300
[Ar!. 6.7J
Power Electronics
For firin g angle delay 0[90°, voltage , and current wave forms ar e shown in Fig. 6.31 (d ). Th e output voltage Vo 15 discontinuous . As Vo made up of ucb. v ue, ubo' ut/,.. .. , tends to become negative at c.ct = 12 0 ~ , 240 e , 360 ~ , FD gets forward biased . Therefore, for each periodi c cycle of 120", output voltage is equal to line voltage for only 90" and for the r emaining 30°, when FD conducts, Vo = O. F or a = 90", conduction angle of SCRs and diodes is seen to be less than 1200 for every output pulse. In other words , conduction angle for both positive and negative group elements is 90 n and for the remaining 30". current completes its path through FD as shown in Fig. 6.31 (d ) for a = 90°. Voltage pulses uab. v b<; , Un are absent from output voltage Vo for this firing angle as well. Withou t FD, .after load voltage Vo reaches zero, a diode from negative group would begin to conduct reducing U o to zero till next SCR in sequence is triggered . For example, at wi = 120", Uo = u cb = 0 and without FO, 03 from negative group would start conducting through T3 from wt = 120" to 1500 when 'SCR Tl is gated. This means that without FO, T3 would conduct for 120° from wt = 30° to 150°, 02 for 90" from cot = 30" to 120 0 and 03 for 3~ '' from wt = 120° to 150 ~ for thi s periodic cycle of 120° extending from cot = 30° to 150°. For firing angle delay of 120°, the voltage and current waveforms are shown in Fig. 6.31 (e). The load current is now assumed discontinuous. For each periodic cycle of .120°, Uo is seen to have three components . When an SCR is gated, thyristor and diode conduct for 60° only. As Vo re aches zero and t e nds to become negative, FO gets forward bias ed and therefore starts conducting for some angle and holds the load voltage to zero. When aU ,the energy stored in ind uc tance is discharged, FO stops conducting and as a result, load voltage rises to load counter e mf E. When Vo = E, n one of the elements of semiconverter bridge is conducting, this is indicated by 0, 0 in Fig. 6.31 (e). It may be s een from above that in a 3-phase semiconverter, SCRs are gated at an in terval of 120:0 in a proper sequence. In a single phase semiconvertt:r, SCRs are fir ed at an inte rl<~ : of 180°. In ord er to obtain full control of the dc output voltage vo, the range of firing angle is:i· A" t o 180". A three-phase semiconverter has the unique feature of working as a ' six-puls~ conuerter for a < 60 ~ and as a three-pulse con uerter for a 2: 60°, a careful obser.·ation of Fig. 6.31 rev eals thi s.
For
3·phase semiconverter, each periodic cycle of output voltage has a periodicity of , 2 0 ~ . AVI'rRga nutput voltaee should. therefore. be calculated over 120° only. For a < 60· , For firing angle l ess than 60· , the output voltage is redrawn in Fig. 6.32 (a ) from Fig. 6.31 (6 ) for some firing angle less than 30° for convenience. In this figure , area abcefda divid ed by :In / 3 would give the average value ofouLpuL volta ge Vo. F or area abcdc, bkc 00' as the origin a nd for area dcefd , take AA' as the origin. Then ~
3
-.
•
V o - 211: (A rc a abcda + Aroa dc cfdl
11 ".
-
(~_au ) V m1coswt . d (Wl).1 = 2"l L [~ _ " l VM1cos wt . d(Wl) + Lv 3V .
Vo = ~ ( 1 + cos a )
or
."
With 00' as the origin, an gle
.
IS
put
(~ •
a } s meas ured to toe left of 00' . therefore minus sign
.OCIOU - ( -"6" - a Ij ;:'- lm l'1 nr'lj', r.11:1. . J:i 5 .l gn 15 . pu t b elore ' n S. 1
.. .(6.55)
Phase Controlled Rectifiers
Voltage u(.!b
=0
at rot
) 01
[Ar!. 6. 7J
= 0 and Vc..: = 0
at wt
=~
in Fig. 6.32 (a J. Therefore, V o can also be
obtained as 3 V O= -2
It
[fn
f3",
1
-3' + u
nl 3
- +0. 3
Vml sin Wl d (wt) +
nl 3
Vmsin wt d (wt )
3 Vrnl
=2n ( 1 + cos 0 ) Rms valu e of output voltage
It
[
v'or : or
for a < 60°, is given by
Von
V. , : -23
... (6.55)
{f'" , ('
_ - -a
r...
) V;. , cos2wtd (wt) + J
V~,
]'12
,
~2
3 - [ .1 rot + ------: ,.si"n wi 1"" 4It 2
.." sin 2wI + ..u..+
]
1
2
I
-(~-. )
/aI[
V;. , cos-0 wt d( wt )}
6 "'0
Vml 2. ..f3 =2"" "'" ~ I 3 + 2 (1 + cas 2 a )
Vo!'
nl 6
- :"'.16
]"2
... (6.56 )
For a ~ 60°. For a ~ 60°, the output voltage waveform is drawn in Fig. 6.32 ( b ) for a firing angle 60 0 < a < 90) for convenienc e. With 0 0 ' as the origin in this figur e, the ave rage output voltage Vo is given by 3 Vo : 2. [Area abedaJ. : -23
It
Voltage
VOl:
=0
at rot
[J'- "[It' "2 -
V ml cos
0
=~ and a
ml
3V (1 + cos O J wt . d(wt ) =-2It
is measured from rot
=r.: / 3
as sho\vn
... (6.5 5)
In
Fig. 6.3:2 (b ).
Therefore, V o can also be obtained as 3 VO = 2/t
J' U
3 V ml
= ~
0 '
V m/ sin
( 1 + cos 0 )
"
~ \.."+-r~- '--.j'X ~ "[ 0 ' I, ¥" A i.
i
wt · d (wt ) ... (6.55)
w,
liJ t
_ _ 21t / 3 _
(a) Fig. 6.32. Ou tput \'altage
(b ) w n'/efor.n ~
for \ 3-pho.:;; e semico nvertcr fo r (e) a
<:
60' .'l.:1d (0 ) c:. > 60'.
302
[Act. 6.7J
Power Eleclronks
It is se en from above that express ion for average ou tput voltage is the sam e for both six-pulse and three-pulse operating modes of a 3-phase semi converter. Rms value of output voltage Von for a > 60° is given by V" = [
2~' t~" 1V~l cos' wt.
= V;' .
..y ~ [
(n _ a ) +
d (WI )
r
~ sin 2" ]'12
.. ,(6.57 )
Example 6.17. (0) .4. 3-phase full converter charges a battery from a three-phase supply of 230 V, 50 Hz. The battery emf is 200 V and its internal resistance is 0.5 .n. On account of inductance connected in series with the battery, charging current is constant at 20 A. Compllte the firing angle delay and the supply power factor. (b) In case it is desired that power flows from dc source to ac load in part (a ), find the 'firing angle delay for the same current.
Solution. (a ) The battery terminal voltage Vo is
But
Vo = 200 + 20 x 0.5 = 210 V 3Vml Vo=--cosa=210V n -1 210 x 1t 474-3" ex =cos 3'12 x 230 = . i) •
For constant load current of 10 = 20 A , Fig. 6.28 (d ) reveals that supply current i.-\ is of rectangular (or square) wave of amplitude 20 A. It is also seen from this figure that iA flows for 120" (or 27t/ 3 radians) over every half cycle of 180e or 1t radian s. Rms value of the supply current I. over
1t
radians is
2l't]112 = 20 ~ _f2 3 = 16.33 A
1 2 I , = [ ;; (20) 3
Rms value of output current, l or = 20 A Power delivered to load = Elo + I~r' r = 200 x 20 + (20)2 x 0.5 = 4200 W Now .J3 V). cos 9 = 4200 W 4200
Input supply pf = 'l3 x 230 x 16.33 = 0.646 lag. (b ) "Vhen battery is delivering power, then
Vo = 200 - 20 x 0.5 = 190 V \'!hen power flows fr om de source to ac load, the 3-ph ase full conv erter then works as a 3·ph ase line comm u cated inv erter.
3 Vml - - cos a = - 190 V n
or
_
a - cos
- ,[ - 190xlt -1-l?7 7?O
3fi x 230 - ... . - .
\
[Art. (j,7]
Phase Controlled Rectifie rs
303
Examp le 6.18. For a 3-phase full converter, sketch the time uari~tlo~,s of input uoltage and the voltage across one thyristor for one com plete cycle for a fir ing angle dela y of (aJ 0" and (b) 30~. For both the angles, find the magnitude of reverse uoltage across thi::; SCR a nd it s commutation time for a three-phase supply uoltage of 230 V 50 Hz. J
Solution_ (0) The three-phase input vo.ltage waveform is shown as vab , v ac, Ubl.' vb" etc. in Fig. 6.33 (a ). A conducting SCR has zero voltage across it. Let the variation of voltage across SCR T l belonging to positive group of Fig. 6.26 be plotted in Fig. 6.33 (a).
"0. "A
1
Q:;OO "ab
'"
' be
"0
' bo
' ,b
'ob
' 0'
"::Ie
"0
0
wI
'00 Ca)
"0' " A.
'"
..=-:CI=300
'"
'00
'be
'bo
0
wI
' ob
C b)
'"
Fig. 6.33. Pertaining to EX3mpie 6.18.
For voltages vab . v ClC ' SCR Tl conducts, therefore voltage across this SCR is VA = V,: - Va = 0 for a peri od of 120", i.e . from Wi = 0 to wt ::: 120° as shown. After Wi = 1 2 0 ~ , SCR T3 conduc ts for 120 0 wi th uk vb« as the outp ut vo ltages. Now cathode of Tl is connected to supply term ina l B through T3 for a period of 120 ~ and its anode to supply terminal A. Therefore, voltage across Tl from wl::: 120" to 240° is VA = Va - Vb' This voltage reverse biases T1 and it is shown as Va b below th e reference line in Fi g. 6.33 (0) . ..\Iter Wi = 240 ~ . fo r voltages U CQ Vcb; SCR T5 conducts for 120 0 and therefore cathode ofTI is conn ected to terminal C th rough T5. Thus, voltage across T1 is VA = L':.! - Ve = Va , below th e r efe rence line for a period of 120 0 from CJJt =240°. At Wi = 21t, T1 is aga in gated and voltage vA = 0 as shown. For firing angle delay of zero degree, each SCR is revers e biased for 240 0 :. Com mutatio n time available for SCR turn-off = circuit
=
~1t
;·rld.
turn~o fftim e. te = ;: ::;ec
x 1000 . x ?_ l't x ~- 0 = .13.33 msec.
4l't
= 3
( b ) For a = 30\ the output voltage waveform and \'oltag{' v...I. ac ross SCR Tl a re ;;l1own !n Fig. 6.33 (b). At wi ::: 150\ when SCR T1 stops conducting, \'ol t::tge ac r053 T1 mu "t foll ow u,,/J cu r ve fo r 1 20~ a~ d i.::i cussed in part (a) a b ove. The r efo re , VA j ump:; fr om zero [0
[An. 6.7]
JIl.J
I' ow ~r
Elcctronil.'S
Fill/si n 30 = 0.5 VItII as shown. A 3 '1'3 r emains on for 120", VA foll ows uu b curve be low the reference li ne for 120°. At cot = 270", when T5 is gated, VA follows IJ lle for 120 0 till '1'1 is gated again, Fig. 6.33 (b ). This figure reveals that each SCR is reverse biased for (2400 - a) . Commutation time avail a ble for SCR turn-ofT = Circuit turn*off time ,
t, =
4J't/ 3 - a
w
=[
sec
! 5~1 x
x 1000 =1167 msec.
In case a ~ 60°, it would be obs e rved that rotc =1t - u . For both the parts , i.e. for a = 0" and 0. = 30", the peak rever se voltage across the SCR is ~ . 230 = 325.22 V. Example 6.19. (a) A 3-phase full·converter feeds power to a resistive load of 10 n. For a firing angle delay of 30°, the load takes 5 k W. Find the magnitude of per phase input sllPply IJoltage.
(b) Repeat part (a) in case a large reactor in series with load renders the load current ripple free.
Solution. (a ) For a resistive load , output current waveform is of the same shape as that of the output voltage wave . It is seen from Fig. 6.28 that for a > 60°, the output vo ltage and output current would be discontinuous for resistive load. However, for a $ 60°, output voltage and current are continuous. Rms valu e of output voltage is already obtained, for u < 60°, in Eq. 16.540 ).
,
V~r
_r;::-
I f = 5000 watts and Vml = \'2
For this example,
" 2" 3 [" -13 ] 2 .v; 3+2cos60
V"
- ._-
=5000 x 10
V, = 188.08 V
or
:. Per phase voltage, (b )
V ph
V,
=13 = 108.591 V
For a constant load current, average loa d current 10 =rms load current,
1or = VRo = ( 3V1tml casu ) ..!. R
,
1-Dr
X
R
ml =[3V -1tcos ex
J1
-R
=5000 W
v, = ~50000 x \122 X 3"cos 30° = 19l.22 V
or
Vph = 110.40 V.
Example 6.20. (aJ A 3-phase semiconverter feeds power to a resisti ve load of 10 n. For a firin g angle d eLay of 30°, the load takes 5 k W. Find the magnitude of per phase i npll t supply voltage. (b) Repeatpart (a) in case load current is m ade ripple free by connecting an ind uctor in series
with the load.
Ph::.sc Controlled Re ctifiers
[Art. 6.7]
305
Solution. It is seen from Fig. 6.31 that for ex < 30°, the output voltage is continuous. For a resistive load, output current is also continuous. R~s valu e of output voltage, for ex < 30 is already obtained in Eq. (6.56). Q
,
2V;~2'.f3 ;{"3 +"2 (1 + cos 60)] = 5000 x 10
... F or
a = 30', -4-
V J = 175 .67 V and Vph = 101.43 V
or (a )
Vo For constant load cu rrent, l or = ave rage load current; 10 =Ii
I
2
or
[
3 Vml xR = - - ( I +cos30') 21t
.,
or
3V
,
1
m =~ (1 + cos ex) R
' -=5DOOW 1
]
10
£n
V, = ,50000 x 3 x 1.866 = 177.44 V and
Vph
= 102.45 V
Example 6.21. Repeat Example 6.15 in case (iring angle delay is 90
Q •
Solution. (a) F ig. 6.31 shows that for ex = 90 0 , the output voltage is discontinuous. For a resistive load, output current is also discontinuous, for ex > 60°, rms value of output voltage is already obtained in Eq. (6.57). : . For ex
= 90
Q ,
~[
4V; n
2 -.
'2 ) + -21 sin 180
1t - -
V, = ~50000 x
or and
V ph
1
1 10
x - 1 = 5000 W
= 298.14 V
= 172 .14 V
(b) For constant load current, l or
Vo =[3V ml =10 = Ii """2n (1 + cos a) ]
• . [3
I~r
xR=
x R1
1
Vrnl (1 + cos 90) ] ' 10 = 5000 W
~
.,
or
0
£,
V, = ,50 DO!} x -3- = 331 ~1 53 V and Vp h = 191.2 V
Example 6.22. Repeat part (a) of Example 6.17 in case 3 -phase full conuerter is replaced by a 3-phase semicoltuerter. . Solution. The battery terminal voltage is
Vo = 200 + 20 x 0.5 = 210 V
But
Vo
3 Vml
=""'2rt" (1 + cos ex) = 210 V -
-{ 210x2,
a-cos l3.'12,230-
1] -6937' -
.
An ex aminati on of Fig. 6.31 rev eals that fo r fir ing angle a> 60 each SCR conducts for 180 - a. So, in this example, each SCR conducts for (180 - 69.37) =110.63 0 • For constant load Q
,
306
P ower Electronics
[M" 6.7J
curr ent of 10 = 20 A, supply cu rrent iA is of squ are wave of amplitude 20 A. As iA flows for 110.63 '01 over eve ry h alf cycl e of 180", t he rm s value of supply current I , is given by
xn]ll2
(20)' 110 .63 =20 - /110.63 180 ' 1 180 =Vol, = 210 x 20 =4200 W 4200 pf =:r:: = 0.6724 lag. , 3 x 230 x 15.68
I
=[!
"n
Power delivered t o load Input supply
=15.68 A
Example 6.23. A 3-phase full conuerter; fed from delta-star transformer, is connected to load RL requiring ripple free load curren t. For a firing angle delay of around 45°, sketch waveforms
{or (a) inp ut voltage u(l b. ua c• ubc etc. (b) load uoltage, load current, thyristor Tl and T4 currents and phase a , b, c cu rrents. In case ac supply is 3-phase, 400 V, 50 Hz, leuelload current = 15 A and a =45°, calculate rec tification effic iency, TUF and input power factor. Sol ution. Th e power circu it diagram of 3-phase full converter, fed from three-phase delta -s t ar trans form er, is shown in Fig. 6.34 (a). etc. is sketched in Fig. 6.34 (b ) with Ua b zero at wt = O. The ma.ximum value of uab , Lla c or u bc isVml ' where Vm l = maximum value of line voltage V I = ..J3 V ,"p ' Note that uab ' u oc ' ubc: etc. are displaced from each other by aD. angle of60°. (a ) The input voltage waveform Llab' Va e'
Ll bc
in
+ A 0---"
TJ
TI
Iv.
T5
I,
" R
i.
A
", L
c o-_--V" T2
Bo----_---' Fig. 6.34 (a ) Pertaining to Example 6. 23 .
For a firing-angle delay of 45 0 , load voltage waveform va is drawn . Load current 10 is constant throughout, SCR Tl conducts when first subscript in load voltage is 'a'. So Tl conducts when load voltage pul ses are uab' Uac' Likewise;T4 conducts for load voltage pulses Ube l Uco ' Source current ia' as shown in Fig. 6.34 (b ), is positive when Tl conducts and negative when T4 ct?!1d ucts . Current ib is shown displaced by 120 0 from ill' Similarly, ic is drawn . Here 10 = 15 A, l or = 15 A, because load current is ripple fr ee. Average value of SCR current, fr om in waveform, is (b )
120
I,
15
ITA =1, x 360 ='3= '3 =5 A Rm s value. of thyris t or current, I n-
= ~ I; x
'*
~~~ =-t =
= B.66 A
Ph:\se Controlled Rec tifiers
v,
JU7
fArl . 6.7] Tl
12
I-a~
:-0'-.1
i: v~ :
l 11a~
:
v'Lv" "-v,, "-v,, "-v,, ",-v" ""v"
o~~
~v" ~u, . ~""
~ ._
' Vrn Lcas 15 ·
iO' .~----~---------II,
wi
wI
c=
wI
wI ;
C
Fig. 6.34
(b)
Waveform for a 3-phase full converter, Example 6. 23.
Fig. 6.35 illustrates th e output voltage waveform for a p-pulse controlled converter. A review of Art. 3.9.3 a nd Fig. 3.34 is also helpful. The origin is taken at 00' , the peak value of the instantaneous output v('lltage va' Firing angle must be measured fr om t he inters ecti on of phase voltage waveforms as shown . T he r efo r e, the limits of i n t eg rati on a re from - (
~-
IX )
to (
( ~-a ) 0'
v,
0 ."
=2
n
I
,
.
~
PO D :
WI
~~~
:. Average output voltage, Vo is Vo
'\
• 11' 1--,. _11 ......'
~ + IX ) as illust r ated in Fig. 6.35. p
WI
Fig. 6.35. Output voltage waveform p-pulse controll ed converte'F
fr(' 1V mp cos p'·a
2
wi. d (wt )
··a
-V mp'
p
I I = " .(L (1'. ) . (!! ) p . Vm o ·
Sin
it
s m Cllt
P
.COSet
C a p •
a )
p
Rrns value of output voltage, V is (II'
. ]112 = ..E..... r; V: cos 2 CJ.U d (wi ) or ". p - ."- -« .1 [ -9 "
V
Jr-: \ . . 0.
for
,'.(6.58)
.lOS
(Art.6.7(
Power Elec trDn ics
, = p. V; 2, ] Vo - [ -+2sin -2, r 21 t p p cos2a
V,, =
or
v.[1 +(1;;
For 3-phase, 6-pulse controlled converter, p
rms va lu e of source current, I .r =
Vo
Here ,
From Eq . (6 .54a ),
=
~ I; x ~~~
3 V ml = -n
...(6.59)
6, therefore, =
cos a =
V,,=-12 x 400 x
~, } cos 2ar
}in (
10
~= 15. ~
=
3 -12 x 400 ,
12.25 A
_ cos 40° = 381.972 V
~ ~+ ~cos 90 ]= 400V 0
P d, = Vo I, = 381.972 x 15 = 5729.58 W
Pa.c = Vor ' lor = 400 X 15 = 6000 W Rectification efficiency
=
~d' = 5~~~'g8 = 0.95493 or 95.493% ~
Input VA = 3 (per phase source voltage) (per phase rm s source 400 = 3 x T3 x 12.25
cur re:!1~)
Pd, 5729.58 067"1 67 - 1% TUF = input VA - ,f3 x 400 x 12.25 = . 0 or .0 P~ 6000 Input pf= inputVA ='J3 x 400 x 12 .25 =0.707 lag.
6.7.4. l\'lulti-pulse Controlled Converters In Fig. 3.31, if diodes Dl ......D6 are replaced by th:yTistors Tl, ..... T6 . a 6-pulse controlled converter, 3-phase M-6 controlled converter or 6-phase half-waue controlled conuerter i s obtained . Similarly, in Fig. 3.42, if twelve diodes are repl ace d by twelve SCRs, a 12-pulse controlled converter or 3-phase 12-pulse controlled converter is obtained. For continuous conduction mode, each SCR conducts for 2 1t radians (or 60 0 ) for 6-puise converter and for ~; S radians (or 300 ) for 12-pulse converte r. In general . for a p-pulse controlled converter, each SCR would conduct for 21t radians . p
__
(6) (')
Vo=m V P -1t
V,, = V. [ 1
=V, [
sin -6
+(:,
)sin
. m
r
cos a = ..:......:...!! 3V 1t cos a
( ~' ) cos 2a
,f3 1 +4~coS2 a ]"'-
For a 3-phase 12- pul:se converter, p = 12, therefore,
V, = Vmp '
(
~2 )sin ( 1~ )cOS" = 0.98816 Vmp cos O'
IAn. 6.S1
Phase Controll ed Rectifiers
301)
1/ 2
v,,~v,[ 1+(~; )sin ( ;; )cos2a]
]1/'
3 =Vl1+;cos2a [ In a 6-pulse converter, 'Nith resistive load, continuous conduction occurs for 0 < a < 60° and di scon tinu ous co nducti on for 60 < a s: 120". The maximum possible firin g angle is therefore, 120°. In a 12-pulse converter, with resistive load. continuous conduction occurs for 0 < a < 75° and discontinuous conduction for 75 < a S 105°. The maximum possibl e firin g angle is 105°. 6.8. PERFORMANCE PARAI\1ETERS OF 3·PHASE FULL CONVERTERS
Here the performance parameters of 3-phase full converter are derived from the va rious waveforms sketched in Fig. 6.36. In a 3- phase system, power-factor angle is defined a5 the angle between phase voltage and phase current. This is the reason for showing the phase voltage Uti, ub, Uc in Fig. 6.36. Line voltage lJab, Uac1 Uk etc. fac ili tate the sketching of ou tp ut voltage waveform uo• so these are also shown in Fig. 6.36. As before, load current is assumed ,. rippl e fr ee. Th e source current is is given by I, (I) ~ I, + II '"
I
C"
t . 2. 3
sin
(nwl + 8")
Here, 1a = 0, because positive and negative half cycles of source current is, about the reference line OJt , a re identi cal in Fig. 6.36. Further, waveform of source cu rr ent i, reveals that
for phase '0', positive pulse is from
( H7nto(a+I~n}
l[rt ,5 s"
all=~
0.+::
Ia [ = n .1t For n = 1,
al =
4:
0
(a+~ )to ( a +
Ia·cos nOJt . d(cot) -
Isinno"'Ill
[ - sin CL.
0.
+ 51t1S
a+~S
and n ogative pulse i5 from
rTl~-.""I: 1a cosncot 0. +
-j sinnwtl
'i]
5; ) ,
7
0.
d(cot)
+ llr-JS ]
a+7::1S
, • n Tt .,. 1 3 I n genera,I all = - -410 . sm n a . sm -3 ..... ror n. = , ,Q.
nn ~
bll
0 ........ . for eo = I , 4. 6 ........ ..
=1. 1t
rr+ ,
:0:-:' IrJ sin ncot. d ( wi ) -
cr.- -
= :~[
I -cosn cot
r,
+::
1a' sin !H.ot d
c.tt-
la +S1tI6 u .'tIS +
I - cosncot
la+~L-::S ] a.,. n: / o
({~t) -j
]
310
Power Elec tronics
[A rt. 6.8 ] U. U,
Ub
o
.u,
lIe
"to '
.;
I - 0. ~ !- Q---J ) -·v~~- ··. '·v~~·· '., " v~~- . ~'
"
,
,'
,....... . . . . . ,
.'
.
"""',': o ..i'" -- 1:.~ ~.. __-"..',-._-;.':C''. "-_+-""'-_--""-_.;, -" _-+,,-_ _ ' --;-;, ' wI
i'lL !. --i.i 0:
u'
•
i '0'
.
1,_-;-_ _ ,"
J
r" ,
- - ' - ---f------c---:c:-
u'
LI
"~nLL ~ ---'-_----;-:---L-~---L---,--_ """WI S
a+ 1f
cr.+...:!!
---1) _' - +-,,----ii---'-.L---;:;c; "iO'~,-_-"I_'_....J."'-_l' w, _
1 . --
,
)
; " -
[ u, Fig. 6.36. Waveforms for 3-phase full converter with
tic!
= 0 when
wt '"' O.
410 • mt = - - cos n a . sm -3 ...... for n = 1,3,5, ..... . nn
=0
... ... for n = 2, 4, 6, .... .. .
nn)'
4 f" . . en =[ [ --;;;. SlIlna. sm 3
410 • mt = - sm nn 3
eo = tan" ( is (t) =
+
[410 . nn )' nn ' cos nil. sm 3
1
and 8n = tan- [-tannal =-na
~: )= tan"
(- tan nIX) = -
mt . L.. -4nnf a . sm. -3 SlIl (nrot -
r
nIX
... (6.60)
flex)
n " 1,3,5,
410 • nn I Jn = T22. nn sm -3 2"2 I , Rms value of fundamenta l current, 1d = l't sin 600
Rms value of nth harmoni c,
-is
=-Ii
I
... (6.61} (I
Phase Controlled Rectifiers
[Art. 6.8]
Rm~ value of total source current, 1$=~ I;
~ ~I0-\1~ 3
1t
X
23
X
311
... (6.62)
DF = cos 91 = cos (- ex) = cos a
fa
1 'I';; ~ -3 ~0 . 955 10 2 ft 3 CDF x DF ~ - cos a~ 0.995 cos a I, I,f6 I, 1t
CDF~ - ~ -. I,x-.
PF ~
... (6.63)
•
HF~THD ~[C~F' - 1 1'" ~[( ~)' -1 ]"" ~0.31084 or 31084% From Eq. (6.55),
V~ ~ V ml
J"3[ •
is.cos 2 a ]"' '\1 2;; "3 + 2"
_1]"' ~ [3. ~, (.!'. ~[6c:s'a (~ + )-1 ~ V~ )' Vo
VRF [ (
21t
';co.2a
. . ActIVe power mput,Pj
=3
V,
,f6
V,
l si_COS 9 1
3Vml
= 3 - T3' ~ , 1
0
I
+ is cos 2<> n' 2 } 9. V~II . cos2a
r
3
,
cos ex
)
(
line voltage, VI
"l3
=
... (6.64)
\
... (6.65)
~ - . - . cos a 10= V'/o
where V$ =
_ 1]112
per-phase source voltage.
Reactive power input, Qi = 3 Vs· 151 . sin 91 = 3.
V,
,f6
.
T3 ·Tt· 1
0,
3Vml
.
sm ex = - . - 10 , sm (X
..(6 .66 a)
3Vml 3Vm1 Vo =--cosaor --~-o 1t 1t cos
But
V
,.
V, . Q i = cos (x ' 10 , sm a. = V o·1o tan
a
Ct .
...(6.66 b)
Example 06.24. A 3-phase full converter delivers a ripp le free load cu rrent of 10 A with a firing angle delay of 45°. TM input voltage is 3-phase, 400 V, 50 Hz. (a) Express tM source current in Fourier series. (b) Find tM DF, CDF, THD and PF. (c) Calculate the active a reactive input powers. Solution. It is seen from Eq. (60GO) that source current for a 3-phase full converter is given by .
l$(t) =
I... -41 nn
0
•
n1t ,
sm -3 sm (nwt - n. o)
""" I, 3, 5
Amplitudes of source current for different harmonics are as under: 410 . n 4 x 10 is _ Sin - = X= 11.03 A and a = 40° For n =1, 1t 3 1t 2 0 4 x 10. For n;= 3, 8m 1t =
3n
Power Electronics
IArt. 6.8J
312
For n = 5,
~x 10 sin 300" = - 2 .205 A
x.
4 x ~ sin 420" = 1.575 A 7x. :. i, (I) = 11.03 sin (WI - 45) - 2.205 sin 5 (WI - 45) + 1.575 sin 7 (WI - 45' ). (b) DF = cos 45 = 0.707 For n = 7,
=
4 I, . • -l6 1.1 = ~ Sin '3 = -;- I IJ
Here
Rms value of source current, fr,o m Eq. (6.62 ), is 1.r = 10 ..
CDF= 1'1
Is
THD
~
=-l6 I, x 10~ 1f=~= 0.955 2 1t .
1t
=[C~F' - 1 )'12 =[( ~)' _1 ]'" =0.31084 or 31.084'k
3 PF = CDFxDF=- x co, 45' = 0.6751 lag
•
(c) Active input power, from Eq.(6.65 ), is
,
3Vm1 ( 3,t2 X400 _ ). P= -.-coso. 10= 1t cos40 xl0=3819.72W
From Eq. (6.66 ),
3Vm1 . Q= --sma . 1. = ~3,t2x400. sm45 x 10=3819.72VAr
Example 6.25 . A 3-phase ful
J
co~uerter, fe
from -Phas:, 400 V, 50 . z' source, is connected to load R = 10 n, E = 350 V and large inductance so that output current is ripple free . Calculate the power ri£liuered to load and inpul pf for ra) firing angiR. of 3~ ' and (b) firing aduance angiR. of 50' .
. 3~ 3,t2x~0 Solution . (a ) For 0.= 30", V" = - - cos a = cos But
n V, = E + I.,R
n
:. Load current
I, = 467 .7;~ - 350 = 11.7734 A
3~'' =
467 .734 V
As load current is constant rms , value of load current I"r = 11 .7734 A. Power delivered to load = EI() + I~r )( R
=350 x 11.7734x 11.7734' x 10 = 5506.82W = V, I , = 467 .734 x 11.7734= 5506.82 W
or
Rms value of source current, from Eq. (6.62 ), is I,r = 10
~= 11.7734. ~= 9.61 3 A
400 Input VA = 3 Vs . / sr = 3 x Ts x 9.613 :. Powe r factoT
=
Power in load = Input VA
5506.82
J3 x 400 x 9.613
Also from Eq. (6.63 ), pf = ~. co, 30' = 0.82697 lag
n
(b) For firing advance angle of 60", CL = 180 - 60 = 120" .•
1..
~ = 3..J2:: 400 co.:; 120" = _ "
270.0 5 V
=
0 82686 1a ' g
Phase Cont rolled Rect ifier s
1M!. 6.91
313
As Vo is negative, this converter is operating as line-commutated inverter. Th e polarity of load emf E must therefore. be reversed. N~
~ = -E+~R
- 270.05 = - 350 + I. R
Load current, I . = 350 -1~70. 05 = 7.995 A Rms valu e of load current, Power delivered by the battery to the ac source through the line-com mutated inverter, p= E10 - /!r R = Vo 10 = 270.05 x 7.995=2159.05 W Rms value of source current,
1$1' =10 ,
~= 7.995 . ~= 6.528 A 400
Supply VA= 3. V•. I" = 3 x Ta x 6.528 _ Power delivered to ac source _
P ower f ae t or -
Also, from Eq. (6.63),
supply VA
-
2159.05
_ 0 47741 u . ao
'1/3 x 400 x 6.528 -
33 ' = cos 120 0 = 0.4775 lag. n n
pf= - cos a
6,9. EFFECT OF SOURCE IMPEDANCE ON TIlE PERFORMANCE OF CONVERTERS For single-phase and three-phase full converters, deri vation of the average output voltages, as given by Eqs . (6.26 ) and (6.54), has been obtained on the assumption that current transfers from the outgoing SCRs to the incoming SCRs instantaneously. This mean s that when incoming SCRs Tl and T2 are fir ed in a s ingle-phase full converter, Fig. 6.10 (0 ), outgoing SCRs T3 and T4 get turn ed off du e to the application of reverse voltage an d the curren t shifts to SCRs Tl and T2 instantaneously. This is possible only if the voltage sour ce has n o internal impedance. Actually, the source does possess internal irnpedance.lf the source impedance is resistive, th en there will be a voltage drop across the resistance and t he average voltage output of a converter gets redu ced by an amount equal to l o.r, fo r a single-phase conve rte r and by 2Io.r, for 3. 3-phase converter. Herelo is the constant de load current and r. is the source resi stance per phase. Since sou rce resistance is usually low, it is assumed that du rat ion of the commutation is very small and the current transfer takes place immediately after the incoming S CRs are fired . Howeve r, a voltage drop caused by source resistance must be taken into consideration as discussed above . rn th e foll o\'.ri ng lin es, the source impedance is ta k en as purely inductive . The load ind uctanc e is _ljsumed large so that output current is virtu ally constant. The source ind ucta:1ce caU.:ies the ou tgoing and incoming SCRs to cQnduct togethe r. During the commutation peri od (when both incomi ng and outgoi ng SCRs are conducting together), the output voltage is equal to the ave rage value of the co nd ucting-phase vol tages . For a single-phase co nverter. the load volt age wi ll be zero and fo r a 3-phase converter, th e load voltage is (va + ub)/ 2 ( ave rage value of the condu ct ing phases a and b). Th e commutation period in seconds, when outgoing and i nc oming SCR3 ar e conducting together , is al s o kn ow n as the o verlap pe riod . T he angular-peri od , dur ir.g which both the incoming and outgoing SCRs ar e condu cti ng, is known as commu ta tion anglt? or overlap angle ~t in degr ees or radians . The efh:·ct of sourc e induc tance !.,> in vestigat ed ir. ~hi3 section ror both single-phase and three-phflse fu ll converter s. h wo.uld be
[Act. 6.9]
314
Power Electronics
observed that the effect of source inductance is (i) to lower the mean output voltage, (ii) to distort the output vo ltage and current waveforms and (iii ) to modify the performance pa ramete rs of the converter.
S.9.1. Single-phase Full Converter The commutation overlap is more predo.rninant in full converters than in semiconverters. In the si ngle- phase full converter shown in Fig. 6.37 (a ), L l is the source inductance. The load current is assumed constant (analysis with pulsating load current is more involved). Fig. 6.37 (b ) gives the equivalent circuit for Fig. 6.37 (a) for analytical purposes . When terminall of sou rce voltage II J is positive in Fig. 6.37 (a ), current i l flows t~oughL " Tl, load and T2; this is shown as ul • L ,. Tl T2 and load in Fig. 6.37 (b). Similarly, when terminal 2 ofuJ is positive, load current i 2 flows through T3, load, T4 ; this is shown as U2' L s' T3 T4 and load in Fig. 6.37 (b).
io
I,
l
fr,
+ TJ
"n
IL,
,
Tl
,
° ". ,.. .'
"
1 ; :-, ...,O#J ,_
b
i,
T4 ,
w
' .
i. ,!., ;'
,.,. -',
"
i
u-;:Y . .:--'
."
'
f' :! ' ,! '
,i ;' - ' """,,:11 :---
,"",,:11 :-
i 011:2 \2 11 ':11: I, ' :
wt
I
i
:: :: 1,1·
' . I,
oP.".c.:..--:,,,.;Q..-"--,0:,:-.4,- --t' ~ 1-t.tT\T2~ ,i-TJTI< -);-
_ Vo + LOAO
,.,
0.-
,,'•
TJ
'
vo{ !.;
I
a
i.......
T2:
A 0
(a)
c
/"
t"..l /'
°
kT2
kT4
0."
l
( n ,TI,) ....... ( l l.n ,n ,n )
'0
~
Fig. 6.37 . (a) Single-phose full converter with source inductance L, (b) its equivalent circuit and (e) typical current and voltage wavcfonns with L, . When TI , T2 are triggered at a firing angle a , the comm utation of already conducting thyristors T3, T4 begins. Because of the presence of source inductance L" the current through outgoing devices T3, T4 decreases gradually to zero from its initial value of 10 ; whereas in incoming thyristors TI, T2 ; the current builds up gradually from zero full value of load current 10 , During the commutation of Tl, T2 and T3, T4 ; i.e. during the overlap angle ~ , KVL - for the loop abeda of Fig. 6.37 (b) gives,
to
1) 1 -
or
LJ
U I - U,
di 1 •
dt
= L,
di2
= 1)2 -
L, dt
(~i -~t')
It is seen from Fig. 6.37 (c) that if VI = Vm sin tilt, then v2 = - VIn sin rot .
L, (~,' - ~:) = 2 Vm sin wi
... (6. 67 )
Phase Controlled Rectifiers
CMt. 6.9]
315
As the load current is assumed constant throughout, il + i2 = 10 , Differentiating this with
respect to t, we get ail di2 -+-=0
... (6.68 )
dt
dt
di l
di2
2V
dt
dt
L~
.
m - - - = --Stnwt
F rom Eq. (6.671,
... (6.691
Addition of Eqs. (6. 68) and (6.69) gives dil
V In
dt
L,
.
... (6.70)
-=-SlDwt
Load current il through thyristor pair Tl, T2 builds up from zero to 11 = 10 during the overlap angle 11; i.e . at cd = ex, il = 0 and at cd = (ex + 11), il = 10 I,
J dil
:. From Eq. (6.7 01,
o
=
V Ltn
,
V ·
or
10
=.,L, [cos CL -
J(a+)l)/ (IJ
sin
ai lll
rot·
dt
cos (CL + Ill]
... (6.71)
It is seen from Fig. 6.37 (c) (middle figure) that output voltage Uo is zero from ex to (ex + Il). Thus the average output voltage Voris given by Vox
V
= -!!! 7t
J (a+~ (a+)ll
sin tot · d(oot)
V
= --.!!!. [cos (ex + 11) - cos (ex + Tt)] Tt ... (6.720 I
. = -Vm [cos CL + cos (CL + Ill] n
2Vm Average value of output voltage at no load, Vo = - cos ex n
2Vm Maximum mean output voltage, Yom. = - n
Eq. (6.72a) can therefore be expressed as
voltage at no load [ (I] V CU'= Maximum mean output 2 COS ex+ cos ex+1l V,m = 2 [cos CL + cos (CL + Ill]
From Eq. (6.71), cos (ex + 11) = cos ex -
... (6.72 bl
00£,
V 10 m
Su bstituting this value of cos (ex + 11) in Eq. (6. 72a ) gives 2 Vm
00£,
2Vm
1t
Tt
7t
V~=--cosa--lo = -- cosCL-2fL . 10 ~
... (6.731
00£ Also, from Eq . (6.71), cos ex = V ~ 10 + CDS (ex + Jl) . Substituting this value of ccs a in Eq. m
(6 .72a ) gives
.. .(6.741
31 G
Powe r Electronics
[Art. 6.91
=
2Vm . cosa
= V",cosa.
...(6.75 a )
0£ [", V m -=0=,:..:. : - [cosa-cos (a+l') I
From Eq. 16.71),
n
For a full-wave diode r ectifier, a
n
=0°
w L~ Io --'--": -Vm 11 - cos 1'1
n
n
Vm
-
:. Inductive voltage regulation
(1 - cos 1')
n = 2Vnmi n =-"--=-= --,--2 V,,/lt ... (6 .75 b)
With the help of Eq. (6.74), a de equivalent circuit for a 2-pulse s ingle-phase full converter can be drawn as shown in Fig. 6.38 (a ). In this figur e, diode D mer ely indicates that load current is unidir ectional. This equi-valent circuit reveals that the effect of source inductance L$ is to present an equ ivale nt resistance of magnitude wL,/1tohms in series with internal voltage of the rectifier, (2 V ml1t ) cos a. The voltage drop due to L , is proportional to 10 and L. Thus as the load current (or source inductan ce) increases, the commutation interval (or overl ap angle) increases and as a consequence , the aver age output voltage decreases as illus trated in Fig. 6.33 (b J. V.
,vm
-- TT cos a:, 10=0 IJ,= 0
wl, ,,-
D
l;m cos
1
Q
I,
J ~)
r"~;~'"
v,
~ 10= -Wl ,
D
W
cos
}
0:
I,
Fig. 6.38. (a) DC equivalent ci rcuit of single-phase full converte r. In single-phase full convEi'i:er, as lo ng as ~ < Tt, the output voltage is given by Eq. (6.73 ). When ).l = Tt, the load will be permanently short circuited by SCRs and the output voltage will be zero because during the overlap angle, all SCRs will be condu cting. An other obser/a tion can be made fr om Fig. 6.37 (c) as follows. If ct = 0, then the mean output voltage can be controlled over ~ < a < 180°. Also, the maximum value of firing angle can be 180 -ll . In pra ctica l ci r cu i~ s, thyristor tak es some tim e to r egain ·its forward blocking capability. Th ~ re fo r e , the maxi mum possible firin g an gle can be 180 -).1 - 0 = 180 - (!! + 0) where (/) / w ) is the thyristor turn-off time including the factor of safety. Here /) is called the
recOI.:eryangle.
I I'
[Art. 6.9 1
Phase Controlled Rectifiers
317
Example 6.26. A single-phase full converter is made to de liver a constant load wrrent. For zero degree firing angle, the overlap angle is 15°. Calculate the overlap angle when firing angle is (a) 30° (b) 45° and (c) 60°. Solution. The de load current, from Eq. (6.71), is given by
Vm
.
I I) :::; - L [cos a - cos (a + ).l )]
Let
'"
III
be the overlap angle for firing angle a l .
.'
10
:::;
Vm Vm - L [cos a - cos (a + 11)1:::; - L [cos W
J
W
(Xl -
cos (a, + IJ. dl
s
or
cos a l - cos (a, + llt):::; cos a - cos (a +)..1) It is given that for a = O, ).l = 15° cos al - cos (a, + ).l,) = cos 0° - cos (0 + 15°) = 1 - cos 15° = 0.03407 or cos (a, + lll) = cos a 1 - 0.03407 (a) For firing angle 0 1 :::; 30°, cos (30 + lll) = cos 30 - 0.03407 :. ~l, = 2.i ~ (b) For 02 = 45°, cos (45 + Ill) = cos 45° - 0.03407 (c) For a3 = 60 cos (60 + ).11) =cos 60 0.03407 : . )11 =2 .23 This example demonstrates that for constant load current, overlap angle decreases as the firing angle is increased. 6.9.2. Three-phase Full Converter Bridge Fig. 6.39 shows a three-phase full-converter bridge with a sour ce inductance LJ in each lin e. The load current is assum ed constant as the ana lysis with puls ating current is quite complicated. In Fig. 6.40 (b) is shown the cond uction of + various SCRs with firing angle a :::; 0 and ove rlap angle )1. = O. In this figure ; T5, T6 conduct up to wt=30 From 0Jt=30 t o 90° (i .e. for 60°), Tl , T6 conduct. From wi = 90 to 150°; T1, T2 conduct and so on. It is seen that only two SeRs conduct at a time , one from the positive group and the other from the negative group. T2 Fig. 6.40 (c) shows the effect of overlap . From wt = 0° t o 30:0 ; T5 , T6 conduct. At = 30°, T5 is outgoing SCR a nd Tl is Fig. 6.39. T.hree-phase ful.1 converl.er wilh incoming SCR and both T5 , T6 be lona"0 to the source Inductance Ls In each line. positive group. As Tl is tri ggered, curr ent through T5 starts decaying while through Tl current begi ns to build up. At wt = 30;) + 11, 15 i",""'Zero whil e It = 10, Ther efore, from wi = 30° to 3D ~ + .u, three SCRs T5, T6, Tl cond uct. After wt = 30° +~ : T6, T1 conduct. At c.ct = 90°, as T2 is triggered , 15 begins to decrease and 12 starts to build up . Therefore, fr om wt = 90 ~ to 90: +.u. thr ee SCRs T6, TI , T2 conduct. At wt = 90° + .u.IG = 0 ilnd 12 = 10 , Aft er wi = 90: + u. only two SCRs T l, T2 conduct. This 3equence of operati on repeats with other SCRs of th e full converter. It may be obs erved from this that when positive group of SCRs arE unde rgoing commutmion. two SCRs fro m the posit i';e groU9 an d on e SCR from the negative group conduct. ..\.{ter th e com mutat ion of positive group is completed ; only tw o SCRs conduct , one fr om th e pO.5 itive grou p and the other fr om the negative group. Sim ii arly, when negat ive group of SCR.5 are un de rgoing comm ut ation, three SCRE cond uct . two fr om th e negati\'e group and one fr om the Q
-
Q
Q
wt
Q
Q
,
•
Q
318
[Art. 6.9]
Power Electronics
positive group and these are followed by !:\va SCRs, one from negative group and one from positive group and so on. Conduction of various thyristors as shown in Fig. 6.40 (c) is as follows: 5-6,5-6-1,6-1,6-1-2,1- 2,1-2-3,2-3,2-3-4,3-4, 3-4-5, 4-5, ,4-5- 6, 5-6 and so on. Vo
va.
vb
T~
,
!
t
k
'm
-oj"
{bl
i
a~O TS , T6
)J-O
ITS
N
T2
T4
T1
TJ
I
I
T2
!%lTTTJ.
~5'T1 Tl
~
T6
j
T2
N.
T6
T2
I
TS
I
T4
~
T4
I
T2
-YI!
~TS.T!
r,
~
T6
group
I
T1
14 ,T6
T2,14
I -tve group
T1
T6
fjT3,TS
A. I'T
T3
n ,T2
~
' 1-6 ,T2
11=12 = ')= 1,,= 15= ' 6 = ' 0
io
1)= 10
..I ~ (el
va. w
" n
,
T1
t"
p
0 ( 0)
ve
T5
16=10
,, )(
X'
~
io
[5= 10
-i
/J
r
-X :
IF10
,
Xi
,,'
group
-tv\!:
wI
X,:
,. ' X ,
, ,
'
'
I,
" ve group. ~ wI
'
'
1-
"
Fig. 6.40. Current and voltage waveforms for a 3-phase full converter showing commutation during overlap.
It is seen that three and two SeRs conduct alternately. tt is also observed from Fig. 6.40 (e) that for 6-pulse converter, there are six shaded areas indicating six commutations per cycle of source voltage. During commutation ofT5 and Tl (transfer of current from T5 to TI ), the output voltage is obtained by taking average of corresponding phase voltages ve and Vg of the positive group. This means that voltage from wt = 30° to 30° + ).L follows the curve
va +
v~
2 - from the positive group;
this is indicated by jk in Fig. 6.40 (a). During commutation ofT6 and T2 , th e ~ ltage waveform from the negative group is Vb; Ve as indicated by m the voltage is
Va
+ Vb 2 as shown by curve op and so
ft.
Similarly, during commutation ofT!, T3;
00.
In Fig. 6.40 , firing angle delay has been
taken as zero just to rughlight the effect ofsource inducta.l'lce. The above treatment is, h owever, applicable for any firing angle delay provided overlap angle is less than 60°. In Fig. 6.40 (a) ; Ua , Vb, Ve are the phase voltages and the output voltage is in be twee n th e hatched portion as 3hown.
319
[Art. 6.91
Phase Controlled Rectifiers
The effect of sour ce inductance Ls is to reduce the average dc output voltage, This reduction is proportional to the triangular (almost) areajk l shown in Fig. 6.40 (a). The average value of this fa ll in output voltage due to overlap is equal to the triangular area j k I divide d by the periodicity of this triangular area which is equ al to x/3. Thus, aver age value of fall in ou tput voltage du e to overlap 3 ~-
J' u
XOL
J' L -d i d(wt)
3 d(wt)~ -
xOSdt
3 L' J )I!W di 3 OJLs Jia . 3w L s oo·-dt= dl= 10 x dt n It Alternatively, average value of faIl in output voltage due to overlap ~--
°
°
_ 3 JI .di . _3oo Ls J lo , _ 3wLs - x/oo Ls dt dt It dt It 10
°
°
Output voltage with no overlap = inter nal voltage of the 3-phase full converter
3 Vml
~ -- cosa.=Vo
•
3Vml
3oo L~
•
•
Output voltage with ove rlap, Vox = - - cos a. ~n
... (6.76 )
10
general, for p-pulse converter, fall in output voltage due to overlap
~L J ' L (dild(wt)~POJL' J0'I.(di ldt 2It 0 dt 2It dt $
_P
-
OJ
J0 I
L,
2It
,
._
dt -
P
OJ
L, . 10 _
?
-
.... x
(J!....) 2 w L s 10 1t
From Eq. (6.58), mean output voltage is ap -phase converter, with overlap, is given by
V~ ~ Vmp(; ) sm(~ }osa-(~} wL, l, 2
"m
For a 2-pulse converter,
Vox =-- coscx -
F or a 3-pulse converter,
Vox =
~~. 10
• 3 {:f. V 2 It
mp
cos
...(6.73 )
• 3 wL, I,
...(6. 77)
2 1t
Ct. -
For a 6-pulse converter, (here replace V mp by Vmt>,
3Vml V ox ~ -It-
cos a-3ooL5" 1
..
0
(6. 76 )
It
An expression, similar to Eq. (6. 72a ) for a single-phase full con ve rter, can be writ ten for a 3-phase full converter as Vml
10 = 2wL~ [cos
.
Ct. -
cos (a + ~l)l
...(6.78 )
In Eq. (6.78), 2 w L; is written in place of ro Ls just bec ause two lines carry current 10 and each line has source inductance Ls ' Substituting t:tt e value of fo fr om Eq. (6.78) in Eq, (6.76), we get -
3V
m'
Vox = - - ' cos it
Ct. -
3
OJ
1t
L, ! 2 v,,/L -
Ws
(c o:;
CL -
cos
(a -
W'
1 -
J20
Power Elect ro nics
[Art. 6.1)) 3 V illI
= ~ [cos CL + COS (a + .u)]
... (6 .790)
= Maximum mean out~ut voltage at no load {cos a + cos (ex + Vo)] =
3 VmI where VOlll = - - for a
"
2Vom
3~phase
...(6.79 b )
{cos ex + cos (a + )J)] full converter.
3V In case overlap angle )J is zero, Eq. (6.79 a) gives V OJ: = --...!!:.!.. cos ex
"
=Vo' This is as expected.
Output voltage for a 3~phase full converter, similar to that given for a in Eq. (6.74), is given by 3 V mI 3 roLl Vo=--cos (cx+)J)+ 10 In a
3~phase
voltage is
v cb
+
2
"
l~pha se
.. .(6.80 )
"
full converter, voltage regulation due to source inductance 3 OJ Ll I , 1 = 1t x .,-;---,--''---;-----; Vo at no load 2 1t f. L , . 10
vob
full conv erter
... (6.81 )
; this is shown accordingly in Fig. 6.41 (a) . In vcb' vob ; first subscripts
c and a indicate the conduction of SCRs T5. Tl together from the positive group and second subscript b indicates the conduction of SCR T6 from negative group. When TI, T6 conduct. the output voltage is vab as shown. For T6 as the outgoing SCR and T2 the incoming SCR, the . average ou t pu t vo It age IS
L'ab
+ b SCrIpts · · 2 L'ac . Th e secon dsu m
Vab, V OC
. d · t e th. e con d uc t Ion · ormica
SCRs 6 a nd 2 tog ether from negative group and first subscript a indicates the conduction of SC R Tl fr om the positive group. The shape of outpu t voltage waveform and the condu cti on of SC Rs for the rema ining part of a cycle can be explained similarly. (e) Fo r firin g a ngl e of 30°, and overlap angle of 30°, the output voltage waveform Va is dr awn in- Fig. 6.41 (b). For obtaining va. it is preferrable to indicate first the conduction ofvariotls S(; R.:5 and then draw the output voltage waveform vo' In Fig. 6.41 (b ), the conducti on of diffe:e nt
311
(Acl. 6.9J
Phase Controlled Rectifiers
wI
..
0
TS,TI
~S ~ T4 ,T6 T6
I
T4
~
Tt
~
~
TS
~Tl ,T3 TI
T6 14,16
T3
TS.T1
~
T1 16,12
~
TS
T' 12.T4
(a)
-=/
~6.rf) _w group S,Tl
~T3 .TS T3
+v~group
Tl
~T6'T1 T1 ~"T4 T4 ~4.T6 T6
W;jS.Tl 15
~3.TS
=11,13
w1
~
T6
T1
P.G.
~~G
llo,r6
(b)
wI
'0'
~PtOkOI i o 0:2.4495A
• wI
(0)
Fig. 6.41. Pertaining to Example 6.27 .
thyris tors is shifted to the right by (l = 30° with respect to their position in Fig. 6.41 (a ) where a = 0°. Now the waveform for uo. as shown in Fig. 6.41 (b), can he drawn easily. The waveform for phase a input current, i.e. la. is shown in Fig. 6.41 (e), As the load is resistive, the waveform of ilJ IS identical to the waveform of voltage Va of Fig. 6.41 (b ). Current ia is positive when Tl is conducting and negative when T4 is conducting. Note that-to without overlap would have four tipples but with overlap, it i3 seen to hav e fi ve ripples in both positive and negative h alf cycles in Fig. 6.41 (c). Ex am ple 6.28. A 3-phase full conuerter bridge is connected to supply !.loUnge of 230 V per phase an.d a fre quency of 50 Hz. The source inductance. is 4 mHo The load curren.t on de. side is constan.t at 20 A. If the load consists of a de source of internal emf 400 V with internal resistance of 1 n, then calculate : (a) fi rin.g an.gle delay and (b) overlap angle in degrees. (d)
Powe r Electronics
[A r t. 6.9J
322
Soluti on. (a ) Converter output voltage = E + 10 R =400 + 20 x 1 =420 V. _ 3-16 · 230 _ 3(2. x 50)4 20 420 7t cos 0 1000 x 7t x
F rom E q. (6 .76 ), or
a= 34.382' :. Firing angle delay is 34.382"
420 = 3-16: 230 cos (a +~) + 3
(b) From Eq. (6.76), or
0+
i2;0~ ~o~ 4 x 20
396 x 7t = 42.6020 3-16 x 230 ~ = 42.602 - 34.382 = 8.22'
11 = cos
-1
:. Overlap angle in degrees = 8.220.
R
=
Exa mple 6. 29. A 3-phase M·3 converter, fed from 3·phase, 400 V, 50 Hz supply, has a'load 1 n. E = 230 V and large L so that load current of 15 A is level.
(a) Calculate the firing angle for inverter operation , (b) If source has an inductance of 4 mH, find the firing angle delay and overlap angle of the inverter. Solution . (a) When a conver ter circuit is working in 3V
ml ---z;t cos a= -
But
Vo = - E + 10 R
230+ 15 x 1 =- 215 V
0= cos
- 1[
- 215 x 2. ] 142 763' 3 x '1/2 x 400 = .
3Vml VQ.t =~ cos
(b) From Eq. (6.77 ),
aD inverter m ode,
3 0. -
L, 10 21t =- E + IaR
00
300 L~ . 10 3 2. =3f L,. I , =3 x 50 x 4 x 10- x 15 = 9 V 3 {2 x 400 _ 2• cos a - 9 = - 230 + 10 =- 215 V
F or a 3-phase converter, cos
(0
cos (139.7 12' + ~)
.
Overlap angle
206 x 2.] - 139 712' 3 'l2 x 400 .
-
- 1[ -
+ Il)
=
a - cos
cos
3f, L,. I ,
0 - -'-,V"':'~ mp
9xif = cos 139.712' - 'l2 x 400
cos 142.22'
Il = 142.22° -139.712"' = 2.508°
Example 6.30. A. 3·phase full converter f~eds a load with ripple free direct current 10 , If source inductance is taken into consideration, show that load current 10 is given by V
10
a
=
,
m =2 ~ , [cos a -
C03 (a
+ !l)]
... (6.78)
S olution. \Vaveform of outpu t voltage : '0 in Fig. 6.40 (0) is sketched with firing angle 0"' and with overlap angle Il. T his figu re shows that when Tl is triggered at a = 0°, cu rrent
.'
..
Phase Controll ed Re clifiers
in T5 begins to decay from Is (= 10 ) and that in T 1 begins to rise from zero, during the overlap angle Jl . Negativ e gr oup SCR T6 is already conducting a current 10 , Thus , during overlap period, equivalen t circuit of Fig. (6.42 ) can be obtained by referring to Fig. 6.39. This equivalent ci rcuit is similar to that given in Fig . 6.37 (b) for a single-phase full converte r.
Q
n L::Z)-''--.f~f'---;--,-~1----l
As T1 and T5 from positive group are conducting together, these got sho r t circuited as shown. KVL fo r the loop Q k l m n , during the overlap period, gives di 1
dt
Va -L3
Fig. 6.42. Equivalent circuit. of 3-phase full converter with conducting SeRs TI, T5 , TG during overlap, Example 6.29.
= U~ - L3
di5
dt
di,
or
diS]
vQ.-v~ =L, [ dt-Tt
F ig. 6.40 (a ) shows that v~ = V",p sin (wi + 120 0 ).
Va =
Vmp sin cot. As u~ lags U
a-
Vc
=V",p Isin we -
Uo
by 2400 , or
v~
Va
by 120::,
300 ) = Vm !. sin (wi _30 0 )
.. .tii)
sin (wt
=..f3vmp sin (wi From Eq. (i), we get
.. .(i )
-
leads
120°)]
di dis Vill i -1- - = - s in (wt- 30') dt dt L,
... (iii )
Load current during overlap period is also assumed constant. i1+is=Io di1
or
dis
Tt+
...(iu)
dt =0
Add ition of Eqs. (iii) and (iv) gives
-didt1 =-Vml sin (cot L3
300 )
... { v )
Fig. 6.4 0 (el, for zero degree firing angle , shows that curren t i1 rises to ' " in overlap angle JJ. . :. Wnen wi = 300 , i1 =0 and when rot = '3 0~ + 11, i1
=II =10
If firing angle is taken into consideration in Fig. 6.40 (e), thEn, -.... hen wt = 30= + a, i j = 0 and when cot =30 + a + Il, il = '} =10 , Th erefore, in order to take intf) account the firing angle CL, the Q
limi t.; cf integrotion for Eq. (v) must be (30' J,
J
,
di l
~ a ) = (, / 6 -
V i" -0-, = i l J~ ~ , l~- +
a ) to \30' + a + u) = ( ~ + a "
r"
si:;, \wt - It / n) dt
~;
324
P owe r Electronics
[Act. 6.101
I(~ · · · 'l/·
1= Vmf I -COS (wt - nI 6) 6 , 2ooL, ( ~ •• Vml
l/.
.
... (6.78)
Ig = 2 w L~ [cos a - cos (ex + jJ)} 6.10. DUAL CONVERTERS
Semi-converters are single quadrant converters. This means that over the entire firing angle range, load voltage and current have only one polarity as shown in Fig. 6.9 (a ). In this figure, Vo and 10 represent, respectively, th e average positive voltage and current of the semi-converter indicating rectification mode and power flow from ae source to the de load. In full converters, direction of current cannot reverse because of the unidirectional properties of SeRs but polarity of output voltage can be r eversed as shown in Fig. 6.9 (b). Thus, a full converter operates as a rectifier in fir st quadrant (both V o, 10 positive) from ex = 0° to 90 and as 0
an inverter (Vo negative but 10 positive) from a = 90° to 1800 in the fourth quadrant . This shows that a fu ll converter can operate as a two·quadrant converter, Fig. 6 .9 (b ). In the first quadrant, power flows fr om ae source to the de load and in fourth quadrant, power fl ows from de circuit to the ac source . 1
loa d
~
r"-.
,tlTIl VTlJ +1 ) A
L
0
A
B
IT!4
~T12
T22i 7T24 A
v.
B
0
- )
T2J "
(b )
(a)
load --A-,.
{ Til A
8
~TlJ
-
~T15frL "
2
'7 T22 )
Vo
126" 124
~
a
C
.f TI'
~
I
,
Vm ,V
T12 \ "
v
, (e)
I '7T2S
•
", "•
T2J
~T21
c
I,
2. ee2
Fig. 6.-:13. (0) Four·qundrant diagram . Non-circulating type (b ) single-phele dual converter and (c) three-ph ase dual converter.
T2I
Phase Con trolled Rectifiers
[Art. 6.10J
325
In case four quadrant operation is required without any mechanical changeover switch, two full converters can be connected back to back to the load circuit. Such an arrangement us ing two full converters in anti parallel and connected to the same dc load is called a dual converter. There are two functional modes of a dual converter, one is non-circulating-current mode and th e other is circulating-current mode. Non-circulating types of dual converters using sin gle-phase and three-phase configur ations are shown in Fig. 6.43 (b ) and (c) r espectively. If full converter marked I, to the left of load circuit in Fig. 6.43 (b) and (c) , is working alone, operation in first and fourth quadrants can be obtained. With full converter marked 2 working alone in Fig. 6.43 (b ) and (c), polarity ofload voltage as well as direction of load current, with respect to converter I , can be reversed. Hence, full converter marked 2 can operate in both second and third quadrants. Thus, a dual converter using tw o full converters can give four quadrant operation as shown in Fig. 6.43 (a ). 6.10.1. Ideal Dual Converter Assume that the dual converter consists of two ideal converters and that there is no ripple in their output voltages. Such a dual converter can be represented by an equivalent circuit shown in Fig. 6.44 (a). Val and V 02 are the magnitudes of average output voltages of converters 1 and 2 respectively. Diodes D1 and D2 shown here in series with the dc voltage sources VOl a~d V 02 indicate the unidirectional flow of current. The current in load circuit can, however, flow in either direction. The firing angles of both the converters are control1ed in such a manner that their average output voltages are equal in magnitude and have the ·same polarity. This can happen only if one converter is operating as a recti·fier and the other as an inverter. The average ou~put voltages for both single-ph ase and 3-phase converters are of the form. VOl = Vom cos a t ... (6.82) and V02 = Yom cos ~ ...(6.83} where, for a single-phase full converter, Yom =(2VII/ n) .. and for a three-phase full converter,
.
Converter 1
,,r-----' 01..., ,,, ,
i TVO'i rL __
l
___ .J
01
+
Conv erler 2
r----- --,,
,,, ,,
,, ,
h
Vo A
I-I
, L_ -
", L_--j Firing anglz con trol 0 Ql + 02 = 180 (~
Vo
, ,, ,
--" 1Cz 1
J----'
Vol
V02
0' BO'
,,
.
,~
0:i
/9ci\\
,
,,' /
"
.. .
J ', , !
...
..,;:
- -- Rec lilicotlon ----- -- Invusion ~
Fi g. 6.44 . (a) Equivale nt circcit of an ideal dual converter (b) Variation of terminal voltage for a n ideal dual converter with firing angle.
Under normal operation, output voltage Val of converter i has upper positive and lower n egative polariti es and out-pu t voltage V02 of converter 2 h as upper negative and lower positi .... ~ polarities in F igs. GA3 (b) and (c). H i3 aS3umed that the two converters have their a .... erage
Power Electron ics
(Art. (dO]
.l2fl
output vol tages equal in magnitud e. Their output voltages would have the same polarity only if polarity of V o:! is r eversed. In other words, their average output voltage Yo can be expressed Vo =
Substitution of VOl and or or
Also
=-Y02
... (6.84)
from Eqs. (6,82 ) and (6.83 ) in Eq. (6.84) gives,
V Dm cos a 1 =- V om cos ~ cos a 1 =- cos ~ = cos (180 -~ ) a 1 + a2= 180° cos al =- cos ~ = cos (180 + 0 = 180 + a,
"1
or
(J.!
V 02
VOl
... (6.85)
av ... (6 .85 a )
As per Eq, (6.85 a), for some value of firing angle O:z, Ct 1 is always greater than 180°. But can never be greater than 180°. Therefore, the solution as given by Eq ,·(S.85) is only possible.
From Eqs. (6.82), (6.83) and (S.85 ), the variation of output voltage with firing angle for the: two converters is as shown in Fig. 6.44 (b), The firing control circuit chang~s the firing angle 0. 1 and 02 in such a manner that Eq, (S.85) is always satisfied. . ' ~ .:: + S.10.2. Practical Dual Converter With the firing angles controlled in a manner that a 1 + ~ = 180 0 and with both the converters in operation, their average output voltages are equal and have the same polarity. On e converter will be operating as a rectifier with firing angle 0. 1 and the other as an inverter with firing angl e (180° - 0. 1), Though their average out put voltages a re equa l, yet their insta ntaneo us voltages VOl and v02 are out of phase in a pra ct ical dual converter. This results in 3 vo ltage difference when the two converters are interconnected and as a consequence, a large circulating current flows between the two converters but not through the load . In practical dual converters, this circulating current is limited to a tolerable value by inserting a reactor between the two converters as shown in Fig. S.45. The circulating current can however, be avoided provided the converters are triggered suitably. In general, a dual co nverter can be operated in th e following two modes. Dual Converter without Circulating Current. With non+circulating current dual converter, only one converter is in operation at a time and it alone carr ies the entire load current. Only this converter receives the firing pulses from the trigger control. The other converter is blocked from conduction; this is achieved by removing the firing pulses from this converter. Thus, only one co nverter is in operation at a time whereas the other converter is idl e. Such an arrangeme nt for the dual' converters is sh own in Fig. 6.43 where there is no reactor in· between the two converters. (a)
Suppose convert er 1 is in opera tion and is supplying the load curr ent. For bl ocki ng co nvertor 1 and switthing on converter 2, first .firin g pu lse~ to con~ert e r 1 are immediately removed or the firing angle of converter 1 is in creas ed to maximum valu e and then its fir ing pulses are bl ocked . With thi s, load curren t would dec;:>.y to zero and then only converter 2 is made to conduct by applying the firing pu ls es to it. Now th: cu rr en t in converter 2 would build up through the load in th e reverse direction. So lo ng as con\' ut er 2 is in op eration , convert er 1 is idle as firing pulses are wi thdrawn fr om it. It should be en::ured that duri ng cha ngeov er from one co nverter to the other, the load current must dec ay ti) zerv. ..l,.fter ene outgoing con'Ierter has stop ped cond uctin g, a del ilY tim e of 10 to 20 m ~e c is in troduced befo re the firing pulses are ::lpplit:d tu :i\vitch u l1 th e inco ming com·ert cr. Thi" time :!:,l~ y i?-:13u r -;? .5 relia ble commut rt ti on of
Phase Controlled Rectifiers
[An. 6. 101
327
SCRs in the outgoing converter. If the incoming converter is triggered before the outgoing converter has been completely turned·off. a large circulating current would flov\,' between the two converters. With non·circulating current mode of dual converter. the load current may be continuous or discontinuous. Th e control circuitry for the dual converter is so designed as to give satisfactory operation during continuous as well as discontinuous load current.
Dual converter with circulating current. In the circulating current mode of dual converter. a reactor is inserted in-between converters 1 and 2 as shown in Fig. 6.45. This reactor limits the magnitude of circulating current to a reasonable value. (b)
~~"~':"=::::;7;::=+1 Reoc\er
~
Til ,
;:
T1J ~
"" Ii'
TI4
T1Z
0 A
.j
~ T22
T24
,j A
", "0,
0
B
if -
.
+
l~+
A B
.'e,l
~/~
_d
-
T2I
d
.
•
;
T2J
•
2. OC 2
(a )
Reactor
It
Til
TIl
II
li+
~r'''. ~ L/2 LI2
TIS
l~+
A B
""
C
I; Tl4
.
T16 ,
Ii
Tl2
0 A
0
+ d
j
T26
J
T24 A B
", ""
II -
-
(- T22
_J
C
T2S
1 T2J
d
d
T21
•
• (b )
Fig. 6.45. Circulating current type dunl converter for (a) single-phase supply and (b ) three-phase 3UllpJy. The firing pulses of the two converlers are. so adjus ted that al + ~ = 180". As for example. if firin g angle of converter 1 is 60°. then firing an gle of converter 2 must be 120°. Therefore, for these firing angles. converter 1 is working 03 a r ec tifier and conver ter 2 as an in .... erter. Though the output voltage at the termina1s of both conVene rs 1 and 2 ha3 th e same ave rage value and also has the s ame polarity. their l03tantane ous output voltage waveforms, hov. .·ever. are not similar as shown by UOl and u02 in Fig. 6.46 (b ). A:::i a con:iequenca otit , circulating current flC.W3 betwee n the two converters. This ci rculating current i3 limit ed by th e reactor. If th e load current is to be reversed. the role of two converters is interc hanged. This means that conve rter
328
Power Electronics
[Art 6.10] v,
(a)
v,, ( b) "'ob
v."
A
/1
/' "
"'OC "....
I .. I ,/
V~.....
VIKl,.....
'1""1" I " I,'
'bl....... ;1
Vcb .. ",
I -' I,' I" /
•"
,,'
...
1 t
I .. '
~h _ ""
"'OC/"
orl
...
I .' "
..
I
I / I,'
WI
v,t ~lood voltog.
e:::::=:: . .
(cl v,
Reactor voltoge
(d )
-------~---------
•
WI
WI
i
-
!I,
,
(I)
WI•
Fig. 6.46. Voltage and current waveforms for a circulating-current type dual converter.
1 is now made to act as inverter by making its firing angle greater than 90° and converter 2 is made to work as rectifier by making its firing angle ~ less than 90 0 such that a 1 + U:! = 180 0 • The nonn al del ay period of 10 to 20 msec, as required in circulating-current free operation, is not needed here . This makes the dual converter with circulating current operation faster. The main disad van tages of this dual converter are as under : (i ) A r eactor is required to limit the circulating current. The size and cost of this reactor may be quite si gnificant at high power levels. (i i) Circulating curr ent gives rise to more losses in th e converters. hence the efficiency and power fac tor are low. (iii ) As the con ver ters have to handle load as well as circulating currents, the thyristors for the two co nverter s are rated for higher cu rren ts. In spite of thes e d rawb o.cks, a dua1 converter with circulatin g current mode is p r e f~rr ;d if load curren t is to be r eversed quite fr equently and a fas t response is desired in the fo ur-qu adrant oper ation of the dual converter.
Phase Controll ed Rectifiers
[Art. 6.IOJ
329
Dual converter operation with waveforms. The operation of the dual converter in the circulating.current mode is described here under the following assumptions : (0 The reactor is loss less. (ii ) The firing angl es of the two converten; are so controlled that a t + ~ = 180°. In Fig. 6.46 (a ), supply line voltages uab. Uac ubc etc. are shown. As an illustrative example for describing the working of a dual converter with wavefonns , let at be equal to 60°. Then. for converter 2. ~ =180° - 60°.= 120°. With a l =60°, the output voltage UOl for converter 1 is indicated by thick line in Fig. 6.46 (a). This output voltage Val is now shown as Vgb in Fig. 6.46 (b ) from wi = 120° to 180°. In this manner, Val is drawn for other intervals of time in Fig. 6.46 (b) . With ~ = 120 0 for converter 2, the output voltage is negative as shown by thick line in Fig. 6.46 (a). A!J the average values of output voltages of both the converters are positive, the output voltage v02 of converter 2 must also be shown positive above the reference line wt . This output voltage vab indicated by thick line in Fig. 6.46 (a ) is now sh own as vila dotted in Fig. 6.46 (b) fr om rot = 180° to 240°. In this manner, vQ2 waveform is drawn positive as uba. u ea . veb etc. in Fig. 6.46 (b ). The load voltage U o is equal to the average value of tne instantaneous converter output voltages VOl and vQ2, i.e. va =
"0= At OJI = 30' ,
"0=
= 60' ,
va =
At OJI
Val
VOI .t vQ2
...(6.86 )
2 V ml
sin 60°+ 0 2
=0.433 Vml
Vml sin 30° + V ml sin 30°
2 0+ Vml sin 60°
2
..
=0.5 Vml
= 0.433 V ml and so on.
This load voltage wavefonn Vo is shown in Fig. 6.46 (e) . Th e re ac tor volt a ge ur is equal to th e difference of converter outpu t voltag es and U02. i.e. U,.
= Val -
V,.
= V ml sin 60° - 0 = 0.866 V ml
...(6.87)
U02
At wt =30°, v ,. =V ml sin 30 - V ml sin 30° =a At Wi = 60°, v r = 0 - Vml sin 60° = - 0.866 VmJ and so on, The variation of reactor voltage Vr is plotted in Fig. 6.46 (d ) Now
where ie is the circul ating current through both the converters and r eactor L . Th e waveform of ie can be drawn from the waveform of v,. as under : At w! = 0°, vr i3 maximum and positive, therefor e slope ofic m ust be maximum and posi tive
di
to satisfy th e relation ur = L d'; ' Thus, ic is shown rising in Fig. 6,46
(e )
with a ma ximum
positive slope..-\t u:t = ",,0", Vr = 0, therefore slope of ie = 0 ; this is possible only wh en ie is di maximu m wi th d~c = O. After wl:: 30°, vr starts becoming negative, the valu e of ie also s tart.s
33U
Power Electronics
[Art. 6. 10]
falling so that diJdt is negative. At 90°) works as an inverter and handles only the circulating current ie' The waveforms for these currents i} and i2 are shown in Fig. 6.46 (f). The expression for the circulating current in the dual converter can be obtained from voltage U r across the reactor. It is seen from Fig. 6.46 (b) that for the time interval (1t/3 + a1) < wt < (n/3 + a1 + n/ 3), the converter output voltages are UOI = uab and u02 = Uhc The reactor voltage Ur , from Eq. (6.87), is U,
= uO l -
U02
=
... (6.88)
Uob - ubc
It is seen from Fig. 6.46 (a ) that
=Vml sin wt . As tlhc lags tlab by 120", it is given by tI~ =V ml sin (oot - 120 tI, = Vml (sin 0Jt - sin (rot - 120 ...(6.89) u, =...J3 Vml sin (rot + n / 6) .
tlah
0
)
0
or
The circulating current giv en by
ic
is obtained from the time integral of reactor voltage
If'
i, = L =
or
ic
=
(a 1 + 1'l/3)/ w
,[3 . Vm l .. r I.I..I.U ,[3 ' V m1 wL
)]
m1 u, · dt= ,f3,v L
J'" (ll
+ rei
3
f' . (a l + 1'l/ 3)/ w
Vr
and is
sin(wt+n/ 6) dt
sin (rot + n/ 6) . d(rot)
... (6.90) .. .(6.90)
(- sin a 1 - cos (0Jt + n / 6)}
It is seen from Eq. (6.90) that the magnitude of circulating current depends upon the firing angle 01 and upon tilt. For any value of firing angle, the peak value of circulating current occurs when Wi =5n/S. This peak value icy is then given by . l ,p
=
,[3vm1
wL
.
[1- SIn all
... (6.91)
Eq. (6.91) shows that peak value of circulating current depends upon the fi ring angle 0. 1, For (Xl = 0, the maximum value of iep is ..J3 V mllwL and for 01 = 900 , icp = O. In Eqs. (6.90) and (6.9 1), Vn:1 is the maxim um value of line voltage. E xampl e 6.31. A 3-phase dual converter, operating in the circulating-current mode, has the fo llowing data:
Per phase supply uoltage =230 VJ f=;5 0 Hz, Calculate the peak ualue of circulating current.
01
=60=,
current limiting reactor, L
=15 mHo
\
..
Phase Con trolled Rectiiiers
(Art.6. 11J
Solution . The peak value of circulating current, for firing angle (6.9!). ', :
SO~IE
= 60°. is given br Eq. "
ra · ,J6 · 230 3 (l-sin60· J:27.7425A. 21tx50 x 15 x lO-
p
6.11.
Qt
331
WORKED EXAMPLES
In this article, some typi cal problems on phase controlled rectifiers are solved. Example 6.32. A single-phase full converter is supplied from 230 V, 50 Hz sou rce. The load consists of R = 10 n and a large inductance so as to render the load current constant. For a firing angle delay of 30°. determine (a) average output voltage (; ) average output current (c) average and rms values of thyristor currents and (d) the power factor. Solution. The waveform s for source voltage v. load current i o, load voltage vo, thyristor cur rent i T! (o r i n) and source current i J (refer to Fig. 6.10) are drawn in Fig. 6,47. u, wI
10
~----~----~--r-~~---++-
o,Lo'-------+.~ , ------tl~'---L~-+.~'~-----+~----~w~ 1
i r -J N O~K!~
Uja~l N o· ::;
Ii
II i, ! i.
ii
o ,
!:T3T4 , ~T1T 2 I I
a 0
••I
- . '-
, ,
10
TJTl.
/
(n ... a )
-1 0
I :, ,
I
t ';
,
T1T2
I.
I
,
( 2nt Q)
I i ,
•
wI
, I , T3T4-n , 'r -
II, i I/ (,,··l "-
:[
wI
"I
i
"I
wI
Fi g. 6.4 7. Pertaining to Exa m pl e 6. 32. (a )
For a single-phas e full conve rter, average output voltage.Vo. Eq. (6.26), is give n by 2Vm
V o = - - cos
n
Q =
Vo
(b) Average output current, f o = R =
2../2 x 230
n 179.303 10
cos
3 0~
= 179.303 V
=17.93 A
(c) It i;i see:1 fro m th e waveform of thyris tor curr ent i n lo r in ) that its aver age value is gi ven by
IT
.\
" 10 17.93 B • • , =I '-,. ) :r =? =- - 0 - = .dQ:}.-"\ - I) '
332
Power Electronics
[Act. 6.11J
Rms value of thyristor current is
_ ,r,.,---'lIT ., =
'I'" n x 2n
10 17.93 =Tz =~ = 12.68 A
..vPo . ~ =10=17.93 A
(d) Rms value of source current, I, = Load power Input power
= Volo = 179.3 x 17 .93 W = V, I, cos ~
For no loss in the power converter,
=Vo 10 179.3x17.93 C08 ~ = 230 x 17.93 = 0.7796 lag
V, I, cos :. Power factor,
~
In general, for a I-phase full converter with ripple free load current as in this example and with no device drops, input power =load power or .., I nput
ViI, cosq.=Volo 2 Vm
1
pf= -n- cos a · 10 x V, . 10
1 212 = 212v. cosa · -=--cosa Tt V, Tt For this example, input pf= 2-.12 cos 30 0 = 0.7796 lag . .n
Example 6.33, In Example 6.32, if source has an inductance of 1.5 mH, then average output voltage (b) the angle of overlap and (e) the power factor Solution. (a) From Eq. (6.73), average output voltage is
cUttrmi~
(a)
2 Vm waL, VOl: = - - cos u- --10 • n = 212 x 230 cos 30' _ 2. x 50 x 1.5 X 10-' x 17 .93 •
n
= 176.614 V (b) From Eq. (6.71),
Vm
10= CJlL [cos a- cos (a+ Jl)]
•
12 x 230 X 10' 17.93= 2nx50 x 1.5 [c0830-C08 (30+1')1
or
I
I ts simplification gives overlap angl e, I' = 32.855 - 30 = 2.855'
(c) Power factor
= Vol o = 176.614 x 17.93 = 0 76791 V, I, 230 x 17.93 . ago
E -x ampl e 6,34 . A 3-phase fully -controlled bridge con ue rter with 415 V supply, 0.04 n resistcnce per phase and 0.25 n reactance per phase is operating in the in.vert:. . . g mode at a firing aciu a.nce an.gle of 35 0 • Calculate the me an generator uoltoge when the current is le vel at 80 A The thyristor uoltage drop is 1.5 V. [IA.S ., 1994 ]
I
1-'
I
Phase Controlled Rectifiers
[Art. 6.11]
333
Solu~ion. Power circuit diagram of a 3-phase full converter reveals that source resistance r f will lead to a voltage drop of210 r,. Two thyristors, one from positive group and another from negative group, conduct together, therefore there will be a constant thyristor voltage drop of 2 V T . The source reactance leads to overlap and its effect is taken care of by Eq. (6.16). By taking into consideration these voltage drops, the average. or mean , output voltage Vo in a 3-phase full converler is given by
V
ax
3VmJ 3wL, =--coso:-21,r - 2 V r - - - Io n ' n
In case 3-phase fun converter is working in the inverting mode, then the load emf E or V, (mean generator voltage in this example) can he obtained from the relation:
3 Vml
.
- n - cos a =-E + 210 r f
3mL,
-t: 2VT + -- Io n
3,j2 x4 15 cos (lBO-35) =-E+ 2x BO x 0.04+ 2 x 1.5 + 3 xO .25 x BO n
or
n
E = 459.022 + 6.4 + 3 + 19.1 = 4B7.522 V :. Mean generator voltage : =E = 487.522 V.
Example 6.35. In Example 6.34, in case load consists of RLE, with R =0.2.n. inductance large enough to make lo:ad current leuel at 80 A and emf E , then find ·the mean value of E for (i) firing angle of 35° and (ii) {iring advance angle of 35°. Solution. (i) When firing angle is 35°, 3-phase full'converter is in the rectifying mod e. Therefore, from Example 6.34, Vax
3V
mJ =E +loR =-ncos 0: -
3wLf 2 10 r, - 2VT - - - I, •
3Vml 3wL, - - cos a =E + loR + 210 rs+ 2VT + - - I ,
or
•
•
3,j2 x 415
_ '. 3 x 0.25 "-''''-== cos 3,,° =E + 80 x 0.2 + 2 x 80 x 0.04 + 2 x 1.5 + x 80 or
•
•
E = 414.522 V. (ii) For firing advance angle of 35°. the full converter is in the inverting mode. From Example 6.34, 3Vm I - - cos a
•
=
3wL - E + loR + 210 r, + 2Vr + - -s 10
•
E = 459.22 + 16 + 6.4 + 3 + 19.1 = 503.522 V. Example 6.36. Fig. 6.48 (a) shows a battery chargin.g circuit using SCRs. The input voltage jn~ neutral to any line is 230 V (rms) and firing angle lor thyristors is 30°. Find the auerage current flowing through the battery. Derive the expression used. S olution. For the parameters given in trus ex ample, th e waveform ofload current is drawn in Fig. 6 .48 (b). When thyri sto r Al is gat ed at a= 30°, it begins cond uction at wt =30 + a = 60°. After its turn·on, when V mp sin ~ = 150 V, thyristor Al gets turned off at wi = ~. Note that h er e ~ is more than 90° as is seen from Fig. 6.48 (b ). Equation governing the con duction period in Fig. 6.48 is or
V ..... sin Wi -E=
ioR
334
Power Electronics
[Art. 6.11)
0" ·
AIV'
o
~
&8 ,
BI...,
~C £. ,
CI "'"
0
'
•
,
'I' , , . ,
wi
I' ! ' . , t : I : I
A
,:/31 11 ' I;' i1i111/J1H I Vmp-E
I5~V
.,
5.n
.
I'
\0
o
~,
: I : , ,i r t -:--TR :' I ''I I I ,,I01 ! " ' ' to r
'.
'1
,I
't'
." ,
., I
I
I
'"
,
I
,
wi
M
N
(al
(bl
Fig. 6.48 . (a) 3-phase half-wave battery charging circuit and (b) its relevant waveforms, Example 6.36.
between the limits of (~ +
(l) and ~
3 '~
10 = - J 21t a+.!!.
> 90°. Thus, average output current is g;ven by
Vmp sin wt. -E
6
R
. d (oot)
m, Icos ",t l a~ + ~/6 - E (~ - a - 30 )J
3 = 2rtR IV 3
= 2rtR IVmp (eos (a + 30) - cos ~
0
~I
- E (~ - a - 30 0 )J
H ere .J2 . 230 sin ~ = 150 V. This gives ~ = 27.47 :1 or 152. 53°, As = 152.53°. This gives th e value of average battery current 10 as under :
~ > 90°, therefore
[,f2.
10 = 2.3 5 230 (cos 60° - cos 152.53°) - 150 (152.53 - 30 - 30) x x 3 = 10. [208.91928J = 19.95 A.
1~0]
Example 6.37. A single-phase semiconuerter, using two thyristors and two diodes as shown in Fig. 6.49 (a), is supplied fro m 230 V, 50 Hz source . The load consists of R = 10 n, E = 100 V and a large inductance so as to render the load current level . For a firing delay angle of 30°, determine (a) a verage output fJoltage (b) average output current (c) average and rms values of thyristor C/lrrents (d ) average and rms values of diode currents (e) input power factor and (fJ circuit turn ·off time. Solution. The wa veforms for vo lt ages and currents are sketched in Fig. 6.49 (b ). When forward-biased thyristor T1 is triggered at firing angl e a. T1D1 start conductmg the constant current 10 , Soon after wt = It. as supply voltage tends to go n egative, diode D2 gets forward biased through D1. Therefo:-e, fr om c.:U = It, loa d current begins to freewheel through T1D2. ThYTistor T2 gets forwa rd biase d after wt = ft . At wt = It + Ct., whe n T2 is turn ed
i,
• (~)",
.l'n
•
.1'"
•
I
R
a
~
"0 b
01
02
-
c
~
L
E: IOOV
(a l
Fig. 6. 49. Pertaining to E:
I
I
Phase Conlr olled Rectifiers
[Art. 6.111
335
on, current 10 begins to fl ow through T2D2 as shown. Soon after wi =27t, as supply voltage tends to go positive, diode Dl gets forward biased through D2. As a result, current flows throu gh T2D l till Tl is turned on at rot = 21t + a and so on. The waveform of output voltage tlo shows that average value of output voltage is given by
Vm
Vo=- (l+cos a)
",
.
"
"
.. .../.
•
" be
"
,
O ~-L~*4'----~~__~~__~,-~ wI
wI
wI
Fig. 6.49. (b ) Voltage and current waveforms pertaining to Example 6.37. (a ) Average
valu e of output voltage
. v'2 . 230
Vo =
(0)
• Vo =-.: ~ 10 R
(1 + cos 30°) = 193. 172 V
193.172 = 100+Iox 10 Average value of ou tput current
10 = 93i~7 2 (c ) It is seen from the wa veforms
=
9.32 A
of thyristor current iTl and diode currem i Dl tnat both
conduc t fo r r:: r adians fo r any val u e of ruing delay an gle. On accou nt of th is , th e circuit of Fig. 6.49 (a) is someti mes call ed symmetrica l configuration for a si ngle-phase semiconver ter.
[Art. 6.11]
336
Power Electronics
Average value of thyristor current
,
I,
9.32
IT . .=I,-=-=--=4.66A 2, 2 2
Rms value of thyristor current
(d) Average and rms value of diode currents are the same as those for a thyristor as discussed in (c) above.
: . Average value of diode current = 4.66 A Rms value of diode current = 6.591 A (e)
Rms value of source current l, .r=
./",,-a
'410-,-
= 9.32...j' -
Rms value of load current
:. Input
~/6)
= 8.508 A
lor =10 = 9.32 A.
Power delivered to load Also
_~
=10 '1----;--,-
=
E I, + ~ X R = 100 X 9.32 + 9.32' X 10
230 x 8.508 x cos $ = Power delivered to load
P{
= 932 + 9.32' x 10 = 0 9202 1
230 X 8.508
.
ag
(f) It is seen from the waveform of vn that circuit tum·off' time is
t
,--,6
e
1t- a =- = 21tx 50 x 1000 rns =8.33 ms, co
Example 6.38. (a) Describe the working of a single-phase dual converter with appropriate woueforms.
Derive expressions for the auerage output voltage and the circulating.current. (b) A single-phase dual converter is fed from 230 V, 50 Hz source. The load is R = 30 nand the current-limiting reactor has L = 0.05 H. For Cl 1 = 30", calculate the peak value of circulating current and also the peak currents of both the converters. Solution. (a ) The circuit diagram of single·phase dual converter is shown in Fig. 6.45 (a ). Singleophase voltage applied across terminals A, B is sketch ed" in Fig. 6.50 (a ) as v.I ' Let the firing angle of converter 1 be Cl i • say around 300 or so. Waveform of output voltage VOl across output term ina ls of converter 1 is show n in Fig. 6 .50 (b). For conver t er 2, ~ =180 - Cll = 1.50 0 and wavefurms of its output voltage v02 is shown in Fig. 6.50 (c). Since v02 is mostly below the reference line wt, it3 ave rage is n egative ..A,.z. per the polarity marking:; of ou tput voltages Va l and va:! in Fig. 6.45 (a ), the average values of output voltages of both the converters m ust be positive. Thus, the wa veform of output voltage V 02 of converter 2 mu:;t be shown posi tiv e above the r eference line rot, this is shown in Fig. 6.50 (d)
J
Phase Controlled Rectifiers
337
[ArI. 6. 111
(a) wI
(b )
Ie )
(d )
"" o "'-7- - -If-?:;--.----+*.;+--+·7·· ~--;-;;271 . lI' i w !
(e )
"' l~ ~; ~
o~--1J.',·,.···[]· ·,.--· [}···.
~()' t :-
,
U
i:
!! !!
i : 'I
.
,-
.
2Vm si'la,
a ,:
(I )
lI'-Cl ,
: '
o ~'-2Y",sin a , !'rr
; :
.
"
I fI ~a l V
:
wI
• 21THl1
V
wI
,:
-'.
1r ~ a , -----;
Fig. 6.50. Waveforms for single-phase dual converter, Example 6.38.
uOI = Vm sin oot and UIr.!
Now
: . Load voltage, to oot = a} and from
=Vm sin oot
= ~+V~ 2 =Vm sin oot as shown in Fig. 6.50 (e l. .Note that mt =1t - a} to wt =Jt, load voltage tlo = 0 Uo
1 :. Average value ofload voltage, Vo = Jt
As per E q. (6.88), voltage
ur
fr om oot = 0
r:t- o.\ Vm sin oot . d (wt ) = -!!! 2V cos a }
across reactor is tlr =
.. (i )
1t
0. ,
Vo l - tI 'll"
Waveform s of Vo l an d
u u2
of Fig.
6.50 (d) reveal that
from oot
= 0 to rot = a} .
from rot
=
ur =- 2 Vm sin rot
a 1 to wt =Jt -al,
tI,=
0
and fr om wt = Jt - a 1 to wi = Jt, ur = 2 Vm sin wt and so on . Wavesh ape of ur is sketched in Fig. 6.50(/) . Note that wh en wt = a l • uf = - 2 Vm sin Ct. l an d at rot = It - CL}. vr = 2 Vm si n CL I . !fie is t he circulating current due to
or
UrI
di then u,= L dte
Powe r Electronics
(Art. 6.11J
338
Th.:! limits of integration for
urI
from Fig. 6 .50 (fJ, are seen to be from zero to 2Vm 2Vm sin rot . dt = wL 1 - cos Wi
lrl/W
i, = L
0
2 Vrn
= CIlL [1 - cos
0. 1
aI '
a /w
Ie, I
J
...!ii>
In case time t is to be included in the i( expression, then
If"- 2 Vm sin
ic = L
I
(a)
t dt
= 2V oor.m (cos tilt -
Maximum value of circulating current icp occurs when cos o
Lcp
(b)
2 Vm = mL [1- cos
rot
cos all
= 1. ... (ii)
0. 1]
From Eq. (ii), part (a) peak value of circulating current i,p =
2!~; ;~005 [1- cos 30'J =5.548
A
Vrn ,J2 x 230 Peak value of load current= R = . 30 = 10.84 A :. Peak value of current in converter 1
= 5.548 + 10.84 = 16.388 A
= 5.548 A
Peak value of current in converter 2
Example 6.39. A semiconductor switch is used to connect a load of 5
n, 0.05 H to a 240 V,
50 Hz supply. Estimate the triggering angle to ensure no current transients. Also indicate the initial triggering angle for the worst transient. (a).
Solution. A semiconductor switch' connecting a load RL to ac supply is shown in Fig. 6.2 Th e solution for the load current io for this converter, given by Eq. (6.7) is repeated here.
.
Vrn .
•.• , -cp) -ysm Vrn . (a-tp) e~ . {-wL R (rot-a) } .. =l,+ll
~o=ysm( ......
H ere magnitude of transient component of load current io is
.
. {R
~. ll=Zsm(a-v)exp - wL{wt -a)
}
In order tha t no current transients occur, when SCR is turned on at a triggering angle n , il must be zero. This is possible only if sin (a - ¢) = 0 = sin 0° or sin 180°. :. Trigg 2ring angle
a
=$ = tan-1 2 1t X 5~ x 0.05 = 72.340
Therefore, wh en trigger ing angle a = $ = 72.34°, no current tr ansients would occur. For wor3t tran3ient current, sin (a - ¢) :. TrigguL"1g a:'lgle
a
= 1 = sin 90"
=90 + ¢ ~ 90 + 72.34' =162.34'
Thuefore, when trigge ring angle a cir::uit.
=
162.34°, worst current transients wOlJld occur in the
Phose Controlled Rectifiers
[Pcob.6J
PROBLEMS
339 .
6.1. (a) Give at least five applicntions of phase-controlled rectifiers . (b ) What is an ideal thyristor switch ? (c) Power flow from I-phase source to load R can be controlled through the use ora thyrislor. Discuss why this method of power flow control is called phase-controlled converter. (d) Give at least two definitions of firing a ngle. lllustrDte your answer with relevant circuit a nd waveforms. .......... 6,2, (a ) A single-phase half-wave SCR circuit feeds power to a resistive load . Draw waveforms for source voltage, load ,toltage, load current and voltage across the SCR for a given fi ring angle 0.. Hence obtain expressions for average and rms load voltage! in terms of source voltage and firin g angle. (b) A r esistive load of 10 0: is connected through a half·wove SCR circuit to 220 V, 50 Hz, single-phase source. Calculate the power delivered to load for a firing angle of 60 ~. Find also the value of input power factor. [Ans. (b) 1946 . 887 W, 0.63421
6.3.
An RL load is fed from single-phase supply through a thyristor. Derive an expressio n for load current in terms of supply voltage, freque ncy, R, L etc. Indicate the ti me limits during which this solution is applicable. For this thyristor-load combination, draw waveforms for load voltage, load cur.rli!nt, sou:-ce current and voltage across the thyristor. (b) An RL load, energised from s ingle-phase, 230 V, 50 Hz source through a single thyristor. h as R :: 10 nand L = 0.08 H. If thyristo r is triggered in every positive half cycle at a:: 75 0 , find curreot expression as a function of time.
(a)
r Ans. (b) 12.023 s in (314 t -
//
68.3°) - 2.3614 e-t:!511 •
6.4 . A single· phase half-wave converter is operated from 230 V, 50 Hz source nne: the load resistance is R :, 12 n. For a firing angle delay of 300 , determin e (a ) the rectificatio:'!. !:fficiE"ncy (b) form facto r (c) voltage ripple factor (d) transformer utili z.:ltion foetor and (el PIV of thyristor. ' (Ans. (a) 36.3 1% (b) 1.66 (c) 1.325 (d) 0.253 (e) 325.22 \'] 6,5 . (a) For 9. single-phase one-pulse controlled converter system, sketch waveforms for load voltage and load current for (i) RL load and (ii) RL load with freewheeling diode across RL. From a comparison of these wavefo rms, discuss the ad v anta~es of using a freew:-tee;· ing diode . . (b ) A battery is charged by a single-ph as~ one-pulse thyristor controlled recti5er. The supply is 30 V, 50 Hz and battery emf is constant at 6 V. Find the resistance to be lnsertec. !:-: series with the battery to limit the charging current to 4 A on the assu mption thnt SCR is triggered continuously. Take a voltage drop of 1 V across the SC R. Derive the exp res.~iQ n used. [Ans (b ) 2.54Si (2;
6.6. (a) A dc battery is charged through a r esistor R as shown in Fi g. 6. 5 (a ). Derive an ex;>ressio:: for the average value of charging current in terms of Vm , E, R, fi.:-ing-:.1 ngle deia;; a. ~~C. Here a> 9 1 where 9 1 can be obtained from Vm si n el :: E. (b ) For an nc sou rce voltage of 230 V , 50 Hz, find the value of ave!'3ge charging cu:-rent r(l~ R = 10 n, E;: 110 V and for firi ng angle delay:: 30~. (c) Derive an expression for the rnL!l value of charging current fo.. r.
pr.
iAns. (a ) Eq. (5. 18), (b) 5.3743 A ~:
:0) [
2~: t(V~ + E' ) (. - 9, - 0) + ~; (sin 20 + si n 29,1- ~ '/~ E .'os a " cos 9, ) ~ ! 591.173 '1Y. 926.175 'IV (d) 0.6855 lag]
:1
Power Electronics
340
6.7. A single-phase one-pulse SCR controlled converter feeds an RL load with a freewheeling diode across the load, piscuss how freewheeling diode comes into play when supply voltage is passing through *ro and becoming negative. Sketcb wavefonns for supply and load voltages, load current, supply current, freewheeling diode current and voltage across the SCR. Derive expressions for the load current as a function of time during conduction WI well as freewheeling periods. Derive also an expression for average load current. ./ . 6.S. (a) Describe the working of a single-phase one-pulse SeR controlled converter with RLE load through the waveforms of supply voltage, load voltage, load current and voltage across . the SCR. Hence derive expression for the load current in tenns of supply voltage, load impedance, firiug angle, load voltage E etc. ( b ) A single-phase one-pulse converter with RLE load has the following data: Supply voltage:: 230 V at 50 Hz, R '" 2 0, L '" 1 rnH, E:: 120 V, Extinction angle ~ '" 220 0 , firing angle a :: 250 • (i) Calculate the voltage across thyristor at the instant SeR is triggered. (ii) Find the voltage that appears across SeR when current decays to zero . (iii) Find the peak inverse voltage for the SCR. Ans. [17.465 V, 329.079 V, 445.27 Vl
Y
B.9. Describe the operation of a single-phase two-pulse mid-point converter with relevant voltage
,
and current waveforms. Discuss how each SCR is subjected to a reverse voltage equal to double the supply voltage in case turns ratio from primary to each secondary is unity. Find the circuit turn-off time provided to each SeR by this converter configuration. 6.10. In a single-phase mid-point converter, turns ratio from primary to each secondary is 1.25. The source voltage is 230 V, 50 Hz. For a resistive load of R ",' 2 0, determine (a l maximum value of average output voltage 'and load current and the corresponding firing and conduction angles, ( b) maximum average and rms thyristor currents, (c) maximum possible values of positive and negative voltages across SeRB, (d ) the value of a for load voltage of 100 V, (e) the value of voltage across SeR at the instant of commutation for a of part Cd). [Hint. (b ) Maximum average thyri,tor current =
2~
J: ';
,in"" . d ["")
etc.]
IAn,. la) 165.63 Y, 82.82 A, a = 0', Y= 180' (b ) 41.41 A, 65.054 A (c) 520.4 Y, 520.4 V (d) 78.025' (e) 509.03 VI 6.11. (a ) A single-phase full converter charges a battery which offers a constant value of E . A resistor R is inserted to limit the battery charging current. Derive an expression for the average charging current in terms of V m' E. R etc. on the assumption that each pair of SeRa is fired continuously in each half cycle. Take Vr as the voltage drop in conducting SCRs. (b) Find the value of R in case battery charging current is 6 A, supply voltage is 40 V, 50 Hz , E", 12 V and Vr = 1 volt.
(c) Find the power dissipated in R. (d ) Find the supply power factoI'. [Hint. Refer to example 6.2.
(b ) 91 ::
sin- 1
i
3
2 · 40 1t in th e denominator]. 1 [Ans_ (a) rrR [2 Vm cos 91 - (E + Vr )(1t - 291))
(d Use Eq. (3.39) with 21t replaced by
(b)
3.994 n [c) 206.446 W (d) 0.9891 Jag!
Phase Controlled RectiI1ers
[Prob. 6]
341
6.12. A single-phase semi-converter delivers power to RLE load with R '" 5 n, L'= 10 mH and E '= 80 V. The ac source voltage is 230 V, 50 Hz. For a continuous conduction, find the average value of output current for n firing angle delay of 50 Ifmain SCR T2 is damaged and open circuited, find the new value of average output current on the assumption of continuous conduction. Sketch the output voltage and current waveforms and indicate the conduction of various components. (AmI. 18.013 A; Tl 01. FD ; Tl 01, FD and so on ; 1.0063 AJ Q
•
'6.13. A single-phase full converter feeding RLE load has the following data. Source voltage = 230 V, 50 Hz; R = 2.5 n, E :: 100 V, Firing angle:: 30°. If load inductance is large enough to make the load current virtually constant, then (a) sketch the time variations of source voltage, source current, load voltage, load current, current through one SCR and voltage across it, (b) compute the average value of load voltage and load current, (c) compute the input pf. [An •. (al Refer to Fig. 6.10 (bl, (bl 179.30 V, 31.72 A (el 0.7796 Jagl 6.14. Repeat Prob. 6.13 in case a freewheeling diode is connected across RLE load.
[Ii ;~~]"']
[Hint. (el Rm, vo1u. oisource current =
[An •. (bl 193.17 V, 37.268 A (el 0.92 lag.)
6.15. Describe the working of a single-phase full converter in the rectifier mode' with RLE load . Discuss how one pair of SCRs is commutated by an incoming pair of SCRs. Illustrate your answer with waveforms for source voltage, E, output voltage and current, source currenl, current through and voltage across one thyristor. Assume continuous conduction. Derive an expression for the average output voltage in terms of source voltage and firing angle. From the voltage differential equation of this converter, show that Vo '" IrJl + E. 6.lS. Describe the working of a single·phase full converter in the inverter-mode with RLE load. Illustrate your answer with waveforms for source voltage, E, load voltage and current, source current, current through and voltage across one 8CR. Assume continuoUs conduction. Find also the circuit turn-offtime . Should the average output voltage be more than E during inverter operation? Discuss. 6.17. A single-phase serruconverter bridge feeds RLE load. Diacuss how freewheeling diode comes into operation and holds the output voltage to almost zero for a given firing angle delay. Sketch the time variations of supply voltage, E, load voltage and current, freewheeling diode current and current through each pair consisting of seR and diode. Find also the circuit turn-off time. Assume the load current continuous. Also, derive an expression for the average output voltage in terms of source voltage and firing angle. 6.lB. (0) Describe how a free wheeling diode improves power factor in a converter system. (b) A separately-excited de motor fed through a single-phase semiconverter runs at a SPeed of 1200 rpm when ae supply vol tage 13 230 V, 50 Hz and the motor counter emf is 140 V. The firing angle delay is 50 Armature circuit resistance is 3 n. Co'mpute the average armature current and motor torque. [Hint. (b) F ind Km '" 1.114 NmlA and proceed] [AllS (bl 110.021 A, 11.164 Nml Q
•
6.19, A single-phase full converter is connected to RLE load. For diJcont inuous load current, draw the source voltage, output voltage, load current end source current waveforms as a fu n::ion of time '..... hen (0 ) exti nction an gle ~ > It (b ) e:ctinction angl e ~ < it with Vm sin ~ < E . EX j:llain how tn:! van ous wave:'arms are obtained and d isCUSJ their DP.tu re.
3-l2
Power Electronics
[Pcob.61
G.20. A single· phase semi converter feed s power to RLE load. For discontinuous load cu rrent, draw t.he source voltage, outp ut voltage, load current, source current and freewheeling d:ode current waveforms as a function of time when ( a ) extinction angle ~ > J[ (b) extinction angle ~ < J[ with Vm sin ~ < E. Explain how various wavefonns are obtained and discuss their nature. 6.21. A single-phase full-converter supplies power to RLE load. The source voltage is 230 V, 50 Hz and for load R = 2 n, L = 10 mH, E = 100 V. For a firing angle of 30 g , find the average value of output current ond output voltage in case the load current extinguishes at (a) 200" and (b) 170°. Derive the expressions used . [Ans. (al 96.2424 A, 192.485 V (bl 106.911 A, 213 .822 VI 6.22. A single-phase mid-point SCR converter supplies constant load cu«eht of 5 A when the triggering angle is maintained at 35°. The input voltage to the converter is 220 V ut 50 Hz. The turns ratio from primary to each secondary is
l
Determine the load voltage and input
power factor. IHint : Primary .4.T, = secondary AT" :. IsJll = 10 . N2 or Is = 10 A etcl IAns. 324.45 V, 0.73741agl
,
6.23. A single-phase full converter delivers a constant load current I". Express its source current in Fourier series and derive therefrom the expressions for the following performance parameters : displacement fac!or, current distortion factor, power factor, total harmonic distortion, voltage rippl!! factor, active (t nd reactive power inputs. 6.24. A I-phose full converter delivers ripple free cunent to RL load with R:; 15 n . The source voltage is 230 V, 50 Hz . For a firing angle of 30" calculate rectification efficiency, voltage ripple factor , displacement factor , current distorti on factor, power·factor, THO, active and reactive powers. fAns. 77.96% , 0.80'1. 0.8 66. 0.90032, 0.7797 lag. 0.4834. 2143.173 W, 1237.362 VArI 6.25. A single· ph ase semi ,cor:ve.-ter delivers a constant load current I". Express its source current in Fourier series a nd .::..:! r :ve therefrom the expressions for the following performance parameters : displacemen t. facto:, current t.:istortion factor, power fact or , total h<:.rmonic distortion, voltage ripple factor, ac:ive anc! r~active power inputs. 6.26. Repeat Prob. €.~4 for a s;n~I~-phase semiconve rter. IAns. S5.22~"", 0.6 139, 0.966, 0.9525 , 0.9201 lag, 0.31973, 2487.6'1 W, 666.57 VArI 6.27. A single-phase .e.sy~!'!'.etrical semiconverter feeds an RL load with R = 10 nand n large L so tha t load cu rre t:~ !S i e~·E:1. Th.:- source voltage is 230 V, 50 Hz. For a firing tingle delay of 30 a , determine (a ) avera~e value of out put voltage and output current, ( b ) average and !'m.s vllh,;es of diode, thyristor and source currents, (c ) input power factor _,lid Cii'Cl..i.it tum-off time. IAns. (t:) 193 .172 -V, 19 32 A I... i n .27 A, 14.756 A; 8.05 A, 12.471 A; zero, 17.64 A (c ) 0.92 ~L10~ . 6.28. For a 3-phase h .. !f·wo. ve diode ~'ectjfier, derive an expression for the ave rage ou tput voltage Vo in te rms of m axim um value of source voltage from line to neu'tr.l l.
Ii this rectifier feed s RL b .:,j with R = 5 na nd L =- 3 mH, find the a verage loa d current for 3-ph a3e input voltage of ·;00 V, 50 Hz. !Ans. 5-1 .011 AI 6.29. (0 )
Phase Conlrolled Rectifiers
[Prob. 61
343
For a three·phase h alf-wave SCR co nverter deli vering continuous output current , derive expres sions for the average ou tp ut vol tage fo r firing a ngle of (i ) 0° < a < 30~ a nd (i i) 30° < a. < 150 A three-phase half-wave SCR converter delivers constant load current of 30 A over the firin g angle range of 0° to 80°. At these two fi ring angles, compute the power delivered to load for an ac input voltage of 400 V from a delta· star transfonner. 0
•
(b )
[AnS:
(a ) Forboth (i ) and (ii ), V o=
3~:mp coso
(b) 8.102kW (e) 1.406S kW]
6.30. A delta-star transfonner feeds power to a load R = 10 n through a 3-phase M-3 converter. The input voltage to converter is 400 V, 50 Hz . Find the power delivered to load for a firing-an gle delay of (a ) 15° and (b) 60°. Derive the express ions used. rAns. (a) 7243.2 W (b ) 4000 W] 6.31. An M·3 converter operates on a 400 V. 3-phase, 50 Hz mains and delivers power to the armature of a dc motor with negligible res istance and infinitely large rea.ctor in the de bus. The transformer has Dy 11 connection with unity phase turns ratio. Back emf is 300 V. Determine the trigger angle . [Hint. Input line voltage to M-3 converter is ff. 400 V] rAns.50. 113 I D
6.32. A 3-phase, 3-pulse converter is c;onnected to RLE load. The source voltage is 3·phas e, 230 V, 50 Hz and th e load current is level at 1(J A. For R::: 0.5 nand L = 2 H , determine (a ) firing angle for E = 134 V and (b ) firing-angle advance for E = - 134 V. [Ans. Ca l 26.472" Cbl 33.824")
6.33. A 3-phase full converler is connected to a resis tive load. Show.that the average output voltage is given by . 3 Vm l
Vo =-.-cosa: and where
Vo = 3 V ml '"
~ml [1 + cos{ o:+i ]
1t
for 0< 0:<"3 for
i
1t
< 0: < 23
maximum value of line voltage.
6.34. For a 3-phase full converter, explain how output voltage wave, for a firing angle of 3 0 ~ , is obtained by using (a ) phase voltages and (b ) line voltages. 6.35. (a ) A resistive load of 10 n is connected to a 3-phase full converter. The load t akes 5 kW for a firi ng angle delay of70 ~ . Find the magnitude of per phase input supply voltage. Derive the expression required for the output voltage in terms of firing angle etc. (b) Repeat part (a ) in case an inductor connected in seri es with the load makes the load current constant. (c) Repeat part (a) in cnse an inductor conn ected in series with the loud mo ke3 the load current contin uous. [Hint. Ca ) First derive an exp ression for the rms value of output voltage,
~
V = Vm !
~(23' -.)+ ~ (,[3 oos 2. - ,nn 2.1]'" [Ans. Ca) 214 .242 V (0 ) 279.55 V (cl 2 13.254 Vi
For a 3·phase semieo nve rter, draw output voltage ..... aveforms fo r a firin g angle delay of 45" '"6.36. indicatin g the conduction of its various on the assumption of con tinuous out;mt e iF~mentJ
curren:. Dis':U35 whether fr eewheeling diode comes in to play or no t. Hence obtain an e:qres s icr: fo r the aver age outpu t vol tage in tenru of ac s upply voltage. fi rin,; angle dele,:.' e"c, by us ing hoth sine and cosine functions for the .su pply voltage. S.37. Ske tch outpu t ~-o lt.n ge wavefor m for a J-phase semiconverter for a firing a ngle delay of ,/ 5' Indicate the con ducti on of var ious elements a nd discuss whether freewheeli ng diode com~.s
~
Power Electronics
[p
3~4
into play on the assumption of continuous load current. Hence obtain nn expression for the average output voltage by using both sine and cosine functions for the supply voltage. 6.38. A 3-phase semiconverter is connected to RLE load. For a firing angle delay of 120~, draw output voltage and load current waveforms in case load current is (a) continuous and (b) discontinuous. For both parts, indicate the conducting elements of the semiconverter during three periodic times of the output voltage wave. Discuss briefly the nature of waveforms obtained . 6.39. A separately-excited de motor fed from 3-phase semiconverter develops a fun load torque at 1500 rpm when firing angle is zero, the armature taking 50 A at 400 V de aod having an armature-circuit resistance of 0.5 n. Calculate the supply voltage per phase. Find also the range of firing angle required to give speeds between 1500 rpm and 750 rpm at full-load torque. 3 Vml
[Hint. For a"" 0 0 and 1500 rpm, - - "" 400, K = 0.25 V/rpm etc.}
V
•
IAn •. 171 .006 V, a. 0' to 86.42')
6.40. A 3-phase full converter thyristor bridge feeds a resistive load R.
(a) Sketch input voltage waveforms for lJab Uac ubc etc. ( b) From (a), sketch the waveform of the output current io for a firing angle of30". (cl From (b), sketch the waveform of input current ia for phase A for a::o 30". Show the duration of conducting thyristors . In case input voltage is 400 V and R = 200 0, indicate the peak magnitude of current ia . lAns. (c) Peak magnitude of ia.:::z 2.828 AJ • 6.41. For a 3-phase full converter, sketch the input voltage waveforms for !Jab !Jac ubc etc and voltage variation across anyone thyristor for onc complete cycle for a firing an'gle delay of (a) 60" and (b) 120'. Find the magnitude of reverse voltage across this SCR and its commutation time for both parts (a) and (b) for a supply voltage of 230 V, 50 Hz. [Ans. (0 ) 325.22 V, JO m sec (b) 281.69 V, 3.33 m sec] • 6.42. A battery is charged from 3-phase supply mains of 230 V, 50 Hz through a 3-phase semiconverter. The battery emf is 190 V and its internal resistance is 0.5 p. An inductor connected in series with the battery renders the charging current of20 A ripple free . Compute the firing angle delay and the supply power factor. 'Ans. 73.263", 0.6521agJ 6.43. (a) A 3-phase full converter is used for charging a battery with an emfof110 V and an internal resistance of 0.2 O. For a constant charging current of 10 A, compute the firing angle delay for ac line voltage of 220 V. Find also the supply power factor. (b) For the purpose of delivering energy from dc source to '3-phase system, the firing angle of the 3-phase converter has been increased to 150". For the same value of dc source current of 10 A, compute the output ac line voltage. IAns. (a) 67.85', 0.36 lag (b) 92.36 VI
6.44. A 3-phase full converter is delivering a constant load current of 50 A at 230 V dc when its input is 3-ph ~s e, 415 V, 50 Hz. If each thyristor h as a voltage drop of 1.1 V when conducting, ca1cul~ta (0 ) the firing angle delay of SCRs (b) the rms currentof SCfu (c) rms source current (d) the mean power loss in each SCR and (e) input pf. (f) In case ac supply has an inductance of 3 mH per phas e, find the new value of firing angle for the same dc power output as before. [Hint : (e) Input ,ower"" output power + power lost in SC Rs
.J3 x 415 x 40.825 x cos ¢ = 230 x 50 + 1.1 x 50 x 6 etc .]
3 IAns. (a) 65.52' (b) 28.87 (e) 40. 825 A (d) 18.33 W (e) 0.3936 lag If) 60.35'1 6A5. (0 ) Discus3 the effect of source inductance on the pcrfonnance of a single-phase full conver te r indica ting cl ea rly the condu ction of vnnous thyristors during one cycl e.
. ... . . .
[Prob. 6J
Phase Controll ed Rectifiers
J .. 5
Derive expressions for its output voltage in terms ofm maximum voltage V m. fi ri ng angle a and overlap angle and (it) Vm. a, L, and load current 10. Here L, is the source ind uctance. Show that the effect of source inductance is to pr esent an equ ivalent r esistance of
wL
- -' 0 in series with the internal rectifier voltage . (b)
•
A single-phase full-converter fed from 220 V, 50 Hz supply gives an output voltage of 180 V at no load . When loaded with a constant output current of 10 A, the overlap angle is found to be 6°. Compute the value of source inductance in henries . IAns. (b ) 4.8084 mHJ
8.46. (a) Show that the performance of a single-phase full converter as effected by source inductance is given by the relation
c.oL, [ 0 cos (a + j,l) = cos a - - y m
(b)
S.47.
(a)
(b )
where the symbols used have their usual me anings. A single-phase full converter is connected to ac supply of 330 sin 314 t volt. It operates with a firing angle a = 7[ / 4 rad . The .totll iload current is maintained cons tant at 5 A and the load voltage is 140 V. Calculate the source inductance, angle of overl a p and the load resistance. [Ans. 17.113 mH , 6.267°, 28 OJ Show that the perfonnance ofa three-phase full converter as influenced by source inductance is given by the relation 2 wL. cos (a + j.l) = cos a - 10 y ml The symbols used have their usual meanings . A 3-phase fully eontrolled bridge converter is fed from a 3-phase 400 V, 50 Hz mains . For firing angle of 60°, output current is level at 25 A and output voltage is 250 V. Calculate the load resistance, source inductance and angle of overlap. [Ans. 10 n, 2.667 mHo 4.8'J
6.48. Describe the effect of source ind uctance on the perfo rmance of a 3-phase full converter with the help of phase voltage wavefonns. Indicate the sequence of conduction of various thyt'ist ors and sketch load current waveforms for both positive and negat ive group of thyristors. Srote the various assumptions made. Derive an expression for its output voltage in terms of supply voltage, source inductance, load curren t etc. 6.49. Repeat Example 6.27 for a firing-angle delay of 450 and overlap angle of 45°. 6.50. A 3·ph e.se full converter is fed from 3-phase, '230 V, 50 Hz supply having per-phase source inductu nce of 4 mHo The load current is 10 A ripple fr ee. (a ) Ca lculate the voltage drop in de output voltage due to sour ce inductance. (b) If de output voltage is 210 V, calculate th e firi ng angle and the overlap period . (c) In case the bridge is made to operate as a 1i ne ,co r'l~utated inverter with cc voltage of 210 V, calculate the firing angle fo r the same load current. [An s. (a) 12 V, (b) 44 .3 7 ~ , 0.3344 ms (d 129.6PI 6.51. Explain how two 3-phase full converters can be connected buck to back to form a circulating cur rent type of dual converter. Discuss its operati on with the help of voltage waveforms across (a ) each converter (b) load and (c) reactor, Take " I = O~. Describe how circulating cU:"l"ent waveform can be obta ined from reactor voltage waveform. If on e of t he two converters is lauded, sketc:': the wave fo rms of the ir load currents 6.52. (a) For n 3-phase dual converter, de:ri:.. ~ an exp ress ion for the C:ri.': ul ::ai ng·current in terr:u at s'Jp pJy voltage, re actor induc tance, fir i:-.g-·angle deby e~ c . Relevant \'ol tage and CI.!:r2:1~ waveforms, needed for this der h·ation . muse be s!< =:ched .
346
[P,ob. 6] (b)
Po'wer Electronics
Two 3-phase full-converters are connected in anti parallel to form a dual converter of the circu lating-current type . The input to the dual converter is 3-phase, 400 V: SO H z. If peak value of circulating current is limited to 20 A, find the value of indue lance needed fo r the reactor for firing' angles of (a) Ct.l '" 30 0 and (b) al = 600 • [An •. (b) (i) 77.97 mH (ii) 20.892 mH]
6.53. A 3-phase full converter, fed from 3-phase, 400 V source, has an output voltage of 450 V de fo r a firing-angle delay or3D a , Calculate the overlap angle and the voltage drop due to overlap. [An •. 6.84", 17.82 VI
Chapter 7
Choppers -----_. __ . ..... ............. _------ ... __ ............•...•.....• ... __ ......... -. .. ..... ... . ....... .. In this Chapter • • • • • • •
Principle at C hopper Operation Control Strategies Step-up Choppers Types of Chopper Circuits Steady State Time-domain Analysis of Type-A Chopper Thyristor Chopper Circuits Multlphose Choppers
"-
........ . . -..... _- .. -.-- ......... -.... __ ........................ __ ....... .••.. ..... _-_ . ....... ..... Many industrial applications require power from de voltage sources. Several of these applications, however, perform better in case these are fed from variable de voltage sources. Examples of such de systems are subway cars , trolley buses, battery-operated vehicles, battery-charging etc. From ac supply systems, variable dc output voltage can be obtained through the use of phase-controlled converters (discussed in Chapter 6) or motor-generator sets. The conversion of fixed de voltage to an adjustable de output voltage, through the use of semiconductor devices, can be carried out by the use of two types of dc to- de converters given below [5}. AC Link Chopper. In the ac link chopper, de is first converted to ac by an inv erter (de to ac converter). AC is then stepped-up or stepped-down by a transform er which is then converted back to de by a diode rectifier, Fig. 7.1 (a). As the conversion is in two stages, de to ac and then ac to dc, ac link chopper is costly. bulky and less efficient.
~c
1
Inverter
Ct
1
~c
1-<)
..,
1
~c
~t
~! (a )
D.C
!
Chopoer
1
DC
~t
Lt (0)
(e)
Fig. 7.1 !e ) .·\C li n:< e:10pp i!r a nd ( b) de choppe r (o r chopper) <.:) Repre3e r,ta:ion of a powe r s emi cond uc tor device.
343
Power Electronics
[Art. 7.1]
DC Chopper. A chopper is a static devi ce that converts fLxed de input voltage to a variabl e de output voltage directly, Fig. 7.1 (b ). A chopper may be thought of as de equivalent of an ae transformer since they behave in an identical manner. As choppers involve one stage conversion, these are more efficient. Choppers are now being used all over the world for rapid transit systems. These are also used in trolley cars, marine hoists, forklift trucks and mine haulers . The future electric automobiles are likely to use choppers for their speed control and braking. Chopper systems offer smooth control, high efficiency, fast response and regeneration.
The power semiconductor devices used for a chopper circuit can be force-commutated thyristor, power BJT, power MOSFET, GTO or IGBT. These devices, in general, can be. represented by a switch SW with an arrow as shown in Fig. 7.1 (c), When the switch is off, no current can flow, When the switch is on, current flows in the direction of arrow only. The power semiconductor devices have on-state voltage drops of 0.5 V to 2.5 V across them. For the sake of simplicity, this voltage drop across these devices is neglected. As stated above, a chopper is dc equivalent to an ac transformer having continuously variable turns ratio. Like a transformer, a chopper can be used to step down or step up the fixed dc input voltage. As step-down dc choppers are more common, the term dc chopper, or chopper, in this book would mean a step-down dc chopper unless stated otherwise.
The object of this chapter is to discuss the basic principles of chopper operation and the more common types of chopper configurations using ideal switches . •
7.\' PRINCIPLE OF CHOPPER OPEltATION
A chopper is a high speed on/off semiconductor switch. It connects source to load and disconnects the load from source at a fast speed. In this m8IUler, a chopped load voltage as shown in Fig, 7.2 (b) is obtained from a constant de supply of magnitude VI ' In Fig, 7.2 (a), chopper is represented. by a switch SW inside a dotted rectangle, which may be turned-on or turned-off as desired. For the sake of highlighting the principle of choper operation, the circuitry used for controlling the on, off periods of this switch is not shown. During the period Ton, chopper is on and load voltage is equal to source voltage VJ • During the interval Tatr, chopper is off, load current flows through the freewheeling diode FD. As a result, load ~erminals are short circuited by FD and load voltage is therefore zero during T off. In this manner, a chopped dc voltage is produced. at the load terminals. The load current as shown in Fig. 7.2 (b ) Yo
v,
\
0+
1
~
------ ..
-------
~T T . . I 10 I i I
,
'sw L ____ JI
+
FD
Y,
I
lOAD
f Vo
,
-T ~ -
~_O::~_~)
,:
£ Toll
I
1 I
---- - -
-
.
I
I
I
w
~
,•
. ..
Fig. 7.2 (0) Elementry ch'1 pper circui t and (b ) output voltage and curren" w;w eionns.
t
[Acl. 7.2]
Choppers
349
is continuous. During T G/\' load current rises whereas during T off. load current decays. From Fig. 7.2 (b), average load voltage Vo is given by VO=T
~T
V' = -~ T V=aV,
+ off T on = on-time; T ot! = off-time T= Ton + Tolf = chopping period Too a =T =duty cycle
(71 ) ....
on
where
Thus load voltage can be controlled by varying duty cycle ex. Eq. (7.1) shows that load voltage is independent of load current. Eq. (7.1) can also be written as Vo=f · Ton ' V,
where
... (7.2)
f= ~ = chopping frequency
7.2. CONTROL STI(ATEGIES
It is seen from Eq. (7.1) that average' value of output voltage Vo can be controlled through a by opening and closing .the semiconductor switch periodically. The various control strategies for varying duty cycle a are as follows: 1. Time ratio control (TRC ) and 2. Current-limit control
These are now described one after the other. 7.2.1. Time Ratio Control (THe). As the name suggests, in this control scheme, time ratio TonlT (as duty ratio or duty cycle) is varied. This is realized in two different strategies called constant {rr!quency system and uariable {rr!quency system as detailed below : 1. Constant Frequency System In this scheme, the on-time Ton is varied but chopping frequency f (or chopping period T) is kept constant. Variation of Ton means adjustment of pulse width, as such this scheme is also caUed pulse· width·modulation scheme. Fi g. 7.3 illustrates the principle of pulse-width modulation. Here chopping period T is constant. In Fig. 7.3 (a ), T on =~ T so that a = 0.25 or ex = 25%. In Fig. 7.3 (b) , Ton =i T so th at a = 0.75 or 75%. Ideally a. can be varied from zero to unity. Therefor e o:.!tput voltage Vo can be varied between zero and source voltage V~. 2. Variable Frequency System
In this sch eme, the ch opping frequency f (or chopping period T ) is varied an.d either (i) on·time Ton is kept cons tant or (ii) off-time TOllis kept constan t. This method of contr olling a is also called frequency-modul ation scheme. Fig. 7.4 illus trates the principle of fre quency modulation. In Fig. 7.4 (a) , TfJ r. is kept cons tan t but T is varied. In the upper diagram of Fig. 7.4 (e ), TfJr. =~ T so ::-.J.': a = 0 25. In the lo;ve:-
diagram of Fig. 7.4 (a ), Tun = ~ T so that a
= 0.75. In Fig. 7,4 (b ), TrJi~ is kept constrmt an d l' is
350
[Act. 7.2)
Power Electronics
•o
( Ton
f-'1
load voltage
- 1/'
V,I--
T olI -
~L-
__
~~
-
____
~~
____- h___ _ t
T--.I
I.
[TOIf
v,
load Yolt('92
.,,-
r-
I-- Too - """'::- T r
,
t
.1
Fig. 7.3', Principle of pulse-width modulation (constant 7).
voL E:l
IT
Toff - --1.
'Is
JI.-
-
Load
T ----1 ./
volto~.
o
-.... .
• t
vo~ V,
-=inn
n n r LOOd voltage
• t
I-T-I
Vl~~~f;fron n
'Is
I-T~
~:~~~f; j.
n0
r t "LOOd voltage
rl
.
Load voltoge
Ton T
-I
.1
• t
varied. In the uppe r di agram of this figure, Ton = i T so that a = 0.25 and in the lower diagram Ton = ~ T so that 0. = 0. 75. Frequency modulation sche-me has some disadv antages as compared to pulse.wid th modulation scheme. Thes2 are as under:
Ch oppers
351
[Mt. 7.3]
The chopping frequency has to be varied over a wide range for the control of output voltage in frequency modulation. Filter design for such wide frequency variation is, therefore, quite difficult. (i)
For the control of a. frequency variation would be wide. As such. there is a p05sibility of interference with signalling and telephone lines in frequency modulation scheme. (iii) The large off-time in frequency modulation scheme may make the load current discontinuous which is undesirable. It is seen from above that constant frequency (PWM) scheme is better than variable frequency scheme. P\VM technique has, however, a limitation. In this technique, Tan cannot be reduced to near zero for most of the commutation cir cuits used in choppers. As such, low range of ct control is not possible in PWM. This can, however. be achieved by increasing the chopping period (or decreasing the copping frequency) of the choppe r. (it)
7.2.2. Current-limit Control In this control str ategy, the on and off of chopper circuit guided by the previous set value of load curren t. These two set values are maximum load current I a.nu and minimum load current I a.lM
/ I o.m• . . ... . 1. ..... . ....
. ·· .. t-_·-,-
0:
..... _ ••• __ . , : ···-r····· - · ·-l ·····---
; l a.rnn
i
.
:
i
:
i
l
\Vhen load current reaches the upper limit l a.nu. ......;:Ton::""TOIl-i. -----..;... T-----o.j : . I chopper is switched off. Now load current freewheels : : : • and begins t o decay e.'(ponentially. When it fans to Vo lower limit l o.mn. chopper is sy..;tched on and load current begins to rise as shown in Fig. 7.5. Profile of . load current shows that it fluctuates between IO./TU and I a.mn • and therefore cannot be discontinuous. Fig. 7.5. Current.limit. control for chopper.
. .
.
Switching frequency of chopper can be controlled by setting I o .nu and l o.mln" Ripple current (= I o.rm - l o.m",) can be lowered and thls in turn necessitates higher switching frequency and therefore more switching losses . Current-limit control involves feedback loop, the trigger circuitry for the chopper is therefore more complex. P\VM technique is, therefore, the commonly chosen control strategy for the power control in chopper circuits.
7.3. STEP-UP CIIOPPERS For the chopper configuration of Fig. 7.2 (0), average output voltage Vo is les5 than the input voltage VJ , i.e. Vo < V,s; this configuration is therefo re called step-down chopper. Average output voltage Vo greater than input voltage V,s can, however, be obtained by a chopper caned step-up chopper. Fig. 7.6 (a ) illustrates an elementary form of a s tep-up chopper. In this article, working principle of a step-up chopper is presente d. l n thi s chopper, a large in duc tor L in se:-ies with source voltage VJ is e33ential a5 shown in Fig. 7.6 (0) . When the ch oppe r CH is on, the closed cu rr ent path is as sh own in Fig. 7.6 (b ) and inductor s tores energy during Tor. period. When the chopper CH is off, <13 ::-;e inductor current cannot die dow n in5ta.'taneo l1..,;1 >~ thi5 curren t is forced to now through th!! diode and load for a time To/;. Fig. 7.6 (c). A3 th e current tends to dec rease, polarity of th e e:r:f in d'Jced in L i~
,.
3-,
Power Electronics
[Art. 7.3]
t,
L
0
, uu u
--~ V
T'
+
,,.
-~
L
-~
l
CHi l :,,
0
t
0
"o'!.r----.---I"
(0)
L
0
+ +
i.
}
-=- Vs
b
CH
is-
+=jj'o
+
-=-
Vo
Vs
:
I---T ---:,
\
!---- T·----I
\
"~I'---+-D-+-;,........JD-+,---o"\ JT~{JI' N
0
+
: II
TOl'\......(Tof j l ,~ Ton ---l'
o
(b)
L
I,
I, v .....::;",
A
0
\
i 0
00 ~ Fig. 7.6 (a) Step-up chopper (h) L stores energy (c) L . dildt is added to V, (d) voltage and current waveforms. r ev ersed as s hown in Fig. 7.6 (cl. As a result, voltage across the load, given by Vo =V, + L (dildt), exceeds the source voltage V,. In this manner, the circuit of Fig, 7.6 (a) acts as a step-up chopper and the energy stored in L is released to the load.
When CH is on, current through the inductance L would increase from 11 to 12 as shown in Fig. 7.6 (d). When CH is off, current would fall from I, to I,. With CH on, source voltage is applied to L i.e. uL =V,. When CH is off, KVLfor Fig. 7.6 (c) gives vL - Vo + V, =0, or vL = Vo - V,. Her e VL = voltage across L . Variation of source voltage v" source curren i" load voltage Va and load current io is sketched in Fig. 7.6 (d). Assuming linear variation of output current, the energy input to inductor from the source, during the period TorP is W ill
= (voltage across L ) (average current through L)
=v.fI;I2)T
TOil
•• ,(7.3)
o"
During the tim e Tolf' when chopper is off, the energy r eleased by inductor to the load is W oIT
=(voltage across L ) (average current =
(Vo - V,)
through
(II; I,).T
L)
off
T off .,.(7.4)
Considering the system to be lossless, these two energi es ginn by Eqs. (7.3) and (7.4) will be equ al.
353
[Art. 7.3J
Choppers
(II I,)
(II I,)
+
+
V, - 2 - TOil = (Vo - V,) - 2 - . Torr
V" TOil = Vo To/f - V" Toff Vo T off = V, (TOIl + T off) = V" T T T 1 Vo=V'r=V' T_T =V, I-a
or
of{
... (7.5 )
'"
It is seen from Eqn. (7.5) that average voltage across the load can be stepped up by varying the duty cycle: If chopper of Fig. 7.6 (a) is always off, a = 0 and Vo =V, . If this chopper is always on, a = l ..and Vo = 00 (infinity) as shown in Fig. 7.7 (a). In practice, chopper is turned on and off so that a is variable and the required step-up average output voltage, more than source voltage . .is obtained.
I,
L
o
v,
i,
v.
o Fig. 7.7.
0. 25
0.5
0.7 5
w
(0 )
Variation of load voltage
~) Vo
with duty cycle (b) regenerative braking of de motor.
The principle of step-up chopper can be employed for the regenerative braking of dc motors. This is illustrated in Fig. 7.7 (b) where motor armature voltage Eo represents V, of Fig. 7.6 (al. Voltage Vo is the dc source voltage. When CH is on, L stores energy. When CH is off, L releases energy. In case Eo/( I- a) exceeds VOl dc machine begins to work as a dc generator and armature current 10 flows opposite to motoring mode. Power now flows from de machine to source Vo causing regenerative braking of de motor. Motor armature voltage Eo is directly proportional to field flux and motor speed. Therefore , even at decreasing motor speeds. reg ene rative braking can be made to take place provided duty cycle and fi~ld flux are 50 adjusted that Eo/ (l - ex) is more than the fIxed source voltage VO' . Exnmple 7.1. For the basic de to dc converter of Fig. 7.2 (a), f?:::oress the folloUJing variables as {unctionsof v., R and duty cycle a in case load is resistive : . (a) A verage output voltage and current (b) Output current at the instant of commutation (c) A uerage and rms free wheeling diode currents (d) R ms val u. e of the outp ut voltage (e) Rms and average thyris tor cu rrents
mE ffectit;e input resistance oftn e chopper.
[Art. 7.3]
354
Power Electronics
Solution. The load voltage variation is shown in Fig. 7.2 (b ). For a resistive load, output or load current waveform is similar to load voltage waveform. (a)
Average output voltage,
Average output current, (b) The output current is commutated by the thyristor at the instant t = T on. Therefore, output current at the instant of commutation is V/R. (c) For a resistive load, freewheeling diode FD does not come into play. Therefore, average and rms values of freewheeling diode currents are zero.
=[TT" .
(d) Hms value of output voltage (e)
. Average thyristor current
v;JI2 =Wi , V,
Ton V,
V,
=T .Ii =a Ii
__ [TT"" . (VR, ]' ]'12 __ 'In r-a . V'
Rms thyristor current
R
({J Average source current = average thyristor current =a . ~
Effective input resistance of the chopper dc source voltage V, . R R· = - - =average source current a . V, a
~=="-'-'=-=----:
Example 7.2. For type·A chopper of Fig. 7.2 (a), de source voltage =230 V. load resistance =10 n. Tafte a voltage drop of2 V across chopper when it is on. For a duty cycle of 0.4, colctl-late (a) average and rms values of output fJoltage and (b) chopper efficiency. Soh,1.tion. (a) Wh~n chopper is an, output voltage is (V, chopper is off. output voltage is zero. (V - 2) T :. Average output voltage =' T on = n (V, - 2)
,
= 0.4 (230 -
2)
volts and during the time
2) = 91.2 V
Rms value of output voltage. V~ = (V,-2~.
[
TT" ]'12 =Wi(V,-2)
= ~ (230 (b)
2) =144.2 V
Power output or power delivered to load,
Po = Power input to chopper, Chopper efficiency
'1' = (141~2)2 =
2079.364 W
2 Pi = V, . f o = 230 x 9;0
= 2097.6 W
= Po = 2079.364 100 _ 99 13 ~~ P, 2097.6 x . ~.
Choppe rs
[Ar!. 7.41
355
Example 7.3. A step -up chopper has in.put uoltage of220 V and output voltage 0(660 V. Ii the cond ucting time of thyristor-chopper is 100)lS. compute the p ulse width of output voltage. In case output-uoltage pulse width is halued for constant frequency operation, find the a verage value of new output voltage. Solution. From Eq. (7.5),
1 660=220--
I-a
2
or
Ton
a="3=T It is seen from Fig. 7.6 (d ) that conducting time of chopper is Ton
=~ T =100 ~s. This gives
chopping period T = 100 x ~ = 150 fi s.
:. Pulse width of output voltage
=T off = T -
Ton = 150 - 100 = 50
When pulse width of output voltage is halved, Toff = 5 0 = 25 2
~s
~s .
For constant frequency operation, T = 150 ~s, T on = 150 - 25 = 125 I..ls T on
125
5
a=T= 150 =6 :. Average value of new output voltage, Vo =220 ~ = 1320 V 1- -
6
7.4. TYPES OF CHOPPER CIRCUITS Power semiconductor devices used in chopper circuits are unidirectional devices; polarities of output voltage Vo and the direction of output current 10 are, therefore . r estricted. A chopper' can, however, operate in any of the four quadrants by an appropriate arrangem~nt of semiconductor devices. This characteristic of their operation in any of the four quadrants forms the basis of their classification as type·A chopper, type·B chopper etc. Some authors describe this chopper classification as class A, class B, ... in place of type-A, type·B .... respectively. In the chopper·circuit configurations drawn h enceforth, the current directions and voltage polarities marked in the power circuit would be treated as positive. In case current directions and voltage polarities turn out to be opposite to those shown in the circuit, these currents and voltages must be treated as n egative. In this section, the classification of various chopper configurations is discussed .
7.4.1. First.quadrant, or Type·A: Chopper
This t}-pe of ch opper is shown in Fig. 7.8 (0). It is observed th a~ chopper circuit of Fig. 7.2 is also type·:\ch opper. In Fig. 7.8 (a), when choppe r CHI is on, va = VJ and current io rows in the arrow direction s ho\....n. When CHI is off , Vo = 0 bu t io in th e load con tinues flowing in the same direction through freewheeling diode FD, Fig. 7.2 (b ). It is thu s seen that average valu es of both load voltage and curr ent, i.e. Vo and 10 ore always positive: this fact is showr, by th e h atched area in the firs t quadr ant of Vo - 10 plane in Fig. 7.8 (b) .
(a)
[Act. 7.4J
.....,.
):,> .... ;1. .. .'
,.
Power Electronics
•
t
,.
'
'----......
"
"
,
CH1,FO
1+
FO
1
I-
lOAC
(b)
(a)
Fig. 7.8. First-quadrant, or type-A chopper.
The power fl ow in type-A chopper is always from source to load. T his f:ho·pper is also called step-down chopper as average output voltage Vo is always less than the input dc voltage V~. 7.4.2. Second-quadrant, or Type-B, Chopper Power circuit for this type of chopper is shown in Fig. 7.9 (a). Note that load must contain a dc source E, like a battery (or a dc motor) in this chopper. . 02
r
"
'I
0
i.,
,, CH2
i
+
,,
tw
I,
L
-~
,
J "
I-
rE
Fig. 7.9. Second-quadrant, or type-B, chopper.
W
When CH2 is on, "0; 0 but load voltage E drives current through L and CH2. Inductance
L stores energy during T,. (; on period) of CH2. When CH2 is off, "0 ; (E+ L ~~) exceeds source voltage V" As a retiult, Qjode D2 is fo rward biased and begins conduction, thus allowing power to flow to the source. Ch9Pper CH2 may be on or off, current 10 flows out of the load, current io is therefore treated as negative. Since Vo is always positive and 10 is negative, power flow is a(wan frOll) load 'to source. As load voltage Vo;(E+L
~;}s 'm ore than source voltage V"
type-.l1 chopper is also called step-up chopper. Both t.ype-A and type-B chopper cOJ::!.figurations have a common negative terminal between their input and oqtput circuits. 7 . 4 . ~. Two-q\l~
j ype·A c h opper, or Type-C Chopper This type of chopper is obtained by connecting type-A and type-B choppers in parallel as show n in Fi g. 7.10 (0). Th e output vol tage Vo is always positive becau se of the presence of freewheeling diode FD across the load. When chopper CH2 is on, or freewheeling diode FD conducts, output voltage Uo = 0 and in case chopper CHI is on or diode D2 condu cts, output voltage Uo = V" The load current io can, however, reverse its direction. Current io fl ows in the arrow direction marked in Fig. 7.10 (0 ), i.e. load current is positive when CHI is on or FD conducts . Load current is negative if CH2 is on or D2 conducts . In other words, CHI and FD operate together as typ e-A chopp er in first quadrant. Likewi se, CH2 and D'2 ope rate together as type-B chopper in second quadrant.
Choppers
[Act,
7.4J
357
02 l
w
~)
Fig. 7.10. Two-quadrant type -A choppe r, or type-C chopper. Average load voltage is always positive but average load current may be positive or negative as explained above. Therefore, power fl ow may be from source to load (first-quadrant operation ) or from load to source (second-quadrant operation ). Choppers CHI and CH2 should n ot be on simultaneously as this would lead to a direct short circuit on the supply lines. This typ e of chopper configurati on is used for motoring and r ege n e r ati~e br aking of de mot ors . The ope rating r egion of this type of chopper is shown in Fig. 7.10 (b ) by hatched area in first and :second quadrants. Example 7.4 . Sketch output uoltage, output current, source current and thyristor (or chopper) cu.rrent waveforms for type-C chopper {or its operati()n in first quadra.nt. Solution. For this example, refer to chopper circuit of Fig. 7.10 (a ). When D2 conducts or CH1 is on , load voltage U o is equal to V, during T OIl' When CH2 is on or FD cohducts, u" = 0 during Tolf' The output voltage waveform Vo is sketched as V, duringTrm anti zero during T ufTin Fig. 7 .11 (a) . Load current io is assumed negative at t = O. It changes from negative to positive value during TOIl a nd from positive to negative value.during T off as shown in Fig. 7.11 (b ).
',' o
b,," -+''''1 I i I--T
;0
, j
,
:· [ (I.mA
v,
I
[
.
t
,
,
:
o~ , ~~ , ~ , +,~--~~~--+-~~_ i l o.m,, : . ,kr1!-+FO \Ki11 ----: FO ~,'!
: ! :--:
02
iT:
: CH2 :02
': ~
. CH2 :
Vl /:
I
V
t
l CX1,
" r---- 2T---------....; Fig. 7.11. Waveform.,
fOi
type-C chop pe r, Exam ple 7.-..
358
[Arl. H]
Power Etectronic3
...
"
"
When v" is positive and io is negativ e, reference to Fig. 7.10 (a) shmys t1:at D2 conducts; when both Ua , io are positive, CHI conducts. What Vo =0 and io positive: FD conducts and further with Uo = a and io negative, CH2 conducts. Source current i.1' with periodicity T , exists when CHI , D2 conduct whereas chopper CHI current exists with a periodicity 2T. Source current and chopper, or thyristor, current wavefonns are therefore, sketched accordingly in Fig. 7.11 (e) and (d ) respectively. It is seen from the wavefonns that average value of both uo• io are positive. Therefore, power flows from source to load. hence type-C chopper operates in first quadrant. In case second qu adrant operation oftype-C chopper is required, average value ofu o should be positive but that of ill must be negative. In Fig. 7. 10 (a), if load current is assumed positive and continuous, then CHI would conduct during T an and FD during T olf. This would result in first quadrant operation of type-C chopper. Chopper CH2 and 02 would. however, remain idle under the assumption of P9sitive load current.
7.4.4. 1Wo-quadrant Type-B Chopper, or Type-D Ch opper The power circuit diagram for two-quadrant type-B chopper, or type-D chopper, is shown in Fig. 7.1 2 (0). The output voltage Va =V J when both CHI and CH2 are on and Va =- V, when both choppers are off but both diodes Dl and D2 conduct. Averf.!,ge output voltage Vo is positive when choppers turn-on time T an is more than their turn-off time T o{f8S shewrdn Fig. 7.12 (c). + 1s
,,.".
CH1:,
-,, : _.,
v,
~ D2
+ --
"
DI ,
¥'--I,
r- -, , iCH2 i : ' .
J
(a)
"o~
(b )
T."
O~T,,::jj ',~"U
lJ I
,,~izi ;. iI i!: :. : o
; ' :. --l CHI CH2t--i CHI CH2'--i .! ,'+. .01D2"';"'-: .. .:
II
: : .: t
::
., ., .' "k-i"f--U--U' °rDv [...5 1
"H~~ Cl ~--..; 00
Cl
r
".
o -lTv'
-
:
,,
:
, ,
II:
:
i
::
,
' , .
-
Too ~ To!!
-
Too I- lo ti
j-vs
,
I
\e ~', i : . : ' I.
0
' \7. \..;
D1D2 ~ ] TD 102-::,
. .'
t
t
'o'H·, CH, .CH2"/1'"
_~(
1
"'/1' (
'r".
I
L.----'t;~
""'J r== PI T~(1" I,
t
(1 t
~
Fig. 7.12 (0) and (b) Two·q uadram type-B chopper, or type·D chopper (c) wavefo rms for Ton > T~ff' Vo is positive, first quadrant operation nnd (d ) wa veforms fo r TOil <.. T ,.r. V is :'lE'gative. fou rth quad rant operation.
Choppers
359
The direction ofload current is always positive, Fig. 7.12 (e), because choppers and diodes can conduct current only in the direction of arrows shown in Fig. 7.12 (a ). Waveform of source current'i. and chopper current iCH1 or iCH2 are al~o sketched in Fig. '7 .12 (c) . As average values of both lIo' ill are positive, chopper operation in first quadrant is obtained and power flows from source to load. Various waveforms for TOIl < Tolf are also sketched in Fig. 7.12 (d) . It is seen that average value of 110 is negative, hut that of io is positive. Thus, fourth quadrant operation of type·D chopper is obtained and power flows from load to source. Average value of output voltage, from Fig. 7.12 (e) and (d ) is _ VJ TOri - VJ • Toff _ T 01\ - Tolf Vo T - VJ ' T (i) In case T Ori> T otr' a > 0.5, Vo is positive 8S in Fig. 7.12 (b) and (c). (ii) In case TOri < T off' a < 0.5-, Vo is negative as in Fig. 7.12 (b) and (d) . (iii) In case TOri = T off' a = 0, Vo = 0, Fig. 7.12 (b ). 7.4.5. Four- quadrant Chopper, or Type-E Chopper The power circuit diagram for a four-quadrant chappel is shown in Fig. 7.13 (a ). It consists of four semiconductor switches CHI to CH4 and four diodes DI to D4 in antiparallel. Numbering of choppers CHI, ... , CH4 corresponds td their respective quadrant operation. For example; for first quadrant operation, only CHI is operated; for second quadrant operation, only CH2 is operated and so on. Working of this chopper in the four quadrants is explained as under: First quadrant : For first-quadrant operation of Fig. 7.13 (a ), CH4 is kept on, CH3 is kept off'and CHl is operated. With CHI, CH4 on, load voltage 110 = VJ (source voltage) and load current io begins to flow. Here both lIo and io are positive giving first quadrant operation. When CHl is turned off, positive current freewheels through CH4, D2. In this manner, both Vo,10 can be controlled in the first quadrant. Note down that type-E chopper operates as a step-down chopper in this quadrant. Second quadrant : Here CH2 is operated and CHI, CH3 and CH4 are kept off. With CH2 on, reverse (or negative) current flows through L, CH2, D4 and E. Inductance L stores ener gy during the time CH2 is on. When CH2 is turned off, current is fed back to source through
diodes Dl, D4. Note that here
(E+ L~:I is more than the source voltage V,.As load voltage
V o is positive and 1 0 is negative, it is sec&nd quadrant operation of chopper. Also, power IS fed
back fr om load to s ource. For second quadrant operation, load must contain emf E as
sho~n
in Fig. 7.13 (a). In second quadrant, configuration operates as a step-up chopper. i,
i,
)... i
CHI :
,
.i
v,
" 01
i,
CH3 i L
,E
~ .. j
~)
~ -;
..
+
OJ
DJ
v,
~VO~
CH2 l~ .J, - t- 02 "
.
..
,
' -l·t;
.
.
."" . ...
CH"/ ;: I 1! ~-
~ , 0 ~, . .~ ~~ 02 CHl.'~:: I :I
T
W
Fi g. 7. 13. Four-quadrant, or Type·E chopper circuit diagram with (a ) load emf E and (b) load emf E reversed.
360
[Act . 7.4]
Power Electronics
T h ir d q u adrant. For third quadrant . ..~ Vo operation, CHI is kept off, CH2 is kept on . I choppe r on '/;P-~~ ~h~~per···. : 4P('ltJ~~.o~ .I~p-dow" chopper and CH3 is operated. Polarity of load emf CH2 ape r.c.1!d '.: CHI ope rated CH2.0ol : L.tornt nfl"l)' • CHI-C H-t an " • E mu st be reversed for this quadrant CH2 -oIT : t hen 0 1-0 4 condllet CH1oa : then CH".02 co ndllci working; this is shown in Fig. 7.13 (b ). When e R3 is on , load gets connected to :-l:!., _ __ __ _ _ .....:!l, sou rce V) so that both v". ill are negative Z clIappen an .1, p-dOUiII chopper I chollper on ",p'.pchopper CH4 operated leading to third-quadr ant operation . When CHS - operlltd CH4-02 : L ttoru utrtY CHl-C HZ: on CH3 is turn e d-off, negativ e current CH3- orr : then CH2· 04 co ndllct CH4 - oIT. then 02, 03 condllct fr eewheel s through CH2, D4 . In thi s E "wned £ revtlu-ed mann er, Vo and i" can be controlled in the third quadr ant. Here chopper operates as Fig. 7.13 Cc) Type-E chopper; oper ation of various a st e p-down chopper operates as a devices in the four quadrants , step-down chopper. Four th quadrant. Her e CH 4 is operated a nd other devices are kept off. Load emf E has' its polarity as shown in Fig. 7.13 (b) for its oper ation in the fourth quadrant. With CH4 on, positive current flows through CH4, 02, Land E . Inductance L stores energy during the time CH4 is on . When CH4 is turned off, current is fed back to source through diodes D2, D3. Here load voltage is negative, but load current is positive leading to the chopper operation in the fourth quadrant. Also power is fed back from load to sour ce. Here chopper operates as a step-up chopper. J'he devices conducting in the four quadrants are indicated in Fig. 7.13 (c) . Example 7.5. Show that for a basic de to de converter, the critical inductance of the fi lter circuit i<; given by .
.. . .
-+____--:-__
..<;~;;(V~'~-~V~o) L- -v -
2fV, Po
where Vo, V" Po and f are load voltage, source voltage, load power and chopping frequency respecti vely, Solution. The critical inductanceL is that value of inductance for which the output current falls to zero at t = T during the turn-off period of the chopper. A typical wavefonn of output current, with critical inductance in the load circuit, is shown in Fig. 7.14 (b). If current variation, fro m zero to 1m " during Ton and from Irru to zero during Torr' is assumed linear, then average value of output current 10 is given by
10 T=
~ 1=
To"
+ ~ 1= T"" =~1= (TM + Toff)=~ 1= T ",
v
r-
• I-T," - J- TO f! ...
L
,
. 'O tt
,
.,
t
'0
~ FD
load
t
.1 (0)
(0)
Fig. 7.14. Pertaining to Example 7.5.
'
[Art. 7.5)
Choppers
361
or 1mz = 2 10 = maximum value of chopper current at t = Ton. It is seen from Fig. T.14 (a}. that when chopper CH is on,
di Vo+L-=V dt • or
or
r
Vo+LTmz=V. M
2/ 0
L-=V -Yo Ton •
L
=
(V,I- Vol To" 2/0
.. .(i)
But average value of output voltage Vo = {To" V,I and output, or load, power Po = Vo 10 . This gives
Po o
and 10=V
Substituting these values of To" and 10 in Eq (il, we get (V. - Yo)
Vi
L =-'--i2'f"V"'.P"o""" 7.5. STEADY STATE TlME.DOMAIN ANALYSIS OF TYPE·A CHOPPER For the type·A chopper of Fig. 7.8 (a) with RLE load, the waveforms for gate signal ii' load current io and load vol tage Uo are as shown in Fig. 7.15 (a ) for continuous conduction and, in Fig. 7.15 (b) fOT discontinuous conduction. In Fig. 7.15 (b), periodic time T is more than that in Fig. 7.15 (a) . The determination of load current expression is useful for knowing (i) the current . profile over periodic time T, (li) the current ripple and (ii i) whether the current is continuous or discontinuous. The object of this article is to study the type·A chopper with RLE load for current variation over T, current ripple and also for the Fourier analysis of output voltage. For RLE type load , E is the load voltage which may be a de motor or a battery. When CHl is on in Fig. 7.8 (a), the equivalent circuit is as shown in Fig. 7.15 (e). For this mode of operation, the differential equation governing its performance is
V.
=R i + L ddi + E
... (7.6)
t
for When CHI is off, the load current continues flowing through the freewheeling diode and the equivalent circuit is as shown in Fig. 7.15 (d). For this circuit, the differential equation is
O=R,'+Ldi+E dt for
(7-, ) ....
Ton < t S T.
Solution of Eqs. (7.6) and (7.7) may be obtained by the use of Laplace transform . It is seen from Fig. 7.15 (a) that initial value of curren t is 1m " for Eq. (7.6) and 1= for Eq. (7.7 ). Therefore, Laplace transform ofEqs . (7.6) and (7.7) is
RI(s ) + L [s/ (s ) -/mol
V -E
= •s
.. .(7.8)
362
[Art. 7.5]
t
Power Electronics
I
l
1
• I
!
I
1
t6
--.. -
L
01
I I
io
Imn
I Vo
I
I
V
•
,
~..J
I
~I' I
I,
I
Vs
rE
t
I
rTon-i
w Fig. 7.15 . Type-A chopper (0 ) continuous load current and
j+ v,
1-
.1
W
(b) Discontinuous load current.
r ~, r l
R
""CI'
I
I
Vo
T~
o
.,,
I
"-
(e)
Imn
)
i
v,
Fig. 7.15. Equivalent circuit for type-A chopper with
L
R
eel CHI on
E
-,
(d)
and (d) CHI off.
I
!-_1-..,I!------'_--'-----_--'-----_ __ 1
~Ton#
~'---H , .~ t =t-T.on ';0
(e)
Fig. 7. 15 (el Pertaining to t' ,
and
From E q_ (7 _8),
E RI(s) + L[sI(s) - 1=] = - s VJI-E L · Imn V, -E Imn I(s ) -s(R+s + =~+--L ) R +s -L L +R _R s· s L S-r L
___ (7_ 9 )
[Art. Hj
Choppers
363
Lapl ace inverse of the above eXpression is i(t) =
for
V.r;E(I_e-~t)+lmne-~t
... (7.10)
OSt S T cw
Similarly, the time-domain expression for curreht from Eq. (7.9) is ... (7 .11)
for
where t' = t - Tan. see Fig. 7.15 (e), so that when t=Tan , 1'=0 t= T. t' = T- TO" = Torr.
and for
The variation of current i(t) from Imn to Irnz for 0 S t 51 TO" can be plotted from Eq. (7.10) and that ofi (t') from 1m:.. to Imn for 0 <: t' S Tafffrom Eq. (7.11). . In Eq. (7.10), at t
=T
M ,
i(t) =1=.
...(7.12) In Eq. (7.11), at( = Toff =T-Tw
i(t')=I~
... (7.13)
where
Eqs. (7.12) and (7.13) can be solved for 1= and 1m. as under: From Eq. (7. 12),
I
=~I_e- (To.IT.)) _ER (I_e-r •./ T.)+1 ernzm . mn
T .. IT•
"
Substitution of Imn from Eq. (7.13) in the above expression gives I
or or
=
m =.; (1-
E E -T.~ ITo -E- e-T. .IT. = V'1 -e- T" I T) --+-e 0
R
+E
R
e-r...lTo ) _
R
R
. e -(T - T... l/ To . e-T... IT.+1
=
~ (1- e- TlTo) + l nu x e- TIT•
lnu(l-e-rlTo) =1(l_e-r..,lJ'o) _ I
e- (T - T••.JIT' .e- T .. I T.
~ ( l_e- T ITo)
T T
= V.[I_e- .l o]_E nu R 1 _ e TIT. R
... (7.14)
Substitution of 1= from Eq. (7.14) in (7.13) gives EE . l m.r =-]l+]i ·e
VIr l_e- ro/ T. ] - (T-T )/T E -(T-T )/ T o+1
(T-T )IT .~
[Art. 7.5)
364
Power Electronics
VJ 1- e- T ... 1T• eT""IT. =Ii 1- e- TIT. e T1T•
V,
Imll=R
eT..,IT. -
1]
eT1T. _l
E
., .
R
E
.. .(7. 15)
- R
In case CHI conducts continuously, then Ton =T and from Eqs. (7.14) and (7.15). ... (7.16) The maximum Irru and minimum 1m " values of load current can be obtained fr om Eqs. (7.14) and (7.15) respectively for given V" R, ex, To. and E. For those who are not familiar with Laplace-transform technique, the following method may be adopted for solving Eqs. (7.6) and (7.7). Eq. (7.6) can be re-written as
~~ =V,-E
Ri +L
where
(R + Lp) i = V. - E d p 5 dt
... (i)
The solution of Eq. (i) consists of two parts, complementary function and the particular integral. Complementary Function: It is obtained from force-free equation (R + Lp) i = O. Its solution is of the type i = AelJt ,
Here p is the root of the auxiliary equation R + Lp "
.
lC,}'.
=Ae
=0 and this root is p =- ~.
-!! t L
Particular Integral : It is obtained from Eq. (i) by putting p = o.
..
Ri=V.-E ,
or
tpJ.
=
V.-E R
Therefore, the complete solution of Eq. (7.6 ) is i;= ipJ . + icy. V -E -!!, = ' +Ae L R ConstantA can be obtained from the initial condition. It is seen from Fig. 7.15 (a ) that at t = 0, the initial current during Ton is i = Imn'
V-E
Imn = 'R
+A
) A=Im,,- V-E J ( R
or ..
i=
V.~E (l _ e-~ f) +Im,, _e- ~'
...{7.10)
This agrees with the s olution in Eq. (7.10). Solution of Eq. (7.7) cnn also be ob tained similarly as in Eq. (7.11).
\
[Art. 7.5)
Choppers
365
7.5.1. Steady State Ripple It is seen from Fig. 7.15 (e) that current pulsates between l nu and f nw The r ipple current (1= -1m .) can be obtained from Eqs. (7. 14) and (7. 15) as follows :
i[ s
l_e-T...I T.
eT...n .- I ]
l nu- 1mn =R 1 - e-TIT• - eTIT-l •
_ V , rl_e- T...IT• (l _e-T...IT.)eT... IT. ] - R l l - e- TIT• - (l_ e-TIT·) . eTIT.
_ V,r (1 - e- T...!T.) _ (1 _ e- T... IT.) elT.. - n I T.]
~Rt
l _e- TIT•
=i['(l_e- T•.IT.) (l_e-'T-T )lT.)] TI T M
1- e-
R
... (7 .17)
•
The ripple current given by Eq. (7. 17) is seen to be independent of load counter emf E. With Ton = ex T and T- Ton = (I-a) T, Eq. (7.17) can be written as
_ -i[' l nu I mn - R Per
. Unit
. npple current
=
(l_e: aTI T.) (l _e-( l - a)TIT.)]
Imx -1mn V IR
- TIT
I-e'
1
=
s
The peak to peak ripple current has maximum value !:Jlnu. when duty cycle a = 0.5 in Eq. (7 .18).
=x
(7.1S) is
~mx ~rom
for co n venience.
-171
1.0
.1. =25 To
t C
0.75
-,• • u
0. 50
2,
.,
R ·c j
nt.t
V ,[(I_eR
O ." )
(l _el _e-.t
O ." ) ]
_ V. [(1_.- 0 .") (l_ e- O.")] - R (l+ e- o·")( I _.- o.,,)
O V .[I _e ."] V. 1 =Ti l +e-o.s..: = R tanh'4 x V. tanh - T
=-
R
4T.
1 L T= -and T =f • R
,:
0. 25
ci. 0. 51
.
0
=
... (7.18 )
•
Eq .
" 6.1
But
e
-
For ex = 0.5 and T I Ta = 5, p_u. ripple current = 0.S4S. For IX = 0.5 and TI T. = 25, pu rippl e current = 1. In this manner, the variation of pu rippl e cu rr ent as a function of du ty cycle IX and r atio TI T, can be plotted as shown in Fig. 7.16 . Its value is maximum when a = 0.5. As L increases, Ta ( = L I R) increases and T I Ta r educes and pu ripple current decreases, Fig. 7.16.
Putting ;
.
(1_e- aT1T. ) ( l _e-(l-Cl)TIT.)
0.25 Duty
Q.5 cycl~.
0.7 5 a _
Fig. 7.16. Per unit ripple current as a fun ction of (l and T I To .
1.0
366
[Arl. 7.5)
Power Electronics 6I
=
In case 4fL » R, then tanh
4~ ==
V,
R
R
4{L
=-tanh-·
:Jr.. Under this condition, maximum value of ripple
current is ... (7.1Ba)
This shows that maximum value of ripple current is inversely proportional to chopping frequency and the circuit inductance. Example 7.6. In the continuous conduction mode of type-A chopper, show that per unit ripple in the load current is maximum when duty cycle is equal to 0.5. Solution. Eq. (7.18) can be used to prove that ripple in the output current, or per-unit ripple in load current is maximum when the duty cycle is equaTto 0.5. Therefore, from Eq. (7.18), ripple current III is _ V, [
t.1- R
l-e
... (i)
-TIT
•
Also per-unit ripple in load current is !J. I (1 _ e- aT/T. _ e- ( 1- a)TI T. + e- TIT.) v,7R = 1 - e- 1'11'.
" .
...(ii)
For obtaining the values of duty cycle for which the ripple in curr.ent is maximum. differentiate Eq. (i) or (ii) with respect to 0 and equate to zero.
~ d
V~ , do
or
(1 -e
-=-
(I..)
- TIT)[O • -e-aTIT•. -
=
-e -'I-a)TIT• . I.. + ~
(1 _ e TI T.)'1
T . e-aTIT._~ ' e-{1-a)TI T. To
~
0] =0
=0
To
or
e-aTIT• = e-(l-o)TIT•
or
aT T -=(l-a)To
0=(1-0)
To
or
This shows that for duty cycle equal to 0.5, ripple in load current is maximum. 7.5.2. Limit of Continuous Conduction In a chopper, if Ton is reduced, Toff increases for a constant chopping period T . At some i'Uw value of Ton. thE;! value of Torr is large and the current i may fall to zero. Since the current i;'l type-A chopper cannot reverse, it stays at zero. The limit of continuous conduction is reached when I",n in Eq. (7.15 ) goes to zero. The value of duty cycle a at the limi t of continuous con duction is obtained by equa ting Imll in Eq. (7.15 ) to zero. Therefore, 1M "
=~e:;;;: ~lll- ~ = 0
[Art. 7.5J
Choppers
367
or
r
or .. .(7.19)
or For giv en E , V" T and Ta ; if duty cycle is a ' as given by Eq. (7.19), then the current is just continuous. If actual duty cycle is less than a' as giv en h ere , th e l oad current wo uld be discontinuous. For some value of Ta I T, say I , value of a' is calculated for various values of m, from 0 to 1. For example, for m = 0.5,
No cl'lopptr operation
0.75
t
0, 50 ,,' = lIn [1 + 0.5 (0'_ 1)1 = 0.6201. For other values of m , a' is computed and E plotted as curve DAB in Fig. 7.17 with T", I T as 0.25 one parameter. For this value ofT",I T ( = 1.0), the OABCO represents continuous conduction region and th e oth er ar ea OABDO as the region of °O~~~~~==t.~--~ O.~7So---~'.0C 0.25 0. 5 di scontinu ous conducti on . In Fig. 7.17 , for m = 0.5 and Tr/T = I , point A gives the limit of Fig. 7.17. Lim jt pC contin uous cQ f!~ q j:tiop con t inuou s co n du cti on with a ' =0.6201. F or for type-A chopper. th ese v a lues of m( = 0.5) and (T. f T ( = I ), if actual duty cycle " is equal to 0.7 (say) , then this point would lie in the re~ o l) OM)(:Q ; tpit value of " would therefore give continuous cpnduction 'for the Joad current. !3trl'ight line ail corresponds to TalT = O. Actually, the chopper operation for the area between (FifJ. 7 . ~ 7)
(i)
the straight line DB and BOD is not possjble as Tal T can never be le~s than
~~ro,
(ii) th e .t raight lin e OB and curve OAB represent. di scontinuous C\lrrent !llQ~Q T. 0<1" < 1. (iii)
curve DAB and DCB represents continuous current mode for
for
~ > 1-
In case actual duty cycle a is equ al to 0.6 (say), then as this point would lie in thf:! reron between the curve DAB and straight line OB. the IOlld current would be discontin\.lQ\ls. C\1ry~~ like DAB can also be drawn for other values of TaI T . One such curve for Tal T = 5 is :jhoWl) ip Fig. 7.17. For this value of Tal T = 5, load current is c:1 iscontinuous for an operatiI'lij' point between st raight line DB and curve marked (Ta/ 1) = 5 ; load current is continuous for ~ operating point between the curve marked (Tal T) = 5 and DCB . 7.5.3. Com pu ta ti on of E xtinction Time (r The exp ress ions for load current obt ained in Eqs. (7.10) to (7.16) apply only when load current is co nti n uous. For Tafl < t < T, the load current may become discontinuous due to lar ge T olf. The ti me tx ' called extinction time and measu red from the instant ! = 0, Fig. 7.15 (b), can be calculat ed as und er :
368
Power Electronics
[Art. 7.5]
As tx is within the time limits of Ton < t < T, Eqs. (7 ,11) and (7.13) should only be used for
computing t x' The value of Imz needed in these equations should, however, be obtained from Eq. (7.12). Thus, in Eq. (7 .12),lmn = 0 at t = tx• i.e. within Ton < t < T . This equation gives lmz as
V -E(l_e- T...1T.) )
... (7.20 )
Imz= ' R
Substitution of this value of lnu in Eq. (7.11) at t'
=t% -
Ton gives i(t' ) = a as
o=-1ll Erl -e-(I-T... VT] + V.-E[l R -e -T" IT] · e-(l-T ~
or
s
lIT
...
E[l _ e- (t" - T...l/ T.]e(ls- Tool/ T. = (V. _ E)(l _ e- T... IT.)
or
e(tz - T .. )IT.
-1
=V.;E (l_e-T..lTo)
This gives the extinction time tx as
t,=T~+Ta~l+ V,;E(l_e- T.. IT.)]
... (7.21)
The average output voltage for the discontinuous current mode as shown in Fig. 7.15 (b) is given by
Vo=
~J: dt=~S:" V.' dt+ J~.. O· dt+ ( "0 '
T~
Edt]
E
=V. T+ T (T-t,) or
... (7 .22 )
Vo = ex V, + (1- i ) E volts.
7.5.4. Fourier Analysis of Output Voltage For continuous load current, the load voltage waveform Vo is as shown in Fig. 7.15 (a). This voltage waveform is periodic in nature and is independent of load circuit parameters. Voltage wave of Fig. 7.15 (a) can be resolved into Fourier series as
._1
where un.
... (7.23)
=value of nth harmonic voltage 2 V, nn
= - - . sin mt a ·sin (n cot + 9n.)
v:
=uV u=T IT and 9 =tan- 1
0.1'
01\
/I
... (7.24)
sin2nnu l-cos2nnu
=tan-l[c~S1tncx ]. sm1tnu
The average value of output voltage Vo can be controlled by varying the duty cycle u. The amplitude of the harmonic voltages, i.e. 2 V. l ntt sin n1t u, depends on n, the order of harmonic and also on the duty cycle u. The maximum value of nth harmonic occurs when sin n1t U = 1 and its value is 2 V,
- - = Jt1l
0.6366V, volts n
...(7.2Sa )
I I!
[Art. 7.5J
Choppers
2 V. ,n nv2
~=
and its rms value is
0.45 V. volts n
369
...(7 .25b)
The harmonic current in the load is given by
.
v,
l,,=Z-, where Z,. is the 19ad impedance at hannonic frequency nf Hz and is given by
Z, = Vii' + (nOlL)' . v V
.
For negligible load resistance R, in = .~r or in"'" n~
--i . This shows that harmonic current n
decreases as n, the order ofhannonic, increases, Another term used fo~ knowing the harmonic content of a waveform, with~ut calculating its harmonic components, isjhe ae ripple voltage Vr It is defined as ... (7.26) In the above equation, Vrlll4l = Vor and Vo are respectively th~ .rms and -average values of the output voltage. It is seen from Example 7.1 that V~ =
va V.
from part (d) Vo = a V, from part (a) Vr = a 2V; = V, ..,Jra-_-a'r,.
and
va. v: -
.:.(7.27)
Ripple factor, defined as the ratio of ac ripple voltage to average voltage is given by V,
=y-
ripple factor
'; ---.-
.
RF=V . -ia-a'
or
,
V. '(l
=~l-a a
TC'; 1 ' -'fa-.
=-
... (7.28)
Example 7.7. ra) For an id
a(V.-E) R
L
RT (Inu: -I".,,) .
(b) For the chopper of part (a), derive an expression for the average current in the freew heeling diode for a continuous load current. (c) From parts (a) and (b), prove that average value of load current 1011 is liven by
Iou=
V,-E R
Solution. For type-A chopper, the output voltage waveform va and the waveforms for load current io, input (or thyristor) current iT and freewheeling diode current i fd are as shown in Fig. 7.18. (a) When chopper is on, the voltage equation for the chopper circuit of Fig. 7.8 (a) is d' RiT+L +E=V.
di
or
R~ ·
diT dt+L dt dt =(Y. -E) dt.
370
Power Electronics
[ArL 7.5)
r-
~Ton
Imn iT
Imn L--+~L--+-4--~ I -+--__ t
I
tId
1m
i
II
I
Fig. 7.18. Pertaining to Example 7.7.
Its average value, Fig. 7.18, is
1 fT~ .
R .T
0
1 f To
"r' dt + T
0
diT (V,-E) fT~ L · dt dt = T 0 dt
1 fl~.
R l TAy + T
I ••
L R l TAy + T (1= - I m .l
or
TM
Ld"r=(V,-E) ""if'
or
=(V, - El a
ITAV =
a(V,-E) L R -RT(Ir=-Imll }
.. .(7.29)
When the fre ewheeling diode is conducting, load voltage is zero. The voltage equation, when FD is conducting, is given by (b)
.
'!..!f!I.
R'fd + L dt +E=O Its average value , Fig . 7.18, is
T 1fT. If di d 1fT R T T 'fd · dt+LT T =d dt+E T T dt=O o. ... t ... or
L
fl...
R . Ifd + T I~ d'fd =- E I
T-TM T
_ L (1= -Im.l E (1- a) ,,, TR .R
or .. .(7.30)
Choppers
[ArL 7.5]
371
(c) Average load current over a complete cycle can be obtained by adding IT AV and lfd from
Eqs. (7.29) and (7.30) respectively.
1.,= =
a(V.-E)
R
L
L
- RT(I~-Im.)+ RT(I~-Imn)-
Vo-E R
To" Vo-E = R a. or
- II II
aV.-E a V.-aE R ' a= R
This will give a maximum value when
_ _ _
• t
• t
\T/__---.
2
=
:: )i------. lol~
The average thyristor current IT is given by
IT=I
R
aV.-E Vo-E R = R . This is the required result.
Example 7.8. For Iype-A chopper, feeding an RLE load. obtain maximum value of average current rating for the thyristor in case load current remains constant. Solution. For constant load current 10, current waveform for thyristor current iT is as shown in Fig. 7.19. Here .
10=
E(l-a)
... (7.31)
• t
Fig. 7.19. Pertaining to Example 7.8
dlT 2aV.-E da= R =0 I
and from this, a = 2~
... (7.32)
•
Therefore maximum value of average thyristor current is obtained by substituting the value of a from Eq. (7.32) in Eq. (7.31).
..
IT~=2:• R · (2~V.-E)=4~'RAmP" • •
Example 7.9. For type-A chopper circuit. source voltage V, = 220 V, chopping period T = 2000 J.1S. on-period = 600 J.lS. load circuit parameters : R = 1 n. L = 5 mH and E = 24 V. (a) Find wMther load current is continuous or not.. (b) Calculate the value of average output current. (c) Compute tM ma.:ri!;num and minimum ualues of steady state output current. (d) Sketch the time variations of gate signal i" load voltage uo• load current i o• thyristor current in freewMeling diode current ifd and voltage across thyristor VT' (e) Find rms ualues of tM first. second and third harmonics of the load current. (f) Compute the average value of supply current. (g) Compute input power, the power absorbed by the load counter emf and the power loss in. the resistor. (h) Compute rms ualue of load current using the results of (b) and (g). (i) Using results of (e). fi n.d the nns value of load current. Compare the result with that obtain.ed in pert (h).
j
.J
372
Power Electronics
. .fMC 7.51
.
."
..'
L 10- 3 . T =-=5 x --=5 X 10- 3 sec a R l ' 3 To = 5 X 10- = 2.5 T 2000 x 10- ' T E 24 - =0.4 ; m =-= =0.11 T. V, 220
Solution. (a)
600
To.
600
a = 2000 = 0.3, T = 5000 = 0.12 .
•
a' = 2.5\n [1 + 0.11 (.0.' -1 )J = 0.13172
From Eq. (7.19) ;
As adual value of duty cycle a ( = 0.3) is more than a ', load current is continuous.
1 = a V, - E = 0.3 x 220 - 24 = 42 A
(b)
o
(e) From
rn;J;
1_.- ]_141 =51.46A
. = 220[ 1 l_ e- 0,4
-1]
220[.0.12 Imn = 1 eo.4 _ 1 -24=33.031A
From Eq. (7.15), (d)
1 0 12
I
Eq. (7 .14),
R
The various waveforms are sketched in Fig. 7.20.
CJ 2.0
'o h
C
4.0
2.6
D
[
I
!
1
,
.
22 0V
!
t in ms·
i~ 1 ° ;' 51.46A .1 1I I lJ.031A
trh
51.':'6 A
r1
I 1
t
t
--IC___.,-
JJ.oJ1A _ _ _--L_ __ L L L
t
ifdj
L._L~ _51_.•_6_A_.L--L~ _ _ _~J_ ' ._OJ_'_A_c_
'T 22 0V
t
tlnUl 05
2.0
2.6
' .0
Fig. 7.20. Perbining to Example 7.9.
•
r
,
[Art. 7.5]
Choppers
373
(e) From Eq. (7.24), rms value of first harmonic voltage is
2V, .
2 x 220 . sm 54° = RO.121 V
V, =...",.- sm (n x 0.3) =:r.::-
. . . 27t
'121t
T
1 10' Here chopping frequency, f = T = 2000 = 500 Hz
=,101'-'+---;;(2c-n-'-x""5'"'0"'0""x""5:-X""."-1000=3o:t):.-, -
..
Z 1 = -JR' + (wI. )'
..
- V, _ 80.121 - 5 0903 A 1,Z, - 15.739762 - .
S · '1 1 lmlary.
15.739762 n
1 = 2 x 220 [sm ' 1080] 1 2 2 .T2'. n ~1'+(2nx500x2x5x10 ')' =1.4983 A
1 - 2 x 220 [in 1620] 1 '-3.,f2 ' n s ~l'+(2nx1500x5x10-')' = 0.21643 A (() From Eq . (7.29), average supply current ,is 1 T,o\V=
0.3 (220 - 24) 5 x 10-' (51.46 - 33.031) 1 lx2000xl0-6
= 58.8 - 46.0725 = 12.7275 A
(g ) Input power = V,I x average supply current =
220 x 12.7275 = 2800.05 watts
Power absorbed by load emf = E x average load current = 24 x 42 = 1008 walts
Pow er loss in resistor (h)
R
=
lor =
2800.05 - 1008 = 1792.05 W
"i!II + Ii + I~ + ~
-J42' + 5.0903' + 1.4983' + 0.21643' = 42.31 A = PR = 1792.05 W =
(i)
Power loss in resistor 10,
=,; 179~.05 =
42.333 A
The results obtained in parts (h) nnd (i) are in complete agreement.
Example 7.10. For type·A chopper, source voltage VI = 220 V, chopping frequency (= 500 Hz. Too = 800 1lS. R = 1 n,L = 1 mH and E= 72 V.
'.
(a) Find whether load current is continuous or not. (b) Calculate the values of auerage output uoltage a7td auerage output current. (c) Compute the maximum and minimum ualues of steady state output current. , (d) Sketch the tim.e uariations of gate signal i" load current tOI load uo~tage uo. thyristdr current iT' freewheel ing diode current iid and uolt?Ce aCrOSS thyristor u7"
Solution. Here
T• = R L = 1 x 10- 3 sec • T=
1 71 = 500 = 2000 J.Lsec
Power Electronics
[Art. 7.5]
374
Ta
-3
7'=f T.=500dO
T
1
=0.5'T. =fT. =2,
m=~= 72 =0327 V, 220 . a
=
TTl!. = f · Tan =,500 X 800 X 10-6 =
0.4,
TM •
1'=0.8 (a)
From Eq. (7.19), a' = 0.5ln [1 + 0.327 (e-2 -l)J = 0.564
As a is less than a', load current is discontinuous. (b) From Eq. (7.21),
t z = 800
X
• t
iT
10-6 + 1 X 10-3
I
In[l + 220 ;72 (l_e- O')] 1.55703 x From Eq. (7.22), =
10-3
7 sec
4
11: = 0.4 x 220 + (1-1 .55703 x 10-3).72 o 2 x 10-3 = 103.95 volts.
(7.20) is
· I I i I I ·
[
U[
As the load current is discontinuous,
/fM = O. The maximum value of current from Eq.
Ii ·
I ·I I. .[
10 = 103.7; -72 = 31.95 A. (c)
I I I Ii!
Fig. 7.21. Pertaining to Example 7.10.
1= = 220; 72 (1- e-
The time variations ofvarions waveforms are sketched in Fig. 7.21.
Example 7.n.An RLE load is operating in a chopper circuit from a 500-volt de source. For the lood, L = 0.06 H, R = 0 and constont E. For a duty cycle of 0.2, find the c/u;pping frequency to limit the amplitude of load current excursion to lOA
Solution. The average output, or load, voltage is given by Vo = aV,
As the average value of voltage drop scross L is zero,
E = Vo = aV, = 0.2 x 500 = 100 Volts. During Ton. the difference in source voltage V, and load emf E. i.e) (V, - E) appears across induc tance L as shown in Fig. 7.22. :. During Ton. volt-time area applied to inductance
= (500 -100) Too = 400 Ton Voll-s ec
,
\
[Art. 7.3]
375
... E
(b)
(a)
Fig. 7.22. Pertaining to
Exa~ple
7.11.
Also, during Tor!> the current through L rises from Imn to Imz. From this, volt-time area across L during this current change is given by
di f'-' •• L · di=L(I=-Im,.)=L · M ° Ldtdt= f°T. "e. dt = fTThese two volt-time areas during TOIl must be equaL
400 To. =L· M 0.06 x 10 TOil = 400 = 1.5 msec
or
Thus, chopping frequency,
f =~ ='; 0/1.
=1.5 0.2 =133.33 Hz. x 10 3
Example 7.12. A series motor used for a rapid transit system is fed through a de chopper. The series motor has total circuit resistance of 20: and inductance of 2 mHo What external inductance should be inserted in series with the armature circuit in' order to limit tm per unit ripple in armature current to 10% for a duty cycle ratio orO.5. The chopping frequency is 1 kHz. Solution. From Eq. (7.18), per unit ripple in current is given by
I/I1Z - Imn V,IR Let
J
=[
= x. Here a = 0.5 andpu ripple = 110 0 = 0.1. Substitution of thes e in the above
o
expression gives
o.1 -or
(1 - ,- °''')(1_ ,-a.'') (1 _ ,- 0''')(1_ , - 0.,,) . 1 _ ,- 0." - "'-"::To; l_e-:r - (I-e o·5X)(I+e o·5.J:) .; -I+e O.5:r
O.I=I-y wherey=e- o.Sr
l+y
or
0.9 -0." Y=--=' 1.10
or
,0" = 10~0 = 1.22222
or x
=TT o
=
0.40134
376
Power Electronics
[Mt 7.5]
T 1 T. = 0.40134 = 0.40134 f
But
r= 1000 Hz
1
T. = 0.40134 x 1000 2
or
L =RT. = 0.40134 x 1000 = 4.983 mH
1'herefore. external inductance that should be inserted for keeping the pu ripple within limits = 4.983 - 2 = 2.983 mHo Example 7.13. A step·down chopper, red from 220V dc, is connected to RL lead with R = 10 nand L = 150 mHo Chopper frequency is 1250 Hz and duty cycle is 0.5. Calculate raj minimum and maximum ual~s of load current (b) maximum value of ripple current (e) average and rms ualues of load current and (d) rms value of chopper current.
Solution. Here
To =
T=
i 7
=
;~o- 3 -1.5 x 10- 3 S
15
1 = 12 50 = 0.8 x 10- 's, T M = aT = 0.4 x 10-' s
Too = 0.4 x 10-' = 0.267 L = 0.8 x 10~ 3 = 0.S33 To 1.5 x 10- 3 • Tg 1.5 x 10- 3 O267
From Eq. (7.14),
220 [1_e- . ] Irru=I2=1o l_e- o.533 =12.478A
From Eq. (7.1S),
Im.=Il= 10
.
261 220 [ eO. - 1] e . =9.S6A
O 533
1
(b) Maximum value of ripple current, M =1, -II = 12.478 - 9.S6 = 2.918 A
fL
V From Eq. (7.18 a), approximate value of maximum ripple = 4 .220 X 10- 3 = 4 x 12S0 x IS = 2.933 A (c) Average .output voltage,
V. = a V. = O.S x 220 = 110 V
Average load current,
I = 110 = 11 A , 10
Also
I, = II ;1, = 12.478 + 9.S6 = 11.02 A 2
Asswning linear variation of current from II to 12, the current during T 0/1 can be expressed as . 1,-11 M lo=I1+-;z.--t=I1+Tt on.
:. Rms value of load current,
on.
,IT. M)' J (I} + T. t . dt
lor = T
on
=i
on
0
on
M ('"(I:+(TM t)' +2I1· T t ) dt O on on
I
1
[Art. 7.6J
Choppers
377
or
Example 7.14. Show that the critical inductance in the load circuit ora step-down chopper is proportional to x (1- x ) where x is ·the duty cycle. Solution. This example is just an extension of Example 7.5. From this example, fr om Eq. (i), the critical inductance L is given by L=
(V~ - Vol Ton
'21
o
•
~
..
Now T rm = a T and Yo = a Vs. substituting these values, we 'ge~~ ~ . ..~ L=
,
V,.T
Here a = x = duty cycle, :. Cr itical inductance, L
V, (1- a) - aT 21
.
V, T
=2T. a (l-a)
.'. L=2T' x (I-x )
•
oc
x (1 - x)
7.6. THYRISTOR CHOPPER CIRCUITS So far a chopper has been shown as a switch inside a dotted r ectangle. Actually, a ch opper consists of main power semiconductor device t ogether with their turn-on and turn-off mechanisms: In low-power chopper circuits ; power tr ansistor s, GTOs etc. are being used widely. In high-power levels, however, thyristors are in common use. The object of this section is to stu dy the thyristor chopper circuits along with their commutation circuitry. The process of opening, or turning-off, a conducting thyristor is called commutation. In de choppers, it is essential to provide a separate commutation circuitry to commutate the main power SCR. It may be recalled that a conducting thyristor can be turned off by r educing its anode current below h olding current value and th en applying a reverse voltage across the device to enable it to r egain its forward bl ocking capability.' There are several ways of turning-off of a thyristor. All these mel.hods differ from one another in the manner in which com mutation is achieved. In de chappell'. commutation circuitry h as passed through num erous innovations. All these commutation circuits can, however, be broadly classified into two groups as under : (a ) F orced Commutation. In forc ed commutation, ext ernal.elements Land C whi ch do not carry the load current continuously, are used to turn-off n conducting thYTistor. Forced comm uta tion can be achieved in the following two ways: (0 Voltage commutation . In thi s scheme, a conducting thyristor is com mutated by the ap plication of a pulse of large r ev:erse vol tage. This r ever se vo lt age is usually applied by
378
[Art. 7.6]
Power Electronics
switching a previously charged capacitor. The sudden application of reverse voltage across the conducting thyristor reduces the anode current to zero rapidly. Then the presence of reverse voltage across the SCR aids in the completion of its turn-off process (i.e. aids in gaining the forward blocking capability of SCR). (ii) Current Commutation. In this scheme, an external pulse of current greater than the load current is passed in the reversed direction through the conducting SCR. When the current pulse attains a value equal to the load current, net .pulse current through thyristor becomes zero and the device is turned off. The current pulse is usually generated by an initially charged capacitor. . An important feature of current commutation is the connection of a diode in antiparallel
with the main thyristor so that voltage drop across the diode reverse biases the main SCR. Since this voltage drop is of the order of 1 volt. the commutation time in current commutation is more as compared to that in voltage commutation. In both voltage and current commutation schemes, commutation is initiated by gating an auxiliary SCR. (b) Load Commutation. In load commutation, a conducting thyristor is turned off when load current flowing through a thyristor either
(0 becomes zero due to the nature of load circuit parameters or (ii) is transferred to another device from the conducting thyristor.
Only three commutation principles listed above are described in what follows though there are numerous other commutation schemes. Chopper circuits with type-A configurat,ion are only studied in this section. For understanding the performance of voltage and current comroutated chopper circuits, Example 7.15 should be studied carefully. Example 7.15. In the circuit shown in Fig. 7.23 (a), capacitor C is initially charged to a uoltage Vo . If the switch S is closed at t = 0, obtain an expression for current i (t) . Sketch time uariation ofuoltage across capacitor and inductor and of the current i (t). ,ndicate also the flow of energy handled by Land C. In case L = 1.6 mH and C = 4p.F, find the time for which current i (t) flows in the circuit of Fig. 7.23 (a).
Solution. When the switch S is closed, KVL for the circuit is .. L Its Laplace transform is
d~~t) + ~
sUes)
f i(t) dt =a
+"clrt [(8)s -
CVo]
-8-
=a
Here negative sign bc";'')re CVol s is used because current i (t) leaves the positive tenninal of C and enter its negative terminal when switch S is closed. Vo
.,
1
[(8) = T 8' + l l LC
or
Vo
T·
1I-fLC 1/.JLC [8' + (l / .JLC )' )
"'0 1 ,where "'0 = VLC [8 + OlOJ Vo . . Us Lapl ace inver se gives ,(t ) = OloL sm "'0 t [(8)
Vo
= "'{)'-'T
,
...(7.33)
•
[Art. 7.6]
Choppers
379
!(t)~ o I
5
.........
,
.!.. ""...
"t =- \10
J~
L
C .
Energy t--lCV; "
L
flow
",=\10
I
'"
TO
"'"
.
2
,~_L'L -, 0 ~
7rW~t
!
L-O
"
--
,,
n
W,t
,
"'-_.J
,
1... Cv 2
0
.1., U m1
'
0
~)
00
Fig. 7.23 (a) Circuit for Example 7.15 and (b) time variations ofi(£), v, and vI.
r
", = ~ c , i(l) . dl - V,
Now
1 V, ~ . = C' wJ,. Jo 510 wot dt - Vo
Vo
1
=LC'3
"',
I
1 -cos wi lo-Vo ... (7.34)
= Vo (1- cos 000 t) - Vo =- Vo cos (wot
where 00; LC = 1 and
00 = ~~C 0
Also
is called the resonant frequency in rad/sec. "L
=L
di(l) dl
VO
=L,. WoL
(cos "'01) . "'0 =Vo cos "'ot
... (7.35 )
.
The time variation of i(t), lie and UL obtained from Eqs. (7.33) to (7.35) is shown in Fig. 7.23 (b). When i(t) ='lm, ue = VL = O. When i (t) becomes zero at wot =n, further conduction stops because diode can't conduct in the reversed direction. Therefore, at Wet =n, capacitor is charged to a voltage Vo but with reversed polarity. When switch S is closed, voltage across L at once becomes Vo, its value is zero at wot = nl2 and - Vo at wet =n. As current i (t) reduces to zero at
Tt,
i
2
We = CV0 again. The flow of
7.6.1. Voltage·Commutated Chopper One ofthe earliest chopper circuits which has been in wide use is the voltage-commutated chopper. This chopper is generally used in high-power circuits where load flu ctuation is not ve ry large. This chopper is also known as parallel-capacitor turn- off chopper, impulse-commutated chopper or classical chopper. Fig. 7.24 gives the power cirC1.1it dingra:n for this type of chopper. In this diagram, thyristor Tl is the main power switch. CO n1 :-:1utation
II
Power Electronics
[Art. 7.6]
380
circuitry for this chopp er is made up of an auxiliary thyristor TA, capacitor C, diode D and inductor L . FD is the freewheeling diode connected across the RLE type load. Working of this chopper can start only if the capacitor C is charged with polarities as marked in Fig. 7.24. This can be achieved in one of the two ways as under: Close switch S so that capacitor gets charged to voltage V, through source V" C, S and charging resistor Re. Switch S is then opened. (i)
Auxiliary thyristor TA is triggered so that C gets charged through source VI" C, TA and the load. The charging current through capacitor C decays and as it reaches zero, Vc = V, and TA is turned off. (ii)
R
~
Vs
L
'Is < R,
<
Fig. 7.24. Voltage-commutated chopper.
With capacitor C charged with the polarities as shown in Fig. 7.24, the chopper circuit is ready for operation. The current ie • iTl • ifd and io are taken as positive in the arrow directions marked. Similarly. the voltages tlc,-uTl. tlTA and tla across C. Tl . TA and load are taken as positive with the polarities marked. Simplifying assumptions for this chopper are (i) load current is constant and (ii) thyristors and diodes are ideal elements. The chopper operation, for convenience, is divided into certain modes and is explained as under: Mode 1. The main thyristor is triggered at t = a and RLE load gets connected across source V, so that load voltage tlo =V" During this mode. there are two current paths as shown in Fig. 7.25 (a). Load currentIo• assumed constant, constitute one path and commutation current ie the other path. Load current 10 flows through source V" main thyristor Tl and load whereas the current ic flows through the oscillatory circuit formed by C, Tl, L and D. The capacitor (or commutation) current first rises. from, zero to a maximum . value when voltage across C is zero at t =t} / 2. As ie decreases to zero, capacitor is charged to voltage (- V,) as shown att = tl in Fig. 7.26, see Example 7.15. The capacitor current changes sinusoidally whereas the capacitor voltage cosinusoidally from t = a to t = t} . This voltage is held constant at (- V$) by diode D. Voltage across TA is (- V,) at t = 0, zero at tl / 2 and V, at t l • this variation is shown as cosine wave in Fig. 7.26. The thyristor current iTl has a peak at t l / 2, because i Tl = ic+Io between t = a and t = t}. At the end of mode I , i.f!. att} ; ic = O. in =10 , tl e =- V" tlTA = V" tla = V.. as shown in Fig. 7.26. n-Iod e II. The conditions existing at t} con tinue during mode II. In other words, for t 1 ~ t ~ t2, ic = 0, iTl =10' tic = - V,s, tlT.-I. = V,s, tla =V" iD = a as shown. Note that during this mode, only main SCR T l is conducting. Mode Ill. When main thyristor Tl is to be t urned off, auxiliary thyristor TA is triggered at the desir ed instant t 2• With the turning on of TA at t!!, capacitor voltage (- V,I) appears across Tl, it is therefore r everse biased and turned off. As the capacitor voltage does the required job of commutating the main thyristor Tl, it is called voltage com mu tated chopper. CUrrent iT1
[ArC 7.6]
Choppers
r
10
T1
.'J=
o
•
C
,
0
TA
t t,
V,
'-0
I,
381
I
e
Vo
a
'~'
• I.
,
10
(a) Mode I, 0 < t < tl 0
+
0
TI
cf =
.,. I
lot
e
'0
~D~
'~u
,
FE -
TA
V,
TI
,•
0
TA
v,
o
10
10
e
0
•,
FD
.• ~ u
-
t2 :5 t < t3 (¢) .M94e. IV, t3 :5 t < T Fig, 7.25. Different modes of volta ge-co'nUn uta-ted chopper.
(e) Mode Ill,
becomes zero at t 2. "After Tl is turned off, capacitor C and auxiliary SCR TA provide the path for load current 10 through VJI , C, TAand the load, see Fig. 7.25 (c). The load voltage is the sum of source voltage and the voltage across capacitor. Therefore, at instant t 2 , load voltage is Vo = V., + V, = 2V, and it decreases linearly as the voltage across capacitor decreases. During this mode , ve = VTI' because capacitor is directly connected across Tl through TA. Ai; the capacitor discharges through the load, ue and Un change from (- V,) to zero at (t 2 + tel. Load voltage Uo changes from 2V., at t2 to V., at (t 2 + tel. After (t 2 + tel, Uc and Un s~lilrt rising from zero towards V., whereas Vo starts falling towards zero. For mode III, t2 ~ t ·St3. Note that Vc and UTI change linearly· from (- V.,) at t2 to V., at t 3 , because load current 10 is assumed constant. Similarly Vo changes linearly from 2V, att 2 to zero at t 3 . Note also that ic =- 10 and iTA =10 during mode III. Mode IV. For this mode, t3::; t < T. At t3, Uc = Un =V" Uo = 0, ic or iTA becomes zero. As ic (or tends to go negative soon after t 3 , thyristor TA is turned off naturally at t z . .As capacitor is slightly overcharged at t 3 , fr eewheeling diode FD gets forward biased. The load current after t3 fre ewheels through the load and FD, see Fig. 7.'2.5. (d ). Note that during fr eewheeling period fr om t3 to T, UTA is slightly negative as C is somewhat overcharged. Durin g this mode, ic= 0, in!:::: 0, i{d=10' UTI = V." vc= Vs+tJ.V, UTA =-tJ.V, vo= 0, iTA = 0. iTA )
At t = T , the main thyristor Tl is triggered and the cycle as described fr om t = 0 to t = T r epeats. Voltage commutated chopper is simple, it has th erefore been used extensively. It, however, suffers from the following dis advantages. (i) A starting circu it is required. - In the re lation i = C( d u/dtl, if i lS const.:l.nt, C is air!!ady constan t, then du/dr
mUH
be cons tan t.
[Act. 7.6]
382
Power Electronics
Load voltage at once rises to 2V, at the instant commutation of main SCR is initiated. Freewheeling diode is therefore subjected to twice the supply voltage. (iii) It can't work at no load. It is because at no load, capacitor would not get charged from - V, to V" when auxilIary SCR is triggered for commutating the main SCR. (ii)
Design Considerations. The values of commutating components C and L can be obtained as under :
Commutating Capacitor C. Its value depends upon the t-.lrn-off time tc of the main thyristor Tl. During the time te. capacitor voltage changes linearly from (- V.) to zero, mode III, Fig. 7.26. It is known that .
lc
du
=C dt
,For a constant load current 10 • the above relation can be written as
V, 1 =C- or C tc
te' 10 C=-V" .
The commutation circuit turn-off time te must be greater than the tliyristor turn-off time tq . Let te = tq + t:.t.
C = .:...(t.'-+....... l>t.:...)_. I""
... (7.36)
V,
Commutating inductor L. It can be designed from a consideration of the oscillatory current established when main thyristor TI is turned-on . The current ie• when TI is triggered, flows through the ringing circuit formed by C, TI, L ; D and is given by
The peak capacitor current
1ep =
V,
mOL
_ = V, .."
•
/c L
This current flows through Tl when it is turned-on. As Tl handles load current as well as 1ep' peak capacitor current should not be too large. It is usual to take lep less than, or equal to, load current l a, i.e . lep S 10 or
L
~ (;;
or V,
J
C
~ S 10 • j7.371
The second method of designing the value of L is from a consideration of the circuit turn-off time for auxiliary thyristor TA.)t is seen from time-variation of UTA in Fig. 7 .26 that turn-off ti me for T.4. is tel and it is given by
. B ut
Choppers
[Mt. 1.61
•
•
i fd
\ .
!
I
, ..
+~I 'T'!.
I.
-1-
~/I ~ I ~J . ' 1/I
,1
t
~,
I. TI
t
•
t
• •
t :
•
t
•
t
t
•
• •
., PO off
Tl on
and 0 on
Fi g . i .26. Current and voltage wave forms for voltage ·commutated chop p~ r.
383
384
Power Electronics
[Act. 7.6]
....
~!l~LC .
t
"
,
2
.
L~(2~IJ ~
or
...(7.38) ... (7.39)
Peak. current through Tl is given by
~
iTlP 10 +
v,.--Jf
... (7.40)
Peak voltage across Tl and TA is uTlp = uTAp = ± V, . CV, Peak current through TA is 'TAp = - t- =10
... (7. 41)
.. .(7.42)
,
Peak voltage across freewheeling diode, u{dp =
Peak diode current,
... (7.43 )
2 VI
iDp=V,
--Jf
... (7.44)
=peak capacitor current. Important features of voltage.commutated chopper. (i) Fig. 7.26 reveals that load voltage is V, from t = 0 to t2 and it varies from 2V, at t2 to zero at t 3' Therefore averag~ load voltage Vo is given by
Vo ~ During the interval is 2 V"
(t3 -
V" t, + 2V, (t, - t,)(1I2) T
.
V,
~ T [T•• + (t, - t,)1
t2), the voltage across C changes from - V, to
Io=C
10
(t, - t,)
.::1
Tor/ = Ton +
i.e . total change
2 CV, or t 3 -t2 = --
2V,
2V,] V, , Vo= TLTol1+4C =rTon where
V I'
2V T C = effective on period o
... (7.45) ... (7. 45a)
Eq. (7.45) shows that effective on period of this chopper is more than the period TOri and is load dependent. Greater the load, less is the effective on period. As average load voltage Vo is dependent on 10 , drooping load characteristics are obtained for this chopper. (ii) Charge on capacitor must be reversed. from + V, to - V, when Tl is turned on. Therefore
m in imum on.period for this chopper is
Minimum duty cycle, Minimum load voltage,
... (7 .4Sa)
VO.mrl = °
m ll .
V~ +
= V~ (a m": -
2V, . 2tc 2T 2fte) = V~ [rtf -J LC + 2ft,1
(..
I, [Art. 7.6]
Choppers
VOmit = f · V, [t l + 2te1
385
.. ,(7.46b )
Maximum on period is (T - 2te). This gives maximum value of duty cycle as T-' 2t
,.. (7,47 )
T <=(1-2(1<)
Q==
Maximum load or output voltage VOnu is given by VO.nu:
=anu . V, +
2V.. · 2te 2.T
=V, (a/7U + 2ft,)
Substituting the value of am,r gives
V,,= = V, [1 - 2(1, + 2(1,! = V, (iii) If main thyristor T1 fails to turn off when TA is triggered, C i.s completely discharged. The commutation, therefore, cannot be resumed in the next cycle and therefore control lost must be r egained by turning off Tl by interrupting the supply. (iv) The circuit cannot be operated at no load.
Example 7.16. A uoltage-commutated chopper feeds power t(f~battery-powered electric car. The battery voltage is 60 V. starting current is 60 A and thyristor turn-off time is 20 JlSec . Calculate the values of the comn:'utating capacitor C and the commutating inductor L. Solution. Let circuit turn·off time te = tq + OJ = 20 + 20 = 40 )is for reli able turn·off of Tl and TA. In Eqs. (7.36) and (7.37), 10 is the maximum load current that the commutation circuitry must be able to commuta~ .
I, f
... From Eq. (7.36), • From Eq. (7.37),
C = 40 x 6~; 10-
6
=40 ~
L ~ 60 60 )' x 40 X 10- 6 = 40 ~H
(
~
1 -6 =16.21 . 40 x 10 A!J per Eq. (7.37), value of L should be equal to, or mor e than 40 ~H , but Eq, (7.39) gives L = 16.21 ~ . Low value of L increases the peak . value of capacitor current, see Eq. (7.44). So higher T1 value of L = 40 ~H should be chosen. + vs' C Example 7.17. A voltage commuta ted chopper !',.. F+ T is shown in Fig. 7.27. With t112 ~ain thyristor Tl conducting, the capacitor C is charged to a source Vs :200V RdOA ... ,~ voltage V, with the polarities as marked. 'u~u From Eq. (7.39),
L
=(2 x 40 x 10-6)' X n
,,.
v,
r;
Now, after the auxiliary thyristor TA is turned on, t112 main SCR remains reverse biased for 100 mi croseconds . Calcula te the value of the commutating component C.
Fig. 7.27. Pertaining to Example 7.17.
Find also the uallU of commutating component L in case maximum permissible current through t112 main SCR is (a) 2.5 times the load current and (b) 1.5 times the peak diode current.
Power Electronics
[Art. 7.6]
386
Solution. When auxiliary SCR TA is turned on, capacitor voltage V, at once appears across main thyristor TI and it is therefore turned off. After it, the load current completes its path through source V" C, TA and the load. The voltage equation for this circuit is iR+
61
idt=V. or
The above equation can also be written as
(RP+~ )q=v,
where
P=~I
Solution of this equation has two components: steady state and transient. Particular integral gives its steady state solution whereas complementary function provides its transient solution.
Pl. Its particular integral is obtained from ~ q = V, qPl. =CV,.
C.F. Its complementary function is obtained from (RP + ~)q = O. ..
qC.F. =M
t
where p is obtained from Rp + ~ = O. This gives p = -
~R
qCP. -_A.-tiRe - tlRC ( qt)=QP1. +qC.F. = CV,+ A e
Therefore,
At t = 0, q =- c v,. Here minus sign is used before C V. ; it is because charging current leaves the positive terminal of C and enters its negative terminal. In other words; C, charged to VJ" now discharges, so negative sign is used before CV,. --CV,=CVJ'+A or A=-2CV,
This gives
q(t)
=CV, -
2CV, e- tlRC
= CV, [1- 2i'tlRC)[
But q.(t) = C,ut (t) = C. un (t), because ut = un. u, (I)
..
= uTdl) =V, [
1 - 2 e- UlRC) ]
.. . (7.48 )
When un reduces to zero after 100 llSec, thyristor T1 is turned off. Thus from Eq. (7.48 ), un =0 =V, 11- z.,-U IRC») lOO x l0'"
or
e (a) Here load cu rrent,
IOC
1
= 2 or C= 14.427!1F V, 10= -
R
Maximum permissible current through the main SeR, from Eq. (7.40) is . V, _/c V, _/c _V, tnp =R'+ V.! ~\fL =2.51f or VJ' -\II = 1.0 R or
4 ., 4 2 L = 9" CR- = 9" (14.42 7)(10) = 641.2 ~H.
Choppers (b )
[Art. 7.6]
387
Peak diode current, from Eq. (7.44), is
iDP=V,~ _Ic . V, _Ic 1.5 V' Y I = In,. '" R + V, 'VI .!V - Ic= V,
or
2
,y t
R
L= C:' = 14.427 x ~O- ' x 100 = 360.68 ~H.
or
Example 7.18. A uoltage·commutated chopper has the following parameters : V. = 220 V, load circuit parameters = 0.5
n. 2 mH, 40 V
Commutation circuit parameters:
L = 20 Iili, and C = 50 J1F Too = 800 ~ec, T = 2000~. For a constant load current of 80 A ; compute the following :
.'. : --:-
, .
(a) Effectiue on period. (b) Peak currents through main thyristor TI and auxiliary thyristor TA
(ej Tum ,offtimes for TI and TA (d) Total commutation interual.
(e) Capacitor uoltage 150 ~ a{teF TA is triggere~. (fJ Time needed to recharge the capacitor to uoltage V,.
Solution. (a) From Eq. (7.45), effe.ctive-on period =TM + 2;" C=800 x 10- '+ 2 (b )
From Eq. (7.40),
~~20 x 50x 10- '
= 1075 ~s.
in,. = 80 + 220'[[= 427.85 A
Since load current is given as constant at 80 A, peak current through TA is 80 A. (c) From Eq. (7.36), turn-off time for Tl is _CV _., _ 50 x 10-, x 220 _ 137 5 tc - I 80 . j.1s,
,
Turn-off time for thyristor TA, from Eq. (7.38), is
,
t"
;r
=.i"LC =~ "20x 50 x 10- ' =49 . 673~
(d)
Total commutation interval
= 2 t, = 2 x 137.5 = 275 ~s (e) Capacitor voltage before TA is triggered is equal to (- V,) , After TA is triggered, capacitor
begins to charge from (- V,). TPerefore, capacitor voltage, after TA is triggered , is given by
1,· t v, =-C
-v'
388
[Act. 7.6]
Power Electronics
where t =time measured fr om the instant TA is triggered v = so x 150 x 10- ' _ 220 = 20 V e 50 x 10 6 (f) Time needed to recharge the capacitor from (- VI) to V, is given by (Total change in voltage) (Capacitance) Load curren t = [V, - (- V,n c = 2 V,· C = 2 x 220 x 50 = 275 10 10 SO ~s.
Example 7.19. An impulse-commutated chopper feeds inductive load requiring a constant current of 260 A The source voltage is 220 V de and the chopping frequency is 400 Hz. Turn-off time for main thrystor is 18 ~. Peak current through main thyristor is limited to 1.8 times the constant load current. Taking a factor of safety 2 for the main thyrisior; calculate the ualues of (a) commutating components C and L and (b) the minimum and maximum output voltage. Solution. (a) Load current 10 = 260 A, V. =220 V For a factor of safety 2, the commutation circuit turn-off time for main thyristor t, = 2 x I S =36 ~s.
From Eq. (7.36),
~ = 36 x 1~;~ x 260
42.545
~F
Peak current through main thyristor, from Eq. (7.40), is 1.SIo = 10 + :.
v,,ff or
220,ff = O.S x 10
L= ( 0.S220 x 260 )' X42.545=47: ~96~H.
(b) From Eq. (7.46a) , minimum value of duty cycle is
"m" = 'f ~LC =. x 400 ~r.4"2"54"5C:x-:-4;;7".5"9'"6 x 10- 6 = 0.0565 Minimum value of output vo~tage, from Eq. (7.46b ), is V O.mn ~V, (am n + 2ft~) = 220 (0.0565 + 2 x 400 x 36 x 10- ') = l S.766 V Maximum value of output voltage is V o.=: = 220 V. 7.6.2. Current-Commuta t e d Chopper
The power-circuit diagram for current-commutated ch opper is shown in Fig. 7.28. In this di ag ram, Tl i s the main thyristor. The other 01 co mp onents , nam ely, auxiliary thyrist or TA, capacitor C, inductor L , diodes Dl and D2 constitute +~commutation circuitry. FD is the freewheeling diode 10 In Tl L+ and R~ is the charging resistor. 02 +~l ,d Simplifying assumptions for the chopper are as 1 1: TA c follows: FO , V, c. '0 •00 (il Load current is constant. t, V" u'--- lfd Cii ) SeRs and diodes are ideal switches. R L (ii i) Charging resistor R ~ is so large that it can ? be treated as open circuit during the commutation interval. Fig. 7.28. Current-commutated chopper.
Fll
:e "e. ,
Choppers
{A" . 7.6]
389
Currents ie' in, i{d and io are treated as positive when these are in the arrow directions marked. Similarly, voltages Ve, vTI, vTA and Vo are taken as positive with the polarities as marked in Fig. 7.28. Like vo1tage~commutated chopper, energy for current~commutation comes from the energy stored in a capacitor. Therefore, first of all, capacitor C is charged to a voltage V, so that energy for commutation process is available. Capacitor C in Fig. 7.28 is charged through source V$ ' capacitor C and the charging resistor Re to a voltage V" After this, main thyristor TI is fired at t = a so that load voltage Vo = V, and load current io = 10 as shown in Fig. 7.30 up to t = t l . With the turning on of TI, commutation circuitry remains inactive. Initiation of commutation process begins with the turning-on of thyristor TA The commutation process for its easy grasp is divided into various modes as follows: Mode I. At time t =t 10 auxiliary thyristor TA is triggered to commutate main thyristor T1. V With the turning-on of TA, ~ oscillatory current ie = ~ sin root [Eq. (7.33)1 is set up in the circuit consisting of C, TA and L as shown in F.i g : 7.29 (a). For the time interval (t2 - t 1), i, and Vc vary sinusoidally through half cycle, Fig. 7.30. During this mode when Vc is zero, ie is maximum though negative. At t 2 , as ie tends to reverse In the auxiliary thyristor TA,· it gets naturally commutated. At t 2• Vc = - V .. as shown in Fig. 7.30 ; i.e. in Fig. 7.28, lower plate is positive and upper plate is negative. During this mode TI remains uneffected, therefore load current and load voltage remain 10 and V.. respectively. Mode 11 . As TA is turned off at t 2• oscillatory current ic begins to flow through C, L , D2 and T1 as shown in Fig. 7.29 (b). Note that after t 2 , ie would flow through T1 and not t hrough Dl. It is because DI is reverse biased by a small voltage drop across conducting thyristor Tl. So after t" i, would pass through Tl and not through Dl. In thyristor TI, ie is in opposition to load current io so that iT! = IrJ - ie' Note that iT! is in the fOIVIard direction through Tl. At t 3 , ie rises to 10 so that iTI = 0 ; as a result main SCR T1 is turned off at ta . Since the oscillatory current through T1 turns it off, it is called current-commutated chopper. During this mode, load voltage remains V, through T1. For this mode , t2 < t < t~. Mode III. As T1 is turned off at t a, ic becomes more than 10 , After t3 , ic supplies load current 10 and t he excess current iDl = ic -10 is conducted through diode DI as shown in Fig. 7.29 (c) and 7.30. The. voltage drop in Dl due to (i, - 10 ) keeps T1 reverse biased for (t4 - t 3) = tc ; this is shown in the waveform for uTl. At t-l ' in case ve exceeds V.I' FD comes into conduction, oth erwise mode IV would follow. During m od:- III, when ie is at its peak value of l ,p
(= :;, }u, =O. Aft« this peak, capacitor voltage ,everses and at t"
upper plate is positive
a nd-lower plate is negative in Fig. 7.28. Mode IV. At t", ie reduces to 10 , as a r esult i 01 =0 and diode D1 is therefor e turned off. After t-l' a constant current equal to 10 flows through source V.!" , C, L, D2 and load and therefore capacitor C is charged linearly to so urce voltage V, at t 5, Fig. 7.29 (d) , So during the tim e (ts - t-l)' it =10 ,
390
Power Electronics
[Art. 7.6] D1
•
,,~
=C3J
r
TA
•
,
i,
C
'0 , 0
OUr"
Y,
+
D2r
TA
+
T1
. 1,
T1
+
i T1 2 (0-l e I,
Y,
0
'0
L At t2
-
R,
-
At t1
1-
(a) Mode I. t1 < t < '2
+
I,
r "
+
Y,
'.~.
t,
'0
, • 0
C~ Y,
0
'.,..
~o--
i,
rj'+
,..,T"..~
-
T1
+
ic .
~=Io
1:1,
.
+
Y,
---JA~
" ~ =C
D2"
i,
~ .~O~
lo=lc+i , ;
tId
FD
-
R,
•
,• 0
'" 0
R, -
<: t3
Dl
~II
L
R, tej Mode III, t3
Dl
10
---JA~
Yei =c
Mode II. t2 < t
(b)
I,
TI i,
•0
L
R,
10
0
U
-
t < t" Cd ) Mode [V, 14 < t < ts (e) Mode V. ts < t < 16 Fig. 7.29. Various m ode~ of current·commutated chopper.
<:
As 01 is turned off at t". tin = tlTA = tic ; this is shown as ab in Fig. 7.30 for tic. tin and tiT..... Now the load voltage 1I0 = V, - Lie = V, - voltage ab at t 4 • This is also marked in the waveform for Llo at t 4. At t s , Lie = Vs' therefore load voltage Llo =V, - Vs ,: 0 at ts. During the interval (ts - t 4 ). Lie incre ases linearly, therefore load voltage Llo decreases to zero linearly during this interval. Mode V. At ts. capacitor C is actually overcharged to a voltage somewhat more than source vo ltage V s' Th erefore, FD gets forward biased and starts to conduct load current 10 at ts. Load voltage Uo is reduced to zero at ts as stated in mode IV. As ie is not zero at t s, the capacitor C is s till connected to load through source V" C, Land D2 ; as a consequence C is overcharged by the transfer of energy from L to C, At t 6. ic =0 and Vc becomes more than source voltage V" Du ring (ts - ls), ie + i fd = 10: With the build up of ifd • ie decays and finally at ts. ie = 0 a.ru:l i{d =10, Commutation process is completed at ts. Total turn-off tim e, or commuta tion in terval, is (fa - t 1)· Att ul v,. is shown as equal taX)' . As D2 is turned off'at ls. v~
=VT.4 = vo ltage xy a t is, Fig. 7.30.
From ts onwa rds , io freewh eels through FD. As ic is zero and D2 is open circuited ; C now discharges through Rc for th e fr eewheelin g inter vaJ of t he chopper, After ls, un r emai ns constant at Vs' because V.f r each es Tl terminals through FD. At t = T, the main SCH. T1 is again triggered nnd the cycl e repe ats.
[Ar t. 7.6]
Choppers
igl~IIII..._ _ _ _ _ _ _ _ _ _ _-!-ICjij]IL-_ ig.j
391
__ "
m
i
..
I
I
..
~1~--~'~-----------T , -----,
. 110
r---r--~~T-~~-~r----t Tolc! turn·off time:
i'd!L__--+__.;-I-c-!-f!-+'---l<~---';i'--+I,- -- -01 I,
i
v,
,o 1
1:;0
TI
on
.I
! 1
y. I
I,
TA
on
i ::;;::,a : I II i I ,S1--j--lly"l b
t2 t] TA n
off off 02 01 on on
t4 ts t, 01 FO
T FO off
off on 02 off
Tl orr
Fig. 7.30. Current and voltage waveforms for current-commutated chopper.
This chopper, developed by Hitachi Electric Co., Japan, is widely used in traction cars. The merits of this chopper are as under : (i) Commutation is reliable so long as the load current is less than the peak commutating current Iep. . . tii) Capacitor is always charged with the correct polarity.
392
Power Electronics
(Art. 7.6]
(iii) Auxiliary thyristor TA is naturally commutilted as its commutating current passes through zero value in the ringing circuit formed by L and C. Design considerations. The value of the commutating components Land C should be so calculated that a reliable commutation is realized for this chopper. The conditions governing the design of L ·and C are as under: (i) Tlie .pe·a k commut~ting current I tp must be more than the maximum possible load current Io . · Thi~ is essential for reliable commutation of main SCR. From Eq. (7.33), the oscillating cur~nt in the commutation circuit is given by
it = V,
~ sin wot =I
tp
sin Wo t
As per the design requirement,
1,p=V.~>1o or
V,
~=xIo
... (7.49 )
where x is greater than 1 and it varies from 1.4 to 3, i.e. 1.4 < x < 3.10 is the maximum possible load current that the commutating circuit has to handle.
%=~
10
. (ii) -Circuit turn-off time tc must be greater than thyristor turn-off time for the main SCR. That is tt = tq + 6/. . It is seen from the current waveform it in Fig. 7.30 that tt = t" - t3
"'oI,=n-26,
or
l cp sin9 t =Io
Also
61=sln.
-1
or
(1 . I 0)=sm
-1
(1)-;
tp
: . Circuit turn-off time for main SCR, 1
t,=-(n-26, ) 000
~[n-2sin-'U:]
t,=
... (7 .50 a)
H(750 b)
The above relation r eveals that as load current 10 increases, turn-off time of main thyristor d~\:reases. Thus for ensuring necess ary turn-off time te, a certain value of ratio (Io/ Icp) must be
maintained. From above,
t, = In - 2 sin- , (lIx)I.JLC rc =
or
Substitution of this value of rc V,
L
in
t, (n - 2 sin-' (lIx)]..fL
Eq. '(7.49) gives
t, :--..,,--,-'--;---. =xIo In - 2 sin-' (l / x )]
.. (7 .51)
[ArL 7.6J
Choppers
V, , to
L~
or
%
, I, [It - 2 sin- (l / x)J
393 ... (7.52)
•
-i:~ 1. [n - 2 sin-' (lI%)J ..fC vL te
From Eq. (7.51),
Substituting this value of {L in the Eq. (7.49) gives V
x I, ~-' [n - 2 sin-' (l / x») C I,
C~=:-:_.::x..:.I''-'.elf-'--,---c
or
... (7.53)
V,[n - 2 8in ' (l / x))
Total commutation interval. The total tum-off time, or the commutation interval, is Fig. 7.30. It can be expressed as a sum of the following components :
(t6 - t 1).
~-I,~~-~+~-~+~-~+~-~
The above four time-components of (ts - t 1) can be obtained as under : (t 2 - t 1) : During (t 2 - tl). waveform of current ie , oscillating at frequen cy
COo. completes one
negative half cycle of1t radians, Fig. 7.30. (t2 - t 1) = time period of half-cycle of oscillating current ~ 2'- ~ rrIiE. "'0 (t 4
-
...(7.54)
t 2 ): For (t. - t 2 ), sine current waveform of ie is examined. At t 2 • ie ~ 0 and at t 4 • ie
attains a valu e 10 after passing through its peak, Fig. 7.30. Angle covered by ie from t2 to t. is equal to (1t - 9 1) radians and as 000 is the angular frequency for i e • (t 4 - t 2 ) is given by (I. - I,)
n-9,
~-- ~
"'0
,~
(n - 9,) ,LC
... (7.55)
(t5 - t.) : The time (ts - t.) can be obtained from voltage considerations across C at t4 and at
Voltage across C at t. = ab = lie at t•. .As this voltage at the instant t. is (90 - 91) ' away from zero crossing of the lie sine wave, ab = V.f sin (90 - 91), Voltage across C at ts =V, :. Increase in voltage across C during (ts - t.) We know that i
=
V, sin (90 - 91)
C ~~. As 10 is constant, I, ~C
--,-Yo:'-"si.::n-:-(9",0,---=-9,,,,)
-,V-,-,
(15 - I,) ~ CV,
or
=V, -
(/5 - I.)
1 - sin (90 - 9,) 1 - cos 9, I, ~ CV, I,
,.. (7.56)
During (ts - t s), current ie is assumed to be 10 cos wot . .As (t6 - t s) is equal to one quarter cycle (lt/ 2 rad.) of a sine wave, (ts - t 5):
1n
n..".
(/ - 1-) = - -~ 6 : l 2 £ ! J0
2
vL C
... (7.57)
394
Power Electronics
[Art. 7.6J
Addition ofEqs. (7.54) to (7.57) gives (ta - t l ) = total commutation interval
5n
= [2
).~
- 61
vLC + CV,
1-cos31 10
,
.
5. ).~ sin 61/2 = 2-61 vLC+2CV, 10 (
... (7.580) ... (7.58b)
Turn-off times. For main SCR, tum-off time from Eq. (7.51) is
t, - t, = t, = (It - 2 61dLC
= I. -
2 sin- 1 (l/%)J ~LC.
Tum-off time for auxiliary thyri$tor, from Eq. (7.55), is t, - t, = t,l = (. - 61dLC =.[. - -!lin-,
.! C~ = .! Lfoo 2
'
V,
or
2
=10--Jf ... (7.59)
The values of Icp and Vep are also the peak ratings of the thyristors Tl and TA. The value of charging resistor Rc is usually taken such that the periodic time T ~ 3CRe. Example 7.20. (a) For a current commutated chopper, peak commutating current is twice the maximum possible load current. The source uoltage is 230 V dc and 17UJln SCR turn-off time is 30 }l sec. For a maximum load current of 200 A. calculate (i) the values of the commutciting inductor and capacitor, (ii) maximum capacitor uoltage and (iii) the pfak commutating current, (b) Repeat part (a), in case peak commutating current is thrice the maximum possible load current. Compare the results obtained in parts (a) and (b). Solution. (a ) He re x = 2, tq = 30 }lsec
Takin g
te =tq + 6 t. !::.t = 30 )J.sec, te =(30 + 30) ).lsec =60 )lsec
=
6
(il From Eq. (7. 52),
L
From E q. (7 .53),
C = 2 x 200 x 60 x ;0-' 230 In - 25in (1))
230 x 60 x 102 x 200 In - 2 sin- 1 (1))
= 16.473 ~H
=49.822 ~F
\
[Art. 7.6]
Choppers
395
(ii) From Eq. (7.59), peak capacitor voltage is
V'P = 230 + 200 ...jr.!-;:-~:-;:;;;;~;--= 345 voits. (iii) Peak commutating current, I,p =xlo= 2 x 200 =400 A. (b) (i)
Here
x= 3,
L=
230 x 60 x 10- ' = 9.342 ~ 3 x 200 [n - 2 sin-I (~)l
C = 3 x 200 x 60 x 10-' = 63.577 ~ 230 [x - 2 sin- I (~)l (ii ) Peak capacitor voltage, V,p = 230 + 200
...j 63.577 9.342 = 306.67 V. . - _. ..
(iii) Peak commutating current,
~
.~
I,p= 3 x 200 = 600 A. It is seen from above that with the increase of x, L is reduced while C is increased, pe* capacitor voltage is reduced but peak commutating current is increased .
Example 7.21. A current commutated chopper is fed from a de source of 230 V. It s commutating components are L = 20 j.1H and C = 50 )1F'. If load current of 200 A is assumed constant during the commutation process, then compute the following : (a) Thrn -off time of main thyristor. (b)
Total commutaticn interIJal.
(c) Thrn. -off time of auxiliary ~Jiyristor. Solution. (a) Peak commutating current from Eq. (7.49) is
_Ic
I,p = v. 'J
t
_IsO
= 230 '\120 .= 363.66 A
!s..
= = 363.66 = 1 8183 xlo 200 '
Turn-off time of main SCR, from Eq. (7.51), is
t, = In - 2 sin- I (1I1.8183)1 " 20 x 50 x 10- 12 - 62.52 ~sec (b)
.-I(
"-I(~)lei - sm
e 1 -- sm
200 363 .66 )-333650 .
Total commutation interval, from Eq. (7.58), is ( 52" -
33,;~~ x ~) " 1000 x 10- 6 + 50 x 10- 6 x 230 1 - CO~;;.365°
= 229.95 x 10- 6 + 9.477 x 10- 6 = 239.427 ~ sec. (e) Turn-off tim e of auxiliary thyristor, from Eq. (7.55), is ("
_a3:;~~ X" )"1000 x 10- 6= 80.931 ~sec.
396
Power Electronics
[Art. 7.6]
7.S.3. Load-Commutated Chopper A load -commutated cnopper is shown in Fig. 7.31. It consists of four thyristors Tl- T4 and one commutating capacitor C. The thyristors Tl, T2 act together 8S one pair and thyristors T3, T4 act together as the second pair for conducting the load current alternately. When Tl, T2 are conducting, these act as main thyristors and T3, T4 and C as the commutating components. Likewise, with the conduction afT3, T4 j these become main thyristors and TI, 1'2 and C as the commutating components. FD is the freewheeling diode across the load. T1 ~r"
T3 ~r"
i,
+ V,
c= ~ VC ,r" H
i,
,
...
T2
-
lfd
FD
I+~,
I,
j~ ~
Fig. 7.31. Load commutated chopper.
Initially, the capacitor is charged to a voltage VJ with upper 'p late negative and lower plate positive as shown in Fig. 7.31. Assumptions made in current-commutated chopper also apply here. The working oftrus chopper can be explained in various modes.as under :
Mode I . With the capacitor C charged with lower plate positive, the load commutated chopper is ready for operation. When thyristor pairTl, T2 is triggered at t = 0, circuit consisting of V." Tl, C, T2 and load shows that load voltage at once shoots to Vo =V., + Vc = 2 V,. Load current now flows from source to load as shown in Fig. 7.32 (0). The capacitor C is charged. linearly by constant load current 10 from Vs at t = a to (- Vs) "at
t,_ When the capacitor voltage becomes (- V.), the load voltage falls from 2 V. to
°
at t I , Fig. 7.33. At t =0, when Tl, T2 are turned on, T3, T4 are reverse biased by capacitor voltage, i.e. at t = 0, un = un = - V,. At t l , un = ur4 = V" i.e. T3, T-4 are forward biased at t l .
Vo
= V, - V.,
=
Mode II. At t l , capacitor C is slightly overcharged, as a result freewheeling diode gets forward biased and load current is transferred from Tl, T2 to FD. From tl onwards, load current freewheels through FD, Fig. 7.32 (b). During (t2 - tl)' ve =- V" Vo = 0, ie = a, i{d = 10• in = iT2 = 0, un = un = V, and UTI = UTI = - !lV., as capacitor is overcharged by a small voltage 6V,. Mode III. At t 2, thyristor pair T3, T4 is tri ggered, load voltage at once becomes Uo =V, + ue =2 VI. Thyristor pair Tl, T2 is reverse biased by Vel this pair is therefor e turned off at t 2. The load current, now flowing through VI ' T4, C, T3 and load charges capacitor linearly from (- V,) at t2 to V, at t 3 , Fig. 7.32 (c). Load voltage accordingly falls from 2 VI at t2 to zero at t 3. During (t3 - t 2); ic = - 10, in = iT"' = 10 but uTI = uT2 = - V, at t2 and VI at t3 ; i.e. thyristor pair Tl , T2 gets forward biased at t 3. Fig. 7.33. At t 3 , capacitor C is somewhat overcharged, FD gets forward biased and therefore after t 3, load current freewheels through FD and load. This is not shown in Fig. 7. 32. 'W hen Tl , T2 are t u rned on at t .. , mode I repe ats .
\
[Art. 1.6]
Choppe rs
..; ~:.;~., TI
391 ~'. ;
T3
C
+ " T2
v,
. I,
(a ) Mode I, 0 < t < i}
(e) Mode III, t2 < t < t 3 Fig. 7.32. Different operating modes of load-comm'UtOteacho,"per.
Design of commutating capacitance. For constant load current 10 , capacitor voltage changes from - V, to V, in time T on. i. e. total change in voltage is 2 V, in time TOIl' Fig. 7.33. I o=C
2 V.
r ,.
c= 10 . T on
or
... (7.60)
2 V. Output voltage. Substitution of T on =
...(7.61 1
2V · C ;
o
. . in Eq. (7.61) gives average output voltage as 2CV 2 · y'! , C · f Vo =V .,. - -' =--,;'~--'-
,
10
10
... (7.62 )
Minimum chopping period. T min= T on :. Maximum chopping frequency. fma:: = -
1
T min
From Eq. (7.60),
1 Ton
-= -
To . - 1 C=_ 2 V, f mlll
I t is seen from the waveform of un . un. un . and uT-4 in Fig. 7.33 that circuit turn-off time fo r each thyristor is 1 j 2V. C V. t~ = 2" T:m = 2" C = 1;:
To
Tot al com mutati on interv al
2 CV = TrM = - ]- '. o
...(7.64 )
.'
...'
•
398
Power Electronics
[ArL 7.6]
. '"",
A load-commutated chopper has the following merits '9Jld demerits. Merits : (i) It is capable of commutating any amount of load cun-ent. No commutating inductor is required that is normally costly. bulky and noisy. (iii) As it can work at high frequencies in the order of kHz, filtering requirements are minimal. (ii)
Demerits : (i) Peak load voltage is equal to twice the supply voltage. This peak can however be reduced by ftltering. . (ii) For high-power applications, efficiency may become low because of higher Bwitching losses at high operating frequencies . (ii i) Freewheeling diode is subjected to twice the supply voltage. (iu) The commutating capacitor has to carry fu ll load current at a frequency of half the chopping frequency. (u) One pair of SCRs should be turned on only when the other pair is commutated. This can be done by sensing the capacitor current that is alternating.
lql . tg2~
,,!~--a--t---------Ir----¥nm~----------~'I !
i,3. I
~_-+___~mL-~ 1 ---1r--+!---~nmL-_"
i'lt:=t==tI=:!!I'LIIt=:tl,=tl==tl=---.1 v'ls~ Vi Li! I><:J ' I I./ . 1
V'~
2V'bJ 2V'N t
i---T~
i'l!. I. l"l : - I i.d
i Tl
~ Ton~
i";T< I
I
-t Tl ,T2 t:
I
I!·
I I ! ; -lot
I
:::J
I p,
I
I
I
P.
I
I
Il l.
W--I T3'''r __ -j II. __ I
~ FD
.1
I
•1 ,L
I
.1
I
.1
t=
•1
I
• t
!/j
v
•t
Fig. 7.33. Voltage and current waveforms for a lo::ad·co mmutated chopper.
,
Choppers
399
[Art. 7.7]
..
7.7. MULTIPHASE CHOPPERS
A multi phase chopper is one that consists of twti'or more choppers connected in parallel. The two-chopper configuration shown in Fig. 7.34 is called a two-pha~e chopper. Similarly, three choppers connected in parallel will constitute a 3-phase chopper. A multiphase chopper may be operated in two modes, uiz. in-phase operation mode and the CHI L phase-shifted operation mode, In the in-phase QOO ~ L~ operation mode, all the parallel connected chopper s ! are on and off at the same instant. In the 10 . • CH~ L + phase-shifted operation mode, different choppers are L___ J I~ OQ ~ I on and off at different instants of time. V, i0 In the two-phase chopper configuration shown in FOI F02 Fig. 7.34, inductance L in series with each chopper is assumed to be sufficiently large in that each chopper . h ch operates independent of each other. Let the load Fig. 7.34. Two-p ase opper. current be 10 and ripple free. For a duty cycle of a = 30%, Fig. 7.35 (a ) shows the in-phase operation of this chopper when both the choppers are on and off at the same instant. The input
..
,
---l1
"
~~b
n
,
H
11:~
lFt :
210~
Q ,
,
. l..-i.T
• t
~T ------'
n
,
: ~T-L
i
I
,
'M'It1 n ::
D D
n " !
"!_ i,- D-l~ : n
,
i
Too
,,
ri
h, '
;~h
,
Ten .
.'
i ,
,
1-1.--4
(0)
• t
' :i : : ', i I ! !
10
• t
:
h n
(b )
Fig. 7.35. Input current waveforms for duty cycle a'" 0.30 for (a ) in-phase operation and (b) phase-shifted operation.
:6
o
i'h DO C ., 1
[
0
I-T~
i'l'-.--'..0 ---,_10-,--0---,-__•
~1r----
.h ' , I,
t
::
I
'
I
U-°+[]-i:-f---LO ---L-,O -'--~t 2~eh ~ H :: n n n n t h
IQt ~ LJ LJ LJ LJ LJ L • t
.. !
, , -l
"T
(0)
, ,
(d)
Fig. 7.35. Current waveform for phase-sh ifted operation for (e ) a = 0.50 and (d) (X '" 0.60.
.t
400
Power Electronics
[Art. 7.7]
current i. obtained by the addition of i l and i2 • is. seen to be doubled as shown. The in-phase ''. operation of multiphase chopper is equivalent to a single--chopper operation. Fig. 7.35 (b ) shows the phase shifted operation for IX =30%. Chopper CHI is on for 0.3 T from t =O. Chopper CH2 is made on such that input cw:rent obtained from i l + i2 is periodic in nature . A comparison of Figs. 7.35 (a) and (b) reveals that for phase-shifted operation, the frequency of input current is doubled and its ripple current amplitude (proportional to Imtu. - I mill ) is halved as compared to the inphase-operation of chopper. In the in-phase operating mode, Fig. 7.35 (a) shows that frequency of harmonics in the input cUrrent is equal to the switching frequency (= 1/7) of each chopper. But in phase-shifted operating mode, Fig. 7.35 (b) shows that frequency of harmonics in the input current is twice the switching frequency ( = l i T) of each chopper. As the frequency of harmonics in the input current is twice the
switching frequency, the size of filter is reduced in the phase-shifted chopper. This shows that phase-shifted operation of multiphase choppers is usually preferred. For a = 50%, the input supply current of phase-shifted operation is continuous and without any ripples, Fig. 7.35 (c). For a = 60%, the supply current is continuous but with a pedestal of half the load current, Fig. 7.35 (d) . A multiphase chopper is used where large load current is required.. The main advantage of this chopper over a single chopper is that its input current has reduced ripple amplitude and increased ripple frequency. As a consequence of it, size of filter for a multiphase chopper is reduced. The disadvantages of a multiphase chopper are (i) extra commutation circuits (ii) additional external inductors and (iii) complexity in the control logic. Example 7.22.A type-A chopper operating at 2 kHz from a 100 V dc source has a load time constant of 6·rns and load resistance of 10 n. Find the mean load current and the magnitude of current rippLe for a mean load uoltage of 50 V. Also, calculate the minimum and maximum ualues of load current. Solution. Lond time constant,
~:::;6 x 10- 3 S i Load resistance, R = 10 n
:. Loaci inductance,
L
Chopping period,
T=
=6 x 10- 3 X 10 = 60 mH 1 71 = 2000 x 1000 = 0.5 ms
Average, or mean, load voltage, Vo =a V, . : . Duty cycle,
Vo 50 a=-=-=05 V, 100 . Ton = 0.5 X 0.5 = 0.25 ms; T olf= 0.25 ms.
As chopping period T = 0.5 rns is much less t han the. load time constant = 6 ms, the current variation fr om minimum current 1m,.. =II to maximum current lmx. = 12, Fig. 7.15 (a ), must be taken as linear. Thus, during Ton period,
I
I
! I
or,
, 'I I
Choppers
[ArL 7.7]
401
... Magnitude of current ripple, Vo al=I,-I'= -L T' fT=
50 -3 ,x O.25 x l0 60 x 10-
= 0.2083 A. Also, average or mean value of load current, 11+12 Vo 10 =-2-=1/ = 50 =5A 10 An examination of Fig. 7.15 (a) shows that maximum value ofload current 12 is given by
1.,=. 1. +tJ.I=1 + VO • T =5+50 x O.25X1O-' =5.104A . o 2 0 2L off 2x60x l()I' 3 Minimum value of load current, !if I, =10 = 5 - 0.104 = 4.896 A.
2
Example 7.23. For the circuit slwwn in Fig. 7.36, show that
and
= Vo
[1_V,
= ~ - :;.
[1- ~:J
+ Vo R .2fL
I
[mi.
V o]
where f =operating frequency of chopper switch S . Assume L to be large enough to ensure linear growth and decay of the current through it and haue continuous current. Solution. The function of capacitor C across load resistanceR is to make the output voltage continuous. In this figure, LC is a filter and FD is a flywheel diode. Inductor stores energy during T o/\ and delivers it during TofT of a cycle. L
R
",
1Fig. 7.36. Pertaining to Example 7.23.
The solu tion up to Eq. (i) in Example 7.22 is the same for this example also. Therefore, Eq. (i) in Example 7.22 gives current ripple as
Power Electronics
[Art. 7.7]
402 .
T Oil
Average load voltage, Vo = a V.. = T V,
Vo Vo TM=VT=fV , ,
or
Toff= T- To. 11: .
Also.
~v.erage load current, 10 =
=7- ~ =7(1- ~:)
.;
It is seen from Fig. 7.15 (a) that
6.1 Vo Vo I_=Io+T=Jf+ 2L T ,,,
= Vo + Vo R 2f£ Similarly.
Imi. =
(1 _V,Yo)
..
~ -;;. (1- ~:)
Example 7.24. For type-A chopper feeding an RLE load, show that ma.;imum value ofrms current rating for the freewheeling diode, in case load current is ripple free, is given by 0.3849
V, (1 _ R
KJ V,
312
Solution. For chopper with RLE load, average load cUlTent 10 is
10=
Vo-E R
aV,-E =
R
Freewheeling-diode current flovis during the period Toff. Therefore, rm s value of freewheeling-diode current, when 10 is ripple freEt, is given by
Ifdr=~Tf
or
10= [T-:,·r [ a VR-E]
=i (aV,-E)('Il-a) = i [a -il-a V,-~I-a
E]
i["a' - a' . V, - , I ~ a
E]
Ifd, =
... (i)
This currentI/d.rw ili have m~urn value when
d I fd,
_1 [1 (2a - 3a') V, +1
da - R 2 or or
~CI?-a'
(2a - 3a') V, E "a'-a' =-~I- a (2a - 3a') -il-a E o'll- a = - V,
E ] _0 2 ~I-a -
Choppers
[Art. 7.7J
403
3cx-2=~ or cx=1(2+~) V, 3 V,
or
Substituting this value of ex in Eq. (i), we get maximum value of rms current rating I'd.rlll of freewheeling diode as under : [(drm
=
i [~ ( ~ V, -E][ ~ ~, 2+
)
1-
(2 +
r
=MH2\~E)V.-E][I- 2~,;.Er
V, +E -3E] [3 V, .3-2 V._E]'" l [2V, - 2E] [V, -E]'/' l . ~ [V _Ei'''. 1 3V, J ~3V,
=.!.[2 R
=
V,
3
=
R
3
R
=.!. ~ V, ff. R 3
=3~3
~3V.
3
'
[1 _Kr/2 V. 12
V R[I-V,Er .V R[ Er"
= 0.3849
1- V,
Example 7.25. Th€ speed of a separately excited dc motor is controlled below base speed by type-A chopper. The supply voltage is 220 V dc. The armature circuit has r" = 0.5.Q and L" = 10 mHo The motor constant is k = 0.1 Vl rpm. The motor drives a constant torque load requiring an auerage armature current 'of 30 A. On the assumption of continuous armature curreTLt, calculate (a) the range of speed control and (b) the range of duty cycle. Solution. For a motor, Vf = Vo =.EG +10 ro The minimum possible speed of dc motor is zero. This gives motor counter emf Eo =0
exV,=Vo=O+lor" a x 220 = 0 + 30 x 0.5 = 15 V
or
15 3 a=220=44
or
Maximum possible value of duty cycle is 1 a V_=E,,+lora 1 x 220 = K.N + 30 x 0.5 220 -15 N= 0.10 = 2050 rpm
or
.1
or Therefore (b)
(a)
range of speed control is 0 < N < 2050 rpm and
the range of duty cycle is
!
< ex < l.
Examp le 7.26. In a battery-powered dc dri ue scheme, a chopper·controlled motor rated at 72 V, 200 A and 2500 rpm is separately-excited at a fiux corresponding ta its full rating. The
[MC 7.7J
404
Power Electronics
current pulsation during acceleration is maintained between 180 A and 230 A. The motor resistance is 0.045 n. while the inductance is 7 mHoThe battery resistance is 0.065 n Neglecting the semiconductor losses, ddermine the chopping frequency and the duty cycle ratio when the speed is 1000 rpm. Draw diagrams to show tM. circuit arrangement and performance during one , UAS., 1988) chopping cycle. Solution. For a separately-excited de motor,
Vt = Vo = Eo + I" Ta 72 : KN + 200 x 0.045 72-9 63 K: 2500 : 2500 V / rpm
or
At 1000 rpm, counter emf of motor
63
.
Eo: K x 1000 ~ 2500 x 1000: 25.2 V From Eq. (7.12),
1=: V'liE
Here circuit time constant, Tp. =
I " .
or
or
(l_e-TMIT')+I~e-T..IT,
~ = o.o:~ ~~.~65 = 0.064 5, R = 0.11 n·
= 230= 72 - 25.2 [1_e- T_ /0 .064
0.11
I7lX
]+ 180 e-T..
IO.064
245.45 e- T",/O.064 = 195.45 Ton = 0.01458 s
From Eq. (7.13),
I~ = - ~ (1- .-T",T, )+ 1=.- T",T.
During the freewheeling period, L 7xl0- 3 _ To : R: 0.045 = 0.1056 s and R = 0.045 n 180=- 25.2 ( 1_e- T"'O. 155~+230e-T"'O. 15" 0.045 790 Torr: 0.15561n 740: 0.010174 s
-.
or Chopping period, Chopping frequency.
T: To. + Torr: 0.01458 + 0.010174: 0.024754 s 1 1 f: T: 0.024754: 40.4 Hz
Duty cycle ratio,
To. 0.01458 Cl =T: 0.024754: 0.588995
= 0.589 or 58.9% Example 7.27. Show by diagram with necessary expLanatio"n a battery-powered de drive scheme using a chopper controlled motor. Sketch the voltage and current waveforms. Define chopping frequency and duty" cycle ratio. In the drive scheme stated aboue, the maximum possible value of accelerating current is 425 A, the lower limit of the current pulsation is 180 A, length of ON period is 14 ms and that of OFF period is 11 ms, the time constant being 63.5 ms. Determine· the higher limit of current pulsation, the chopping frequency and the duty cycle ratio." [LAS., 19891
Choppers
[Art. 7.7J
Solution.
Maxim~
405
value of current, 1111% = 425 A
Lower limit of current pulsation = 180 A . :. Minimum. value of current, I",n = 425 -180 = 245 A 14 Here Ton = 14 ms, Totr= 11 ~s. T = 25 ms, Ta = 63.5 ms, a = 25 = 0.56
I
From' Eq, (7.12),
= V' -E ( 1 _e- T..1T• ) +1
nu .
R
V -E
425 = 'R
or
nan
(1- e- I4163 .5
· e- T..1T• ..
)+ 245 '
.
e - 14/ 63 .5
V -E 'R =1175.023A
or
For higher current pulsation, Ton is ke.pt 14 ms. For duty cycle ratio a less than 0.56, the current pulsation will be more. For a =0.5, the current pulsation is the highest, .So with a = 0.5, Ton = 14 rna, from Eq. (7.12), we get ~ 425 =1175.023 (1 - eor
14/83 . ~
+ Imn e-14/63.5
1= =239 · 9956" 240 A Higher liinit of current pulsation ' = 425 - 240 == 185 A Ton. 14 Chopping period, T = - = - = 28 rns a 0,5 1 1 Chopping frequency, f=-= = 35,714 Hz T 28x10 3 Duty cycle ratio, a=0,50
-
Example 7.28. Fig. 7.37 (a) shows a chopper circuit operating at 100 Hz with Kd = 0.5. The load. current at steady state is continuous but uaries between'3 amps and 10 amps. S ketch the waueshape of
(a) the load current iL (b)
the current if through tlu!. freewlu!.eling diode DF
(c) tM current ie throUgh tlu!. commutating capacitor.
(GATE, 1994; lA.S ., 1997]
Solution. The circuit of Fig. 7.37 (a ) is redrawn in Fig. 7.37 (b) where the positive directions for ie• Ve, iL • vL' iTHI and if are indicated: Source voltage V, charges commutating ~0 1
~
v,
...
4
,..;..., '
..
\,
~1
OF
~!\l
.,
TH7/-'
02
01
~ TH I
V,
~
OF il
vl
~ -I
(b )
(a)
Fig. 7. 37. Pertaining to Example 7.28.
--
- v,
l-
'f
02 L
\,
, 1' TH2
[Art. 7.7)
406
Power Electronics
capacitor C to voltage + V, tht'ot:gh D2, L , C, load parameters R . L and souree voltage V" The working of this chopper is divided into certain modes as under : Mode I : Main thyristor THI is turned on at t =0. load voltage tiL =V, and load current iL begins to build up from 3 A. Capacitor voltage remains dOlT.'lant. The waveforms for Uc . UL . iL and i THl (= ill arc shown in Fig. 7.39. The circuit operation during this mode is shown in Fig. 7.38 (a ).
~
.
il
.,..-'l
+
TH1
II
c~
Y,
L
- v,
~
v,
+1
-'
il (b)
At t2. U,=-Vs
ul=Vs
il l
Mode It 0 < t <: tl
(a )
TH2
C
iL"i. TH1
~:r'
I
-1
t,
THl
Y,
+1
I
i THl
Mode II, tl
'l
j
+
01
..
Y,
OF
At tJ.ic= i.L and i TH1 =O
(C)
Mode 1I1,
t2
~L
< t < ta
(d)
f
uL=O
~
v,
:
...
02
t
At t,; ic=O. u(=Vs
Mode IV, t3 < t < t ..
Fig. 7.38
Mode II : Au.xilhuy thYristor TH2 is turned on at tl to commutate main thyristor THl. With TH2 on, ic begins to flow through C, L, TH2 and THI. During this mode, i~Hl = iL + ie. At time t2 ; ic =0, Uc =- V s' uL = Vs- As ic tends to reverse soon after t 2• TH2 is turned off. Capacltor current during this mode is given by Eq. (7.33). i.e. V.
~ . sin 000 t.
Mode 111 : Al:1er TH2 is till'ned off at t 2 • reversed current i c now begins to flow through C, TH1. D2 and L . Note that during this mode, i THI =iL - i e . When ic rises and becomes equal to iL , 1 ;'.'-11 decays to zer o and thyristor THI is therefore turned off at time t J • This means that Sou!"ce voltage is disconnected from load RL at tao As iL tends to decay from 10 A. DF gets forward biased by L Eventually,
!L
~;
and load current begin s to freewhe el through R , L and DF after t3'
dec ~ys to 3 A at is. At t3 , capacitor voltage
ue is still n egative as shown in Fig. 7.39.
I
1
[Art. 7.7J
nm I I
., : !vsft. sin ~ot
....... "'.
407
•
•
I
L
I
I
•
I
n--llllr-lY.., Modes :: :, I ' , ' , I, l OA
I
I, TH2
on
,
,
,
t2 tJ
I,
ts . TH1: : on OF ott
TH2THI D1 02 pff off 02 OF
on on 0' on
oH
I
Fig. 7.39. Pertaining to chopper circuit of Example 7.28.
Mode IV: After THl is turned off at t3. ic rises through the oscillating circuit formed by Dl, D2, L and C. Finally. at t", ic = 0 and Ve = V" Soon after as if: tends to reverse, diodes DI, D2 get turned off. Capacitor C charged to proper pol9l'ity and voltage is now suitable for next repeat cycle initiated by triggering TH l on at t a" During the . interval (t5-t~. lo ad current freewheels through DF, +~>--r-_';"0l--' R and L and decays from 10 A at I, to 3 A at Is as shown in I
t,.
Fig';:':'Ple 7.29. In Ihe circuit shown.in Fig. 7.40, S is an ~Vi
ideal semiconductor switch, V" is the de source voltage and Vo is the load emf. Show that this circuit configuration can be made to,operate as a step-up or step-down. chopper.
}L Fig. 'l.40. Pel""...aining to txample 7.29.
~~v.
Power Electronics
408
Solution. In Fig. 7.40, when switch S is closed, current in circuit V.I' S, L rises from 11 to 1'2 during time T Ori as shown in Fig. 7.6 (d) . di
L dt or
1,-1,
=UL
arL. -T-=V.
L. L1I=V... Tf)(I
'"
.. .(i)
where 12-11 =M, current ripple.
When switch S is opened, current begins to flow in L, Vo and diode D, and the current falls from 12 to I I during time T otr. 12 -II L. r--=V,orL . M=V,. To(! o(!
...(ii)
From Eqs. (i) and (ii), we get L . M =V• . Ton = Vo ' Tolf. or ... {iii)
It is seen from Eq. Wi) that when '(:( < 0.5, circuit of Fig. 7.40 operates as a step-down chopper. In case a> 0.5, this circuit would operate as a step·up chopper.
PROBLEMS 7.1. (a) Discuss the main types of de choppers. Which of these is more commonly employed and why? Enumerate the applications of de choppers. (b) Describe the principle of de chopper opera,tioD. Derive an expression for its average output voltage. . (c) A 120 V battery supplies RL load through a chopper. A freewheeling diode is connected across RL load having R = 5 nand L =60 mHo Load current varies between 7A and 9A. Calculate time ratio Ton/Torr for this chopper. [Hint. (c) 10 =8 A, Cl = 1/3 etc.) .
[Ans. (c) O.~l
7.2. (0 ) What is time ratio control in dc choppers? Explain the use of TRC for controlling the output voltage in choppers. ( b) What is current limit control? How does it differ from TRC ? Which of these control strategies is preferred over the other and why? 7.3, (a) Describe the principle of step-up chopper. -Denve an expression for the average output voltage in terms of input voltage and duty cycle. State the assumptions made . (b ) A step-up chopper has output voltage of two to four times the input voltage. For a chopping frequency of 2000 Hz, determine the range of off-periods for the gate signal. (Ans. (b) 250 ~, to 125 !IS]
7.4. (n) Wh a t is .meant by step-up chopper? Explain its operation. Sketch the input voltage, input current, output voltage and output current waveforms. State the various assumption made. How can a step-up chopper be used for the regenerative braking of de motors? Discuss.
,
Choppers
[Prob.7)
409
opera'tiig'~om 230 V dc supply. Compute the average value of load voltage for a chopping "frequency of 2000 Hz. (b ) 287.5 V[
( b) A step-up chopper with a pulse-width.of100 IlS is
.
. [An•.
7.5. (a ) What is a dc chopper? Describe the working of type-B chopper. Does it operate as a step down or step-up chopper ? Explain. . \ (b) A d e battery is to be charged from a constant dc so~rce of 220 V. The dc battery 15 t o be charged from its internal emf of 90 V to 122 v. Th~ battery has mternal re!istance of 1 n . For a constant charging current of 10 A, comput!l~the range of duty cycle. . 1Ans. (b) 0.4545 to 0. 61 7.S. For type-A chopper , express the following variables in te s of V" R , 10 and duty cycle a In case load inductance causes the load current 10 to remain nstant at a value 10 = VoiR Here V.f is ~e source voltage. (a ) Average output voltage and current. (b ) Output current at the instant of commutation. (c) Average and nns values of freewheeling diode cuiTent. (d ) Rms value of the output voltage. (e) Average and rms values of thyristor current. ~ketch tbe time variations of gate signal ig , output voltage tlo, output current io, thyristor current iT and freewheeling diode current ifd' . [Ana. (a) a V" VoiR (b).YoIR (e) (1- a) 10• ..)(1- (1) 10 (d ) ..Ja V, (e) a 10• ..Ja 101
7.7. Draw the power circuit diagram for a type-A chopper. Show load voltage waveforms for (i) a =0.3 and (ii) a =0.8. For both tbese duty cycles, calculate (a ) the average and rms values of output voltag.e in terms of source voltage V... (b) the output power in case of resistive load Rand (e) the ripple factors. [Ans. (a ) (i) 0.3 V, . 0.5477 V, (ii ) 0.8 V" 0.8944 V,
(b ) 0.3 V;/R, 0.8 V;/R 7.8.
(a ) A chopper controls power given to an
("b )
(e)
1.5275, 0.5.1
R-L load. 'For T1Ta. =Q, derive an expression for
the value of duty cycle a below which the per unit value of minimum load current falls • below x per unit of V.,IR. A chopper bas the following data : T·1 0 00~,R.2n, L . 5mH Find the duty cycle a so that per ucit value of minimum load current does not fall below (i) 0.1 and (ii ) 0.3 of V~ /R.
[Hint. (a) Use Eq. (7.15) witb E. O. Here x .
~;~]
[An•.
(a )
a.
~ In [1 + x (e Q -
1)1 (b) 0.12, 0.344]
7.9. (a ) A chopper, fed from a 220-V dc source, is working a t a frequency of 50 Hz a nd is connected to a n R-L load of R =5 nand L =40 mH oDetermine the value of duty cycle at which the mi nimum load ctuTent will be (i ) 5 A (it) 10 A (iii) 20 A and (iv ) 30 A. (b) For tbe values of -a obtained in (a ), calculate t be corresponding values. of maximum currents and th e ripple factor s. [Ans. (a ) 0.328, 0.50582, 0.7222, 0.86 184 (b ) 26.823 A, 1.4313 ; 34.3996 A, 0.9884 ; 40.055 A, 0.6202 ; 42.3767 A. 0. 40041 7.10. (a) A de chopper fe eds power to an RLE load with R = 2 n , L '" 10 mH and E = 6 V. If this chopper is operating at a ch op ~ i n g fr eque ncy of 1 kHz a nd with duty cycle of 10% from a 220-V d e source, compute t he m~im u m and minimum curre nts ta ken by the load .
410
Power Electronics
[Prob. 7]
(b) A d e chopper is used to control the speed oCa separately excited dC ,motor . The de supply
voltage is 220 V, armature resistance ro = 0.2 n and motor constant Kg <> "" 0.08 VI rpm. The motor drives a constant torque load requiring an average armature current of 25 A. Determine (i) the range of speed control (ii) the range of duty cycle. Assume the motor current [I.A.S., 19901 to be continuous. IAns. (a) 9.016 A, 7.0367 A (b) Duty cycle : 1144 < a < 1; Speed: 0 < N < 268.5 rpm]
7.11. Sketch output voltage, output current, source current and thyristor current waveforms for type-C chopper for its operation in second quadrant. Indicate the conduction of various devices . For ripple free load current, discuss the operation of type-C chopper. ' 7.12. Describe the working of type-D chopper with appropriate waveforms to demonstr9.te its operation in first as well as fourth quadrants. Indicate clearly the range of duty cycle for which it operates in first and fourth quadrants. 7.13. Describe the working of type-E chopper with relevant circuit diagrams and its operation in all the four quadrants . Record in each quadrant, the brief account of its working as step-up or step-down chopper. 7.14. Write voltage equations governing the performance of type-A chopper duri~g TOIl and Toff periods for RLE type load. Hence obtain therefrom expressions for the maximum and minimum currents taken by the load on 'the assumption of continuous output current. 7.15 . A step-down chopper is fed from 230 V dc and its duty cycle is 0.5. Calculate rms value of output voltage for fundamental, second and third harmonic components. Hence. express tbe o~tput voltage as a function of Fourier series. . [Ans. 103.552 V, zero, -34 ..52 V. uo "" 115 + 146.423 sin cot - 48.81 sin 3 (LIt} 7 .16. For type-A chopper connected to RLE load, write the basic voltage equations and derive the expressions for the maximum and minimum values of load current in terms of source voltage V" R, E etc. Hence show that the expression for per unit ripple in the load current is given by (1_e-aTIT.) (1:"'e-(1-a)TIT.1 (l_e-!IT.)
where T"" chopping period. a = duty cycle and To. "" LIR 7.17. A type-A chopper feeds RLE load. For low value of Ton. limit of continue us conduction is reached when load current during Ton < t < T falls to zero. Derive an expression for this load current from basic voltage equations and hence obtain therefrom that the duty cycle a' at the limit of continuous conduction is given by
at:: TnT In [1 +.!. (e TIT• - 1)] V, where Vol =source voltage, Tn = LIR and T::: chopping period. 7.18. For type-A chopper feeding an RLE load, find an expression for the duty cycle for which the average current rating of freewheeling 'diode would be maximum . .A3sume constant load current:-
[ Ans.~ ( I + ~,}a > ~ ]
7.19. A battery with its terminal voltage of 2t)Q V is supplied 'With power from type-A chopper circuit. The output voltage of the chopper consists of rectangular pulses of2 ms duration in a n overall cycle time of 5 ms. Internal resistance of the battery is negligible. Calculate : (0) ripple factor ( 0 ) averag' and rms values of output voltage (c) nns value of the fundamental com ponent of output voltage (d ) ae ripple voltag' IAns. (a) 1.2247 (b ) 80 V 126.49 Y (e) 85.63 Y (d ) 97.98 V]
Choppers
[Prob.7J
411
7.20. A type-A chopper feed s power to RLE load with R '" 1.5 n, L '" 6 mH and E '" 44 V. Other data for this chopper is as under : Source voltage = 220 V dc, chopping frequency = 1 kHz, Output vol~age pulse duration =400 ~sec . Repeat parts (a ) to (i ) of Example 7.9. [Ans. (a ) (1' - 0.221, current is continuous Cb) 29.33 A Ce) 33.764 A, 24.975 ACe) 2.496 A, 0.7719 A, - 0.1716 A, if) 11.781 A, (g) 2591.82 W, 1290.52 W, 1300.5 W, Ch) 29.445 A Ci ) 29.445 AI . 7.21. A 500 kW dc series motor used in a high-speed train is controlled by a chopper circuit. The inductance of the armature and series field windings is augmented by an external inductance . The dc source voltage for the train is 1000 V. The duty cycle varies from 0.15 to 0.9. Find the range of total inductance (its maximum and minimum values) in the armature circuit in terms of chopping period in case the amplitude of armature current excursion is limited to 25 A. [Ans. 3.6 T to 10 T henriesl 7.22. Describe a voltage-commutated chopper with relevant current and voltage waveforms as a function of time. The chopper operation may be divided into certain well-defined modes. Enumerate the various simplifying assumptions made . Show that effective on-period for this chopper is load dependent. Find also the minimum permissible on-period in terms of comm':ltating parameters. 7.23. A voltage-commutated chopper delivers power to RLE load for which R ::: 0 and L = 8 mH oFor· a chopping frequency of 200 Hz and dc source voltage of 400 V, find the chopper duty cycle so as to limit the load current excursion to 40 A. [Ans. 0.8 or 0.21 7.24. (a) Derive expressions for computing the magnitude of commutating components C and L for
a voltage-commutated chopper. Relevant voltage and current waveforms may be drawn to assist these derivations. Discuss the considerations on which design value of these components depend. (b ) Describe the various important features of a voltage-commutated chopper, such as effective on period, minimum on-period etc. 7.25. Draw t he power circuit diagram for a current commutated chopper. Explain the working of this chopper by dividing its commutation-process interval into some well -defined modes. Show distinctly the total turn-off time, turn-off times for mtin and auxiliary thyristors in the relevant waveforms drawn . 7.26. Ca) Discuss the conditions governing the design of commutating components Land C for a current-commutated chopper and hence obtain expressions for these parameters in terms of source voltage , load current, circuit turn-off time etc. The current and voltage waveforms needed for obtaining the expressions of Land C may be sketched. Cb) A current-commuta ted chopper has the following data : Source voltage = 220 V dc ; Peak commut ating current = 1.8 times the load current; Main SCR tq = 20 ).Is ; Factor of safety =2 ; Load current = 180 A. Determine the values of the commutating inductor and capacitor, maximum capacitor vol tage a nd the peak commuta ting current. [Ans. 13.832 ~H,30.002 ).IF, 342.22 V, 324 AJ 7.27. The com m ut a ting compon ents for a current -com mutated chopper are C =40 ~F a nd L = 181lH. DC so urce voltage is 220 V and load current is constant a t a value of 180 A duri n&" the commu tation interval. For this chopper, calculate (a) circuit turn-off time for main thyristor, ( b ) circui t turn ·off time for auxiliary thyristor, and (e) total commuta tion interval. Derive th e expr essions used in parts of Ca), (b ) and (c ). [Ans. (a) 53.12 ).is (b) 68. 71 ).is (c ) 203.1i s usl
7.28. Ca ) Dis cuss lhe worki ng of a load-commut a ted choppe r wit h rel ev an l '1 olta ge a nd current wav efo rms. Show vol tage variation across each ptl!r of SCRs as a function of time .
,.
.'
412
Power Electronics
[Prob.7J
I I
Derive an expression from which the value of com mutating capacitor oethia chopper can be computed. (b ) A load-commutated chopper, fed from 230 V de sow"u, bas a constant load current of 50 A. For a duty cycle of 0.4 and ~ chopping frequency of 2 kHz, compute (i) the average output voltage, (ii) the value of commutating capacitahce, (iii ) circuit turn-off ~e for one thyristor pair and (ill) total commutation intervaL (Ans. (b) 92 V ; 21.739j.lF ; 100 IlS ; 200 ~J
I
7.29. What is a multipbase chopper? Bring out clearly. with appropriate waveforms, the difference between the in-phase operation and phase-shifted operation of a multiphase chopper. Hence show why phase-shifted operation is always preferred. Enumerate the merits and demerits of multiphase choppers . 7.S0. Describe a three-phase chopper with appropriate circuit diagram ~ ..Draw the input current waveforms for both in-phase and phase-shifted operations of this chopper on the assumption of ripple free output qurrent. Suitable values of duty cycle may be chosen to illustrate your answer. 7.31. (a) A chopper circuit drives an inductive load from a 200 V dc supply. Given the load resistance of 40 n, the average ftlad current of 30 A and the operating chopper frequency of 400 Hz, compute the ON and OFF periods of the chopper. . ' (b ) A separately-excited dc motor is supplied from a 60 V dc source through a fixed frequency chopper. The rated speed is 900 rpm and tbe rated current 30 A. Armature circuit resistance is 0.25 n. Find the duty cycle ~atio of the chopper at rated motor torque for a . speed of 300 rpm ignoring current palsations. (Ans. (a) 2.083 ma, 0.417 ms (b) 0.4iS7] .
is
7.32. (a ) A chopper circuit as shown in Fig. 7.41 (a) is inserted between a battery, V, .= 150 V and a load resistance R =10 n.·The circuit turn-off time for the main thyristor Tria 110 ~s and the maximum ' pem:lissible current through it is 30 A. Calculate. the values of the commutating components L and C. lIAS .• 1995] T1
c ' 1:. V,
'"
10
v,
.J'
TA
R
400V
"""" •~ . . '0 Cb)
Ca )
Fig. 7.41. Perta.inipg to Problem 7.32. (b)
fig. 7.41 (b) shows the circuit schematic of a chopper driven separately-excited dc motor. The single-pole double-throw switch operates with a s...."ttching period T of 1 ma. The duty ratio of the .switch (Ton/ 7) is 0.2. The motor may be assumed 10ss1es9, with an annature inductance of 10 mH oThe motor' draws an average current of 20 A at a constant back emf of 80 V, under steady state. (i) Sketch and label the voltage waveform Vo (t) of the chopper for one switching period (ii) (iii )
Sketch and label the mo.tor CU1'Tent i., (t ) for one switching period T Evaluate the peak-to-peak curren t rippl e of the motor.
LAn •. Ca ) L =1.587 mH, C =15.87 ~F
,
Choppers
413
Minimum current = 16.8 A, maximum current =23.2 A, peak·to·peak current = 6.4 Al 7.88. In Fig. 7.42. (a), the ideal switch S is switched on and off with a switching fr equency f = 10 kHz. The circuit is operated in steady state at the boundary of continuous and discon· tinuous conduction, so that the inductor current i is as shown in Fig. 7.42 (b ). Find (b) on·time Ton of the switch and (b ) the value of the peak current lp[GATE, 2002J (b )
sl _ 5
C
y
'I ~ 100~H ~~ [
500V
ON
lOFF
...+. . .. _._._:...
_ lp
,
0 ~>+'~-----4'--~~~-,~Ton---:-Tot!~
(a)
t
(b)
Fig. 7.42. Pertaining to Problem 7 . 33. [Hint. L . lp =- 100. TOil during Ton. etc. See Example 7.291
IAns. (a).8(! .33 ~s (b ) 83.33 Al
Chapter 8
Inverters ................•.......•.•........•....•............. . ........ .....................•............ ~
~
In this Chapter • • • • • • • • • • •
Single-Phase Voltage Source Inverters : Operoting Principle Fourier Analysis of Single-Phase Inverter Output Voltage Force-Commutated Thyristor Inverters Three Phase Bridge Inverters Voltage Control In Single-Phase Inverters Pulse-Width Modulated Inverters Reduction of Harmonics in the Inverter Output Voltage Current Source Inverters · Series Inverters stngle-Phase Parallel Inverter Good Inverter
............... ...•....• ...•.•...•...••...........•••.....•••.••..•...•. .............•...•.• ~
~
~
~
...
~
A device that converts de power into ac power at desired output voltage and frequency is called an inverter. Some industrial applications of inverters are fo r adjustable-speed ac drives, induction h eating, stand by air-craft power supplies, UPS (uninterruptible power supplies) for computers, hvdc transmission lines etc. Phase-controlled converters, when operat ed in the in verter mode, are called line-commutated inverters, Chapter 6 . But line-commutated inverters require at the output terminals an existing ac supply which is used for their commutation. This means that line-commutated inverters can't function as isolated ac voltage sources or as variable frequency generators with dc power at the input. Therefore, voltage level, frequency and waveform on the ·ac side of line-commutated inverters cannot be changed. On the other hand, force commutated inverters provide an independent ac output voltage of adjustable voltage and adjustable frequency and have therefore mu ch wider applications. In this chapter, force-commutated and load commutated invert ers are described . The dc power input to the inverter is obtained from an existing power supply network or from a rotating alternator through a rectifi·e r or a battery, fuel cell, photovoltaic array or magneto hydrodynamic (MHD) generator. The configuration of ac to dc converter and dc to ac inverter is called a dc-link converter. The rectification is carried out by standard diodes or thyrist or converter circuits discussed in Chapter 6. The inversion is performed by the metl-~ s discussed in this chapter. Inverters can be broadly classified into two types; voltage source inverters and current source inverters. A voltage-fed inverter (VFl), or voltage-source inverter (VSl), is one in which the dc source h small or negligible impedance. In othe words, a voltage source inverter has stiff de voltage source at its input terminals. A current-fed inverter (CFI) or current-source inverter (CSl) is fed with adjustable current from a dc sour ce of high impedance, i .e. from a stiff dc current source. In: a CSI fed with stiff current source, outpu t current waves are not affected by th e load.
:
415
[Art. 8.1]
Inverters
In VSIs using thyristors, some type of forced commutation is usually required; however, load commutation is possible only if the load is underdamped. In VSIs usin g GTOs, switching-off is achieved by applying a negative gate-current pulse. VSIs using transistors, like BJTs, PMOSFETs, IGBTs or SITs, can be turned off by the control of their base current. Switching-off of the devices with the help of their gate or base currents is called self-commutation .. So the self-commutated inverters using GTOs and transistors do not require additional commutation circuitry as needed in thyristor-based inverters. This reduces the complexity and cost of the self-commutated inverter circuits and at the same time, enhances the reliability of their operation. From the viewpoint of connections of semiconductor devices, inverters are classified as under: . 1. Bridge inverters
2. Series inverters 3. Parallel inverters The object of this chapter is to describe the operating principles of both single-phase and three-phase inverters and to present their elementary analysis. As before, switching devices are assumed to possess ideal characteristics. Since bridge inverters are more popular, more emph asis is given to their description. 8.1. SINGLE-PHASE VOLTAGE SOURCE INVERTERS: OPERATING PRINCIPLE In this section, operating principle of single-phase voltage source inverters is discussed. 8.1.1. Single-phase Bridge Inverters ' Single-phase bridge inverters are of two types, namely (i) single-phase half-bridge inverters and (ii) single-phase full-bridge inverters. Basic principles of operation of these two types are presented here. Power circuit diagrams of the two configurations of single-phase bridge inverter, as stated above, are shown in Fig. 8.1 (a) for half-bridge inverter and in Fig. 8.2 (a) for full-bridge inverter. In these diagrams, the circuitry for turning-on or turning-off of the thyristors is not shown for simplicity. The gating signals for the thyristors and the resulting output voltage' waveforms are shown in Figs. 8.1 (b) and 8.2 (b) for half-bridge and full-bridge inverters respectively. These voltage waveforms are drawn on the assumption that each thyristor conducts for the duration its gate pulse is present and is commutated as soon as this pulse is removed. In Figs. 8.1 (b) and 8.2 (b), ig1 - ig4 are gate signals applied respectively to thyristors T1-T4.
[
i"l~_..,
"
i92t,-__ .
..L_ _
is
Ys 2"
l
,,
01
Vo
Vs
~
,,
to
2"
02
(a)
Vs
-2
_
~J..~
-T
,
~¥ .1 (b)
Fig. 8.1. Single-phase half-bridge inverter:
.. f
+-_~!L,.'--_.. ,,, ,
- n---+- T2--:- Tl --:-- T2--i
0
T
_
,,
1"
lIs
,
.~
!,r2i
t
I.
416
[Art. 8.1]
Power
Ele~cs
~'
Single-phase half bridge inverter, as shown iIi Fig. 8.1 (a) , consists of tWo seRs, two diodes and three-Wire supply. It is seen from Fig. 8.1 (b) that for 0 < t ~ T12, thyristor T1 conducts and the load is subjected to a voltage V/2 due to the upper voltage source V/2 ~ At t = T 12, thyristor Tl is commutated and T2 is gated on. During the period T12 < t ~ T, thyristor T2 'conducts and the load is subjected to a voltage (- V/2) due to the lower voltage source Vs/2. It is seen from Fig. 8.1 (b) that loadvoliage is an alternating voltage waveform of amplitude V/2 and of frequency liT Hz. Frequency of the inverter output voltage can be changed by 'controlling T . The main drawback of half-bridge inverter is that it requires 3-wire dc supply. This difficulty can, however, be overcome by the use of a full-bridge inverter shown in Fig. 8.2 (a). ,It consists of four SeRs and four diodes. In this inverter, number of thyristors and diodes is '
twice of that in a half bridge inverter;'This, however, does not go against full inverter because
the amplitude of output voltage is doubled whereas output 'power is four times in this inverter '
, as compared to their corresponding values in the half-bridge inverter. This ,i s evident ITom
Figs. 8.~ (b) and 8.2 (b). ' '
"
iT'
.
[
~
L
~t
t
i g3,ig4
1
.I
Vo
roo
V's
l.. s T1
03
0
02
- Vs
6
T/ 2
*3T/2 ,T
2Ti'
Vs
~!1-+T3+T1 ' T2 T4 ' 12 +T3-j T4
(a)
(b) Fig. 8.2. Single-phase full-bridge inverter.
For full-bridge inverter, when TI, T2 conduct, load voltage is Vs and when T3. T4 conduct , load voltag,e is - Vs as shown in Fig. 8.2 (b) . ~requency of output voltage can be controlled by varying the periodic time T . ' I'n. Fig. 8.1 (a), thyristors TI, T2 are in series across the sourCe; in Fig. 8.2 (a) thyristors Tl, T4 or T3, T2 are also in series across the 'source. During inverter operation, it should be ensured that two SeRs in the same branch, such as TI, T2 in Fig. 8.1 (a ), ,do not conduct simult aneously as this would lead to, a direct short circuit of the source. '
For a resistive load, t wo SeRs in Fig. 8.1 (a) and four SeRs in Fig. 8.2 (a) would suffice, because load current io and load voltage Vo would always be in phase with each other. This, however, is not the case when the load is other than resistive. For such types ofloads, current io will not be in phase with voltage Va and diodes connected in antiparallel with thyristors will allow th e current to flow when the main thyristors are turned off. As the energy is fed back to the de source when these diodes con duct, these are called fee dback d iodes. In Fig. S.l (a ), DJ, D2 are feedback diodes and in Fig. 8.2 (a) , Dl, D2, D3, D4 are feedback diodes.
.. I
r
Inv,e rters
[Art. 8. 1]
41 7
8.1.2. Steady-state Analysis of Single-phase Inverter Figs. 8.1 '(b) and 8.2 (b) reveal that load voltage waveform does not depend on the nature of load. The load voltage is given by Va ' for half-bridge inverter, Vo = ~ ..... .. 0 < t «T12
Va
=- 2 ....... T 12 < t < T
and for full-bridge inverter,
Vo = VB
........ 0 < t < T 12
= - Va ....... T 12 < t
The load current is, however, dependent upon the nature of load. Let the load, in general, consist of RLC in series. The circuit model of single-phase half-bridge or full-bridge inverter is as shown in Fig. 8.3 (a). In this circuit, load current would finally settle down to steady state conditions and would vary periodically as shown in Figs. 8.3 (c) to (f). It ,i s seen from these waveforms that ' and
io =± 10 .......... at t = 0, T, 2T, 3T,..... .
io = ±lo .......... at t = T12, 3T12, 5T12, ......
The voltage equation for the circuit model' o(Fig. 8.3 (a) for l}~lf-bridge inverter and for 0< t < TI2 is given by ; ,-:. -..-,
Va
d
l'f'todt . +'Vel
' . io to + L dt + C
2" = R
...(8.1)
For full-bridge inverter, replace V,/2 by Va in Eq. (8.1). In this equation, Vel is the initial voltage across capacitor at t = O. For T 12 < t < T, or 0 < t' < T 12, the voltage equation for half-bridge inverter is
f
Va = R'to + L dt' di + C 1 ~'d' - 2" t + V c2
...(8.2)
and for a full-bridge inverter, replace ( - V/2) by (- Vs ) in Eq. (8.2). In this equation, Vc2 is the initial voltage across capacitor at t' = O. Differentiation of Eqs. (8.1) and (8.2) gives 2
d io R d io 1'. dt 2 + L dt + LC to =0 2
and
d io R d io 1. dt'2 + L dt' + LC to
=0
The solut.ion of these second order differential equations can be obtained by using initial conditions as specified above. Components constituting the load decide the nature of load current waveforms. F or a full inverter, the rectangular output v oltage waveform is shown in Fig. 8 .3 (b). For this inverter, v ariou s current waveforms for different load characteristics are cirawn in Fig. 8.3 (c) t o (f) . The nature of t h es e current waveforms.. is briefly di cu ssed in what follows: R load. For a resistive load R, load CUITe t wav eform io is iden tical with load voltage waveform Vo and diodes D1-D4 do not come into conduction, Fig. 8.3 (c). RL and RLC overdamped loads. The load current wav eforms for RL and RLC overriamped loads are shown in Figs. 8.3 (d) and (e) respectively. Befor e t = 0, thyristors T3,
418
Power Electronics
[Art. 8.1]
a ~
-------
.
--~ .
T12 I
to 10
I
I
i
2T
3T/2
T ( b)
ir---:"
I
o -10 ---_._ --
to
' - T1,T2 --L T3.T4+ T1 ,T2 (e)
10
.
R load
--L- T3,T4 I Components I '}conductlng r I RL Lo.ad ..'
O~~~~~~--~--~--~~~~+---------~t
-10
to 10 O~~--~~,~--r-~--~~~--~------~ I .
-10
,I,
,:
i03, i T3,T4 0,)T1,T2 i03,1 T3,T4 .!}' Components i 04 : I02' : 04' I co nd ucting i. i (e) I I . I
i.o
I
I
I
.
,
I
:RLC underdomped
I
01 02
T3 T4
:I i'
b3}Compon~nts 04 conductrng
(1)
Fig. 8.3. Load voltage and current waveforms for single-phase bridge inverter.
T4 are con ducting and load current io is flowing from B to A, i.e. in the revers~d direction, Fig. 8.2 (a) . This current is shown as - 10 at t = a in Figs. 8.3 Cd) and (e). After T3, T4 are turned off at t == 0, current io cannot change its direction immediat ely because of the nature of load. As a r esult, dic....es Dl, D2 start con du cting after t = 0 and allow io t o flow against the supply voltage V s, As soon as Dl, D2 begin to conduct, load is subjected to Vs as sh own. Though Tl, T2 are gated at t =0, these SeRs will not turn on as these are r everse biased by voltage drops across diodes D1 and D2. When load current through DI, D2 falls to zero, T1 and T2 become forward biased by source voltage V s' T l and T2 therefore get turned on as these are gated for a period T 12 sec. N ow load current io flows in the positive direction from A t o' B. At t = T 12 ; T I , T2 are turned off by forc ed commut ation and as load current cann ot rev erse immediately, diodes D3, D4 come int o onduction to allow the fl ow of current io after T/ 2.
I
419
[Art. 8.1]
Inverters
Thyristors T3, T4, though gated, will not turn on as these are reverse biased by the voltage drop in diodes D3, D4. When current in diodes D3, D4 drops t o zero; T3, T4 a!"e turned on as these are already gated. The conduction of various components of the full-bridge inverter is shown in Figs. 8.3 (d) and (e) . RLC underdamped load. The load current io for RLC underdamped load is shown in Fig. 8.3 (f). After t = 0 ; Tl, T2 are conducting the load current. As io through Tl, T2 reduces to zero at t l , these SCRs are turned off before T3, T4are gated. As Tl, T2 stop conducting, current through the load reverses and is now carried by diodes D 1, D2 as T3, T4 are not yet gated. The diodes Dl, D2 are connected in antiparallel to Tl, T2 ; the voltage drop in these diodes appears as a ,reverse bias across Tl, T2. If duration of this reverse bias is more than the SCR turn-off time t q, i.e. If (T12 - t l ) > tq ; Tl, T2 will get commutated naturally and thertfore no commutation circuitry will be needed. This method of commutation, knows as load commutation, is in fact used in high frequency inverters used for induction heating. In single-phase bridge inverters shown in Figs. 8.1 (a) and 8.2 (a), thyristors are shown as switching devices. Note that basic inverter operation is not dependent on the particular semiconductor device used. It means that if npn transistors (or GTOs, IGBTs etc) are used as switching devices in place of thyristors as shown in Fig. 8A~ normal inverter operation is , obtained. The operating principle of an inverter using transistors, Fig. 8.4, can be described merely replacing T (for thyristor) by TR (for transistor) in Figs. 8.1 (b), 8.2 (b) and 8.3 (c to f). '
01
03
01
+ Ys
02
i.o
Vo LOAD
04
D2
(b) (a) Fig. 8.4. Single-phase (a) half-bridge and (b) full-bridge inve.rters using transistors. Example 8.1. (a) A single-phase full bridge inverter is connected to an RL load. For a dc source voltage of Vs and 'output ,frequency f = 1/ T, obtain expressions for load current as a function of time for the first two half cycles of the output voltage. ,(b) Derive also the expressions for steady-state current for the first two half cyCles. (c) For R = 20 nand L = 0.1 H , Qbtain current expressions for parts (a) and (b) in case source voltage is 240 V de and fre quency of output voltage is 50 Hz. ' Solution. (a) For the first half cycle, Fig. 8.3Cb), i.e. for 0 < t < T 12, the voltage equation for RL load is
.
dio
Vs =Rlo +L dt
... (8.3)
Its Laplace transform, with zero initial conditions, is , V ' " _S =R 1(8) + Ls . 1(s) =1(8) [R + Lsl s Its time solution is, ...(8.4)
for
420
Power Electronics
{Art. 8.1]
This is the expression of current as a function of time for the first half cycle from the instant of switching in with io(t) = O'at t ~ O~ ' , At t
=T 12, current io(t) of Eq. (8.4) becomes the initial value for second half cycle. ' io(T12) = ~S(l
..
-e- ~)
... (8.5)
For second half cycle, time limit is from TI2 to T or 0 < f < TI2 where f voltage equation for RL load during second half cycle is ,' - Vs
. dio to + L df
=R
=t - T12. The ...(8.6)
Its Laplace transform, with initial current io (TI2) given by Eq. (8.5), is Vs - - s =1(s) [R +Ls] - io' (TI2)· L Vs L . io(T12) l(s) =- s(R + Ls) + -R""'::+:":"L-s---:"
or
V(
io(! ') =-
Its time solution is
;
=-
1 -e
- ~ t ,), - ~ t' L I + io (T 12)e L
zt 1 .;;.e-~t')
V(
V (
L+ ;
_RT)
RT) -
V ' V( ~ t' ~o(t ') =- ; + 2 - e 2Le L
R
or
-~t'
1-e 2L e L
... (8.7)
for
0< t' < T12. Eqs. (8.4) 'a nd (8.7) give the transient solution for load current for first and second half cycles respectively. (b) Under steady-state conditions, at t =0, io(O) =- 10 , Fig. 8.3 (d). Under this condition, Laplace transform of Eq. (8.3), is V ---!. =l(s) [R + Ls] + L 10 s Its time solution is At t == T 12, io (t)
io(t)
m
=
~(
_!i
1- e
L
t)
-
10 e
_!i
t
L
... (8.8)
=10 Fig. 8.3 (d), therefore from Eq. (8.8)
~0(TI2) = 1 = i(1- e-~)- Io. e-~ 0
RT
VS
or
1 - e-
10=1[ '
2L
RT
... (8.9)
1 + e- 2L Substituting t his value of 10 in E q. (8.8), gives .
£o(t)
V s(
=R
RT
R
--
R -II) Vs l - c 2L - - t 1- e - Rt e L
R
- 1+ e 2L
... (8.10)
Eq. (8.10), gives the steady-state solu .on durin g the first h alf cycle, i. e. for 0 < t < T 12. /
[Art. S.l}
Inverters
421
For second half cycle, at t = T12, io (TI2) =10, Fig. 8.3 (d). Under this initial condition, Laplace transform of Eq. (8.6), is
V
.
- ~ = 1(s) [R + Ls] - L 10
Its time solution is
io(t)
=-
=-
~ (1- e-~ t')+ 10 e-~
i( R s'
R
-L 1~ e
t'
t') + -Vs . 1 -
RT -e 2L
R
R7' . e
L t'
...(8.11)
R- I +e
2L
Eq. (8.11) gives the steady-state solution during the second half cycle, i.e. for 0 < t' < TI2 where t' =t-TI2. .
R
(c) Here L
1 1 RT = 200, T = 7 = 50 = 0.02 sec., 2L = 2.
Expression for transient current during the first half cycle from Eq. (8.4), is io(t) = 22~0(I_e-200t)= 12 (l_e- 200t } Expression for transient current for second half cycle, from Eq. (8.7), is . (t') - -240 - +240(2 - -e- 2) e- 200 t'
lO
20 20 ="':'12 + 22.376 e- 200 t' . 240 l-e- 2 From Eq. (8.9), 10 = 20 . _ 2 =9. 139 Amps. . l+e Steady-state current for the first half cycle, i.e. for 0 < .t < T 12, is obtained from Eq. (8.10) as io(t) = 12 (1 - e- 2OO t) _ 9.139 e- 200 t = 12 - 21.139 e- 200t for 0 < t' < TI2
For the second half cycle, steady state current from Eq. (8.11), io(t')
~~
=- 12 (1- e- 200 t') + 9.139 e- 200 (
' .'
=-12+21.13ge-200t' forO
q
R dt + C =Vs
or "
f
=Rio + ~ io dt .
... (8.12)
Its Laplace transfonn, with zero initial conditions, 's ~ Vs R [s Q(s) - q(O)] + C =-;Vs
or
Q (s)
Its tim e s olu tion is,
q(t)
1
=Ii . s(s + l iR e)
=CV
.
(1 _ e- tlRC )
Power Electronics
[Art. 8.. 1]
422
'Vc(t)
or
i
Also,
(t)
o
=qg) =Vs (1 - e- tlRC)
...(8.13)
dv (t) V =C _ c_ = dt .. . R
...(8.14)
e- tiRC
_5
Eqs. (B.13) and (B.14) give the transient solution for vc(t) across C and io(t) throughRC load during the first half cycle . . At t = T12, vc(t) ofEq.(B.13) becomes the initial value ofthe second half cycle,
..
vc(T12)
~ Vs [1 -e- ~c]
...(8.15)
Voltage equation for RC load for second half cycle is - Vs =Rio+ dq . q . R-+-=- V dt' C S
or
~ f io dt' ...(8.16)
Its Laplace transform, with initial voltagevc(T12) given in Eq. (8.15), is
.. ~ . Vs
R[s Q(s) - C· v c(TI2)] + C =- -. "
s .
.
or
Q(s)
=-
R· . C vc(T12)
C . Vs s(RCs + 1) +
.
s+ RC
Its time solution is
q(t')
=- C V,( l- e- t'IR
)
+ C ve(T12) e-t'IRC T
t'
v c( t') =- Vs ( 1 - e- RC) + ~s ( 1 -t~ -
or
t'
2RC )
e - RC
=- Vs + Vs (2 - e- 2RC Je- RC Vs ( = - - 2 - edt R Voltage and current solutions in the first helf cycle are given by Eqs. (8.13) and (8.14) and in the second half cycle by Eqs. (8.17) and (B.18).
Also
ioW)
=C d
vc(t')
Under steady state working, the waveform for voltage Vo across capacitor and load current io through RC are as shown in Fig. B.5. (b)
...(8.17)
T)e-
2RC
r
At t = 0, vc(O) = - Vo. Therefore, Laplace transform of Eq. (8.12) under this condition is R [8 Q(s) + CVO] +
or or
[ . 1]· . ..
I
F
wt
.wt·
Vs Q(s) Rs + C . = -; - CR Vo .
. Vs
...(B.1B)
wf
Qg) =:s
Q(s) = R
t'
RC
Fig. 8.5. Pertaining to Example 8.2.
1·
'LI)-( sls +RC)
8
CVo 1 + Re
J I
Inverters
[Art. 8.1]
=CVs (1- e- tlRC ) - C Vo e- ti Re Ve(t) = VB (1 - e- tIRC) - Vo e- tl RC or At t = T12, vc(TI2) = V o, Fig. B.5, therefore, from Eq. (8.19), Its time solution is
q(t)
T
T
...(8.1'9)
.'
~ Vo e- 2RC
ue(T12) = Vo =Va (1; e- 2RC ) 1- e- 2RC
or
423
.Vo = Va
... (8.20)
---T-
1 +e- 2RC T
1-e- 2iiC
Frem Eq. (8.19),
T
t .
e RC
... (8.21 )
1 + e- 2RC t
.
.
Fromahove equation, £o(t) = C
2 Vs
dUe (t) ~
=R
e- Rc
.
... (8.22)
T
1+e
.
2RC
.
.Eqs. (8.21) and (8.22) give the steady-state solution for vc(t) and load current ioCt) for the first half cycle, i.e. for 0 < t < T 12. . .' For second half cycle, at t =T 12, Uc(T12) =Vt). Under this condition, Laplace transform of Eq. (8.16) is ~ . V R[sQ(s)-CVo]+ C
=---;
.
or
.
CVo
1
Vs
Q(s)=-
R '~1 +
1
s s ,+ RC i
Its time solution is or
q(t') = -
Uc(t')
=-
S
+ RC
+ CVo e-t'I RC Vs ' (1- e-t'lRC) + Vo e-t'I RC
CVa (1- e-t'I
=_V .
s
)
(1 _ ~~) + V 1e-
'
s
e-
T 2RC
T
1 + e- 2RC
Also
. (t') = C
£0
e - ~~ . .
...(8.23)
.
d v (t') c dt t'
2VS
=-R
- RC
e
... (8.24)
T 1+ e- RC
Eqs . (8.23) and (8.24) give respectively the steady-state solution for capacitg£ voltage uc(t' ) an d load current io(t') for second half cycle, i.e. for 0 < t' < T 12. . (c)
1
1
Her eR=20n. C=501J,F. RC=1000, T=,=0.002
From Eqs. (8.13) and ( 8.14), the transient expressions for the first half cycle are ve(t) = 240 (1- e-lOOOt)
an d
io(t) = 12 e- 1000 t)
From Eqs . (8. 17) and (8.18), the transient expressions for the second h alf cycle are vc(t') = - 240 + 240 (2 - e- 2) e- lOOO t' =_ 240 + 512. 48 e- lOOOt'
.
424
Power Electronics
[Art. 8.2] iO(t')
an d From Eq. (8.20),
Va
=-
12 (2 - e- 2) e- 1000 t'
= 240
1 -e
l+e
=_ 22.376 e- 1000 t'
-1
. _ 1 = 110.91 volts.
For the first half cycle, steady state voltage and current, from Eqs. (8.21) and (8.22) are vc(t) = 240 (1 - e- 1000 t) _ 110.91 e- 1000 t = 240 _ 350.91 e- 1000 t . and
. 2 x 240 e-lOOOt -1 = 17.545e-lOOOt 20 l+e For the second half cycle, steady state voltage and current, from Eqs. (8.23) and (8 .24) are vc(t') = - 240 (1 - e -1000 t') + 110.91 e- 1000 t' ;::: _ 240 + 3qO.91 e- 1000!'
io(t)=
io(t') = - 17 .545 e- 2000 t'
and
Example 8.3. A single-phase bridge inverter delivers power to, a series connected RLC load with R =2 n and oiL = 10 .0. The periodic time ,T = 0.1 msec. What value of C should the load ·· have in order to obtain load commutation for the SCRs. The thyristor turn-off time is 10 ~sec. Take circuit turn off time as 1.5 tq.Assume that load current contains only fundamental component. . Solution. The value of C should be such thatRLC load is underdamped. Moreover, when load voltage passes through zero, the load current must pass through zero before the voltage wave, i.e. the load current must lead the load voltage by an angle 9 as shown in Fig. 8.6. Recall Vo Fund. comp oof voltage r( the phasor diagram for RLC series circuit. From . . Fund. compo of this phasor diagram, . .,1---...... . current XC-XL tan 9=--R-=
Now
or or
9 w
"",
wt
Here Xc> XL as the current is leading the voltage. Now (9 / w) must be at least equal to circuit turn-off time,i.e. 1.5 x 10 = 15 Ilsec. ..
.
,~
Fig. 8.6. Pertaining to Example 8.3.
.
- = 15 x 10- 6 sec
103
0.1 9 = 21t X 104 x 15 x 10- 6 = 0.9424778 rad = 54° . X -10 tan 540 = c ,. 2 f = - = 104 Hz
Xc = 12.752764 =
1
4
21t X 10 xC
C =1.248 IlF.
8.2. FOURIER ANALYSIS OF SINGLE';PHAS~ INVERTER OUTPUT VOLTAGE The output voltage va is shown in Fig. 8.1 (b) for a single-phase half-bridge inverter and in Fig. 8.2 (b) for a single-phase full-bridge inverter. These output or load voltage waveform s do n ot depen d on the n atu r e of load. Voltage waveshapes of Figs , 8. 1 (b) and 8.2 (b) can be r esolved int o Fourier series as under : Va = n
I"" _ 2V n1t
_5
=1, 3, b . .... ..
sin nwt volts
" .(8 .25)
[Art. 3.2]
Inverters
425
for single-phase half-bridge inverter and vo =
4mt Vs sin nwt volts
"'" ~ .
... (8.26)
n = 1, 3, 5, ...... for single-phase full-bridge inverter. Here n is the order of the harmonic andw = 2rif is the frequency of the output voltage in rad/s. . The load current io can, therefore, be expressed as
..
"'"
4 Vs Z sin (nwt - n) Amps
~
io =
n7t·
n = 1,3,5, ... ... where Zn = load impedance at frequency n.f
n
. ~[R'+Ln:-_n!f)' and phase angle n is
n = tan-
1
I
...(8.27)
r
... (8.28)
nwc] rad
.. :(8.29)
R .....
The output, or load, current at the instant of commutation is obtained from Eq. (8.27) by putting rot = 1t. Its value is . io = 10 at rot = 1t rad In case 10 > 0, forced commutation is essential. If10 < 0, no forced commutation is required and load commutation, as described for RLC underdamped load in Art. 8.1.2, can be relied upon. If 101 = rms value of the fundamental component ofload current, then the fundamental load power POI is given by = I~1 R = VOl 101 cos <1>1 = rms value of fundamental output voltage. POI
where
VOl
The fundamental output power POI does the useful work in most of the applications (e.g. electric motor drives). The output power associated with harmonic current does no useful work and is dissipated as heat leading to rise in load temperature. Examp le 8.4. A single-phase half-bridge inverter has load R = 2.0. and dc source voltage Vs
2:=115V (a) Sketch the waveforms for Vo, load current i 01 ' currents through thyristor 1 and diode I and voltage across thyristor TI . H arm onics other than fundamental com ponent are neglected. Indicate the devices that conduct d uring diffe rent intervals of one cycle. (b) Find the power delivered to load due to fundamental current. (c) Check whether forced co mutation is required.
Solution. (a) The fundamental component of output voltage, from VOl
. 2 Vs . = - - sm wt 7t
X
The rms value of t his voltage, V Ol = 2rc
~o = 103.552 V
q. (8 .25), is
426
[Art. 8.2]
Power Electronics
and the load current, 101==
~l == 103 552==51.7_76A
2
The fundamental frequency component of load current is i Ol == 51. 77 6 ,[2 sin rot t 2n
The waveforms for the various voltages and currents are shown in Fig. 8.7. For resistive load, diodes do not come into conduction, therefore iDl is zero. When Tl cond,:!cts, vTl == O. When T2 conducts, vTl =Vs as shoWn.
....
{
t
(b) Power delivered to load
== I~l R == (51. 776)2 x2 == 5361.5 watts
When ·Tl is conducting, power to load is
\' - delivered by upper source and when T2 is on,
-t
Fig. 8.7. Pertaining to Example 8.4.
lower source delivers power to load. \'8 =-·1 2 8
Power delivered by each source Here Is
= average value of fundamental component of source current over one 1 = 21t
flt.o'i2
1n
.
1 01
-
sm rot . d (rot)
cycle.
,[2101 =-1t
=,[2 x 51.776 =23.304A
1t
Power delivered by each source Power delivered by both the sources
=115 x 23.304 =2679.96 watts
=2 x 2679.96 == 5360 W
P ower delivered by both the sources is equal to that consumed by the load. (c) As the diodes do not conduct, forced commutation is essential.
Exa mple 8.5. For a single-phase full-bridge inverter, \'8 = 230 \' dc,T = 1 ms. The load
consists of RLC in series with R
= 1 0.,
wL
=6 n
and
:c
= 7 n.
(a) S ketch the waveforms for load voltage Vo> fundamental component of load current i Oll source CUrrent is and voltage across thyristor 1. Indicate the devices under conduction during differe nt intervals o.f one cycle. (b) F ind the. p ower delivered to load d ue to funda men tal component. (c) Check wh ether forced commutat ion is required or not. Take thyristor turn-off time as 100 !lS.
Solution. (a) The load voltage waveform in Fig. 8.8.
va and its fwndam ental component VOl are shown
427
[Art. 8.2]
Inverters
Rms value of load voltage, from Eq. (8.26), is 4Vs 4x230 VOl =7tT2 = 7t T2 = 207.1 V
Vo 230V
Rms value of current, .
.
101 .
,
o~'-"-;----~~'-~~'~----------~-'~
VOl
=Z
T/2 --_
- 1
"
T t
/
,/
VOl
= [R' + ( ooL = 2
2~7.1
[1 +(-1)2J .
. -1
~ )']'12 1/2
= 207.1 = 146.46 A
~
XL -::-Xc· ~1 ° R = tan (~1) = - 45
Or-~----~:~~------~~+-~t-
.,
. ,i 230V ---------_....:.!----+-------. 1)T1
The fundamental component of current a function of time is iOl = ~ 101 sin (rot.,...
iOl
as
. ~ tc
O ~------~'~~--------~~~
T/2
= ~ 2%:1 sin (rot +45°)
.1
-
-
I------T....,..
I
-----J
Fig. B.8. Pertaining to Example B.5.
= 207.1 sin (rot + 45°)
Load current iOl and source current is are plotted in Fig. 8.8 and the conducting components are also indicated. .
2
(b) Power delivered to load
=I~, R = (2~1 Jxl = 2i.445 kW
This must be equal to the power Ps delivered by the source.
where Is = average value of the fundamental component of source curren.t =! 7t
J1t0 ~ 101 sin. (rot + 45°) d (rot)
= 207.1 [_ cos (oot + 45°)] It =207.1 [2 cos 45°] = 93.23 A . 0
7t
P~
7t
=230 x 93.23 =21. 443 kW
Fig. 8.8 re~als th at UTI is negative for some time before T3, T4 from circuit turn-off time can be' obtained . (c)
ar~
t riggered. Thus
.
".
.or
7t
ro tc = '4 1 T
. tc ="4
.
. 2' =0.125 ms = 125 ~s
.A.s voltage drop in diodes Dl, D2 rever se biases TI, T2 for 125 ~ s , which is more th an the thyrist or turn-off tim e of 100 )ls, no forc ed commutation is required.
428
[Art. 8.2]
Power Electronics
Example 8.6. A single-phase full-bridge inverter is fed from a de source such that fundamental component of output voltage is 230 V. Find the rms value of thyristor and diode currents for the following loads: (b)R='2n, xL=sn, Xc=6n.
(a)R=2n
Solution. (a) Rms value of fundamental component of load current
230
101
=2= 115 A
Fundamental component of output voltage VOl is shown in Fig. 8.9 (a). For resistive load, load current waveform i Ol and thyristor current iTl handled by Tl are also shown in Fig. 8.9 (a). For R load, diode does not conduct, therefore diode current iD1is zero. From the waveform of in, rms value of thyristor current is IT!
" 2 ]"112 =[ 2n1 J"0lt (1m sin rot) . d (rot) " [ ]1/2
=2lqn
J:
(1- cos 2 rot)· d (rot)
=I; = 1l~ ..J2 = 81.33 A .
Rms value of diode current, (b)
IDl
. .':.
=0
Rms value ofload current, 101 = 2
230
2V2=81.317A [2 + (8 - 6) ]
Phase angle,
CPI
=tan- 1
XL-XC
R
=45°
For R = 2 n, XL = 8 n and Xc = 6 n, the fundamental component of load current lags the output voltage by 45°, this is shown in Fig. 8.9 (b). The thyristor-current waveform i Tl and Vo
Vs 0
\)0
--.f VOl ,
"
,
,,
•• /
Vs
~
,, ,,
-v~
,
.
,.,
1...
wt
'-
1101
0
, .-.-
Vol
,,.
""
,,
,
wt
" ....... '"
iol
0
,
,. ,
wt
wt 1
i Tl 0
wt
7r.
iD~t
I 2lf
I I
,, I
rr:
wt
I
2fT
wt
..
27T
7T
(a)
(b) Fig. 8.9. Pertaining to Example 8.6.
wt
[Art. 8.2]
Inverters
429
diode-current waveform i Dl are also shown in Fig. 8.9 (b). It is observed from this figure that rms value of thyristor current is
ITl =[2~ J:""
(1m
..
sin rot)' . d (rot)]", .
31t
=2~1<[lrot-Sin:rotl:
112
] =0.47S75 1m
=0.47675 ~..J2 X 81.317 =54.818 A . Rms value
of diode current,
IDI
=[i" S:/4
12
(1m
Bin rot)' . d
(rot)J
x...f2 x 81.317 = 17.328 A As the load current io does not change from positive to negative at an angle rot < 1t, no time = 0.15070251m = 0.1507025
is available for SCR to turil. {)ft; forced commutation is therefore essential. '- . Example 8.7. A single-phase full-bridge inverter has RLC load of R =4 n, L =35 mH and C = 155~. The de input voltage is 230 V and the output frequency is 50 Hz. (a) Find an expression for load current up to fifth harmonic. Also, calculate (b) rms value of fundamental load current~ (c) the power absorbed by load and the fundamental power, . (d) the rms and peak currents of each thyristor, (e) conduction time of thyristors and diodes if only fundamental component is considered. Solution. (a) From Eq. (8.26), an expression for output voltage is
4 Va . 4Va .
4Vs . ' vo = sm rot + -3 sm 3 cot + -5 sm 5 cot •
1t
1t
>• •
.7t
.' 1 · '3 1. 5 ] = 4 X7t230 [ sm rot + '3 smoot + 5' sm rot = 292.85 sin 314t + 97.62 sin (3 x 314t) + 58.57 sin (5)< 314t)
Load impedance at frequency n.fis
x ~ x10-' xn - 21< x501~'155 xn 1
Zn. =4 +j [ 2lt 50 35
=4+j(l1n-~)n Zl = "'4
2
+ (11- 20.54)2 = ~ 0 .345 n
~, = tan- 1
and
Z,= cjl3 =
an d Similarly,
11 - :0.54) = _ 67.25'
4'+(ll X3_
20 54
:i
] = 26.4S n
tan- 1 [ 33 - 20. 54/ 3] = 81.3
Zs = 51.05 nand cjl,5 =85.50
0
430
Power E lectronics
[Art. 8.3]
Load current; from Eq. (8.27), is given by .
Zo
292.85. ( t
= 10.345 sm
(t)
+
6'7 25°) . 97.62 . (3 t . . + 26.46 sm (t)
81 3°) -
..
. 58.57 . (5 + 51.05 sm
.
r.,+ _ UJl,
85.50) , ..
' .= 28.31 sin (314 t + 67.2'5°) + 3.689 sin (3x314t -
81.3°) . . +1.1473 sin (5 x 314t - 85.5°)
(b)
Rmsvalue of fund'amentalload current, . I . . I7LI 28.31 2002 A I Ol-~-~- . ~2
'12
,
(c) Peak load current . 2
2
2
1m = "28.31 + 3.689 + 1.1473 = 28.572
Rms load current
= 28;52 = 20.207 A
Load power
= (20.207)2
Fundamental load power, (d)
2 .
POI
Peak thyristor current
X
4 = 1633.3 W
= 1~1
X
R = (20.02)2 X 4 = 1603.2 W
= 1m
=:=
28.572 A
Rms value of thyristor current = (e)
A
..... ""
28.~72 = 14.286 A
Fundamental component of current is i OI = 28.31 sin (314t + 67.25°). .
.
This current leads the fundamental voltage component by 67.25°. This means that diode conducts for 67.25 0 and thyristor for 180 - '67.25 = 112.75° :. Conduction time for thyristors 112.75 x 1t . ;: 180 x 314 = 6.267 ms Conduction time for diodes
67.25 x 1t = 180x 314 = 3.738 ms,
In case SCR turn-off time is less than 3.738 ms, load commutation will occur and no forced commutation will be required under the assumption of no harmonics. . 8.3. FORCE·COMMUTATED THYRISTOR INVERTERS
For low-and medium-power applications, inverters using transistors, GTOs and IGBTs are becoming increasingly popular; However, for high-volt age and high~current applications, thyristors are more suitable. In voltage fed inverters, thyristors remain forward biased by the de supply voltage. This ent ails th e use of forced commutation for inverter circuits using thyristors. A3 stat ed earlier, forced commutation r equires a precharged capacitor of correct polarity to turn-off an already conducting t hyristor. A large variety offol'ced commutation circuits h ave been described in the technical1it erature. Here popularly used McMurray technique will be described leading to the
,
.
[Art. 8.3]
Invert ers
431
discussion of two types of force-commutated inverters, viz modified McMurray inverter and McMurray-Bedford inverter. These are now described in what follows: 8.3.1. Modified McMurray Half-bridge Inverter Load commutated voltage-source inverter has been discussed in the latter part of the Sect ion 8.1.2. It is shown there that for obtaining load commutation in VS!, the load circuit must be underdamped, i.e. capacitive reactance of the 'load must be more than its inductive reactanC8. The object of this section is to describe modified McMurray half-bridge inverter which is a current-commutated VS!. · Fig; 8.10 shows a single-phase modified McMurray half-bridge inverter. It consists of main thyristors Tl, T2 and main diodes DI, D2. The commutation circuit consists ·of auxiliary thyristors TAl, TA2, auxiliary diodes DAl, DA2; damping resistor R d , inductor L and capacitor C. Three-wire dc supply is required andac load is connected between terminals A and B as shown in Fig. 8.10. The function of capacitor is·to provide the energy required for commutating the main thyristors. Inductance Ll is for limiting dildt to a safe value in m ain and auxiliary thyristors. As an auxiliary thyristor is used for commutating the main thyristor, this inverter . is also known as auxiliary-commutated inverter.
I I
I
02
~
I
I I
...... _--------------------------- -
Vo
I I ______ ________.AI
+
LOAD
Fig. 8.10. Power circuit diagram for a single-phase modified McMurray half-bridge inverter.
In the ·original inverter circuit given by McMurray, elements DAl, DA2 and Rd were not present, hence the circuit of Fig. 8.10 is now commonly known as the modified McMurray inver ter. As three-wire dc supply is required, .a s in Fig. 8.1, the term "half-bri dge" is added t o this inverter of Fig. 8.10. . The following simplifying assumptions are made for this inverter: Load current r emains constant during the commutation interval. (ii) SCRs and diodes are ideal switches. (i ii) Inductor L and capacitor C are ideal in that they have no resist ance. (i)
The operation of this inverter of Fig. 8.10 may be subdivided into various modes as follows: Mode I : yristor T1 is conducting a constant load cU11'ent 10 , i.e . i Tl = 10, Capacitor C is already charged to a voltage Vs with right h and plat e positive because of the commutation of previously conducting thyrist or T2. In this mode, the equivalent circuit of this inverter is as shown in Fig. 8.11 (a). With Tl conducting, commutation circuit is passive in this mode.
432
[Ait. 8.3]
Power Electrouics
Mode II. Auxiliary thyristor TAl is triggered at t = 0 to turn offthe m'ain thyristor Tl. As TAl is fired, capacitor current ic starte building up through resonant circuit consisting of L, Tl, TAl and C. Voltage drop across Tl reverse biases Dl, current ic can therefo:re flow only through Tl and not through Dl. As 10 is constant, an increase in ic causes a corresponding decrease in iTI so that iTl =10 - ic (KCL at node B) . Waveforms ofi c' iTl> 10 and Vc are shown in Fig. 8.12 and equivalent circuit is as shown in Fig. 8.11 (b). Att l • ic rises to 10 and therefore i TI = O. As a result, main thyristor Tl is turned off at t l . Mode III. Mter t l , as resonant current ic exceeds 10, the excessive current ic - 10 = i Dl circulates through feedback diode Dl, Fig. 8.11 (c). The voltage drop across Dl reverse biases .Tl to bring it to forward blocking capability. When capacitor voltage Vc discharges to zero, resonant current ic rises to peak value lcp as shown in Fig. 8.12. After attaining l cp ' ic begins to decrease and in so doing, C begins to get charged in the reverse direction. At t2, ic falls to 10 , In case Vc is somewhat more than source voltage Vs at t 2 , diode D2 gets forward biased and starts conducting. In such a case, mode IV is absent; otherwise mode IV follows. Mode IV. Mter t 2, as ic tends to fall below 10, diode current i Dl becomes zero and Dl, therefore, stops con.ducting. Constant load current 10 continues flowing through V/2, TAl, C, L and load as shown in Fig. 8.11 (d). Load current charges capacitor C linearly with reverse polarity and at t3 , Vc is somewhat more than Vs. Mode V. At t 3 , as V:; becomes slightly more than Vs' an examination of Fig. 8.10 reveals that diode D2 gets forward biased and thus an alternate path for 10 is provided. Load current 10 is now shared by resonant circuit and D2. Current through D2 flows through lower source Vs/2, D2 and load. After t 3 , ic begins to decrease whereas iIJi starts building up so th.at the sum of ic and iD2 i3equal to 10 , i.e. ic + iD2 = 10 (KCL at node B), Fig. 8.11 (e). The supply voltage Vs. through D2, is n ow impressed across the resonant circuit. As current ic is falling from 10 to zero, the energy stored in L is transferred to C and as a consequence, capacitor is overcharged to a peak voltage V mat t 4 • . The main thytistor T2 is usually given the trigger pulse 1t --JLC seconds after thyristor TAl is fired, i.e. between the interval t3 and t 4. But T2 will not turn on because of the reverse bias applied to it by the voltage drop in D2.
Mode VI. At t 4, iD2 rises to 10 and at the same time ic falls to zero. As ic tends to reverse, TAl is turned off at t 4 • Now Vc > Vs , capacitor C therefore discharges through R d , DAl, source voltage Vs' D2, L an d C, Fig. 8.11 (f) . Note that current ic is now negative as it is flowing opposite to its positive direction. For a constant load current 10 , KCL at node B gives iD2 = ic + 10. During this mode, i D2 is more th~ 10 , Here the question arises: how can diode current iD2 in D2 be established and deliver load current 10 when sour ce voltage Vs opposes i D2 ? It is the overcharging of capacitor C to a voltage higher than Vo that causes the establishm ent of diode current i D2 • The cirC"Jit traced by ic is usually critically damped so that Vc gradually reduces to Vs' At ts. ic becomes zero, Vc = - Vs and iD2 = 10, This is the end of mode Viand also the end of commutation interval. The volt age drop across Rd and DAl applies a r everse bias across TAl and completes its commutation process.
[Art. S.3]
Inverters
I~
Vs
433
r 1~Io T1 ~ . ld~.
r
-L TA1b
12
I
I
Vs'
I
'"
oIlLOADI
I.
1.0 -
- "0 (b) Mode II,
Mode I, t < 0
. (a)
lO1=tc- I o
10
~LI~Al We ~c
~-----~I------~-'~II .
A't~
A T
~~B0_'....,
2
nfl.rlf'..
,
.A
10 ,
lc
0
+ -
+
tI l
ic < 10
Tl: b
01
L
rI ..
ic
B
10
10 . L.O.A
'- "0 (c) Mode
+
III, ic > la,
(d)
tl < t < t2
Mode IV, ic = la, t2 < t < t3
01
10
10 10
I
LOAD
:+
'10 -
'. (e) Mode V, ic < 10 and ic + iD2 = 10' t3 < t < t4
A
LOAD + Vo if) Mode VI, iD2 > 10' t4 < t < t5
B
~ ~02 ' LOAD + '10 (g) Mode VII, ic = 0, iD2 =i o, t5 < t < ts
Fig. 8.11. Operating modes of modified McMurray inverter of Fig. 8.10.
Mode vn, After t s, the circuit model is as shown in Fig. 8.11 (g). The decreasing load current io = iD2 becomes zer o at t s, Main thyristor T2 is already gated on during t4 ~ t3 interval, i.e. 7t ;/LC seconds after TAl is fired . But it will not get turned on at this moment because ofthe reverse bias applied to it by voltage drop in D2 due to current im . At t 6, iD2 = 0 and T2 is no longer reverse biased. Therefore, after t a, source voltage applies a forward bias across T2 and the trigger pulse already applied t o it turns it on. Load is now subjected to negative current as desir ed. Note that load wa already subjected to negative voltage through D2 at t3 at the commencement of its conduction.
[Art 8.3]
434
Power Electronics
After t 6 , capacitor charged to voltage - Vs (i.e. with the left plate positive) is ready for the ne;xt commutation process . The commutation process from T2 to D1 is identical to that described above.
Design of commutating circuit components. For modes II and III of the modified half-bridge McMurray inverter of Fig. 8.10, the circuit parameters that come into play are Capacitor current ic
,
, I ~--~--~----~~----~~---7~~~--~~t
I
_._.t--I
on
on
t6;
t2 tJ I
I
I
I
t
I
I
I
I
.:
ill ~ TIl!---ll --t- I . 01 02 TAl off on off .
i
T1
T2
off
~ -----t-'2J1-l}
OA 1
Modes
02
. off
off
DAi
T2
: tired on
on
Capacitor voltage I·
Vc
'
I t::O ~(1!...-tl)--~_ :
-Vs
000
I
..
, I
~ Totat
commutation interval - - -.. . . ;.
Fig. 8.12. Voltage and current waveforms for inverter of Fig. 8.10.
Land C only. Therefore the commutation, or capacitor, current ic during these two modes is given by
: = V-/c· ' (00 t s -\J L sm (00 t = Icp sm
Lc
where and From Fig. 8. 12, at
Icp
=Vs ~ L
.. .(8.30)
1 (00 = -vLC
t l , ic = 10 = 1cp sin 1 sm . tl=--·
o
(00 tl
-1[-10]
...(8.31 )
Icp
(00
Fig. 8. 12 also reveals that (00
or
tz
=7t -
(00
=1t -
tl
sin- 1 (lollcp )
sin- l (lol l cp)] Ther efor e, circuit turn-off time for main thyristor T is
t z = 1/ (00
[7t -
tc = tz - tl = ~ [It - 2 sin- 1 (J,, /I cp )] (Do
. .. .(8.32)
Inverters
(Art.
~.3)
435
This time r;~ must be greater than the thyristor turn-off time t q . In practice, this condition can be realized by several different combinations of C and L as shown in Fig. 8.13 (a). The current commutation pulse ie that gives the requ:ired turn-off time with the minimum amount of capacitor energy
~ CV2 ,
is the optimum pulst::. This can be licbieveC1 froni. Eqs. (8.31) and
(8.32) as under.
2
or Let
I :!:j!;:
10
= x so that
(I '\'
t
... 110 -.WOe - =-1t - SIn I-
From Eq. (8.32),
. 2
'llc
.
I
p)
.
10 . 1t wotc) Wo tc -=sm(---- ·=cos·Icp \. 2 2) 2 Wo tc 1 cos - - =-
2
tc :JLC
or
...(8.33)
x
1
...(8 .34)
= 2 cos- (l/x) = g(x)
heX)
L-------=-"1.s=----- x· (a)
(b)
Fig. 8.13. Pertaining to the choice of Land C for the inverter circuit of Fig. 8.10.
The energy W that the commutating 'capacitor should provide for commutating the main thyristor is
W =!C~=! U02 2 a 2 Substituting the value ofVs =
From Eq. (8.34), "LC =
2 cos
I~p ~ from Eq. (8.30) in the above relation, we get 1 _ rL 1 W ='2 CVs . Icp -Yc = 2 -,fLC . Va Icp
t _~
( l/x)
. Substitutin g this value of-vLC in the above relation,
we get
w= l
tc ' Vs . Icp 2 2 co - 1 (l/x)
=
tc' Vs . x 10 4 cos- 1 (l / x)
436
[Art. 8.3]
P ower Electronics
Here the product Vs 10 tc has the dimensions of energy. In normalized form, the above relation is
w =
Vs 10 tc
x
4 cos- 1 (1 / x)
.. .(8.35 )
;:::. h(x)
On plotting hex) against x from Eq. (8.35), it is found that normalized commutation energy hex) has a minimum value of 0.446 when x = 1.5, Fig. 8.13 (b) From Eq. (8 .34),
g(x) = 2 cos-
1
(1 / 1.5) = 1.682
The design is carried out on the basis of worst operating conditions which consist of minimum supply voltage Vmn and maximum load current l orn' :. From Eq. (8 .30),
V mn
~ = lcp =x l orn = 1.5 lorn
-{f"
1.5 lorn L Vmn ~,..,,-;:::;tc tc "LC=-=-
or From Eq. (8.34),
=
g(x)
...(8.36) ...(8.37)
1.682
Multiplication of Eqs. (8.36) and (8.37) gives 1.5 tc 10m C = 1.682 V mn
-vr
= 0.892
tc lorn Vmn
.. .(8.38)
Vmn C = 1.5 10m
From Eq. (8.36),
...(8.39)
Multiplication of Eqs. (8.37) and (8 .39) gives tc' Vmn
L
tc Y mn
= 1.682 x 1.5 1 =0.3964 ~
...(8.40)
0m
For critical damping of the resonant circuit consisting of R d , L, C in series in Fig. 8.10,
~--l - (Rd - )2=0 LC
2L
From above, the value of resistance th at gives critical damping is
Rd=2~ ·
...(8.41)
8.3.2. Mod ified McMu r r ay full-bridge in:,,;erter A single-phase modified McMurray full-bridge inverter is shown in Fig. 8.14. The number of thyristors, diodes and other component s in full-bridge inverter is double of those in half-bridge inverter of Fig. 8.10. The operation oHhis inverter du ring commutation process is similar to that described in the previQus section for a single-phase half-bridge inverter. For example, for mode I, thyristors T1, T2 aTe conducting and th e load current completes its path through source V s' .T1 , load, T2 and back to source. For mode II , TAl and TA2 are triggered together for commutating main SeRs T1 and T2. For mode III , commut ating current ic in both the circuits goes beyon d load current ]0 so th at T l and T2 are turned off and so on.
[A rt. 8.3]
In verters
437
Fig. 8.14. Single-phase modified McMurray full-bridge inverter.
8.3.3. McMurray-Bedford half-bridge inverter Power circuit diagram of a McMurray-Bedford half-bridge inverter is shown in Fig. 8.15. It uses less number of thyristors and diodes as compared ar---------~----~----~ to modified McMurray half-bridge inverter. The number Tl
of capacitors and inductors is, however, large. This ~ . 01 Cl inverter, Fig. 8.15, consists of main thyristors T1, T2 L1
Vo + and feedback diodes D1, D2. Commutation circuitry LO AD consists of two capacitors C1, C2 and magnetically 9 10 10 L2 ....,,, coupled inductors L1, L2. Actually L1 and L2 constitute Vs 02 C2 one inductor with a centre tap so that Ll =L2 =L. The '2 TZ inductance of this centre-tapped inductor is usually of the order of about 50 ~H. The inductor is wound on a core with an air gap so as to avoid saturation. The value Fig. 8.15. Single-phase
McMurray-Bedfored half-bridge
of capacitance for t h e two capacitors is the same, i.e. C1 inverter with L1 = L2 = L
= C2= C. The inverter of Fig. 8.15 is a voltage-com mutated and C1 = C2 = C.
VSI. . .In a branch consisting of two tightly coupled inductors in series with two thyristors, if one thyristor is turned on, the other con.ducting thyristor gets turned off. This type of commutation . is called complementary commutation. As a consequence, inverter shown in Fig. 8.15 is alsl) known as complementary-commutated inverter. The simplifying assumptions are the same as for the inverter discussed in previous section. The working of this inverter can be
expla~.ned
in different modes as follows :
Mode I. In this mode, thyristor T1 is conducting and upper dc 30urce supplies load current 10 to the load, Fig. 8.16 (a). Ai; the load Cllrrent is almost constant, voltage drop across commutating inductance Ll rroportiOnal to Ll.
~:}s negligible. With zero voltage drop across
Ll and T1, voltage across C1 is zero and voltage across C2 is Vs because point g is now effectively connected to point a through L1 and T1 8:'!'d lower plate of C2 is connected to point f . The equivalent circuit for this mode is as shown in Fig. 8.16 (a). In, this figure, voltage of node g with respect to poi.ntf is Vs ' The potential of points d and c is the same as that of point gwith respect to f . In other words, the potential of all the three nodes g, d, c with respect to point f is Vs; this is shown in Fig. 8.17.
Mod e II. At t = 0 -, thyristor T2 is triggered t o initiate th e commutation of Tl. With the turning-on of T2, point d gets connected to e or f, i.e. to th e negative supply terminal. Voltage across C1 and C2 cannot change in st ant aneously, th erefor e a volta ge Vs appears across L2. As L1 and L2 are magnetically coupled, an e u al volt age is induced across L1 with terminal c
438
Power Electronics
[Art. 8.3] a .
~ 'Is
2
-...
i...
T1
I
Io
I
C1
+
L2
Vs
L2
-Vs
i
(a.)
L2
C2
• T2
.
9
Mode I, before i = 0
.. Io
2
(b)
Im/2
b
--,
T
i
)!
I
%-.l
~C1
~2
vs-L
10 /2 Io' 1e2 10/2 C2 J:t ils
'I
~rroll
L2
~
I 10
~T
:
•
I
L2
.
10/2
Im/2 (d) Mode II, t < tl
•
I
C1_ Vs
2
2
to
17f
J 10
(e)
I (
L2
02
tL 1
..
1m
1~;
Mode III, at t1
~o
II 0
'Is
1
I
ioJ02)
to
T
rm
I
"sl,.
1 1
Vt.-L
...Io!2
~OA
Mode II. t =0 +
(c)
Mode II . at t = 0
10/2
T
10 -
10
!I
10
- "0+
l! V, .I.
'lsI
~
g
- "0+ . C2
0
L1
"TI
c.J.
10
L
o
b
T +
~
10
r f
a
b
to=~2 J
L2
02
1
r
if) Mode IV
Fig. 8.16. Different modes for the inverter of Fig. 8.15.
positive. Voltage un across terminals of thyristor Tl can be found oy applyincr KVL for the
. loop c b a (e d c in Fig. 8.16 (h), th erefore, Un -
V V -t-;f
+ Vs + Vs
=0
This shows that point c is positive with respect to b by Vs volts, i. e. Tl is subjected to a reverse voltage of - Vs ; it is therefore turned off at t = 0 + . Load current 10 flowing through Tl and L1 is at once transferr ~ d to L2 and T2 so as to maintain constant mm((proportional to L1, I ) in the centre-tapped in ductor as per the constant flux linkage th eorem . Cu ent directions
Invuters [Art. 8.3) --------------------------------~----------------~
439
for iel , ic2 are shown in Fig. 8:16 (c). KVL for the loop consisting of Cl, C2 and .t he source V, for this figure gives ., C1
~1 Ji el dt + ~2 Jie2 dt -
VB (voltage across C2) + V, (source voltage) -= 0 or iel =ic2 '
=C2.
KCL at node g in Fig. 8.16 (c) gives iCl+ i e2=lo+10
or This shows capacitor C1 and C2 both carry current 10 at t =0 + . Half of i el flows through load and the other half through L2 ; same is true for i e2 • Fig. 8.16 (c). Now C1 is getting charged from zero voltage and at the same time C2 is getting discharged from VB at the same rate. As c~tpacitor C2 is placed across L2, an oscillating current is set up in the closed loop formed by C2, L2 and T2. After one-fourth of a cycle, this oscillating current rises (initial value 10) to a maximum value of 1m in L2 and T2 and at the same time Vc2 acroSs C2 falls to zero. At this moment, i.e. one-fourth of a cycle after the "instant T2 is triggered, KCL at node g, Fig. 8.16 (d), gives " i c1 + i c'}. = 10 (load current) + 1m (eurrent in L2 and T2)
or This is shown in Fig. S.16 "(d). The variation of i el , i c2 ' in and io from t:= 0 to tl is shown in Fig. 8.17. With the turning on of T2, voltage of node d drops to zero whereas that of node cshoots up to 2Vs with respect to node fat t =0 + , Fig. 8.16 (b). From t =0 to tIl voltage of node g drops to zero in one-fourth of a cycle of Vs cos 000 t. Similarly, voltage of node c reduces from 2Vs to zero at tl in one-fourth of a cycle, Fig. 8.17. Thyristor Tl is reverse biased until voltage of node c falls to Vs' commutation time for Tl is therefore tc as shown in Fig. 8.17. For the circuit consisting of C2, L2 and T2, ringing frequency is 1
<00
Periodic time
= ~LC 1 fo
27t
To == - = tl
<00
_~
= 27t -vLC
7t_~ = -To =--vLC 4 2
Here L is the inductance of L2 and C is the capacitance of C2.
Circuit turn-off time tc < one quarter of a cycle
or
or
tc < tl t <~ c
2 ,/LC
440
P ower Electronics
IArt. 8.3J
.. t
I
tD2f
1
f
.
i T2
.
gated
I
t02 Current in 02 I
b
i
----~--~~~----~~-=~--~--------~
I
t c l=i c2.
I I I
I
rr:-r 11
1m
0 I.....o-.-IT TI off
T2 on
t1 • I
02 on
:m:
.t
.ti. ,
T2 off .
tJ
N
, : r ; [ - - } M-:-des ;
02 off T2 on
Fig. 8.17 . Voltage and current wave-forms for the inverter of Fig. 8.15 ..
Mode TIl. At t l , capacitor Cl is charged to supply voltage Vs and therefore no current can fl ow through C1 , i.e. ic1 = O. After one-fourth of a cycle from t = 0, i .e. at t l , Vc2 = O. Just after t l , (/0 + ImY 2 through C2 tends to charge it with bottom plate positive. As a result, diode D2 gets forward bi ased at t l . Thus, now entire current (Io + 1m) is transferred to D2 so ~hat both i cl = ic2 = 0 just after t l but i D2 = 10 + 1m ; this is shown in the equivalent circuit of F ig. 8.16 (e). Diode current iD2 =10 + 1m is also shown in Fig. 8.17. The energy stored in in uctor L2 at tl is dissipated in the closed circuit made up of L2, T2 a nd D2. At time t 2 • t his en ergy i~ entirely dissipated, therefore current iT2 decays to zero and as a result, T2 is tu rned off at t 2• Fi 8.17. Sometimes , a small resistance is inclu ded in series
Inverters
[ rt. 8.3]
441
with the diode to speed up the dissipation of stored energy in L2. As iT2 decays from 1m at tl to zero at t 2, iD2 also decays from 10 +lm at tl to iD2 = io =ab at t 2. Load current also decreases from 10 at tl to io = iD2 =ab at t 2. Mode IV. When the current in through L2 and T2 has decayed to zero, the equivalent circuit is as shown in Fig. 8.16 (f>. A load current io= iD2' still continues flowing through the diode D2 as iD2 during (t3 - t 2) interval, Fig. 8.16(f). Mode V. Finally, load current io through the diode D2 and load decays to zero at t3 and is then reversed. As soon as io, equal to iD2 . tends to reverse, D2 is blocked. The reverse bias across T2, due to voltage drop in D2 no longer exists. Therefore, thyristor T2 already gated during the interval (t3 - t 2) gets turned on to carry the load current in the reversed direction as shown in Fig. 8.17. The capacitorC1, now charged to the source voltage V s ) Fig. 8.16 (e), is ready for commutating the main thyristor T2. The magnitude of commutating circuit parameters Land C, for minimum trapped energy, is given by .
and
Vs
tq
... (8.42)
L=2.351om '. I t C = 2.35 o~ q
...(8.43)
s
= thyristor turn-off time. lorn = maximum load current to be commutated.
where and
tq
8.3.4. McMurray-Bedford Full-bridge Inverter A single-phase McMurray-Bedford full-bridge inverter can be realized by connecting two half-bridge inverters as shown in Fig. 8.18. The various components required are double of those in the half-bridge inverter of Fig. 8.15. The working of this inverter is similar to that described for half-bridge inverter in the previous section. For example, for mode I, thyristors TI, T2 are conducting and load current flows through Vs ' T1, L1, loa~, L2, T2. Voltage across Cl. C2 is zero but C3, C4 are charged to voltage Vs' For initiating commutation of Tl. T2 : thyristors T3. T 4 are triggered. This reverse biases T1" T2 by voltage - Vs' these thyristors are therefore turned off and so on.
D1
03
C3
L3 s
LOAD i.o - v + o
i.o
04
D2
L2 C2
T2
Fig, 8.1S . Sin ."le-phase McMurr ay-Bedford full-bridge inverter.
442 ~.4.
[Art. 8.4]
THREE. PHASE
Power Electronics nn.IIJ(~~
INVElrt'EltS
For providing adjustable-frequency power to industrial applic~tions, three-phase inverters are more common than single-phase inverters. Three-phase inverters, like single-phase inverters, take their dc supply from a battery or more usually from a rectifier. . A basic three-phase inverter is a six-step bridge inverter. It uses a minimum of six thyristors. In inverter terminology, a step -is· defined as a change in the firing from one thyristor to the next thyristor in proper sequence. For one cycle of 3600 , each step would be of 60° interval for a six-step inverter. This means that thyristors would be gated at regular intervals of 60° in proper sequence so that a 3-phase ae voltage is synthesized at the output terminals of a six-step inverter. Fig. 8.19 (a) shows the power circuit diagram of a three-phase bridge inverter using six thyristors and six diodes. As stated earlier, the transistor family of devices is now very widely used in inverter circuits. Presently, the use ofIGBTs in single-phase and three-phase inverters is on the rise. The basic circuit configuration of inverter, howeverl remain~ unaltered as shown in Fig. 8.19 (b) for a three-phase bridge inverter using IGBTs in place of thyristors. A large capacitor connected at the input terminals tends to make the input de voltage constant. This capacitor also suppresses the harmonics fed back to the dc source. . In Fig. 8.19(0:) inverter using six thyristors, commutation and snubber circuits are omitted .for simplicity. It may be seen from Figs. 8.1 and 8.19 that a three-phase bridge inverter consists of three half-bridg~ inverter::: arranged ~ide by side. The three·ph8.~e load is assumed to be star connected. In Fig. 8.19 (a), the thyristors are numbered in the sequence in whkh they are . triggered to obtain voltages V"b, Vbc' veil at the output terminals a, b., C of the inverter. .
v5
-
c
(a) Fig. 8.19. Three-phase bridge inverter using (a) thyristors
(b) (b )
IGBTs.
- There are two possible patt erns of gating the thyristors. In one pattern, each thyristor conducts for 180° and in the other, each thyristor conducts fO T 120°. But in both these patterns, gating signals .are applied and removed at 60° intervals of the output voltage waveform . . Therefore, both the s ~ modes require a six step br' dge inverter. These modes of thyristor conduction are described in wh at follows: 8.4.1. T h r ee-phase 180 Degree Mode VSI In t he three-phase inverter of Fig. 8.19, each SCR conducts for 180° of a cycle. Thyristor pair in each arm, i.e. Tl , T4 ; T3, T6 and T5 , T2 are turn9d on with a time interval of 180°.
"l
"
[ rt. 8.4]
Inverte rs
443
It means that T1 conducts for 180 0 and T4 for the next 1800 of a cycle. Thyristors in the upper group, i.e. T1, T3, T5 conduct at an interval of 120°. It implies~hat if T1 is fired at wt = 0°, then T3 must be fired at wt =120 0 and T5at wt =240°. Same is true for lower group of SeRs. On the basis of this firing scheme, a table is prepared as shown at the top of Fig. 8.2C. In this table, first row shows that T1 from upper group conducts for 180°, T4 for the next 1800 and
r-- 180
I
LT5 00
Steps {II
0
-I-
T1
160<>----,
;
I
I
T4
I
T2
T1 T2
T5
60<> 120~ 1800 21.0° 300 0 3600 60 0
I IT : m !17 !
7
I T3
r
T5 T6
120 0 180 0 ?1.0° 300' 360 0
i it I I I IT i m : N.
:
~
: 11 :
5.6,1 ,6.1,2' 1.2.3 '2.3."'3.4,5: 4.5.6: 5.6.' : 6.1.2: 1.2.3,'2.3,,,: 3.".5: ",5.6i}Cho,,~ucting ' I I " I " , t yrrstors , I I ' " , I ,
r vt ' ,
v
I
I
:
,
'
,
,
I
I
I
,
I
I
!
00:
I I
I (a)
~I
I,
I
"
I I
I ' I
, '
I I
~
o
I
,
:
,
I
_
w~
Y b,
C)---==>-~-t--...!...L~-=-+.. . ,. . ";--"---+-.....L.-";---lf---+--~ wt
1_ 2;5
I
I -L
r ,
=\'00- Y bo
'10
I
Vc:.
i I
(b)
I
,
I
I J
, =vbo-vco
I I
~120'
1
I,
I
,, 120~
,
,
I
I
!; J
I
,
-Vs
I
:
I
I
,
v'.,
~.,
I
,I
I
I
,
,
I I
I
,I ,
I
I
j.
I
I
,
, ,,
I
,,
I I I
,I
,
,, r
,.,
wt
I
,I I
.' WI I .
,
I
, , . II ,
! I
,
-:r
1
,
, , , I
2.,,
I
I
I
I
3.,,
1
,I :
~
-V
I
51
Fig. 8 .20 . Voltag,= waveforms fer I SO ' m od'? 2-oh?se VSI.
1..,,
Ult
1444
[Art. 8.4]
Power Electronics
then again T1 for 1800 and so on. In the second row, T3 from the upper group is shown to start conducting 1200 after T1 starts conducting. After T3 conduction for 180 0 , T6 conducts for the next 180 0 and again T3 for the next 1800 and so on. Further, in the third row, T5 from the upper group starts conducting 1200 after T3 or 240 0 after T1. After T5 conduction for 1800 , T2 conducts for the next 1800 , T5 for the next 180 0 and so on. In this manner, the pattern of firing the six SCRs is identified. This table shows that T5, T6, T1 should be gated for step I ; T6, T1, T2 for step II ; T1, T2, T3 for step III; T2, T3, T4 for step IV and so on. Thus the sequence of firing the thyristors is T1, T2, T3, T4, T5, T6 ; T1, T2 ... . It is seen from the table that in every step of 60 0 duration, only three SCRs are conducting-one from upper group and two from the lower group or two from the upper group and one from the lower group. ,
The circuit models for step I-IV are shown in Fig. 8.21. During step I, thyristors 5, 6, 1 are conducting. These are shown as closed sWitches and non-conducting SCRs 2, 3, 4 as open switches in Fig. 8.21 (a). The load terminals a and c are connected to the positive bus of de source whereas terminal b is connected to the negative bus of de source, Fig. 8.21 (a). The load voltage is vab = vcb= V" in magnitude. The magnitude of line to neutral voltage can be obtained as under : During step I, 0 ~
rot
0
< ~, Fig. 8.21 (a), thyristors conducting 5, 6, 1.
,. - -V,-Z-3 - -2 .V, Z
Current,
1-
Z+"2 The line to neutral voltages are
. Z
and
V,
Vao
=vco ='1'2 =3
vOb
. 2 Vs
= 'I Z = -3 ~
o
The above line to neutral voltages may be written as
va\)
=vco =3
and
2~
vbO = -
3 ' These
voltages are shown in Fig. 8.20 (a) during step 1. For t~e next step II, the line to neutral voltages are as under:
I
0
During step II, Current,
~s rot < 231t , Fig. 8.21 (b), thyristor conducting 6,1,2. 2 Vs ,. -- 3 Z 2-
.
or
2 V"
. Z
.
Vs
vaO
='2 Z =3"":J VOb =VOc ='2"2 ="3
VaO
=3'
Vs
2Vs
V bO
=VcO =- 3'
These output voltages are plotted in Fig. 8.20 (a) . In this manner, th e variation of phase voltages vao, vbo' vco as obtained in Fig. 8.21 up to step IV and similarly for other steps, is plotted in Fig. 8.20 (a). It is clear that for each cycle of output volta ge of each phase, six steps are r equired and each step h as a duration of 60 0
•
0
[Art. 8.4]
Inverters
Step I
445
c
0-----:. .- - - - . : ; - --.... vs +
T
a
'----.--l----f~
b
2Vs
2Vs
z
-3
T
o-----'--.....J -- -_____ ._1":" t5
(a)
Step II
0-60° ; 5,6, 1 closed.
Vaa
=Veo =V s /3
Vba
=-
Vob
v---........---~--.,...,
= - 2Vs/3
a +
-----. - --or + z
2Vs
"3
r--~o-'-"---I- ~ Vs z z
Vs
T
"3
CF"~-----1-+-_ _----,
c
b Voo
(b)
Step III
60-120° ; 6, 1,2 closed.
= 2
vba
o---~---~----~
__ J
V/3
=Vco =- Vsl3 a
b
o----l--.-.------:,. -- . "l' + + Vs
5
z
2
"'3
"-=---i0~---I-.--.~ ~
b
~
z
o z
3
0-----'----'----------.1 C
c '--------' closed.
vaa
(c) 120-180° ; 1, 2, 3
=vbo =Vs/3
vea = -
Step IV
2Vs/3 b
3
---------·f·+
+
5
2Vs - 3
2
a
b
-
r----+---.. Vs
--t ~
!!. 3 o---~--__l ___ l 3'
a
c
Vba = 2 V s /3 (d) 180-240° ; 2, 3, 4 closed. Vao = veo = - V/3
Fig. 8.21. Equivalent circuits for a 3-phase six-step 180 0 mode inverter with a balanced
star-connected load.
The line voltage vab = v at) + Vab or Vab = Vao - Vba is obtained by reversing to Vao as shown in Fig. 8.20 (b). Similarly, line voltages vbe =vba - vco and plotted in Fig. 8.20 (6).
vba Vca
and adding it =Vca - vaa are
446
[Art. 8.4]
Power
Electro nic ~
The three rows in the top of Fig. 8.20 also indic ate the pattern of gating ~igrial waveforms. At wt = 1t, when iC/l is removed, Tl is turned off and simultaneou.:;ly i ';4 is applied to turn on T4. Similarly, at rot =21t / 3, when ig6 is cut off, TS is turned off and at the same instant ig'J i~ applied to turn on T3. Same is true for other thyristors. b
I
b
It is seen from Fig. 8.20 that pha.3e voltages have six steps per cycle and line voltages haVe one positive pulse and one negative pulse (each of 120 0 duration) per cycle. The phase as well as line voltages are out of phase by 120°. The function of diodes Dl to DS is to allow the flow of currents through them when the load is reactive in nature. The three line output voltages can be described by the Fourier series as follows: Vab
4V. n1t ' . I -' cos -S sin newt + 1t/ S) n1t
=
.. (8.44)
n = I, 3, 5, Vbc
4Vs
= n=
Vca
II,.3,. 5,-n1t
= .I n = I, 3, 5,
For n
n1t
cos -S sin newt -1t/ 2)
... (8.45)
:~,; cos nS1t sin n( wt + 5S1tJ
.. .(8.4S)
= 3, cos 3s1t = O. Thus, all triplen harmonics are absent from the iinevoltages as given
by Eqs . (8.44) to (8.4S). T he line voltage waveforms shown in Fig. 8.20 represent a balanced set of three-phase alternating voltages . During the six intervals, these voltages are well defined. Therefore, these voltages are independent of the nature ofload circuit which may consist of any combination of res istance, inductance and capacitance and the load may be balanced or unbalanced, linear or nonlinear. . F ourier series expansionofline to neutral voltage
v
(J()
=
I
n=6k±1
where
Vcw
2VS .. --smnoot n 1t
in Fig. 8.20 is given by ...(8.47)
k =0,1,2, ...
For a lin'e ar star-connected balanced load, phase or line currents can be obtained from Eq. (8.47) . E xpression s similar to Eq. (8.47) can be written for vbo an d vco by replacing wt by (oot - 120°) and (oot ~ 24CO) respectively. In Fig. 8.21, load is assumed star connected and three phase and line voltages are obtained as shown in Fig. 8.20. For a delta connected load also, phase or line voltage waveforms Vab' Vbc' vca as shown in Fig. 8.20 would be obtained directly. Therefore, for a llnear delta-conne ct~ load, phase and line currents can be obt ained from Eqs. (8.44) to (8.4S). From E q. (8.44), r ms value of nth component of line voltage is VLf'. =
4 Vs
n1t
T2 n1t cos 6
...(8. 48)
Rms valu e of fund amental line voltage,
1t 4 Vs V L1 = -:rn-=~ cos S = 0.7797 Vs 'i . 1t
... (8.49)
[Art. 8.4]
Inverters
447
It' is seen from line voltage waveform vab in Fig. 8.20 (a) that line voltage is Vs from 0° to 120°. Therefore, rms value ofline voltage V L is 2 /3
VL = [
~Jolt
,1/2
V;d(WX)j
_
rr
=\f~
V.. =0.B165V"
...(B.50)
Rms value of phase voltage Vp is
VL ~ VP=~=3V~=0.4714 Vs
...(8.51)
Rms value of fundamental phase voltage, from Eq. (B.47) , is 2V.'I V L1 V p1 = T2X = 0.4502 V" = ...J3
... (8.52)
8.4.2. Three-phase 120 Degree Mode VSI The power circuit diagram of this inverter is the same as that shown in Fig. B.19. For the 120-degree mode VSI, each thyristor conducts for 120° of a cycle. Like 180° mode, 120° mode inverter also requires six steps, each of 60° duration, for completing one cycle of the output ac voltage. For this inverter too, a table giving the sequence of firing the six thyristors is prepared as shown in the top of Fig. 8.22. In this table, first row shows that Tl conducts for 120° and for the. next 60°, neither T1 nor T4 conducts. Now T4 is turned on at we = 180C and it further conducts for 120°, i.e. from wt = 180° to wt =300°. This means that for 60° interval from CJ)t = 120° to CJ)t = 180°, series connected SCRs T1, T4 do not conduct. At we = 300°, T4 is turned off, then 60" interval elapses before T1 is turned on again at we = 360°. In the second row, T3 is turned on at CJ)t = 120° as in 180° mode inverter. Now T3 conducts for 120°, then 60° interval elapses during which neither T3 nor T6 conducts . At UJt = 300°, T6 is turned on, it conducts for 120° and then 60° interval elapses after which T3 is turned on again. The third row- is also completed similarly. This table shows that T6, Tl should be gated for step [ ; T1, T2 for step II ; T2, T3 for step III and so on. The sequence of firing the six thyristors is the same as for the 180° mode inverter. During each step, only two thyristors conduct for this inverter - one from the upper group and one from the lower group ; but in 180 0 mode inverter, three thyristors conduct in each step. . . The circuit models for steps I-IV are shown in Fig. 8.23, where load is assumed to be resistive and star connected. During step I, thyristors 6, 1 are conducting an d as such load termin al a is connected to the positive bus of de source wh ereas terminal b is connected to negative bus of de source, Fig. 8.23 (a). Load terminal c is not connected to dc bus. The line to neutral voltages, from Fig. 8.23 (a) are Vs vao =2'
or
Vs ubo=-2
and
V co
Vs
V ob
=2
=0
These voltages are shown in Fig. 8.22 (a) during step I of 0° - 60° . F or step II, thyristors 1, 2 conduct and load voltages are va~ = V/2, Vco = - V/2 and Vbo = 0, Fig. 8 .23 (b) ; these voltages are plo tted in Fig. 8.22 (a). This procedure is follo wed for obtainin g lo ad voltages or the remaining steps and these phase voltages are then plotted in Fig. 8.22 (a).
448
Power Electronics
[Art. 8.4]
0'
Step~1 I cond~cting{, 6 1 thyristors
I
'
,
va I
60' 120' 180· 240' 300· 360' 60' 120' 180' 240· 300· 360·
im
I In :
IiI:
: II 117 : 12: i YI I 1II : 17 : 12: I
'I 12 I 2 3 I' 34 ' I. 5 I 5 6 I 61 ,I 1,2: 2,3 " 3,4 ,1 4 ,51 .5,6 l6,1
, '
I"
,
I
1
l-12,0-i
'
I
' I ' ",
'I
'
",
I
,
,
,
,I
:
:
::
1
1
1
1
, ,
O~-'--~I--~--r-~~~~~~--~~--~--L-----~ W.t
(a)
010+
Vb
: I
~
Veof o
: ~.
.1.....11-+--....
. . . . v/
,"T T
',I
.
I
1
1
,
I.
I I
--4-.-..1
,-I
l-120.--.1 ,
I
_I:
I
I
.
~I_-+I_.....I
,I
1
C:t I I
1
I
, 1
wt"
wt
..
wt
I
I 120·4-.j
II· " :
(b)
I
L
I I
I
I:
I
.\
wt
.
:"
O~-+-L~--+---~~--~--4---L-~---L--J---+-----w-t~'
>,
Fig. 8.22. Voltage waveforms for 120 mode six-step 3-phase VSI.
0
. vab =Vao
The line voltages
vbc
- vbo
=vbo -
vco
and
are also plotted in Fig. 8.22
(b).
It is seen fr om Fig. 8.22 that phase voltages have one positive pulse and one negative pulse (each of 1200 duration) for one cycle of output alternating voltage. The line voltages, however, have six steps per cycle of output alternating voltage. As stated before, the three rows in the top of Fig. 8.22 indicate the pattern of gating signal waveforms. .
The merits an demerits of 120-degree mode inverter OVer ISO-degree mode inverter are as follows : In the 1800 mode inverter, when gate signal ig1 ' cut-off to turn off T1 at rot =1800 , gating signal igl\ is simultaneously applied to turn on T4 in the same leg. In practice, a comm t ation interval must eri t between the removal of ig 1 and application of ig 4 , ' because (i)
Jf,.
In v~rte rs
[Ar t. 8.4]
otherwise dc source would experience a direct short-cir<;uit same leg.
ough SCRs Tl and T4 in the
Step I +
a
+ R
·-·l~;
o .-- .
R
'Is
T
0------'
b
c (a) 0-60
0
~-----' ;
6, 1 closed
Voo
== V/2
Vbo
=-
Voo
=
Vs/2 and
lIeo
=0
V/2 - .V/2 and
IIbo
=0
= Vs/2 == - Vs/2 and vao
=0
Step II ·
2 b
C~-----=-'
(b)
60-120° ; 1,2 closed
Veo =
Step III 5
+
2 b
R (c)
c '----------" 120-180° ; 2,3 closed
Vbo veo
Step IV b
3
+
R
'Is
2"
o . -
a
'Is
R
2" a
C'---~---'
(d)
180-2400
;
3, 4 closed
449
Vbo
== Vs / 2
vao=~V/2
and
v eo =O
F ig. 8.23. Equiv al ent circuits for a 3-phase six-step 120 0 mode inverter with balanced star-connected resistive load.
450
Power Electronics
[Art. 804]
. This difficulty is overcome considerably in 120-degree mode inverter. In this inverter, there is a 60° interval between the turning off of T1 and turning on of T4. During this 60° interval, T1 can be commutated safely. In general, this angular interval of 60° exists 'between the turning-off of one device and turning-on of the complementary device in the same leg. This 60° period provides sufficient time for the outgoing thyristor to regain forward blocking capability. (ii) In the 120° mode inverter, the potentials of only two output terminals connected to the dc source are defined at any time ofthe cycle. The potential of the third terminal, pertaining to a particular leg in which neither device is conducting, is not well defined; its potential therefore depends on the nature of the load circuit. Thus, the analysis ofthe performance of this inverter is complicated for a general load circuit. For a balanced resistive load, the potential of all the three terminals is, however, well defined. This is the reason load is assumed resistive in Fig. 8.23. For a balanced delta-connected resistive load, the line voltages as shown in Fig. 8.22 (b) are obtained directly.
The Fourier analysis of phase voltage waveform Vao
= fI
Similarly,
Vbo
of Fig. 8.22 (a) is
2 Vs n1t. L -cos -6 sm n (rot + n/6) nn
... (8.53)
nn · . L -2Vs cos -6 sm n (rot -n/2) nn
...(8.54)
L
... (8.55)
= 1, 3, 5
= fI
Vaa
= 1, 3, 5, .. ....
nn . ( 5nJ v co = ~ -2Vs' cos - 6 sm n wt + -6 n1t fI = 1, 3, 5, ......
. The Fourier analysis ofline voltage waveform vab of Fig. 8.22 (b) is
and
vab
= ~ ~
3· V n7t
8
. ( nJ sin n rot + '3
... (8.56)
fI =6k±1
where
k::: 0,1,2,3...
Similar expressions for
Vbc
and
Vca
can also be written.
Rms value of fundamental phase voltage, from Eg. (8.53), is 2Vs n V p1 = ~ . cos '6 = 0.3898 V s
...(8.57)
Rms value of phase voltage, 2
. V
1
= .n- J0 p
21t/ 3
[
1/ 2
V
[--.2!] .d (wt)]
='\1 - ~. '3
2 .V _8
2
=~ = 0.4082. Vs . '16
V
.. .(8.58)
ms valu e of fundamen tal line voltage, from Eq. (8 .56), is
3Vs V L 1 =~ =0.6752 Vs '12 1t
=.In '1 3 Vp1
... (8 .59)
Rms value of line voltage,
VL
= -J3vp =~= 0.7071 Vs
.. .(8.60)
Inver ters
[Art. 8.4]
451
Example S.S. A three-phase bridge inverter delivers power to a resistive load from a 450 V de source. For a star-connected load of 10 Q per phase, determine for both ('1) 180 0 mode and (b) 1200 mode, (i) rm s value of load current (ii) rm s value of thyristor current . (iii) load power.
Solution. For a resistive load, the waveform of load current is the same as that of the applied voltage. In view of this, waveforms of phase-load current and thyristor current are as shown in Fig. 8.24 (a) for 180 0 mode operation and in Fig. 8.24 (b) for 1200 mode operation. io 2Vs !
3R Or-----~----~~----~
__
wt
O~--~~------~-------
wt
~)
~)
Fig. 8.24. Pertaining to example 8.8 (a) 180 0 mode (b) 120 0 mode:
Upper waveform of Fig. 8.24 (a) shows that rms value of per-phase load current I()r is given by (a) 1800 mode :
. J2 1]1/2 ( [ l(::;J3(J =[-[x 4 (3: 45:;3 :l~ 3' .
1
lor
.
=
Vs
3 X 10
1t
+ 2 3' +
X
7t
2Vs
Rms value of thyristor current is
+
X
3
X
10
.
7t
Vs
X
2
3'
:= ,/450
= 21.213 A
.~
112
l-J:..~ [{ Tl - 2x ( 3 x 10
2
2
1
21t [2 X 450J XS+ 3x 10
x~}
]
= --./ 225 := 15 A
Power delivered to load
= 3 I;rR = 3 (--./450)2 X 10 = 13.5 kW
(b) 12.Q m ode: Upper waveform in Fig. 8.24 (b) gives rms value of per-phase load cur rent l or as under:
.
n
.; I or = [ 1 ( 2 450 X 10
X
Rmsvalue of thyristor current,
I" Lo ad power
=[ix ~ 2;;~oI
327t]l/2 := '1. ~ 337 .5 = 18.371
r
x 2;
= 12 99 A
= 3 l ;r R = 3 (~ 337 .5 )2 x 10 = 10. 125 kW
452
Power Electronics
[Art. 8.S]
8.5. VOLTAGE CONTROL IN SINGLE-PHASE INVERTERS AC loads may require constant or adjustable voltage at their input terminals. When such loads arefed by inverters, it is essential that output voltage of the inverters is so controlled as to fulfil the requirement of ac loads. Examples of such requirements are as under: (i ) An ac load may require a c'o nstant input voltage though at different levels. For such a load, any variations in the dc input voltage must be suitably compensated in order to maintain a constant voltage at the ac load terminals at a desired level. (ii) In case inverter supplies power to a magnetic circuit, such as an induction motor, the voltage to frequency ratio at the inverter output terminals must be kept constant. This avoids saturation in the magnetic circuit of the device fed by the inverter. The various ~~thodsfor the control of output voltage of inverters ar e as under: (a) External control of ac output voltage External control of dc input voltage: ' (c) In ternal control of inverter. (b )
The first two methods require the use of peripheral components whereas the third method requires no peripheral components. These methods are now briefly discussed.
8.5.1. External Controlof ac Output Voltage There are two possible methods of external control of ac output voltage obtained from inverter output terminals. These methods are: . (a) AC voltage control (b) Series-inverter control
These are now discussed briefly.
(a) AC voltage control: In this method, an ac voltage controller is inserted between the output terminals of inverter and the load termin als as shown in Fig. 8.25. The voltage input to the ac load is regulated through the firing angle control of ac voltage controller. This method gives rise to higher harmonic content in the output voltage; particularly when the output voltage from the ac voltage controller is at low level. This method is, therefore, rarely employed except for low power applications.
Constant de voltage
Inverter
AC voltage controller
Controiled ae vo Itage
AC load
Fig. 8.25 . External control of ac output voltage.
-
(b) Ser ies-invert er contr ol : This meth od of voltage control involves the use of two or . more inverters in series. Fig. 8,26 (a) illustra tes h ow the outpu t voltage of two inverters can be . summ ed up with the h elp of transform ers to obtain an adjust able output voltage. In this figur e, the inverter output is fed to two transformers whose secondaries ar e connected in series . Phasor su m of the two fun damental voltages V Ol ' V02 gives the r esultant fundamental voltage Vo as shown in Fig. 8.26(b). H ere V o is given by 2 Vo =[ V 2Ol + V 02 + 2 V Ol ' V 02 cos 8J1I2
t
Inverters
[Art. 8.5J
453
Constant de voltage
(a) (b) Fig. 8.26. Series inverter control of two inverters.
It is essential that the frequency of output voltages VOlt V 02 from the two inverters is the same. When e is zero, Vo =VOl + V 02 and for e =1t, Vo =0 in case VOl =V 02 ' The angle e can be varied by the firing angle control of two inverters. The series connection of inverters, called multiple converter control, does not augment the harmonic content even at low output voltage levels. .
8.5.2. External Control of de Input Voltage In case the available voltage source is ac, then dc voltage input to the inverter is controlled through a fully-controlled rectifier, Fig. 8.27 (a) ; through an uncontrolled rectifier and a chopper, Fig. 8.27 (b); or through an ac voltage controller and an uncontrolled rectifier, Fig. 8.27 (c). If available voltage is dc, then dc voltage input to the inverter is controlled by means of a chopper as shown in Fig. 8.27 (d). . Input voltage-control techniques shown in Fig. 8.27, in which de voltage input to inverter is controlled by means of components external to the inverter, has the following main advantage. Constant Fully controll ac voltage ed rectifier
-
Controlled
Filter
d c voltage
Inverter
I Controlled a c voltage
(a)
Constant oc voltage
Uncontrolled rectff ier
-
Chopper
f-
Filter
Cont rotled Inverter d c voltage
Controlled ac voltage
(b)
AC voltage Constant ac voltage co ntroller
r-
-
Uncontrolled rectifier
-
F ilter
Controlled
Inverter
dc voltage
Contro lled ac voltage
(e )
Consta nt de voltage
Chop per
Filter
Cont roUed de v oltage
I ver ter
Controiled QC voltage
(d )
Fig. 8.27. External contr ol of dc input volt age to inverter ; (a ), (b ) and (d) with de sour ce 0 the input.
(e )
with ae source on the input
~S4
Power Electronics
[Art. 8.6]
(i ) Output voltage waveform and its harmonic content are' not affected appr eciably as the invert er output voltage is controlled through the adjustment of dc input volfage to the.inverter.
This method of voltage control, however, suffers from the following disadvantages : (i ) The number of power converters used for the control of inverter output voltage v dries from two to three, Fig. 8.27. More power-handling stages result in more losses and reduced efficiency of the entire scheme.
(ii) For reducing the ripple content of dc voltage input to the inverter, filter circuit is required ir.. 9.11 types of schemes shown in Fig. 8.27. Filter circuit increases the cost, weight and size and at the same time reduces efficiency and makes the transient response sluggish.
(iii ) As the dc input is decreased, the commutating capacitor voltage also decreases.
.
This has the effect of reducing the circuit turn -off time
(t ~ C ~) for the SCR for a constant
load current. Therefore, for a large variation of output voltage for a constant load current, control of dc input voltage is not conducive. This difficulty can, however, be overcome by a separate fixed dc source for charging the commutating capacitor, but this makes the scheme costly and complicated.
8.5.3. Internal Control of In verter Output voltage from an inverter can ·also be adjusted by ex.ercisin~a control within the inverter itself. The most efficient method of doing this is by puise~widtn" inodulation control used within an inverter. This is discussed briefly in what follows: P ulse w idth modulation control. In this method, a fixed dc input voltage is given to the inverter and a controlled ac output voltage is obtained by adjusting the on and off periods of the inverter components. This is·the most popular method of controlling the output voltage and this method is termed as p ulse-width modulation (PWM) control. Th e advantages possessed by PWM technique are as under : (i) The output voltage control with this method can be obtained without any additional components. (ii ) With this method, lower order harmonics can be eliminated or minimised along with its output voltage control. As higher order harmonics can be filtered easily, the filtering requirements are minimised. . The main cl!isadvantage of this method is that the seRs are expensive as they must possess low turn-on and turn-off times. PWM inverters are quite popular in industrial applications, these are therefore discussed in detail in the next section.
8.6.
PUL~E.WIDTH
MODULATED INVERTERS
.
PWM inverters are gradually taking over oth er types of inverter s in indU$trial applications. PWM t echniques are ch aracterised by const ant amplitUde pulses. The w'dth of these pulses is, however, modulated t o obtain inverter ou tput voltage control and to reduce it s h armonic content. Different PWM techniques are as under: (a ) Single-pulse modul ation (b) Multiple-pulse modulation (c) Sinusoidal-pulse m odulation. In PW f inverters, for ced commutation i essen tial. The three PWM techniques l'sted above differ from each other in the harmonic content in th eir respective 0 tput vo t ages . Thus, choi e of a particular PWM t echnique depends upon the permissible harmonic cont ent in the inverter output voltage.
[Art. 8.6]
Inverters
455
In industrial applications, PWM inverter is supplied from a diode bridge rectifier ana an LC filter. The inverter topology remains the same as in Fig. 8.2 (a) for a single-phase inverter and in Fig. 8.19 for a three-phase inverter. But now the devices are switched on and off several times within each half cycle to control the output voltage which has low harmonic content. In the following lines, the basic principles ofPWM techniques for single-phase inverters are illustrated and then the methods of obtaining such output voltages are considered.
8.6.1. Single-pulse Modulation The output voltage from single-phase full-bridge inverter is shown in Fig. 8.28 (a). When this waveform is modulated, the output voltage is of the form shown in Fig. 8.28 (b). It consists of-a pulse of width 2d located symmetrically about 1t/2 and another pulse located symmetrically about 31t/2. The range of pulse width2d varies from 0 to 1t ; i.e. 0 < 2d < 1t. The output voltage is controlled by varying the pulse-width 2d. This shape of the output voltage wave shown in Fig. 8.28 (b) is called quasi-square wave . Fourier analysis of Fig. 8.28 (b) is as under:
nt
7t/ 2
. 2 bn =
d + ) (7t / 2-d)
. 4Vs [. n1t. ] Vs sm nc.ot . d (cot) = n1t sm 2 sm nd
.. .(8.61)
Positive and negative half cycles of Vo in Fig. 8.28 (b) are symmetrical about 1t/ 2 and 37t/2 respectively. In addition, these half cycles are also identical. As a result, coefficient an = O. Thus the waveform of Fig. 8.28 (b) can be described by Fourier series as
~
4Vas m . n7t . nd' - sm smnwt n1t 2 n=I,3,5 . 4V ' ' Vo = ~ [sin d sin rot - ~ sin 3 d sin 3 rot + % sin 5 d sin 5 rot .. .... ] vo::; L.
or
... (8.62) .. .(8.63)
When pulse width 2d is equal to its maximum value of 1t radians, then the fundamental component of output voltage, from Eq. (8.63), has a peak value of 4 Vs ...(8.64) v0 1m =-; For pulse width other than 2 d = 1t radians, the peak value offundamental component, from . 4 Vs . Eq. (8.63) , IS - - sm d. 1t
I
t
Vs~----"'"
2'TT
7T
1
-Vs (a)
I I I I
I
0.75
I I I 1
"onm
wt
-
"01 m
0,50
I
I
I
I I
I..
I
I
0,25
I I
2'11'
wt 45' 90' 135' 180' Pu ls e Width (2d) in d~ gr e~ s _ (b)
(c)
Fig. 8.28 , (a ), (b ) Single-pulse m odul ati on (SPM) (c ) Harmonic content in SPM ,
456
Power Electronics
[Ar t. 8.6]
Ifnd is made equ al to 7t or d
=~n or ifpulse width is J1lade equal to 2d = 27t, Eq. (8.62) shows . . n
t h at nth harmonic is eliminated from the inverter output voltage. For example, for eliminating 7t third harmonic, pulse width of 2d must be equal to 23 = 120°. The peak value of nth harmonic, from Eq. (8.62), is. 4. Vs
vonm
From Eqs.(8.64) and (8.65),
.
=-n7t- sm nd
Vonm
sin nd
VOlm
n
...(8.65) ...(8.66)
In Eq. (8.66), note that vOlm is the peak value of the fundamental component of square voltage w.avefrom of width 2d = 7t. The ratio as given by Eq. (8.66) is plotted in Fig. 8.28 (c) for n ~ 1 (plot of sin d), n = 3 (plot of sin 3d/3), n = 5, 7 for different pulse widths. It is seen from these curves that when fundamental component is reduced to 0.5 for 2d = 60 0 , the amplitude of third harmonic isj sin 90 = 0.33. When fundamental component is reduced to about 0.143, all the three harmonics (3, 5, 7) become almost comparable to the fundamental. This shows that in this method of voltage control, a great deal of harmonic content is introduced in the output voltage, particularly at low output voltage levels. The rms value of output voltage, from Fig. 8.28 (b), is
Yo,.
=
[V; 7t.2d]1/2=V.., [~dll/2 7t
...(8.67)
8.6.2. MUltiple-pulse Modulation This method of pulse modul a tion is an extension of single-pulse modulation. In multiple-pulse modulation (MPM), several equidistant pulses per half cycle are used. For simplicity, the effect of using two symmetrically 'spaced pulses per half cycle, Fig. 8.29 (a), is investigated here. In this figure, pulse width is taken half of that in Fig. 8.28 (b), but their amplitudes are the same. This means that rms values of pulses in Figs. 8.28 (b) and 8.29 (a ) are equal to that given in Eq. (8.67). For the waveform of Fig. 8.29 (a), Fourier constants are as under: bn
. =:;2 flt0 Vo sin nc.ot . d(rot) 2
=:;
f
(y + d/ 2)
.
(y- dl2 ) Vs . sm nc.ot . d(rot) . 2
The use of fact or 2 in the above expression accounts for the two pulses from 0 to 8.29 (a
wt 2.".
(a )
(b)
Fig. 8.29. Symmetrical tw o-pulse modulation pertaining to MPM.
7t
in Fig.
[Art. ' 8.6]
In verters
I
Y d12 4V~1 . -nd - cos n wt y...- d/2 = -8Vs. SIn ny sm bn -- n n' n1t 2
As in Fig. 8.28
(b), an = ~9
457 ... (8.68)
in Fig. 8.29 (a) also.
Therefore, the wavefornf of Fig. B.29 (a) can be described by Fourier series as
Vo
B Vs. . -2 nd sm . nrot =L -sm nysm n1t
n
= 1,3.5
8 Vs ,[ .
or
(B .69)
.. ,
. 3d. . 5d . ] '2 sm wt + '1. 3 sm 3y . sm 2" sm 3 wt + 5'1. sm 5y sm 2" sm 5wt + ...
. d.
va = ---;- sm y sm
...(B.70)
The amplitude of the nth harmonic of the two-pulse waveform of Fig. 8.29 (a), from Eq. (B.69), is
B Vs . . . nd v =- -smny · sm ~ n n1t . 2
... (B.71)
Eq. (B.71) shows that magnitude of Vn depends upon y and d. This expression also shows that when Y= ~ n or d = 21t, n nth harmonic can .be eliminated from the output voltage. But this has the effect of reducing the fundamental component of output voltage. For example, take pulse width 2d = 72° for single-pulse modulation of Fig. B.28 (b). Then, from Eq. (B.65), the peak value of fundamental voltage component is 4V . .
v01m =__s sin 36° = 0.7484 Vs ,
1t
For two-pulse modulation and pulse width d = 36°, y in Fig. 8.29 (a), is
or in general.
Y= 180 - 72 + 72 = 540 3 4 1t-2d 2d 1t-2d
y-
- N+1
d
+- +2N- N+1 N
...(8.72)
Eq. (8.72) is valid in case pulses of equal width are symmetrically spacer\. Here N is the number of pulses per half cycle. Eq. (8 .72) can also be obtained by referring to Fig. 8.29 (b). For N pulses per half cycle, th ere are (N + 1) intervening equidistant spaces, each of width 91 as shown in Fig. 8.29 (b). Note th at for these equidistant spaces, va = O. Total width of these (N + 1) equidistant spaces ~ (N + 1) 91 ~ (1t - width of N pulses) = (1t -2d) . 1t-2d 91 = N + 1
or
Fig. 8.29 (b) shows that 92 = h alf of the pulse width = ~. This figure also reveals that y~
9 1 + 82
1t-2d
or
d
y= N+ 1 + N
Fe
val e of fundam ental voltage component, fr om E q . (8.7 1), is v0 1m
8 V. = ~ sin 54 sin 1B" = 0.637 Vs,
.. .(8.72)
458
Power Electronics
[Art. 8.6]
It is seen from above that fundamental component of output voltage is lower (0.637 V s ) for two-pulse modulation than it is for single-pulse modulation (0.7484 Vs )' It can be shown that for more number of pulses per half cycle, the amplitudes oflower order l?armonics are reduced but those of some higher harmonics are increased significantly. But this is no disadv!'l.ntage as higher order harmonics can be filtered out easily. The symmetrical modulated wave shown in Fig. 8.29 (a) can be gener~ted by comparing an adjustable square voltage wave Vr of frequency ro with a triangular carrier wave Vc of frequency roc as shown in Fig. 8.30 (b). This comparison is done in a comparator, Fig. 8.30 (a) . In Fig. 8.29 (a), there are only two pulses per half-cycle but in Fig. 8.30 (b), there are four Carrier signal Yc.freq we Reference signal Yr. freq. W
Tr iangular wave
i
lIfe
r wt
Trigger Trigger pulse generator pulses to SCRs
Comparator
(a)
, .. ,, .,'I :'.',' .., ·, : 2T i- r- . , , ,, ,· ,, ,
.
, ,
'10
Square wave '
I
Ys r-
;
1
,.
I
I
.. .
I
I
I I I
:
,, ,,,,
, ,,, I
I
I
I
: ,I , :,
,
I ,,, ,,, ,
I
I
I
I
,
1, I ,, ,
,
I
I
~
,
I
1 (c)
-
.... .....
I
I I I
II :;1""
,, Ult
,
I
1T:
!:
I
.,, I
..... .. :: 'I'l'
. (b)
Fig. 8.30 . (a ) Pertaining to multiple-pulse modulation (MPM) (b ) Output voltage wavefonn with MPM (c) Vc and Vr shown on a larger scale.
pulses ' per half cycle. The triggering pulses for' thyristors are generated at the points of
intersection of t he carrier and referen ce signal waves. The firing pulse.. so generated turn-on
the SeRs so that output voltage Vo is available during the interval triangular voltage wave
exceeds the square modulating wave shown in Fig. 8.30 (b) . In this figure, fc and f are the
frequencies in Hz for the carrier signal and reference signal respectively. This figure reveals
that
A= ~ and ~f = 1t and the number of tr igger pulses is ~ x i = 4. In general, the number of
pulses generat ed per half cycle can be determin ed from Fig. 8.30 (b ) as under: .For t riangular carrier wave, pulse width :;:
A.
For square r eference wave, width of h alf-cycle
=
if'
... Number of pulses per half-cycle, N = Number of hill-t ops per half-cycle, N = Length of h alf-cycle of squar e reference wave Width of 0 e cycle oftriangular carrier wave
.... .... \
,[Art. 8.6]
Inverters
N
or
= 1/ 2 f == fc ;:: we life
2f
459
... (8,73)
2w
Note that N in Eq . (8. 73) must be an integer. The pulse height of the reference, or modulating, signal can be controlled within the range 0 < V r < Ve and pulse width the range 0 <
Z; varied in
Z;; < ~ by adjusting the magnitude Vr of the reference square wave. The pulse
width is 2dl N on the ass umption of same rms voltage as in single-pulse modulation. In Fig. 8.30 (b), pulse width 2dl Nis given by
~~(~-2x] A general expression for the pulse width can be obtained by sketching the first cycle of carrier signal on a larger scale as in Fig. 8.30 (c). From this figure, pulse width, in general, is given by
. where
x,
~~ (t-2x)
defined in Fig. 8.30
(c) ,
Ve 1t1 2N
or
... (8.74)
.
is Vr
=-;
n Vr 2N Ve
x=_ ·
From Eq. (8.74), the pulse width is
~ ~ (~-~~)=(1-~)~
..
(8.75)
In MPM method, lower order harmonics can be eliminated by a proper choice of 2d and y. But the rms voltage in Figs. 8.28 to 8.30 is the same, i.e.
_ (2d]1I2
Vo r - Vs
1t
This means that if lower order harmonics are eliminated, the magnitude of higher order harm onics 'l<'l ould go up . But this is not a disadvantage, as higher order h armonics can be filtered out by the use of filters at the output terminals of the inverters. 8.6.3. Sinusoidal-p ulse M odu lation (SPWM) In this meth od of modu lation , several pulses per h alf cycle r e used as in the case of multiPle-pulse modulation (MPM). In MPM, the pu se width is equal for all the pulses. But in SPWM, the pulse width is a sinusoidal function of th e angular position of th e pulse in a cycle as shown in Fig. 8.3l. For r ealizing SPWM, a h igh -fr equency triangular carrier wave Vc is compared with a sinus oid al refer ence wave v r of t he desir ed fr equency. The in t ersection of Ve and vr waves det ermin es the switching inst ants and commu tation of the m odul ate ~ pulse . In Fig. 8.31, Vc is the peak valu e of triangular carrier wave and V,. th at of . e refer ence, or mod'ulating, signal. The carri er and referen ce w av ~s ar e mixed in a r;ompar a tor as in Fig. 8.30 (a). When sinusoidal wav e has magnitude high er th an the triangul ar wave , the comparator outpu t is
:- ':' '.\~ ' .
~
..
460
Power Electronics
[Art, 8.6] Reference wove, freq. f Carr ier wove, treq. Ie
wt
(a) Carrier wave,freq . fc Reference wave, freq. f
wt 277
f---+--+-'-, 1/2 f ~"""""'-+~...-..I , 110
I
Vs
wt 2rr
(b) Fig. 8.3l. Output voltage waveforms with sinusoidal pulse modulation.
high, otherwise it is low. The comparator output is processed in a trigger pulse generator in such a manner that the output voltage wave of the inverter has a pulse width in agreement with the comparator output pulse width. When triangular carrier wave has its peak coincident with zero of the reference sinusoid, there are N =
~fPu1ses per h alf cycle; Fig. 8.31 (a) has five pulses. In case zero of the triangular
wave coincides with zero of the reference sinusoid, there are (N - 1) pulses per half cycle; Fig. S .31 (b) has
(;r-
1} i. e. four, pulses per half cycle. .
The ratio of V,l V c is called the mo dulation index (MD and i t controls the hannonic content of the output voltage waveform. The magnitude of fundamental component of output voltage is proportional to MI, but MI can never be mor e than unity. Thus the output voltage is controlled by varying MI . Harmonic analysis of the output modulated voltage wave reveals that SPWM has the following important features : (i) For MI less t han one, largest h armonic amplitudes in th e output voltage are associated \Vith harmonics of or der fel t ± 1 or 2N ± 1, where N is the n umber of pulses per h alf cycle. Thus , . by increasing he n mber of pulses pe a1f cycle, ilie order of dominant h armonic fre quency
Inverters
[Art. .8.6]
~ 61
can be raised, which can then be filtered out easily. In Fig. 8.31 (a), N ::; 5, therefore harmonics of order 9 and 11 become significant in the output voltage. It may be noted that the highest order of significant harmonic of a modulated voltage wave is centred around the carrier frequency fc [in Fig. 8.31 (a),{c ::; 10]. It is observed from above that as N is increased, 'the order of significant harmonic increases and the filtering requirements are accordingly minimised. But higher value of N entails higher .switching frequency of thyristors. This amounts to more switching losses and therefore an impaired inverter efficiency. Thus a compromise between the filtering requirements and inverter efficiency shoUld be made. (ii) For MI greater than one, lower order harmonics appear, since for MI::' 1, pulse width is no longer a sinusoidal function of the angular position of the pulse. In addition to the three PWM techniques discussed above, there is another PWM technique called mUltiple-pulse modulation with selective reduction (MPMSR). In this technique, the number of M pulse positions in each quarter cycle are so selected as to reduce or eli!llinate M harmonics from the output voltage waveform [6]. This PWM technique will, however, not be discussed here. 8.6.4. Realization of PWM in Single-phase Bridge Inv erters The output voltage waveforms shown in Figs. 8.28 to 8.31 reveal that output voltage from an inverter is V s, zero or - Vs' Such waveforms can be realized in single-phase inverters as under: . Single-phase full.bridge inverte r. In the inverter of Fig. 8.2 (a), when + Vs is to be obtained in the positive half cycle, thyrist ors T l , T2 are turned on. For obtaining - Vs in the negative half cycle, thyristors T3, T4 should be turned on. For zero output voltage, i.e. if the load . is to be short~circuited ; then Tl, D3 or T3, Dl from positive group; or T4, D2 orT2, D4 from negative grou p should conduct depending upon the direction of load current. This means that for obtaining zero output voltage at the end of each pulse, one of the two conducting SeRs . should only be turned off. Under t his strategy, only one thyristor n eed be turned on for obtaining the next voltage pulse. Switching on and commutation of thyristors should be so arranged as to utilize the thyristors symmetrically. Let us illustrate this with an example. Suppose output voltage ofpulse width 21t1_3 radians is to be obtained in each half cycle. This pulse width is symmetrically placed as shown in Fig. 8.32. The waveform of load current io is (a)
Vo
Vs
A
2'lT -3
'TT
-
6
B
2'"
"3
I
: io 0
UJt I
( a) ( b)
jT2.~~ T"T2 n.OJ~TI .72
I
J
I
'
",;Tl,03 ;T3.01!--- n , T4 ~:i2,OIoiT.4.02~ TJ ,T"
Fig. 8.32. C nduction of various components
fOT
I
-.:T4.o2:I T2,O"f.. , I
--;13.01:11.031.,.
single-phase bridge inverter of Fig. B.2
(a).
462
Power Electronics
[Art. 8.6)
assumed as sketched in Fig. 8.32. It is obvious from these two waveforms that from B to C ; Tl, T2 should conduct and from E to F ; T3, T4 should be on. From C to D, Vo = 0 but current io is positive. Therefore, from C to D ; either Tl, D3 orT2, D4 should conduct, this is shown in Fig. 8.32. From D to E, Vo = 0 but io is negative. Negative current with zero output voltage can exist only if T3 or T4 together with one diode are on. When T3 is on, then T3, Dl should conduct and with T4, D2 should conduct, Fig. 8.32. From F to G, Vo =0, io is negative. For this ; T4, D2 or T3, DI shoul4 conduct. From D to E, if T3, DI conduct, then now T2, D4 must conduct in order to utilize the thyristors symmetrically. In case T4, D2 conduct from D to E, then T3, DI should conduct from F to G. From G to H ; T2, D4 or TI, D3 conduct as shown. The conduction from G to H is similar to that from A to B. It may be observed from Fig. 8.32 that, during one cycle, each thyristor conducts for 1500 and each diode for 30 0 • (b)
Sin gle-phase half.bridge inverter. In single-phase half-bridge inverter of Fig. 8.1
(a), zer o value of output voltage cannot be obtained. The output voltage can either be
V/2 or - V/2. In Fig. 8.33, V/2 from A to B is obtained with TI on, from B to C with T2 on, from C to D with T1 on and so on. For obtaining a symmetrical waveform for output voltage in Fig. 8.33, intervalAB =interval DE; intervalBC =interval EF and so on. The output voltage can be controped through the adjustment of width 2d. .,
.....
o A
r ~ 2"
• 'IT
_.
8
_..... _.... __.
"--T
r-
2d i'-
C
E
0
F
wt
G
~
t.-...
-J
Fig. 8.33. Output voltage waveform obtained through PWM in half-bridge inverter.
Example 8.9. A single·phase bridge inverter, fed from 230 V dc, is connected to load R = 10 nand L = 0.03 H . Determine the power delivered to load in case the inverter is operating at 50 H z with (a) sq uare wave output (b) quasi-square wave outp ut w ith an on·period of 0.5 of a cycle and (c) two symmetrically spaced pulses per h alf cycle with an on-period of 0.5 of a cycle.
Solution. In order t o calculate t he power delivered t o load fairly accurately, harmonics up to seventh may be considered. (a )
Square-wave output: From Eq. (8.26), rms value of fun dament al voltage is V
01
= nT2 4Vs = 4 x 230 =207 10 V 1t . T2 .
Load impedance at fun damental frequency is Z 1 = [10 2 + (21t
X
50
X
0.03)2]112 = 13 .7414 .Q
[Art. 8.6]
Inverters
463
. 207.10 101 = 13.7414 =15.0712 A 4 x 230
V03
= 3 x 7t X T2= 69.035 V
Z3 = ...J102 + (27t
and
X
50 x 3 x 0;03)2 = 29.9906 n
69.035
Similarly,
103
= 29. 9906 = 2.302 A
1
=
05
1 07 -
920 x 1 5 x 7t X T2 ~102 + (21t x 50 x 5 x 0.03)2
= 0.8598A
920 x 1 - 0 4434 A 7 x 1t X T2 ~102 + (21t x 50 x 7 x 0.03)2 - .
Rms value of resultant load current,
2
2
2
2
10= [ 101+103+ 1 05+ 1 07
J1/2
Power delivered to load = 1~ R
[15 .07122 + 2.302 2 + 0.85982 + 0.44342] X 10
= 2333.76 W
(b) Quasi-square wave output: For quasi-square wave or single-pulse modulated wave, use Eq. (8.62), where pulse width, 2d = 0.5 x 180° = 90° or d = 45°. From this equation, rms value of fundamental voltage is . =
4V
VOl =
.
~ sin d = 47t~ ~o sin 45° = 146.423
101
= 13.7414 =
V03
= 3 4xx1t 23&sin 3 x 45° = 48.8075 V X .
10.6556 A
48.8075
Similarly,
146.423 V
103
= 29.9906 = 1.6274 A
105
= 5~: :3&- sin (5 x 45°) x 48. 1; 324' = -
•
•
·0
0.6079 A
. (7 450) · 1 - 0 313- A 107 -- 7 4x230 x 1t x ,,'2 sm x x ,~6 . 727 - - . ::>
2 2 Power delivered to load =(10.6556 + 1.6~7 42 + 0.607g2 + 0.3 135 ) x 10
= 1166.58 W
(c) F or two symmetrically spaced pulses per half cycle, use Eq. (8.69). For t his equation , d . . .180 - 90 45 2 = 0.5 x 180 = 90° or d = 45° and from Eq. (8.72), Y= 3 +"2 = 52.5°. From Eq. (8.69), rms value of fund ament al voltage is . d 8Vs . 8 x 230 . 52 50 . 45 125 755 V V·01 = ~ sm y sm 2" = 7t . T2 sm . sm """2 == .
125.755 1 01
= 13.7414
= 9.1515 A
.
. (- 2 5 3) . -sm . V 03 = 3 8x x1t 230 X fi . sm D . X
(45 2 x 3)
=
48 .8
~V
D
464
Power Electronics
[ArL 8.7]
103
48.815 . = 29.9906 = 1.6277 A
105 ::; 107
Power delivered to load
::
:~ sin (52.5 x 5) sin (22.5 x 5) x 48.1~324 =1.575 A
= 7 ~:~3~ sin (52.5 x 7) sin (22.5 x 7) x 66 .~27 =0.0443 A = (9.1515 2 + 1.62772 + 1.5752 + 0.04432)
X
10
= 888.82 W.
8.7. REDUCTION OF HARMONICS IN THE
I~RTER
OUTPUT VOLTAGE
There are several industrial applications which may allow a harmonic content of 5% of its fundamental component of input voltage when inverters are used. Actually, the inverter output voltage may have harmonic content much higher than 5% of its fundamental component. In order to bring this harmonic content to a reasonable limit of 5%, one method is to insert filters between the load and inverter. If the inverter output voltage contains high frequency harmonics, these can be reduced by a low-size filter. For the attenuation of low-frequency harmonics, however, the size of filter components increases. This makes the filter circuit costly, bulky and weighty and in addition, the transient response of the system becomes sluggish. This shows that lower order harmonics from the inverter output voltage should be reduced by some means other than the filter. Subsequent to this, high frequency component from this voltage can easily be attenuated by a low-size, low-cost filter. The object of this section is to study these methods of reducing low-order harmonics from the output voltage of an inverter. ·
8 .7.1. Harmonic Reduction by PWM I t has alr eady been discussed that when there are several pulses per half cycle, lower-order h armonics are eliminated. Fig. 8.34 illustrates output voltage waveform that can be obtained from a single-phase full- bridge inverter. This waveform can also be obtained from a single-phase half-bridge inverter, but then the amplitude of voltage wave would be V/2. The waveform of Fig. 8.34 needs ten commutations p er cycle (= 360°) instead of two in an unmodulated wave. The voltage waveform of Fig. 8.34 is symmetrical about 1t as well as 1t/ 2. Vo
2
3,
Vs
0
4
~
~ 7
."/2 .
8, 9
10 ,
wt
3Tr/2
- Vs
Ct2 ...J
Fig. 8.34. Harmonic reduction by PWM in single-phase inverter. As this voltage waveform has quarter-wave symm etry, an = O. 4 bn = -1t V s JCOl _ sin n wt . d (oot) -
o
f2 sin Cl. !
1 - 2 cos na~ + 2 cos nCl.2 }
nrot . d (wt)
1
+ f~/2 . sin n wt . d (oot) Ct z .
...(8. 76) ...(8 .76 )
Inverters
[Art. 8.7]
465
If third and fifth harmonics are to be eliminated, then from Eq. (8.76),
4Vs [1 - 2 cos 3 a l b3 = -
3
1t
= 4Vs [1- 2 cos 5a l + 2 cos 5~] == 0
b
and
+2 ~os 3 a 2] =0
5
5 1 - 2 cos 3a.l + 2 cos 1 - 2 cos 5a l + 2 cos
1t
or and
I
3~
=0 5~ = 0
The above two simultaneous equations can be solved numerically to calculate a l and ~ under the condition that 0 <
and
I':
= 0.3867 Vs
. The amplitude of the fundamental component for these values of a l and a 2 is
i
4Vs b1 =[1 - 2 cos 23.62 + 2 cos 33.304] :;:: 1.0684 Vs
n.
The amplitude of the fundamental component ofunmodulated output voltage wave is I~
bl
I
4Vs
.n
=- 1t
=
1.27324 Vs
In terms of the fundamental component of unmodulated voltage wave, the amplitude of 7th, 9th and 11th harmonics are respectively 24.78% (= 0.31555 x 100/1.27324),40.86% and 30.37% but third and fifth harmonics are eliminated from the inverter output voltage wave . The amplitude of. the fundamental voltage is 83 .91% or 0.8391 times the amplitude of fundamental component of unmodulated voltage wave. Thus, with this method of harmonic reduction, inverter is derat ed by (100-83 .91) 16.09%. Another disadvantage of this method is that there are additional eight commutations per cycle and this leads to more switching losses in the thyristors. 8.7.2. Harmonic R educt ion by Transformer Connections Output voltage fro m two or more inverters can be combined by means of transformers to get a net output voltage with reduced harmonic content. The essential condition of this scheme is that the output voltage waveforms from the inverters must be similar but phase-shifted from each other. Fig. 8. 35 (a) illustrates two transformers in series. Their output voltages, VOl from inverter 1 and V 02 from inverter 2, are shown in Fig. 8.34 (b). Here V 0 2 waveform is taken to h ave a phase shift ofn/ 3 rad ' ans with respect to VOl waveform as shown. The resultant output voltage Vo is obtained by adding t he vertical ordinates of VOl and V 02 . It is seen that Vo has an amplitude oL2Vs from wave
Vo
j to
is a quasi-square wave .
71:,
~ t o 2n and so on . Note that sh ape of the
output voltage
466
Power Electronics
[Art. 8.7]
1:1 2
t
Inverter I
vol
I ... :!'t--
DC
L 0 A 0
Input
'
Vs
.
/
·,· ,
/ I I
,,I
I
I I
,
,, ,
I
,
·
Vo
I
:
,
: 4rr ')
rr
,,I
, /
/
/
'.
.,.2Vs
..,. rr: 3~
1 wt
,
I
. /
vo2
IT
,
;
/
3 t -v 5~' .
t
Inverter
.
, ,/ ,I
,,,
.,: ;
wt
2rr '" 7rr
3"
1:1
Fig. B.35. (a)
- 2Vs (a) . ' . . (b) . Harmonic reduction t-v transfonner connections (b) Elimination .ofthir1::l ;. ,.." and other triplen harmonics. ....~, :.. '< ~ ;t:::::..:. ,.. ". .. .• 4 .
The Fourier analysis of waveforms VOl and v 02 gives
-
vOl
4V [ '. 1sm ' 3rot + 5' 1sm ' 5rot + 7 1sm ' 7rot + = --;8m rot +"3
U02
=
$.
--;-lr~nn. ( 4V$
rot -
.
...
:~
.
]
l
. 3( 1t) 1 . 5 ( rot - '31t 1j '37t) + '31 sm rot - '3 + 5' sm
+~ sin 7 ( ~ +] Wi -
The resultant voltage
Vo
is
)
= vOl + v 0 2
~ 4:" €[sin (ott -~)+ i sin (5oot +~J+ ~ sin (7 o~ - ~)+]
..
18.771
The expression for resultant voltage Vo as given above can be obtained from VOl and analytically or graphically. The summation by graphical method is carried out as under: .
V02
An examination of the expressions for VOl andv02 reveals that for the fundamental frequency; V 02 lags VOl by 60°, this is shown in Fig. 8.36 (a) . The resultant of VOl and V 02 mw~t be ..J3 times VOl (or V 02 ) and at the same time, the result~nt lags VOl by 30"'. Net value of fundamental frequency voltage. must, therefore, be associated with fi sin (rot - 7t/6). For third Vol
..
...
Vo2
Freq . W (a)
Vol
\
Freq.3w
Freq.5w
Freq.7w
(b)
(c)
(d )
Fig. 8.36. Pertaini ng to the summ ation of first, third, fifth and seventh harmonic voltages.
.
Inver ters
467
[Art. 8.7]
harmonic, V 02 lags Val by 180°, Fig. 8.36 (b), their resultant is therefore zero. For fifth harmonic, Vo2 lags Val by 300° or V02leads Val by 60°, see Fig. 8.36 (c) ; its resultant is V3 times VOl or V 02 and it leads VOi by 30°. Thus, the resultant of fifth harmonic voltage must be associated with ,f3 sin (wt + 1t/6). Similarly, resultant of seventh harmonic voltage must be associated with ,f3 sin (rot - 1t/6), Fig. 8.36 (d) .
. It is seen from Eq. (8.77) that third and other triplen harmonics are eliminated from net output voltage wave. The amplitude of fundamental component ofu o is 4Vs _In V olm == -"V 3 1t
In case output voltages UO I , U02 from invert~rs 1 and 2 has no phase shift, then amplitude of the fundamental voltage wave is 8 V/7t. This shows that with phase shift, the amplitude of the fundamental voltage is 4~s {3 x .
8~ = ~ times the amplitude of fundamental voltage v.ith no s
phase shift. With this method ofharnionic reduction, there is thus a derating of
8Vs
12"" I -
4 {3
-; Vs ' 3 I (~ . 8V I 100 == 1 - 2) 100 == 13.4%
'
s
1t
.
in their net output power so far as fundamental component is concerned. The degree of . derating with this method is, however, less than that obtained in PWM harmonic reduction method . .. The disadvantage of this method of harmonic reduction is the need for more number of inverters and transformers of similar ratings .
8.7.3. Harmonic Reduction by Stepped-wave Inverters In this method, pulses of different widths and heights are superimposed to produce a resultant stepped wave with reduced harmonic content. Fig. 8.37 illustrates two stepped-wave inverters fed from a common dc supply. The two transformers used have different turns ratio from primary to secondary. In this figure, the turns ratio from primary to secondary is assumed three for tran~former 1 and unity for transformer 2.
vo
,+
1:3
I
Inverter
-
I .
I
trV 111001
t
Inverter II
· ~W I 10 i j
IVoZ
iI I
, :1
Fig. 8.37. Harmonic reduction by stepped wave inverters.
wt'"
7T
(a)
vo 2AI 1
i
;
V~ II
" Olp vo~
I i
•
1
l
'"!2
I:
I
CJ -r;
I
I
b
wt ()
:
r---:---.'- ._ . _ .-:- ~ ~
'--.r-
-+--~~---,:
-"-i - I - o - - - - lis ! ~
TOI
0
:
(I
: 4 Vs I
I ,
'71
w 't . (c)
Fig 8.:38 . Wa veforms fo r stepped-w ave inverters
468
[Art. 8.8]
Power Electronics
The inverter I is so gated that its output voltage is vOl as shown in Fig. ~.38 (a). During the first-half cycle, output voltage level is either zero or positive. During S' ~cohd'half cycle (not shown in the figure), the output voltage would be either zero or negative .',This output voltage waveform is given the name two-level modulation. For inverter II, the triggering is so arranged as to give output voltage V02 as shown in Fig. 8.38 (b). It is seen from V02 waveform that the level of output voltage is positive, negative or zero during the first half cycle, this inverter has therefore three-level modulation. The resultant output voltage from a series combination of inverters I and II is obtained by superimposing the waveforms of Figs. 8:38 (a) and (b). This summation depicted in Fig. 8.38 (c) shows that the amplitude of output voltage is 4 Vs and waveform has four steps. Fourier analysis of Fig. 8.38 (c) would give harmonics whose amplitudes would depend upon the values of d l , d 2 , d 3 , d 4 and amplitude of vo. By a proper choice of these parameters; third, fifth and seventh harmonics can be eliminated or attenuated considerably and the fundamental component optimised. Note that it is the three-level mod ulation of second inverter that helps in achieving the required wave-stepping of the resultant output voltage waveform. It is seen from the waveform of Fig. 8,38 (c ) that this waveshape is more nearer to sinusoidal wave.
8.8. CURRENT SOURCE INVERTERS So far, voltage-source inverters have been discussed. In these inverters, input voltage is maintained constant and the amplitude of output voltage does not depend on the load. However, the waveform of load current as well as its magnitude depends upon the nature of the load impedance, In the current-sour e inverters (CSls), input current is constant but adjustable. The amplitude of output current from CSI is independent of the load. However, t h e magnitude of output voltage and its waveform output from eSI is dependent upon the nature of load impedance. The dc input t o CSI is obtained from a fixed voltage ac source through a controlled r ectifier bridge, or through a diode bridge and a chopper. In order that current input to CSI is almost ripple free, L-filter is used before CSI. A eSI converts the input dc current to an ac current at its output terminals. The output frequency of ac current depends upon the rate of triggering the SeRs. The amplitucl!e of ac output current can be adjusted by controlling the magnitude of dc input current. A eSI does not require any feedback diodes, whereas these' are required in a VS!l:. Com mutat ion circuit is simple, as it cont ains only capacitors. As power semi-conductors in a CSI have to withstand reverse voltage, devices such as GTOs, power transistors, power " MOSFETs cannot be used in a eSI. The CSIs find their use in the following applications: (i) Speed control of ac m otors (ii) Induction heating
(iii) Lagging VAr compensation (iv) Synchronous motor starting. In this section, basic principles of single-phase eSI are consider ed fir st. Then single-phase forc e-commutated and a uto-sequential commutat ed CSIs are described.
8.8.1. Single-phas e CSI with Ideal Switches A singl e-phas e CSI with ideal thyrist ors is shown in Fig. 8 .39 (a ). Here a thyristor i:;; ass um ed an ideal swit ch vvith zero comm utation time. Positi ve directions for load voltage Vo and
Inverte rs
[Ai-t. 8.8]
469
Current input to C51
0 to
r 0
I
,T/2
t
~T I
-I
I
Tm: nT4' TlT2 TlT4:, "
Vo
1+
.
:IE
I-
i
i (a)
(b)
Fig. 8.39 . (a) Power circuit diagram and
(b)
waveforms for an ideal single-phase CSI.
load current io are indicated in Fig. 8.39 (a ). The source cOilsists of a .voltage source E and a large inductance L in series with it. The function of high-impedance reactor in series with volt age source is to maintain a constant current at the input terminals of CSI. In other words, dc input current I to CSI is constant as shown in Fig. 8.39 (b). In Fig. 8.39 (a ), when Tl, T2 are on, load current io is positive and equal to I. When T3, T4 are on, load current io is negative and equal to - I as shown in Fig. 8.39 (b). The output frequency of io can be varied by controlling the frequency of triggering the thyristor pairs Tl , T2 and T3, T4. It is seen from Fig. 8.39 (b) that output current io is a square wave of amplitude equal to the dc input curr ent I . Assume that load consists of a capacit~r C. It is known for a capacitor that dvo
io = C dt As io is constant,
slope~tO must be constant over every half cycle. This slope is positive
fro m zero to T 12 and negative from T 12 to T. On this basis, waveform of load voltage V o is drawn in Fig. 8 .39 (b ). The input voltage to the CSI, i.e. Vin = Vo when Tl, T2 conduct and Vin = - Vo when T3, T4 conduct. Note that 'j ~veshape of vin can be drawn by referring to the waveform of vo' For f= l i T as the fr equency of output voltage Vo or current, input voltage vin. has a frequency of2f, this is revealed by an examination of Fig. 8.39 (b ). The dc current 1, input t o CSI. is always unidirectional. If average value of vin is positive, power fl ows fr om source to load . In case average value of v in is negative, pow er flows from load
to s()urce, i,e. regeneration of power takes place. In a practi cal inverter, the load current waveform is not a square wave owing to t he fact that rise and fall of current can 't be in ~tant an e o us as sh own in Fig. 8.39 (b ). On account of finite commutation time , a pr actical inverter h as finite times for the rise an fall of curre t.
470
[Art. 8.8]
Power Electronics
eSIs may be load or force commutated. Load commutatioIi is possible when loadpf,i.s.....; . ' leading. For lagging 'pfloads, forced-commutation is essential. Here single-phase eSIs using' . ,.... forced commutation are studied. Use of commutating capacitor is an important featur~ of force-commutated eSIs. . . 8.8.2. Single-phase Capacitor-Commutated CSI with R Load As stated above, all force-commutated eSIs need capacitors for ihejrcommutatioIi. Here the term 'capacitor commutated' is used just to distinguish this eSI from other eSIs. Power circuit diagram for a single-phase eSI with resistive load R is shown in Fig. 8.40 (a). The source for this inverter is a constant but adjustable dc current source. Capacitor C in parallel with the load is used for storing the charge for force-com mutating the SCRs. The thyristors TI to T4 are the four power switches. These SeRs are gated in pairs; T1, T2 together by gating signals ig1 . ig2 and T3, T4 by i g3 , ig4 as shown in Fig. 8.41. Positive directions for load current ia and load voltage va are marked in Fig. 8.40 (~). . . I "t'
T3
t
I
8
Yin
T2
;"1
0: ~
.
IT/2
b
i.e
(b) C
--------I" tc
A
---l~
C
1
ve= '10
___ J. B
a
d ~)
(e)
"1
T1 ,T2.1.TJ.T4-l- T1 ,T2-l- fl,T ,J
(a)
I
I F
IT
d
"1
1
------- +
C
-:'l~ vc=vo
VI
-.=.-~c
b
~
Fig. 8.40. (a) Power circuit diagram of 1-$ CSI with R load (b) AC output current waveform Equivalent circuit of Fig . (a) for 0 < t < TI2 and (d) Equivalent circuit of Fig. (a) for TI2 < t < T.
Before t = 0, let the capacitor voltage be Vc = - Vl, i.e. capacitor has left plate negative and right plate positive in Fig. 8.40 (a). When T1, T2 are gated at t = 0, th'e capacitor voltage Vc reverse biases conducting thyristors T3, T4; these are therefore commutated immediately. The source current I now flows through TI, parallel combination of R r.n d C and through T2. From zero to T 12, iTl = iT2 = I, output current iac = I ; capacitor voltage Vc changesJrom - V l to Vl through the charging ofC by current i c ' Note that here load voltage Vo = Vc' Thus, the waveform Va Vc
of ia = R = R h!'ls the same nature as that of vc' see Fig. 8.41 . When T3, T4 are gated at t = T 12, Uc = Vl r everse biases TI, T2 ; these are therefore turned-off immediately. The source C Trent now flows through T3, parallel combinati on oiR, C and T4. From T 12 to T, i T3 = i T4 = + I but iac = - I . The variation of ac current iac is shown in Fig. 8.40 (b).
Un der stea dy state oper ation of the CSI, various current an d voltage waveforms are sketched in Fig. 8.41. At t = 0, when the capacitor is charged with volt age Vc = - VI' then
[Art. S.lS]
Inverters
.
U
471
VI
o = Vc = - VI and load . current io = - -R1 = -11 ' From t = 0 to T I Z, capacitor charges frum
. T v - VI to VI Therefore, at t = 2' io = ~
v
V
= ; = Rl = 11' This is uu, = Vo from t = 0 to TI2 whereas V ifL = - Vo from T/2 to T.
shown in Fig. 8.41. Input voltage
It is seen from Fig. 8.40 (a) that when Tl, T2 are conducting for 0< t < T/2, currents it' io are leaving the node A and current I is entering the node A. Therefore, equivalent circuit for
Fig. 8.40
(a)
for 0 < t < ; may be drawn as shown in Fig. 8.40 (c). KCL at node A in Fig. 8.40
or at node b in Fig. 8.40 (c) gives
io + ic =1. or g
i.;,i '[1-------'------.
c
• t
I I
ig3,tg41L____~=====L
____.t:====::I.___,___ L
.I
. lT1.lT20t
• t
I
[ ,.t :l_===t_~-T===~~______~J, ________~ I ______~L-~ ~ T1 ,T2 -.:.-- T3, T4
.,. !.
T1,T2
.1.
T3,T4
--J
Vo,10
Or---------~~------~~r_------~--------~~~
Vin
0
-V,
-
I
vT1 .
I
VT2
r
+V1 0 ~--------~~~--~--------~r_~~--~--t
-v, t
\
T1 T2
J
t
T3 TL.
t
T3 T4
Fig. 8.41. Current and voltage waveforms for single-phase CSI with R load .
(a)
472
Power Electronics
[Art. 8.8]
At t = 0, io = - II' therefore, ic =I + II' Just before T 12, io =II' therefore ie =I - II' This is shown in Fig. 8.41. Just after T 1 2, when T1, T2 are off and T3, T4 are conducting, currents io, ic win continue flowing in the same direction. AB such, all the three currents io, ie' I enter the node B in Fig. 8.40 (a). This gives the equivalent circuit of Fig. 8.40 (d). KCL at node B of this figure gives ' io + ie + 1=0 or ie = - I - io At t = (T12) + , io =+ I}> therefore, ic =- I - II = - (I + II)' This is shown in Fig. 8.41. At t = T -, io =- Iv therefore, ic = - I + II = - (I - II) and so on. Voltage across thyristor T1 is zero when T1, T2 are on. At T 12, when T3, T4 are turned on, u Tl = VT2 = - ve = - Vo = vinfrom T I 2 to T . These waveforms are sketched in Fig. 8.4l. The nature of the waveforms of ie and Vc in Fig. 8.41 can also be verified by the relation '. ie = C . dv/dt. At t = 0, dv/dt is positIve and high, therefore, ic is positive and high. At t = (TI 2) -, dv/ dt is reduced, likewise ie is also lowered. At t = (TI 2) +, du/ dt is negative and pronounced and so is ic as shown. Analysis. From 0 < t < T 12, equivalent circuit for the CSI of Fig. 8.40 (a) is as shown in Fig. 8. 40 (e). The capacitor is initially charged to a voltage - VI' Traversing the closed path a bed a, we get .. .(8.78) Differentiation of Eq. (8.78) with respect to time gives
R~: +~ = ~ or (RP + ~}o = ~
...(8.79)
Complementary function of the solution is obtained from force-free equation . .
rp+ ~)I,p=O or P=- R~ I cp == Ae- tIRC
For particular integral, put p =
io
°in Eq. (8.79)
C
I
= C or
io =1
Thus, complete solution for load current io, from Eq. (8.79), is ) to. = PIC . .+ .F . == 1 + A e- tlRC ... ( 8.80 Under steady state operation, the load current at t = 0, from Fig. 8.41, is io = - II' Therefore, from Eq. (8.80), - 11= I+A or A = - (l +/ 1)
io =I - (l + II) e- tlRC or for
.. .(8. 81) io = / (1- e- tI RC ) - II e- ti R C 0 < t < T 12. As only steady solut ion is desired, current io at t = T 12 becomes 11. Substitution of these
values in Eq , (8.81 ), gives
[Art. 8.8]
Inverters
1 - TI2 RC]I 11. = I [ l+e - e _ T I 2 RC
or
=I if 2 ~C »
473 .. .(8 .B2)
1 or T» RC
Substitution of 11 from Eq. (B.B2) in Eq. (B.B1), gives
.io =1 (1- e- _I{ll+e ~ e~~:;~]. eio = I[ 1 ~ 2 1 + eX:-(~;/2 RC)] tI RC
or
The output voltage vo, or capacitor voltage
)
Ve
ti RC
...(B .B3 )
is given by
vo=v, =R io=R+- 2 1 +
e,;-(~R;/2RC)l
The turn-off time te provided by the circuit to each SCR is obtained from the condition that when t = t e, Vo = V c = ioR = 0. Therefore, from the above equation, Vo
exp (- t/RC)
or
=Vc =ioR =RI[l- 2 .
e-t~;~RC]
1+e
=7
°
=21 [1 + exp (- T 12 RC)l
tc= RC ln[1+exp ' (~TI2RC)l
or
... (B.B4)
The average value of the input voltage Yin is obtained from the equation 1 Yin = TI2 .
=
fT12. 0
loR dt
~ I R {o12[121 +ee-~~]dt . 2RC .
or
[1- 4ftCT (1-+
expo (- TI2RC) )~ expo (- TI2RC) U
When input power Yin . I is positive, power is delivered to the load.
V.
In
Design considerations.
(i)
=IR
1
... (B .B5)
It is seen from Eq.
(B.B4) that as T is reduced (e.g. with T = 0, te = 0), or
inverter fr equency is increased, tc reduces. But the circuit commutating time tc should not be less than the SCR turn off time t q . This means that there is an upper limit to the inverter frequency beyond which inver ter SCRs will fail to commutat e. When T is large or i n ert e r fre q'u enc y ( = 1/T) is low, the plot ofio, or vo, versus time t, from Eq. (8.83), becomes fl atter as shown by dotted cu rve in Fig. 8. 42. As this curve shape is near er 0 a squ ar e w a ve , it can be inferred t ha t fo r low inv ert er (ii)
Fig. 8.42. Waveforms for I-phase CSI with R load.
474
Power Electronics
[Art. 8.8]
frequencies, inverter has square wave output for load current io or load voltage Va, When T is small, or inverter frequency is high) waveform of va' ori~ is shown by full-line curVe in Fig. 8.42. As this full-line curve is closer to a sine wave, it can be said that for high inverter frequency, CSI has sinusoidal waveshape for output load current or load voltage. (iii) Square-wave current . It has been found that fo.r obtaining ·square wave of the load
T
.
current, 2RC > 5.00.
If tq is the tuin-offtime for the SCRs used in theCSI, then from Eq. (8.84), tlJ = RC In
2 . . _ 5:: RC In 2 = 0.69 RC 11- e .
_ . tq C - 0.69 R
or
...(8.86)
T For 2RC = 5 or for T = 10 RC, maximum frequency,
1
{max =
.1
T = lORe
Substituting the value of C from Eq. (8,86),
= _1_ . 0.69 R t:max 10 R t C
= 0.069
t II
(iv) Sinusoidal wave output. For obtaining sinusoidal waveshape for load curre!}t, frequency domain analysis shows that capacitive reactance Xc at 3 times the minimum . frequency {min should be less than R12, i.e. Xc
or or or
R
-
2
.
at 3 {min S ~
1 1
L
-
-
C ~ 0.106 R . (min
...(8.87) .
S.S.3. Single-ph ase Auto~sequential Commutated Inverter (I-phase ASCI) Out of the force-commutated current source inverters, auto-sequential commutated inverter is. the ost popular. Though thr ee-phase ASCI is t h e u niver sal choice in industrial applications, hete only single-phase ASCI is presented so as to highlight the basic concepts of . this type of inverter.. . . . . Fig. 8.43 (a) shows a single-phase bridge inverter with' auto-sequ ential commutation. A const ant current source I feeds the loa d which is assumed here an in ductance L for simplicity. Thyristor p airs TI, T2 and T3 , T4 are alternatively switched t o obt in a n early square wave loa d current. Two commutating capacitors, one CI in the upper half and the other C2 in the lower half are connected as shown. In Fig. 8.43 (a), diodes Dl t o D4 are connected in series with each SCR to prevent the commutation capacit ors from discharging into the load. The
[Art.
Inver ters I
~ . 8]
475
I
I
I
I
(a) (6) Fig. 8 .43 . (a) Single-pha8e bridge auto-sequential commutated inverter with L load
(b)
Mode I.
inverter output frequency is controlled by adjusting the period T through the triggering circuits . .. of thyristors. The operation of this inverter- can be explained in two modes as under.
Mode I. In the beginning, i .e. before t::: 0, assume that T3, T4 are conducting and a steady current 1 flows through the path T3, D3, L, D4, T4 and source 1 as shown in Fig. 8.43 (a). The commutating capacitors are assumed to be initially charged equally with the polarity as shown in Fig. 8.43 (a ), i.e. Vel::: V c2 = - VeO ' This means that both capacitors Cl and C2 have right hand plate positive and left hand plate negative . . At time t = 0, thyristors TI, T2 are gated. The thyristor pair T3, T4 is turned-off by the application of reverse capacitor voltage V eO' Now pair TI, T2 conducts c,,!rrent 1. The path for this current I is through TI, CI; D3, L, D4, C2 and T2 as shown in Fig. 8.43 (b). Both the capacitors will now begin charging linearly from - Vco by the constant current I . The diodes DI, D2 remain reverse biased by VeO initially. The voltage vDl across DI, when it is forward biased, can be obtained by traversing-closed path abeda as vDl+ VeO
C~2 f I dt= °
Note that voltage across L is zero because of constant current !. .. .(8.88)
t1•
As the capacitor charges, voltages VDl across DI rises linearly. Eventually, at some time reverse bias across DI vanishes, vDl becomes zero and diode DI starts conducting. An
equation, identical to Eq. (8.88), holds good for diode D2 also. Actually, DI and D2 start conducting at the same instant t l. The time for which DI, D2 remain r everse biased is obtained from Eq. (8.88) by equating vDl = O.
or
.. .(8.89)
476
P ower Electronics
[Art. 8.8]
The capacitor voltage vel = Ve2 = Ve appears as reverse voltage across thyristors T3, T4 when TI , T2 are gated. The value of Ve is given as vel
2J I dt.
= Ve2 = Ve = - VeO + C
. vel = Ve2 = v~ (t l ) =- VeO + ~ I tl
Its value at time tl is
Substituting the value oftl from Eq. (8.89) in the above expression, Vol
=V,2= v, (t,)= -
V", + ~ ( ; V
dl)=0
This means that voltage across CI, C2 varies linearly from - VeO to zero in time t l . Mode I ends when t = tl and Ve = O. Note that tl is also the circuit turn-off time for thyristors of Fig. 8.43.
Mode II. Diodes D3, D4 are already conCl~cting, but at t = t l • diodes DI, D2 get forward biased and start conducting. Thus, at the end of time tv all four diodes DI, D2, D3 and D4 conduct. As a result, commutating capacitors now get connected in parallel with the load as shown in Fig. 8.44 (a). For simplicity in analysis, Fig. 8.44 (a) is redrawn as shown in Fig. 8.44(b) where two parallel capacitors Cl and ~ are combined into one as C (= C / 2 + C/ 2). In this figure, KCL gives 1+ io = ie (= icl + ic2 ) .
.
.
.
Lcl =L~2' Lcl = Lc2
As
dio I L 4-dt C
For this figure, KVL gives,
Ji ·. dt=O e
J(I + io) dt
dio 1 L dt + C
or
1 .
="2 Le
=0
2 d i o io I L-+ -=-2 dt C C
or
...(8.90)
For solving Eq. (8.90), the initial conditions at t =0 are . dio . io = I and dt = 0 I
I
c 01
1
I
i.e
03
t
to
04
C2= ~
01
I
-.
t
Cht
T1
~to\"
L
~
04 ~ i04 ·te 2to 02
02
1
-=--- t e2
ic2 C2=
1
I (a )
03 ~\03Kic'
(b )
(c)
Fig. 8.44. Equivalent circuits for I -ph ase ASCI with 10 d L.
t
llO2 T2
[Art. 8.8]
Inverters
477
It should be noted that inEq. (8.90), time t is measured from the instant Dl, D2 begin conduction, i.e. time t for mode II is measured from the instant mode I is over. In Eq. (8.90), for particular integral
I
los
C=-C or ios =- I and for complementary function,
[LPi + C1).to = 0 2 1 Lp +-=0 C
or
2
This gives
P
1 = - LC =-
p
=±jwo
000
= ~LC
or
2·2 2 000 = J 000
1
where
iot =Aeiwo t
+ Be- jwot
The total solution for the current is obtained by adding ios and iot
io(t) = ios + i ot = _ 1 + Aeiwo t + Be- jwo t t = 0, io =I
At
... (8.91)
1=-I+A+B A +B = 21
or At t
= 0,
dio dt
... (8.92)
. .
= 0, from Eq. (8.91); dio
.
. joooBe- JWo t = 0
dt = jooo Ae'wo t jWo(A - B) = 0
or or
(A - B) = 0
... (8.93)
From E qs. (8.92) and (8.93), A = B = 1 From Eq. (8.91),
. ic(t)
Now capacitor current,
+e=- 1 + 21. [eiwot . 2
jWot ]
·
=1[2 cos 000 t -
ic = 1 + io = 2 1 c~ 000 t
1]
.. .(8.94) .. (8.95 )
The voltage across capacitor is
I f lc' ' dt = oooC 21 . sm Wo t
vc = C This expression
foy V c
can als o be obtained as Vc
But
2
. d io dt
= VL = L
CD o =
1 LC or
.
=L ooo . 21 · sm 000 t 1 (OoC
LCD o = -
... (8.96)
478
Power Electronics
{Art. 8.8] .
21.
...(8.96J
Vc =-C sm 000 t 000
~ . ...JLC . sin 000 t =21 . ...JLIC
Also
Ve:;::
From Eq. (8.95),
ie1 :;:: ic2 :;:: %ic:;:: I cos wot
sin
...(8.98)
The existence of various currents i C1 ' i C2 , i Dl etc. is marked in Fig. 8.44 are given by .
and
.. .(8.97)
t
000
(c).
Diode currents
i D3 = i Cl :;:: I'cos wot } fi .. .. or 0 < t < t2 tDl =1 - LCl =1(1- cos wot) . .
As current iCl tends to reverse', diode D3 prevents its reversal. Similarly, diode D4 prevents
the reversal :nf CUITE!,J:ltj,C2. ... From.theinitiati()llofmode II, a time t2 must elapse for the current iC1 to become zero. This time .t2 be o~tained by equating iCl to zero.
cru:
iC1 = I cos Wot2 :;:: 0 1t
cos wOt2 = cos _. 2
or
1t
or
.. .(8.99)
t2 = .2000
The capac,:itor voltage Vc at the end of mode II, L e. at t
=t2 = 21t
,
can be obtained from Eq.
000
(8.96).
21 v =-
waC
c
Load current at t :;:: t2 is io:;:: 21 cos %- I = ~ I. This shows that load current ha~ reversed from + I to - I during mode II of t2 duration. . .
.
Th~ waveforms for vc,io,i c1 • iD1 etc. are plotted in Fig. 8.45. It i~ seen from the waveform for
Vc
that capacitor voltage changes by
2Vco
during each commutation interval.
Th e total commutation interval te from Eqs. (8.89) and (8.99) is given as . C 1t t c =t 1 +t 2 =2I - V co + 2(00
... (8 .100)
·
It is seen from Eq. (8.96) that vc ismaximum and equal to
Veo
when wot = 1':/2
21 woC C
V c;:;;' -
Fr om Eg. (8.89),
C 2l 1 V :;::-·-=-= ...JLC co 21 WoC 000
t =1
21
..
.
Therefo re, commutation interval from E q. (8.100) is . .. t
7t '\ (. 1t '\ =(I -001+ -2(.Oo! 1 = l+- I"L C 2)
r;;-;::;
c
\.
0
J
.
... (8.101)
{Art. 8.8]
Inverters 1 \9 't9 2
...47?
1
i 9 3 ,19 ,,[
I
T
,I
~t
F
.. t
· ,
I· :, r-~----.---~~~----~--~4-~---------+~--~t ,
·, ,.,
I!
: .
: ...Ao,
I; I:
.. ;
;.-.I'---------l'~
,
H-~-------L---~~~~~~---~I~~------------~~----.t
,,
tc..i
Ii ·
I:
.1';'
!---n,T2
+
Tl
13 ,T4
T3
·T2 fired
,! :'
-_.+0-1.+ ___- 'T1 .T2 ------l: . Tl
· T4fir~d
T2 firtd
Fig. 8.45. Voltage and current waveforms for I·phase ASCI.
At the end of total commutation interval (t 1 + t 2), the steady input current I flows through Tl, DL load L, D2 and T2. This constant current continues to flow till the next commutation process is initiated by gating SeRs. T3 and T4. Example 8.10. A single-phase autosequen.tial com.m.utated current source in.verter feeds a load R. Describe its working with appropriate circuit and wave{r>rms. Find also the circuit , turn-off time for the thyristors.
Solution. For the circuit diagram, replace L by R in Fig. 8.43 (a ). In other words,.Figs. 8.43 and 8.44 are being used here to describe the working of this inv,e rter with R as load in place of L. Worki ng of this inverter is explained in two modes as before. M ode I : Initial1y, it is ar-sumed that T3, T4 are conducting. A constant current I is flowing through the path T3D3 R D4 T4 and the source 1. 9apacitors are initially charged so that Vet = vc2 = - YeO' Voltage across load is IR = YeO '
When Tl T2 are turned on at t = 0, T3 T4 ar e tu rned off by the reverse voltage Ve O appearing a cross t hem. Current I now starts flowing through Tl Cl D3 R D4 T2 . Capacitor voltages Vel and ve2 start rising from - V eO towards zero. If Vzn is the forward voltage drop across Dl, then KVL for the loop a bed in Fig. 8.43 (b), (with L r eplaced by R). gives v D1
+ VeO
-
J
~ r dt- IR = 0
480
[Art. 8.8J
Power Electronics
or
V Dl ::::
But
VeO
~ I f dt
-VeO+ IR +
= IR,
:.
Diode D1 will start conducting when D1 begins conduction, then vDl =0 =
~I
~ I f dt.
VDl
=
vDl
becomes zero. If t1 is time after which
t1
vDl
= 0 and
= 0 or t1 = O. This means that when T1 T2 are turned
on, T3 T4 are turned off immediately and diodes D1 D2 start conducting at the same time. In other words, all the four diodes start conducting as soon as T1 T2 are turned on. Mode II: With all the four diodes on, the equivalent circuit is shown in Fig. 8.44 (a) which is simplified to that given in Fig. 8.44 (b), (with L replaced by R). In this figure, KCL gives 1+ io = ie KVL for the loop consisting of Rand C gives
R io + ~ R io + ~
f ie dt =0
f (I + io) dt = 0
dio
or
io
I
Rdt+C=-C
Its particular integral gives the. steady-state solution as is =- I. For transient part of the solution, proceed as under: . ",: ;~~,.• (RP +
~}o = ° or
P=-
..
"t. =A e-tIRC
Total solution for load current is
io
R~
= is + it = - I + A e- tl RC
When t = 0, in'tial current through R is io = I, :. A = 2 I :. Load current,
io
Capacitor current, '
.
=- I + 2 I e- tlRC =1(2 e- tl RC I
"e =
1)
2' I - tlRC + £0 = e
"e1
.
. 1 . ' I -tiRC = Le2 = '2 £e = e '
Diode D1 current,
"D1
.
' =1(1 '- e- 'tlRC) =I - "el
Capacitor voltage
Vel
1 = C/2
Vel
= _ 21 e- tl RC , (- R C) + k C =..: 2IR e- tlRC + k =- 2 VeO e- tlRC + k
or
f'
"cl '
When t = 0, initial volt age across capacitor is ..
Ve l
.. .(i )
.
Vel
.. .(ii ) .. .(ii i)
f
dt = C 2 I e- tI RC . dt '
=- VeO ' this gives k =VeO'
=V eO [1- 2 e- tl RC]
.. . (i v )
When T l T2 are on ; v el. vc2 appear as r everse bias across T3 , T4 respectively. Therefore, circuit turn-off time tc for T3, T4 (01' for any thyristor in Fig. 8.43 wit h R as the load) can be obtained from Eq, (iv) by pu tting Vel = 0, '
[Art. M.S]
Jnvetters
4S1
i9l,ig2t i9
30i 4 9
1
'?
I T/2
I'
L
L
1
T
.."'.
.4
~
t
T
Vel. Vc2
'.
II
I
I I I I
I
I
tO __l ~""!\I!I
)
I I ' I
.:
I
t03 ______ ~ __
I I
I
~ ~' : !/
' ...
I
t e l=tc2 I
I
T1
T2
T3
r1
T4
T2
T3 T4
Fig. 8.46. Pertaining to Example B.10.
a = VeO or
[1- 2 e VRC ]
te =RC In 2
.. .(v)
From above, waveforms for Vel from Eq. (iv), io from Eq. (i), iel = ie2 from Eq: (ti) and iD1 from Eq. (iii) are plotted in Fig. 8.46. It is seen from Eq. (iv) that vel will change from - VeO at t = 0 to + VeO after an infinite time. But it is usual to take that in time 4 RC, vel varies from - VeO to . + VeO as shown in Fig. 8.46. I
Example 8.11. A single-p hase auto-sequential commutated CSI is fed from 220 V de source. The load is R = 10 n. Thyristors have turn-off time of20 ~s and inverter output frequency is 50 Hz. Take a factor of safety of2. Determine suitable valu e of source induc tance assuming a maximum current change of 0.5 A in one cycle. Neglect all losses. Find also the vaLues of commutating capacitors.
Solution. Time of one cycle, T
=-1 =-1 sec f 50
di a. 5A :. Rate of change of current; dt = -""1' = 0.5 x 50 = 25 A/sec
I
A short circuit at the load terminals of th e inverter puts the most s evere conditions on the source. S o the value of source inductanc e must be obtained from these considerations.' V = L di S dt
482
Power Electronics
[Art. 8.9]
:. Source inductance,
L = 2225° = 8.8 H
From the' previous example, from Eq. (v), circuit-tum-off time is given by
tc =RC In 2 20 x 2 x 10- 6 = 10 . C In 2
or
• 40 x 10- 6 C= 10ln2 =5.77IlF .
or
8.9. SERIES INVERTERS .Inverters in which commutating components are permanently connected in series with the load are called series inverters. The series circuit so formed must be underdamped. As the current attains zero value due to the nature of the series circuit, series inverters are also classified as self-commutated inverters or load-commutated inverters. These inverters operate at .high frequencies (200 Hz to 100 kHz), the size of commutating components is, therefore, small. These inverters are used extensively in induction heating, fluorescent lighting etc . . The object of this section is to describe single-phase series inverters.
8.9.1. Basic Series Inverter The circuit diagram for a basic series inverter is shown in Fig. 8.47. It consists of load resistance R in series with commutating. components Land C. The values of Land C are so chosen that the series RLC circuit forms an underdamped circuit. Two thyristors Tl and T2 are turried on appropriately so that T2 L output voltage of desired frequency can be obtained. When thyristor Tl is turned on, with T2 off, current i starts building up in the RLC circuit, Fig. 8.48. As the circuit is Load underdamped; the load current, after reaching some peak value, decays to zero at point a, Fig. 8.48. At point a, as the load Fig. 8.47. Basic series inverter. current tends to reverse, SCR Tlis turned off. After instant a, some minimum time tq . min must elapse for T1 to regain its forward blocking capability. This minimum time is given by _
where and
(J)
(0,.
-'qmio
= ou tput frequency In rad/sec = circuit ringing frequen cy in rad/sec.
n
(7-
=: - ;, =%
... (8.102)
In Fig. 8.48, time interval between the instant Tl is turned off and the instant T2 is turned on is indicated by Toft = ab, where Torr> tq . min' After thyristor T l has comm utated, upper plate of capacitor attains positive polarity. Now when T2 is t urned on at instant b, capacitor begins to d.1s charge and load current in the rever sed direction builds up to s ome peak negative value and then decays to zero at instant c. After this, time Toft = cd must elapse for T2 to recover. At d, T 1 is again turned on and the pr ocess repeats. In this manner, c is con verted to ac with the
Inverters
[Art.
~.I)]
o is2tl
c ~1
i o •i
________~[J ~~________________~[J~______
i
r--- 11 /w --~ri
I
I
°
.
a;
I I
E
_V
ro
:
: :
~ 1(/W(~,~
~/2
-----r~.L---------~~-----~-: :0
. "T : ~1t_~:
.!
: --j
.
__-----+-J
;--11/w ."( To:!
-y:. . .
;
C
Tolf.
!
---.~-t----+-,-.,.----L- .-r----\:
1 ,
cOL I VL
tI. I
.
Vs TVe o
L!r--~~~~~--~--~--Mr--~°1
-'.
~-. -
t
~
Mode 1---; : :
Mode 1lI--i Mode II :
O~--~---*~----------~~-------~~-----L--t
.; Fig. 8.48 . Current and voltage waveforms for basic series inverter of Fig. 8.47 .
help of series inverter. In Fig. 8.48, TOff = ab, or cd, is called circuit turn-off time or dead -zone time. Th e capacitor stores charge during one half cycle and releases the same amount of charge during the next half cycle. As a consequence, the positive half cycle of current is identical with the n egative half cycle ofload current. In a practical series inverter, P9sitive and negative half cycle may not be sine waves. 8.9.2. Ar.. ysis of Basic Series Invert er The operation ofthis inverter is described in three different modes as below_
Mode I. With T2 off, the equivalent circuit of basic series inverter of Fig. 8.47 is as 5hown in Fig. 8. 49 (a). In this figu r e, the capacitor Cis assun.. . ed to be initially ch arged to voltage VCIJ with lower plate positive. Mode I begins with the turning on ofthyristor TI. Thus, with Tl on , .KVL for th e closed circuit of Fig. 8.49 (a ) can be written as .. (8.1 03)
[Art. 8.9]
484
Power Electr onics
Its Laplace transform is
[
1] = Vs + Veo
Is R + Ls+ or
--"---=
sc
s Vs + Veo I(s) = L .
The roots of S2 + R s + _1_ = 0 are s L LC
1 2
s
R +L
s:-
.. .(8.104)
1 LC
I(.!l J2_~
=_ll. + ... 2L - 'J
2L
LC
2
~ ) . - LIe] must be negative, i.e.
R)21 24L
2L - LC < 0 or R < C
As the circuit is underdamped, [(
(
s =-
R S= 2L
where If Wo =
~lc' then wI' = ~w~ - S2 I(s)
Let
~ ±J'{~ -(it ... /1
and wr = 'J LC -
or Wo =
=
J
(
2 2i RJ
~ w; + S2 . Therefore, from Eq. (8.104) ,
Vs + Veo [ 1 ].. L (s + S- j wr ) (s + S + j (D,.)
1 A B
(s..l.. S- jwr) (s + S+ jw r) = s+ S- jWr + S + S+ j wr
1
From above,
-1
A=-.-andB=. 2Jwr 2) wr
Its Laplace inverse is Here
=-1; ±j ro,
S= ~
.. .(8.105)
is called damp ing factor,
U: = ~lc 0
is r esonant frequency in rad/sec , wI' is
known as circuit ringing frequency in r ad/sec and W = operating, or output, frequency in Tad/sec =2 1t{. . The plot of this current of' Eq. (8 .105) shows that load current i (t) will be zero when
wI' t =1t or t = ~ sec. This is shown as oa = 1t1 wr in Fig. 8.48. W
· . d o·f T Ime perlO
' , OSCI'11 atlOn
oa
1t = =-w,.
-.J
1t
R
2
11LC - ( 12L )
.. .(8 .106 )
[Art. 8.9]
Inverters
•
Output frequency,
f=
T1
-.-
1
2(oa + TOff)
L
T2
f=~ ·· ____~~l________
or
485
... (8.107)
27t +2T :V1/LC - (R I 2L)2 off R It is seen from Eq. (8.107) that output frequency can be controlled, by varying (i) Toff'
(b)
(a) it depends upon power semiconductor device used, (ii) load resistance R, (iii) L o~ (iv) C. In Fig. 8.49. Equivalent circuit of Fig. 8.47 with (al
any case, output frequency f caIlnot be more Tl on, 1'2 off, (b) Tl off, 1'2 on.
than the circuit ringing frequency
1
fr = 27t . :VIlLe _\RI2L)2 Hz, i.e.f < (,.
The voltage across inductance L, from Eq. (8.105), is
di
VL = L -d = L . t
Va + Veo L
V+V = s eo. e- f,t
1 - ~t . - [e . (J)r" cos Wr
[(J)r
Wr
cos
{J)r
'1Wr + ~
wI'
eo _
wI'
_ .
- (Vs + V eo)'
where
~t
. '
]
2 cos
(J),t -
~ . ~ sm wl't
1
00,. - ':I
"~-~
V+V a
-
t - ...,. e . sm wrt .. ..~
t - ~ sin 00/1
= Vs + Veo e- ~t .1'1Wr2 + ~ 2 [ W ./ 2r =
):
(J)r
e- c;t ~W? + ~2 [cos'l' cos wrt - sin 'l' sin Wl't] 000 (J)r
-f,t
.e
...(8.108)
cos (wrt + 'l') .
"'0 =resonant frequency =--J",~ + ~2 and 'If =tan-I ( ~ }
see Fig. 8.50.
The voltage across capacitor C is
ve = Vs - vR - vL
. =Vs - Vs + Veo . R L e-c;"-t sm wI' t (J)r
But
Fig. 8.50 . Pertaining to wI" ; and WO o
486
Power Electronics
[Art. 8.9] .W == Vs - (Vs + Vco ) e- C,t 2W [2 r
sin'll sin w,.t.,+1:;08 , . wrt cos'll - sin wI". t sin 'Ill .
...(B.109)
radians, i.e. when O)rt = 1t, i(t) Eq (8.109), is given by After
1t
vel (t ;: 1t/ Wr )
=0, therefore capacitor voltage vel at t =1t/wl"
' = V.:l. = V~ - (V.• + or v co) e
. ".)t/ ro
W
'.
0
0)1'
cos
(
1t -
from
'II) .
...(8.110) .. Vel = Vs + (Vs + Yeo) exp [":"1t ~/O),.l Mode II. During this mode, both SeRs T1 and T2are off. Therefore, Vc = VV cl ' i = 0, VL = 0 dUI'ingmode II. Mode III. This mode begins with the turning on of thyristor T2. Equivalent circuit for this mode is shown in Fig. 8.49(b). Time origin for this mode is taken when T2 is switched on. Therefore. KVL for the circuit of Fig. 8.49 (b) is
J'
1 R-'.+ L -di dt + -Cl' dt = Vc 1
... (8.111)
I
Initial voltage Vel causes the capacitor to have upper plate positive. Therefore, with the turning on of T2 , the current in load R is reversed as desired. It is seen that Eq. (B.111) is identical with Eq.(B.103). Their solutions must, therefore, be identical. Thus, solution of Eq. (8.111), from Eq. (B.105), is given by ' . V el ' e-~... sm . wl't t.() t = --L-' ... (B112) . wr .
Peak value of positive current, as given by Eq. (8 .105) would be equal to the peak value of the negative current given by Eq.(B.112) only when Vel = Vs + Yeo' The waveforms for IQad current io or i, voltage, ve across capacitor, VL across L and source current i., are sketchedin Fig. 8.4B. As stated bef.ore, dead. zone ab (. = ~:.
.
W
~
~
) must be greater .
than the thyristor turn-off time t q . It is seen that source current is flows only during the positive half cycle of load current. Drawbacks. The basic series inverter shown in Fig. B.47 is very simple. It, however, suffers from the following drawbacks. 1. For a load power, the load current is taken from the supply during the positive half cycle only. This has the effect of increasing the peak current rating of the de source. 2. Since the source current flows during the positive half cycle only, its harm onic ' content is much pronounced. 3. The m aximum operating frequency of the inverter is limited because this frequency to has t o be less than the circuit ringing frequency. 4. For output frequency much lower than the circuit ringin g frequ ency, the load voltage wavefor m gets distorted considerably due to the increase d duration of th e dead zone. 5. }\mpl'tude and duration of load current flow in each half cycle depends on the load circuit parameters. Therefore, this inverter suffer s from poor output regulation. 6. In this in verter, commuta ting components have to carry the load curren t con t' nuo'USly, therefore high ratin g of these components is necess ary.
, )
[Art. 8.9]
Inverters
487
Out of the drawbacks enumerated above; 4, 5 and 6 are inherent in all types of series inverters. However, the drawbacks listed at 1, 2 and 3 can be mitigated by modifying the configuration of the basic series inverter. This is discussed in what follows.
8.9.3. Modified Series Inverter Fig. 8.51 shows a modified series inverter in' which the drawback of limited operating frequency is overcome. It consists of two thyristors T1, 1'2, de source voltage Vs' load R, capacitor C and mid-tapped inductor. Actually, inductor is one but it has a centre tapping so that inductance of both halves are equal, i.e. L} = L 2 . Let T1 be switched on so that load current i builds up through R and then decreaRes. When current i is nearing zero value, ue across C is less than Vs + V eo ' Fig. 8.48 and at this instant, turn on SCR 1'2. Since current i is decaying, the polarity of emf induced in L} would be asiudicated in Fig. 8.51. Since L}, L2 are mutually coupled, voltage across L2 has induced voltage with polarity as shown. Now KVL for the loop consisting of T1, Vs' R, C, L} ~md T1 gives uTl - Vs + iR drop + Vs + Veo - emf in L} (say 0.5 V eo ) = 0 UTI
=(- 0.5Vco -
f.ig. 8.51. Modified series inverter with coupled .... inductors.
iR drop)
Since uTI is negative, T1 is reverse biased, as a result T1 is turned off by UTI' This shows that when 1'2 is turned on, even before load current has reached zero, T2 gets turned off. At the same time, when 1'2 is turned on, it gets forward biased by a voltage uT2 =, emf in L2 + (Vs + V eo ) + iR drop. In basic series inverter of Fig. 8.47, ifT2 is turned on before current in T1 vanishes, there would be a dead short circuit of source voltage Vs; this will eventually damage the inverter. However, in modified series inverter of Fig. 8.51, there is no danger of any short circuit. This important feature permits the modified series inverter of Fig. 8.51 to operate at frequency higher than the circuit ringing frequency.
8.9.4. Half-bridge Series Inverter
I
In the circuit of Fig. 8.51, the load draws power from de source in positive half cycle only, even though its operating frequency limit is surmounted. This drawback of intermittent power flow can be overcome by half-bridge series inuerter shown in Fig. 8.52. This inverter consists of two thyristors, mutually-coupled inductors L} = L 2 , two capacitors C 1 = C2 = C, de source Vs and load R. In this inverter, power is drawn from de source during both the half cycles as explained below. Assume that capacitor C2 is initially charged to voltage Veo
with the lower plate positive, Fig. 8.53 (a). Since C1 and C2 are in
series across the battery, (voltage across C 1 + voltag8 across C 2 )
Fig. 8.52, Half-bridge se ries m ust be equal to battery voltage '\.-78' Capacitor C 1 would, therefore,
inverter. be charged to voltage Vel = Vs + Veo with upper plate positive.
488
Power Electronics
[Art. 8.9)
Let the thyristor Tl be turned on at t == O. With Tl on, two current. paths i l and i2 are established in Fig. 8.5Q (a) . One path is for i l (shown dotted), it consists of driving voltage V , + Veo (net voltage in the loop of i 1) , Til L1I Rand C2 = C. The second path is for i2 (shown full line), it consists of driving voltage Vs + V co (net voltage across C 1), T 1 , L11 Rand C l = C. Since these two paths for i1 and i2 are identical, currents i1 and i2 must be equal, i.e. i l =i 2· Since current i l is driven by de source · and i2 by capacitor C 1; 50% of the load current ill (= i l + i2 = 2 i1 = 2i 2) is drawn from de source and the remaining 50% from the capacitor discharge. 11
r------ ......-----,
r----- ....... - ------ --- -- --- .-- -- .... --,
:
:
i1
is
i,
:
I
• il
.
+
tl ........ ____._ - - - - -
ttl :
.J
(a) Fig. 8.53. Half-bridge series inverter
(a)
initially
Vel
=Veo
(b) and (b) initially
lle2
=Yeo '
At the end of positive half cycle at a, Fig. 8.54, load current io becomes zero, T1 is therefore, turned off. Now the voltage across C 1 is Veo and that across C2 is Vs + Veo with the polaritiei? as indicated in Fig. 8.53 (b). When T 2 is turned on at instant b, Fig. 8.54, currents i1 and i2 are established as shown. LoadcllTrent io (= i1 + i 2) now get reversed as desired in an inverter. Current i1 (shown dotted) is driven by voltage V$ + Veo and the path it trllverses is C1> R , L 2 , T 2 , Vs and C l = C. Current i2 (shown full) is driven by voltage Vs + Veo and the local loop at traver~es is made up of C 2, R , L 2 , T2 and C 2 = C. Since the paths negotiated by currents i1 and i2 are identical and the driving forces ara also equal, i l = t2 . As before, 50% of the load power comes from the dc source and the other 50% from the capacitor discharge. ' -. v e2'
In Fig. 8.54 are sketched the waveforms for voltage across C 1 as V el' voltage across C 2 as load current i o' source current is or i1 and voltage across thyristor Tl as un. Waveforms
for i . = i z = ~ io are also shown in this figure, It is seen from the waveforms that dc source delivers current i" or power, to load during both the half cycles, as a conseql.l~nce (i ) ripples in source current get reduced and (ii) peak current rating of dc source is also curtailed, Examp le 8.1 2. In a self-commutated SCR circuit, the load consists of R = 10 12 in series w it h com mutating component s of L = 10 m E and C = 10 flF. Check whether the circ uit will com muta te by itself w hen triggered from zero voltoge condition on th e capacitor. Wh at will be t he voltage across the capaci tor and induc tor at the time of commu ta tion ?
Inverte r s
[Art. 8.9]
tg~ 6
~
t
,
\g~ t
489
n
!l
t
tic 1
T 10
Vs+V cO
t
Ve2
t
t
,"
~
~Tl
.
: ~T2------i ! - Tl - - - : r ; .i .: ,
VTl
"
t
i ,
t
-VeO ...L
Fig. 8.54. Waveforms for half-bridge series inverter of Fig. 8.52.
" d a lso di Fm dt at t =0
. .
Solution. Here As R? <
~,
-3
R2 = 10 x 10 = 100 . 4L = 4 x 10 x 106 = 4000 , C
10 X 10-
the circuit is underdamped. Therefore, the series circuit will commutate on
its own when triggered from zero voltage condition on the capacitor. Also
S=.!i. = 10 x 1000 = 500 2 x 10
2L
Wo =
1 " r;-;::r 3 1fC = -v 10 = 3.1623 )(10 rad/sec
=..j~ - S2 =[ 10 2.5 10SX =3.1225 10 =tan- (~J = tan- 3. 1225 500 J=9.097° 'II 10 7
wr
12
X
-
1
1
(.Or
(
X
X
3
3
rad/sec
490
Power Electronics
[Art. 8.9]
The load current is zero, i.e. SCR will commutate when
(Or
t
= 7t
7t
or
t = 3.1225 x 1000 = 1.006 ms
St = 500 x 1.006 X 10- 3 = 0.503 From Eq. (8.108), the voltage across L at the time of commutation, with vL
VeO
= 0, is
VeO
= 0, is
= V. 3.1623. -0 :503 (180° 9097°) s 3.1225 e cos +. = - 0.60472 Vs
, From Eq. (8.109), the voltage across C at the time of commutation, with ' .
Vc
'I
= Vs[ - e
- 0.503
3.1623 (180° 9 097 0 ) ] x 3,1225' cOS - .
;; 1.60472 \18
The voltage across :apacitor can also be obtained as under. Note that commutation : . ' ,
VR
= 0 at the time of
'
Ve
=Vs -
vR - VL
= Vs -0- (:- 0.60472) = 1.60472 Vs
F rom E q.(8. 105) , WI'th V cO = 0 , di dt = (0Vsi '[ e - ~t(Or COS
t -" e - c,t. . SIn
, ' J! (Or
(Or
t] ,
r
, =- = , (-dil dt L Vs
Vs = 100 V Als ' 10x10- 3 s
=0
,
,
Example S .13. Calculate the outpu t frequency of a series inverteF with the following. parameters: Inductance L Torr = 0.2 ms.
=,6 mH,
capacitance C = 1:2 microfarad, load resistance R = 100 ohms,
If the load resistance is varied from 40 to 140 ohms, find out the range of output frequency. [I.AS., 1986]
Solution. From Eq. (8.106), the time period of oscillation is given by 7t
~l/LC -
,
(RI2L) 2
7t
112
103 x 106 _ 1002 x 10 6 ] [ 6 x 1.2 ' ,' 4 x 36
=
. 1t
= 0.377
ms
1000 (8.333)
..
Fr0l'!l Eq. (8.107), the output frequency, , 103 f= 0.377 x 2 + 2 x 0.2 = 866.55 Hz.
When R = 40 n, output fre quency
, 2
=
r10
l
1
= 1046 .2 Hz.
1t 3
x 106
1eOO x 10 6 x 1.2 4 x 36
112
6
]
+ 0.4
X
10-
3
[Art. 8.9]
Inverters When R
491
= 140 n, output frequency
=
1
2 = 239.8 Hz 3 1t 112 + 0.4 X 106 3 2 10 X 106 _ 140 x 10 ] [ 6 x 1.2 4 x 36 .
:. Range of output frequency
= 239.8 Hg to 1046.2 Hz
Example 8.14. In a single-phase series inverter, the operating frequency is 50kHz q,nd the thyristor turn-off time tq = 10 J.1S. Circuit parameter are : R =3 n, L = 60 1lR,t; ;7.5W and' Vs =220 V dc. . .' . ' '" Determine (a) the circuit turn-off time and (b) maximum possible operating frequency, assuming a factor of safety = 1.5. Solution. (a) Operating frequency,
W=
R
21t X 5000 rad/s
3xl~
.
; = 2L = 2 x 60 = 0.025 x 10 1 (t)o = =.JLC Circuit ringing frequency, wr
106 =~60 x 7.5
6
. 6
=0.0471 x 10
=-Jw~ - ;2" =106 -J0.04 7i 2 -
.. ~~
0.025 2
= 0.03992 X 106 ~ 0.04 X 106 rad/s
From Eq. (8.102), the circuit turn-off time (= time of dead zone) is
1t 1 . 1t
t =---=-~- -0-.0-4';";:':-1-0-:-'~ = 21.46 )J.s C (t) wr 21t X 500 As circuit turn-offtime (= 21.46 IJ.S) is more than the SCR turn-off time, the series inverter would operate satisfactorily. when the dead zone is just equal to tq x factor of safety = 10 x 1.5 = 15 ),lS, the maximum possible operating frequency would be obtained. From Eq. (8.102), we get (b )
1t 1t t q x. 1.5=-- . (t) ' Wr 1t 15x 10- 6 = 1t . ' 21t . fmax 0.04 X 106 or f max = 5345.5 Hz. Example S.15. In Example 8.14, initial voltage across capacitor is 80 Vand output current is assumed sin usoidal. Calculate the p ower delivered to load, and average and rms values of thy ristul current. Solution. It is seen from Fig. 8.48 or 8.54 that maxim:u m value of load current across 1t
wh en
Wr t 1
.
1t
="2 or when t1 = 2 wr ·
1t
6 = 32 .27)J.S 2 x 0.04 x 10 F r om Eq. (8.105), t he m aximum value of load cu rren t Iom:r would occur when sin
t1 =
I o .mx =
·
Vs + Veo _ ~ t L e l
wr ·
Wit
= 1.
492
Power Electronics
[Art. 8.10]
22~ + 80 6 exp [- 0.025 x 106 x 32.27 x 10- 6] 0.04 x 10 x 60 x 10. == 55.79 A == peak value of thyristor current
=
55 .79 = 39 46 A T2 --;n o .
= lo.m.x ==
:. Rms value of load current
== (39,46)2 X 3 == 4671.27 W Load power Since thyristor current has a periodicity of 21t radians, rms value of thyristor current == lo.m.x == 55.79 == 27 895 A
2
2
.
Assuming lossless converter system, average value of source current, I
Average thyristor current,
sA
== 4671.27
220
ITA == ISA ==
2
= 21 233 A .
21.233 == 10.617 A . 2
8.10. SINGLE-PHASE PARALLEL INVERTER
.
The basic inverter circuit for a single-phase parallel inverter, utilizing capacitor for its commutation, is shown in Fig. 8.55. It consists of two thyristors T1 and T2, an inductor L, an output tr ansformer and a commutating capacitor C. The transformer turns ratio from each primary half to secondary winding is assumed unity. The output v oltage and current are Va and ia respectively. The fu~ction of L is to make the source eur rent constant at la. Positive directions for voltages and currents are marked in Fig. 8.55. During the working of this inverter, capacitor C comes in parallel with the load via the transformer. That is why it is called a parallel inverter.
R
T2
-
Fig. 8.55. Single-phase capacitor commut ted parallel inverter with centre-tapped transformer.
The operation of this inverter can be explained in some well-defined modes as under:
Mode I : In this mode, thyristor Tl is conducting and a current flows in the u pper half of primary windin g. Thyrist or T2 is off. This current estabHshes m agnetic flux that 1inks both the halves of prim ary win din g. As a result, an emf Vs is in duced across upper as well as lower h alf of primary win ding. In oth er wotds, total voltage across prim ary winding is 2Vs' This volt age charges the commutating capacitor C to a voltage of 2Vs with upper plate positive as
[Art. 8.1 0]
Inverters L.
-
i.o
T1
-
L
493
10
Io
~Vs
R
Io!
R
--
-
I~ tlo_______Io~__~~ 10
10
10
10
(b)
(a) Fig. 8.56. (a) Mode I, t < 0 ; (b) Mode II, t = 0 + L
T1
to
.
Ig
L
i.o
~
10 R 2 Vs
10
~)
r
Vo
+:
R
, I
I
j
I I
I I
L__
+
.. 10 ~)
Fig. 8.56. (c) Mode II, tl S t < TI2 (d) Mode III, just after t = T I2 .
shown in Fig. 8.56 (a)~ Thyristor T2 is forward biased through Tl by the capacit'or voltage 2VII • Eventually, a steady state current 10 flows through VII' L , Tl and upper half of primary 'fvinding. During this mode, Vo =V s' Vc = 2Vs, io =10,vTl =0, ic =0, iTl =10 as shown in mode I in Fig. 8.56(a) .
.Mode IT : At time t = 0, thyrist or T2 is turned on by applying a triggering pulse to its gate. At this time t = 0, capacitor volta ge ~Vs appears as a reverse bias across T l, it is therefore turned off. A current 10 begins to fl ow through T2, lower half of primary winding, Vs and L as shown in Fig. 8.56 (b). At the sam e time, capacitor voltage 2Vs is applied across the total transformer prim ary and a capacit or curren t - i.is established. Negative sign. before ic means that current ic flows opposite to its positive direct ion assumed in Fig. 8.55 . Before T2 is on, i.e. at t = 0 -, mmfin upper primary winding is Io N l and zero in the lower primary winding. Soon after T2 is on, i.e. at t = 0 + , mmfs linking both upper and lower halves cannot change suddenly. Therefore, at t = 0 ) - ic = 10 such th at m mf in the lower half remains zero and mmf in the upper half is equal t o mmf at t = 0 ~ . After t = 0 + , capacitor C discharges and current ic is su ch that it supplies th e load current io and balances the primary and secondary ampere turns of the tr ansform er. Capacitor current continues flowing till capacitor has charged from + 2Vs to -:- 2Vs at time t = t l . Load voltage also changes from Vs at t = 0 to - Vs at t ;::;; t l • Fig. 8.56 (c) and Fig. 8.57.
494
[Art. 8.10]
Power Electronics
Mode III : When capacitor has charged to - 2V s with upper plate negative and lower plate positive; SCR Tl may be turned on at any time. In Fig. 8.56 (d ), Tl is triggered at t = T 12. Capacitor voltage 2Vs applies a reverse bias across T2, it is therefore turned off. After T2 is off, capacitor starts discharging as shown in Fig. 8.56 (d). Current i.: is now positive . Mmfs in theuppet lind lower halves remain unchanged from their values just before T 12. When l c decays to ·zeto. u,,!:::+ 2Vs , Vo = V s' iTl =10 = V/ R, Fig. 8.57. igl
t
om ... am i92+~.I ~~mJ ~:L-____~__~___________ ~m~____________~_ 2Vs
--.,
!
,
I
I
, I
I
'I
I .
~tc ~! I ,
- - - i' 2Vs
, . - - -.......
I I
I I
;
I
I
.
; 1,
I
I
: 1
1 I
1
I I
·
I
-~-.--ll
__--<.; ...... -t--]]i -. --~•...;: , 4-1
:
I
! :
:
~'
I
I '
I~ I
-1 0 , ! ......0----'--
T2 - - - - t - - + - - T1
I
:
--~"WOIt"''''TI-L--
T2 ......,-.-
O~--~--~--------~-----L------~~+---------~--~_ .'
I .
i
O ~--~-~~----~--~--------~~-L-~------~----
t= T/2
T1 ON
L:T T2 ON
T1
-
ON
Fig. 8.57. Waveforms for currents and voltages pertaining t o singl e-phase para llel i nverter.
l
Inverters
(Art. 8.10]
4'95
The waveforms for uc ' uTl, Uo etc are shown in Fig. 8.57. At t = 0 + , when T2 is turned on; uTl =- 2Vs, ie =-10' iTl = 0 and iT2 =10 as shown. AB the turns ratio from whole primary to secondary winding is 2, the load voltage has half the amplitude of capacitor voltage. However, the load voltage has the same waveform as the capacitor voltage; see Fig. 8.57. .
8.10.,1. Analysis of Parallel Inverter In Fig. 8.56 (b), the secondary mmf must be balanced by the primary minf at any time t. It is seen from this figure that net current in the lower primary half is (10 - iJ upward and in the upper primary halfis ic downward. Secondary mmf =mmf in the lower primary half + mmf in the upper primary half.
io N2 = (10 - ic) Nl Load or secondary current,
ie Nl
N
io = (10 - 2 ie) N l 2
.
But
Zc
dUe
= C dt
io =(10 - 2 C Also, load voltage Uo
~')~:
... (8;113 )
= ioR. This voltage when referred to whole primary is
ioR
uC=N (2N l )
... (8.114)
2
or
Its particular integral is, Its complementary function is, l
__n_t VC.F.
=A e
4RC 2
2IOR -~ct UC = - - 2 - + Ae n H ere :, = R
(~:
J
... (8.115)
is the load resist ance r eferred to on e h alf of the primary winding. lois dUe
the load current on the primar y side when dt = O.
496
[Art. 8.10]
Power Electronics
:. 210 (resistance referred to one half of primary winding) = 2Vs
2IOR -2-= 2Vs n
or
-
With this, Eq. (8.112) becomes 2
V
At t
=.
0, ve
= 2Vs ,
:.
c
= 2Vs +A e
-~t
4RC
2Vs = 2Vs + A. But this gives redundant solution.
At t = T 1 2, voltage across capacitor becomes opposite to what it is at t expressed as [V e at t
= 0] =-
= T 1 2]
[v e at t
n
2Vs + A
or
2
T
=- 2Vs -A e- SRC =
A
or
= O. This fact can be
-4V
:2 s
.
1 + e- n TISRC
2e-ntl4RC
I
ve= 2Vs 1-
2
2
[
[n2T]
J
...(8.116)
n T I 1 + exp - 8RC , I
1.
. I ·cIrcmts, " F or practlca exp - 8RC «
.. .(8.117 ) U,=2v,[1-2ex+ ~~] In Fig. 8.57, Vc at t = 0 is 2Vs ' But in Eq. (8.117); at t = 0, ve =- 2Vs' In order that Eq. (8.117) is compatible with the waveform for ve in Fig. 8.57, we re-write Eq. (8 .117) as U,
=2V, [2 exp (- ~~
)-1]
..(8.118)
It is seen from the wavefonn of Ve and vTI in Fig. 8.57 during the interval t = 0 to t = t1 that UT1
=- u, =2 V, [ 1 -
exp (-
~E1J. Circuit turn-off time t, is the time taken by
from its initial value of - 2Vs'
~erefore,
UT1
to become zero
n2tc]
or or
vT1=2Vs [ 1_2e- 4RC =0 n2 t __e = In 2
4RC
t = 4Re · In 2
n 2
e
...(8 .119)
Also, commutating capacitance, C=
n 2 ·t e 4R · In 2
..(8 .120)
[Art. 8.11]
In verters
497
Eq. (8 .119) reveals that for a given value ofload resistance, the value of capacitor should be so selected as to give adequate circuit turn-offtime tc for the off-going thyristor. In Eq. (8.119), t c > toff' where toffis the thyristor turn-off time. It is seen from Eq. (8.120) that ifload resistance is small, size of capacitor required is large. Example 8.16. In a single-phase capacitor commutated parallel inverter using two thy ristors and a center-tapped transformer, the source voltage is 220 V dc. The centre-tapped transformer has a turns ratio from each half primary winding to secondary winding of 3 : 1. For a load resistance of 20 n, find the value of capacitor C to obtain 20 ~s turn-off time on the thyristor. Assume the inductor L large and transformer ideal. Take factor of safety as 2.
Solution. From Eq. (8.118), v, = 2V, [ 2 exp (Here n
':::t)-1]
= ~ . Circuit turn-off time tc = 2 x 20 = 40 ~s
'The circuit turn-off time is obtained when
Vc
reduces to zero from 2Vs ' This gives
2
n tc C = 4R In 2 1 40 X 10- 6 - 9' 4 x 20 In 2
.
= 0.0815 ~F .
8.11. GOOD INVERTER Here, some of the requirements of a good inverter are enumerated. 1. Its output voltage waveform should be sinusoidal. 2. Its gain (ae output voltage/de input voltage) should be high. 3. Its output voltage and frequency should b~ controllable in the desired usage. For examp1e, inverter must be capable of keeping Vl f constant for some applications . 4. The power required by its controlling circuit should be minimum. 5. The semiconductor devices used in the inverter should have minimum switching and conduction losses 6. Its overall cost must be minimum without sacrificing its reliability. 7. Its working life must be long. 8. It should produce minimum electromagnetic interference (EM!). 9. The size of t h e filters r equir ed should be small. Many of the points listed above, apply equally well to other types of power-electronics converters. 'PROBLEMS
~
8.1 . (a ) What is an inverter? List a few industrial applications of inverters. (b) What are line-commutated inverters? How do they operate? Explain the difference between line-commutated and force-commutated inverters . . (c) What are the two ma 'n types of inverters? Dis tinguish betw een th em eJ!:plicitly . 8.2. (a ), Describe the working of a s' ngle-phase half-bridge inverter. What is its m ai n drawback? Explain how this drawback is overcome. Discuss how out put power in single-phase full -bridge invert er becomes four ti mes t he power h andled by a single-phase half-bri dge inverter.
498
P ower Elect r onics
[Prob . 8]
What is the purpose of connecting diodt:s in antiparallel with thyristors in inverter circuits ? Explain how these diodes come into play. 8.3. A single-phase full-bridge inverter may be connected to a load consisting of (a) R (b) RL or RLC overdamped (c) RLC underdamped. For all these loads, draw the load voltage and toad current waveforms under steady operating conditions . Discus-s the nature of these waveforms. Also indicate the conduction of the various elements of the inverter circuit .. Is it possible for this inverter to have load commutation? Explain. 8.4. For a single-phase full bridge inverter, Vs = 230 V dc, T = 1 ms. The load consists of RLC in (b )
~eries with R
~
= 12 . Q " wL
=8 Q ~ we = 7 Q .
the wavefonns for load voltage vo, fundamental component ofoutput current i 01 , source current is and voltage across thyristor 1. Indicate the devices that conduct during different intervals of one cycle. Find also the rms value of fundamental component of load current. (b) Find the power delivered to load due to fundamental component. (c) Check whether forced commutation is required or not. [Ans. (a) 132.586 A (b) 21094.8 W (c) It is required.]
(a) Sketch
..
8.5. (a) Write Fourier series expression for the output voltages and currents obtained from single-phase half-bridge and full-bridge inverters. (b) A single-phase bridge inverter is fed from 230 V dc. In the output voltage wave, only fundamental component of voltage is considered. Determine the rms current ratings of an SCR and Ii diode of the bridge for the following types of loads:
R
2Q (ii) roL = 2 Q Find also the repetitive peak voltage that may appear across a thyristor in parts (i) and (ii) . . [Ans. (i) 73.211 A, zero A, 230 V (ii) 51.776 A, 51.776 A, 230 Vl (i)
=
8.6. (a) A single-phase full bridge inverter is connected to a dc source of V,. Resolve the output voltage waveshape into Fourier series. (b) A single-phase full-bridge inverter delivers power to RLC load with R = 3 Q and XL = 12 Q. The bridge operates with a periodicity of 0.2 ms. Calculate the value of C so that load commutation is achieved for the thyristors. Turn-off time for thyristors is 12 !lS. Factor of safety is 2. Assume the load current to contain only the fundamental component. [Ans. (b) C = 2.148 !IF.] 8.7. A single-phase full-bridge inverter feeds power at 50 Hz to RLC load with R = 5 Q, L = 0.3 Hand C = 50 )IF. The dc input voltage is 220 V dc. (a) Find an expression for load current up to fifth harmonic. Also, calculate: (b) power absorbed by the load and the fundamental power, (c) the rInS and peaJt currents of each thyristor, .(d) conduction time 'of thyristors a nd diodes if only fundamental component were considered. [Ans. (a) 9.036 sin (rot - 80.72°) + 0.357 sin (3 wt - 88.90°) + 0.122 sin (5 wt - 89.38°) (b) 204 .54 W, 204.12 W (c) 9.044 A, 4.522 A Cd) 5.513 ms, 4.487 ms] 8.8 . Describe modified McMurray half-bridge inverter with appropriate voltage and current waveforms. The total commutation interval may be subdivided into certain well-defined modes ..,for the purpose of explaining its operation. 8.9. (a ) For a single-phase modified McMurray half-bridge inverter, find an expression th~ves . the circuit turn-off time for the main thyristor in terms of load current, peak capacitor current etc. Discuss how comm utatin g circuit components can be designed on the basis of minimum commut ation energy The r elevant voltage and current waveforms must be drawn. (b ) A single-phase modifi ed McMurray inverter is fe d by a de sou rce of 230 V. The de source volt age m ay flu ctuate by ± 25%. The load current durin g commutation may vary from 20
· Inv erters
[Prob.8)
499
to 80 A. If thyristor turn-off time is 20 !lsec, calculate the values of C and L. Use a factor
of safety of 2 .
Also, obtain the value of resistance that gives critical damping.
(Ans. (b) 16.547 !J.F, 34.1895 !J.H, 2.875 OJ 8.10. Describe McMurray-Bedford half-bridge single-phase inverter with relevant voltage and cur rent waveforms. The working of this inverter may be explained in certain well defined modes . Enumerate the simplifying assumptions made. 8.11. Discuss the principle of working of a three-phase bridge inverter with an appropriate circuit diagram. Draw phase and line voltage waveforms on the assumption that each thyristor conducts for 180 0 and the resistive load is star-connected. The ~quen.ce of firing of various SeRs should also be indicated in the diagram. ~ 8.12. Repeat Problem 8.11 in case each thyristor conducts for 120°. 8.13. Repeat Problem 8.11 in case ioad is delta-connected. 8.14. A star-connected load of 15 n per phase is fed from 420 V dc source through a 3-phase bridge inverter. For both (a) 180 mode and (b) 120 mode, determine (i) rms value of load current (ii) rms value of thyristor current (iii) load power. (Ans. (0) 13.2 A, 9.333 A, 7840 .8 W (b) 11.43 A, 8.083 A, 5879.02 W] 0
0
8.15. (0) What is the need for controlling the voltage at the output terminals of an inverter? Describe briefly and compare the various methods employed for the controi of output voltage of inverters. (b) For the series-inverter control of voltage, two single-phase inverters are connected in series . Each inverter has output voltage of 400 V and each transformer has primary to s~copdary turns ratio of 1/2 . Calculate the resultant output voltage from this scheme in case firing angles for the two inverters differ by 30° (Ans: (b) 1545.48 V] 8.16. (0) What is pulse width m odulation? List the various PWM techniques. How do these differ from each other? . (b) For a single-pulse modulation used in inverters, show that output voltage can be expressed as U0
. n1t. . . 2, -4Vs . sm -2 sm nd sm nwt n1t
=
n = I , 3, 5
. where 2d is the pulse width.
U
Sketch the variation of
Onm
u01m
.
as function of pulse width 2d. Here .
Uonm
is the peak value
of nt h harmonic component and vOlm that of the fundamental component. (c ) A single-phase full bridge inverter has rms valu e of the fundamental component of output voltage, with single-pulse modulation, equal to 110 V. Com pute the pulse width r equired a nd the rms value of out put voltage in case dc sou r ce volt age is 220 V. [An8 : (c) 67.471° , 134.693 V1
8.17. For a single-phase bridge inverter, source voltage is 230 V de and the load is series RLC with R = 1 n, wL = 2Q and
:c = 1.5 n. The out put voltage is controlled by single pulse modulation
a nd the pulse width is 120°. Determine the magnitude ofrms values offundamental, third,
fifth and seventh harm onic components of th e output current.
Also, find the power deliver ed to load .
( Ans: 160. 43 A, zero, 3.6786 A, 1.8537 A, 25754; 7 WJ
500
Power Electronics
(Prob. 8] 8.18.
(a) For
the symmetrical two-pulse modulation shown in Fig. 8.29, prove that (i) the magnitude of nth harmonic voltage is
• nd
8V . -;;ts sm ny sm 2 1t-2d d Y=n+l+N
(ii)
( b)
where N = number of pulses per half cycle. Describe how multiple-pulse modulated wave can be generated from carrier and reference waves. Hence show that (i)
we
number of pulse per half-cycle, N = 2w'
(ii )
pulse width,
~=
[1- ~) . ~
where Vc and Vr are the amplitudes of carrier and reference signals respectively. 8.19. Explain sinusoidal-pulse modulation as used in PWM inverters. Discuss the conditions under which the n umber of pulses generated per half cycle are
~f or (~f
-I}
Here fc and f are the
frequencies of carrier and reference signals respectively. 8.20. A single-phase bridge inverter feeds power to a load of R =12 Q and L = 0.04 H from a 400 V de source. If the inverter operates at"a frequency of 50 Hz, determine the power delivered to load for (a) square wave operation (b) quasi-square wave operation with an on-period of 0.6 of a cycle (c) two symmetrically spaced pulse per half cycle with an on-period of 0.6 of a cycle. [Ans. (a) 5285 .56 W (b) 3400.96 W (c) 2706.34 WI 8.21. ( a ) For harmonic reduction in single-phase inverters, two identical transformers are used in series. If their rectangular output voltage waveforms are shifted from each other by 120°, t hen sketch these voltage waveforms and their resultant waveform on the ass ~p~ tion that transformer secondary voltages oppose each other. Find also an expressiori'fl}~ the net output '/oltage as a function of time. Hence find the percentage derating of the inverter so far as its fundamental powercomponent is concerned. ...... ~" (b) A single-phase full bridge inverter has waveforms for its output voltage Vo and output current io as shown in Fig. 8.58. Explain and indicate the devices that conduct in the various intervals throughout the cycle.
'Is ---,__--.., O
'"/2
.
.
~~ 2,"~ 6 : 3 :
2",
I
2 ' :...J'TT \. 'Tr : ' 6:T~ I
:
j
~..,......-..
wi
I
l
,~
f
:, :
(-Vs ) I
i
I
wt
Fig. 8.58: Pertaining to Problem 8.21 (b ).
[Ans.
(a) Vo = 4; s ( b)
J3[
sin ( Vi +
~)+ ~ sin (srot - ~) + ~ sin (7wt + ~}.J 13.4%]
T 4D2, DlD2, T1T2, T1 .3, T2D4, D3D4, T3T4, T3Dl, T4D2 etc
or T3Dl, D1D 2, T1T2 , TI D3, T2D4, D3D4, T3T4, T4D2, T3D! et c.
~,
[P rob.8]
Inverters
501
8.22. The stepped wave output voltage waveforms of Figs. 8.59 (a) and ( b) are to be obtained by the series cascading of two stepped-wave inverters . Describe how these wave-forms can be . realized. Illustrate your answer with appropriate waveforms through the use of two-level or three-level modulations for Fig. 8.59 (a) ; and only two-level modulations for Fig. 8.59 (b) .
_--..-- ·3V5 ··2V5
_·Vs
O~L-______~-_~~~~~~~__w~t+ ~/2
~
wt
"-----~ dl ~~-~ O~~------~------~~~+ ~/2 7T
(a)
(b)
Fig. 8.59. Pertaining to Problem 8.22.
8.23. (a) A single-phase CSI is fitted with ideal SeRs. Describe its working when its load is a capacitor C. Show that the frequency of input voltage to eSI is twice the frequency of . triggering the thyristors. (b) A single-phase eSI (with ideal switches) has the following data:
1= 30 A, f =500 Hz, Load capacitance = 20 j.l.F
For this inverter, calculate
. (i) the circuit t~rn-off time.
(ii) the peak val ue of reverse voltage that appears across thyTistors. [Ans. (b) 500 !-IS, 750 Vl 8.24. (a) Describe a single-phase capa.citor-commutated CSI connected to load R with the help of its power circuit di agram and waveforms for gating signals, load ·c urrent, capacitor voltage and current, input voltage and voltage across one thyristor. ' From the equations governing its performance, show that load current is given by
(b)
i =1[1-2 exp.(-tlRC) ] o 1 + expo (- TI2RC) A single-phase capacitor-commutated eSI connected to load R has the following data: R =40 n, C = 50 j.l.F, f = 500 Hz, Source cunrent =40 A For this CSI. (i) obtain an expression for tbe outpu t current as a function of time and find its value at t = 0 and t = T12, (ii) find the circuit turn-off time, (iii) comp ute the average value of in put voltage and the power delivered to load. [Ans. (b ) 40 [1- 1.245 exp ol~ 500 t)l, - 9.8 A, 9.8 A; 438. 14j.l.s ; 32.52 V, 1300.8 Wl
8.25. Describe a sin.gle-ph ase a to-sequential comm utated eSI with L load . Write appropriate expressions governin its performance and prove therefrom that total circuit turn-off time for this' inverter is given by
tc =
(1+ ~ ) -JLC
Waveforms for gating signals , capacitor voltage and current and load current should also be sketched . Fin d also ..ne circuit turn-off time for each thyristOJ~. ( ns. tc for aa h th yris tor = ~LCl
502
Power Electronics
[Prob. 8]
8.26. A single-phase ASCI has the following data ; Load inductance, L = 4 ).iH, Source current = 20 A Time during which four diodes conduc~ = 10).is
For this CSI, determine
(a) the value of commutating capacitan ce, (b) the total commutation interval (c) the maximum capacitor voltage and (d ) the circuit turn-off time for each thyristor. [Ans. 10.132 ).iF, 16.366 ).is, 25 .133 V, 6.366 ).is] 8.27. In a I-phase ASCI, with load L, SCRs T3, T4 are conducting a constant current 1= 10 A. If Tl and T2 are turned on at t = 0 to force commutate T3, T4 ; find the time required for the load current to fall to zero. Load L = 10 ).iH and commutating capacitance C = 6 ).iF. Find also the total commutation interval and the circuit turn-off tim~ for each of the SCRs . (Ans. 15.858118 , 19.913 ).is, 7.7461ls1 8.28. A single phase auto-sequential commutated inverter is used to deliver power to a load of R = 12 n from a 240 V dc source. If the inverter output frequency is 60 Hz, thyristor turn-off time 15 ).is and factor of safety 2, then determine the suitable values for source inductance and the commutating capacitors. Neglect all losses and assume a maximum current change of 0.4 in one cycle. . (Ans. 10 H, 3.6067 ).iF1 Derive the formula used for determining C. 8.29. (a ) Describe the working of a single-phase .senes inverter with appropriate circuit and waveforms . ( b ) For this inverter, derive an expression for the output frequency in terms of circuit parameters and Toft'
8.30. In a self-commutated SCR circuit, the load consists of an inductance of 12 mH, a capacitance of 8 ).iF and a resistance of 15 n all connected in series. Check that the circuit will commutate by itself when triggered from zero charge conditions on the capacitor. Calculate the voiltage across the inductor and capacitor at the time of commutation . Find also
J
di at t = 0 (d t '
[Ans. It will comm utate, - 0.538 V s , 1.538 VSI 83.33 Als]
8.31. (a) A single-phase quasi-square wave inverter operates on a 24-V battery and it is required to feed power to a230-V, 50 Hz load. PMOSFETs are used as switches. Draw the .power circuit diagram and explain its operation (b) For a heater load of 1 A at 230 V, find the rms value of fundamental component of inverter output current in case inverter of part (a ) operates with (i ) square-wave output and (ii) quasi-square output with 1200 pulse width. Treat the transformer ideal. [Ans. (b) (i ) 10 .643 A (ii) 12.29 AJ
8.32. Draw the circuit diagra m of a series inverter and indicate the need for an optimum time margin. Also, indicate the m erits and demerits of this inverter . 8.33. Enumerate the various lim itations in a basic series inverter. Explain h ow the limitation of operating frequency i~ surmounted in this inverter. 8.34. Discuss the method of over coming the intermit tent power fl ow in a basic series inverter. Ill ustrate your answer with relevant circuits and waveforms. 8.35. A single-phase modified series inverter with coupled inductors has the following data ; Vs = 220 Vdc, R = 2 n, Ll = L2 = 40 ).lR, C = 8 ~, operatin g frequency = 5 Hz, tq = 12 IlS. Determine (a) circuit t urn-off time and (b) the maximum possible operating freque ncy, taking a factor olf safety of 2. · [Ans. (a ) 37.17 ).is (b ) 5758.2 Hz] 8.3S. In Prob. 8.35, initia l voltag,e across capacitor is 60 V and ou tput current is assumed sinusoidal. Calculate the load power, an d aver age and rms values of thyristor current. tAns. 4075. 6 W ; 18.525 A, 31. 916 AJ
Inverters 8.37. (a) Calculate the R = 100 n, (b) Calculate the L = 5 mH,
[Prob.8]
503
maximum possible frequency of a series i!.lverter~;ith the following data: L = 0.05 H, C = 10 )IF . , output frequency of the series inverter if the parameters are: C = 0.2 IlF, R = 200 n, T off =- 0.2 ms. [Ans. (a) 159.15 Hz (b) 2442.36 Hz]
8.38. Describe the working of a single-phase parallel inverter with relevant circuit and waveforms. ~.39. A single-phase parallel inverter delivers power to a resistive load through a centre-tapped ideal transfonner. Derive expression for the capacitor voltage on the assumption of constant source current. Hence obtain therefrom an expression for the circuit turn-off time. 8.40. Discuss the requirements of a good inverter.
Chapter 9
AC Voltage Controllers -_ ..... __ ... __ ..........................................•. -_ .................
--_. __ ......•........ _-
In this Chapter • • • •
Principle of Phase Control Principle of Integral CyCle Control Single-Phase Voltage Controllers Sequence Control of AC Voltage Controllers (Transformer Tap Changers) ....... .. ................... __ ................ __ .......................... ...... .. .. __ ....•....
----
-
AC voltage controllers are thyristor based devices which convert fixed alternating voltage directly to variable alternating voltage without a change in the frequency. Some of the main applications of ae voltage controllers are for domestic and industrial heating, transformer tap changing, lighting control, speed control of single·phase and three·phase ac drives and starting of induction motors. Earlier, the devices used for these applications were auto-transformers, tap-changing transformers, magnetic amplifiers, saturable reactors etc. But these devices are now replaced by thyristor-and triac·based ac voltage controllers because of their high efficiency, flexibility in control, compact size and less maintenance. AC voltage controllers are also adaptable for closed-loop control systems. Since the ac voltage controllers are phase-controlled devices. thyristors and triacs are line commutated and as such no complex commutation circuitry is required in these controllers. The main disadvantage of ac voltage controllers is the introduction of objectionable harmonics in the supply current and load voltage waveforms, particularly at reduced output voltage levels. The object of this chapter is to study single-phase ac voltage controllers so far as their principle of working and gating signal requirements are concerned. Their use in transformer tap changers is also considered. For regulating the power flow in ac voltage controllers, control str.ategies are of two types : 1. Phase control 2. Integral cycle control These are no\,.. described in what follows.
9.1. PRINCIPLE OF PHASE CONTROL Principal of phase control with respect to phase-controlled rectifiers has already been described in Art. 6.1. Th e basic philosophy of phase control in single-phase voltage controllers is the sam e. Here, switching device is so operated tha t load gets connected to ac source for a part of e3.ch half-cycle (or a part of each cycle) of the input voltage. For ill ustrating the principle of phase control in single-phase voltage controllers, consider th e power circuit diagram oi Fig. 9.1 (a ). It consists of one thyristor in antiparalI el with one diode . This circuit configuration is called single-phase half-wa ve voltage controller . "Then SCR is for-.. . a rd biased dur ing posi tive half cycl e, it is turned on firing angle a. Load voltage at once
AC Voltage Controllers
505
[ArL 9.1) v. T1
+ "v Vmsinwt
0
""
Cl
I.
+ 0'
,.1
-
I.
Vm
/
,,
w
1"/ 2
,
f-n
01
I I,
I I
,,
I -------i
R
21'J, '
W
-
21'J / '
W2
wI
1 :"-TI -
, .-
I
w
a. ~ "'/2
~
~)
Fig . 9.1. Single-phase half-wave ac voltage controller (a) Power-circuit diagram and (b) voltage and current waveforms.
V jumps to Vm. sin n, likewise load current becomes ;
sin a. The waveforms of load voltage flo
and load current io are drawn in Fig. 9.1 (b). Thyristor get turned off at wt =1t for R load. After wt = n, negative half cycle forward biases diode Dl . therefore Dl conducts from rot = n to 2n. Note that only positive half cycle can be controlled, nega~i.ve h alf cycle cannot be controlled. As such, single-phase half wave voltage controller is also called single·phase unidirectional f)ol tage controller. Fig. 9.1 (b) reveals that positive h alf cycle is not identical with negative half-cycle for both voltage and current waveforms. As a result, de component is introduced in the supply and load circuits which is undesirable. From Fig 9.1 (b). rms value of output voltage is given by
'12
~ : [ 2In (" II! sin' ",I d (cot) ] • .2 r" II! [ v,,: II! 4n J. (1 - cos 2 cot) d (Olt) : 4n (2n -
V
V,,:
~m
[H
(2n-a)+ Sin22a})'12
a) +
sin 2a ] 2 ... (9.1)
Rms load current, AV,e rage value of output voltage is, Vo =
or
V
Vo =~ 2n I-coswt I"a
Vm
j
;1t (Vm. sin rot . d (Wi)
: - (cos a- I ) ... (9.2) 2n Power circuit diagram of single·phase full,wQf)e uoltage controller is shown in Fig. 9.2 (0). It consists of two SeRs connected in antipara11el. During positive halfcycle. Tl is triggered at firing angl e a, it conducts fr om wt = a to n for R load. During negative half cycle,T2 is triggered at wt = 1t + ex, it conducts from rot =n + a to 2n. Voltage and current waveform s are sketched in Fig. 9.2 (b ). This figure reveals that positive half-cycle is identical with negative h alf cycle fo r both voltage and current wavefonns. The power circuit of Fig. 9.2 (a), therefore , introdu ces no direct com ponent in the supply and load circuit. This circuit is thus more suited to prClc tical
506
Power Electronics
[Art. 9.1J
Vo I
(0
+
+ 'V Vm sln wt
wI
2.
T1
,,
R
vol
I
,i
II
i'--T2--;i
'0
T2
:
I I
"
w
WI
W
Fig. 9.2. Single-phase full·wave ae voltage controller (a) Power-circuit diagram and (b) voltage and current waveforms.
circuits than single-phase half-wave circuit. In this chapter, therefore l only full-wave ac voltage controllers are described. Single phase full-wave ac voltage controller is also called single phase bidirectional voltage-controller. Fig. 9.2 (b) shows that rms value of output voltage is
~ ~ ( V~ sin' 001 . d (001) ]""
V =[
Sin2rot .,.or = ~ 21t 100I _ 2 or
V
~.~ iz:-[
H
l"
1
a
(n - Il) + Sin 2 ex } JI2 2
... (93)
Average value of output voltage would be zero. It has been stated above that ac voltage controllers are phase-controUed converters. The principle of phase control has been illustrated in Figs. 9. 1 and 9.2. In these figures, the phase relationship between the start of load current and the supply voltage is controlled by varying the firing angle. Recall a similar statement made in Art. 6.1 for phase-controlled rectifiers. A1?J the controlled output in Fig. 9.1 and 9.2 is ac, these are called phase-controlled ac voltage controllers or ac voltage controllers. E xample 9.1. A single-phase half-waue ac uoltage controller feeds a wad of R =20 n with an input uoltage of 230 V, 50 Hz. Firing angle of thyristor is 45°. Determim (a) rms value of output voltage (b) power delivered to load and input pf and (c) average input current. Solution . Here V. = 230 V, Vm = ,J2 x 230 V, ex = 45' =~, R = 20 (0 ) From
Eq. (9.1), rms value oflaad voltage is
V~= ~m [H (2n - ex) + Sin22a } = (b)
r
'i'2 ~ 230 [ {~ ( 2n _ ~ ) + sin2 90 } V~
Rms value of load current, Ion = If =
Load power,
n
P, =f;, x R
224.682
20
= 11.2341'
11.2341 A x 20
= 2524.1 W
r
= 224.682 V
AC Voltage Controllers
[ArC 9.1]
507
Rms value of source curre.nt, IJ = rms value ofload current, lor = 11.2341 A
VA = V, .l, = 230 x 11.2341 VA
:. Input Input pf
. Po V o,.· 10,. Also mpulp!= VA = V 1
V
= Vor
*[ ( ~{
2n -
V or
1 =V· From Eq. (9. 1),
"or
r·J
: . Input pf =
' l or
J
. '2}]'12 =mput pf.
Vo r 1 1 sma V, = 2 [ { (2. - ex) +
T2 ;;
~ ) + 8in290 }
r
= 0.9770 (lagging)
(c) From Eq. (9.2), average output voltage V, =
Average input current,
I
,f2 x 230 2n (cos45-1)=-15.17V '.
= V, = _ 15.17 =._ 0 7585 A ON R 20 .
-
Negative value of average input current is due to the fact that current during positive half cycle is less than that during the negative half-cycle.
Example 9.2. A single-phase full-wave ac voLtage controller feeds a load of R = 20 n with an input voltage of 230 V, 50 Hz . Firing angle for both the thyristors is 45°. Calculate (a) rms value of output voltage (b) load power and input pf (c) average and rms current of thyristors.
Solution. Here V, = 230 V, Vm = -J2 x 230 V, ex = 45' =~, R
=20 n
(a ) From Eq. (9.3), rms value of output voltage is V,,= 230 [
H(
n-
~ )+8in290' } )'12 = 219.304 V
V" = 219.304 (b) R ms va Iue 0 f Ioa d current, 1or = If 20 = 10.9652 A Load power,
P, = I;" x R = 10.9652' x 20 = 2404.71 W
Rms value of source current, IJ
= rms value of load current, l or = 10.9652 A
:. Input pf
= 2302404.71 095351 . aggmg. x 10.9652 =.
Also,
From Eq. (9. 3),
". t f l it sin 90 0 9-3VV = mpu p = [~{ ( 1t -"4 ) + 2 }]"' = . " .:> J
(c)
Average thyristor current, ITA =
2~ £: Vm sin Wi, d
(wt)
508
[Art 9.2]
Power Electronics
Vm
= 2 !tR (1 + cos Il) .f2 x 230
= 2. x 20 (1 +cos 45°) =4 .418 A
Rms value of thyristor current, 1'1> = =
[-..!.., r~ 2!tR
•
sin' ·wI . d (wi)
~[ ~{(.-a)+ Sin221l}
=";: ::~O[ ~
{( 1t-~)+
]"'
r
sin2900} It is seen from the expression of current In- that it can be re-written as I T,
r
= 7.7524 A
=Vm [V,,]= 'l2R V" = 219.304 =7 736A 2R V, 'l2 x 20 .
9.2. I'H1Nnl'LE 0 .. INTEGIl.\L CYCLE {'ONTIWL
It is stated above that ae voltage controllers are phase-controlled devices. The principle of phase control is illustrated in Figs. 9.1 and 9.2. In these figures, tbephase relationship between the start of load current and the supply voltage is controlled by varying the firing angle. As the controlled output is ae, these are called phase-controlled ae voltage controllers or ae voltage controllers. In industry, there are several applications in which mechanical time constant o~ thermal time constant is of the order of several seconds. For example, mechanical time constant for many of the speed-control drives, or thermal time constant for most of the heating loads is usually quite high . For such applications, almost no variation in speed or temperature will be noticed if control is achieved by connecting the load to source for some on-cycles and then disconnecting the load for some off-cycles. This form of power control is called integral cycle control. So integral cycle control consists of switching on the supply to load for an integral number of cycles and then switching off'the supply for a further number of integral cycles, Fig. 9.3. u, wI
,
;'9 1
,
,,
~T1 ~T2-:
wt
m
. :'
..
. ,' . m
\0
L..-.___
OO - - - - --o<.!_._ _ 011
\.
on _ __ _ _ . ~ l ._
Fig. 9.3. Waveforms pertaining to integral cycle control
off
wt
AC Voltage Controllers
[Art. 9.2)
509
The principle of integral cycle control can be explained by referring to Fig. 9.2 for a single·phase voltage controller with resistive load. Gate pulses i,l' i(l turn on the thyristors Tl, T2 respectively at zero-voltage crossing of the supply voltage. The source energises the load for n (= 3) cycles. When gate pulses are withdrawn, load remains off for m (= 2) cycles. In this manner, process ofturn·on and turn·offis repeated for the control of load ~ower. By varying the n~b er of n and m cycles, power delivered to load can be regulated as desired. The waveforms for source voltage II,. gate pulses and output voltage 110 are shown in Fig. 9 .3 for n = 3 and m = 2. Power is delivered to load for n cycles. No power is delivered to load for m cycles. It is the average power in the load that is controlled. In literature, integral cycle control is also known as on-off control, burst firing, zero-voltage switching, cycle selection or cycle syncopation. For sinusoidal supply voltage, the nos value of output voltage V"," can be obtained as under :
~. [Jo v! sin2 oot d (oot), for first on-cycle 2rt
V; =
.
perla city
'. V! sin2 Olt · d (ool), for s~~ond on-cycle + ...... +
+ J0
f:' v!. 8in2 001 . d (001), for nth on.cYcle] For n on-cycles and m off-cycles, the periodicity = (n + m) 21t radians, see Fig. 9.3.
V~ =.[ 2. (::+ m) f:"v!. .in' 001 · d(001»)",
..
~
]112
~n:m) f~ (1- 2(01) d (Olt) V ~ =~ .,j n =v,j n =V {Ii" -v2 n+m ' n+m = [4.
or
COB
...(9.4)
I
where
V, = rIDS value of Bource voltage
and
k
=_n+m n_ is the duty cycle ofac voltage controller.
Rms load current,
V" l or =R
Power delivered to load
= V:, = ~
(_n_)=k · ~
R n+m
R
R
... (9.5)
= rms value of load current, lor
Rms value of input current, I,
Input VA = V, (rms value of source current)
=V
J •
Input VA
X
:. Input pf
I~
V" =V.. . lor = V.. . R
p f = power delivered t o lo ad
=V:, R
R = V" V,, Vor V,
="'1/ n +n m
={Ii"
... (9.6)
510
Power Electronics
{ArL 9.2J
As each thyristor conducts for 1t radians during each cycle of n on-cycles, the average value of thyristor current is given by
1 2n
ITA =
S' 1m sin oot . d for first on-cycle + 1 S' 1m ' sin rot · d 21t for second on-eycle + ...... + 1 S ' 1m ' sin d (rot), for nth on-cycle} 2ft (rot),
0
(IDt),
0
rot ·
0
=
n f"l . Sinwt . d(wt) 2n (n + m) 0 m
=lm
. _ n_=k
1t
. 1m
.. .(9.7)
1t
m+n
Similarly, rms value of thyristor current is
1TR = [2n (::... m) = 1m
2
~
S:1~ sin' wt · d (wt)f = 1m ..fk
n
n+m
... (9.8)
2
Integral cycle control introduces less harmonics into the supply system , the supply undertakings therefore insist upou the consumers to use integral-cycle method for heating loads and for motor-control drives. AC voltage controllers with on-off control have specific applications as discussed above. Phase-controlled ac voltage controllers are, however, more common. A!J such, phase-controlled ac voltage controllers will only be discussed and analysed in what follows :
Example 9.3. A single-phase "oltage controller has input "o/toge of230 V, 50 Hz and a load of R = 15 n For 6 cycles on and 4 cycles off. determine (a) rms output voltage, (b) input pf and (c) auerage and rms thyristor currents. Solution. (a) From Eq. (9.4), rms value of output voltage is
~
V =V n = 230 or 'n+m (b)
From Eq. (9.6), input pf
Also power delivered to load
=..fk =
~ n+m n =~ 6 6+4
=t'. R= Dr
~ 6+4 6
= 0.7746 lag.
V!, = 178.157' =2116W R 15
Input VA
=230x23~5"t6=2731.74 VA
:. Inputpf
2116 =2731.74 = 0.7746 lag
(c) Peak thYTistar current, 1m =
= 178.157 V
23~5-12 = 21.681 A
From Eq. (9.7), average value of thyristor current,
ITA = k 1m = 0.6 X 21.681 = 4.1407 A n
"
I
[Art. 9.3]
AC Voltage Controllers .
511
From Eq. (9.8 ), rms value of thyristor current, I
_ 1m' -Ik _ 21.681 x ,fQ.6 _ 8 397 A 2 2 - .
TR-
9.3. SINGLE· PHASE VOLTAGE CONTROLLERS
Fig. 9.4 shows four possible configurations of single-phase ac voltage controUers. Fig. 9.4 (a ), similar to Fig. 9.2 (a ), uses two thyristors connected in antiparallel. The trigger sources for the two thyristors must be isolated from one another because otherwise the two cathodes would be connected together and the two thyristors would be out of circuit as shown in Fig. 9.4 (b) . Thus, no control of the output voltage would be possible. r-_ _-."In~Triggcr
+ T1
-
source
T1
'iLt. +
i. 12
+ , V'
o _
T2
0 0
A
+
L
0 0
.0 A
_
1-
(ol
(bl
"
T2
0'
01 .
'" v,
, '0 5 A
-0
(el
to +
+
!
.0 A
_ 0
+
!
Vo _ 0
(d l (el Fig. 9.4. Single-phase ac voltage coa tr ollers.
Singl e-phase full-wave volta ge controller can al so be re alized by using two thyristors TI, 12 and tw.o diodes DI. D2 as shown in Fig. 9.4 (c). Since cathodes ofTl, T2 are connected together, only one triggering circui t can serve the purpose. TI, Dl conduct during positive half cycle and T2. D2 during negative half cyde. As two devices conduct at the same time, there are more conduction losses and, therefore, reduced efficiency ofth e circuit Bl'Tangement of Fig. 9.4 (e ).
512
[An. 9.3]
Power Electronics
Scheme shown in Fig. 9.4 (d) employs four diodes and one thyristor. For this circuit, there is no need for any isolation between control and power circuits. This scheme, therefore offers a cheap ac voltage controller. The voltage drop in the three conducting devices (two diodes and one thyristor) will. however, be more than in Fig. 9.4 (a) and consequently its efficiency would be low. The circuit shown in Fig. 9.4 (e ) uses one triac. This configuration is suitable fo r low·power applications where the load is resistive or has only a small inductance. The triggering circuit for the triac need not be isolated from the power circuit. 9.S.1. Single.phase Voltage Controller with R Load Fig. 9.5 (a) shows a single·phase voltage controller feeding power to a resistive load R. As stated before, two thyristors are connected in antiparallel. Waveforms for source voltage ul • gating pulses i,l, i,2, load current io• source current i l , load voltage uo, voltage across Tl as un and that across T2 as Un are shown in Fig. 9.5 (b) .
~
w.t
wt \92
wt
, :, I (" ... q) a
!
I
211' I (2'Ir +el) I ,
!'
. ,i
.i
i
"!'
wt
i--n 1i!rT2-.jI ir-T1 -!I
,
wt
wt
t. +
v.l
v,
R
Voltagridrop a 'fo5s: TI l ' ,..... (l)
wt
w
W
Fig. 9.5. (a) Single-phase ac voltage controller with R load (b ) Voltage
and current waveforms fo r figure
(a).
,
AC Voltage Controllers
[Art. 9.31
513
Thyristors Tl and T2 are forward biased during positive and negati ve half cycles r es pectively. During positive half cycle, Tl is triggered at a firing angle ex. Tl :'ita:ts conducting and so urce voltage is applied to load from ex. to n. At 7[, both uo, io fall to zero. Just after !t, Tl is subjected to reverse bias, it is therefore turned off. During negative half cycle, T2 is triggered at (!t -r- ex). T2 conducts from n + ex to 2Jt. Soon after 2Jt, T2 is subjected to a r everse bias, it i::; therefore commutated. Load and source currents have the same waveform . From zero to ex, Tl is f9rward biased, uTI = u$ as shown. From ex, Tl conducts, uTI is therefore about 1 V. After r.:, Tl is reverse biased by source voltage, therefore (In = (I" from Jt to 1t + a. From It + ex to 2n, T2 conducts; Tl is therefore reverse biased by voltage drop across T2 which is about 1 to 1.5 V. The voltage variation (lTI across SCR Tl is shown in Fig. 9.5 (b ). Similarly, the variation of voltage (lTI across T2 can be drawn . In Fig. 9.5 (b), voltage drop across thyristors Tl and T2 is purposely shown just to highlight the duration of reverse bias across Tl and T2. Examination of this figure r eveals that for any value of ex, each thyristor is reverse biased for n / oo sec. There is thus no restriction on the value of firing angle a. Firing angle can, therefore, be controlled from zero to 1t and rms output voltage from V, to zero. Here Vs is the rms value cf source voltage. .:. Circuit turn·off time,
'c= -W• sec.
Harmonics of output quantities and input current. It i::; seen fr oll). Fig. 9.5 (b ) that waveforms for output quantities (voltage Vo and current io) and input current i J are non·sinusoidal. These waveforms can be described by Fourier series. As the positive and negative half cycles are identical, dc component and even harmonics ~r~_absent. The output voltage (10 can be represented by Fourier series as under : ... (9.9) n " 1. 3, 5, .. .
where
an =
and
bn
The load voltage
Ua
~
r ".
••
2f
=-
(la
.. .(9.10)
(Olt). cos nOlt . d (W I)
.
(rot). sm nwt. d (wi)
...( 9 .11)
•• during the first half·cycle is
ua =Vm sin wt ....... a < cot < it Substitution of ua = Vm sin wt in Eqs. (9.10) and (9.11) gives On
2V f" sin wt. cos nwt. d (wt) =~
•
•
=Vm
r•
= Vm
[
•
cos (n + 1) a-I _ cos (n - 1) a- I n +l
It
and
b'l
[sin (n + 1) cot - sin (n -1) WlJ. d (wt )
n -l
1
V f" sin wt. sin mot , d (wt) = 2 ---.!!!.
• • f" [cos (n -1) wt - cos (n
Vo.' =-
•
•
'+
1) (O.:J . d (oot )
... (9.121
Power Electronics
[Art. 9.3J
514
=Vm 1t
[Sin n+1+ a_sin n-1 .a] 1)
(n
..
(n -1)
(9.13)
where Vm = V2 V, and V, = rms value of source voltage. For obtaining Eq. (9.12), note that for n
= I, 3, 5 .. .cos (n + 1) rt = 1 and cos (n -1) 1t =1.
The amplitude of the nth harmonic output voltage Vnm and its phase 4ln are given by , , + b'II. and-'+'11. = tan- 1an Vnm = .-van b
...(S.14a)
o
Vom Inm = R = nth harmonic load current
and
... (S.14b )
For fundamental frequency, i.e. for n = 1, V1m and ¢II cannot be obtained from Eqs. (9.12) to (9.14), because these become inJeterminate for n = 1. This difficulty can, however, be overcome by putting n= 1 in Eqs. (9.10) and (9.11) and substituting the value of Vo = V m
sin
CJJi.
.
2f'
and
.
Vm [COS2a-1] 2
a 1 ='1t a Vm smwt · cos rot · d(rot)='1t
... (S.15)
hI =~ (V sin2wt ·
...(S .16)
m
d(oot)
= ~m [Sin22a + (rt-a)]
From the coefficients a 1 and b 1, the ·peak. value cif the fundamental V 1m and its phase ¢I I are given by '
frequ~ncy
voltage
. ..
V1m =[ai + bil l 12 =
11m
and
•'+'
I
:m [{ cos 22a - 1}'+ {Sin22a +(n - a)}r VIm
='If" =
...(S.17a) •
amplitude of fundamental component of load or source current ...(9.17b )
cos 2a - 1 ] =t an- 1 -a1 = t an-, [ b sin 2a+ 2(1t-a)
.. .(S.18)
i
When ac voltage controller is used for the speed control of a single-phase induction motor, only fundamental component is useful in producing the torque. The harmonics in the motor current merely increase the losses and therefore heating of the induction motor and reduced efficiency. Harmonics also cause more noise and vibrations in the motor. For heating and lighting loads, however, both fundamental and harmonics are useful in producing the ac controlled power. In such applications, rms value Vor of the output voltage should be known. It can be obtained from va= Vm sin wt as follows:
'12 V~= [~(~Sin'Oltd(wt) ] =
and
~; [~ {(" - a) +~ sin 2a}r
V" = rms value of load. or source current l or = If
f ... (S.lSa) ... (S.lSb )
[ArL 9.3)
AC Yolt:lgc Controllers
515
The' average power P delivered to load of resistance R is
1. ]
p,V!.v!.[ =I~R =R= 2nR (n -
a) +"2 sm 2a
~
1 sin2a] = nR [ (n-a)+"2
... (9.20)
Maximum power PmtNi is delivered'to load when· a = O.
V'
P~ =-' R
This gives
P:
=
~[(7t -a) +~ sin 2~] = per unit power.
In terms of harmonic components, . ) P =' R'" (101 + 12 03 + I'05 + ...... =
~ (~I + ~3 + ~, + ...... )
Fig. 9.5 (b ) shows that source current waveform is identical with load current waveform. This mean,s that expressions for both load and source Currents for the appropriate harmonics
are the same. Power Factor. Assuming that source voltage re;nains sinusoidal even though non-sinusoidal current is drawn from it, the power factor is given by
p{ = where
Real power _ V,II cos 4'1 = II . cos 4'1 ... (9 .21) Appar~nt power V, · Irm.t Irm.t 11m II ;;; T2 = rms value of fundamental component of load or source current
given by Eq ..<9.17 b) I rm . ;;; l or ;;; nns value ofload or source current, Eq. (9.19 b), $,
=phase angle between
V, and I I. Eq. (9 .18)
Another expression for pf can be obtained as follows : Real power delivered .to load
=
~r
Apparent power delivered to load = V, . lrml V~
= V, ' lor = V, . If
V'.,.IR V., pi= V" Vor' R =-V, From Eq. (9. 19a ).
pi=
i: =[*
{(n- a)
+! sin 2a}]'" =[per unit power)'"
... (9.22 )
The marimum value of rms output voltage and cunent occurs at a. = 0 and are given by V. and V, IR respectively, Eq. (9. 19). For a =0, harmonics are absent , these are th erefore also the maximum values of fundamental r ms voltage and current.
516
Power Electronics
(Art. 9.3J
,
Example 9.4. A single-phase voltage con troller fee ds power to a resisti ve load of 3 n from 23() V, 50 H z StJ/lrce. Calculate : (a ) the maximum values of auerage and rms thyristor currents for any Ii-ring angle a . rb) the minimum circuit tum-off time for any firing angle a, (c)
the ratio of third-harmonic uoltage to fundamental voltage for a
J'
=
(d) the maximum ualue of di I dt occurring in the thyristors, (c) the angle a at which the greatest forward or reverse uoltage is appLied to either of the
thyristors and the magnitude of these uoltages. Solution. (a ) It is seen from Fig. 9.5 ( b ) that current through thyristor flows from 0 to for the first cycle of 21t radians . Therefore, average thyristor current from Example 9.2 is Vm ITA = 21tR (1 + cos oJ
Its max.imum value occurs when a current is I
TAM
= O. Therefore,
1t
maximum value of average thyristor
=Vm = ,J2"X230=34 ,5 12A 1tR Tt x 3
Rms thyristor current, from Example 9.2, is
ITR
=[ 21.f~ ( ; sin J d(WI)r ~n [(n - ex) + ~ sin 2a WI
r
= ;
Its maximum value occurs at a = O.
I
TRM
ITR_V
ALia
I r,wM ITR.\1
= -J22 xx3230 = 54 .211 A
= Vm 2R
Vm
rtR.
It
= 2R ' Vm ="2 = ~ x 34.512 = 54.211 A
for uTl. uT2 in Fig. 9.5 (b ) show that for any value of firing angle a, the circuit tl!.r!l-off tir=J. e is a lways proportional to 7t: radians. (0 ) \Vaveforms
:. ~.
' IT t'l me = -n CI' rCU.l't turn-o
CIJ
.'
1 = 1 = V. 0 1 sec = 10 m-sec. = 2nItr= -r 2 2 x 50
_ .:rd h nrmoni c,""irom Eq. (9.12 ), aJ
=Vm rcos 240' 1t 4
b. = Vm J
-1_ cos 120' d f rom E q. (913 2 -1 ]= 0 .37-,) Vm 1t an . ),
[-Sin240::' _ ;;in 120=]- =0.64952 V",
It
4
2
Th e ampl itud e of third harmonic voltage, from Eq. (9.1 4), is
y"~ m
.'~ b'3JI .-' : 07 Vm .0- -;-
= la l +
It
AC Voltage Controllers
[Art. 9.J)
517
The amplitude of fundamental frequency voltage from Eq. (9.17a) is
,
..
:m[{Sin2120 +_~] + {COSl~O' -l}
VIm
=
V 3rn
0.75 = 2.63634 = 0.2845
VIm
, ]1F2 =2.63634 ; "
(d ) As there is a sudden rise of current from zero to Vwt. I R sin a at firing anglE':: a , di / dt is infinity. . (e) The wavefonns of un , Un in Fig. 9.5 (b ) r eveal that the greatest forward or reverse
voltage would appear across either of the thyristor s when a these voltages is
=~
or a > ~ . The magnitude of
V:'n = ..f2 V...
9.3.2. Single-p hase Voltage Controller with RL Load Fig. 9.6 (0) shows a single-phase voltage controller with RL load. In Fig. 9.6 (b) are shown waveforms for source voltage [J" gate currents i,l and iTl . load and source currents io and i.. , load voltage uo. voltage uTl across SCR Tl and voltage un across thyristor T2. V~sinwt
v,
... -
--
+ vT1 =
,,
TI
is
to T2
+
+
- v12"
v, 'V
Vo
!
1 .,,
~"
t92t
r
', 1 •
'
Vtn
/ ;
iT
·
lJ.
:
l :
\'Tl
,
.'
,.
u: "r2Hr-G-i=!
I
wt
.
•
.
I : .
~ Y~ :
.i"~ :
I"
, ' ,:, :''
:' =9qrj.!
=.J.J-dk.
(c }C! SO
a:t •
.I"~.
i:
I
I :
2": :
•
. , )l " o. .... y..;
w~ -
; (1T +cr )
::: .,
To
t
, ,
/~
}'1
wt •
: :-n -T · . ;--T2f t-TI T ! : i': i I .:, : ,"
;: I ::
;,-,lI ',
.. ..: ;
Q-Y-~
.'
I
I
: :{ 2/T-t cr
o
,
·
wt
io.i,
v
1T
I i'C2lT1'crl
...
R
(a )
11 / 2
·,
•
191
io
",
2.
I
(bJc > itI
Fig. 9. 6. S ingi!! -pr..13e vol tage controller with RL load.
wI
-..1--;w ;;;-,•
518
[Art. 9.3]
Power Electronics
During zero to 11:, TI is forward biased. At o::t = a , TI is triggered and io =in starts building up through the load At n, load and source voltages are zero but the current is not zero because of the presence of inductance in the load circuit. At ~ > 11:, load current reduces to zero. Angle p is called the extinction angle. After n, TI is reverse biased but does not turn off because io is not zero. At P only, when io is zero, TI is turned off as it is already reverse biased. After the commutation ofTl at~, a voltage of magnitude Vm sin ~ at once appears as a reverse bias across TI and a!; a forward bias across T2, Fig. 9.6 (b). From Pto 11: + a , no current exists in the power circuit, theretore, Vo = 0, vn~= - vJ and vT2 = vJ' Th}'l;stor T2 is turned on at (n + a) > p. Current io =in starts building up in the reversed direction through the load. At 211:, v, and Vo are zero but in = io is 'not zero. At Cn + a + iJ, iT2 = and T2 is turned off because it is already reverse biased. At (n + a + r>, Vm sin (n + a + iJ appears as a forward bias across TI and as a reverse bias across T2, Fig. 9.6 (b). From (n + a + y) to (21t + a ), no current exists in the power circuit. As before, Vo = 0, vn = v, and vT2 = - v" At (2n + a), TIts turned on and current starts building up as before. When Tl conducts, voltage drop across it appears as a reverse bias across T2. Similarly, when T2 conducts, un appears as a reverse bias across Tl. It is seen from Fig. (9.6 b) that waveform Vn is obtained after inverting waveform un. Waveform un ~hows that Tl is reverse biased for n radians. Same is true for thyristor T2. Therefore, circuit tUrn-off time tc for each SCR, for any firing angle a, is
°
tc = n/ ro sec. The expression for load current io can be obtained as under: The KVL for the circuit of Fig. 9.6 (a), when TI conducts, gives
. . dio uJ=Vm sm rot=Rto+L dt ... a < Wi < ~ The solution of this equation is of the form V io =~ sin (rot - 41) + Ae-{RIL)t ... (9.23) Z where Z = [R' + (",L)'] "' and $ =tan- 1 (ml.IR) Constant A can be obtained from the initial condition according to which io = at mt = a i.e. at t = a/co. Therefore, V .0= ; sin (a-41) + Ae- Ro.ILf4)
°
or
A
=_ Vm sin (a _ 41) eRIlILw
Z Substitution of this value of A in Eq. (9.23) gives io as
io= ' ; [Sin (cot-$) -sin (a - $) e:cp
{i (: -I]J
.(924)
It is seen from Fig. 9.6 (b) that io= 0 again at cot = ~ or at I = ~ /w . Substitution of thi s condition in Eq. (9.24) gives
sin (~- =sin (a - exp[i (a: ~] $)
$).
.. (9. 25)
AC Voltage Controllers Extinction angle
~
[Art. 9.3J
519
can be obtained from Eq. (9.25).
The conducti'Jn angle y during which current flows from angle ex to angle Y=
~
~
is given by ".(9.26)
-
For a given value of load phase angle ex, angle~. 1800'~~::;:~~~~;;;;----'-1 is determined for various values of ex from Eq. (9.25) I~ and thus a relationship between y and ex can be realized from Eq. (9.26). For the various values of y and Ct, curves shown in Fig. 9.7 are obtained for 'I different values of ¢I. Note that phase angle c;. cannot exceed 90° .
t
It is seen from Figs 9.6 (b) and 9.7 that as
through T2 flows from n + a to n + ex + y. T2 remains off from n + cx + y to 2n + cx. At 2n + cx, T1 is turned on. With progressive decrease in a, y may become equal to n. Under this condition, when y is just equal to n, Tl will be on from a to 'it + a and iTl flows from cx to 7t + ex. Further, T2 will be on from 1t + a to 2n + ex and current iT2 flows from 1t + a to 21t + cx. Thus,
90'
og '----,"='=_~,___+_;;_~~__::.,.__:! 30°
60°
90°
120·
150·
180· .
CI_
Fig. 9.7. Y versus a curves for ac voltage contI-oller of Fig. 9.6 (a).
I
0 to ex -T2 conducts .. (A) a to (n + ex) -Tl conducts (n + a) to C2n + ex) -T2 conducts.and so on This shows that load current will never become zero for any segment of time and therefore for all the time, load is connected to source. Thus, for y =1t, the load voltage is equal to sinusoidal source voltage provided the voltage drop in thyristors is neglected. Under these . conditions. load behaves as ifit is being fed directly by the ac source .. What is the value of ex for )Vhich y =7t and load is directly connected to ac source? For this, consider that RL load, with load phase angle ¢I, is connected directly to ac source. Under steady state, the load current will be a sine wave and lag behind the voltage wave by an angle <\l as shown in Fig. 9.6 (c). The current is positive from $ to 7t + $ and negative from n + ¢I to 21t + <\I. Fig. 9.6 (e). Ifit is required to obtain the current waveform of Fig. 9.6 (e) through the operation of Fig. 9.6 Cal, then of power ci' when y= n,
from
I
oto c;. - T2 conducts to (. +~) -Tl conducts (n + 41) to (27t + ¢I) -T2 conducts and so on ~
... (B)
A companson of expressions (A) and (B) reveals that when ex =¢. y = rr.: . This can be verified by referring to Eqs. (9.25) and (9.26). When
sin (~ - ex) = 0 = sin 7t ( ~ -
or From Eq. (9.26)
Y=~-
This shows that for l-phase BC voltage controll er, w aveforms of Fig. 9.6 ( b ) a re applicabl e only when ~ and that of Fig. 9.6 (c) for a ~ ~.
520
Power Elec tronics
[Art. 9.Jj
Ope r at ion with as $. Assume that ac voltage controller is working under steady state with n=$. From zero to $, T2 conducts and fr om $to(1t+¢\}. Tl conducts; fr om It + $ to 2n: + $, T2 conducts and so on as shown in Fig. 9.8 (b) . When T2 conducts, Tl is r everse biased by voltage drop UT2 in T2. This r everse bias across T } is shown as Un from 00 to ~ , 1[ + $ to 2 It + $ and so on. Similarly, when Tl conducts, T2 is reverse biased by voltage drop in Tl , this reverse bias is shown as un from ¢I to 1t + $, 2n: + $ to 3n: + ¢I and so on. Voltage drop across T1 is shown as Un in Fig. 9.8 (c) and that across T2 in Fig. 9.8 Cd). la ) O <-~------~c-c------ctc-,------~~------ Wl
~ TZ
f------ T1--+-----:::::::::-;-T2: ---'---+---Tt ---'--;-- T2
o a : 9: : (rr+o ) lor T2
Ibl
:, i Oe-~,------+-~;-----~~,~----~~~---W I ! alorTl ~o-i !(2TT .. a)tOrTl tI :---- rr ----"'--;1 !
Ic)"To~ : # ~ ac:¢
(0)
hi
or
\112
ri
U
u
. .
.
-
;
L
WI
F
wI
I ('It-It )
J u frt
!'I j
J~ {r..-a\'d: : ... t ]
Fig. 9.8. Single-phase vol tage controller with RL load and n S 6.
Not let firing angle a be made less than (I. When Tl is triggered at a < $, Fig. 9.8 (c ), T1 will not get turned on because it is reverse biased by voltage drop 1iT2 in T2 which is cond ucting current i T2 . Tl will get turned on only at $ when in = 0 and reverse bias due to voltage drop un in T2 vanishes. Now T1 will conduct from ¢I to 1t + ~ . T2 will he triggered at an angle (n: + a) < (n: +!jl) as shown in Fig. 9.8 (d ). As T1 is conducting, a voltage drop Un in Tl will apply a r ever se bias across T2, as a result T2 "Will not get turned on at it + a; but only at n + 41, when iT} = o. Now T2 will conduct from n +!jl to 2n +!jl and so on . This shows th at load voltage and load current waveforms will not change from what these are at a = 41. Thus the reduction of a below 4> is not able to control the load voltage and load current. The ac output power can be controlled only for a> (D . Note that for as $, " remains equal to it . Thu s the control range of firing angle in phase·controlled ac voltage controllers is e < a < 1SQ:> . Gating s ign a l requirements. For R load as in Fig. 9.5, thyristor Tl stops conducting at n:, 72 is now forward biased after n. When T2 is triggered at 1t + a, it gets turned on as it is already forward bi ased by source voltage. Thus pulse gating is s uitable for R load as shown in Fig. 9.5. Pulse gating is, however, not sui table for RL loads. The r eason for this can be explain ed by referring to F~g. 9 .9(a ). At a, Tl is fired and the current grows as shown. At Jt + a. T2 is fired. As T1 is st1!l conducting, voltage drop across T1 reverse biases T2 at!t ... a , T2 is therefore not turned on at n: + a. At (a + y) , in decays to zero nnd T1 stops conducting. as a result T2 gets forw ard bia:::ed but the gate pulse i&'Z applied to T2 at it + ct is already zero and th er efore T2 does not get turned on . At (2n: + a ), gate pulse is applied to Tl, it gets turn ed on because it is a!r~ady forw
-.
[MI. 9.3]
AC Voltage Controllers
521
v, 0
io~t H ig~t
i
i
I
i
i r~ I I ~ I I ~ I, i ! II::"" I ,' '
I
(21'1'+u)
;
'
i,
wt
I
:i:i ,
I
I
'
I
I
wI "
, •
wt•
~1'1'
n
fzl'l' +~ ~1'1' I
I
"' , i" :,
II
wt"
I
I
(0)
I
i
JlDDDD
(2'11+w
('II'+u) (2",+ct) ( Cl+V)
(b )
I ,. I:
o~~~·~t~~~~~~~~~
wI•
3'1f
•
wt (d)
Fig. 9.9. Types of gating signals
Fig. 9.9. (a) Single-phase voltage controller with RL lC'lad with pulse gating.
(b) pulse gating (e) continuous gatip.g (d ) high-frequency carrier gating.
waveform due to the conduction of Tl alone. This half-wave rectifier operation of the ac voltage controller is undesirable. This difficulty can, however, be o·...eroome by applying a continuous gate single to the SeRs Tl and 1'2 so that when iTl becomes zero at (ex + 1), 1'2 gets turned on due to the presence of continuous signal. A continuous gate signal is shown in Fig. 9.9 (c). The duration of '" continuous gate signal should last. for a period of (n - ex)/ w seconds. Strictly speakmg, sustained gate-pulse may not last from ex to 7t as shown. Fora =$ or a > $, agate pulse of narrow widthi5 sufficient to trigger the SCR In case a < $, miniml,l1l1 ,..;dth of gate pulse should be equal to $ plus the angle required for the current to reach latching current value.
.
In practice continuous gating is undesirable as it leads to more heating of the SCR gate and at the same time, it incr eases the size of the pulse transformer. The technique that mitigates the above disadv antages of continuous gate signal and ensures thyristor tum on is to use a train of firing pulses from a. to 1t as shown in Fig. 9.9 !d) . This type of signal is also term ed as high-frequency carrier gating. Example 9 .5. A singLe-phase-voLtage controller is employed for controlling the power flow from 230 V, 50 Hz source into a Load circuit consisting of R = 3 nand ooL = 4 n. Calculate (a) the con trol range of fi ring angLe, (b) the maximum' value of rms load current, (c) the ma:"Cimum power and powe r factor, (d) the maximum ualues of auerage and rms thyristor currents, (e) the maximum possible ualue ofdildt that may occur in the thyristor and (f) the conduction angle for \X =0° and a = 1200 assuming a gate pulse of duration 7t radian.. Solution. (a) For controlling the load, the minimum value of firing angle a. = load phase angle , 4> =tan- !
~ = ~an- l ~ = 53.13°. The maximum possible value
of a is 180e .
: . Firin g angle control r an.ge is 53.13° ~ a ~ 180°. (b ) The maximum value of rIDS load current l or occurs when a = ¢> = 53.13 ~ . But at this value of firing an gle, the power circuit of ac voltage controller behaves as if load is dir ectly connected to ac source. Ther efore, maximum value of rms load current is 230 230 T" = ~R' + (wL)' = + 4' = 46 A.
'13'
522
Power Electronics
[Art. 9.3J (c)
=I~ R = 46 2 X 3 = 6348 W
Ma:Gmum power
t'", R 46 x 3 Power factor = V I = 230 = 0.6 . •
(d ) Average thyristor
0
current is maximum when a = ¢I and conduction angle r =1t. From Fig.
9.6 (c ),
1
fa.. V;
I TAM = 2. a
sin (wt-$) d(wt)
=Vm = 1tZ
-<2x230 =20707A 1t X ~32 + 42 . .
Similarly, maximum value ofrms thyristor current is
ITm=~~J:" {; Sin(wt-af =
d(wt)]",
~ = ~: ~30 = 32.527 A
dio (e) Maximum value of di occurs when a
=¢I . From Eq. (9.23 ).
-
..
"
dio w · V", dt = Z cos(wt-$)-O.
Its value is maximum when cos (oot - ¢I) = 1
.
(~:
L
= -<2:230 52. x 50 = 2.0437 x 10' A/sec.
(f) For a = 0· , Fig. 9.7 shown that conduction angle y i. 180·. For a= 120· and ~ = 53.13· , Fig. 9.7 gives a conduction angle of about 95°.
Example 9.6. For the circuit shows in Fig. 9.10 (a), sketch the lJ)aueforTn$ of output uoltage and current for the following ualues of firing angles. (a) Only T2 is triggered at"!t =0, 21t, 41t etc. (b) Only Tl is triggered at'oot =6, 2rt, 41t etc. (c) T2 i s triggered at Wi = 0, 21t, 4Tt etc. But Tl is triggered at wi = ex. 2rt + a. 4rt + a and so on. Take a around 40°.
Solution. (a ) At w =0°, supply voltage is passing through zero and becoming positive. Therefor e, when th yristor T2 is triggered at wi = 0, 21t, 41t etc. it gets turned on and load voltage U o equals source voltage u•. At Tt, 3rt etc, as source vo:tage tends to become negative, T2 is turned off as load current io is zero. Lo ad voltage Uo = 0 from 1t to 2n etc. as shown in Fig. 9.10 (b·il. For R load, current waveform is identical with voltage waveform . (b ) This part is sim il ar to part (a ), except that the voltage amplitude is now -12 · 230 volts, F ig, 9.10 (b·il). (c) At cd = 0, 2rc , 4/t etc. when T2 is triggered, it gets turned on and Uo =-12.115 sin wt . At wt .; (X , 2it + a, 4it + a. etc. when fonv8l'd bi as ed thyristor Tl is trigger ed, it gets turned on. But
wh en Tl gets on, a voltage equ al to ..f2. 115 sin Ct appea rs as r everse bi as across T2, it is th erefoTe turned off a t Wi = a, 2it + a e t c. T hus . fr om wi = 0 to a , Uo follows th e cu rve
[Arl. 9.4J
AC Voltage Controll ers
(i)
§~
... 'V
-
-
I
(il)
-.
R
! I
Yot
wi
"
Yo (Iii)"
1
-
• wt
I
...
;;,...
115\1
f
- -· /2.230V
io
115\1
...
Y,
.""
.
,!~,. I I i
•
Yo
T1
...
YOh O~J2·1I5V
523
~~
'"\ {2'Ir -tal
"
,.
wi
(b)
(a)
Fig. 9.10. Pertaining to Example 9.4 .
..f2 . 115 sin rot but from cot =a to x, Vo follows the curve .,[2 · 230 sin rot a,s shown in Fig. 9.10 (b-iii). From 1t to 2ft, Vo = O. From 21t to 21t + a, Uo = v2.115 sin rot and from 2ft + a to 3rt, Uo = -.f2. 230 sin CJ:t and so on. As the load is resistive, load current waveform is identical with
load voltage waveform. 9.4. SEQUENCE CONTROL OF AC VOLTAGE CONTROLLERS (TRANSI'OMIER TAP CIlANGERS)
Sequence control'of ac voltage controllers is employed for the improvement ofsysterp. pf and for the reduction of harmonics in the input current and output voltage. Sequence control of ac voltage controllers means the use of two or more stages of voltage controllers in parallel for the r egulation of output voltage. The term 'sequence control' ""leans that the stages of voltage controllers in parallel are triggered in a proper sequence one after the other so as to obtain a variable output voltage with low harmonic content. The object of this section is to describe two-stage as' well as multistage sequence control of voltage controllers. A single-phase sinusoidal voltage controller is also explained. 9.4.1. Two-stage Sequence Control of Voltage Controllers A two-stage sequence control of ac voltage regulators employs two stages in parallel as shown in Fig. 9.11 (a). The turns ratio from primary to each secondary is taken as unity for convenience. This means that for source voltage Vs = V m sin cot , VI = V2 =V m sin rot and sum of two secondary voltages is 2Vm sin rot. The load may be R or RL. For both types of loads, jor obtaining output voltage control from zero to rms value V, use only thyristor pair T3, T4 in Fig. 9.11 and k eep T1 , T2 off. For zero output voltage, a is 180 0 for T3, T4 and for V, C( is zero. For output voltage control fr om V to 2V, a for thyristor pair T3, T4 is always zero and for thyristor pair TI, T2 ; a is varied from zero to 180D • The main adv antage of two-stage sequence control oj ac voltage controller over single-phase full-wave ac voltage controller is the reduction of harmonics in the load and supply currents.
{Acl. 9.4J
Power Electronics Tt
+
'0 +
(b>
(0)
Vo
12,2v
I!v
-- '- ' -
(c>
Fig. 9.11. (a ) Two-stage sequence controlled ae voltage controll er (b) R load
(c) RL
load.
(a) R esis ta nce load. For r esistance load, the load current waveform is id entical with output voltage waveform . When thyristor pair T3, T4 is in oper ation with Tl, T2 off, then the output voltage and current waveforms are as shown in Fig. 9.5 (b ). The rm s val ue of output "oltage under this operation is given by Eq. (9. 19 a ). When both pairs T l, T2 and T3, T4 are in operati on, then firing angle fo r T3. T4 is always zero whereas firing angle a for pair T1. 1'2 is varied from 180= to zero for obtaining ou tput v\lltage from V to 2V. The output voltage, when thyristor T3 is triggered at wi = 0, follows u2 =Vm sin wt curve. When SCR T1 is triggered at Wi = Ct, voltage lit r everse biases T3, it is therefore turned off. ..:\Iter th is. T 1 begi ns conduction and the ou tput voltage j umps from U2 to (u l + u2) and foll ows 2 \ ~~ si n wt cunre. At wt = ro, output voltage and current are zero. At this instant, T4 is triggered and outp ut voltage follows V m sin Wi curve. At wi =1t + ex, when fonvard biased SCR T2 is triggered, T4 is r eve rse bia::ied by Vm s:n a. it is therefore turned off. When T2 begins c o ndu~ti on, output voltage foll ows 2 Vm sin mt curve as sho'wn by the negati': e half cycle in Fig. 9.11 (b ), In thi s fi gure, output current ' . .·avefo rm to is shown identical with output villtage w a \' efo rn~ L'a.
AC Voltage Controllers
[MI. H )
When buth pairs 1'1. 'f2 a~a·T3~'T4 ate in use, or for the two·st age sequ encE' control of s.inglfo -pha."i e uc voltage controllH, the output \·oltage vQ wavefo rm is shown in Fi~ . 9.11 { bl. From 0 to a, V follows Vm sin wt curve and from a to rt, V adopts to 2 Vm sin. wt curve. Rms value of this output voltage V can now be obtained as u nder. . Q
Q
Q
V., = [
~ { ( V~ sin' rot . d (rot ) + ( 4 V~ sin' Wi . d (~) } ]
v!. = ~ { ; or
( (1- cos 2 \OJI) d (wI)
l/1
~ 2 v!. (1 - cos 2 rot ). d (WI ) }
V~ = [~ ( o._Sin22o. )+ 2;' ( ._c:+sin22o. )r'
... (9.27)
Vw Rms value of load current, IQr =i f
It is also seen from Fig. 9.11 (b ) that T3 conducts for a degrees from lilt Therefore . rms value of current for thyristor T3 (or T4 ) is I n, = [
1 ., 2rtR-
r V~
sin' wI. d (rot ) ] "'
Q
=V2Rm [ ~.. ( 0. -
Sin 2 0.
)]"2
2 Al so, thyristor Tl conducts for (rt - a) radians from wt =a to wi of thy ristor Tl (or T 2 ) is given by I TlR
=OQ to lilt =a .
[~r 4 ~ sin' WI . d (rot ) ]'12 2 JtR a Vm[1;; ( • sin22o. ) ]'12 =/f
...(9.28)
=rt . Therefore , rms current
=
- 0.+
... (9. 29)
Wh en thyristors Tl , 7'2. T 3, T4 are in use, upper second ary winding handles both cycles of top SCR pair; whereas lowe r secondary winding has to handle both cycle of top and bottom pairs. :. Rms current rating of upper secondary, I, = ,fi X I T" and rm s current rating of lower secondary. I, = i(,fi x IT")' + (,fi. In ,)' ) ,12
... (9.30) ... (9.3 1)
If voltage rating of upper and l'1!Yer second ary windings ar e equa l to V" then tran:lfor mer rating = V, (11 + 13 ) VA . Exampl e 9.7. A two-stage sequence controlled single-phase ac voltage controller is feeding a load of R =20 n The .'iOllrCi! milage is 230 V. 50 Hz and turns ratio from primary to each t ra ns/orm .:r seconda ry is /lfII !.'" ,""or CU.'D"sl age seqllent.:e conu()i, the {irin g eng le of" upper thyristors is 60~. Ca lculatot (a ) rms ua lue oj outpu t uoltage (b) rms ualue of current fo r upper thyristors (c) rms ualue of current fo r lower thy ristors (d) transformer VA rating and (e) i npu t power factor.
Solut ion. Here V, = 230 V, Vm
=
{2 x 230 V, a = 60", R = 20 n
526
Power Electronics
[Art. 9.4J (a) From Eq . (9.27), rms value of output voltage is
v
- or
(b)
(!'3. _sin2120 ) + 2 x 2
= [ 2 X 230' 2n
X 1t
230' (
r
1t
_!'3+. sin 2120' )]'12 = 424.94 V
From Eq. (9.29) rms value of current for upper thyristors is 230 lTV '= {2 ;0 [
~ ( n - ~ + sin ~20' )
= 14.585 A
(e) From Eq. (9.28), rms value of current for lower thyristors is
1 = {2 x 230 ,.". 2 x 20
[! (!'. _ n
.
1/ 2
sin 120" )] 2
3
= 3.595 A
(d ) Rms current rating of upper·secotidary,
1, = {2 x 14.585 = 20.623 A Rrns current rating oflawer secondary, from Eq. (9.31) is
3.595)')'12 = 21.244 A
1, = [({2 x 14.585)'+ ({2'x Transformer rating = V, (II + 13)
'" 230 (20.623 + 21.244) = 9629.41• VA I ...
.v.;,.
(e) LoadPower'P'= R= :. Input pf
.
2
'
",
.,
42494 20 =9028.7W ,
P, 9028.7 .' . . : = VA = 9629,41 = 0.9376 (laggmg)
..
RL load. When pair T3, T4 alone is in operation, then the waveforms of output voltage and current are as shown in Fig. 9.6 (b) for a > ~ and in Fig. 9.6 (c)for a S ~ . For obtaining output voltage from V to 2 V, firing angle for T3, T4 is always zero whereas firing angle for Tl, T2 is varied from 1800 to zero ..It is assumed that during positive half cycle, firing pulses for Tl, T3 last from mt =0 to rot = 7t and during neg~tive half cycle, firing pulses , for T2, T4 extend from 'rot = '1t to 2Tt, ' During positive half cycle, T3 is conducting and a voltage v2 =..J2 V sin mt is applied to the load, At rot = a, when Tl is triggered, T3 is tunied off by reverse voltage VI and output voltage jumps to VI + Vz = 2 Vm sin rot as shown. At rot = n, (VI + Vi> reaches zero but output current io is not zero because of the presence of L in the load. Thus, Tl continues conducting until wt =~ . where io decays to zero and Tl. already reverse biased by (ul + is turned·oft'. Thyristor T4. already gated at wt = 1f, starts conducting lowering the voltage to V2 as shown. At rot =1t + a, T2 is triggered, v l turns off T4 and output voltage in th e n egative half cycle jumps to (u l + vz) as shown, At wt = 21t, (Vt + (2 ) reaches zero but io is n ot zero because of L, At wt = 1t +~, io reach es zero. T2, already reverse biased by (V I + is turned off and T3 already gated at wt = 21t is turned on lowering the voltage to Vz at wt =1t +~ . At wt = 2n + n, already gated Tl turns on and T3 turns off and output voltage jumps from V z to (VI + IIi> in t he positive half cycle. The output voltage and output current waveforms are shown in Fig. 9.11 (e).
vv.
vv,
1:
,I"
[Act. 9.4J
AC Voltage Controllers
527
9 .4.2. Multistage Sequence Control of Voltage Controllers Multistage sequence control of ac voltage controllers is employed when it is desired to have harmonic content lowe:r than that in a twrrstage sequence control. Fig. 9.12 shows the power circuit for n -stage sequence control of voltage controllers. In this figure, the transa I v former has n secondary windings. Each • • J secondary is rated for v.ln where v, is the oA" source voltage. Voltage of terminal a with n resp ect to 0 is v,. Voltage of tenninal b is , b (n-1) vs/n ,,I . ~. (n - 1) v!ln and so on. If voltage control ,, from vdO = (n - 3) vsln to vco = (n - 2) v,ln v, ,, -,,is required, then thyristor poor 4 is fired at ,,I , a =0 0 and the firing angle of thyristor pair ' 1 3 c ( I 3 is controlled from a = 00 to 180 0 whereas I I • + all other thyristor pairs are k ept off. ~ Similarly, far controlling the voltage fram Vs 'V ubo = (n - 1) u,ln to v(lO = u., thyristor pair I 4 I . d ( 2 is trigger ed at a = 0 0 whereas for pair 1, , , I . ,,, ·a is varied from 0 0 to 180 0 by keeping the ,,I , remaining (n - 2) SCR pairs off. Thus, the ,, , ,, load voltage can be co ntrolled fr om ,, ;?> ,, v,ln to u, by an appropriate control oftrig.,I n v, /n gering the adjacent thyristor pairs. • The presence of harm onics in the 0" 6 n output voltage depends up on t1;1e A 0 magnitude of voltage variation. If this voltage variation is a small fra ction of the o total output voltage, the hannonic content Fig. 9.12. Multistage sequence control of in the output voltage is small . For ae voltage controllers. _ example, for voltage control' from (n - 2) u/n to (n - 1) 1:J,In, ifvaltage variation is v ,ln« (n - '1) u"ln, then the harmonic content in the output voltage would be small. 9.4.3. Single-phase Sinusoid al Voltage Controll er For obtaining continuous voltage control over a wide range with low harmonic content and improved pf, a multistage sequence voltage controller must have quite a lar ge number of stages . . This is, however, an expensive proposition. An alternative to th is, with less number of stages and called single-phase sinusoidal voltage controller is usually employed; this is shown in Fig. 9.13. This voltage controller has one primary winding and (n + 1) secon dary windings, i.e. singlephase sinusoidal valtage controller employs (n + 1) stages; 0, 1,2, 3 ...n. Th e top secondarywind* ing, numbered A, is called vernier winding. It s rating is u volts. The voltage ratings of the remaining n windings are in geometric progressi on with a r atio of 2. Thus, if v is the volta&e rating of secondary numbered I, th en voltage ra ting of secondary numbered 2is 2 u, that of numbered 3 is 4 v (= 23 - 1 . v), that of numbered 4 is 8 U (2 4 - 1 . u) and that of numbered n is 2" - 1 . ~. volts. The power circuit of Fig. 9.13 uses two sets of thyristors for n windings. Th ese are Tel, TC2, .. ... called con trol thyristors and TB1, TB 2..... called by -pass thyristors. Vernier wind ing h as two pairs afthyri s ~o rs marked TeA and TBA. Thyristors pertaining ton s tages, i.e. Tel , TC 2, ... ,
, ""
J, , ~
~
,
2""
""
J2
"" ""
.
.I,
I
,
""
""
..
528
Powe r Electronics
[A r t. 9.4]
Ten o.nti TB1 , TB2 ,... , TBn are either on or off throughout a cycle. This mean s that con t rol and
by- pass thYTi5tor:. :.lre made to act as switches which remain either on or off during a cycle. When co ntrol SCR pair for any stage is on and its by-pass thyristor pair is off, then voltage of that stagt: would appear across the load and a load current would flow accordingly. On the other hand. if cont rol SCR pair is off and by-pass pair is on for any stage, then this particular stage would be by-passed and will not contribute any voltage across the load. T hus, with an appropriate series combination of secondaries from 1 to n (without vernier winding), the load voltage can be varied from u to (2 11 - 1) u in discrete steps of u. This special featur e of choosing any series combination of secondary w indings is, however, not available in the multistage sequence control of thyristors in F ig. 9.12. ", As stated above , an addition al stage -4., is employed as a vernier to TCA pe rmit continuous control of voltage TBA a' from zero to u. It is a phase-controlled .l seco ndary wind ing. Th is winding contributes harmonics to line and load I ~'" currents. The hannonic content would, TCl however, be much lower because TB l a' co ntribution of voltage by vernie r J winding to the load voltage is only a , s~al1 fraction of the total load voltagE:. . ", I
~ ~Ok t
II
The operation of the power circuit of Fig. 9.13 can now be explained fo:.- a r esistance load. Let it be required to vary the load voltage from 10 L' to 11 L'. For obtaining this voltage range, stages 2 and 4 together with vernier winding are r€quired . For stages 2 and 4, by-pass SCRs are kept off but their control SCRs are kept on all the time. For the remaining stages from 1 to 11, all by-pass SCRs ar e kept on \vhile their control seas are kept off all the time. With this, the circuit of Fig. 9.13 reduces to that shown in Fig. 9 .14 (a ).
+
'" ·s -
o
, _ 3
~o
~4
1
f
J
~2 Y o
If
!
Y J
r
TC2
.,...
J
I
.YO
TC3 0'
J
TC4 J
""
~ TB3 ,
If
L
0 A
n .(2 -1) •
0
l!
I
oJ'
0'
TB2
~ TB 4
If
, ,,
During the pos itive half cycle, ,Jo TC n thyristor T3 is turned on at wi := 0" 2 •
; ~n L I
~
-.-
AC Voltage Controll ers
[PrOD. 91
529
T1
• T3 T2
... L
a A o
~ Fig. 9. 14. Operation of voltage controller of Fig. 9.13 .
W
PROBLEMS 9.1. (a) What is an ac voltage controller? List some of its industrial applications. Enumerate its merits and demerits . (b) What are the control strategies for the regulation of output voltage in ac voltage controllers ? Discuss the merits and demerits of the control strategies listed above . (c) What are the advantages and disadvantages of unidirectional as well as bidirectional controllers? Which one of these is preferred and why? 9.2. (a) Draw the possible configurations of a single-phase voltage controller and compare them. A single-phase voltage controller using two SeRs in anti parallel must have its trigger sources isola ted from each other. Why? Explain with a suitable diagram . (b) Describe the principle of phase control in single-phase half-wove ac voltage-controller. Derive expressions for the average and rms value of output voltage for this voltage controller. 9.3. A single-phase unidirectional voltage controller is connected to a load of R =- 10 n. Input voltage is 230 V, 50 Hz. Firing-angle delay is 30°. Determine (a) nos value of output voltage (b) average and rms values of thyristor current (c) average and rIDS values of diode current and (d) input power factor. IAns. (0) 228.3 V (b) 9.6585 A, 16.146 A (e) 10.352 A, 16.261 A (d ) 0.9926 Jag) 9.4. (a) Discuss the principle of phase control in single-phase full-w ave ac voltage co·ntroller.Derive expression fo r the rms value of its output voltage. (b) A single-phase full-wave ac voltage controller has a load of R = 5.Q and the input voltage is 230 V, 50 Hz. If load power is 5 kW, find (a) firing-angle delay of thyristors and (b) input power fa ctor. [Hint : a: - Sin 2O: = 1.657. By tri al and error a. =92.5°] 2
IAns: (a ) 9'2.5° (b) 0.6874 lag]
9. 5, (al Describe the principle of burst firing for a single-phase ac voltage controller. Derive an expression for the nTIS value of output voltage. (b ) A single-phas e ac voltage controller uses burst-firi ng con trol for heating a loa d of R = 5 n with an input voltage of 230 V, 50 Hz. For a load power of 5 kW, determi::e (i)
530
[Prob. 9)
Power Electronics
the duty cycle (ii) input power faclor (iii) average a nd rms thyristor currents. Derive the expressions used . [Ans. (b) 0.4726, 0.687Jllag, 9.785 A, 22.3575 AJ
'"
9.8. (a ) For a single-phase voltage controller feeding a resistive load, draw the waveforms of source
(b)
voltage, gating signals, output voltage, source and outp'ut currents and voitage across one SCR. Describe its working with reference to the waveforms drawn. Analyse the output voltage waveform of part (a) into various harmonics with Fourier series and find expressions for the . a~plitude of nth harmonic Vnm and its phase ~/.
9.7. (a) For single-phase voltage controller, connected to a resistive load, analyse the output
voltage waveform into various harmonics using Fourier series and find ~xpressions for the amplitude of fundamental voltage component VIm and its ph~ ¢II ' ( b) A 230-V, 1 kW electric heater is fed through Q triac from 230 V, 50 Hz source. Find the load power for a firing angle delay of 70". Derive the expression used for the voltage . . [An~. (b) 713.414 watts]
9,8.
(a)
For a single-phase ac voltage controller feeding a resistive load, show that power factor is given by the expression:
[~ {(n _a) +~ sin' 2a} ]112 (b )
A single-phase half-wave ac voltage controller, using one SCR in a ntiparallel with a diode, feed ~ 1 kW, 230 V heater. For a supply voltage of230 V, 50 ·Hz, find the load power for a firing-angle delay of (i) 0" (ii) 180" (iii) 70". IAns. (b) 1000 W, 500 W, 856.707 WJ
9.9. (a) Compare the merits of controlling the heater power by a triac using integral cycle control . to .. over the phase-angle control. (b) A heater load is controlled through a triac from a single-phase source. Detennine the firing angle delay when the power is at CO 50%, (ii) 70% of its maximum power. Derive the expres sion used . (d A he ater load is controlled by means of single-phase ac voltage controller. Dctennine the firing angle delay, when the controlled power is at (i ) 50%, (ii) 70% of its maximum power. Derive the expression used. [Ans. (b) 90",71.5" (c) 180", 99"1
9.10. (a) Define the term 'p8wer factor'. Derive its expression for a single-phase voltage controller feeding a resistive 'load circuit and show that (b)
Power factor =: [per unit power )112 For the circuit shown in Fig. 9.15, do the following : (i) Sketch the waveforms for two cycles of supply voltage, supply current, load voltage and load current: fO:r a firing angle of about 45" tor- the t WO ' thyristors. (ii) For 230 V, 50 Hz as the supply volt· age, find the power consumed by load in case a = 60" and R "" 10 O. Derive the expression used for power.
T1
01
03
+ R
T2 04
02
t
Vo
I
Fig. 9.15. Pertaining to Prob. 9.1 0 (b ).
(ii i) In case diode D1 gets open circui ted, draw the load current waveform and calculate the power delivered to land. IAns. (b) (ii) 4255.8 W (iii) 2127.9 WI
9.11. A single-phase voltage controller with resista. nce load has the foll owing data : Supply mains: 230 V, 50 Hz, R = 4 O. Calculate : (a ) the firing angle a at which the greatest forward or revers e voltage is applied to either of the thyris tors and the magnitud e of thes e voltages;
AC Voltage Con trollers
[Prob. 9]
531
the greatest forward or reverse voltage that appears acr oss either of the thyris tors for firing angles of 60° and 120 ~ ; (c) the TInS value of fifth harmonic current and its phase for ci -= n1 4.
(b)
[Ans.
(a)
a. -= ~ or at a >~, 325.27 V
(b)
281.691 V, 325.27 V
(c)
27 .284 V, - 63.435°J
9.12. (a) A single-phase voltage controller, with two thyristors arranged in anti parallel, is con-
nected to RL load . Discuss its working when firing angle is more than the load pC angle. Illustrate your answer with waveforms of source voltage, gate singals, load and source currents, output voltage and voltage across both the thyristors. Hence derive an expression for the output current in terms of source voltage, load impedance, firing angle etc. (b) A single-phase voltage controller has the following data : Source voltage '" 230 V at 50 Hz, Load = 0 +j 4- n. Calculate: (i) the control range of firing angle, (ii) the maXimum value of rms load current, (iii) the maximum value of average and rms thyristor currents, (iv) the maximum,value of dildt that may occur in the thyristors, (v) the value of conduction angle for a = 90° assuming gate pulse width of n radians. . [An•• (b) 90" < a < 180" ; 57.5 A; 25.88 A, 40.66 A ; 2.5546 x 1d' AI,ec ; 180"J 9.13. (a) For a single-phase voltage controller, develop a relation between conduction angle yand
9.14.
9.15.
9.16.
9.17.
9.18.
firing angle a and plot their variation as a function of load ph<\se angte 41. Under what conditions conduction angle y becomes equal to 1t ? (b) Discuss the operation of a single-phase voltage controller with RL load when firing ungle a is less than, or equal to, load phase angle 9. Hence show that for a less than 9. output voltage of the ac voltage controller cannot be regulated. (cl For a single-phase voltage cOJ;ltroller, discuss how pulse gating is suitable for R load and not for RL load. Hence show that high-frequency carner gating is essential for RL loads . (a) Des cribe the working of a two-stage sequence control of voltage controllers for R load. What is the advantage of this controller over single-phase full-wave voltage controller? (b) Derive an expression for the,rms: value of output voltage when both stages are in operation. A two-st.f!ge sequence cont.olled single-phase ac voltage controller ,is connected to a load R = 10 n and input voltage is 200 V, 500 Hz. Turns .atio from primary to each transfonner secondary is unity. The firing angle of the upper group of thyristors is 30" for the two·stage control. Calculate (a) the rms value of output voltage (b) the rms value of current for upper thyristors (e) rm3 value of current for low er thyristors (d) transformer VA rating and (e) input powe. factor. (a) Describe the working of a two-stage sequence control of voltage controllers for RL load. (0) Distinguish between two-stage and multistage sequence control of voltage controllers. What are the advantages of multistage over two-stage sequence control? Describe mul-. tistage sequence control of voltage controllers. Descri be a single-phase sinusoidal voltage controller with vernier winding. What are the functions of controlled and by-pass thyristors? Discuss how output voltage waveform from 7 v to 8 v can be obtained from this voltage controller . .' A single-phase bidirectional oc voltage controller delivers a load power of 50 kW with an efficiency and power factor of 0.9 and 0.85 respectively. The input voltage is 230 V, 50 Hz. Determine the maximum poss ible ratings of (0) average and rm s thjTistor cUJ"!"ents and (b) repetitive voltage of the thyristors. [Ans. (0) 121 .92 A, 200 .93 A (c) 325.22 \11
Ch apter 10
Cycloconverters
... -........ ..... -.•.••••......••.•......• ••••. .....• ..................•....................... -.~
~
In this Chapter • • • •
Principle of Cycloconverter-Operation Three-Phose Half-Wave Cycloconverters Output Voltage Equation for A Cycloconverter Load-Commutoted Cycloconverter
•••••••••••••• _ ••••• m •••••••••••••••••••••• •• •
•
•• •
•••
~
•••••••••••••••••••••••••••••••••••••••••••••
A device which converts input power at one frequency to output power at a different frequency with one-stage conversion is called a cycloconverter. A cyc1oconverter is thus a one-stage frequency changer. Basically, cyc1oconverters are of two types, namely: (i) step-down cycloconverters
and
(ii) step-up cycloconverters .
. In step-down cycloconver t ers, the output fre quen cy fo is lower than t he supply frequency fs, i.e. fo < f~· In step-up cycloconverters, fo > f s · The operating principles of step-down cycloconverters were developed as far back as 1930. At that t ime, mercury-arc rectifier was used as a cyc1oconverter for converting three-phase 50 Hz supply to single-phase 16 ~ Hz supply for use in ac traction. system in Germany. A single-phase series4l1otor, when operated at a l ower frequency, gives better operating characteristics. In the United States, a cyclocon vert er comprising 18 thyratrons was employed to drive a 400-HP synchronous motor for several years in Logan power station [61. The cycloconverter systems at that time did not find widespread use only because early systems were not technically attr~ctive and economically viable. Wit h the advent of high-power thyristors, cycloconverters are again becoming popular. At pr esent, t he applications of cycloconverters include the following: (i) Speed control of high-power ac drives
Indu ction h eating
. (i: i) Static VAI Com pensation
(iv ) F or convert i:U'g variable-speed alt ernat or voltage to constant frequen cy output volt age for us e as pow er supply in aircraft or shipboards. (i i )
The general use of cycloconverter is to provide either a variable fr equency power from a fixed input fr equency pow er (as in a c mot or speed control) or a fix ed fr equen cy power from a variable input frequency power (as in ai cr aft or ship bo r d power supplies or win d generators).
of
The obje this cb apter is t o present both 8i gle-ph ase an d three-phase cycloconverters at an introductory level.
Cychconverters
[Art. 10.1]
io~. PRINCIPLE OF.CYCLOCONVERTER OP.ER:AT~ON - -:"';'~~'... '
.'.
533
.
In this section, basic principle of operation of step-up as well as step-down cycloconverter is presented. Single-phase to single-phase cycloconverter, though seldom used in practice, is considered here for describing the principle of operation of both the types of cycloconverters. A single-phase to single-phase device of the mid-point type is shown in Fig. 10.1 (a) and of the bridge type in Fig. 10.1 (b). With the help of this figure, the basic principles of both types of cycloconverters are described here.
r------- --------.,
r---- ------ -- - ---...,
I
: I
I
:
I I I I
+
"s
'V
O~~--o&~---.!;yA
I
,
I
J ------------7p- converter
(a)
.
I
,
~-------.-----
N converter
-J- J
(b)
Fig. 10.1. Single-phase to single-phase cycloconverter circuit (a) mid-point type and (b) bridge type.
10.1.1. Sin gle-phase to Single-phase Circuit-Step-up Cycloconverter For understanding the operating principle of step-up device, the load is assumed to be resistive for simplicity. It should be noted that a step-up cycloconverter requires forced commutation. The basic principle of step-up device is described here first for mid-poin t and then for bridge-type cycloconverters. 10.1.1.1. Mid -point cycloconverter. It consists of a ingle-phase transfonner with mid tap on t h e secondary winding and four thyristors. Two ofthese thyristors PI, P2 are for positive group and the other two N1, N2 are for the negative group. Load is connected between secon dary winding mid-point 0 and terminal A as shown in Fig. 10.1 (a). Positive directions . for output voltage Vo and output current io are marked in Fig. 10.1. In Fig. 10.1 , during th e positive half cycle of supply voltage of Fig. 10.2, terminal a is posit ive 'with r espect t o terminal b. Therefore, in this positive half cycle, both seRs PI and N2 are forw ard biased from eDt = 0 t o wt =11:. As such SCR PI is t urned on at wt = 0° so th t load voltage is pos ' tive with terminal A positive and 0 n egative. The load voltage n ow foll ows the positive en velope of the supp ly vo ~ag e, Fig. 10.2. At instant {J.Jt 1 P I is force commutated nd forward-bias ~d t h yristor N2 is t~Tled on so that load v oltage 's egative v.-ith t ermL--l· 1 0 positive a.."'1 dA negative. Tbe lo ad, or output, voltage now trac~s the negative envelope of the supply voltage , Fig. 10.2. At cot 2 , N2 is force commutated an 1 PI is t rued on, the load voltage is now positive an d foll ows the positive envelope of sUPP~J-' voHage, Fig. 10. 2. After rot = it, terminal b is posit ive with r espect to t erminal a ; both s e Rs P2 and Nl are th er efor e forward biased from wt ::; 11: t o 211:. At wt = 7t, N2 is force commutated an d fonvard bi ased SCR P2 is turn ed on . At rot == 2~ + fs
21
P 2 is forc e commutated and forward bi".3c SCR N1 is turned on . n this
fo
mann er, thyristors P I , T2 for first half cycle ; P 2, Nl in the second half cycle nd so on are
534
[Art. 10.1]
Power Electronics
P2 .
I
SUPPIY voltage envelope
~-----------------~=~ --------~ 1s Fig. 10.2. Waveforms for step-up cycloconverter. . '. switched alternately between positive and negative envelopes at a high frequency. As a result, output voltage of frequency fo; higher than the supply frequency fs' is obtained. In Fig. 10.2, fs is the supply frequency and fo is the output frequency. Also fo = 6 fs in Fig. 10.2. 10.1.1.2. Bridge-type cycloconverter. It con's ists of a total of eight thyristors, PI to P4 i.e. four for positive group and the remaining four for the negative group. When a is positive with respect to x in Fig. 10.1 (b), i.e. during the positive half cycle of supply voltage of Fig. 10.2, thyristor pairs PI, P2 and N1, N2 are forward biased from rot = 0° to rot = n. When forward biased thyristors PI, P2 are turned on together at rot = 0°, the load voltage is positive with respect to xin Fig. 10.1 (b), forward.biased thyristors PI, P2 are turned on together at rot = 0° so that load voltage is positive with terminal A positive with respect tv O. Load voltage now traverses the positive envelope of supply voltage, Fig. 10.2. At rot l , pair PI, P2 is force commutated and forward biased pair N1, N2 is turned on . With this, load voltage is negative with terminal 0 positive with respect to A. Load voltage now follows the negative envelope of . source voltage, Fig. 10.2. At rot 2 ; N1, N2 are force commutated and P1,P2 are turned on. The load voltage is now positive and follows the positive envelope of source voltage. Mter rot =n, thyristor pairs P3, P4 and N3, N4 are forw ard biased, these can therefore be turned on and force cDmmutated from rot = 1t to rot =21t. In this manner, a high-frequency turning-on and force commutation of pairs PI P2, N1 N2and pairs P3 P4, N3 N4 gives a carrier-frequency modulated output voltage across load terminals.
In Fig. 10.2 conduction ofthyristors PI, P2 and N1, N2 for mid-point cy~loconverter of Fig. 10.1 (a) is only shown. It is fairly easy to indicate the conduction of thyristors PI to P4 and N I . to N4 in Fig. 10.2. 10.1.2. Sin gle-phase to Single-phase Circuit-Btep-down Cycloconverter . . .
A step-down cycloconverter does not re quire forced comm u tation . It requires phase-controlled converters cOlmected as shown in Fig. 10.1. These converters need only line, or natural, commutation which is pr ovided by a c supply. Both-mid-point and bridge-type cycloconverters aTe described in what follows : . 10.1.2.1. Mid-poi nt c ycloconverter. T 's type of cycloconverter will be described both for discontinuous as well as continuous load current. The load is n ow assumed to consist of R and L in series. (a) Discon tin"llOUS l oad current. When a is positive with respect to 0 in Fig. 10.1 (a ), forward biased SCR PI is triggered at rot ~ Ct. . With this, load current io s tarts building up in the positive directi on from A to O. Load current io becomes zero at wt - ~ > 1t bu t less than (n + a ), Fig. 10.3 (c ). Thyristor PI is thus n aturally commutated at rot = ~ wh ich is already reverse
Cycloconverters
[Art. 10.1]
535
.,' tilt
(a)
(b)
wt
(c)
wt
Fig. 10.3. Voltage and current waveforms for
step-down cycloconverter with discontinuous load current.
, biased after 1t. After half a cycle, bis positive with respect to O. Now forWard biased thyristor P2 is triggered at rot = 7t + Ct. Loadcurr~nt is again positive from A to 0 and builds up from zero as shown in Fig. 10.3 (c) . At cbt = 1t + ~, io decays to zero and P2 is naturally commutated. At 27t + Ct, PI is again turned on. Load current in Fig. 10.3 (c) is seen to be discontinuous. After four
positive half cycles of load voltage and load current, thyristor N2 (after P2, N2 should be fired)
forward biased, it starts
is gated at (41t + Ct) when 0 is positive with respect to b. AS N2 ,conducting but load current direction is ,reversed, i.e. it is now from 0 toA . Mter N2 is triggered,
load current bUil~s,(p i'Ilthe n gative direction as shown in Fig. 10.3 (c). In the next half-cycle,
o is positive with re'sp-ect to a but before Nl is fired, io decays to zer o and N2 is naturally commutated. Now when Nl is gated at (51t + Ct), io again builds up but it decays to zero before thyrist()f N2 in sequence is again gated. In this manner, four negative half cycles ofload voltage and load current,equal to the number of four positive half cycles, are generated. Now PI is again triggered to fabricate further four positive half cycles of load voltage and so on. For , discontinuous load current, natUral commutation is achieved, i.e. PI goes to blockin g state before P2 is gated' and so on. '
is
In Fig. 10.3, mean output voltage and current waves are also shown. It is seen f rom this " •figure that frequency of output voltage and current is fo =
1
{s.
(b) Contin uous load current. When a is positive with re :,~e ,..t t o 0 in Fig. 10.1 (a ), PI is , tr iggered at rot = Ct, positive output voltage appears a cross bad and load current starts building up, Fig. 10.4 (c). At rot = 1t, supply and load voltages a e zel'C'I. .Aft er wt =1t, PI is r everse biased. , As load current is continuous, P I is n ot t urned off at cot = Te . "Vhen P2 is triggered in sequence at 1t + Ct., a reverse voltage appears acr oss PI, it is therefor e turned o' f by nat ural commutation. When PI is commutated, load current has built up to a valu e equal to RR, Fig. 10.4 (c) . With the turning on of P2 at (n + a) , output voltage is again pos'tive as it was with PIon. As a consequen ce, load current builds LIp further than RR as shown in Fig. 10.4 (c) . At (21t + a ). when PI is again t u rned on, P2 is naturally commutated and load current through P I builds up
536
(Art. 10.1]
Power Electron ics
beyond RS as shown. At the end of four positive half cycles of output voltage, load current is RU. When N2 is now triggered after P2, load is subjected to a negative voltage cycle and load current io decreases fr om positive RU to negativeAB (say) as shown in Fig. 10.4 (c). Now N2 is commutated and Nl is gated at (51t +a). Load current io becomes more negative thanAB at (61t + a), this is because with Nl on, load voltage is negative. For four negative half cycles ot output voltage, current io is shown in Fig. IDA (c). Load current waveform is redrawn in Fig. IDA (d) under steady state conditions. It is seen from load current waveform that io is symmetrical about rot axis in Fig. IDA (d). Thepositive group of voltage group and current wave consists offour pulses and same is true for negative group of wave. One positive group of pulses along with one negative group of identical pulses constitute one cycle for the load voltage and load current. The supply voltage has, however, gone through four cycles. The output frequency . . is, therefore, fo = fs in Fig. IDA.
i
wt
wt
tUt
wt
Fig. 10.4. Voltage and curren t waveforms for
step-down cycloconverter with continuous load I.!urrent.
10.1.2.2. Bridge-type cycloconverter. The operation of bridge type cyclo convert er shown 'n Fig. 10.1 (b) can be easily explained for both discontinuous and continuous load currents . The voltage and current waveforms would again be as shown in Fig. 10.3 for discontinuous load cu rrent and as in Fig. lOA for continuous load current. The explanation of bridge-type cycloconverter is left as an exercise to the reader..
Examp e 10.1. (a ) A single·phase bridge type cycloconverter fee ds a load R. For an output frequency equal to one-t ird of the inprtt frequency, sketch output voltage wavefo rm for a fi ring angle of about 30°
[Art. 10.1]
Cycloconverters
537
(b) Derive an expression for the rms output voltage in part (aJ.
Solution. (a) For this example, refer to Fig. 10.1 (b) which is the circuit diagram for single-phase bridge-type cycloconverter. The source voltage waveform Vs and load voltage waveform Vo are sketched in Fig. 10.5. The half cycles of source voltage are numbered 1,2, 3, ..... for the sake of convenience.
w!
wt
Fig. 10.5. Voltage waveform pertaining to Example 10.1.
During positive half cycle 1, thyristors PI' P 2 are forward biased, these are therefore triggered at a . PI> P 2 conduct from a to 1t because of resistive load. During negative half cycle ,2 of source voltage; P3, P4 are forward biased, these are therefore turned on at 1t + a and so on. During negative half cycle 4;N3, N4 are forward biased, these are, therefore; turned on at rot
=3 1t + a
and so' on. For output frequency fo
=~ fs'
three positive half
cycle~ _are
obtained
first and then three negative half cycles are secured as shown in Fig. 10.5. Waveform of mean output voltage Va is also shown in Fig. 10.5. For one half- cycle of low-frequency fo' there are 1
three half cycle of source freque~cy fs' we have fa =3" f s ' . . It is seen from output voltage waveform, consisting of three positive and three negative half cycles , that rms value of output voltage is given by
Var = [
~ ( V! sin
1/2
2
rot d (CIt) ]
•
,
This expression is similar to that leading to Eq. (9.3) . Therefore, from E q. (9 .3), rms value Vor is .
:. ,
Vm
Vor=~ [.
n1 (
1t-Ct.+
mn . 2a 2
ll'
'
1/ 2
, .. .(10 .1)
Example 10.2. A single-phase bridge-type cycloconverter has input voltage of 230 VJ 50 z a n d loa d of R =10 .n. Output frequency is one-third of input frequency. For a fi ring angle delay of 30°, ca lcu late (a) rms value of output voltage (b) rms current of each con verte r (c) rm s current of each thyristor and (d) inp ut power factor. Solution. Her e Vs
.
(a )
,
= 230 V,
V m = ..f2 x 23 0 V, R
=10.0, a::: 30° =~. 6
From Eq. (10.1), we get
V or =
'1'2 T2 x 230 [ 1. ( _ ~6 + sin260 1t 7t
0
1l1l2 =2')... 6 .c 6 V -J
538
[Art. 10.2]
Power Electronics
Rms valu e ofload current, . (b)
. .' For a n output frequency of fa,
v
= ~ = 226.66 = 22 67 A or. R 10 . . . ..' 1 ' each converter conducts for 2 fo =1t radians, with a
.I
periodicity of 21t radians. ". .' . ' " . . lor ' . 22.67 . :. Rms value of current for each converter. Ip = T2 = ~ == 16.03 A (c ) Each thyristor handles r m s current for 1t radians with a periodicity of 2 1t rad.
. t or :. Rnis va1ue 0 f current C'lor each thyns (d ) Rms source current,
--!-t-2 -.J2 -_ 120r --' 22.67 -- 11.335 A 2 Is = lor
= 22.67 A
Input VA = 230x 22.67 VA, Load power = 22.67 2 x 10 22.67 2 X 10 = 230 x 22.67 = 0.9856 lag . .
:. Input pf
10.2. THREE-PHASE HALF-WAVE CYCLOCONVERTERS The object of this section is to consider how single-phase low-frequency output voltage is fabricated from the segments of 3-phase input voltage waveform. Then three-phase to three-phase cycloconverters are described. 10.2.1. Th r ee-phase to Single-phase Cycloconverters ' . For converting three-phase supply at one frequency to single-phase supply at a lower frequency, th basic principle is to vary progressively the firing angle of the three thyristors of a 3-ph ase h alf-wave circuit. In Fig. 10.6, firing angle at A is 900 ; at B firing angle is somewhat less than 90 0 , at C the firing angle is still further reduced than it is at B and sci on. In this . manner, a small delay in firi.~g angle is introduced at C, D, E, F and G. At G, the firing angle is zero and the mean output voltage, given by Vo = V dacos
voltage
wt
Fig. 10. 6. Fabricated an mean output voltage waveforms for a single-phase cycloconverter .
540
[Art. 10.2]
P ower Electron ics
load current, two three-phase half-wave converters must be connected in antiparallel as shown in Fig. 10.8. The conve:rter circuit that permits the flow of current during positive half cycle of low-frequency output current is called positive converter group. The other group permitting the flow of current during the negative haH cycle of output current is called negative con verter group. For a three-phase to single phase cycloconverter, schematic diagram is shown in Fig. 10.8 (a) and basic circuit configuration in Fig. 10.8 (b). This figure uses two 3-phase half wave converters in anti-parallel, the positive group for the conduction of positive load current and the negative group for the flow of negative load current. Examination of Fig. 10.7 reveals that when output current is positive (above the reference line (Ot), positive converter conducts. Under this condition, positive converter acts as a rectifier when output voltage is positive and as an inverter when output voltage is negative. When output current is negative, the negative converter conducts; under this condition, negative converter acts as a rectifier when output voltage is negative and as an inverter when output voltage is positive. It can thus be inferred, in general, that one of two component converters in Fig. 10.8 would operate as rectifier if the output voltage and current have the same polarity and as an inverter if these are of opposite polarity.
Negative group
IG Reactor - Grol,lp
'tGroup
1-phase ac load
l-phase ac load
- - - - 4 I - - - - - Neutral
---+---- Neutral (b )
(a)
Fig . 10.S. Three-phase to single-phase cycloconverter (a) schematic diagram and
(b) basic circuit configuration with IG reactor.
Fig. 10.8 is almost similar to Fig. 6.45 for a dual converter where two phase-controlled converters are connected in antiparallel. As in a dual converter; in Fig. 10.8 also, both the component converters belonging to one phase can be phase-controlled simultaneously to fabricate th e output voltage. Though the output voltages of the two converters in the same ph ase h ave the s m e average value, their output voltage waveforms as a function of time are, however, different and as a result, the e will be a net potential difference acro s the two converters of Fig. 10.8 (a). This net voltage would cause a circulating current in the two converters , this l ~ simil ar to a dual converter. This circula ting current can be avoided by removing the gating signals fr om idle converter or can be limit ed to a low value by inse rting an intergroup r eactor (IGR) between the positive and n egative group converters as shown in Fig. 10.8 (b ). In or der th at the average valu e of the two co verters are equal in magnitude and opposite in sign, th e sum of their firing angles ill st be 180 In other words, if ap and an are the firing angles for positive and n agative group converters respectively, then these firing angles should be so controlled as t o satisfy the relation a p + an = 180°. 0
•
Example 10.3. Describe a thre e-phase to single cycloconverter fe eding RL loaG and w ith Ol'- tpu t frequency one·fourth of input frequency. Ill ustrate y our answe r by shouing ~m e cycle of
lOW-freq uen cy output voltage. In dic ate clearly the triggering of uariou-s thyristo s.
541
[Art. 10.2]
CyclocODverters 3
S olu tion. It is seen from Eq. (10 .2) that progressive variation in triggering angle required is given by ~ x 120 = 30°. Power circuit diagram for a 3-phase three-pulse converter is given in Fig. 10.9 (a). Firing angle are (Xl for thyristor T l , ~ for T2 and (X3 for T 3. In Fig. 10.9 (a), thyristor Tl is triggered at
= ~ fs is obtained.
o
oLr> II
M
(;
O'l
g
o
>
For sketching negative half cycle of low frequency output voltage, T2 is triggered at
Co ell,...;
I-<
~o-
;>
~
S ell
;::
)<
8~
o11
-6
_ _-+)<..
::l 0
~
~
...c: ~
c;S
..c:
.
0
I-<
..c:
E-<
542
[Art. 10.2]
Power Electronics 3-ph~e----.-----~~--____~____~~______~______~___ . fs
P
L----+_....J
N
p
N
p
N
(a)
A
3-phas~ ~~~----~---------.--------~~~----.--------, supply =B--f-.-----+-.......----......;,..-+-......-----+~.__----_+--._----r...,
fs
~C_4--+-~--+-~~.__---+--+~.__-+--~~--_+--+_~--+__r~
.
(b)
Fig. 10.10. 3~p-hase to 3-phase cycloconverter employing 3-phase half-wave circuits
. (a) schematic diagram and (b) basic circuit arrangement.
'I
This scheme, shown inFig. 10.11, employs thirty-six thYristors and is called 6-pulse, 3-phase
to 3-phase cycloconverter. In this circuit, each phase group consists of a 3-phase dual-converter
with two IGRs. The load ph ases, shown in st ar in: Fig. 10.11, mu st not be interconnected. Ifit
is done, then positive group of one output phase and negative group of other output phase would
be joined together through IGRs without load impedance which is undesirable. In case it is . essential to interconnect the load phases in star 0 delta, then each phase group,is supplied separately from t hree secondary windings 8 1, 82 and S3 of a three-phase transformer as sho'vVn in Fig. 10.12. In this arrangement, as individu al phase groups are isolated from each other on . the input side, the interconnection of load phases in st ar or delta is permissible. The magnitude of output volt age in a 3-phase bridge cir cuit of Fig. 10.11 is double of that in the IS-thyrist or cir.cuit of Fig. 10.10 (b) . In case volta ge an d current ratin gs of all the SeRs in Figs. 10.10 (b) and 10.11 are identical, then t otal VA rating of bridge circ.uit wou ld be double of that of the I S-thyristor circuit. Three-phage bridge cir cuit gives a smooth variation of output voltage, but its control circuit is complex and expensive.
Cyc1oconverters
.·. 3-phas
.
543
[Art. 10.3]
A B
SUPpIY~C~ · ~~~---+~~r-+-~~~---+~~r-r-+-~4---~~
1s
3-ptlase load '-----------~----..........,----------'
Fig. 10.11. Six-pulse, 3-phase to 3-phase cycloconverterusing 36 thyristors . and with isolated load phases. ..
~
i~; Z;Z
~{ ~ { 2{
3-'3
input
fs
fl ~'7 } ..Y 4 ~ ~
l 4f ~
) '1) '1)"
~
~
j
S1
S2
•
. ~--
IIJ I
~
~
."0.
I
I
I
~ I
I
JTr' 53 .J'llr'.
....rrf\
I I
,,'1 ~ '" IGR.2~ 2~ l JJfo
) ~) ' ) '1
"
IGR
1.
~ l ( 4(
~
4
1"11'"
~~
[
J' }~;~ IGR ~ it;. 'f/0 ~~
fo
Fig. 10.12. Three-phase bridge cycloconverter using 36-thyristor s and with non-isolated load ph ases .
10.3. OUTPUT VOLTAGE EQUATION' F~ it t;YCLOCONVERTER
~':_~:' .~
i
In this s ection, emf expressions for the line-commutated phase-contr olled eye oconvert er s are dis cussed.
[Art. 10.3]
544
Power Electronics
A cycloconverter is essentially a dual converter but so operated as to produce an alternating output voltage. Each converter in a cycloconverter works as a phase-controlled converter with a varying firing angle. Vo
In a 3-phase half-wave converter, each
1.
1t
phase conducts for 23 radians of a cycle of
.
,(~~Q) I
I I
I
21t radians . In general, for an m-phase half-wave converter, each phase conducts
I I
I
for 21t radians in one cycle of 21t radians, m
wt this is shown in Fig. 10.13. Actually, an expression for the average output voltage is already obtained in Eq. (6.58), Art. 6.7.4. Here it is derived again for the sake of Fig. 10.13. Output voltage :navef?nn for m-phase continuity. With time origin AA' taken at half-wave converter wIth finng angle c. the peak value of supply voltage in Fig. 10.13, the instantaneous phase voltage is
v = V mp cos rot =..f2 V ph cos rot where
Vph
=~ ~2 = rms value of per-phase supply voltage.
It is seen from Fig. 10.13 that conduction takes place from - ~ to ~ for a . m m firing angle
a,
the conduction is from (- ; +
a)
to (; +
a)
= 0°.
For any
Thus average value of output dc
voltage Vd , equal to the average height of shaded area in Fig. 10.13, is Vd
mJ;l;;+a) = -2 ( Tt --+a It
)
V mp
cos rot· d(wt)
Tt) = V mp [(m). sm - cos a Tt m
m
For zero fIring angle delay, the average value of direct voltage V do
= V mp
(m) sin~ =..f2 1t m
... (10 .3a)
I
Vph (.m)sin
lTt :
~ m
Vdo
is given as ... (10.3b)
In an actual cycloconverter, the firing angle is gradually varied. For any firing angle, the . output phase voltage at any point of the low-frequency voltage wave is equal to V do cos a on the assumption of continuous conduction. · I n Fig . 10.6, Vdo . cos a = 0 when a = 90° at A and Vdo . cos a = V do when a = 0° at G. Neglecting the voltage fluctuations superimp osed on the mean output voltage in F ig. 10.6 or
Fig. 10.7, the low-frequ ency output voltage waveform is sketched in Fig. 10. 14 ((1,) . It is seea
from t his fi gure th a t peak value of low- fre quency (If> output volta ge for zero firing angle 18
V d o' Since If output voltage varies sin soidally, its fund am ental rms value per phase is given by
. or
... (10.4)
Cycloconverters
[Art. 10.3]
545
The effect of source inductance leads to v d commutation overlap as discussed in Art. 6.9.1. If firing angle is zero, the output voltage wt waveform would be as shown in Fig. 10.14 (b)
due to overlap angle in a single-phase full (a)
converter. At the same time, in the inversion v
mode with a > 900 , the maximum value of firing Vol 271 angle cannot be more than 180 - 1..1. as shown in 0 wt wt Fig. 10.14 (c). This shows that on account of --l~'l-71-~ -1 iJ ~ commutation overlap and thyristor turn-off (b) time, the firing angle range in a cyc1oconverter (c) is 1..1. < a < (180 - 1..1.). Fig. 10.14 (a) Low-frequency output voltage 0
o.~(\
wave form of a cycloconverter. Single-phase
This implies that, in practice, the firing full-converter output voltage wavefonn when angleap of the positive converter cannot be zero, overlap angle is Il and (a) a =0, rectifying mode see Fig. 10.14 (b) and at the same time, and (b) a = 180 -Il, inverting mode. maximum value of firing angle
:r[
VPh . (
:
}in(: J]
.
·
.(106)
As a mn is always greater than zero, the voltage reduction factor, r, is always less than unity.
Eq. (l0.6) gives the rms value of the per-phase output voltage for a 3-phase to 3-phase or 3~phase to sin gle-phase cycloconverter employing m-phase half-wave circuits shown in Fig . 10.8 or 10.10. Note t hat these converter circuits consist of two converters, one 3-pulse positive group converter and the other 3-pulse negative group converter. Eq. (10.6) is also applicable for 3-phase to 3-phase or 3-phase to sin'gle-phase cycloconverter employing 6-pulse bridge converter of Fig. 10.11 or 10.12, but then m is equal to the number of pulses, i.e. m.= 6. Since bridge convert ers of Fig. 10.11 or 10.12 employ -phase full converters, Vph in Eqs. (10.4) and (10.6) mll.-ct be replaced by line to line voltage Vi as done in 3..phnse full converters . E x a mple 10.4. A 3-ph ase to single-phase cycloconverter employs 3-p ulse positive and negative group converters. Each con verter is supplied from delta / star tran sformer with per p hase turns ratio of 2 : 1. The supply voltage is 400 V, 6 0 Hz. The RL load has R = 2 Q and at low outp ut freque ncy, roof.. =1.5 n In order to account for commutation overlap and thyristor turn-off time, the fir ing angle in the inversion mode sh ould not exceed 160°, Compute (a) the valu.e of the fun dam ental rm s outp ut voltage.
Power Electronics
(Ar t. 10.3]
546
(b) rms output current and (c) output power.
Solution.
(a)
Per phase input voltage to transformer = 400 V.
Per phase input voltage to converter, 400 . . Vp h=-=200V 2 . Voltage reduction factor, r = cos (180 - 160):: cos 20° For 3-phase 3-pulse device, m ~ 3. From Eq. (10.6), the rms value offundamental voltage is
VM = cos 20 [200 (b)
Rms output curr.ent
(c)
Output power
· .
(~} sin ~] = 155.424 V
= 155.4242 [_ tan- 1 1.5 ~22 + 1.5 20 lor = 62.17 1- 36.87° Amps.
= I;r B = (62.17)2 X 2 = 7730.22 W.
Example 10.5. Repeat Example 10.4 in case 3-phase to I-phase cycloconverter employs 6-pulse bridge converter.
Solution. (a) Per phase input voltage to converter = 200 V . Line voltage input to bridge converter = 200..J3V Voltage reduction factor, r = cos 20° For 3-phase, 6-pulse device, m = 6. From Eq. (10,6), the rms value of output voltage is
V" ~
cos 20 [ 200 ;13(~ )sin ~] ~ 31084 V
This example demonstrates that output voltage in a 6-pulse bridge converter employing 36 thyristors is double of that in a 3-pulse half-wave converter using 18 thyristors. (b) R ros outpu t current
=
I t - 1 1.5 121310.84 + 1.52 - an 2
= 124.34 (c )
Rms output power
1- 36.87° Amps
=(124.34)2 x 2 =30920.88 W.
This example shows that output power handled by a 6-pulse bridge converter is four times the power handled by a 3-pulse converter. Exam ple 10.S. In a siT~gle-phase cycZoconverter arrangement for changing the frequency . of the supply voltage to a load, as show(l in Fig. 10.15 (a), the load terminal L should be connected- to (a) A (b) B (c) C (d) D or E.
Solution . A careful examination of Fig. 10.15 (a ) r eveal that this figure is the same as that of I-phase to i-phase mid-point cycloconverter of Fig. 10.1 (a) ..Thethyristor are marked Pl , P2, N1, N 2 in Fig. 10.1 (a) and on the same patt ern, these are marked asPI, P2, Nl, N2 in Fig. 10.15 (b ). In Fig. 10.1 (a), load terminal A is conn ectoo directly to PI, P2, N1, N2 . But, in Fig. 10.15 (b), load terminal L must be connected to mid-point B of the intergroup reactor. Therefore, here the answ er is (b). In Fig. 10.15 (a), if IG eactor is not used, then load terminal L m ay be connected to A , Bar C as shown in Fig. 10.15 (c). Fig. 10.15 (c) is exactly ident ical to Fig. 10.1 (a).
Cycloconverters
[Art. 10.3
547
E
(b)
(a) E
,
Fig. 10.15
(a) (c)
(c)
Pertaining to Example 10.6 (b) Load terminal L connected to B 1-phase to 1-phase mid-point cycloconverter redniw.tl.
Basic principle of working of I-phase to I-phase mid-point cycloconverter can also be . explained with the help of Fig. 10.15 (b) if reactor is used and with Fig. iOJ:5 (c) if reactor is not used.
Example 10.7. A six-pulse cycloconverter, fed from 3~phase, 400 V, 50 Hz source, is delivering a load current of 40 A to a 1- phase resi.stive load. The source has an inductance of 1.2 mH per phase. Calculate the rms value ofload voltage for firing angle delays of (a) 0° and (b) 30°.
.
Solution. The object here is to first calculate Vdo of Fig. 10.14 (a) and r . Vdo and then its rms value. (a) Firing angle ex = 0°. Peak value of output voltage, fo.r a 3-pulse converter, from Eq. (fO.3 b), is
Vdo~Vmp(~ )sin (~ ) . For a 6-pulse converter, V mp must be ·replaced by V ml =-.J2. VI. Vdo ::: -.J2 x 400 x ~ sin ~
Reduction in voltage ·dueto source inductance, from Eq. (6.76), for a 6-pulse converter is
- _ SeaL _31
1t
.
0
. :. Peak value of output voltage,
Vo.mx =.J2 x 400 x ~ s.in 30° _ 3 x 21t >< 50 : 1.2 x 10- 3 X 40 = 525.71 V : . Rms value of load voltage =
525.71
,12
= 37 1.79 V
548
Power Electronics
(Art. 10.4]
(b) Firing angle a = 30.0.
Peak value of output voltage, from Eq. (10.3 al, is V d o = ,f2 x 400 x ~ sin ~ . cos 30° Reducationin voltage due to overlap is the same as for a
= 0°
:. Peak value of output voltage 3
V
. = ,f2 x 400 x ~ sin ~ x cos 300 _ 3 X 27t X 50 x 1.2 x 10- x 40 = 453.33 V O.mx 7t 6 7t
Rms value of output voltage =
.
453.33
12 .= 320.60 V
10.4. LOAD·COMMUTATED CYCLOCONVERTER
.
A step-up cyc1oconverter discussed previously requires forced-commutation. An additional circuitry for force-commutating the thyristors in a cycloconverter is, therefore, essential. The process of forced commutation, however, does not depend upon the source or load voltage. In the step-down cycloconverter discussed in the previous sections, the phase-controlled cycloconverter rely on natural commutation for their operation. The natural, or line, commutation is provided by the supply voltage. In these cycloconverters, the output frequency fo is less than the input supply frequency fs' A load-commu tated cycloconverter differs from the force-commutated and line-commutated cycloconverters discussed so far. In load-commutated circuit, the thyri tors are commutated by the reversal of the load voltage. This implies that the load circuit must have a generated emf that should be independent of the source voltage. The most usual example for such a load is wound-field or permanent-magnet synchronous machine. For such loads, the load frequency may be equal to, or greater than, the source frequency and for both these cases, thyristors will be naturally commutated by the reversal of the load circuit emf.
PROBLEMS
.
10.1. (a) What is a cycloconverter ? Enumerate some of its industrial applications. (b) Describe
the operating principle of single-phase to single-phase step-up cycloconverter with the help of mid-point and bridge-type configurations. Illustrate your answer with appropriate circuit and waveforms. The conduction of various thyristors must also be indicated on the waveforms. 10.2. Describe the basic principle of working of single-phase to single-phase step-down cyclocon verter for both continuous and discontinuous conductions for a bridge type cycloconverter. Ma!'k t he conduction of various thyristors also. 10.3. A single-ph ase to sing:-e-phase mid-point cycloconverter is delivering powe to a r esistive load. The supply transformer has turns ratio of 1 : 1 : 1. The frequency ratio is fol f s = 1/5. The firi ng angle delay for all the four SeRs are the same. Sketch the time variations of the following waveforms for a: =0° and a: =30°. (a) Supply voltage (b) Output current and (c) Supply current. Indicat e the conduction of various thyrist ors also. 10.4. A s.in gle-phas e to single-phase bridge-type cycloconverte is fed from 230 V, 50 Hz source and has a load R ::: 20 n. Output frequency is one-fifth of in pu t frequ ency. For a firing angle delay
Cydoconverters
549
[Prob. 10]
of 45°, calculate (a) nns value of output voltage (b) rms current for each converter (c) rms current of each thyristcr and (d) input power factor. Derive the expression for rms output voltage for this cycloconvel'ter. [Ans. (a) 219.3 V (b) 7.775 A (c) 5.483 A (d) 0.9535 lag]
Pl
10.5. A single-phase to single-phase cycloconverter of Fig. 10,16
, 1
is used for obtaining an output frequency of '3 times the input frequency. Turns ratio from primary to upper secon
dary is 111 and to lower secondary is 1/2. The thyristors are
so triggered as to obtain a symmetrical output voltage
, waveform. Sketch the output voltage and output current waveforms for a resistive load for ,firing angle ex = 0° and , a =30Q• Indicate the conduction of various thyristors.
Fig, 10.16. Pertaining to Example 10.5.
10.6. For the circuit of Fig. 10.16, source voltage is V m sin rot. Derive expression for the rms value of output voltage (a) for firing angle a = 0 and (b) for firing angle ex. . " ['. Hint: (a) V"r
[ 1
=
r
2·'
.
m 31t o 6V sm rot. d
1
. ('
«(j}[t)
J
"
etc.
sin 2 ex
[ 1[
Ans.(a)Vm
(b)Vm:;t
1t-ex+
2
\]Vz
1
J
10.7. In Prob. 10.6, source voltage is 230 V, 50 Hz and load is R = 20 n, Find the power delivered lAns. (a) 5290 W ( b) 5136.01 WJ , to load for (a) ('). = 0° and (b ) ex = 30°. 10.S. Cycloconverter circuit of Fig. 10.16 is used for obtaining on output frequency to = ~ x sourc e 'frequency (s' Turns ratio from primary to upper seconc.ary is 111 and to lower seconda;:y lin. Derive expression for the nns value of output voltage for firing angle a . For. Vs = 230 V, 50 Hz; n = 3; R = 20 n and ex = 45°. find the load power. ,...
Il An
, 1/2
'r;'"
r!
S.V m L
n,2
6
+ 2 (I _ 1t
\
n
a+
sin 2a 2
1 ~ JJ'
(3
I
8814 k'r W .,, ~:) J
10.9. Describe h ow single-phase low-frequency output voltage can be fabricated from the segmer:.ts of 3-phase input-voltage waveform through the use of a 3-phase half-wave circuit? Show a complete cycle of low-fr-equency outputs voltage. In case load current lags the low-frcquer.cy output voltage . discuss the operation of positive and. negative group phase-controlled c')n verters. '10.10. (a) Discuss, why 3-phase to 1-phase cyc1oconverter requires positive and negativ'2 ; :-cup phase-controlled converters. Under what conditions, the groups work as inve:~ e;:-s or rectifiers ? How shoul - the firing angles of t he two converters be controlled ? ( b ) Describe 3-phas e t o 3-phase cyc1oconverter with relevan t ci rcuit arrangements using 18 SeRs an d 36 SCRs . What are the advantages of 3-phase bridge circuit cyc1oconverler over IS-thyristor device ? 10.11. (a ) S how that the fundam en tal rros value of per-phase output vol age of lo'w-frequency fo r
an m-pulse ycloconverter is given by
Vor= VPh ( : )sin
H ence express
Y o!"
(~ J
in terms of voltag e reduction fact or.
550
Power EJectronics
[Prob. 10]
(6) A 3-pulse cyc1oconverter feeds a single-phase load at 200 V. Estimate the value of the supply voltaee. Derive the formula used. . [Ans. (b ) Vph = 241.85 V] 10.12. A 3-phase to single-phase cycloconverter employs a 6-pulse bridge circuit. This device is fed from 400 V, 50 . Hz supply through a delta/star transformer whose per-phase turns ratio is 3/1. For an output frequency of2 Hz, the load resistance is roo L =3 n. The load resistance is 4 n. The commutation overlap and thyristor turn-off time limit the firing angle in the inver sion mode to 165°. Compute (a) peak value ofrms output voltage (b) rms output current and (c) output power. [Ans. 301.215 V (b) 42.605 L - 36.87° A (c) 7260.74 Wl 10.13. Repeat Prob. 10.12 in case 3-phase to I-phase cydoconverter employs 3-pulse positive and negative group converters. [Ans. (a) 150.602 V (b) 21.301 L - 36.8T (c) 1815 Wl 10.14. A 3-pulse cyc1oconverter, fed from 3-phase, 400 V, 50 Hz supply is delivering a load current of 30 A to a I-phase resistive load. The supply has an inductance of 1 mH per phase. Calculate the rms value of load voltage for a firing angle of (a) 0° and (b) 45°. [~8. (a) 187.83 V (b) 152.76 V]
/
Chapter 11
ome
Appl~cations
.•............•..... ......... .....•...••.•...•.....• .•.•..•............•............. .. .......... _.
~
~
In this Chapter • • • • • • •
Switched Mode Power Supply (SMPS) Uninterruptible Power Supplies High Voltage DC Transmission Static Switches Static Circuit Breokers Solid State Relays Resonant Converters
..................•... .•.•......•..•........•.................•....... -- ••..•....................•• Several applications of power electronics are listed in Art. 1.2. Study of all these applications will be a voluminous task. Even then, some of these applications described in this chapter will be of interest to the reader.
11.1. SWITCHED MODE POWER SUPPLY (Sl\'IPS) With advances in electronics, need for dc power supplies for use in integrated circuitE' q C~) and digital circuits has increased manifold. For such electronic c'r cuits, NASA was the first to develop a light-weight and compact switched mode power supply in the 1960s for use :n i t~ space vehicles . Subsequently, this power supply became popular and presently, annual production of SMPSs may be as high as 70 t o 80% of the total number of power ~upplies produced. At this juncture, a question may arise that controlled dc supply can also be obtained frO'll phase-controlled rectifiers. Then why go in for SMPS ? An ac to dc rectifier operates a: supply frequency of 50 (or 60) Hz. In order to obtain almost negligible ripple in the dc output voltage. physical size of filter circuits required is quite large. This makes the dc power supply inefficient, bulky and w eighty. On the other hand, SMPS works like a dc chopper. By operating the onJoff switch very r apidly, a c ripple frequency rises which can be easily filtered by Land C filter circuits which are small in size and less weigh ty. It may therefore be inferred tha.t it is the requirement of small physical size and weight that h as led to the wide spr ead use of SMPSs. As stated a bove, SM PS is bas ed on the chopper p L~ ipl e . The ou tput dc voltage is controlled by varyin g the duty cycle of chopper by PWM or F'M techniques . Th e circuit configurations used for SMPS can be cl assified into four broad categories; namely flyback, pushpull, half bridge and full-bridge. In SMPS circuits discussed here, PWM technique is used for th e inverter. The ou tput of the inverter is then converted to dc by a diode rectifi er. As the inverter is made to oper ate at
. very high frequency, the ripples on the dc ou tput voltage can be filt ered out easily by using
small fi lter components . If the switching devi ces are po\ver transistors, the chopping frequen cy
is limited to 40 kHz. For power MOSFETs. t h e chopping fr equency is of the order of 200 Hz:
Power Electronics
[Art. 11.1]
552
as a result, size of the filter circuit and transformer decreases leading to considerable savings. At such high frequency, ferrite core is used in trans~ormers. The four categories of SMPSs listed above are now discussed briefly. 11.1.1. Flyback Converter The circuit configuration for flyback converter is shown in Fig. 11.1. It consists of a power MOSFET Ml, transformer for isolation purposes, diode D, capacitor C and load. An . uncontrolled rectifier converts ac to de output which is fed to flyback SMPS as shown in Fig . .11.1. o io When power MOSFET is + +. turned on, supply voltage Vs is applied to the transformer c primary, i.e. VI = Vs' A corres ponding voltage V2' with the polarity as shown in .Fig. 11.2(a), is induced in the Uncontrolled transformer secondary, i.e. redifier M1 Va . v 2 = N N 2 · As V2 reverse bIases I
diode D, equivalent circuit of Fig. 1l.2(a) is obtained. Filter . ' Fig. 11.1. Flyback SMPS. capacitanceC is a.s smned larg~ enough so that capacitor voltagev c (t) = load or output voltage Vo is taken as almost constant. When Ml is turned off, a voltage of opposite polarity is induced in primary and secondary windings as shown in Fig. 11.2(b). Voltage'across transformer V ' . . . S secondary is L'2 =- - Vo = - N N 2• Diode D is forward biased' and starts cond cting a current I
i D · As a result, energy stored in the transformer core is delivered partly to load and partly to charge the capacitor C . . D
+
+
c
+
-
iD
c
V1 + Vs
I M1
+
U~o, (a)
(b )
Fig. 11.2. Flyback SMPS equivalent circuitd uring (a)
Waveforms for
VI '
To n.
and
(0)
Torf'
v2, transformer magnetizing curren t im and dioqe cu r ent
in Fig. 11.3. During the time M l is on,
VI
= Vs' ~2 =;, . N 2 • For
iD
are shown
magnetiz ' g current, it is
~
assumed that. transformer core is not demagnetized complet ely at th e end of periodic time
553
[Art. 11. 1]
Some Applications
T = Ton + T off. In other words, it means that transformer magnetizing current at t = 0 is not zero but has some positive value Imo. Therefore, during Ton' magnetizing current rises linearly from its initial value Imo to Iml at t =Ton. With the rise of im during Ton' magnetic'energy gets stored in the transformer core. The variation of im as shown in Fig. 11.3 can be expressed as under: . Vs "m(t)=Imo+rt ...... O
where L At t
= transformer magnetizing inductance, H . '\ Vs tm (Ton):;:-: ImI =Imo + L . Ton
= Ton'
...(11.2)
When M1 is turned off, the emfs induced in primary arid secondary windings are reversed as shown in Fig. 1l.2(b). Diode D is now forward biased. A current in transformer secondary winding begins to flow through D. As this current iD or magnetizing current im reduces from ImI to Imo at t = T, transformer core energy is delivered to load. · During T off, Ml is off and V2 =- Vo. This voltage when referred to primary is
.. Vs
,
= T,
1
~
Or-L-;---,---+----+------;--t~
~
1
. Ton - N2 . NI . L
I
Imo
. t to
Nl .
Substituting the value of ImI from Eq. (11 .2) in the above expression, we get .
:,"
or----+-------...;------;.....----'---+-
N: . L(T - Ton)
tm (1) = Imo + L
t '
= - N~ N 1. The fall of
...(11.3)
, V im(T) = I m1 -
,..
V
VI
i
2
.
.~·f~· '
current im during Toff can be expressed as under: . V . im (t) =lml - NO Nl . (t - Ton) ..... Ton < t < T At t
... (11.1)
I
Im~
,
..
."
::
_.
:
t
I
:
ot?~~~~Io ~ f..,Ton-L- TO
T - Ton)
ff-+
~'--T ----l
.. .(11.4)
Since the net energy st or ed in core over periodic time T is zerq,
Ton L T o!1 '--l
t
Fig. 11.3. Waveforms for flyback converter SMPS.
or
or :: Load voltage, where a
N2 =N ' transformer l
.
, a . Vs . Ton
a · Vs . k
Vo = T - Ton = l- k rns ratio from seconda y to primary
... ( 11.5)
554
[Art. 11.1]
and
k =T
Ton
Power Electronics
' duty cycle of flyback converter.
It is seen from Fig. 11.2 (b) that open circuit voltage across Ml is . Vo Vo Voc = V I + Vs = N . NI + Va = - + Va 2 a . · ~ ·k .~ From Eq. (11.5), Voc = 1 _ k + Vs = 1- k
...(11.6)
Eq. (11 .3) gives current on primary side of the transformer. This current, when referred to secondary side, is equal to diode currenti D . N1 ..
iD(t)=im(t)'N
2 [ =NI N Iml -
.
==
1.
Vo
N . NI . L
(t -
.
1
Ton)
22
I
V a . a.L
ml_ - 2 0
(t - T )
... (11 .7)
all
Flyback converter offers simple SMPS and is useful for applications below about 500 W. 11.1.2. Push-pull Converter SMPS with push-pull configuration is shown in Fig. 11.4. It uses two power MOSFETs Ml and M2 and a transformer with mid-taps on both primary and secondary sides. As in flyback converter, an uncontrolled rectifier feeds push-pull SMPS . Inductor L and capacitor C are the filter components. .
+
Uncontro ll ed rect ifier
M2
Co ntr ol cir cuit
Fig. 11.4. Push-pull SMPS.
When Ml is turned on, Vs is applied to lower half of transformer primary, i. e. vI = Vs' As V . . a result, voltage v2 = N S N 2 is induced in both the secondary windings .Voltage V z in the upper .
.
I
h a lf secondary fo r war d bias es diode D l, therefo re loa d v olt a ge Vo i s giv en by Vs
VO=N N2 = aVs ' I
When M2 is turned on,
V
v2
..
=- N : N2 is induced
forward biased and Vo
555
[Art. 11.1]
Some Applications VI
=- VB is applied to upper half of primary winding. Consequently, .
...
in both the transformer secondaries. As
=a Vs
V2
is negative, diode D2 gets
as before.
This shows that voltage on primary swings from + Vs with Ml on to - Vs with M2 on . Power MOSFETs Ml and M2 operate with duty cycle of 0.5. When Ml is off, the voltage across Ml terminals is Voc = 2V". As both Ml and M2 are subjected to open-circuit voltage of 2Vs, this configuration is suitable for low-voltage applications only. 11.1.3. Half-bridge Converter The circuit for half-bridge SMPS configuration is shown in Fig. 11.5. It consists of an uncontrolled rectifier, two capacitors Cl and C2, two power MOSFETs Ml and M2, one transformer with mid-tap on the secondary side, two diodes Dl and D2 and filter components Land C.
+
1~ C1
1\111
...
D1
". "
+ Vs
_2
LOAD
Vs
Uncon1 rolled rectifier
~ 2
C2
1\112 1
Con1rol circuit
2
Fig. 11.5. Half-bridge SMPS.
voltage across each .of the two VTwo capacitors Cl and C2 have equal capacitance, . therefore . . is -28 • When Ml is turned on, voltage of Cl appear s across transformer primary, i.e .
.
.
and voltage induced in secondary is
.
V
v2
VI
=. Vs 2
. .
== ~ . N2 and diode Dl gets forward biased. When M2 1 .
Vs is turned on, a reverse voltage of 2 appears across tr ansform er ~imary from C2, i.e. . VI
=.
:s
and voltage induced in secondary winding is
-
Vz
=- ;;; 1
N 2, t herefore diode D2 gets ~
~
~
.:.
forwar d biased. This means that tr ansfor mer primary voltage swings fr om - ry to + n' Average output voltage, however, is
556
Power Electronics
[Art. 11.1]
When Ml is off, open circuit voltage across Ml terminals is V oc.= Vs. When M2 is off, as . before Voc =Vs' For h .v . de applications, half-bridge converter is , therefore, preferred over push-pull converters. For l.v. de applications, push-pull SMPS is preferred due to low MOSFET currents. 11.1.4. Full-bridge Converter
The circuit arrangement for a full-bridge SMPS is shown in Fig. 11.6. It consists of an uncontrolled rectifier, four power MOSFETs, transformer With mid-tap secondary, two diodes and LC filter circuit. As in all the previous circuits, the.function of control circuit is to sense the output load voltage and to decide about the duty ratio of MOSFETs . .+
~
M3
4~
M2
01
1 ..
u ncontro ll ed rect ifier ...
[
.
L
02
.
1
Control circuit
2
3
4
Fig. 11.6. Full bridge SMPS.
\ When power MOSFETs Ml and M2 are turned on together, voltage V s appears across transformer primary, i.e. forward biased and Vo is reversed, i.e. .
VI
Vl
=V~
.
= aVs' When M3
= - Vs and v2 = -
V
and secondary voltage
V2
=N 1 S
•
N 2
=. c.i V s'
Diode Dl gets
and M4 are turned on together, the primary voltage
V S N N2 1
= - a Vs'
.
.
Therefore, diode D2 now begins to conduct
and the output voltage is again. Vo = a Vs' The open circuit voltage across each MOSFET is VDC = V s' Of all the four configurations of SMPSs, ful -bridge converter operates with minimum voltage and current stress on the power MOSFET. It is th er efor e very popular for high power applications above 750 W. The overall size of SMPSs is dependent on its operating fr equency. Use of power transistors is limited to approximately 40 to 50 kHz. Above this operating frequency, power MOSFETs are used up to about 200 kHz. . . . The main advantages of SMPSs overcoriventioriallinear power supplies ar a under: the sam e power rating, SMPS is of smaller size, lighter in weight and possesses h igher effi cien y because of its high-frequency operation. . SMPS is less sensitive to input volt age variat ions.
(i ) For (i i )
[Ar t. 11.2]
Some Applications
557
The disadvantages of SMPS are as under: . (i ) SMPS has higher output ripple and its regulation is worse .. (ii) SMPS is a source of both electromagnetic. and radio interference due to high fre quency switching. (iii) Control of radio frequency noise requires the use of filters on both input and output of SMPS. . The advantages possessed by SMPSs far outweigh their shortcomings. This is the reason for their wide-spread popularity. and growth.
11.2. UNINTERRUPTIBLE POWER SUPPUES There are several applications where even a temporary power failure can cause a great deal of public inconvenience leading to large economic losses. Examples of such applications are major computer installations, process control in chemical plants, safety monitors, general communication systems, hospital intensive care units etc. For such critical loads, it is of paramount importance to provide an uninterruptible power supply (UPS) system so as to maintain the continuity of supply in case of power outages. Earlier UPS systems were based on an arrangement sh9wn in Fig. 11.7. This scheme is usually called rotating-type UPS. This arrangement consIsts of DC motor-driven alternator, the shaft of which is also coupled to . a diesel engine. The three-phase mains supply, after rectification, charges a dc battery-bank and feeds the dc motor as well. The uninterruptible UPS to Critical loads
I 3
¢
supply mOins
Rectifier
T
Batter y Bank
--'
Fig. 11.7. Rotating-type UPS system based on de motor/alternator set .
power supply needed is taken from the alternator output terminals. When main supply fails, the diesel engine is run to t ake over the load. Starting of the diesel engine takes 10 to 15 seconds. During this period, the battery-bank is able to maintain the alternator speed through the dc motor and the flywheel, thus giving a no-break supply to the critical load. At present , however, static UPS systems are becoming popular up to a few kVA ratings. Static UPS systems ar e of two types; namely short-break UPS and no-break UPS. In short -break UPS, the load gets disconnected from the power source for a short d r ation of the order of 4 t o 5 ms . In n o-bre ak UPS, load gets continuous uninterrupted supply fr om the power source. These are now discussed briefly. Short-break l.JPS. In situ ations where short interruption (4 to 5 IDS) in supply can be tolerated, the short-break UPS sh own in Fig. 11.8 is used. In this system, main ac supply is r ectified to dc. This de outpu t fro m the r eetifi r charges the batteries and is also conver1:ed to
558
[Art. 11.2]
Power Electronics
ac by an inverter, Fig. 11 .8; M er pjissing through the filter, accan be deliver~d to load in casQo,-'" .~ '~','~ /~'!''t'! ';' normally-off contacts are closed. Under normal circumstances, normally-on contacts are clos'e<;i" , ' and normally-off contacts are open and the main supply delivers ac power to the load. At the same time, t h e rectifier supplies continuous trickle charge to batteries to keep the'mfully " charged, In the event of power outage, normally-off switch is turned-()n and the batteries deliver ac power to critical load through the inverter and filter. A momentary interruption in NO ~~I~~_ ~_~ ~ ' I,
'I
: I
r---------------------~I Power flow
I
" I
I I
=»
Static : ' \ ' ; transfer ~---- -----~ switches '
,
----~----,
I
,Main, Rectifier ac o-~~~ supply ac- d c
I
~~~
Inverter dc- ac
Filter
~--~y----~
,N ormally off
! I
I
r
I
,
N~~~~~~-;-F~~
• •:'10
Fig. 11.S. Short-break static UPS configuration.
the supply (4 t05 ms) to the load can be observed in case lamps and fluorescent tubes are a part' of the load. When normally-on switch is opened and normally-off switch is turned on, lamps will have a transient dip in their illumination whereas the fluorescent tubes will be off momentarily and then get turned on. When the main ac supply appears, critical load gets connected, through normally-on swit ch, to the supply mains. Again, a momentary interruption in the illumination is noticed. The arr angement shown in Fig. 11.8 is also referred to as stand-by power supply.
No-b reak UPS. When a no-break supply is required, the static UPS system shown in Fig. 11.9 is used. In this system , mainac supply is rectified and the rectifier delivers power to maintain ,required ch arge ' on ,the batteries. Rectifier also supplies power to inverter continuously which is then given to ac-type load through filter and normally-on switch. In case of main-supply failure, batt eries at once take over with no-break of supply to the critical load. No dip or discontinuity in the illumination is observed in case of no-break UPS. Th is ' configuration of Fig. ,11.9 has the following additional advant ages: ,The inverter can be used to condition the supply delivered to load. (ii) Load gets protected from transients in the main ac supply. (iii ) Inverter output fr equency can be maintained at the desired value. (i)
In case invert er failur e is detected, th e lo ad is switched on to the main ac supply directly by t ur ning on the n ormally-off static switch and opening the normally-on static switch. The transfer of load fro m inverter t o main ac supply t akes 4 to 5 ms by static transfer switch as compared to 40 t o ,50 ms for a mechanical contact or. After inverter fault is cleared, uninterruptible power supply 's again rest o ed to the load t hr ough the normally on switch. Th e batteri es are n ow echarged from the main supply by adjusting the charger at maxim um ch arge rate so that batteries are char ged to their full capacity in the shortest possible time.
'\
..
[Art. 11.3]
Some Applications
r-------- -,
I
.
.-
I
I
I
I
I
'
~~~-------------.----------------~----~-t1 Power flow
".
. '.
Normally 'OFF '
I
I .
:
. Static ~ : transfer. \ L ________ "
switc~es
. Main Rectif ier Inverter ac o---40--i ae-de ~......~ dc-ae supply
559
\
. .
~--------iI
. I I
I
:
I
......
Filter ~-
t·'------v~--~ Normo li yon
'"--l1G-.......
I
Crit ic al . load
I
L.- Normally 'ON'
IL ________ 4 I
Fig. 11.9. No-break UPS configuration.
I
,.
The standby batteries in the UPS system are either nickel-cadmium (NC)or lead-acid type. NC batteries have the following advantages : . (a) Their electrolyte is non-corrosive . . (b) Their electrolyte does not emit an explosive gas when charging. (c) NC batteries cannot be' damaged by overcharging or discharging, these have there fore longer life. Cost of NC batteries is, however, two or three times that of lE~ad-acid batteries; The time period for which a battery or a battery-bank can deliver power to load through inverter at the required voltage level depends upon (i) the size of the batteries and (ii ) nature of the load. 11.3. HIGH VOLTAGE DC TRANSMISSION It is well known that electric power generated in power plants is transmitted to the load
centre on three-phase ac transmission line . However, for bulk power transmission over long distances, high voltage dc (HVDC) transmission lines are preferred. HVDC transmission possesses the following advantages over AC transmission system: (i) In HVDC transmission system, one or two conductors and smaller towers are required as against three conductors and tall towe.rs in ·AC transmission system . . HVDC transmission, therefore, costs less . .(ii) Fault clearance in HVDC is fast er, therefore DC transmission system possesses improved transient stability. (iii) Size of conductors in DC transmission can be reduced as there is no skin effect. (iv) Two AC systems at different fr equencies can be in ter connected thro gh HVDe transmission lines . . (v) For power transmission through cables, HVDC is pr eferred as it requires no charg ing current and the reactive power. The addition al cost of convertin g and inverting equipments makes HVDC~ransmission I u n economical for low -p ow er supply ov er short dist ances. However, for l arge-power tra smission over long distances, HVDC turns out t o be economical. As a result, HVDC links are being used worldwide at power levels of several gigawatts with the use of thyristor valve*. 1
" The te rm "thyris tor valve n , used on HVDC systems , denotes a number of thyristors connected in series and p.:.rallel to get the required voltage and curren t ratings.
560
[Art. 1l.3]
Power E lectronics Ld
C
o
N ' V E
~~--..,....jR
T .
o R
B
· Hh
B
. . . AC
..
AC
syst~m
Iii ler
Ld
.
Fi.g. 11.10. Basic layout for HVDC transmission system .
Fig. 11.10 shows the basic layout of an HVDC transmission system. Two AC systems A and,B are interconnected by the DC line. If power flows from A to B, converter A then operates as a rectifier and B as an inverter. Reverse power flow from B toA is also possible with B acting as a rectifier and A as an inverter. AC filters reduce the current harmonics generated by the converters from entering into ac systems. DC filters and smoothing inductors Ld reduce the ripple in the de voltage. Both converters A and B have 12-pulse configuration. The centre-point of converters A and B is earthed with one line, or pole, at +kV and the other line, or pole, at -kV with respect to .earth for a ± kV system. With both the ends earthed, the power how can be maintained with +kV line and the ground or with -kV line and the ground. 11.3.1. Types of ·HVDC Link · There are two basic types of HVDC transmission systems. These are m onopolar link and bipolar link. Monopolar or unipolar link shown in Fig. 11.11 (a) offers the simplest arrangement. It uses a single conductor which has either positive, or negative, polarity. It is preferred to have . Rectifying
Inverting
Return current
I
I
L _______ ~---~----------~
(a)
,---...,,+ 3 phase o--....-ooC:?II':~-t
1----------1
1-<4~t---:)
AC
AC S';l s tem
3 phase
\L.-~J---:~Tt1-()
o-H-+-qDo--i
Retu r n cu r r ent Rect i fy ing
Fig. 11. 11 . Types of HVDC lin
( b) (a )
I nverting
monopolar link and (b ) bipola link.
system
Some Applications
[Ar t. 11.3]
561
negative polarity for the single conductor as it produces less radio interference. The return path is provided by ground or sea. The return current through ground or sea leads to higher condudion losses, electrolytic action and large potential gradients. , In bipolar HVDC transmission, two conductors are used, one is pdsitive and other is negative with respect to the ground as shown in Fig. 11.11 (b). As stated before, the neutral points are grounded at both the ends. As the positive and negative conductors carry equal currents, there is no earth current. In case one line is opened due to fault, the other conductor and the ground will fonn unipolar link and half the rated power can be transmitted untill the fault is cleared. It is obvious from above that bipolar system ofHVDC is more reliable than the unipolar or monopolar system. As such, HVDC bipolar link is more commonly employed. A typical bipolar HVDC arrangement is described in what follows. 11.3.2. Bipolar HVDC System Three-phase 6-pulse phase-controlled converters were discussed in Chapter 6. Current harmonics generated on the ac side and the, voltage ripple produced on the dc side of the
AC
AC
System
Filter
AC Filter
A
B
Ld
Ld
(a)
+
+
3- phose
AC
tJdl
Supply
"
System
Ud =V d1 +Vd2
,
-
+
,i (b )
(b) Fig. 11.12. (a) Schematic diagram for bipolar HVDe sys m Twelve-pulse converter obtained by connecting two six-pulse converters .
562
[Art. 11.3]
Power Electronics
converter used in HVDC system must be reduced. This is achieved by using twelve-pulse converter oper ation which requires the use of two six-pulse converters fed from delta-delta and delta-star transformers as shown in the schematic diagram of Fig. 11.12 (a). The details of the two converter connections are shown in Fig. 11.12 (b). The delta-star bridge gives six-pulse output voltage Vdl whereas the other delta-delta bridge also delivers six-pulse output voltage Vd2 ' The two secondaries, one in star and the other in delta, cause a displacement of 30° in the two six-pulse output voltages such that a twelve-pulse output voltage is obtained from Fig. 11.12. It is seen from this figure that the two six-pulse converters are connected in series on the dc side and in parallel on the ac side. The series connection on the dc side helps to meet the high-voltage requirement in HVDe systems. When the power flow is from system A to system B, converters A operate as rectifiers . The average value of output voltage of converters A is as under : Average value Ofvdl = average value ofvd2 3Vml 3wLs or V d1 =Vd2 =- .- cos a - - - - Id 1t
1t
where V ml = maximurn value of line to line input voltage to each of the two six-pulse converters Ls = transformer leakage inductance per ph ase referred to the converter side Id = dc current a = firing angle delay. The average voltage · available across both the series-connected converters, i. e. de link voltage is given by 6Vml 6wL s 6 .. .(11.8) Vd = 2Vd1 =2Vd2 =- - cos a - - - Id =- [Vml cos a - wLs I d ] 1t
1t
1t
11.3.3. Control of HVDC Converters Positiv e and negative poles are operated under identical conditions. Therefore, HVDC system of Fig. 1l.12(a) can be ·represented on per pole basis as in Fig. 11.)3 (a). In this figure, 3wLs
-'f-
-.
Rci
+ J II . ~ COS(l Vd 71 -3 - [ \Iml cos '/ 1f
..
- wL slct ] Rec t ifi er
In verte r
~--------~------~
(a) (b) Fig. 11. 13. (a) Representation of Fig. 11.12 (a ) and (b) its equivalent circuit.
A and B ar e six-pulse con verters. Further, it is taken that converter A is operating as a r ectifier and B as an inverter . Current Id is then given by
Vda
Id
-
Vdb
= - --=R=-d-
wh ere Rd = resista. ce of on e transmission line conductor .
Some Applications
[Art. 11.3]
563
In practice, one converter is made to control the transmission line voltage and the other to control the current I d . The inverter is usually made to operate at a constant margin angle . y = 7t - a - ~ from commutation considerations. It, therefore, follows that inverter is assigned the job of controlling Yd' The current Id and therefore power level is controlled by the rectifier.. At constant extinction angle y, the inverter de voltage is given by
1=;;3 [Vml cos
3·roL s Id Vdb ="[3Vml -7t- cos y- -7t-
y- w Ls Id ]
and the rectifier output voltage Vd is . . 3 Vd = Vda =Vdb +Id Rd =- [Vml cos y- wLs I d] + IdRd 7t 3 Vd=-[VmICOSCl.-wLsId]
Also,
7t
... (11.9a)
... (11.9b)
'
Eq. (11.9) leads to the equivalent circuit ofHVDC system as shown in Fig. 11.13(b).
Example 11.1. Two six-pulse converters are used i'n bipolar HVDC transmission system. The ac systems are 3-phase, noooy, 50Hz. The input transformers have a leakage inductance of 10 rpH per phase. The current in de line is 300A. The inverter marginal angle is 20°. Resistance of each de transmission line is 1 n. Calculate firing angle of the rectifier, its output voltage and de link voltage. Solution . It is seen from the equivalent circuit of Fig. 11.13 (b) that
3Vml 3roL s . Id 3Vml 3roL s . Id
- - cos a = - - cos y + Id . Rd 7t
7t
7t
7t
7t
cos a = cos y + -. _ . I d · Rd
3Vml
or
.
7t
= cos 20° + 3 'f2 x 11000 x 300 x 1
or
a= 16.283°
3Vml 3wLs . Id Rectifier output volt age, Vd = - - cos a - -.....::....--=.
7t
7t
= ~7t ['1'2 x 11000 x cos 16.283 .=
DC link voltage
0
-
27t x 50 x 10 x 10- 3 x 300]
13357.2 V
= 2Vd = 2 x 13357.2 = 26714.4 V = 26.7 14 kV.
Example 11.2. Two six-pulse converters, used for bipolar HVDC transmission system, are · rated at 1000 MW: ± 200 kV. CalculaU! the rms current and peak reverse voltage ratings fo r each of the thyristor valves. Solution. The dc transmission voltage
= 200 + 200 = 400 kV
Direct current in t h e transmission lines ,
3
I = 1000 x 10 = 2500 A d 400 . It is seen fr om the working of a 3 phase full converter that eac 120 for a periodicity of 3600 • 0
thyristor conducts for
564
P ower Electronics
[Art. 11.4]
:. Rms current rating of thyristor 20 =Id -360 =2500 3Vml - - cos (X = 200 kV Also,
1Wn
~ . A - = 1443.38 3
It
For extreme cases, V·
or
ml
3V
(X
ml = 0° and -It=200 kV
= 2003 x It. ..=209.44 k V
Since there are two SeRs conducting simultaneously in a six-pulse converter, the peak reverse .voltage across each thyristor valve = 209.44 ;: 104.72 kY. 2 11.4. STATIC SWITCHES A switch having no moving parts is called a static switch. Power semiconductor devices which can be turned on and off within a few microseconds can be used as fast-acting static switches. For high-power applications, thyristors are being used as static switches whereas for low-power applications, power transistors are preferred. Static switches are now replacing mechanical and electromechanical switches because of several advantages listed below : (i) On time of a static switch (SS) is of the order of 3 /ls, it has therefore very high switching speed. (ii) SS has no moving parts, its maintenance is therefore very low. (iii) SS has no bouncing at the time of turning on .
(iu) SS has long operational life.
In static switches, however, attention must be paid to leakage current during their off periods.
~) (a)
~)
~
Fig. 11. 14. Single-phase swi tches electromechanical (b ) triac and (c) two-thyristors in anti-pardUel.
An electromechanical switch sh own in Fig. 1l.14(a) is actu ated by magnetic coil or plunger. The static swit ches us.:ng t riac and two thyristors in anti-parallel are shown in Fig. 11 .14. (b) and (c) respectively. Static switches are now bein g used for relays, circuit breakers, fuses, fl ashers, UPS, automobile blinkers etc. . It may be' observed that cir cui t of ~'ig. 1l.14(c) is the same as that of Fig. 9.4(a) for single-phase ac voltage controller . But these two circuits are operated differently. Not e that static means changel ess. In other words, this implies th at static switch merely connects a load
(Art. 11.4]
Some Applications
565
to the supply or disconnects a load from the supply. Static switch does not change or control the power deJivered to load as it is done in a single-phase voltage controller. In static switches, the semiconductor switches are turned on at zerO-:crossing of load! current, whereas it is not so in single-phase voltage controllers. Static switches can also be used for latching, current and voltage detection, time-delay circuits, transducers etc. Static switches are of two types : ~i) ac switches and (iir-dc switches. If the input is ac, ac SSs are used and for dc input, dc SSs are used. Switching speed for ac switches is governed by the supply frequency and turn-off time of thyristors. For dc static switches, the switching speed depends on the commutation circuitry and turn-ofrtime of fast thyristors. AC switches may be single-phase or three-phase. Static switches are discussed in what follows: 11.4.1. Single-phase AC Switches The circuit diagram of a single-phase ac switch is shown iIi Fig. 11.15 (a). Here two thyristors are connected in anti-parallel. For resistive load in Fig. 11.15 (a), the waveforms for sour e voltage v S' triggering pulse i gl for Tl and pulse ig2 for T2, load voltage Vo and load current io are shown in Fig. 11.15(c). Note that Tl is triggered at rot = 0°, rot = 2lt, .... ,·. and T2 is triggered at rot =It, 3lt ..... when the load current waveform is passing through zero. For RL load, outpur- . or load current io lags Vo by load power-factor angle 4> =tan- I
~.
For RL load, Tl must be
triggered at rot = 4>, 2lt + and so on, Fig. ll .15(d). A triac TR can replace two anti-parallel t hyristors as shown in Fig. 1l.15(b). For triac, only one pulse ig = i gl = ig2 in each half cycle will be required. \)s
T1
0
wt
+ tg
(a)
7r
wt
: wt I
I I I I
(b)
I
:
I
•
•
: mlg~
O~~~----+---~---
wt (c)
(d)
Fig. 11. 15. Circuit diagram for single-phas e a c switch (a) using two thyristors (b ) using one triac (c) waveforms for R load (d) waveforms for RL load .
566
[Art. 11.4]
Power Electronics
AB load current waveform io is a sine wave, it can be expressed as
io ::: 1m sin rot Rms value ofload current i o, or source current is, in Fig. 11.15 (c) and (d) is 1
lor::: Iar =[ ~
JIt0 1m2 ~m . 2rot· d (rot)- ]112 ::: T2 1m
... (1110) .
Note that each thyristor carries current for 180 0 for each cycle ofload current. Therefore, . average value of thyristor current is 1 [
ITA::: -2 It
. . . 1m sin rot . d(rot) ::: 1m 0 . It
...(11.11)
and rms value of thyristor current, .
ITr=[21It~I~Sin2cOt . d(rot)]
112
_ I;
... (11.12)
A hi-directional switch, using four diodes and one thyristor (or a triac or a GTO) is shown . in Fig. 11.16(a). This circuit, in performance, is similarto that ofFig.. 1l.15(a). The waveforms for source voltage vs, triggering pulse ig for thyristor T, load current io or source current is and thyristor current iT are shown in Fig. 1l.16(b). It is seen from this figure that rms value of thyristor current, 1/2 I 2 ... (11.13) I Tr ::: [ ~ I! sin rot . d(oot) ] , =~
*
and average value of thyristor current, 1 ~ 21m ITA = - L 1m sin rot· d(oot) = 0
It
... (11.14)
It
i. g
wt R
+ v
n
Fig. 11.16. (a) Single-phase ac bi-directional swi tc
2"
(6)
(a ) (b)
its waveforms.
wI
Some Applications
[Art. 11.4]
567
11.4.2. DC Switches As st ated before, in dc switches the input voltage is dc. Power semiconductor devices used in a dc switch may be transistors, thyristors or GTOs.
When thyristor is usee , it must have forced commutation circuitry as an integral part of dc switch. One such circuit giving the principle of operation of a dc switch is shown in Fig. 1l.17(a). Here T1 is the main thyristor and TA is the auxiliary thyristor. Capacitor C is charged to source voltage Vs with lower plate positive. When T1 is on, normal load current 10 flows . from source to load through Tl. For breaking the dc circuit, auxiliary thyristor TA is turned on. Capacitor C at once applies a reverse voltage across T1 turning it off at once. After this, load current flows as shown in Fig. 11.17(b). Capacitor now gets charged from + Vs to - Vs and current through TA falls below its holding current to turn it off. Subsequently, freewheeling . diode takes over and load current eventually decays to zero. For further details, refer to Figs. 7.24 to 7.26.
T1
+
+
c
FD
TA
TA
(b)
(a) Fig. 11.17.
Single~pole
thyristor dc switch.
Sin gle-pole dc switches using a transistor and a GTO are shown in Fig. 1l.18(a) and (b) respectively. In Fig. 11. 18(a), when forward base curr ent is applied to transistor, TR gets turned on and dc voltage Va appears across load. When base current is removed, TR gets turned off and load current falls to zero. A freewheeling diode FD is necessary for the inductive load.
TR
+
+
Bose drive
circuit
(a) (b) Fig. 11.18. Single-pole (a) transistor dc switch (b ) GTO de switch.
In Fig. 11.18 (b ), gat e tur n-off' thyristor is turn ed on by a short positive gate pulse as in ordinary thyristors . \Vhen r equired, GTO can be turned off by a short n egative pulse applied to its gat e t ermin als. Note th at a GT O r equires no for ced-commutat; on eire itry.
11.4.3. Design of Static Switches The design of static switches involves the determ 'n a t' on of voltage and current r atings of power semicondu ctor devices employed. Their design is illust r ated with an example.
568
Power Electronics
[Art. 11.5]
Example 11 .3. A single-phase ac switch of Fig. 11.16 (a) is used in between a 230-V, 50 Hz source and a loq,d of 2 kW at pfof O.8lagging. Determine the voltage and current ratings of (a) the thyristor and (b) diodes of the bridge. Take a factor of safety of two.
Solution . (a) Peak value ofload current,
= 2000 'i'2 = 15 37 A
I
230 x 0.8
m
.
Rms value of thyristor current, from Eq. (11.13) is
_1m - 15.37 - 10 87"A I Tr-T2-~. Average value of thyristor current, from Eq. (11.14), is ITA
= 21m = 2 x 15.37 = 9.785 A 1t
Maximum value ofrms current for SCR
1t
= 10.87 x factor of safety.
"
= 10.87 x 2 = 21.74 A
Maximum value of average current for SCR = 9.785 x factor of safety. = 9.785 x 2 = 19.57 A For the configuration of Fig. 11.16(a), thyristor is always on, therefore it is subjected to almost zero inverse voltage. However, PlV for this SCR may be taken as V m = -.J2 x 230 = 325.22 V. So choose a thyristor with PlV = 325.22 V, maximum rms on-state current 21.74 A and maximum average on-state current = 19.57 A. (b) Each diode conducts for 1800 for a periodicity of 360 0 • This gives maximum value of diode current
=I; == 15237 = 7.685 A
and max imum value of average diode curren t =
I;
= 15.37 = 4.892 A. 1t
Diode also experiences zero inv€!rse voltage under ideal conditions. So diode PIV may be selected as for the thyristor. So choose a diode with PlV = 325.22 V, maximum rms on-state current = 7.685 x 2 = 15.37 A and maximum average on-state current = 4.892 x 2 =9.784 A. '11.5. STATIC CIRCUIT BREAKE~S
"
"
Static circuit breaker s are semicon duct or-based circuits capable of providing a fast and r eliable interruption to a continuous cucrent. Static circUit breakers are of two types; static ac circuit breakers and static dc circuit breakers. High-current circuit breakers employing thyristors are now discussed briefly. 11.5.1. Stat~c AC Circuit Breakers Static a c switch can be made to ope. ate as a stati ac circuit breaker, In Fig. 11 .19 (a) is shown a simplified circuit configuratio for static ac circuit breaker an d Fig. 11.1 9 (b) gives relevant voltage and current wa', eform.s. As in static switches, thyristor s 1 and 2 in Fig, 11.1 9(a ) axe turned on at th iIlBtant load current is passing through zer o. For breaking the circuit, th e trigger ing pulse is ',vi thdrawn . For example, at rot =41t + <1>, if t riggering pulse ig 1 is not app it:d to TI, it will n ot get turned on. T2 is already off just before rot = 4n + ~ . Therefore,
[Art. 11.5]
Some Applications
569
o~----~----+-----+-----~----~
wI
T1 lo
R Vs
l
+
Vo
(\J
I
tg
I
~JPul,e tran,'or mecs
L
(a)
Fig. 11.19. (a) Static ac circuit breaker and
(b)
(b)
its relevant voltage and current waveforms.
the continuity of the circuit is broken. So when turn-off command is received by the control circuit due to some system fault, the gating pulse is withdrawn from Tl or T2 and eventually the circuit is broken. In case turn-off command is received just after 31t + <\>, load current will be broken only at 41t + <\>, i.e. a delay of 1t r adians or half-cycle is a must. If turn-off command is received at the instant (31t + <\» < rot < (41t + <\», even then the circuit is broken at the instant rot ;: 41t + <\> only. This shows that maximum time delay for breaking the circuit is one half-cycle i.e. : seconds after turn-off command is accepted by the control circuit due to some exigencies in the system. 11.5.2. Static DC Circui t Breakers A simple arrangement of static dc circuit. breaker is shown in Fig. 11.20. This circuit is similar to that shown in Fig. 5.4(a), pertaining to class-C commutation. As stated before, when input voltage t o a circuit consisting of thyrist ors is dc, forced
commutation is essential for t urning off a thyristor. For complete
+ analysis, refer to section 5.3. Here only brief discussion is given.
ed on, load voltage becomes
When main thyristor Tl is t equal to source voltage Vs an d c~ a citor C begins to char ge
Vs through the cir cuit Vs , R 2 , C and TI . Eventually capacitor C gets T2 charged with r ight h and plate positive. For breaking the circuit,
auxili ary thyristor T2 is t rned on. Capacitor voltage V c at once
applies a reverse voltage Vs across SCR Tl and turns it off. After
Fig. 1l.20.
Tl is for ce commutated, capacitor will ch arge from + Vs to - Vs Static dc ci r cuit breaker. , through the circuit V s , load, C and T2. When C is fu lly charged t o - Vs (l eft and plate posi tive) , current through load will be zero and at the same time cur ent through R2 will be less th an the holding current of SCR T2 . As a result, T2 w ill get turne off naturally. From this, the v alue of R2 an be determined.
570
[Art. 11.6]
Power Electronics
Example 11.4. For the static de circuit breaker shown in Fig. 11 .20, the supply voltage is
200 V de and load current requ ired is lOA. SCR T l has a turn-off time of 20 ~ and SCR T2
has a holding current of 5mA. Find the values of parameters R 2 , C and load resistance. Take
a factor of safety of 2. Solution. Load resistance, RL = 200 = 20 n 10
Holding current determines the value of R 2 . Therefore,
200
R2 = 3 = 40 k n. 5 x 10 It is seen from Eq. (5 .9b) that voltage vTI
=-
ue
VTI
across T1, after T2 is turned on, is given by
= Vs
[1- 2e- tIRC ]
... (11 .15)
The turn-off time te for T1 can be obtained from Eq. (11 .15) by calculating the period during which uTI falls from - Vs to zero. Therefore, from Eq. (11.15), o = Vs [1 - 2 exp (- tel RL . C)]
or tc =RL C In 2
C=
6
tc =20x2xlO- =2.885)lF.
RL In 2 20 x 103l n 2
11.6. SOLID STATE RELAYS AC and dc static switches can be used as solid st ate relays (SSRs) in ac and dc circuits respectively. In ac circuits, thyristors or triacs are used wher eas in dc circuits, transistors are preferred. Solid state rel ays have no contacts or moving parts. These are now being used extensively and are replacing the conventional contact-type electromagnetic relays in applications like control of motor drives, resistance heating etc.SSRs need electrical isolation between control circuit and the load circuit by means of optocouplers or pulse transformers . Ph ot o- diode
+
~-------\----~ I
•
I
V1
I
:-------\--i_
Photo t ra nsi'stor
R
I I
.
I
:
I
I I
I
I
I
I
II
___
II
I
. I
I
---
I
I
! ILE D I
I
I
I
I
I I
pLED
I
I
I
:... __________]
: I
I
I
I
__ -J
.
Optc coupl er
(a)
fJ)
Fig. 11.21. Optocouplers using (a) a photo-diode and (b ) a photo-transistor.
An optocoupleT consists of infra-red light emittin g diode OLED) and a photo-diode or a ph oto-transistor. An optocoupler h aving ILED an d phot o-d 'ode is sh mvn in Fig. 11. 21 (a ). A shor t puls e V1 applied to ILED will cause it to emit light on to ph ot o-diode which will t hen begin to conduct in the reverse direction as shown . An optocoupler using photo-tr ansistor is shown in F ig. 11.21 (b). As before, a short pulse V 1 applied t o ILED will throw light on t he base of phot o-trans istor an d turn it on. As photo-tr ansistor is m ore sensitive t h an a ph oto-diode, optocoupler s bas ed on opto-t ransistors are m ore comm on .
[Art. 11.6]
Some Applications
571
11.6.1. DC Solid State Relays A dc solid state relay using opto-coupler for isolation purposes is shown in Fig. 11.22. When control pulse V c is applied to ILED, it emits light and t urns on the photo-transistor. The current output from the photo-transistor acts as the base current for transistor TR. Consequently TR is turned on and source voltage Vs is applied to load. When control pulse Vc is absent, TR gets turned off and load voltage is zero.
+
'L ·-·-·-·-·7··-·~ ~-------------r · --.J Optocoupler
Fig. 11.22. DC solid-state relay using an optocoupler.
11.6.2. AC Solid State Relays Fig. 11.23 shows two basic circuits for ac solid-state relays. Fig. 11.23 (a) uses a pulse transformer for isolation purposes and in Fig. 11.23 (b), isolation is provided.by an optocoupler. When control signal appears across the primary of pulse transformer, its secon dary applies a triggering pulse to turn on the triac. As a result , circuit is completed through us' load and triac and therefore, source voltage is applied to the load. Solid state Relcy
Solid state Relay r---------------~
:,
.----+-'---,
r----------),-- - - - - , i 0 pta cou pie r .------4-........~-__, I . r----------"1
II I
j
r---t-........-----,
:,
+ lis
('\j
I I
,
\
Control : signal "t-
AC voltage
,,
,, ,L. ___ _ __________ : I
I
I
I I I
,,, ,
sourc~
~----------~
~-------- - ~--- -- -- ----------~
(a)
Fig. 11. 23. A'C solid-state relays using (a) a pulse transformer and
(b)
(b) an
optocoupler.
In Fig. 11.23 (b), control signal turns on the photo-transistor. If the ac supply has upper terminal positive as sh own in Fig. 1l .23(b), the current 'lY'i ll flo w thro ugh R , D 1, photo-transistor,D2, triac gate and source. This current will turn on the triac an d load gets energised by source voltage Us ' The function of R is to limit th e tl O\ V of gat e cur rent of tri c. If low r t ermin al of ac s pply is positive, the current will fl ow throu gh triac g at~1 D3, photo transistor, D4, R and source Us ' Triac gets turned on and source voltage is applied to load.
572
[Art. 11.7]
Power Electron ics
11.7. RESONANT CONVERTERS In SMPSs discussed in Art. 11.1 and in the PWM inverters described in Chapter 8, the swit ching devic'es arc made to turn-on and turn-off the entire load current at high dildt. The devices handling high dil dt also experience high-voltage stresses across them; due to these two effects, there are increased power losses in the switching devices. In case size and weight of the converter components is to be reduced, switching frequencies are increased. At these high frequencies, switching losses and high~voltage stresses are further aggravated. Another major drawback of high dildt and high dvldt caused by rapid on and off of the switching devices is the electromagnetic interference. The shortcomings enunciated above can be minimised if each switch in a converter is turned on and off when the voltage across it and/or current through it is zero at the instant of switching. The converter circuits which employ zero-voltage and/or zero-current svvitching are called resonant converters. In most of these converters, some form of L-C resonance is used, that is why these are known as resonant converters. In this section, resonant converte r s employing zero-current switching (ZCS ) and . zero-voltage switching (ZVS) are described. 11.7.1. Zero-Current Switching Resonant Converters There are two types of ZCS resonant converters, L-type and M-type: Both of these circuit topologies use L and C as a series resonant circuit; in addition L also limits dildt of the switching current. Here first L-type and then M-type ZCS resonant converters are presented. 11.7.1.1. L-type ZCS Resonant Converters. An L-type ZCS r esonant converter is sh own in Fig. 11.24. The switching device S in the figure can be a GTO, thyristor, BJT, power MOSFET or IGBT. At low kilohertz range; GTO, thyristor, transistor or IGBT is used whereas for megahertz range, power MOSFETs are pr eferred. Inductor L and capacitor C near the dc source Vs form a r es onant circuit whereas L 1 , C 1 near the load constitute a filter circuit. Direction of currents and plarities of voltages as marked in Fig. 11.24 are treated as positive.
s
Y
r-
L
T
Ll
tL
ic -+ Uc
'----.......v ------' Resonant -:ircuit
Fig. 11.24. L-type
c
10
o
Cl
'----.......v,....----~ . Filt er circui t
zero -current~switching
resonant converter.
The circuit of Fig. 11.24 is initially in the steady stat e with constant load current 10 , Fiiter inductor Ll is relatively large to assume that current io in L1 is almost constant at 10 , Initially, switch S is open; r esonant circuit parameters have iL = 0 in L and Vc =0 across C and the load current 10 freewh eels through the diode D. F or the sake of onvinience, work ing oftrus converter is divided into five modes as under. For all these modes, tim8 t is taken as zero at the beginning of each mode . Mode I . (0 ~ t ~ t 1 ) . At t = 0, swit ch S is turned on. As 10 is fr eewh eeling thro ugh diode D, voltage acro ss ideal di.ode VD = 0 an also V c = 0 Fig. 11 .25(a ). It implies that source voltage
. '..\
.
574
[Art. 11.7]
Power Electronics
s
s
L
L
i.e
c
D
lo-----.J
<4
(a)
S
Vs
Mode I, iD ::: 10 - iL
(b) Mode II,
tL
L
C
Vs
..
r-+
- i.e =10
i.e
10
Vc 3
To
iL = 10 + ic
r.
V
D
tD 10
10
(c) Mode III, iL ::: 10 - ic (d ) Mode IV, - ic ::: 10 . (e) Mode V, iD =10
Fig. 11.25. Equivalent circuits for the operating modes ofL-type ZCS resonant converter.
.
II
r m -- ZoS
---- ------- --- -- ---------
10 l--+----+---+-+_ T - - -- - -.....:
1
1
t
I
1
I
I . I I
I· ,
; - . - - t 2- - :I tJ IL-t, --...;
:I
:I
I
I
I
1
, I
I
I
,
I
I , . I
I I I
t I
r-ts-"
I
I
~,'][~ m ~IY · ---...... '._y~I
iI
ic
;.
I: I
I
I
•
I I I
, I ,
I
I
i
:
I
I
I
t
I
I
I
I
I
o ~~---+--~~~---.--~----~~--~----~---.
t
liT
0,
I
:,
I
,
S on
S
oH
Fig. 11.26. Wavefo rms for L-type
5
on
zes resonant converter.
t
!
f
[Art. 11.7]
S me Applica tions
575
Mode IV. (0::; t ::; t 4 ). As switch S is turned off at t = 0, capacitor begins to supply the load current 10 as shown in Fig. 11.25 (d). Capacitor voltage at any time t is given by vc = Vc3 -
~ Jic . dt
As magnitude of capacitor current ic =10 is constant,
. 10 vc=Vc3- C t
.. .(11.19)
This mode comes to an end when Vc falls to zero at t == t 4 • . '. 10 . or 0 = V C3 - C t4 or
t _ C· VC3 4 10
... (11.20)
At t=0,vr=Vs-Vc3 and at t=t4 ,vr=Vs -0=Vs as shown in Fig. 1l.25(d). As 10 is constant, capacitor discharges linearly from V C3 to zero and vrvaries linearly from (Vs - VC3) to Vs as shown in Fig. 11.26.
Mode V (0::; t ::; t 5). At the end of mode IV or in the beginning of mode V, capacitor voltage Vc is zero as shown in Fig. 11.26. As Vc tends to reverse at t = 0, diode D gets forward biased and starts conducting, Fig. 11.25(e). The load current 10 flows through the diode D so that iD =10 during this mode. This mode comes to an end when switch S is again turned on at t = t5 ' The cycle is now repeated as before. Here ts = T - (tl + t2 + t3 + t 4)· The waveforms for switch or inductor current Lv capacitor voltage vL' diode current i D , capacitor current ic and voltage across switch S as VT are shown in Fig. 11.26. It is seen that at turn-on at t = 0 (0 ~ t ::; t l ), switch current iL = 0, therefore switching loss v~L = 0. Similarly, at turn-off at t3 (0 ::; t ::; t 3 ), iL = 0 and therefore v-J,L = O. It shows that the switching loss during Vs turn-on and turn-off processes is almost zero. The peak resonant current 1m = Zo must be more than the load current 10 , otherwise switch current iL will not fall to zero and switch S will not get turned off. The load voltage Vo can be regulated by varying the period t5' It is obvious t h at longer the period t 5 , lower is t he load voltage. 11.7.1.2. M-type ZCS R esonant Converter. An M-type ZCS r esonant converter is before, Land C form the r esonant circuit and L l , C 1 the filter circuit. sho\'/D in Fig. 11.27. Capacitor C is connected across the series combination of switch Sand L; but in L-type convert er, C is connected across diode D. Working of this converter can be divided into fi ve modes as for the L-type resonant converter. The time origin t = is redefined at the beginning of each m od e.
°
M ode I (0 ::; t ::; t 1) . Prior to m ode I , i.e. before switch S i turned on at t;::: 0, load current 10 fr eewh eels throu gh diode D. Also, voltage Vc = Vs before S is closed . At t = 0, swi ch S is turned on. Now current iL begins t o develop throu gh Vs . L an d D as shown in the equivalen t circui t of
576
-[Art. 11.7]
Power Electronics
+
Ve
-
i.e
C S
i o=10
iL L
10 L1
tD
D
C1
~ol
R
'-----y----'
F i I ter circuit
Fig. 11.27. M-type zero-current-switching resQnant converter. . . V · this mode, Fig. 11.28 (a). It is seen that iL = t rises linearly as in L-type converter. Also, diode
i
current iD
= 10 -iL
Att=t 1• 1 ,L
or
tl
=-0 -
... (11.21)
Vs
At t = t I , iL =10 and iD =10 - 10 = 0, diode D is therefore turned off. Voltage capacitor stays at Vs through D. -t
Vs -
Vc
across
+
_ Vs
C S
s
tL
L
iD
~ (0)
v
.
s. ~
Mode I, iD =10 - iL
(b) Mode II, iL
-
Ve
=10 + ic
+
(c)
Mode III,
+ Vs
iL = 10
- ic
_
C
Ve = V C3 aU =0, iL = 0 Fig. 11.28. Equivalent cir cuits fO T -type
(d) Mode IV,
(e) Mode V, iL
°
=
zbs resonant converter.
Mode II (0 S t S t 2). Afte D turns off, load current 10 flows through Vs and L as shown in Fig. 11 .28(b). Also, C and L fo m a resonant circuit where the urrent ici given by
Some Applications
[Art. 11.7]
ic = Vs and capacitor voltage At t;: O. Vc;:
VSI
Vc
1f
is given by
sin wot = 1m sin wot
Vc
= Vs cos wot <-
ic = O. iL = 10 , When wot2 =1t or t2 =
Vc =- Vs. ic = 0 and iL =10 ,
577
:0 =
1t
...JLC. cap a citor voltage
.
t
= ;.ic=Im.iL=Io+Im and vc=O.
Whent=2: 0
During this mode, iL =10 + ic =10 + 1m sin wot and capacitor gets charged from Vs to - Vs as shown in Fig. 11.29.
t
I I
T
t
I
I
lc Modes \
I
I
I
I
:
I
I
1 l--lI-,----~IIIl ~N
\
I
I
I \
-/ 7~ I I
I I
I
:
• I I
I
I
I
I
t
Fig. 11.29. Wavefonns for M-type ZCS resonant converter.
Mode m (0 ~ t ~ t s). Equivalent circuit for this mode is given in Fig. 1l.28(c). Various waveforms for M-type ZCS r esonant converter are shown in Fig. 11.29. It is seen that capacitor voltage at th e end of second mode is negative, i.e. Vc = - Vs' During mode III, the capacitor voltage is given_by Vc
and
= - Vs cos COo t, ic
=1m sin root
iL= l o-ic= l o- l msinooot At t = t 3, switch current iL falls to zero as in L-type converter. Also at t = t 3• Vc
=- V, cos Wo ts =- VC3 and ic =lm sin w(h
_1 [10)
= 10 so that iL = O. This gives
_1[10)
-'[7'c sm . ...(11.22) - =~1JL; COo 1m 1m Mode IV (0 ~ t ~ t~. In the previous mode, as iL falls to zero and t end to reverse, switch 8 is n aturally turned off. In this mode, th erefore, S remains off and the equivalent circui t d . t s= -1sm
578
Power Electronics
[Art. 11.7]
Fig. 1l.28(d) applies. Load current 10 flows through Vs and C. At t charges C from - VC3 at t = to Vs at t =t", Fig. 11.29. Therefore, dv V + VC3 Io=C- =1 s .~ dt t"
°
=0, Vc =-VC3' Current 10
V s + VC3 t" = C ~-::I;-o--
or
.. .(11.23)
At t 4 , Vc = Vs' Actually, at t", C is somewhat overcharged with left hand plate positive and consequently diode D gets forward biased . . Mode V (0::;; t ::;; t 5 ). At t = 0, diode D starts conducting and 10 freewheels through D as shown in Fig. 11.28 (e). SwitchS is open and voltage Vc stays at V" through D. Switch current iL remains zero as S is open. At time t = T, switch S is again turned on and the cycle repeats. 11.7.2. Zero-Voltage-Switching Resonant Converters A zero-voltage-switching (ZVS) resonant converter is shown in Fig. 11.30. It consists of diode D1 and capacitor C connected across the switch S. AP. in ZCS converter, ZVS resonant converter has L, C as the 'resonant circuit components and Lv .Cl as the filter circuit .components. The function of resonant capacitor C is to produce z~ro voltage across the switch S. Diode D2 provides a free wheeling path to load current 10, As the name suggests, the switch S in ZVS resonant converter is turned on and off at zero-voltage across the switch. S
L
. I
\'0= 0
~--~~~
~--~----~
D2
~--'I
'y~---'
Resonant circuit
Filter circuit
Fig. 11.30. Zero-volt age-switching resonant converter.
The working of this converter can be divided into five modes with equivalent circuits as shown in Fig. 11.31. As before, the time origin t = is redefined at the beginning of each mode. Load,cur rent 10 is assumed constant and filter,inductor current io is also taken to remain level at 10 as filter inductor is relatively large. Initially, switch S is on and conducting 10 - Therefore, inductor current iL =10 and initial voltage across capacitor Vco = 0. Mod e I (0 ::;; t ::;; t l ). At t = 0, switch S is turned off. From the equivalent circuit of mode I, Fig. 11.31(a), it is seen that constant current 10 flows th rough Va. C and L . As a re~lt. voltage across switch S or C builds up line8!ly from zero t o V s at time t = t 1 . Diode D2 is off. As the capacitor is charged fr om zero to Vs, capacitor voltage Vc is given by
°
I =C dv o
dt
or A c tim e t
= tIl
... (1 1. 0, 4)
Some Applications
[Art. l1. 7J
°
579
°
N.ote that voltage across diode D2 is vD2 =Vs at t = and vD2 = at t = t l . Also, at t =0, Vc =0; therefore switch S is turned off at zero voltage as required.
Mode II (0 ~ t ~ t 2 ). At t == 0, actually capacitor is somewhat overcharged, i.e.. Vc > VS; th,erefore diode D2 becomes forward biased. Now a resonant current iL is set up in series cir,cuit Vs' C, Land D2, Fig. 11.31(b), where iL is given by i; =10 cos Wo t. The capacitor voltage Vc is given by Vc
where V m = 10
=Vs + Vm sin wot
... (1l.25)
~ = 10 Zo and Zo = ~ is the characteristic impedance
of the circuit in
ohms. The peak switch or capacitor voltage Vpk occurs when wot = 1t/2 or t = %~LC and its .value is V pk = Vs + Vm = Vs +Ioz,o At t =t 2, iL =- 10 where wot2 =1t or t2 = 1t ~LC and capacitor voltage is Diode D2 current is given by iD2 = 10 - 10 cos wot.
At t = 0, iD2
Vc
= Vs,
i
=0, at t = ~LC, iD2 =10 and at t = t2,iD2 = 210 ,
It may be observed from the waveforms that a ZVS resonant converter is the dual of ZCS .-resonant converter. .
Mod e III (0 ~ t $; t s)' Initially, i.e. at t = 0, Vc = Vs and iL = - 10 , With time t reckoned zero from the beginning of this mode, capacit or voltage is given by S
,iov,N
.~
(a)
Mode I, iL = 10
(b)
~il
L
~
Uc
-
(c)
Mode III,
ir=Io
s
01
02
(d)
Vs
Mode IV, iL = -hs + L t
(e)
Mode V, i T =
iL ==
Fig. 11.31. Equivalent circuit for ZVS resonant converter.
j
i02
02
~
\V ,
Mode II, it == 10 sin root
L
+
IoVs
L
10
v a == Vs
10 I
~
at t == 0
*
580
Power Electronics
[Art. 11.7]
vc =Vs - V m sin wot iL = - 10 cos (Oot iD2 =10 - 10 cos root
and so that
At time t = t 3 , Vc = O,iL =IL3 and
o=Vs -
-IL3 . This gives
V m sin Wo t3
t3 ~ 'LC Sin-l(~J~
or
iL
iD2 =10
... (11.26)
At the end of this mode, i.e. at t = t 3 , Vc = 0; as a result reverse bias across Dl vanishes and begins to flow through Dl.
Mode IV (0 ~ t ~ t 4). During this mode, capacitor voltage is clamped to zero by diode Dl conducting negative current i L . As soon as antiparallel diode Dl begins to conduct at t = 0, gate drive is applied to switch S. The inductor current iL rises linearly from - l L3 to zero. At this instant, reverse bias of Dl vanishes and already gated switch S turns on. This shows that . switch S turns on at zero voltage and zero current. Mter this, current rises linearly to 10 in the circuit formed by Vs' S, L and D2. The linear variation of current from l L3 is given by
O L-+---~----~~--------~--~;-----r---~~--
iL 10 O~~--~~--~--~~--4---~--+-~----~----~~--
S
s :
. of f
off ' :
"T
,i--""T"--;
o~.-----~--
I
J
tI
i02
.
tl -
J
- -\2
'::
__
--~~--+---~--~~-I I
----tJ l--
:: I
-
:
__~____+-~___
14_++-
I
,, , - - - - -,- Vo :
- -
.
-
MOd ~S~ I ~JI ---'.I ill;--N - ..........-'l _
Fig. 11.32, Wavefonn fo r ZVS resonant converter.
[Probe 11]
Some Applications .
~L
At t
=t ...
.
VS =- I L3 + L t
. V
iL
581
.
=10 =-h3 + LII t ... This gives
t, '" (10 + 1L3)
(~, J
. . C11.27)
Diode current iD2 =10 + i L . At t = 0, iD2 = 10 + l L3 andtime t' = t..,'iD2 = 0. During modes II, III and IV, diode D2 is in conduction, therefore vD2 = 0, as shown in the waveforms of Fig. 11.32.
Mode V (0:5 t:5 t 5). At the end of mode IV, or in the beginning of mode V at t = 0, iL r~aches 10 and therefore diode D2 turns oft'. Switch S continues conducting 10 as shown in Fig. 1l.31(e). Note that voltage vD2 =Vs during this mode.' Mode Vends at t = ts when switch S is turned off again at zero voltage. The cycle now repeats as before. The various waveforms for these five modes are now sketched in Fig. 11.32. It is seen from these waveforms that for a ZVS resonant converter: (i) switch, or inductor, current is limited to 10 (ii) average value of output voltage Vo can be controlled by controlling the interval t5' This shows that average power delivered to load can be controlled by regulating the output voltage Vo for a given load current 10 ,
11.7.3. Compar ison between ZCS and ZVS Converters In
v zes, the switch is required to handle a peak current of 10 + ZS. .
V
For natural turn-off, ZS
0
0
must be more than 10 , There is, therefore, ~ upper limit to the value of load current in zes converters. In ZVS, the switch is r equired t o withstand a peak voltage of Vs + loZo. This shows that peak switch voltage is dependent on the load current 10 , A wide variation ofload current would need large voltage across the switch. As peak voltage across the switch is .a dominating factor, ZVS converters are used only for constant load applications. In general, ZVS is preferred over zes at high sWitching frequencies, primarily due to . internal capacitances associated with the switch. PROBLEMS
'. . '
.
"
11.1. (a) What is SMPS ? Give its operating principle and indu strial applications. (b) List t he various types of SMPSs, Describe SMPS with a pushpull configuration. 11.2. Describe flyback SMPS with relevant equivalent circuits and waveforms. Derive the various . expressions for voltages and currents involved. 11.3. A flyback SMPS supplies a load of 40A at 5V. The source voltage is 240V dc a d the transfonner initial magnetizing current is 0.4 A. The power MOSFET is operating at a freq uency of 50 kHz with .l duty cycle of 0.4, Detennine the transformer turns ratio from primary to secondary and ita inductance. Assume ideal components and no ripple in load voltage . .. 'nd al the open-circuit vol age acr oss the semicon duct or device . [ADs. 32, L = 0.571 mH, 400 V1
582
[Prob.11]
Power Electronics
11.4. Describe SMPSs using half-bridge and full-bridge configurations. Enumerate the advantages and disadvantages possessed by SMPSs. 11.5. What is an UPS? Give its industrial applications. Describe rotating-type, short-break static and no-break static UPS configurations. Why are nickel-c'ldmium batteries preferred over lead-acid type batteries in UPSs ? 11.6. (a) Give the merits and demerits ofHVDe transmission system over ac transmission system. (b) Describe both types of HVDC links with relevant circuits.
Derive the equivalent circuit of an HVDC system.
11.7. (a) Two six-pulse converters are used in bipolar HVDC transmission system. The ac systems are 3 phase, 11 kV, 50 Hz. The input transformers have a leakage inductance of 8 mH per phase. Resistance of each transmission line is 0.8 O. The inverter marginal angle is 18° and rectifier firing angle is 15°. Calculate current in dc line, rectifier output voltage and dc link voltage. (b) An HVDC transmission system, using two six-pulse converters for bipolar transmission, is rated at 1000 MW, ± 250 kV. Determine the mis current and peak reverse voltage ratings for each of the thyristor valves. [Ans. (a) 276.07 A, 13684.3 V, 27.368 kYJ?! 1154.7 A, 130:9 kV] 11.8. (a) What is a static switch? List the merits ofstatic switches over mechanical switches. (b) The
circuit of single-phase ac voltage controller is the same as that used for single-phase ae switch. Discuss how these two differ from each other. (c) Describe single-phase ac switches using (0 one triac and (ii) bidirectional switches. Derive average and rms values of currents for the semiconductor devices used .. 11.9. (a) Describe single-pole dc switches based on (i) a thyristor (i i ) a transistor and (iii) a GTO. (b) A single-phase ac switch, using two thyristors in antiparallel, is inserted between 230 V, 50 Hz source and a load of 10 kW at a p{ of 0.8 lagging. Determine (i) the voltage and current ratings ofthyristors and (ii) the firing angles of thyristors. Take a factor of safety of 2. [Ans. (b) (i) 650 .44 V, 76.85 A, 48.924 A (ii) 36.87° and 216.87°] 11.10. (a) Vlhat is a static circuit breaker? Describe static ac as well as static dc circuit breakers. . (b) A
static dc circuit breaker of Fig. 11.20 has input voltage of 220 V dc and load current of 5 A. Thyristor T1 has turn-off time of 15 :Ils and thyristor 2 has holding C'lrrent of 6 rnA. Find the values of parameters R 2 , C and load resistance. Take a factor of safety of 2.5. [Ans. (b ) 36.67 k 0, 1.231lF, 44 0]
11.11 . (a) What are solid state relays? How is electrical isolation obtained in these relays? (b) Describe dc solid state and ac solid state relays with relevant circuit .diagrams.
11.12.
What are resonant converters? Give their advantages over PWM controlled converters. (b) Describe M-type z es resonant converter with relevant circuits and waveforms.
n .1S. (a) Give the advantages and limitations of zes reson ant converters.
(b ) Describe L-type ZCS resonant converter with r elevant circuits and waveforms. 1 1.14. (a) Give the principle of ZVS resonant converter. (b) Describe a ZVS resonant converter with approp:-i ate circuits and waveforms.
1 , .15. (a ) Give the advan tages and limitat ions of Z S r esonan t converters .
(b ) Wh at is the difference between L -t ype a nd M-type zes converters? (c) Compare zes and ZVS converters . (a)
,:
.'
. _. '.'
.
C apter 12
.Electric Drives
......
~
••••.•.•..••....••.......•.••••.....•••.....•••.....•......•.•......••••.•.......•....
In this • • • • • • • • • •
~
...... -.
~hapter
Concept of Electric Drive DC Drives Single-phase DC Drives Three-Phase DC Drives <:;;hopper Drives A. C. Drives Induction-Motor Drives Speed Control of Three-Phase Induction Motors Synchronous Motor Drives Some Worked Example
......•...........•.............•.............• ....... ........................
.•..•....•
,~~~
~
......
In this chapt er, first the concept of electric drive is given and then dc and ac drives are described. The object of this chapter is not to discuss electric drives exhaustively but at an introductory level. 12.1. CONCEPT OF ELECTRIC DIUVE
In many of the industrial applications, an electric motor is the most important component. . A complete production unit consists primarily of three basic components; an electric motor, an energy-transmitting device and the working (or driven) machine. An electric motor is the source of motive power. An energy transmitting device delivers power from electric motor to the driven machine (or the load); it usually consists of shaft, belt, chain, rope etc. A working machine is the driven machine that performs the required production process. Examples of working machines are lathes, centrifugal pumps, drilling machines, lifts, conveyer belts, food-mixers etc. An electric motor t ogether with its control equipment and energy-transmitting device forms an electric drive (10). An electric drive together with its wor king mac . e constitutes an electric-drive system (1 . A ceiling-fan motor with it s speed regulato but withou t blades is an example of electric drive. Other examples of eledric drives are : a food-mixer without food to be processed, a motor and conveyer-belt without any material on its belt. Some examples of electric-drive systems are : a ceiling-fan mot or with r egulator and also with blad es, a fo od-mixer with foo d to be process ed, a motor and convp.yer-b elt wit h material on its belt and so on. Fig. 12.1 shows an electric drive syst em. The electric drive, consisting o electric mot or, its power controller and ener gy-transmitting shaft is also indicated in Fig. 12.1. A m odern electric . drive system using a fee dback loop is illustrated in Fig. 12.2. In this chapter, el ectric drives controlled wough power-electronic converters are only des,cribed.
584
Power Electronics
[Art. 12.2]
-, Main Power Source
I I ' I I
Power Controller
Motor
I
Fill
I I I
I
Working Machine
L I ______________________________ ~I
-Fig. 12.1. An electric-drive system. Main Power
Source
Command
)----~
Power Electronic Converter
L.--..,.-~---4
Motor
Wor king Machine
Rotor position or
speed sensor 1 - -......- _......
Fig. 12.2. Block diagram for a modern .electric drive system using power electronic converter.
Electric drives are mainly of two types : dc drives and ac drives. The two types differ from each other in that the motive power in de and ac drives is provided by dc motors and ac motors respectively. 12.2. DC DRIVES DC motors are used extensively in adjustable-speed drives and position control Their speeds below base speed can be controlled by armature-voltage control. Speeds above base speed are obtained by field-flux control. As speed control methods for dc motors are simpler and less expensive than those for ac motors, dc motors are preferred where wide-speed control range is required. applicatio~s.
Phase-controlled converters provide an adjustable dc output voltage from a fixed ac input voltage. DC choppers also provide dc output voltage from a fixed dc input voltage. The use of phase-controlled rectifiers and dc choppers for the speed control of dc' motors have revolutionized the modernindustria1 controlled systems. The dc motors used in conjunction with power-electronic converters are de separately excited motors or de series motors. These motors \. ]1, therefore, be studied her e. Depending upon the type of ac sourc~_ or the method of voltage control, dc drives are classified a;5 under: 1. Single-phase dc drives 2. Three-phase de drives 3. Chopper drives. First the basic operating characteristics of dc m otor aTe presented and t hen three speed control strategies as mentioned above are described.
J
[Art. 12.2]
Electric Drives
585
12.2.1. Basic Performance Equat ions of DC Motors Equivalent circuit and basic performance equations for a separately-excited dc motor and a dc series motor are presented in what follows . (a) Separately-excited dc motor. The equivalent circuit for a separately-excited dc motor coupled with a load is shown in Fig. 12.3 (a) under steady-state conditions. The load torque TL opposes the electromagnetic torque Te. For field circuit, Vf=If · rf ...(12.1) For armature circuit, V t =Ea + Ia r a Motor back e.m.f. or motor armature e.m.f., Ea = Ka ~ rom = Km rom Te =Ka ~ la =Kmla Also, Te =D rom + TL where V t =motor terminal voltage, V la = armature current, A
=field flux per pole, Wb Km =Ka ell =torque constant, NmlA or, emf constant, V-sec/rad r a = armature circuit resistance, n ~
rom = angular speed of motor, rad/sec
rf =field circuit'r esistance, n D =viscous friction constant, Nm-seclrad.
Electromagnetic power, P = rom . Te watts
From Eq. (12.1), Ea =Km rom,=V t - lara
V t - lara V t - lara or rom = Km = Ka ~
... (12.2)
It is seen from Eq. (12 .2) that speed can be controlled by varying (i) armature terminal voltage V t , known as the armature-voltage control and (ii) the field flux <», known as the field-flux control. Base speed is defined as the speed at which motor runs under rated armature voltage, rated field current and rated armature current. Speeds below base speed are obtained by armature-voltage control. During this con trol, armature current and field flux (or field current) are k ept constant so as to meet the torque demand. So the armature voltage control meth od is also termed as constant-torque drive method because motor torque Te =Ka <» Ia remains almost constant.
+
+
T
-
o---------~
-(~'(' l lfvf
1
+
Eol
(a) (b) Fig. 12.3. Equivalent circuit of a (0) separately-excited de motor and (6) de series m otor.
[Art. 12.2]
586
Power Electronics
Speeds above base speed are obtained by varying the field current or field flux and by keeping V t and Ia constant at their rated values. As flux decreases, speed increases so that motor e.m.f. Ea remains almost constant, Consequently, field-flux control method is also called . constant-power drive method as power P =Ea Ia remains substantially const ant. The variations of '{e' P, la' If' and Vt against speed are shown in Fig. 12.4 for a separately-excited dc motor_ I
II I·
i i
Speed
!
~
Armature voltage ,Vt .
·~-·--A;"~;-~;:;~~ent,Ia
Ia
:~:-':j~:.~~:-.:::+::.- . . . ~ . . . . . ~Field flux,;
cP
If
.A-
I
. •/. '-vvJ t
..............................C
'I
;/ .
--...
'I
'
.
...
Field current,It
~
... ...
... ....
O~~----------------~--------------~~-----------Speed 8ase I speed
Leon stant torque drive
__
• I' Constant power--t drive
Fig. 12.4. Characteristics of a separately-excited de motor. (b ) DC series motor. For a series motor, field winding in series with the armature circuit . . . is designed to carry the rated armature current. Fig. 12.3 (b) gives the equivalent circuit of a dc series motor driving a load torquef~.
Forthe armature circuit in Fig. 12.3 (b),
V t = Ea + la (ra + rJ . T, =K"u cl> l a For no saturation in the magnetic circuit, = CIa
.. Also Fr om E q. (12.3),
OT
speed,
T, =KaC1a2 =k1a2
Ea =Ka cI> rom =Ka C,Ia rom =k 10, rom
V t = k la rom + la (ra + r.)
Vt =10, [k rom + (r0,+ r,)]
Vt -Ia (ra+ rs)
(t,)m
=
kI
a
V,
=k la where
... (12.3)
(ra + r,,) k
rs =serie&-field resist ance, 0
,~ = Ka,C =a constantin Nml A2 or in V-siA. Tad,
.... (1 2 04)
587
[A r t. 12.3]
Electr ic Drives Te,P
Pow er, P
I /~
~---------------/.
.
I
i
i
i i ./
Bose speed .
Speed
ii 10
Armature terminal voltage)Vt
~.-.-.-.-.- . -
. I
Armature cu rrent)I a
~ Constant
torque drive
..I.
Speed
Constant power drive ·
Fig. 12.5. Characteristics of a dc series motor.
For speed control up to base speed, armature terminal voltage V t is varied with fa k ept . constant. Therefore, P (= V t fa) varies linearly and torque Te = k f~ remains constant. For speeds above base speed, series field flux is decreased by the use of diverter or tapped-field control and Ia is kept constant. Therefore, torque Te = Ka <1> Ia decreases but power P = Ea fa remains substantially constant. Speed control of de motors, when fed through single-phase or three-phase converters, is now studied in what follows. 12.3. SINGLE·PHASE DC DRIVES Fig . 12.6 illustrates th e general circuit a rrangement · for the sp ee d control of a separ ately-excited dc motor from a single-phase s our ce. The firing angle control of converter 1 regulates the armature voltage applied to dc motor armature. Thus, the variation of delay angle a l of converter 1 gives speed control below base speed. The variation of the firing angle u2 of converter 2 installed in the field circuit gives speeds above base speed. At .low values of a l for converter 1, armature current may become discontinuous. The discontinuous armature current causes (i) more 10sse in the armature and (ii) poor speed regulation: It is usual to insert an inductor L in series with the armature circuit to reduce the ripple in the armature current and to make the armature current continuous for low values of motor speeds . Depending upon th e type of power-electronic
L
Armc tu r e Field
1 ph o se
SO l.J i Ce "
Fig. 12 .6 . Ge eral ci r cuit anan ge m ent for s in gl e-phase de drives .
588
Power Electronics
[Art. 12.3]
cpnverter used in the armature circ1J,it, single-phase d~ drives may be subdivided as under: 1. Single-phase half-wave converter drives 2. Single-phase semiconverter drives 3. Single-phase fhll-converter drives 4. Single-phase dual converter drives. The converter drives listed above are now briefly discussed. In all these types, it is assumed that armature current Ia is constant. . ..
12.3.1. Single-phase Half-wave Converter Drives A separately-excited dc motor, fed through single-phase half-wave converter, is shown iii Fig. 12.7 (a). Motor field circuit is fed through a single-phase semiconverter in order to reduce the ripple content in the field circuit. Single-phase half-wave converter feeding a dc motor offers one-quadrant drive, Fig. 12.7 (b). The waveforms for source voltage us, armature terminal voltage ut • armature current i a• source current is and freewheeling diode current ifd are sketched in Fig. 12.7 (c). Note that thyristor current iT = is: The armature current is assumed ripple free . Such t~e~ of drives are used up to about i kW dc motors. T
('\J
FD
Us
=V m sin w t
D2
D1
DCI
wt
•
wI
wt
o ,
I d~ ~
H
o
'
(b ) (c) Fig, 12.7 . Single-phase half-wave converWT drive (0) C1Tcuit diagram ( b ) quadrant diagram and (c ) waveforms .
!
wI
I • 4rr w I
J
.[Art. 12.3]
Electric D rives
589
For single-phase half-wave converter, average output voltage of converter, Vo == armature terminal voltage, V t is given by Eq. (6.1) as Vo
=V t = Vm 21t (1 + cos a) ,
for 0 < a < 1t
... (12.5
)
where V m = maximum value of source voltage. For single-phase semiconverter in the field circuit, the average output voltage is given by . Eq. (6.28) as
Vm
V = - 1t (1 + cos al) . f
for 0 < a l < 1t
... (12.6)
It is seen from the waveforms of Fig. 12.7 (c) that rms value of armature current, Iar= Ia rms value of source or thyristor current, Isr=
1t-a'J1I2 ~ =Ia (21t ~ Ia21t-a
... (12.7)
rms value of freewheeling-diode current, I fdr
=.. . II 21t + ex =I "a
21t
(1t + exJl/2 21t
.. .(12.8)
a .
Apparent input power = (rms source voltage) (mis source current)
=Vs' Isr
Power delivered to motor =E~a + Ia 2 . r a =(Ea + Ia ra) Ia =V t . la
2
EJ a +Ia ra V t · Ia
... (12.9) =-:-=--:-- Vs . Isr VB . Isr E xample 12.1. A separately-excited dc motor is supplied from 230 V, 50 Hz source through ' a single·phase half-wave con trolled converter. Its field is fed through i-phase semiconverter with zero degree firi ng-angle delay. M otor resistance ra = 0.7.0. and motor constant = 0.5 V-sec / rad. For rated load torque of 15 N mat 1000 rpm and for continuous ripple free currents, determine (a) firing-angle delay of the armature converter (b) rms value of thyristor and freewheeling diode currents (c) input power factor of the armature converter.
Solution . (a) Motor constant =0.5 V-sec/rad =0.5 NmJA =Km
But m otor torque, Te = Km 1a .
Input supply
pf =
(.
:. Arm ture current Motor emf,
= 15 = 30 A 0.5
E o =Km . (Om
=0.5 X 21t x601000 = 52.36 V
For 1-phase half-wav e converter feeding a de motor,
Vm
V t = 21t (1 + cos a) = Eo + Ia ra
Vt =
or ••
CL
,(2 X 230 . ' 21t (1 + cos a) = 52.3·6 + 30 x 0,7 = 73.36 V
=
X 21t -1] = 6533'6 cos- 1 [73.36 '12 x 230 .
Thus, firing-angle delay of converter 1 is 65.336°
0
590
Power Electronics
[Art. 12.3] (b ) Rms
value of thyristor current, from Eq. (12.7), is I
Tr
=I a (l 1t 21t - a ]U2 =30 (180 - 65.336 JU2 = 16 931 A =I I 360 . sr
Rms value of free Ii'lheeling-diode current, from Eq. (12.8), is
= I ( 1t +
I fd r
(c)
aJ112 = 30 (180 +360 65. 336J1/ 2 = 24 766 A .
21t
a
From Eq. (12 .9), input power factor of armature converter V t ' Ia
73.36 x 30
.
=Vs . l SI" = 230 x 16.931 = 0.5651 lag. 12.3.2. Single-phase Semiconverter Drives A separately-excited de motor, fed through two single:.phase semiconverters, one for the armature circuit and the other for the field circuit, is shown in Fig. 12.8 (a). Both converters 1 a~ 2 are connected to the Same single-phase source. This converter also offers one-quadrant + T11
T12
T22
T21
a
FO
'---+----. b 0 12
0 11
\,fd
D21
D22
(a )
I
I
: Tl1011 !
I
I
I
ia
'I.
wI
•
Ii
~FO: T120lt , I I I
'
wt
wI tTl
WI
(b) F ·g. 12.8. Single-ph ase semiConverter drive (c circuit diagr
and
(b )
waveforms .
[Ar t. 12.3]
Electric Drives
·591
drive and is used up to about 15 kW de drives. The waveforms for currents and voltages are sketched in Fig. 12.8 (b) on the assumption of ripple free armature current. Load voltage . waveform for Vo =VI is the same as shown in Fig. 6.11 (b). For a single-phase semiconverter, average output voltage, from Eq. (6.28), is given by V . Vo = VI = ~ (1 + cos a) ... (12.10' a) 1t
Vm
. Vf = 1t (1 + cos a1)
For field circuit,
.. .(12.10 b)
It is seen from the waveforms in Fig. 12.8 (b) that rros value of source current,
I" = I. [ "
~"
f'
... (12.11)
.
1/2
rmsvalue of freewheeling-diode current, I fdr =Ia [ rros value of thyristor current, IT. = I. ( " ;"" VI·la
Input pf
r'
i]
.. .(12.12)
..
'
... (12.13) .,"
,", .
-,
A single-phase semiconverter is also called single-phase half-controlled bridge converter.
E a m p le 12.2. A separately-excited dc motor, operating {r'o m a single-phase half-controlled bridge at a speed of 1400 rpm, has an input voltage of 330 sin 314t and a back emf 80 V The SCRs are fired symmetrically at 0: = 30° in every half cycle and the armature has a resistance of 4 n. Calculate the average armature current and the motor torque. Solution. For a single-phase semiconverter feeding a separately-excited motor, Vo
Vm
=V t =-11:
(1 + cos a) = Ea + Ia ra
330 - (1 + cos 30°) ·= 80 + I a' 4 1t
196.01 = 80 + Ia . 4 :. Averag e ar ma ure curren t, la = 1 96 .0~ - 80 = 29.003 A
E =K
Motor e~f,
a
m rom
=K
m
21t
x 1400 60
_ · 80 x 60 Km = 211: X 1400 = 0.546 V-slrad or 0 ..,46 Nm l A.
or :. Motor torque,
Te = Km Ia = 0.546 x 29 .003 = 15.836 Nm.
Example 12.3. Th2. speed of a .15 hp, 220 V, 1000 rp m de series m otor is controlled using a i-phase ha lf-con trolled bridge converter. The combined arm ature (m d field resistance is 0.2 . n. Assuming continuous and ripple free motor current and speed c 1000 rpm aRd k = 0.03 ~m/amp2, determine (a ) motor current (b) motor torque for a fir ing on ale a = 30°. AC voltage
i8 250 V Derive
any formu la used.
(lA .S:, 11)91 )
592
[Art. 12.3]
Power Electronics
Solution.
Refer to Fig. 12.3 (b) for a dc series mot or and Fig. 12.8 for a single-phase semiconverter.
From Eq. (12.3) for a dc series motor,
V t =Ea +Ia (ra + r,)
Motor torque, Te = Ka Ia' For no saturation, = CIa
.. Te=KaCIa2=kIa2 where k is a constant in Nm/amp2. Ea =Ka rom =Ka CIa rom =k Ia rom Also ' Constant k in the expressions for Te and Ea is the same. (a) From above, V t =Vo:: Ea + Ia (ra + r,) . V
or V t =Vo =~ (1 + cos (X) =Ea + Ia. (ra + rs) =kIa rom + Ia (ra + r,)
7t -v2: 250 (1 + cos
300)'~ 0.03 Ia x 2it X6~000 + 0.2 Ia
209.97 = 3.3416 Ia
. . 209.97 ,
: . Motor armature current, Ia = 3.3416 = 62.84 A (b) Motor torque,
Te =k Ia 2 = 0.03 (62.84)2 = 118.466Nm.
12.3.3. Single-phase Full Converter Drives Two full converters, one feeding the armature circuit and other feeding the field circuit of a separately-excited dc motor, are shown in Fig. 12.9 (a). This scheme offers two-quadrant drive, Fig. 12.9 (b) and its use is limited to about 15 kW. For Tegenerative braking ofthe motor, the power must flow from motor to the ac source and this is feasible only if motor counter emf is revers.ed because then eaia would be negative. Note that direction of current cannot be reversed; as SCRs are unidirectional devices. So, for regener ative breaking, the polarity of ea must be reversed which is possible by reversing the direction of motor field current by making
, delay angle of full converter 2 more than 90° . In order that current in field winding can be .
reversed, the field winding must be energised through single-phase full converter as in Fig.
12.9 (a) . , 2V ' ;..(12.14 a) For the arm~ture converter 1, Vo = V t = 7tm cos (X for 0 < (X < 7t For the field converter 2,
2Vm Vf = - cos (Xl 1t
...(12.14 b)
for 0 < 0.1 < 7t
From the waveforms in Fig. 12.9 (c), it is seen 'that
WT
7t 1a2 . -1t = 1a
rms value of source current,
I 8r =
rms value of thyristor current,
1 = 1 2. ~ Tr
[a
I - ]112 =~
...(12,15)
'12
27t
. Vt .10 2Vm I a · -.f2 From Eq. (12.9)) input supply pf = V ,I = cos ct . V . I
3
sr
1t
m
a
2-v2 = - - cos a It is seen from Eq. (12.16) that input pf dep ends on the fIring angle a8sU:ffiptions of const ant arinature current.
" .(12.16) CJ.
only under th e
Electric Drives
[A rt. 12.3]
593
+ TIl
T13
T23
T21
T22
T24
.-----....... a
L..---4_-..... b
T12
(XI
(a)
wt
wt
wI
Tr
21'(
371
wt
..
(b)
(c) Fig. 12.9. Single-phas e full converter drive (a) circuit diagram (b) two-quadra nt diagram and (c) wavefonns.
Example 12.4. A separately-excited dc motor dri ves a rated load torque of 85 Nm at 1200 rpm. The field circuit resista nce is 2 00 n and armature circuit resistance is 0.2 n. The field wind ing, connected to i -phase, 400 V source, is fed thro ugh 1-phase full converter with zero degree fi ring ang le. The armature circuit is also fed through another fu ll converter from th'! same i-phase, 400 V source. With mag net ic saturation neglected, the motor constant is 0.8 V-sec /A-rad. For ripple free armature and field currents, determi ne (a) rated armature cu rrent (b) firing-angle delay of armatu re converte r at rated lo ad (c) spe ed regu lation at full load (d) inp u t pf of the armat ure con verte r and t he d rive at rated load. Solution. (a) For field converter, firL. g-an gle delay = 0° :.
F I· e Id. v oIt age,
V
f
2V =~ =2 1t
2 x 400 7t
= 360V
594
[Art. 12.3]
Field current,
Power Electronics
V 360 I =.:J. = = 1.8 A f rf 200
With magnetic saturation neglected, $ =K1 If Ea = Ka 4> rom == Ka Kl If' rom = KIf' rom'
where K has the units of V-sec/A-rad.
Te = Ka 4> Ia = Ka Kl If . Ia = KIf ' ~a 85 = 0.8 x 1.8Ia 85 Rated armature current, Ia == 0.8 x 1.8 = 59.03 A
Similarly,
(b)
2Vm VI = Vo = - - cos a =Ea + Iaro =K If' rom + lara
Here
1t
2~ 1tx 400 cos · a ~ 0 .8 x 1.8 x 21t X601200 + 59 .03 x 0 .2
or
= 180.96 + 11.81 == 192.77 V a = 57.63
or
0
(c) At
the same firing angle of 57.63'\ motor emf at no load, Eo = VI = Vo = 192.77 V = K If rom 0
:. No load speed, or
. Ea .. 192.77
W mO
=KIf = 0.8 x 1.8 =
13387 · d/ . ra sec
N = 1278.35 rmp Speed regulation at full load _ No load speed-full-Ioad speed full-load speed
=(1278·;:0~ 1200)x 100 = 6.53%. (d)
Input pf of the armature converter == VI' Ia
Vs . Iar
= 192.77 x 59 .03 = 0 48191 400 x 59.03
.
ago
Also, from Eq. (12. 16), input pf of the armature convert er 2~ 2~ =- cos a == - cos 57.63 1t 1t
0
== 0.48191a g
Rms value of current in armature converter, Iar =10 = 59.03 A
Rms value of current in fiel d circuit, 1fT = I f = 1.8 A
Total
rID S
current t aken from the source, 1ST
I pu t
=~laT2 + I fr2 == '-159. 032 + 1.82 = 59.06 A
VA = Vs ' Isr
= 400 x 59 .06
Electric Drives
(Art. 12.31
595
With no loss in the converters, total power input to motor and field
=Vt . 10. + V f · If = 192 .77 x 59.03 + 360 x 1.8 = 12027.2 watts Power input in W 12027.2 0 50911 = Input in VA =400x59.06=· ago
Input pf of the drive
Example 12.5. In Examp(e 12.4, the polarity of the counter emf is reversed by reversing the field excitation to its maximum value. Calculate (a) delay angle of the field converter (b) delay angle of the armature converter at 1200 rpm to maintain the armature current constant at 50 A and (c) the power fed back to the supply during regenerative braking of the motor. Solution. (a) The field voltage is reversed to its miuimumvalue of 360 V. 2Vm .. Vf = ~ cos (Xl = - 360 V 0 or (Xl = 180
(b) With field current reversed, motor emf Eo. is also reversed. Vo = V t = - Eo. + 10. ra 2V . ~ cos (X = - 180.96 +50 x 0.1 = -175.96 V . 1t or
(X
=
cos
175.96 x 1t] =-11925"0 2T2 x 400 . '*
-1 [ -
(c) Power fed back to the ac supply
=Vt . 10. = 175.96 x 50 = 8798 watts Example 12.6. A 220 V, 1500 rpm, 10 A separately-excited dc motor has an armature resistance of 1 ohm. It is fe d from a single-phase fully-controlled bridge rectifier with an ac source voltage of 230 V, 50 Hz . Assuming continuous load current, compute (a) motor speed at the fir ing angle of 30° and torque of 5 Nm (b) developed torque at the firing angle of 45° and speed of 1000 rpm. (GATE, 1996 ) Solution. Under rated operating conditions of the separately-excited dc m otor, Vt=E a + Iara =Km rom +10. ' ra 21t x 1500 or 60 + 10 xl = 50· 1t . Km + 10 220 = Km .
Km =
:. Motor const ant, (a) For
220 -10 . 501t = 1. 337 V-s /rad or 1 .337 N mlA.
a torque of 5 Nm, m otor armature curren t,
5
10. = 1.337 = 3.74 A
The equ ation givin g t he operation of converter-motor is
Vo =V t =Eo. + 10. ra
2Vm
- - cos a =Km · (j)m +10. ra
1t
2..[2 X 230
- - - - cos 30 0 = 1.337 rom + 3.74 X 1 1t
or
rom =
21t · N
~
179.3 - 3.74 1.337 .
= 131 .31 ra d/s ec
= 131.31 r ad/sec
~
596
[Ar t. 12.3]
Power Electronics
Motor speed = (b )
131.31 x 60 27t
= 1253.92 rpm
For ex = 45°,
450 = 1 337 27t x 1000 I 2 {~f x 230 7t cos . x 60 + a 146.4 = 140.01 + Ia
X
X
1
1
Ia = 6.~9
or Motor developed torque,
=6.39 A Te =Km Ia =1.337 x 6.39 = 8.543 Nm
Example 12.7. A 220V, 1000 rpm, 60A separately-excited dc motor has an armature resistance of0.1 n. It is fed from ·a single.phase full converter with an ac source voltage of230V, 50Hz. Assuming continuous conduction, compute (a) firing angle for rated motor torque at 600 rpm (b) firing angle for ra~ed motor torque at (-500) rpm (c) motor speed for
C(
= 150° and halfrated·torque. .
Solution. Under rated operating conditions of the motor,
V t =Ea +Ia ra =Km wm +Ia ra 2:0 = Km 27t
or
=[2!:0 1-0~O] x 60 = 2.044 y-s/rad or 2.044 Nm / A.
K" (a)
For rated motor torque, armature current = 60 A Vo 2..f2" ~ 230 cos
or
C(
= V t =Km wm + Ia r a = 2.044 21t ~0600 + 60 x 0.1 = 134.43 V _
C( -
(b)
x6~000 + 60 x 0.1
cos
-1
[134.43 x 7t] _ 49 -120 2T2 x 230 - .D
At (-500) rpm, 2..f2" x 230 7t cos
= 2 044 21t (-60500) + 60 x. 01
c(.
= - 107.024 _ C( -
(c )
~
,",os
- 1
+ 6 =- 101.024 V
rl 2101.024 '12 x 230x 1t] -_ 119 .27AO '±
At h alf-rated t orque, mo tor armature current
= 1. x rated current = 1. x 60 = 30 A 2
..
2
2"2 x 23 0 cos (1 50°) =2 .04 4 x wm + 30 x 0.1 1t
- 17 9. 30 = 2 .044 wm + 3
- 182.3°
com= 2. 044 :. Spe ed,
=-
89.188 rad/sec
-= _ 89. 188 x 60 = _ 85 27t
.
683
rpm.
(Ar t. 12.4J
Electric Drives
597
.i
12.3.4. Single-phase Dual Converter Drives A single-phase dual converter, obtained by connecting two full-conv e.'ters in anti-parallel, is shown fe eding a separately-excited dc motor in Fig. 12.1() (a ). Its use is limited to about 15 kW dc drives . It offers four-quadrant operation, Fig. 12.10 (b ). For working in first and fou r th quadrants, converter 1 is in operation. For operation in second and third quadrants, converter 2 is energised. Four-quadrant operation demands that fi eld winding of the motor is energis ed from a single-phase, or three-phase, full converter. . 2V For c onverter 1 in operation, V t = ~ cos al for 0 ~ a 1 ~ 1t 1t 2Vm For converter 2 in operation, V t = - cos a 2 for 0 ~ ~ ~ 1t 1t where U 1 + ~ = 1t 2Vm Fur field converter, V , =--cOSU3 forO~a3~1t I 1t
v!
I
Re verse Re g . bra Kin g Con v. 2
Forward motori ng Conv. 1
-10
Ia
T21 ~~:"':"':".;",;j Co nv 1
Revers e motoring
Forward Reg . brak ing
-Vt Us
.
(b )
(a )
Fig. 12.10. (a) Single-phase dual converter feeding a separately-excited de motor (b ) four-quadrant diagram. .
Note t hat in Fig. 12.10, (i) Converter 1 wit h Ul < 90° operates the motor in forward motoring mode in quadrant l.
(ii) Converter 1 with u 1 > 90° and wit h field excitation rever sed oper ates t he motor in
forward regen erative braking mode 'in quadr ant 4. (iii ) Convert er 2 with a 2 < 90° operates the motor in reverse motoring mode in quadrant 3. (iu) Converter 2 with u 2 > 90° and with field excitation reversed operat es t he mqtor in
reverse regen erative braking mode in quadrant 2. ' E DC DIDVES I ,» ... . THHED!jDIIAS ~~ ~ ........
I
'
r; • '. •
-
'.
•
. . ' ...~ .~.
•
:~ ~
Lar ge d c motor drives ar e always fed t ough three-phase converters for th eir speed control. A t.1rree-ph 8 controlled converter fe eds power t o the armat ur e circuit for obtaining speeds belO'N base speed. Anoth er three-ph ase controlled converter is insert ed in th e fi eld circu 't for gettin g speeds above base speed.
598
[Art. 12.4]
Power Electronics
The output frequency of t hre e-phase converters is higher than those of single-phase converters. Th erefore, for reducing the armature current ripple, the inductance required in a three-phase dc drive is of lower value than that in a single-phase dc drive. As the armature current is mostly continuous, the motor performance in 3-phase dc drives is superior to those in single-phase dc drives. The three-phase dc drives, as in single-phase dc drives " may be subdivided as under: 1. Three-phase h alf-wave converter drives 2. Three-phase semiconverter drives
3. Three-phase full-converter drives 4. Three-phase dual-converter drives These converter controlled dc drives are now described one after the other. Armature current is assumed ripple free for convenience.
12.4.1. Three-phase h a lf-wave converter drives; Fig. 12.11 (a) illustr ates a 3-phase half-wave converter drive consisting of two converters and a separately-excited dcmotor. The armature circuit of the motor is fed through a 3-phase half wave converter whereas its field is energised through a3-phase se.miconverter. This converter offers one-qu adrant operation Fig. 12.11 (b) and may be used up to about 40 kW motor ratings. Two-quadrant operation can also be obtained from three-ph ase half-wave converter drive in case motor field winding is energised from single-phase or three-phase full converter. iT1
A T1
La
13
B
A
+
T2
FO
C
8 C
Vo= lJ I
T3
01
02
03
N
(a) VO'V I
,
"0
VI
:-- Tl I
..... I
I I I
I
T2
I IL l" I I
I
-
I
I I
I I
T3
"1I 4
I I
wt
I
Tl--{ I
10 wt
10
iAh1
ooo_---- 21T.- - ---J
(b )
..
(c) Fig. 12.11 . Three-phase hal f-wave converte drive (a ) ci cuit diagram (b) -quadrant diagram (c) wave orms .
wt
-
Elec ric Drives
[Art. 12.4]
599
.1
For a 3-phase half-wave converter, average value of output voltage or armature terminal voltage, from Eq. (6.50), is ...(12.17)
where V ml = maximum value of line voltage and ex is the firing angle for converter 1. The voltage expression of Eq. (12.17) is valid only for continuous armature current. For three phase semiconverter, the average value of field voltage, from Eq. (6.55), is given by .. .(12.18)
A three-phase half-wave converter drive is not normally used in industrial applications as it introduces dc component in the ac supply line. It isseen from the waveforms of Fig. 12.n (c) that rms value of armature current, Iar =Ia rms value of phase or line current, _ f1 - - -1 2 2n ...(12 .19) lsr = \J Ia 3"' 2n = l a -'I '3
_,r-
. ... .. - ,.
average thyristor current,
...(12.20)
rms thyristor current,
... (12.19)
12.4.2. Three-ph ase Sem iconver ter Drives The circuit diagram for a 3-phase semiconverter feeding a separately-excited dc motor is shown in Fig. 12.12. The field winding of the motor is also connected to three-phase semi converter. This drive offers one quadrant operation and is used up to about 115.kWratings . . On the assumption of continuous and ripple free armature current, wavefor ms for line current iA and thyristor current iTl are sketched in Fig. 12.13 for firing angle ex =30° and also for ex =90°. An examination of these waveforms would reveal that (i) for firing angle ex ~ 60°, each thyristor co?ducts for 120° and (ii) for 60° < a < 180°, each thyristor conducts for (180° - ex). Note that freewheeling diode comes into conduction only when firing angle of 3-phase semi converter is mor e than 60° i.e. when ex> 60°. The waveforms of Fig. 12.13 reveal that freewheeling diode conducts for (a - 60°) in case load, or armature, current is continuous. Also, conduction angle of thyristor + conduction angle offreewheeHng diode = 120°, when armature current is continuous . As armature current is ripple fr ee, rms value of armature current, lar =la. It is also seen from Fig. 12.13 a unde : For a 0;;,.60°, rIDS value of supply line current, iA is given by,
I ="\ !I2 2n .l fi r 'f a 3 n
=fa ~
... (12 .21)
and r ms value of th yrist or curr ent iTl is given by
-fT 21t 2
~]1I2
ITr- L~ a 3 '2n
_ - il
-la 'J 'S
.. .(12 .22)
600
[Ar t. 12.4J
Power Electronics
to
T12
T13
T23
T22
T21
LA
A<'>-_-'0
t-----oA
B <.>----+---.... b
t----+---
FO
C o----+----~--~~c 012
011
~---r--~-----oC
013
023
022
021
~ I
~
~~1
Fig. 12.12. Three-phase semiconverters feeding a separately-excited dc motor.
wt
I T3L
°0
io
I
Tl
T2 03
I
"
T1
13
I
01
1
I
02
13'
T2
03
I
o.
01
I
02
f 10
(Y)
d
wt
tA
..
wt
tTl
r---
Each SCR conducts for 120 for Ct< 60°
t-- 120 °-.,
120°-.,
wt
to~tI~------------~~-----------------tfo wt
I I
o
o
en
"
d
~1 ~~~9O0~~~~.-~~
=90"..f'""'TTI I-- 90 LA rl.__ I ____ .J l.Lc L.' ;.. C(
0
::
LLlliJ 2;r
:
I- 90o ~
I I
-o 8
~I ~~ ~ .1 j-' 90° -1
______
__
____, ________
~-p... wt
- - - - - - --""1
Each SC R condu ct s tor ( SO- cd fo r Ct >6 0"'
Ia )... r<>--------- 271 - -
-
-
- -
-...;
Fi g. 12.13 . Voltage and curren t waveforms for a
three-pha se s emi con ve rt er drive of Fig . 12.12 .
.
wt
FOT
601
[Art. 12.4]
Ele r ic Drives
60 0 < a <: 180 0 , Tms value of supply line current iA is given by
Isr = [ Ia
2
( 180 - a 112 180
]~
r
=Ia
~ 180 - Ct. 180
. ...
and rros value of thyristor current iTl is given by
IT, =I.
(18~6~
... (l2~23)
(12.24)
Q(
From above, it is obvious that average thyristor current is ~ I. for a < 60· and ( 18~6~ Q( I.
J
for 60 0 < a < 180 0 • For 60 0 < a < 1800 , freewheeling diode has rms value of Ia of Ia;
~ a 1~~0
and average value
a- 60
120'
3V
.
ml =V t = ~ (1 + cos a) for 0 < a < It
For converter 1,
Vo
For converter 2,
3Vml Vf =~ (1 + cos ( 1) for 0 < <11 < It
...(12.25 a) ...(12.25 b)
E x ample 12.8. The speed of a separately-excited de motor i~ c-ontrolled by means of a 3-phase semieonverter from a 3-p hase, 415V, 50 Hz supply. The motor constants are: inductance 10 mH, resistance 0.9 ohm and armature con stant 1.5 Vlradls (NmIA). Calculate the speed of this motor at a torque of 50 Nm when the converter is fired at 45 0 • Neglect losses in the converter.
Solution. Armature constant, Km
= 1.S V/radls or 1.S NrnJA.
Motor torqae, Te = Km Ia = SONm
SO 100
:. Motor armature current, Ia = 1.S = 3 A
The equation for the converter-motor combination is
3Vml (1 + cos a) = Ea + lara = Km ffim + Ia ra
rn-
~.
-
3 ~ X 41S a 100
21t (1+coS4S)=1.5x(Om+sxO. 9 478.3 = 1.5 (Om+ 30 478.3 - 30 COm = 1.S = 298 .867 r ad/s
or or
2 {;' = CO m- = 298 .867 r ad/s 6
:. Motor speed,
N = 298.8;: x 60 = 2853.97 r pm
Example 12.9. A 600V, 1500 rpm) BOA sep arately·excited dc motor is fe d through " th ree-phase sem icon. verter from 3-phase 4 00V supply. Motor armature resi tance is 1 Q. Armature current is assumed c nstant. (a) For a fi ring angle of 45 0 at 12 00 rp m, compu e the rms values of source and thy ristor currents, average v..zlue of t hyri tor cu rren t and the input sup ly power factor. (b) Repeat part 'a ) for a firing ang le of 90 0 at 700 rp m.
602
[Ari.
1 2.4~
Power Electronics
Solu tion. Under rated operating conditions,
,
Vt = Ea + Ia r a = Km Wm +Lara . 600 = Km 21t x6~500 + 80 xl 520 x 60 Km::: 21t x 1500= 3.31 V-s/rad (or Nm/A).
or Motor constant (a)
For the converter-motor combination, 3Vml V t = 21t (1 + cos a) =Ea +Ia ra = K,;,. wm + Ia ra
x
3 ..f2 400 (1 .450) = 3 31 21t x 1200 I . 21t + cos . x . 60 + a
X
1
461.01 = 415.95 + Ia
:. Armature current,
Ia
= 45.06 A
From Eq. (12.21), rms value of source current,
-" '; ..•.
Isr = Ia ..jf= 45.06 ..jf= 36.791 A
From Eq; (12.22), rmsvalue of thyristor current,
1 - .;.. l ' .
= Ie; T3 = 45.06 T3 = 26.015 A
Average value ofthyristor CU~-tent 1 = 3' x 45.06 = 15.02 A
Input supply power fact&
(b) V t = 3
V . Ia
461.01 x 45.06
=ra. tVa .Iar =ra x 400 x 36.791 =0.815 lag
~: 400 (1 + cos 90°) = 3.31 X 21t ~~OO + Ia X 1
270.05 = 242.64 + Ia
:. Armature current, Ia = 27.41 A
Rms value of source current,
15r = 27.41 ' ..jf= 22.38 A
Rms value of thyrlatorcurrent, 17r= 27.41-k = 15.825 A
Average value of thyristor current
= 27341 = 9. 137 A
Input supply power. f.actor
= 270.05 x 27.41 = 0 4774 1
x 400 x 22. 38 ' ag,.
ra
12.4.3. Three-phase F ull-converter Drives The circuit diagram, consisting of one three-phase ful converter in the armature circuit and another 3-phase (or I -phas e) full converter in the field circuit, is as shown in Fig. 12.14. It offers two-quadrant drive and is used up to about 1500 kW drives. For regenerative purposes , th e polarity of coun ter emf is reversed by revers 'ng the field excitation by makin;:: the firing-angle delay of converter 2 more than 90° .
j
Electric D ives
{Art. 12.4]
6D]
to + T13
T1 5
tA A c::r-...-.... a
T25
T23 A
V o= 1.)\
B 0---+---'" b
B
C o---+--~--... c
C vf
T22
l)CC 2,0(, Fig. 12.14. Three-phase full converters feeding a separately-excited de motor.
For converter 1 in the annature circuit, the average output voltage, from Eq. (6.54), is given by 3Vml Vo = VI =- - cos ex for 0 ~ a ~ 7t ... (12.26 a) 7t
For converter 2 in the field circuit, 3Vml Vf =- - cos a l for 0 1t
~
ex l
~
...(12.26 b)
1t
where V ml = maximum value of line voltage. Voltage and current waveforms for a = 30 0 and for constant armature current are sketched in Fig. 12.15. It is seen from this figure that each thyristor conducts for 120 0 for continuous arm ature current. This gives rms value of armature current 'ar I = I a· . rms value of source current, from iA waveform,
~r =
o
I
I
iA
~
•
,'
' c< :300
2
Ia'
':'
1:-i ~1200~ 1
21t
1 x~
3
!
._
12
=Ia \j 3
... (12.27)
10
I I
I
I
I
:
! I.
I
:
wt
I
I
rr4 3..1.4 5~
,
'
I
'
I
o ~--~~~~~-,~~~~----~~----~
__________ cu t
ir~t
1"- 120°--1
I
I..
f 10
I
I
110
I
wt ..\ Fig. 12.15. Voltage and current waveforms for firin g angle of 30° for a three-phase full-converter drive of Fig. l2 .14. 2n
•
604
Power Electronics
[Art. 12.4]
r ms value of thyristor current, from iTl waveform,
ITr =
~
2
21t
1
Ia ' :3 X 21t
_
fl
.. .(12.28)
Ia -\J '3
==
and average value of thyristor current, 21t 1 IT.".... == Ia . -3 x ~ . =-31 I a . 21t
.
..:\12.29)
It may be observed in Fig. 12.15 that source current iA is positive when flrst subscript with voltage is a, as in Vab, va{;' Similarly, source current iA is negative when second .subscript is a, just as it is in vba' vca' On this basis, source current waveforms for phases B and C can also be sketched. Example 12.10. A 100 HV; 500 V, 2000 rpm separately-excited dc motor is energisf.d from 400 V, 50 Hz, 3.phase source through a 3-phase full converter. The voltage drop in conducting thyristors is 2V The dc motor parameters are as under .
r4
=0.1 n,
Km
=1.6 V-slrad,
La
=8 mHo
Rated armature current = 2 10 A. No-load armature current = 10% of rated current. Armature 'current is continuous and ripple free . (a) Find the no-load speed at firing angle of 30°. (b) Find the firing angle for a speed of2000 rpm at rated armature current. Determine also the supply power factor. (c) Find the speed regulation for the firing angle obtained in part (b). ' Solution . (a) The motor terminal voltage, V o == V t == 3..J2 ; 400 cos 30° = 467.75 V
Also
V t =Ea+ I ar a+ 2
467.75 = K m com+ 21 x 0.1 +2
or
:. No-load motor speed (b)
= 467'~~6- 4.1
rad/sec or 2767 .2 rpm.
At rated armature current and at 2000 rpm, Vo
= VI =Km . rom + Ia ra + 2
or
3~ 1tx 400 cos Ct -_ 1•6 X 21t X602000' + 210 X 0 . 1 + 2 -- 358 .1 V
or
X 1t a == cos-1 [358.1 3<\f2 X 400 = 4847° .
1
Rms value of source current , from Eq. (12.27 ) is
I!r == Ia : . Supply pf
~= 21 0 ~== 171.46 A
VI ·fa 358. 10 X 210 = \f3vs 'lsr -13 X 400 x 171.46
= 0.633 lag
At rated load , sp eed is 2000 rpm, armature terminal voltage V t == 358 .1 V and firing angle is 48.47° . At this firing an9'le, if rated load is r educed t o zer o, then (c )
Va = Vt
=35 S.1 == K m . (Dm + 21 x 0.1 + 2
[Art. 12.4]
. Electric Drives
605
rom = 358.11.6 4.1 ra d/ sec or 21128 . rpm
or
= 2112.8 - 2000 x 100 = 5.64% 2000
:. Speed regulation
Example 12.11. A 230V, 1500 rpm, 20 A separately-excited dc motor is fed from 3 -phase full converter. Motor armature resistance is 0.6 n. Full converter is connected to 400 V, 50 Hz source through a delta-star transformer. Motor terminal voltage is rated when converter firing angle is zero. (a) Calculate the transformer phase turns-ratio from primary to secondary. (b) Calculate the firing angle delay of the converter when (i) the motor is running at 1000 rpm at rated torque and (ii) the motor is running at (- 900) rpm and at half the rated torque'.
Solution. (a) For zero degree firing angle, motor terminal voltage is rated i.e. 230 V. Therefore, ..
3..J2V
----'-z cos 0° = V t = 230 V 1t
. 230 x 1t .
Vz = 3 T2' x 1 = 170.34 V
or
.-.'.
Here V z is the line voltage. Per-phase voltage on transformer star side is
vph = 170.34 T3 =98 .35 V Per-phase voltage input to tra:'1sformer delta
= 400 V
:. Transformer phase turns' ratio from primary to secondary 400 . = 98.35 = 4.067.
(b) (i) At 1500 rpm ,
Ea = V t - la ra = 230 - 20 x 0.6
= 218 V
. 218
At 1000 rpm, motor emf . = 1500 x 1000 = 145.33 V
For this motor emf, armature terminal volta~e at rated torque is
Vt
=Ea + 1 0 ra = 145.33 + 20 X 0:6 -= 157.33 V
3Vm1
But
- - cos Ct. = 7t
-
or
Ct. -
Vo = V t = 157.33 V cos
- 1[
157.33 x 1t ] _ 4" 840 3 ..J2 x 170.34 - o.
1 (ii ) t half the rat ed torque, armature current 1a = 2 x rated current
=1:2 x 20 = 10 A 3 oFf x 17 0. 34 1t
cos
Ct.
=-
900
15 00 x 218 + 10 x .6 = - 124.8 if 1
o
-
Ct. -
124. 8 x J- . ,) " cos . r L3 '12 170.34 - 12... 861 - 1
-
1t
.
606
[ rt. 12.4]
Power Electronics
E xa m ple 12.12. A 230 V, 10 kW; 1000 rpm separately-excited dc motor has its armature resistance of 0. 3 n and fi eld resisiance of 300 n. The speed of this motor is controlled by two 3-phase full converters, one in the armature circuit and the other in the field circuit and both are fed from 400 V, 50 Hz source. The motor constant is 1.1 V-s / Arad . Armature and field currents are ripple free. (a) With field converter setting to maximum field current, calculate firing angle for the armature converter fo r load torque of 60 Nm at rated speed. (b) With the load torque as in part (aJ and zero degree firing-angle for armature converter, speed is to be raised to 3000 rpm. Determine the firing angle of the field converter.
Solution. field voltage,
(a)
F or maximum field current, firing angle offield converter is zero. Therefore, V = 3Vml f
Field current, M~~e~
= 3-../2 x 400 =540.1 V.
1t
1t
I = 540.1 = 1 8 A f 300 . ~=~.~
With no saturation, _• =Kif' :. Ea =Ka KIf' (Om where k is a constant in V-s/A.rad Motor torque, Te =Ka Ia == Ka KIf' Ia = k If' Ia or 60 = 1.1 x 1.8 x Ia 60 Ia = 1.1 x 1.8 = 30.30 A :. Motor current, For the motor converter,
=k If' (Om
3Vm1
Vt =Vo = - - cos a = Ea + Ia ra =k If ' 1t
(Om
+ Ia ra
3...[2 ~ 400 cos a = 1.1 x 1.8 x 21t ~~OOO + 30.30 x 0.3 = 216.435 V :. Firing angle of armature converter, (J.
(b)
=cos-1 [216.435 x 7t] =66376 3 '12 x 400 .
0
_
With zero degree firing angle of the armature converter,
3-..J2 7tx 400 cos 00 =-1.1 X- I fX
27t
I = 540.1 - 9.09 f 345.58
or :. F iel d voltage, -
3000 30 30 0 3 60 + . x .
X
=1 "" 366 A'
Vf =If · rf= 1.5366
.0
x 300 =
3--12 x 400 7t
cos
Ctl
:. Firing angle of field converter,
a
_ l -
- 1
cos
[3003xTx1.5366 x 1t]_31 406 400 .
• 0
EX811,lple 12.13. In a speed controlled dc drive, the load torque i 40 Nm. At time t = 0, the operation is under steady state and the speed is 500 rp m. Under this condition at t =0+, the g enerated torque is instantly increased to 100 Nm . The inertia of the drive is 0.01 Nm sec 2 l rad. The friction is
negligible.
Electric Drjyes
[Art. 12....]
607
(a) Write down the differential equation governing the speed of the drive for t> O. (b) Evaluate the time taken fo r the speed to reach 1000 rpm. [GATE, 1998]
Solution. (a) At t = 0, steady state exists and therefore, generated torque, Te = TL, load torque In general, the dynamic equation for the motor-load combination is generated (or motor) torque = inertia torque + friction torque + load torque doo m Te =J dt + Doom + TL
or
As friction torque is zero, Doom of the drive at t > 0, as
=O. This gives the differentiai equation, governing the speed doom
Te =J ---cJ:t + TL doo m 100 = 0.01 dt + 40 (b)
From Eq. (0,
. doom
.. .(i)
60
dt = 0.01 = 6000 doom dt = 6000
or Its integration gives,
...(ii)
Initial speed at t = 0 + r em ains 500 rpm. Therefore 00
= 21t X 500 = 100 1t rad/sec
60 6 1001t 1 0 = 600 0 x - 6 - + A
mo
From Eq. (ii),
rom
t
Final speed
or A
-1t
= 360
1t
= 6000 -
. '!: '
360·
_ 21t x 1000 _200 1t d/ rom 60 6 ra sec 2001t
1t
1t
t = 6000 x 6 - 360 = 360 sec = 0.0873·sec :. Time t aken for the speed t o reach 1000 rpm =0. 0873 sec. Example 12.14. A dc motor dri ven from a (ully"con trolled 3-phase con verter shown in Fig. 12 .16 dra ws a de current of 100 A with negligi ble
100A
ripp le. ~ (~)
Sketch the ac line current iA for
one cycle.
(b) Determi ne the 3 rd and 5th ha rmonic componen ts of the line cu rrent as a percentoge of the fundamenta l current. [GATE, 1. 998)
Solution . (a ) The ac line current iA for on e cycle is sketched in Fig. 12.15 fOT a fIr ing angle ct u nder the assumpti on of negligible ripple in the armatur e curren t Ia = 100 A.
3 phase
inp ut tA
A 0 -_ _----",
B o-
-t-----...
c c---+--~--...
Fig. 12.15. Partainini to Ex mple 12 .14.
608
[Art. 12.4]
Power Electronics
(b ) Th e line current iA shown in Fig. 12.15 can he expressed in Fourier series as
0041
. tsn
• = tAn ='L" -n1ta
n =
mt. t '- no. ) cos -6 SIn ( nro
'.
,
I, 3, 5,-
Rms value of the nth harmonic line current is given by
41a
1&X
Ian == :.r2 . n 1t COB '"'6=
2 ..f2 . 1a
.
. n 1t
. cos
n1t
6"
Rms value of fundamental current,
I a1 =
2..f2Ia -J6 1t . cos 30° = -1t I a
Rms value of third-harmonic current,
I s3
Ia
= 2..f2 31t cos 90° =0
Rms value of fifth-harmonic current,
"
2-./2 Ia -J6
Is5 = 51t cOS 150° =- 5 1t ICI
From above, third harmonic current as a percentage of fundamen tal current = 0% and fifth harmonic current as a percentage of fundamental current
Is5
= l sI x 100 = -
-..f6 Ia
1t
51t x <\16. Ia x 100 = -
Exam p le 12.15. A dc motor d riven from a 3-phase full converter shown in Fig. 12. 17 draws a dc line current of 60 A with negligible ripple. (a) Sketch the line voltage vab taking it zero-crossing and becoming positive at rot = O. Also, sketch the line current iA for one cycle for 0.= 150°. Indicate also the conduction of devices. Thyristor current iT should also be sketched. . (b) Calculate average and rms values of thyristor current.
,. 20%. 10=60 A
iA
A cr-__-
T3
T5
..... a
B O----I-------..b
C o----+------+--..... c
T2
Fig. 12.17. Three-phase full converter feeding a dc motor, Example 12.15.
(c) Compute power factor at the ac source .. (d) If motor constant is 2 .4 V-sec / rad d.nd armature circuit resistance is 0. 5 n, calculate th e m otor speed.
Solution.
(a ) Note that for line voltages Va b' Uac• Ubc' V ba etc.
and with
12.18, firing angle a for thyristor T1 must be measured from wt
vab
as shoVlm in Fig.
= ~ . Accordingly, Ct = 150
0
is
measured from th e instant rot = 1t/3 in Fig. 12.18. Motor current 1a = 60 A is shown constant in Fig. 12.18. At ex = 50° ) Tl is turned on . So voltage Uab will send constant current f a t hr ough TIrTS. Thyristor T1 will con duct fOT 120°; for
Electric D rives
[Art. 12.4]
609
O~----~----------~----------------~~--------~w~t~
i:[
!60 A
~
wt
"A '
,
I
'
wt
~120o~
°1
J
~ T4T3 ~TJ ~ - - 120·
-r-:
T1T6
:
-t- TlT2 --ii .
:
I
160A
I
7t /6
~'"
'It
I
wt
271
.
Fig. 12.18. Waveforms for Example 12 .15.
the first 60°, Tl, T6 con d ct t ogether. For the next 60°; Tl, T2 conduct together as shown in Fig.
12.18. Voltage Vbo == - Vab will cause T3, T4 to conduct for 60° and Vca will force T5, T4 to conduct for the next 60° as shown. Note that voltages Vbo' Vca will cause line current iA to be negative whereas for vo.b' v ac ' line current iA is positive. Thyristor current iT through Tl will flow only .when iA is positive. (b)
Average thyristor current,
60 = -Ia =-= 20 A 3 10
3
60 I Tr = T3 = -J3 =: 34 :642 A
Rms thyristor current, (c)
ITA
.
Rms value of source current, lsI' =10.
_ f2 _ 12 .'V '3 =60.\/ '3 A
Power delivered to motor Power factor at a c source
= '13 . Vs . Isr =
[
3-J2 V s 7t
1
cos 150° x 10.
= ~7t cos 150 = -
{3
-f3 Vs . 10. 12
0.827.
Minus sign for th e power factor merely indicates t e system to be in the inversion mode. (d )
v:o = Vt = 3,[2 7tx 400 cos 1-0 b
0
=
K m wm + 60
X
0 .5
Power E lectronics
[Art. 12.5]
610
=
(Urn
- 467
.
73~
2.4 ·
30
·
.
.
.
= - 207.39 rad/sec or
. . . .. -1980.43rpm.
The motor is in the ·regenerative braki~g mode with emf Ea r eversed ·from its motoring mode polarity. 12.4.4. Three-phase Dual Converter Drives The schematic diagram for a 3-phase dual converter dc drive is shown in Fig. 12.19 . Converter 1 allows motor control in I and IV quadrants whereas with converter 2, the operation in II and III quadrants is obtained. The applications of dual converter are limited to about 2 MW-drives. For reversing the polarity ofmotor generated emf for regeneration purposes; field · . circuit must be energised from single-phase or three-phase full converter. .
.
.
.
.
.
+
1 phose or 3 phose I .e .
Fig. 12.19. Three-phase dual converter controlled separately-excited de motor . .
When converter 1, or 2, is in operati,on, average output voltage is . 3Vml . Vo= V t = cos a l for 0 S; a l S; 1t
... (12.30)
With a 3-phase full converter in the field circuit, . 3V ml Vf = - - cos af for 0 S; afS; 1t
... (12.31)
1t
1t
In case circulating current-type dual converter of Fig. 6.45 is used, then as per·Eq. (6.85),
a l + a 2 = 180 12.5. CHOPPER DRIVES
0
. .
..
,
.'
,
When variable dc voltage is to be obtained from fix ed dc voltage, dc ch opper is the ideal choice. Use of chopper in traction systems is now accepted all over the world. A chopper is inserted in between a flxed voltage de source and th e de m otor annature for its speed control below base speed. In addition, chopper is eas' ly adapt abl e for regenerative brakin CT of de mot ors and thus kin etic energy of the drive can be r eturned to the d e source. This results in overall energy sa'ling which is 'the most welcome feature in transp ortation systems r equiring frequent stops, as for example in rapid transit sys tems ChoppeT drives are also used in battery- operated ve 'des wher e energy savin g is of prime importance .
Ii'
Electric Drives
(A rt. 12.5]
611
Though choppers can be used for dynamic braking and for combined regenerative and dynamic con trol of dc drives, only the following two control modes are described in what follows . 1. Power control or motoring control.
2. Regenerative-braking controL Both the chopper control methods are now described. In addition, two-quadrant and four-quadrant chopper drives are also discussed. 12.5.1. Power Control or Motoring Control Fig. 12.20 (a) show:s the basic arrangement of a dc chopper feeding power to a dc series motor. The chopper is shown to consist of a force-commutated thyristor, it could equally :,:rell be atransistor switch. It offers one-quadrant drive, Fig. 12.20 (b). Armature current is assumed continuous and ripple free. The waveforms for the source voltage Vs' armature terminal voltage VI = vo, armature current ia, dc source current is and freewheeling-diode current ifd are sketched in Fig. 12.20 (c). From these waveforms, the following relations can be obtained: ... (1 2.32)
.
where
Ton ex = duty cycle = T
f = ch{)pping frequency = ~
and Chopper
ts
~------..,
+
io=io
I
I I
,
:
+
1.. ______ ...
V'1
V~~Vt kaT
FD
Vs
rVs
t•
(1-alT
·1· {j
,
tvs
II ,
I
I
:
tfd
I
I I I I
to=1.o
L. ; ,I,
t
10 ;
( a)
t Vo=Vt
1.s
Ton
i
Tof1
r :
I I
- 10
10=10 -Vt
(b )
ttd1 f.--
, I
t ,
OJ lo 0
,
T ---l (c)
Fig. 12.20 D.C . Chopper for Svries motor rive (a) circuit di agram
(b) quadrant diagr am and (c) waveforms.
[ t
•
612
Powe Electr onic;s
[Art. 12.5]
Power delivered to motor = (Aver age motor voltage) (average motor current)
,-..'
=V t · 1=a· V s ·1a a
Ton Average source current = - T . I a Input power to chopper
= a· I a
= (average input voltage) (average source current) = Vs' ala
For the motor armature circuit,
or
VI = aVs =Ea +la (ra + rs) =Km ' rom +10 (ra +rs)
aVs - 1a (ra + rs)
... (12.33) vo: Vt Wm = ---K=---=-
1,--
m
It is seen from Eq. (12.33) that by varying the duty cycle a of the chopper, armature terminal voltage can be controlled and thus speed of the dc motor can be regulated.
So far, armature current ia has been assumed
ripple free and accordingly, waveforms in Fig. 12.20 are sketched. Actually, the motor armature current will rise during chopper on period and. fall during off period as shown in F ig. 12.21. The current expressions during on and off periods are obtained in Chapter 7 on choppers. By referring to this choppe r, armature current la(t) during on period, from Eq. (7. 10), is given by
l~
D
n _!--COo'
Q'---'-----"--_........_"--_........
:;
!! ,I :, ,
,
II
1
,
,. OL-__~-L--;--~-~-~-l- Ton-l Totf~: t is + ' !
h 'm'(C]
o
:
I
: ,(1-ex)T ~
itd
l mx
•
n !
ia (t) Here R L
=-
...
t
:
i :
... (12.34) The armature current during the off-period, from Eq. (7.11), is given by
t
,
t Fig. 12 .21. Wavefonns for d e chopper drive of Fig. 12.20
~ (l-e-~t)+lmx ' e-~t
(a).
... (12 .35)
ra (armature resis tance) + rs (series-field resistance)
= La (arm ature inductance) + Ls (series-fi e d in ductance)
=
Under st ea dy-state operating conditions,
V t = aVs = E a +Jefl·
Example 12.16. A de serie' motor is fe d fro m 600 V de source through a chopp er. The de motor h as the follo w ing param eters: Ta
= 0.04 .0,
r3
=0.06 Q,
k
= 4 X 10- 3 Nm / amp2
Th e auerag armature current of 300 A is ripple free. For a chopper d uty cycle of 60%, dete rmine .' (a ) inp ut p otuer fro m the source (b ) mo tor speed and (c) motor t rque .
~
[Art. 12.5]
Electric D ives
Solution.
(b)
613
Power input to motor
= V t . Ia = a Vs . Ia
= 0.6 x 600 x 300 =108 kW.
For a dc series motor, aVs =Ea + Ia R =k fa Wm + Ia R 0.6 x 600 = 4 x10- 3 x 300 X Wm + 300 (0.04 + 0.06) 360 - 30 Wm = 1.2 . = 275 rad/sec or 2626.1 rpm (a)
Motor torque, Te =k I/ = 4 x 10- 3 X 300 2 =360 Nm. Example 12.17. The chopper used for on-off control of a dc separately-excited motor has supply voltage of 230V dc, an on- time of 10 m sec and off-time of 15 m sec. Neglecting armature inductance and assuming continuous conduction of motor current, calculate the average load current when the motor speed is 1500 rpm and has a voltage constant of Ku = 0.5 V /rad per [I.AS., 1985) sec. The armature resistance is 3 n. (c)
Solution. Chopper duty cycle Ton
10
a = Ton + Torr = 10 + 15 = 0.4 For the motor armature circuit, VI = aVs = Ea + Ia ra =Km . wm + lara
0.4 X 23.0 = 0.5 :. Motor load current,
fa
21t X
1500 60 + Ia
X
X
3
=92 - ~5 X 1t = 4.487 A
Example 12.18. A de chopper is used to control the speed of a separately-excited de motor. The dc supp ly voltage is 220 V, armature resistance ra = 0.2 n and motor constant Ka = 0.08 VI rpm. T hi motor drives a constant torque load requiring an average armature current of 25 A Determine (a) the range ofspeed control (b) the range ofduty cycle a.Assumed the motor current to be continuous. [I.AS., 1990) Solution. For the motor armature circuit, V t = a Vs = E a + I ara As mot or drives a constant to qu e load, motor t orqu e Te is constant and therefore arm ature current remains constant at 25 A. Minimum possible motor peed is N = .0. Therefore, a x 220 = 0. 08 x 0 + 25 X 2. 0 = 5 5 1 a = 220 = 44 Maximum possible motor speed corresponds t o a = 1, i.e. when 220 V de is directly applie ,and n o chopping is done. .. l x220 =0. 08x N+ 25 x O. 2 22 0 - 5 or N = 0.08 = 2687.5 rpm
,
:. Range
44 < a < 1.
0
speed control: 0 < N < 2687.5 rpm and corresponding range
0
duty cycle:
614
Power Electronics
[Art. 12.5]
Examp le 12.19. A separately-excited dc motor is fe d from 220 V dc source through a chopper operating at 400 Hz. The load torque is 30 Nm at a speed of 1000 rpm. The motor has ra = 0, La =2 mH and Km = 1.5 V-sec/rad. N eglecting all motor and chopper losses, calculate (a) the minim um and m ax imum values of armature current and the armature current excursion,
the armature current expressions during on and off periods. Solution. As the .armature resistance is neglected, armature current varies linearly between its minimum and maximum values. Te 30 Average armature current, 1a = Km = 1.5 = 20 A ( b)
(a:
Ea =Km' rom = 1.5 x
Motor emf,
21t X 1000 60
.
= 157.08 V
Motor input voltage,
T = 1. = _ 1_ = 2 5 ms f 400 . .
On-period, Ton. = a T = 0.714 x 2.5 = 1.785 ms
Off-period, Torr = (1- a) T = 0.715 ms
During on-period Ton.. armature current will rise which is governed by the equat:.:m,
Periodic time,
o+ L
dia dt + Ea = Va
=3146 A I
dia _ Va - Ea = 220 - 157.08 dt L 0.02
or
s
dia Ea = - 157.08 = _ 7854 AI dt = - L 0.02 s With current rising linearly, it is seen from Fig. 12.21 that During off period,
dia dunng . Ton 1m:r. =1mn + (dt
J Ton X
=lmn.+3146 x 1.785 x 10- 3 or 1mx = 1mn. + 5.616 For linear variation between 1mn and 1mx. average value of armature current I +1 I = mx mn. = 20 A a
or
...(it )
1mx = 40 - Imn 1m" = 17.912 A. : . Armature current excursion =lmx - Imn = 22.808-17.912 = 5.616 A (b ) Armature current expression during tum-on,
.
.. .(i)
2
Solving Eqs. (i ) and (ii), we get 1mx = 22.808 A and
~-- -
(di
a
. g Ton La(t) = Imn + dt dunn
= 17.192+ 3146t
J
X
t
for O ~t ~ TO Tl
Ele,ctr ic Drives
615
[Art. 12.5]
Armature current expression during turn-off,
1
£a(t) = Imx .+ (dia dt during Toff x t = 22.808 - 7854
t
for 0 :$ t
:$
Toff
Example 12.20. Repeat Example 12.19, in case motor has a resistance of 0.2 n for its armature circuit. Solution. (a) From Example 12.19, annature current, Ia = 20 A and motor emf, Ea = 157.08 V; source voltage, Vs = 220 V. For armature circuit, aVs = Vo
=V t =Ea + Ia ta = 157.08 + 20 x 0.2 = 161.08 V
= 1~~.g8 = 0.7322 Ton = aT = 0.7322 x 2.5 = 1.831 ms a
T off = T - Ton
= 0.669 ms, ~ = 0~022 = 10
During T on from Eq . (12.34), armature current is . () = 220 - 157.08 (1- -lOt) . I -lOt t 0.2 e + mn . e
ta
=314.6 (1 At t
e- lOt ) + Imn . e- IO t
= Ton = 1.831 ms, cu rrent become Imx. This gives ia (t)
.. .(i) .
=Imx =5.7079 + 0.98187 Imn
During T off, from Eq. (12.35), armature current is
\
' . It )=-157.08(1_ -lOt) I . -lOt Za ~. 0.2 e + mx e = -785.4 (1- e-
At t
=0.669 ms,
ia(t)
lOt
)
+ Imx ' e-
=Imn. This gives
ia(t) = Imn = - 5.237 + 0.99331mx
Solving Eqs. (i) and (ii), we get
Imx = 5.7079 + 0.98187 (- 5.237 + 0.9933 Imx)
= 0.5658 + 0.9753 Imx
0.5658
Imx = 0.0247 = 22. 907 A .
or
Imn =- 5.237 + 0.9933 x 22 .9;O~ == 17. 516 A .
: . Arm atur 'e current excursion .:..~ .. =Im;c - Imn =229 07 - 17.516 = 5.39 A
.
(b)
/
10t
.Arm atur e current expression during turn-on p e lod is
ia (t) = 31 4.6 (1- e- 1Ot ) + 17.516 e- lOt
Armature current axpression during turn-off period i~
ia(t) =- 78 5.4 (1 - e- lOt) + 22.907 e- 10 I
...(ii)
616
[Ar t. 12.5]
Power Electro nics
12.5.2. Regenerative-Braking Control In regenerative-braking control, the motor acts as a generator and the kinetic energy of the motor and connected load is returned to the supply.
V-E
During motoring mode, armature current la = t a, i.e. armature current is positive and . ra the motor consumes power. In case load drives the motor at a speed such that average value of motor counter emf Ea (= Km . wm) exceeds Vt, Ia is reversed and power is delivered to the dc bus. The motor is then working as a generator in the regenerative braking mode. The principle of regenerative braking mode is explained with the help of Fig. 12.22 (a), where a separately-excited dc motor and a chopper are shown. For active loads, such as a train going down thehill or a descending hoist, let it be assumed that motor counter emf Ea is more than the source voltage Vs' When chopper CR is on, current through armature inductance La rises as the armature terminals get short 'circuited through CR. Also, vt = 0 during Ton. When chopper is turned off, Ea being more than source voltage Vs' diode D conducts and the energy stored in armature inductance is transferred to the source. During Toff' vt = Vs' On the assumption of continuous and ripple free armature current, the relevant voltage and current waveforms are shown in Fig. 12.22 (b). With respect to first quadrant operation as offered by motoring control of Fig. 12.20 (a), regenerative braking control offers second quadrant operation as armature terminal voltage has the same polarity but the direction of armature current is reversed, Figs. 12.22 (a) and (c). From the waveforms of Fig. 12.22 (b), the following relations can be derived: The average voltage across chopper (or armature terminals) is Vt
Torr =r' Vs = (1 -
... (12.36)
a) Vs
iJ~------,o'T""!I-O-
. t
t.O
+
+
D
iT t
..
r- -, I
CH I
I
i
tJ T
L .. J t
... •
(a )
(b)
Fig. 12.22 . Regener ative braking of a separately-exci ted de motor (a) circuit diagram (b) waveforms (c) quadrant d 'agram .
(c)
[Art. 12.5]
Electric Drives
617
Power generated by the motor
=V t . Ia = (1 Motor emf generated,
a) Vs ·Ia
Ea = Km wm = V t + Ia Ta = (1- a) Vs+Ia ra
... (12.37 )
Motor speed during regenerative braking, Wm
=
(I-'a) Vs + lara
'
Km
dia When chopper is on, Ea - lara - La -dt = 0
or " di With chopper on, L a must store energy and current must rise, i.e. d; must be positive or ... (12.38)
or With chopper off, (Ea - lara) must be more than Vs for regeneration purposes and therefore [Vs - (Ea - Ia r a)J must be negative. This is possible only if current decreases during off period, di ' i.e. in the above expression must be negative.
d;
[Vs - (Ea -lara)] ~ 0 - (Ea - laTa) ~ (- V s)
or
(Ea - Ia Ta)
~
Vs
... (12 .39)
Eqs. (12 .38) and (12 .39) can be combined to give the conditions for controlling the power during regenerative braking as
o ~ (Ea -
l aTa)
~
Vs
... (12.40)
Eq. (12.40) gives the conditions for the two voltages and their polarity for the r egenerative braking control of dc separately-excited motor. Minimum brakin g spee is obtained when E a - IaTa = 0 or
- Km
:. Minimum bra k ing speed
Cl)mn
= Ia ra
Cl)mn
Ia ra =K
.. :(12 .41)
m
Maximum possible braking speed is obtained when Ea-1aTa =Vs
:. Maximum br aking sp ed,
... (12 .42)
[Art. 12.5]
61 8
Power Electronics
Th us regenerative braking control is effective only when motor speed IS less than more than Cl)mn ' This can be expressed as
w mx
and
. . . ' V s + I ar a I a ra Therefore, the speed range for regeneratIve brakmg IS K : K or (Vs + Ia r a) : Ia r U ' m
m
Regenerative braking of chopper-fed separately-excited dc motor is stable, it is therefore discussed here. DC series motors, however, offer unstable operating characteristics during regenerative braking. As such, r~generativebraking of chopper-controlled series motors is difficult.
Example 12.21. A dc chopper is used for regenerative braking of a separately-excited de motor. The dc supply voltage is 400V. The motor has ra = 0.2 fl, Km = 1.2 V-s I rad. The average armature current during regenerative braking is kept constant at 300 A with negligible ripple. For a duty cycle of 60% for a chopper, determine (a) power returned to the dc supply (b) minimum and maximum permissible braking speeds and (c) speed during regenerative braking. Solution. (a) Average armature terminal voltage, VI
=(1 -
a) Vs
=(1- 0.6) x 400 =160 V.
Power returned to the dc supply = V'a = 160 x 300 W = 48 kW (b )
From Eq. (12.41 ), minimum braking speed is Cl)mn=
Ia · ra 300x O.2 Km , = 1.2 = 50 radls or 477.46 rpm
From Eq. (12.42), maximum braking speed is
rom%=
Va + I a . r a 'Km
400 + 300 X 0.2 =
1.2
= 383.33 radls or 3660.6 rpm (c)
When working as a generator during regenerative braking, the generated emf is
Ea =K m rom = VI + lara. = 160 + 300 X 0.2 = 220 V :. Motor speed,
<.Om
220 = 1.2 radls or 1750.7 rpm
2. .S. Two-quadrant 'Chopper Drives Motoring control circuit for chopper drives offer only first -quadrant drive, because armature voltage and arm ature current remain posi ive over the entir·e range of speed control. In r egenerative braking, second-quadrant drive is obtained as armature terminal voltage rema' 8 positive but direction of armatur e current is reversed , In two-quadrant de motor drive, both motoring mode as well as regenerative braking mode are carried out by one chopper configuration. On stich circuit is ShovID in Fig, 12.23 (a ) whic consists of two choppers CHI, CH2 and two diodes DI, D2 and a separately-excited de motor.
[Art. 12.5]
Ele tric Drives
619
+
D2 CH2,D2
CH1, D1
-1 0
(a) (b) . Fig. 12.23. Two-quadrant dc chopper drive (a) circuit diagram and (b) two-quadrant diagra.m.
Motoring mode. When chopper CHI is on, the supply voltage V, gets connected to armature terminals and therefore armature current ia rises. When CHI is turned off, ia free wheels through Dl and therefore ia decays. This shows that with CHI and Dl, motor control i11: fir!!t quadrant is obtained. . Regener t ive mode. When CH2 is turned on, the motor acts as a generator and the armature current ia rises and therefore energy is stored in armature inductance La' When CH2 is turned off, D2 gets turned on and therefore direction of ia is reversed. Now the energy stored in La is returned to dc source and second quadrant operation is obtained, Fig. 12.23 (b). In this figure, first-quadr ant oper ation of dc motor is sometimes called forward-motoring mode and second-quadrant operation as forward regenerative· braking mode. 12.5.4. Four-qua drant Chopper Drives In four-qu adrant dc chopper drives, a motor can be .made to work in forward-motoring mode (first quadr ant), forward regenerative braking mode (second quadrant), reverse motoring mode (th ird quadrant) and reverse regenerative-braking mode (fourth quadrant). The circuit shown in Fig. 12.24 (a) offers four-quadrant operation of a separately-excited dc motor. This circuit consists of four choppers, four diodes and a separately-excited dc motor. Its operation in the four quadrants can be explained as under: . Forward motoring mode. During this mode or first-quadrant operation, choppers CH2, CH3 are kept off, CH4 is kept on whereas CHi is operated. When CHI, CH4 are on, motor voltage is positive and positive armature current rises. When CH I is turned off, positive armature current free-wheels and decreases as it flows through CH4, D2. In this manner, controlled motor operation in first quadrant is obtained. F orward regenerative-brak ing mode. A dc motor can work in the regener ative-braking mode only if mot or generated emf is made to exceed the de source voltage. For obtaining this mode, CHI , CHS and CH4 are kept off whereas CH2 is operc.,Ted . When CH2 is turned on, negative armature current rises thr ou gh CH2,D4, E a , La' ra ' WhenCH2 is turned off, diodes Dl, D2 are turned on and the motor acting as 9. gener ator r eturns energy to the de source, This r es Its in crward r egenerative-braking mode in the second-quadr ant. .. Reverse m otorin g m ode. This ope ating mode is opposite to forward mot oring m ode , Choppers CHI, CH4 are kept off, eH2 is kept on wh ereas CH 3 is operatzd. When CH3 and CH2 are on, armature gets conn ected to source voltage Vs so that both armature volt age V t and armature current ia are nega tive. As armature current is reversed, motor t orqu e is rever sed and ~ onse quently motoring m ode in third quadrant is obtained. Wh en CH3 is turned
620
Power Electronics
[Art. 12.6]
Vt
=Vo
CH2operat E:'d CHl op NatE:'d CH1, CH 3,C H4oft CH2 J CH3 off Forwarded Forwarded Reg . Motoring 8raking
+
-1a CH3 operated CH1.CH4 off Reverse Motoring -VI
(a)
CH 4 opNated CH1,CH2,CH3 off Reverse RE:'g Braking
(b)
Fig. 12.24. Four-quadrant de chopper drive (a) circuit diagram and (b) four-quadrant diagram.
off, negative armature current freewheels through CH2, D4, Ea La' r a; armature current decreases and thus speed control is obtained in third quadrant. Note that during this mode, polarity of Ea must be opposite to that shown in Fig. 12.24 (a). Reve rse R egenerative-braking m ode . As in forward braking mode, reverse regenerative-braking mode is feasible only if motor generated emf is made to exceed the dc source voltage. For this operating mode, CHI, CH2 and CH3 are kept off whereas CH4 is operated. When CH4 is turned on, positive armature current ia rises through CH4, D2, raJLa' Ea' Note that in this mode also, polarity of motor emf Ea must be opposite to that shown in Fig. 12.24 (a). When CH4 is turned off, diodes D2, D3 begin to conduct and motor acting as a generator returns energy to the dc source. This leads to reverse regenerative-braking operation of the dc separately-excited motor in fourth qu adrant. . Note that in Fig. 12.24 (a), the numbering of choppers is done to agree with the quadrants in which these are operated. For example, CHI 's operated for first quadrant, .... , CH4 for fourth quadr ant etc. . 12.6. A.C. DRIVES
.
Primarily, electric drives can be divided into two groups, dc drives and ac drives. DC drives have alre ady been discussed in this chapter. Now ac drives are described at their introductory level only. Advantages and disadvantages of ac drives with respect to dc drives are as under : Advantages of ae drives (i) For the same ra ting, ac m otors are lighter in weight as compared to dc motors. (ii ) AC motors r equir e low maint enance as compared to dc motors. (iii) AC m otors are les8 expensive as compared to equivalent de motors.
(iu ) AC motors can work in hazardous areas like chemical, petrochemical et c. wher eas
dc motors are unsuitable for such environments because of commutator sparking. Disadvantages of a c drives (i ) Power converters for the control of ae motors are more compl ex.
(i i) P ower converters fOT ac d ives are more expensive. (iii) Power
converters for ae drives generate h arm onics in the supply system an load cir cuit . As a resul t, ac motors ge t derated.
[Art. 12.7]
Ele tric Driv es
621
The advantages of ac drives outweigh their disadvantages. As such, ac drives are used for several industrial applications. In general, there are two types of ac drives: 1. Induction motor drives 2. Synchronous motor drives.
These are now described in what follows .
12.7. INDUCTION-MOTOR DRIVES
,
Three-phase induction motors are more commonly employed in adjustable-speed· drives than three-phase synchronous motors . Three-phase induction motors are of two types, squirrel-cage induction motors (SCIMs) and slip-ring (or wound-rotor) induction motors (SRIMs). Stator windings of both types carry three-phase windings. Rotor of SCIM is made of copper or aluminium bars short-circuited by two end rings. Rotor of SRIM carries three-phase winding connected to three slip rings on the rotor shaft. When 3-phase supply is connected to three-phase stator winding, rotating magnetic field is produced . The speed of this rotating field, called synchronous speed, is given by ' 120 f1'
Ns = - p - rpm
Also , where
, cos;::
fl COl
or
2fl
ns = p
rps
.. .(12 .43)
41tf1 2co 1 p - = p rad/ sec
... (12.44)
= supply frequency in Hz,
= supply frequ ency in r ad/s
P = number of stator poles. Rotor cannot attain synchronous speed. It must. run at a speed N r less than N s , where' . Nr=Ns (l-s)
where
N r = rotor speed in rpm com = rotor speed in radl.s . Ns -Nr COs - wm s=shp= =--Ns Ws
Analysis and p e. onnance . Per-phase equivalent circuit of a three-phase induction motor is shown in Fig. 12.25 . In this circuit , r 2 = rotor resistance referred to stator, X 2 = rotor leakage reactance referred to stator. r l + j Xl is the stator leakage impedance andXm ;:: magnetizing reactance. In this
(a)
(b)
? 'g. 12 .25. P er·phase equivalent circuit of a three-phase indu ction motor referred to s a
OT.
622
Power Eledronics
[Art. 12.7]
figure, core-loss resistance Rc is not shown. But, for determining the shaft power or shaft torque, the core loss must be taken into consideration. In case stator impedance drop is assumed negligible as compared to tenninal voltage Vl' Xm can be moved to stator terminals to get the simplified equivalent circuit of Fig. 12.25 (b). From this per-phase circuit model, stator current!l and rotor current 12 can be calculated. Once 11' 12 are known, the performance parameters of a 3- phase induction motor can be determined as follows : From Fig. 12.25 (b),
VI 12 = ~-----.:---'~--r2
rl
+ -s + j
(Xl
.:.(12.45)
+ X2)
Air-gap power (power transferred from stator t o rotor through the air gap),
P
g
=3 12 r2 2 S
... (12.46)
Rotor ohmic loss = 3122 r 2
Developed power in rotor, Pm =Pg - rotor ohmic loss
. =31; r, (1 ~ . Pm 2 Developed rotor torque,Te =- =312 r 2 . wm
8/ "
=~ .1 W
s
Also, '
1-:- SJ
-- . S
.. .(12.47)
. (1-) 1
W6
S
2 r2
2
.. .(12.48)
S
P T =J
... (12.49)
W,
e
Substituting the value of 12 from Eq. (12.45), we get 2
3 T, =
V1
OO'[rl + ';
r2
J
+ (Xl + x,)'
... (12.50)
S
Motor power input, P = Pg + stator core loss + stator copper loss
Output, or shaft, power, Psh =Pm - fixed loss (friction and windage loss) .
Psh
Motor efficiency, 11 =P Output, or sh aft, torque, Tsh
P
P
=-romsh = (Os ( 1sh-
... (12.51)
s)
Slip at which maximum torque occurs is given by m
8 =
R
r2
...(12.52)
+ (Xl + x2i
Substituting this value of Sm in Eq. (12.50), w e get an expression for maximum t orque as 2 1
3 s [ r1 + ~r12V+ (Xl + X2)2 T e.m = 2w
1
.•. (12. 53)
The ma."{imum torque, also called p ull-ou. t or breakdown torque, is independent of r otor resistance. H owever, sm is directly raportional to atar r esistance.
• 623
[Art. 12.7]
Electri Drives
It is seen from Eq. (12. 50) that if tQ.ree-phase
induction motor is energised from fixed voltage· at a
. c onstant frequency, motor t orque is a function of t he slip. ' For -different values of slip, the speed-torque characteristic of a three-phase induction in()tor is plotted in Fig.l2.26. In motoring mode under ' normal rUnI:l,ing (8 < 8 m ), as stator current is not high, the air'-gap flux remains substantially constant and torque increases with increase in slip from zero to sm' After maximum torque Tem at slip 8m, the slip increases, stator current rises much more than the rated current, air-gap flux decreases and therefore torque decreases with an increase in
slip.
In case rotor resistancer 1 is neglected, which is usually true in lar g e induction motors, the expressions for slip and torque are as under:
Speed
o
-:::----------~-;r~~~ --1- .... Ttm~-'"
,
Forward motoring
I
\-Stator
\CU"'( .
as
I
slip
Torque
-Reverse plugging
'... --.--------------'" - I.-
... (12.54)
Fig. 12.26. Speed-torque characteristics Of a three -~hase induction motor ..
... (12.56) hat .. .(12.57) 8
'Sm
8m
S
-+ If S < s m' then
Te
2 ·8
2 (ros-{J)m)
.,.(12,58)
or or Motor speed,
.. .(12.59)
It is seen from Eqs . (12.58) and (12. 59) that for small v alues of slips, m otor .t orqu e is proportional to slip E q . (12. 58), an motor speed decreases as t orqueTe = load torqu e TL • . . increases, Eq. (12 .59). From Eqs . (12.54) an d (1 2.59), ... (12 .60)
Th is expression shows that drop ' r esistance.
speed r"om no 10 d t o full load dep end s on rotor
624
[Art. 12.8]
Power Electronics
12.8. SPEED CONTROL OF THREE-PHASE INDUCTION MOTORS Three-phase induction motors are admirably suited to fulfil the demand of loads requiring substantially a constant speed. Several industrial applications, however, need adjustable speeds for their efficient operation. The object of the present section is to describe the basic prin ciples of speed control techniques employed to three-phase induction motors through the use of power-electronics converters. The various methods of speed control through semiconductor devices are as under: (i) Stator voltage control (ii) Stator frequency control (iii) Stator voltage and frequency control (iv) Stator current control (v) Static rotor-resistance control (vi) Slip-energy recovery control. Methods (i) to (iv) are applicable to both SeIMs and WRIMs whereas methods (v) and (vi) can be used for WRIMs only. These methods are now described in what follows. 12.8.1. Stator Volta ge Control It is seen from Eq. (12.50) that motor torque Te is proportional to the square of the stator supply voltage. A reduction in the supply voltage will reduce the motor torque and therefore the speed of the drive . If the motor termina~ voltage is redu ced to KV'1 where K < 1, then the motor torque is given by 3
Te = -
ro.
.
(KV1l 2
~ I + :'J + (Xl + X,)'
'2
... (12.61)
.
s
For the purpose of varying the voltage applied to a 3-phase induction motor so as to achieve a speed control, a 3-phase ac volt age controller is us ually employed. Fig. 12.27 (a) shows a three-phase ac voltage controller feeding a three-phase induction motor. By controlling th e firing angle of the thyristors connected in antiparallel in each phase, the rms value of the stator voltage can be regulated. As a consequence, motor t or que and thus speed of the drive is controlled. In Fig. 12.27 (b ), for load torque T L , a is the operating point at rated voltage and Speed A
0-----'" A B
8 0--
-.
Rated vol ta g e
C o-----... T2
(a )
O ~~
_ _ _ _ _ _ ___
(b)
Torqu e
Fig . 12.27 . (a ) Three-phase ac voltage contr oller feeding a 3-phase inductio n motor (b) Speed-torque characteristi cs as effe cted by :~ator volt age control.
[Art. 1_.8)
Electric Dr ives
OA is the motor speed. For reduced stator voltage (K = 0.5), b is the operating point and OB is the reduced motor speed for load torque T L . This m ethod is suitable for motors having large value of Sm' For 10w-2lip motors, the range of speed control is very narrow.
Stator-voltage-c ontrol method offers limited speed range . It is u sual to use 3-phase voltage controllers. Their use, however, introduces pronounced harmonic contents and input supply power factor for the voltage controller is quite low. These are, therefore, useri for low-power drives like fans, blowers and centrifugal pumps requiring low starting torque. For these types ofloads, the load torque is proportional to speed squared and input current is maximum when slip s = 1/ 3, this is proved in Example 12.22. Example 12.22. (a) A three-phase SCIM drives a blower-type load. No-load rotational losses are negligible. Show that rotor current is maximum when motor runs at a slip s = 1 / 3. Find also an expression for maximum rotor current. (b) If three-phase SCIM runs at speed of (i) 1455 rpm and (ii) 1350 rpm, determine the maximum current in terms of rated current at these speeds. The 1M drives a fan and no-load rotational losses are ignored.
Solution. (a) The torque required by a blower-type load is proportional to speed squared.
= (1- s) Pg are negligible, Pm = power required by load.
Mechanical power developed in motor, Pm As no-load rotational losses
(l-s ) P g =TL
or 312
2 r2
S
· Wm
s) = T L . rom
(1 -
1 _ [rom. TL . s]1/2 2 - 3r2 (1- s)
or Bu t rom
= ros (1 ~ s) and TL = k
(J)m 2.
...(i )
Substituting these in (i), we get,
2 1 =[ros(l-S).TL.S]1/2 =[ (J)s·k,wm2 ·S ]1/ . 2 3r2 (1 - s) 3r 2 _
S .
- rom [
-
or
[ S . k . ros ]112 k . ros 1/2 _ 3r (1 s) ro 3r s ] 2 2
12 =...fS (l - s) [
k ·w 3]112 s 3r2
...(i i )
The slip at which rotor current 12 becomes maximum can be foun d by obtaining
~2 and
equ ating it to zero.
['k OJ 3]1I2+{S (-1 ) [-kWs 3] =0
1 1
s dI 2 - = - ' r= (1-s ) ds 2 'l( S 3r z
or
l- s
-
2
- =s
or
3 T.~
·
1
s= -
3
.. .(i i i>
626
[Art. 12.8]
PO'dtr
This SDOYVS. that 12 is maximum at a slip of putting s =
l
in Eq,
l
The maximum value of 12 is obtained by
(ii) . 1112
1 ..
fl. ~ kCJ)s 3 = ~ , CJ) [k CJ)s] -\J '3 3 [ 3r2 J 9 s r 2
2 · mx
For 1455,
I, =..fS (1- s) [k
full-lo_~9
...(iv)
~;,.'r'
slip SI
= 1500 - 1455 = 0 03 1500 .
1 From above,
112
=-
(b) From Eq. (ii), (i)
Electronics
2 , mx _
12 , r
-
_f12 _fT2 .-\J~ . 3 3 _ 3 3 _ 2 291 'fs'; (1 - SI) - ~0,03 (1- 0.03) - . -\J~
Here 121' is the rated, or full-load, rotor current. (ii)
"
For 1350 rpm, full-load slip .;" 1500 - 1350 _ 0 1 1500 -.
SI -
12 , mx
~~
-[-2,-r - -'0-.-1~(-1~_-0L-.1-)
".
I
= 1.352
12.8.2. Stator Frequency Control By changing th e supply frequency, motor synchronous speed can be altered and thus t orque and speed of a 3-phase induction motor can be controlled. For a three-phase induction motor, per-phase supply voltage is VI =..J2 1t (1 Nl
... (12,62)
.
Etectric Drives
[Art, 12,8]
627
. ., 47t fl 2Wl . ;Synchronous speed, Ws = -p = p rad/s
Te =
Motor torque,
ros .12 -;r2
.3 ·
2
... (:;.2.63)
Slip,
Here f2 angw2 are the rotor frequencies in Hz and radls respectively. Substituting the value
W
.
of slip s = ~ in Eq. (12.63), we get WI '
2
3P V 1 . WI Te = - · ? 2w l r2 ' OJi 2 --2-
W2
r2
. 2
+ (:)1 (l1 + Z2) . ','
~
2
3P V 1 'W 2 = --2' 2 2 2 . r2 2Wl r2 + W2 (ZI + [2)
Slip at which maximum torque occurs is given in Eq. (12. 54) as
... (12.64)
r2
... (12.65)
s mT--+Xl
X2
Rotor frequency in radls at which maximum torque occurs is given by WI . r2
w2m
r2
= SmT ' WI = WI (Z 1 + Z)2 = -Z-Z 1+ 2
Note that w2m does not depend on the supply frequency WI' Substituting r2 in Eq. (12.64) gives maximum torque T e.m as 2 3P V~ . W2m . (l1 + l2) T e.m = --2 . 2 1 2 2 l Z 2 2Wl W:!m (l1 + 2) + W2m ( 1 + 2) v12 3P = 4W12 . ZI + l2 '
Eq. (12.66) indicates that T
em
' Wl
2
T em
3P
= w2m' (l1 + Z2)
... (12.66)
is inversely proportion al to supply-frequency squared . Also,
V/
= -'-4 II + Z2
At given source volt age VlI 3{ . Zl:':2 is constant, therefore, T em
.
w/ is also constant. As the
operating frequency WI is incr eased, Tem . 0012 remains constant but maximum torque at increased frequency WI gets r educed as shown in Fig. 12.28. Supposing rat ed frequency for a motor is 50 Hz and Te . m = 100 Nm, If the mota is n w oper.a ted at 100~, then 100 (2
1t
x 50)2 = (new maximum '\
Power E lect ron i s
[A r t. 12.8]
628
torque) (21t x 100)2 or the maximum torque at increased frequency of 100 Hz is 25 Nm. Such type ofIM behaviour is similar to the working of dc series motors. With constant voltage and increased-frequency operation, air-gap flux gets reduced; therefore, during this control, IM is said to be working in field-weakening mode. Constant voltage and variable frequency control of Fig. 12.28 can be obtained by feeding 3-phase IM through three-phase inverters discussed in Chapter 8.
S peed ~
2.5
~
(a ) the motor speed at rated load (b) the slip at which maximum torque occurs and
/
• W 2 =consta nt
T
e.m .
1
Q. III
Example 12.23. A 3-phase, 400V, 15kW, 1440 rpm, 50 Hz, star-connected induction motor has rotor leakage impedance of0.4 +) 1.6 n. Stator leakage impedance and rotational losses are assumed negligible. If this motor is energised from 120 Hz, 400V, 3-phase source, then calculate
2 .0
;/' Tem =constan t , , I I
o~~----~
_________
Torque Fig. 12.28. Speed torque characteristics of a 3-phaseIM with stator frequency control with constant supply voltage.
(c ) the maximum torque .
Solution. Full-load torque, Te .(l (a)
..
=p
=
41t x 120
4 = 120 1t rad/ s .
= 0.4 +) 1.6 x
Rotor impedance at 120 Hz
T
20 150 = 0.4 +) 3.84 =) 3.84 n
= 99.472 = _3_ .. (400/ ~)2 x
~
(l
120 1t
.
(3.84)2
0.4
S
1 0.4 ( 400 ')2 s = 401t x 99.472 x T3 x 3.8i = 0.1157 .
: . Full-load slip, 1111 J.V. otor spee d a t rate dId oa
-_ 120 x 1.20 (1- 0.1157) -_ 3183.48' rpm 4
Slip at which maxim wn torque occurs is given by s
(c)
:m = 21t x6~440 x 15000 = 99.4 72 Nm
At 120 Hz, synchronous speed,
41t fl Ws
(b )
=
From Eq. (1 2.5 6), wit
Xl
m
= 0,
= xr2 = ~ = 0 1042 3.84 . 2
m aximum torque,
Te m = ~. V 1 = _3_. (400/ -J'3Y Ws 2x2 120 1t 2 x 3.84 2
= 55.262 N m.
12.8.3. Stator Voltage an d F r equ ency Con t rol FOT
a 3-phL.s e 1M , st ator vo tage per ph ase is given by
VI = -.f2 1t fl
. N phl '
~ . k wl
" .(12.66)
Electr ic Drives
[Art. 12.81
629
It is seen from above equation that if the ratio of supply voltage ' \ to supply frequency fl is kept constant, the air-gap flux remains constant. From Fig. 12.25 (6 ) and Eq. (12.50 ), the starting torque is given by
3
T e.st = -
.
(Os
"1
2
2
(r1 + r 2) + (Xl + X2)
2 . r2
. 20)1
As (r1 + r2) « (xl ~ X2) and (O~ = P' we get
".(12.67 )
From Eq. (12.56), maximum torque is given by
T
::::~ .
e.m Ws
2
V1 2 (xl + x2)
3P
V12
= 2(01
. 2 . 001 (l1
+ l 2)
3:[::J !
... (12 .68)
=
Il I, Eq. (12 .68) shows that if V1 l wl, or air-gap flux <1>, 1:S kept constant, the maximum to rque remains unalt ered . Eq. (12,67) indicates that starting tor que i inversely proportional to supply frequ ency Wv even if air -gap flux is kept constant. At low values of frequ encies, the effe ct of resistances cannot be neglected as compared to the reactances . This has the effect ofreducing the magnit u de of maxim um torque at lower frequencies as shown in Fig. 12.29 . In practice , at low frequencies , the supply voltage is increased to ma' ntain the level of maximum torque. This meth od of speed control is also called volts / hertz control. Sp ~"d I.J <.J W If stator r esistance is neglected, then from Fig. 12.25 (6 ), ws 1> 52:' 53 - 54 the slip at which maximum torque occurs is given by r2 s =-- m Xl + X2 r2
=---(01
A s the supp ly frequency maxim um torqu e incr eases.
(ll (01
+ l2)
... (1 2.69 )
is r educed, the s lip at
In Fig. 12.29, load torque TL fo r a certain load is also shown. It is seen from this figure that as both voltage and \ . fr equen cy are varied (usu ally b elow their r ated values ), ~s peed of the drive can be controlled. The centrol of both voltage and fre qu ency can be carried out (so as t o keep
7V
const an t) 'through t h e use of thr ee-ph ase invert ers or
Torque
Fig. 12. 29. Speed-torqu e
characteristics of a :' -ph ase with 'lo1ts/hertz contr ol
....
mI
630
[Art. 12.8]
Power Electronics
cycloconverters. Inverters are used in low and medium power drives whereas cycloconverters are suitable for high-power drives like cement mills, locomotives etc. Variable voltage and variable -frequency can be obtained from voltage-source inverters . Four such circuit configurations are shown in Fig. 12.30. In Fig. 12.30 (a), three-phase ac is converted to constant dc by diode rectifier. Voltage and frequency are both varied by PWM inverter. The circuitry between the rectifier and the inverter consists of an inductor Land capacitor C, called rilter circuit. The function of filter circuit is to smooth dc input voltage to the inverter. This Circuitry in between rectifier and inverter is called de link. In Fig. 12.30 (a) , -regeneration is not possible because of diode rectifier. Also, inverter would inject harmonics into the 3-phase ac supply. In Fig. 12.30 (b), three-phase ac is converted to dc by diode rectifier. Chopper varies the de input voltage to the inverter and frequency is controlled by the inverter. Use of chopper reduces the harmonic injection into the ac supply. Regeneration is not feasible in the scheme of Fig. 12.30 (b). Fig. 12.30 (e ) uses a 3-phase controlled rectifier, dc link consisting of Land C and a force-com mutated VSI. Voltage is regulated by controlled rectifier and frequency is varied within the inverter. Here regeneration is possible if three-phase full converter is used. Regeneration is also feasible in the scheme shown in Fig. 12.30 (d ). It uses a 3-phase dual - converter, L-C filt er and inverter. Level of dc input voltage tothe inverter is regulated in dual converter whereas frequency is varied within the VSI inverter. 3 phasE'
PW M In vertert-----t
ac supp ly
(0 ) Chopper
3 phasE'
ac
Inverter :------i
supply
(b)
L 3' ph.ase
ac sup pl y
f~
In v ert er r--------; r-------~----~
Dual c-onver ter
Cd) _ Fig. 12.30. Three-J5hase induction motor speed control through voltage source inverters.
Electric Drives
[Ar t. 12.8J
631
It may be observed from above that voUs/hertz control crffers speed control from standstill up to rated speed ofIM. This method is similar to the armature-voltage control method used for the speed control of a dc motor. Example 12.24. A 3-phase, 20 k W, 4-pole, 50 Hz, 400 V delta connected induction motor has the following per phase parameters referred to stator: rl
= 0.6
n, r2 = 0.4 0,
Xl =X2
=1.60
Its magnetizing reactance is neglected. If this motor is operated at 200V, 25 Hz with DOL starting, calculate (a) current and pfat the instant ofstarting and under maximum torque conditions; compare the results with normal values, (b) starting and maximum torques and compare with normal values. Solution. The subscripts n and d are used here for normal operation and for reduced-voltage reduced-frequency operatiqn respectively. (a) Starting current: 112
= 119.31 A
+ (3.2)']
(b)
Starting torque: .T
= en
Ws
= Ted
l... 12 r 2 n
S
5~n (119.31)' (Oj4)= 108.75 N m
3 2 0.4 = 251t (106) X T
=171.673 Nm
It is seen fro m above that at reduced-voltage reduced-frequ ency oper ation (i) starting current decreases, h owever, starting torque becomes more, (ii ) power factor at st arting is improved, therefore power input to motor is more. It can be inferred from abov e tha t with reduced-voltage and reduced-frequency (keeping V I f or flux constant) oper ation, th e per form ance ofIM at starting is improved. Maximum torqu.e. The slip at which maximum torque occurs is given by
~ -- Ii'a. 62, + ,'3.22
632
Power Ele tronics
[Art. 12.8] Smn=
I
~ .4
\i 0.6 + 3.2
2 =0.123 400
(i) current gets reduced whereas pf is improved under maximum torque conditions, (ii) the maxi mum torque , however, gets reduced. and (iii) maximum torque occurs at higher value of slip. 1~. 8.4 .
Stator Cur ren t Cont r ol The developed torque and therefore th e speed of a 3-phase 1M can also be controlled by stator-current control instead of stator-voltage control. The behaviour of a motor with stator-current control is differ ent frem that obtained with voltage-source inverter. Consider a const ant current flfed into the stator windings of a three-phase 1M. Fig. 12. 31 (a ). So far as r ot or current 12 ic:: conc erned, stator leakage impedance (r1 + jx 1) plays no role. Therefore, the effect of (rl + jXl ) can be omitted when studying constant-current mode of motor operation. However, (r1 + j x 1) does influence the magnitude of applied stator voltage VI ' For stator curr ent I v th e r otor current 12, fr om F ig. 12.31 (a ), is given by ... (12.70)
[Ar t. 12.8]
Electric Drives
633
Speed Ws~~~~~~~~----
---==::.-",.-;' ~~
Torque at rated 11 without saturation
+ VI
-1--+---1
O
W-~'-L
_________________ Torque
(0) (h) Fig. 12.31. (0) 3-phase I.M. per-phase equivalent circuit (b) speed-torque curves with stator-current control.
The internal, or developed, electromagnetic torque Te is
T
!:!
e
1.. . 122 '2 (j).~
3
= ro•. (: '
-. I
8
(I 1 X"i
. . '2
J
+ (X2 + X,,.)'
.. .<12.71)
,
Invoking the principle of m aximum power tr ansfer in the equivalent circ'I,lit of Fig. 12.31 (a.) with '1 + jXl = 0 and with constant cur rent 11, we get
or
S
m
'2 =--m x 2 +X
... (12. 72)
Eq. (12.72) gives the slip at which maximum t orque occurs when 3-phase 1M is fed from .
,
current-controlled source. Maximum torque Tell! is obtained by substituting 2.. .
Sm
=X2 + Xli! in Eq.
(12 .71).
...(12.73)
Also, ... (1 2 . i4)
It is se en fr om Eq . (12. 74) t at maximum torque is (i) pro portional to stator-curr ent squared, (ii ) independent of supply frequency fl and (iii) independent of rotor resista n e
634
[ r t 12.8]
Power EJ ctron ics
The starting torque, from Eq. (12.71), is (II' Xm)2
3
T e . st =-· Ws
2 r2
+ (X2 + X m)
2 ·
~ .. (12.75)
r2
The speed-torque characteristics for different stator currents are shown in Fig. 12.31 (b). For comparison purposes, speed-torque curVe at rated voltage is also sketched. At rated current II = 1.0, starting torque is very low as compared to that obtained with rated voltage VI == 1.0. At starting, slip s = 1.0, r2 + jX2 is quite low in magnitude producing almost a short-circuiting effect leading to very low current through Xm and therefore very low air-gap flux and low stator voltage. , r As a consequence, starting torque is quite small. As speed rises or slip falls, ~ + j x 2 rises, current s
throughXm increases and as a result, stator voltage, or air-gap flux rises leading to higher torque. With no magnetic saturation, torque rises to quite a high value as shown by dotted line. In practice, the saturation will limit the peaking in maximum developed torque as shown by solid curve for II = 1.0. The speed-torque curves for II = 0.5 and 2.0 are also shown in Fig. 12:31 (b). A constant current for the 3-phase 1M can be obtained from a 3-phase current source inverter (CSI). The advantages of torque and speed control by CSI. fed 1M are (i) fault current level control and (ii) current input is almost unaffected by motor parameter variations. The disadvantages of urrent-fed drives are (i) generation of unwanted harmonics in the system and (ii) torque pulsations. Fig. 12. 32 shows two circuit configurations for current-fed inverter 1M drives. In Fig. 12.32 (a), 3-phase controlled rectifier gives out controlled dc voltage. Inductor L converts this voltage to constant current. CSI regulates the output frequency and ther efore the torque and speed of 3-phase 1M. In Fig. 12.32 (b), uncontrolled de voltage is r egulated by chopper which is then conv81ted t o current source by inductor L. As before, CSTthen controls the torque and speed of three-ph ase I. M.
I
3 phase
ac supply
L
t---,----~--ll
Current source in v er ter t---../
(a)
3 ph a se
Cu rrent
ac
so ~ e
supp l y
inver t er
~-,./
(b) Fig. 12 .32. (a) 3-phase 1M speed control through current-source inverter.
Example 12.25. A 400 V, 4-pole, 50 Hz, 3-phase, star-conn ected induction m otor has ~1 = 0, Xl = X 2 = 1 .0, r 2 = 0.4 0 , Xm = 50.n, all referred to stator. This induction motor is fed from (i) a constant- voltage source of 231 F pe p hase a'nd (ii) a constant-current source of 28A.
Electric Drives
[Ar t. 12.8]
635
For both parts (i) and (ii), calculate (a) the slip for maximum torqu.e (b) the starting and maximum torques (c) the supply voltage required to sustain the constant current at the maximum torque. Solution. The equivalent circuit for this motor is shown in Fig. 12.33 (a) . Its Thevenin' equivalent circuit is drawn in Fig. 12.33 (b) where Xe and Ye are given by Xe = 1 ~150 = 0.9804 n
Y~ = 2315~50 == 226.5 V
and
0.9804
In
1.Q.
n.
,n
0.4
0.4
s
5
(b)
(a)
Fig. 12.33. Induction motor per-phase equivalent circuit, Example 12.25.
It is seen from the equivalent circuit of Fig. 12.33 ( b ) that the slip at which maximum torque occurs is given by (a) (i)
sm
(ii)
=
1.~8404 = 0.202
For constant-current operation of 1M, slip Sm is given by Eq. 12.72.
..
smT
(b) (i)
= 1 ~~O = 0.00784
. . 41tfl
Synchronous speed,
(Os
For constant voltage input, .. Te st
=p =
4rt
X
4
50
=50rt rad/s
is given by
Te . st
3
=ro .12st s
2
; r2
226. 52
3.
= 50rt · ·0.4-? + 1.98042 x 0.4 = 96.012 Nm T
3
y2
(Os
2 (x2 + Xe)
=_. e·m
e
3 (226.5)2 . - 501t 2 (1.9804) = 247. 3'7 Nm (ii)
For con stant CUl'Tent input , Te. st from Eq. (12.75) is · 3
(28
50)2 _ 2 xO .4 ==5 .7D6 Nm 1t 0.4 + 51
Te. s t = SO- .
X
2
636
Power Electrorlics
(Art. 12.8]
Maximum torque, from Eq. (12 .73) is 3 (28 X 50)2 Te m = 50n' 2 x 51 = 366.993 Nm It is seen from above that for constant-current mode, Te.st is much smaller (5.756 Nm) as
compared to Te at for constant-voltage mode (96.012 Nm) . But maximum torque for constant-current mode is much more (366.993 Nm) than its value for constant-voltage mode (247.37 Nm). Also the slip at which maximum torque occurs is very low (0.00784) for constant· current mode as compared to its value for constant-voltage mode (0.202). (c) At maximum torque, sIlIT:: 0.00784. From the equivalent circuit of Fig. 12.33 (a), the magnetizing current 1m is given by
Supply voltage required to sustain constant current of 28A is given by V 1 :::: ..J3 (19.806) (50) =1715,2 V This shows that maximum torque at low value of slip necessitates a large value of source voltage which is much higher than the rated source voltage, But acquiring such a large voltage is not feasible. Thus, a large value of maximum torque at low slip is not a practical possibility. In actual practice, saturation occurs and the magnitude of Xm is reduced. As a consequence, the value of maximum torque under constant-current mode has a much lower value than that computed here. Summ ary of Characteristics of Adjustable-frequency Ind uction-motor Drives. Speed-torque characteristics of a three-phase induction-motor drives depend upon the methods of control techniques employed. For different stator frequencies , a family of speed-torque characteristics as shown in Fig. 12.34 can be obtained. These char acteristics can be subdivided into three regions; constant-torque region, constant-power region and high- speed series-motor region. A summary of these characteristics is reviewed here. Per unit spt'ed
r
w1=3,0 High speed series motoring region
2< w,< 3 I
~ Wl=2.0 ~-,
\/ em ex
I
l
t
UJ I<
l
H -....,.C-.__\'\
\
-
WI
Me x m . a Il_
- - - -.
~.
-r
:::onstant power r€'g io n
torque
w, =1. 0 h~=~::::::=---):---------t Const an t torque r egio n
Va I
,
Maxm . a llow ab l e torque
~
Torq ue
Fig. 12.34. TyPical speed-to qu~ characteristics of a 3-phase ind uction m otor with variable voltage and variable-freq uency power supply .
[Art. 12,8]
Electric Drives
637
Constant-torque region. As explained before, this region of constant torque can be obtained by voltslhertz control as shown in Fig. 12.29. In the low-frequency range of speed, the effect of stator resistance is compensated by a boost in the stator voltage as shown in Fig. 12.35. In this region, stator current is kept constant at its rated value. Power, equal to the . product of constant torque and speed, varies linearly with speed as shown. Slip frequency remains constant during this region.
t
I . r
Constant power, P ._L._·.~
.
I---I-l--~t.-~.....;----.....;:"....,.....--......j,"'-
/
/
/p
.
.
'-.
-'-_
! -~~
-- -
... / ' ,_----1 / ----7----7--- T- S, ip frequency
•
_~;
J
~/
•
!
W2
.
I
,
2
~ Constant
torque
ConSlant power .-----t-reg I on I
region
3 per unit speed series I speed motoring region --t
---k- High
Fig. 12.35. Stator voltage, current, slip-frequency, torque arid power variation with speed for speed-torque characteristics of Fig. 12.34.
Constant-power region. When maximum speed, called base speed, is attained in the constant-torque r egion, stator voltage reaches its rated value. Motor speeds beyond base (or rated) speed are obtained by keeping stator voltage constant and lowering the stator frequency. Torque in a 3-phase induction motor is given by
Te
=
~
CPower input to rotor)
(Os
3
=w·Ezl2cos92 s
At low rot or frequency, cos 82
(=
_I
r
'1'2
_'2 '2 +(21tf2i 2 s )
3
... (12.7 6)
T e =- · Ezl 2 (Os
E2
But ro tor current, 12 =- - '2
.
-+J S
· At sma 11 S1iP':"
T2
S
» X 2'
Jis almost nearer to unity.
X2
Th"18 gIves I 2 =- sE,)
'2
638
Power Electronics
[Art. 12.8]
Substituting this value of 12 in Eq. (12. 76), we get
3 sE/
T =-· W~
e·
'2
3
. WI'
= 3P 2'2
~
3P
E22
002
=Ws
E22
--r; = 2001 . WI . --r;
. ( E2J2 . 00 WI
...(12.77)
2
Emf induced in rotor at stand-still, E2
= ~1t fl
. Nph2 . ~. kW2
. WI
. = T2 . Nph2 . ~ . kW2 E
This gives air-gap flux, ~ Cc ~ WI
From Eq. (12.77),
Te
00
3P 2 2r . ~ . ~
2
or If stator
imped a~ce
-:
Te =K . ~2 . 002 is neglected, then
VI
...(12.78)
=..J2 1t fl . Nphl .4> . kwl
WI
=T2 . N phl . ~ . kwl :. Air-gap flux, From E q. (12.78),
VI
~oc WI
Te
oc
K[ V1J2. ~ WI
,I
2( ,
Vi . 00 =Kl . ~ VI ~ J Te == KI 2" W2 y,ll
or
OOl
1
...(12.79)
1
Eq. (12.79) gives approximate value of motor torque for a 3-phase 1M working in the low-slip range. It shows th at if 1M is operated at const ant stator voltage VI and constant rotor frequency (02' the motor t orque is inversely proportional to supply-frequency squared. In case slip (or rotor) frequency ~ is incr eased linearly as WI is increased to obtain high-operating speed
(=2;' rad/ s} then :: i. constant and therefore T, . "'1=K . V,' [~ ]remainS constant
giving a constant power charact eristic. During- const ant power operation, stator voltage is kept con stant bu t stator fr equency is raised. As a result , stat or flux decreases or air-gap flux is weakened . In view of this, constant-power mode of a 3-phase 1M is also called rield -wea.~e ni ng m ode of an 1M. The upp er limit of constant-power mode is reached when maximum working value of rot or frequency is r eached. High-sp eed seri es-motor in g region. From Eq. (12.79), T e::;; Kl '
V/ '2002 00 1
... (12 .79)
[Art. 12.8]
.Electric Drives
639
After th e constant-power region, high-speed series-motoring region is obtained. In this region, stat or voltage VI and rotor frequency 002 are maintained constant at their maximum valu es. It is seen from Eq. (12.79) that under constant VI and 002 operation of a 3-phase 1M, output torque Te is inversely proportional to supply frequency squared or Te varies inversely as speed squ ared. In other words, Te . remains constant and series-type characteristics are thus obtained. As this region corresponds to high-speed series-type characteristics, operation under constant VI' ~ and variable WI is usually referred to as high-speed series-motoring region. In Fig. 12.34, speed-torque characteristics at different stator frequencies in the constant-torque, constant-power and high-speed series-motoring regions are shown. The maxim'um torque, indicated by dashed line, is constant below base speed and decreases inversely with speed above base speed and upto WI =2.0. In the constant-torque and constant-power regions, maximum allowable torque is shown somewhat lower than the maximum or breakdown torque Tem just as a matter of precaution because inverter current carrying capability is l~mited. Maximum allowable torque is indic_ated by solid line in the constant-torqu e region and by solid-curve in the constant-power region. E xample 12.26. Is it possible to obtain Ward-Leonard type ofcharacteristics from a 3-phase induction motor? Discuss. Motor field Arm . voltage Solution. Word-Leonard system of speed control control control for a separately-excited dc motor gives torque-speed and power-sp eed charact eristics as shown in Fig. 12.36. i Constant i Constant I power P lJ A comparison of Figs. 12.35 and 12.36 shows that torque Te Ward-Leonard type of characteristics as obtained in a I. I dc motor can be obtained from a 3-phase induction i ,/ motor. i / From base speed down to zero speed, the speed / /p control is obtained by (i) armature voltage control in a / dc motor and (ii) stator voltage and frequency control I I / in a 3-phase indu ction motor. In both, the flux is kept constan t by keeping (i) field current constant in a dc 0 2 per unit speed motor and (ii) V I f constant in a 3-phase 1M. Armature current in a dc motor and stator current in 1M are kept Fig. 12.36. Torque-speed and constant at their rated values. In both, constant torque power-speed characteristics for and variable power charact eristics are obtained from Ward-Leonard system of speed control. zer o to b ase speed. Above base speed, the speed control is obtained by field-weakening method in both dc and ac motors. Constant power and variable torque characteristics are obtained in both types of motors as shown. Arrnat ur.e current in dc motor and stator current in 3-phas e 1M are kept constant. In dc m otor, armature voltage is k ept constant whereas field flux is weakened by decreasing the field current. In 1M, stator voltage is ke pt constant and air-gap flux is weakened
w?
.r--------...t.
I
00
by increasing stator frequen cy but by keeping ~ constant. WI
12.8.5. Stati c Rotor-resist ance Contro In a slip-ring indu ction motor (SRIM), a 3-phase variable resistor R 2 can be inserted in th e rotor circuit as shown in Fig. 12.37 (a) . By varying the rotor circuit resis tan ce R2> th e motor torque can be controll ed as shown in Fig. 12.37 (b). The star ting torqu e and starting current can also be varied by controlling the r otor cir cuit resistan ce, Fig. 12.37 (b) and (c). The dia6.dvantages of his m ethod of speed control a e : (i) r educed effi ciency at low speeds ( ii ) speed
640
[Art. 12,8)
Power Electronics
Speed
3 phose supply 0 - - 11
o (a)
o
Torque
Stotor current
(b)
(c)
Fig. 12.37. Three-phase 1M speed control by rotor resistance (a) circuit arrangement
(b) effect on developed torque (c) effect on stator current.
changes very widely with load variation (iii) unbalances in voltages and currents if rotor circuit resistances are not equaL In spite of these, this method of speed control is used when speed drop is required for a short time, as for example in overhead cranes, in load equalization etc. The three-phase resistor of Fig. 12.37 (a) may be replaced by a three-phase diode rectifier, · chopper and one resistor as shown in Fig. 12.38 (a). In this figure, the function of inductor Ld is to smooth en the current [d' GTO chopper allows the effective rotor circuit resistance to be varied fo the speed control of SRIM. Diode rectifier converts slip-frequency input power to dc at its output terminals. 'tVhen chopper is on, Vdc ~ Vd = 0 and resistance R gets short-circuited. When chopper is off, V dc = Vd and resistance in the rotor circuit is R. This is shown in Fig. 12.38 (b). F.lom this fi gure , effectiv e external resistance Re is Re
Tall
where k = T
=
R· Tof[ . R (T- Ton) T = T =R (1- k)
.. .(12.80)
= duty cycle of chopper.
3-phase supply
Resistance
R
o --1 Ton :--Toff -i l--- T - - - l I
I I
I I
J--...-.____I ~~1 IIId R . ~T _____l
(a)
(6 )
Fig. 12.38. (a) SRIM control by static variation of external rotor resistance ( b) wavefonns pertain' ng to Fig. Ca l.
t
Electr ic Drives
[Art. 12.8]
641
alysis of induction mot or with ch opper control. The equivalent circuit for 3-phase 1M, diode rectifier and chopper circuit of Fig. 12.38 (a) is as sh~wn in rig. 12.39 (a). Ld
+
f
Re=R(1-k)
I
1-
__,1 =-1
(a) Ld
-
+
f
I1
Vdc
Re=R(1-k)
( b)
Fig. 12.39
(a).
Equivalent circuit for Fig. 12.38
(a), (b)
its approximate equivalent circuit.
If stator and rotor leaka~e impedances are neglected as compared to inductor L d , equivalent circuit of Fig. 12.39 (b) is obtained. Stator voltage VI when referred to rotor circuit gives slip-freauency voltage as VI '
5 .-
NI
.
N2 = s a VI
=s E2
where E2 =rotor induced einfper phase at stand-still VI = stator voltage per phase rotor effective turns per phase, N2 . a= t t ffi t' t h N = per phase turns ratio from rotor to stator. s a or e ec lve urns per p ase, I Voltage 5 E2 = 5 a VI' after rectification by 3-phase diode bridge appears as Vd (rectifier output voltage).
..
. 3Vml
Vd = - - = n
3·...J3Vmp n
3...J3 _'" =- . '12 . sa VI =2.339 saY l n
. .. (12.81)
where Vmp = maximum value of phase voltage = .J2 sa VI Total slip power = 3 s Pg • For n o losses in the rectifier, this must be equal to Vid '
.. P~r-phase
35 Pg =Vd1d
developed power, Pm = (1 - 8) Pg
Vd1d
= (1 -
Also,
s)
3s
Pm= Te . Wm = Te . W3 (l- s)
From Eqs. (12.82 ) and (12.83), we get
Vd l d (1 ..,. s) =T . ro (1 3s
e
s)
...(12.82 ) . .. .(12.83 ) "
642
Power Electronics
[Art. 12.8]
I d--
or
T e . CJ)s • 38 Vd
Substituting the value of Vd from Eq. (12 .81), we
ge~
3· s Te . Ws
= 2 .339 . sa V I = 1.2826 TL = 3Te , ld
Te' Ws
a VI
... (12.84 a)
Load torque where Te = motor developed torque per phase. Id
T
·
L =-::-=-=-=----,;--;2.339 a VI
.. .(12.84 b)
Eq. (12 .84) shows that inductor current Id is independent of motor speed. Assuming inductor to be ideal, dc voltage at the rectifier output, Vd = Id . R (1 - k). Vd = 2.339 s a VI = Id . R (1- k) I d · R (1- k) s = -.;....~::-----=-:;;2.339· aVI
wm = Ws (1- s)
_ [ _ Id . (1 - Ws 1 2.339 . a VI
From Eq. (12.81), Slip, Motor speed,
... (12.85)
R k)]
... (12.86 a)
R k)]
Id . (1 N r =Ns [ 1 - 2.339 . a VI
Also
Substituting the value of Id from Eq. (12 .84 b) in Eq. (12.86 a ), we get W m
=W
. T
s[
·
R
(1
-l~)l
L 1 - --=--=----(2.339 . a V I )2 CJ)s •
...(12.86 b)
For fixed value of duty cycle, speed falls as load torque TL is increased. Each diode in Fig. 12.38 (a) conducts for 120°. Th e waveform of rotor current i2 is shown in Fig. 12.40. For a ripple-free output current I d , it is seen from Fig. 12.40 that
271
wt
rms value of rotor curr ent, 12 =
~1/· 23 ~ = ~. Id 1t
.
...(12.87)
Fig. 12.40. Rot or current waveform
for ripple-free I d .
Rotor c rrent r eferred to stator ,
11 =N2 12 = aI2 =a . Id . - (2 NI ~3
...(12 .88 )
F ouri er analysis of the waveform in Fig. 12.40 can be obtain ed from Eq. (8.61 ).
2 b1 = 1t
5lt/ 6 .
2
lt/ 6
1t
J
Id sm wt · d (rot) =-Id
1- coa
This gives fundamental compon ent of rotor current as
2..[3
=Id
f1tl 6
rot I
It/6
7t
I
Electric D r ives
[Art. 12.8] :~
.. . (12.89) Fundamental component of r otor current r eferred to stator
N2
a
-..f6 . a ·l
... (12.90) d Nl 7t Example 12.27. A 3-phase, 420V, 4-pole, 50Hz, star-connected SRIM has its speed controlled by means of GTO chopper in its rotor circuit. The effective phase turns ratio from rotor to stator is O.B. The filter inductor makes the inductor current ripple free. Losses in the recti{zer, inductor, GTO chopper and no-load losses of the motor are neglected. Load torque, proportional to speed squared, is 450 Nm at 1440 rpm .. (a) For a minimum motor speed of 1000 rpm, calculate the value of chopper resistance R. For the value of R obtained in part (a), if the speed is to be raised to 1320 rpm, calculate. (b) inductor current (c) duty cycle of the chopper (d) rectified output voltage (e) efficiency in case per-phase resistances for stator and rotor are 0.015 nand 0.02 n respectively. Solution. Per-phase stator voltage,
420
VI =T3 = 242.5 V
III = -
Load torque at 1000 rpm, Synchronous speed,
.TL Ws
121 =
.121 = -
1000)2 =450 ( 1440 = 217.01 Nm
=
Minimum motor speed, wm . mn =
27t x 1500 60 = 50 7t rad/s 27t X 1000 60 = 104.72rad/s
From Eq. (12.84 b), the inductor current,
. 217.01x507t
Id = 2.339 x 0.8 x 242.5 =75 .122 A
For minimum motor speed, duty cycle k = 0 in Eq. (12.86 a).
Id . R (1 - 0)]
. _
. • {Om· mn = (Os [ 1 2.339 a V (a)
l
75.122XR] 104.72 = 50 7t [ 1- 2.339 x ·0.8 x 242.5 , This gives
R = 2.0134 n
(b) New load t orque at 1320 rpm, TL
1320)2 . =450 ( 1440 =378.125 Nm
Inductor current from Eq. (12.84 b) is 378 .125 x 50 7t .. ~ Id =2.339 x 0.8 x 242. 5 = 130.890 A 27t x 1320 . (c) (Om = 60 = 138.23 rad/s . F rom E q. (12 .86 a ),
138.23= 501t [ 1 _ 130.8.95 x 2.0134 x .(1 -2. 33 9 x 0.8 x 2 .2. 5 = 50 7t [1 - 0.5808 (1- k )]
:. Duty cycle of chopper, k = 0.7934
k) ]
644
Power Electronics
[Art. 12.8] ' = 15001320 81 IP, s 1500
(d)
=012 .
From Eq. (1.2.81), Vd = 2.339 x 0.12 x 0.8 x 242.5 = 54.452 V
(e) Power loss in chopper resistance =Vd Id = 54.452 x 130.895 = 7127.5 W
Also, power loss in chopper resistance
=Ii· R (l-k) =(130.895)2 x 2.0134 (1- 0.7934) =7127 W From Eq. (12.87), inductor current referred to rotor,
_12
_12
..
12 =·V 3 . Id ='13 x 130.895 = 106.88 A
Total rotor ohmic loss :;:: 312 2 r2 = 3 x 106.882 x 0.02 = 685.4 W
From Eq, (12.88), stator current,
11 =al2 = 0.8 x 106.88 = 85.504 A _
Total stator ohmic loss = 3112rl =3 x 85.5042 X 0.015 =329 W
Power output =TL' <.om = 378.125 x 21t x601320 =52268.25 W· = 52268.25 + 7127.5 + 685.4 + 329 = 60410.15 W
Power input
Efficiency =
~~!~~:~~ x 100 = 86.52%
Example 12.28. A 3-phase, 400 V, 50 Hz, 960 rp m, ·star- connected S RIM has the following per-phase parameters referred to stator: r1=0.10, r2 =O,080, X1 =X2=0.3 0,Xm =Q. Per-phase turns ratio from rotor to stator = 0. 7 Speed of this motor is controlled by a GTO chopper in its rotor circuit. For a speed of 800 rpm, the inductor current is 110 A and the chopper resistance is 2 n. Calculate (a) the value of chopper duty cycle (b) efficiency for a power output of 20 kW and for negligible no-load losses. (c) the input power factor.
Solution.
(a)
Per-phase voltage =
. Synchronous speed, F rom Eq. (12.86 a), • I
W
= 230.9 V
Ns = 1000 rpm [ llOx 2 (1- k) .. ] 800 - 1000 1- 2.339 x 0.7 x 230.9
This gives chopper duty cycle, k =0.656 (b) Powe loss in chopper =Ii· R (1 - k) = 1102 X 2 (1 - 0,656) = 8324.8 W Rms value of rotor current r eferred to stator, fromEq. (12.88), is
.
.
-
_12
~ f2
11 = a ' Id' '1 3 = 0. 7 x HOx '1 3 =62.87 A
Power loss in stato and rotor resistances =3 (62.87)2 x (0 ,1 + 0.08) =2134.4 W ./ Power input = 20,000 + 8324.8 + 2134.4 = 30459.2 W 20,000 Efficiency = 30459.2 x 100 = 65.66%
~
Electric Drives (c) From Eq.
[Art. 12.8]
645
(12.90), fundamental component of rotor current refe~red to stator is ...[6
-.f6
7t
7t
III =-. a ·Id = - . (0.7) x 110=60.04 A Power input Input
\ =...f3 x 400 x 60.04 x cos 9 = 30459.2 W
30459.2 .
pf= x 400 x 60.04 = 0.7323 lag
ra
12.8.6. Slip-Power Recovery Schemes In chopper method of speed control for SRIM, the slip power is dissipated in the external resistance and it leads to poor efficiency oithe drive. However, instead of wasting the slip power in the rotor circuit resistance, it can be conveniently converted by various schemes for the speed control ofSRIM. Two important slip-power recovery schemes are static Kramer drive and static Scherbius drive. These are now discussed in what follows. 12.8.6.1. Static Kramer Drive. The circuit configuration for static Kramer drive is shawn in Fig. 12.41. The slip-frequency power from the rotor circuit is converted to dc voltage which is then converted to line frequency and pumped back to the ac source. As the slip power can flow only in one direction, static Kramer drive offers speed control below synchronous speed only. 3-phase supply
V, N2
°1=
N1
L..-"""-'T"""1--'
V2
c---t----t----. Slip power
Diode rectifier
Line commutated . invertE'r
Fig. 12.41. Static Kramer drive.
The slip power from ro tor is rectified to dc voltage by diode bridge. Inductor Ld smoothens the ripples in the rectified volt age Yd' This voltage Vd is then converted to ac volt age at line frequency by line- commutated inverter and fed ba'ck to 3-phase supply. As the power fl ow is fr om r otor c'rcuit t o supply, static Kram er drive offers constant-torqu.e dri ve. As stated befor e, this sch eme offers speed control below synchronous speed only. Simplified torque and speed expressions for th is drive can be derived as follows . Rot o voltage per phas e = SE 2 where E2= per pha&e rotor e .ro .f. at st andstill s = slip an d
Power Electronics
{Art. 12.8]
646
Voltage sE2 is ectified to Vd by diode bridge. Uncontrolled output voltage of diode rectifier, from Eq. (6.54), with a = 0, is _ 3 x maximum value of input line voltage V d ·It
=
3 -.f2 (..J3 sE 2) 1t
3 ...rtf =--sE 2 1t
N2 E 2 =VI N =a VI
But
I
effective rotor turns per phase, N2 a. - effective stator turns per phase, NI VI = supply voltage per phase. 3...[6 Vd = -_. saVI ,= 2.339 saYI
where
1t
.. .(12.91)
For three-phase line-commutated inverter, average dc output voltage (with no transformer) IS
Vdc = ~
3..[6
7
VI . cos
a
= - 2.339 VI . cos ex
.. .(12.92)
Here negative sign is used to confinn to the firing angle range. At no lo ad , e ectric torque Te is negligible and dc link current Id is almost zero . Consequently, the two direct voltages of Eqs. (12.91) and (12.92) must balance. Thus, 2.339 saVI
or
= - 2.339 VI cos ex 1 a
Slip,
s = - - cos a
... (12.93)
If a = 1, slip s = - cos ex. For a = 90 0 , s = 0 (speed synchronous) and for ex = 180 0 , s = 1 (speed zero). This shows that no-load speed of the motor can be controlled from near stand-still to full speed as firing angle a of line-commutated inverter is varied from almost 180 0 to 90 0 • During the analysis, we proceed from 3-phase stator voltage VI to de voltage Vd and also from 3-phase voltage V I to three-phase transformer, to the line-commutated inverter and to Vdc:
In practice, rotor circuit volt age is less than stator voltage and therefore, a < 1. Thus, a 3-ph ase transformer is often r equiTed between the ac supply network and the inverter in or der t o step down the supply voltage t o a level th at is appropriate t o Vd . Let t he transform er turn-r atio be aT where per phase output voltage of transfor mer as input voltage t o invert er, V 2 per phase supply voltage, VI
aT = --------------------.-------.---~--~~----------------
Note th at turns r atio CXr is usually ess than unity. AC volta ge across inverler t erminals, V 2 = aT ' VI ' The inverter dc voltage Vde • from Eq. (12.92), is given by Vdc = -
=-
3.,,[6
316
----;t . V2 ' cos a =- 1t ' aT' VI cos ex 2.339 . a-r ' V l cos a
... (12.94)
[Art. 12.8]
Electric Drives
6-47
...:.:~~:~~,':J'~~.'! .-:.
With the use oftransformer, from Eqs. (12.91)
2.339 saV1 = - 2.339 aT'
an~ (12.~4),
.
i-\ cos a
.
we get'.
. aT slIp, s:;;: - - cos a
or
... (12.95)
a
In order to develop motor torque, a rotor currentI2 is required and the rectifier rotor voltage Vd must force a current Id against the inverter dc voltage V de' When the motor is loaded, speed, falls and the increased rotor voltage sE 2 can overcome the voltage drops in the rotor windings, in the dc link circuit and the converters.
If resistance of the rotor circuit and inductor Ld are neglected, then total slip power,
3sPg 3s . (Os . Te
= Vdc . Id = Vdc . Id
T
= Vdc·ld
or
e
3s·
... (12.96)
Ws
Substituting in Eq. (12.96), the valu es of s from Eq. (12.93) and V de from Eq. (12.92), we get
Te
=
2.339 VI . cos a· Id 1 3· _. a cos a· wS
2.339 a VI . Id
=- 3 - .
Ws
... (12.97)
When a transformer is used in Fig. 12.41, then substitute in Eq. (12.96), t he values of s from Eq. (12 .95) and Vdc from Eq. (12.94) and we get
Te =
2.339 . aT ·V1 cos a . Id 2.339 a VIId - - 3 - ' Ws aT 3 . - . cos a . (0 a s
... (12.98)
An examination of Eqs. (12.97) and (12.98) reveal that these equations are valid whether a transformer is used in the static Kramer drive circuit or not. It is observed from Eq. (12.97) or Eq. (12.98) that steady state torque is
link curr ent, Id (ii) proportional to stator s upply voltage, VI (iii) proportional to effective r otor to stator turns' ratio, a
and (iu) inversely proportional to synchronous speed, (Os'
The 'dc link current Id is given by (i) pr oportional to dc
Vd - Vdc
= resistance of de link inductor, Rd Vd = Vdc + I d · R d = 2.339 sa1/1 Id
or Slip,
V dc + ld . Rd,
s = -=--::::-::-~-=2.339 a VI - 2 .3 39· aT ' V 1 cos a 2.3390V1
=---::-=-
~~;----
2.339 aV1
Power Electronics
[Art. 12.8]
648
Motor speed rom is given by
=
Cl)s
(1 - s)
=
Cl)s
[
1
Id . R d
aT
1 + ~ cos a - 2.339 . aV
...(12.99)
I
Under steady state, from Eq. {12 .98), totaHorque 3Te is given by 3Te =T L
aV ·I
I d =2.339 - ro s
CI), •
or
TL
...(12.100)
I d = -=-=c=-::---;-; 2.339 a VI Substituting this value of Id in Eq. (12.99), we get rom
= Cl)s
[1
=
+ aT cos a a
K=
(2.339, aVI )
1..
=w, [ 1 + ' ; COS"
2'
TL]
... (12.101) ... (12.102)
Rd 2
F r om Eq. (12.101) or (12 .102), the no-load speed of the drive is given by
w'"
Rd
+:T cos a - K TL1
Cl)s .
where
Cl)s •
(2.339 a VI)
(12.103)
From Eq. (12.1 01), the speed-torque characteristics of static Kramer drive shown in Fig. 12 .41 are plotte d, for different fi ring angles, in Fig. 12.42. It is seen that these characteristics are similar to a separately excited dc motor with armature voltage control. Static Kra mer drive systems are us e d in large p ower-pumps a n d · com press or type loads wh ere speed control is within n arr ow range and below synchronous speed .
Speed
r _ ___:: cx : 90 •
t
o
p .u. torque
Fig. 12.42. Speed-torque char acteristics of sta tic ~amer drive for open loop system ..
Example 12.29. e, 3-ph ase, 42 0 V, 4 -p ole, 50 Hz, star-con nected SRIM has its speed controlled by m eans of sta tic Kramer drive. T he effective phase turns ratio from rotor to stator is 0.8 and transformer h as phase turns ratio from l.v. to h.v. as 0.4. T he inductor current is ripple fre e. Losses in diode rectifier, inductor, inverter and transformer are neglected. The load torque is p roportional to speed sq uared and its val ue at 1200 rpm is 450 Nm. For a motor operating speed of 1000 rpm, ca lculate.
(a) rotor rectified voltage (b) inductor current (c ) delay angle of the inverter (d) efficiency, in case inductor resistance is 0.01 n and per-phase resistances for stator and rotor are 0.015 nand 0.02 0 respectiuely.
Eledric: Drives
[Art. 12.8]
649
~-------------"'-----":"'------""'---=--------.,
(e) For the firing angle obtained in parl (c), tlu load torque is increased to 500 Nm, fi'nd the motor speed. Solution. (a) Per-phase stator voltage,
. 420
VI =-ra= 242.5 V 1500 -1000 1 1500 ='3
Turns ratio from rotor to stator, a = 0.8
From Eq. (12.91), rectified voltage,
Vd = 2.339 saVl
1
= 2.339 x '3 x 0.8 x 242.5 = 151.26 V
Slip,
s=
J
(b) Load torque at 1000 rpm, T L
=450 x (i~~~ =3T, =312.5 Nm
. Synchronous speed,
=-
47t{l
00
• P From Eq. (12.100), inductor current is
=
47t X 50 4
=50 7t rad/s
00, . TL 50lt x 312.5 Id = 2.339 aV 1 = 2.339 x 0.8 x 242.5 = 108.18 A
(c )
From Eq. (12.95 ),
.
a.r. s=-- · COS
1
or
0.4
'3 =~ 0.8 cos a a (d)
= H)= C08-
1
131.81°
Power fed back to supply
=Vdld =151.26 x 108.18 = !l6363.31 W =TL . OOm =312.5 x 27t x601000 =32724 .92 W =Ii .Rd = (1 08.18)2 x 0.01 = 117.03 W
Power output Loss in inductor
~Id =~x 108.18 =88.33 A
Rotor current,
12 =
Rotor ohmic loss Stat or current,
=3122 rz = 3 (88.33)2 x 0.02 =468.1 3 W
I I = a 12 =0.8 x 88. 33 = 70.664 A
-
Stator ohmic loss Power input
= 3 (70.664)2 x 0.015 = 224. 70 W
= 32724.92 + 468.13+ 224.70+ 117.03= 3353 .78 W
32724.92
= 33534.78 x 100 =97.58%
Efficiency (e)
From Eq. (12.101), we get motor speed a under: w m
= 50lt 11 + 0.4 cos 131.81 L
0.8
= 104.121 radl ~
or
0 -
l
S07t X 0.0 x 5 00 (2.339 x 0.8 x 242.5)2.J
994. 3 rpm .
-.;:."
+.
650
[Art. 12.8]
Power Electronics
Exam p le 12.30. A static K ramer drive is used for the speed control of a 4-pole SRlM fed from 3-phase, 415 V, 50 Hz supply. The inverter is conn ected di rectly to the supply. If the motor is required to op erate at 1200 rpm, find the firing ad vance angle of the inverter. Voltage across the open-circuited slip rings at stand-still is 700 V. Allow a voltage drop of 0.7 V and 1.5 V across each of the diodes and thyristors respectively. ln~uctor drop is neglected. Solution . Rot or induced emf at stand-still = 700 .V (line) _ 700 V E 2-Ts
.. Slip,
= 1500 - 1200 = 0 2
1500 DC voltage across the diode rectifier is s
.
Vd = 3-.[6 sE2 _ 2 x 0.7 = 3-./2 x 0.2 x 700 _ l.4
It
·
It
Inverter dc voltage is given by 3-./2 x 415 ] Vdc = [ .. cos a - 2 x 1.5
It
With no voltage drop in inductor, Vdc = Vd
..
_ 3-./2 x 415 cos a + 3 = 3-./2 x 0.2 x 700 _ 1k
It
or
a
=
cos
,' ;. : .... .
It
- 1 [ - 184.6379 x It] = 109 240
3'12 x 415
.
:. Firing advance angle of inverter = 180 - a = 180° - 109.24° = 70 .76° E xample 12.31. Repeat Example 12.30 in case there is an overlap angle of 18° in the rectifier and 4° in the inverter. Solution. Average output voltage for a 3-phase full converter, with overlap, is given by Eq. (6.79 a), which is reproduced here, for convenience. 3Vml Vox =-2- tcos a + cos (a + ~)]
It
For uncontrolled rectifier bridge, de output voltage is . 3-.[6 sE2 . Vd = 21.[ [cos 0° + cos (0° + 18°)]- 2 x 0.7
= 3..J6;It~2fs 700 [1 + cos 18°] . Similarly, inverter de voltage is V"" ==
W::
(cos (x+ cos (0: +
_[{3~2: 416
- 1.4 = 183.012 V
~)} - 2x 1.5]
(cos a+ cos (Ct.+ 40)} -
3]
= - 280.18 [cos a + cos (a + 4°)] + 3 With no-voltage drop in induct or, V dc = V d
or
- 280. 18 [cos C( + cos (Ct. + 4°)] + 3 = 183.012 V cos Ct. + cos (a.+ 4°) = - 0.6425 cos Ct. + cos Ct. . cos 4° - sin Ct. sin 4° = - 0.6425 1.9976 cos CI. - 0.07 sin a = - 0.6425
...(i)
[Art. 12.8]
Electric Drives
Note that
A cos a - B sin a
-1
9 -- t an
where Eq. (i) gives 1.9976 cos
=..JA 2 + B2
(l -
651
cos (a + 9)
B
A
0.07 sin (l ='11.9976' + 0.07' [cos ( (l + tan- 1
1~~~6]
=- 0.6425 1.999 cos (0.+ 2°) =- 0.6425 a= 106.75° Firing angle of advance of inverter = 180 - 106.75° = 73.25°. Example 12.32. Using the data ofExample 12.30, find the voltage ratio of the transformer to be interposed between supply and the inverter for a minimum speed of 1200 rpm. 3..f6 sE2 3...f6 Solu tion. Here Vd = and Vdc =- -- . aT' VI . cos a
or or
1t
For an ideal inductor, 3-16 sE 2
1t
Vd = Vdc
3..f6 =- -. aT . VI . cos a
VI
. cos a 1t 1t E2 Minimum speed means maximum slip. Here s would be maximum in magnitude when firing angle of the inverter a ;: 180°. aT' VI s E2 0.2 x 700 140 .. s= or aT:;: - - = 415
E2 Vi 415 : . Tra n sformer voltage ratio per phase from h .v. to l.v. is 415/140 or aT = 0.3373.
..
or
s:;: -
aT'
12.8.6.2. S tati c Sche-rbius Drive. In st atic Kramer drive, speed of SRIM ,can be control led below synchronous speed only. For the speed control both below and above synchronous speed, static Scherbius drive scheme is used. There are two possible configurations to obtain such a drive; these are (i ) DC link static Scherbius drive and (b) cycloconverter static Scherbius drive. These ar e discussed briefly as under. .. . •..'~ " DC link S cherbius Drive. In subsynch ronous speed control of WRIM, slip power is removed from the rotor circuit and is pumped back into the ac su.pply. In supersynchronous speed control, the additional power is fed into the rotor circuit at slip frequency. The circuit diagram of Fig. 12.43 allows both subsynchronous and supersynchronous speed control. It consists of one WRIM, two phase-contr olled bridges, smoothing inductor and a t ransformer as shown.. For subsynchr onous speed control, bridge 1 h as firing an g~ e less than 90° whereas bridge 2 has firing an gle more than 90°. In other words, bridge 1 work6 as rectifier and bridge 2 as line-commentated inverter for subsynchron ous mot or control. The slip power fl ows from rotor c!!cuit t o bridge 1, bridge 2, tr ansformer and t o th e suppl~ ~ For su persynchronous motor control, bridge 1 is m ade to work as lin e-commut ated inverter
with firing angle mor e than 90° and bridge 2 as a r ectifier with firing angle less than 90°. The
power fl ow is now from the supp y t o transform er, bridge 2, bridge 1 and to the rotor circuit .
N ear synchron ou ~ speed, slip fr equency emfs are ins ffici ent for n u lfal commut ation of thyris tors. This difficulty can, h owever, be Dv e come by sing forced commut ation. Thus, the pro 'sion ofboth sub synchron ous and supersynchronous speed oper ation complicates the static converter sys tern and nullifi es the advant ages of simplicity and economy wh ich are inher entin a pur ely subsynchr onous drive . In addition, static Scher bius drive is expensive than s tu tic Kramer drive because s 'x diodes are r eplaced by six tn)7is tors and their contr olled circuit ry. It o fer s con s tant torqu e-drivE; scheme.
652
{Art. 12.9]
Power Electronics
3 phase supply
tS-------I---I~
"Bidir~ctional slip pow~r,!sPg
Phase controll~d bridge 1
Phas~
controlled bridgt" 2
Fig. 12.43. DC link static Scherbius drive.
Cycloconverter Scberbius Drive. The dual controlled converter system used in de link Scherbius drive is now replaced by one phase-controlled line-commutated cycloconverter as shown in Fig. 12.44. Such schemes are used for very high-power pumps and blower-type drives. Cycloconverter per mits the slip-power flow in either direction and the machine can, therefore, be controlled in both subsynchronoUB and supersynchronous ranges with motoring ,and r egeneration features: As the slip power is either returned t o, or taken from, the supply mains, cycloconverter static Scherbius drive offers constant-torque drive scheme. 3 phase
supply
i?- ' ~ r~
P
N
8 idirt"c l i ona l slip '
flow, :': sP9
p ow~r
Fig. 12.44. Cycloconverter static Sch'erbiuB drive.
i 2.9. SYNCHRONOUS 1'IOTOR DRIVES
'.
'.
Synchronous motors have two winding ,one on the stator is three-phase armature winding and the other on the rotor 's the field win ding. The three-phase winding on its stator is similar
,
Electric Drives
(Art. 12.9]
653
---
to the 3-phase winding on the stator of a 3-phase 1M. Field winding is excited with de and it produces its own mmf called field mmf. Three-phase stator winding carrying three-phase balanced currents creates its own rotating armature mmf. The two mmfs combina together to pr oduce resultant mmf. The field mmf interact s with the resultant mmf to produce electromagnetic torque . A synchronous motor runs always with zero slip, i.e. at synchronous speed given by Eq. (1 2.43). Power factor of synchron ous motors can be controlled by varying its field current. .For the speed control of synchronous motors, both inverter and cyclonverters are employed . The various types of synchronous motors are : (i) Cylindrical rotor motors (ii) Salient-pole motors (iii) Reluctance motors (vi) Permanent-magnet motors.
These are now described briefly in the following lines.
12.9.1. Cylindrical Rotor Motors These motors have uniform air gap. The per-phase equivalent circuit for a cylindrical-rotor synchronous motor is given in Fig. 12.45 (a). I n this circuit, E,= excitation voltage =~. 1t . f cI>, . Nph . kw, V t = armature terminal voltage, ra = armature resistance, Xs = synchronous reactance and Z, =...Jra2 +X,2 is called the synchronous impedance.
__.. . _-
VI _-------/ Z s
,I
I
~
(a) (a)
" 10
(b)
·Fig. 12.45. Cylindrical-rot or synchronous motor equivalent circuit and (b) its phasor diagram ata lagging pf load.
It is seen from the equivalent circuit th at
V t = Ef+la
or
. Z,
=Vt - E[ =Vt _ Er
I a
Z,
Z, Z,
.
Eq. (12.104) shows that armature current 10 is the difference of two currents
.. .(12.
V4)
V
E
_ I
Z~
an d ;!-,
Zy
lagging b~hind their respective voltages by impedance an gle 9z as shown in Fig. 12.45 ( b). Here impedance angle 9z is given by - S, = tan
1
(~)
.
[Art. 12.9]
654
Power Electronics
., . ., Pow r input to the motor is given by P im = V t [Component of la in phase with V t ] In the above expression, subcripts i and m denote input and motor respectively. It is seen from Eq. (12.104) and Fig. 12.45 (b) that the component of la in phase with V t is
~ ] Vt [ Z, cos 9z - Zs cos (0 + 9z ) .
.
•
.... .
.
]
P. im = V t [~ Zs cos 9z - ~ Zs cos (0 + 9z) Vt 2
=Zcos9z s
V t ' Ef Z cos (0+9 z )
... (12.105)
s
r
.
Now cos 8z = Za and 8z = (90 - etz). Substituting these in Eq. (12.105), we get
s
V/ Z,
Ef · Vt •
{
P im = '72" ra - ~ cos (0 - etz) + 90°
Ef · V t . =- sm (0 - N)
Z,
--z
Power output from the motor, p o .m It can be proved similarly that
=Ef
}
V/ + -Z2 . r a
... (12.106)
II
[component of lain phase with Ep
_ Ef · V t • E/ Pam - -Zsm (0 + (lz) - --"'2' ra
,. Z,
Here Pom i8 the developed power and shaft power = Pam - rotational losses.
Developed t orque,
P
Te =~
ro,
=-1
ro,
[E~ Z.V sin (0 + t
s
(lz) -
...(1 2.107)
E2 . r a]
~
Z.
If armat ure resistance is neglected, then
E[' V Pim=Pom = X t sino
. ...(12.108)
s
1 EC V t ~o ros X,
~d
~=- .
where
ros =
1
...(12.109)
= synchronous speed in rad/s .
~d
0 = load, or power, angle
The torque versus load angle characteristic for a cylindrical-rotor synchronous motor is
sh own in Fig. 12.46 (a ). F or stable operation of synchronous motor, the load angle 0 should
never exceed 900 • Power factor of a synchronous motor' depends on the field current . The variation of arm ature current with respect to fi eld current for different loads on the synchronous motor is shown in Fig. 12.46 (b). As the shape of these curves resemble the letter V, these are called V-c urves of a synchronou motor. Note that V-curves are obtained for const ant shaft load and fo const ant termin al volt age. Unity pf rYe is shown dotted. For low values of fi eld current (under-excit ation), synchronous m otor operates at a lagging pf. As t he field current is increased, it would start operating at unity pf (nor mal excitation) . If the field current is still increased beyond the unity pf point (over-excitation), synchronous motor begins to operate at a leading
¢
~
[Art. 12.9]
Electr ic Drives Ia
Pull out
Torque
Rated
655
Half rated torq ue
No load
Lead ing pf -
unity pf
(a)
(b)
Fig. 12.46. Cylindrical-rotor synchronous motor (a) torque- angle characteristic and (b) its V-curves.
The power given by Eqs. (12.107) and (12.108) and torque of Eq. (12.109) have per-phase values. For 8 = 90°, pull-out power P 171.% and pull-out (or maximum) torque T e , m are obtained. Ef · V t ...(l. 2.110) Pmx=----X s
T
and
1 E =_._ - fX-' Vt e'm (0
S
... (12.111)
8
For fixed field excitation, the excitation emf Ef is directly proportional to supply frequency. The synchronous reactance Xs is also directly proportional to frequency. This shows that
iE
in s
Eqs. (12.110) and (12.111) is independent of frequency variation. If supply voltage V t is varied in V ' , proportion to frequency so that Vtlf or _t is constant, then pull-out torque, Eq. (12.111) remains (Os
constant. Pull-out power P 1TIX = Te ,m x (Os , however,
rises linearly with speed as shown in Fig. 12.47.
At base speed «(Os = L~), rated voltage and
rated frequency are reached. Beyond base speed,
rated voltage is kept constant, but inverter
frequency can be increased to obtain higher
operating speeds of synchronous motor.
Wit h constant field excitation, increase in , E frequen cy k eeps Xf const ant as stated befor e. s
, With supply voltage remaining con stant above o 1.0 2.0 per un it base sp eed, pull-out power, Eq. (12 .110), rem ains ~ Cons ta n t torquetConstant power __ frequency 1 '
const an t, but pull-out t orqu e = - . P 171.% falls Fig. 12.47. Variation of pull-out torque, pull-out ,
(Os
in versely with rise in speed as shown in Fig. 12.47.
power, terminal voltage and armature current with frequency for a synchronous motor.
Example 12.33. A 3-phase, 400 V, 50 Hz, 6 pole, star· connected round -rotor synchronous m otor has ZJ =0 + )2 n. Load torque, proportional to speed squared, is 340 Nm at rated synchronous speed. The speed ofthe motor is lowerrd by keeping VI f constant and maintaining unity p f by field control of the moto . For the motor operation at 600 rp m, calculate (a) supply
656
Power Electronics
JArt. 12.9]
voltage (b) the armature current (c) the excitation voltage (d) the load angle and (e) the pull·out torque. N eglect rotational losses. . Solution. (a) For 600 rpm, the supply frequency, P · N. f= 120 As
6 x600 120 =30 Hz.
~ is constant, the supply voltage Vt is given by Vt _ 400 30 - 50
(b)
=
or
V t =30x8=240V
Load torque at 600 rpm, 600 J2 T L =340 [ 1000 I
. =122.4 Nm
P = Te . <.Os = TL . <.Os = 122.4 x
Power output,
As rotational losses are neglected, .,f3 Vt Ia cos 9
21t x 600 60 = 7690.62 W
=P
ra7690.62 x 240 x i =18.50 A
:. Armature current, Ia = (c)
Ef
30 Xs = 50 x 2 = 1.2 .Q
240 P er-phase supply voltage, Vt = T3 = 138.57 V
Per-phase armature current, Ia = 18.50 A From the phasor diagram, excitation voltage is
xi
Ef = ..JVt 2 + CIa = ..J138.572 + (18.5 X 1.2)2 = 140.34 V per phase Line value of excitation voltage = .,f3 x 140.34 = 243.07 V (d) From the phasor diagram of Fig. 12'.48, we get load angle S as I::
u (e)
T
e.m
Fig. 12.48. Phasor diagram pertaining to Example 12.33.
=t an- 1 (11 XsJ=t an- 1 [18.5 x 1.2) =9'.100 V 138.57 . t
=1- . E[· Vt =~ <.Os
Xs
207t
140.34 x 138.57 1.2
=773 77 N .
m.
12.9.2. SaJien t-pole Motors The armature winding on the stator of a salient- pole motor is similar to that of a cylindrical-rotor motor. Fie d winding on the rotor is a concentrated winding on the salient poles . In this type of motor, the air gap is not uniform. For analysis purposes, its armature current is resolved into t NO components, called direct-axis current Id and quadrature-axis current I q . Likewise, there are two reae.ances, Xd =d-axis synchronous reactance and Xq = q-axis synchronous reactance.
657
[Art. 12.9]
Electric Drives
The phasor diagram of a salient-pole synchronous motor with negligible armature resistance is shown in Fig. 12.49 (a). It is seen from this pha"sor diagram that
Vt=Ef+}ldXd+}lq Xq . Id:::: Ia sin (9 - 0), Iq = Ia cos (9 - 0) Also. V t sin B = Iq . Xq =Xq . Ia cos (9 - B) =Xq Ia [cos 9 . cos B+ sin 9 sin B]
sin 0 [Vt - Ia Xq sin 9] =Xq . Ia cos 9 . cos 0
... (12.112) .
tan 0 =
Ia Xq cos 9 ... (12.113) V t - I a . Xq sin 9 Power P is given by p:::: Vt cos 0 (current along q-axis) + Vt . sin B (current a~ong d-axis) = Vt . cos B. Iq - Vt . sin B. Id ...(12.114) But Id Xd =Vtcos 0 - Ef Vt cos B-E or Id =-~X=-d-....!..f ..
Substituting these values of Id•Iq in Eq. (12.114), we get E ·V V/ ( 1 1 ). . P= f sino+- - - - sin 20 Xd 2 Xg Xd
q-axis
l
... (12.115)
I
I
Iq Xq :
Te . Resultant torque
(a)
(b)
Fig. 12.49. Salient-pole synchronous motor, ~ (a ) 'ts phasor diagram and (b) its torque versus load a ngle characteristics, . ' . E ·V ' . t f The power in Eq. (12.115) has two components. The first component Xd sin 0, similar
to that of round-rotol" motor, i3 caned el ectromagneti c power. The other component
~
2
(i. -i.)
sin 2 S is call ed the reluctance power, as it is present due to different reluctances
'along direct a..-ds and quadrature axis .
.
.
[Art. 12.9]
658
Power Electronics
Developed t orque Te is given by 2
l· 1
Vt (1 . 1 sm28 T =P- = 1- [E(' Vt sino+---X e
(Os
Xd
O)s
2
Xq
d
.. .(12.116)
As for power, the first and second components in Eq;· (12.116) are called espectively the electromagnetic torque and the relu ctance torque. . Th e torque Te versus load angle 8 characteristics are drawn in Fig. 12.49 (b) with the help ofEq. (12. 116). It is seen that torque is maximum at a load angle less than 90°.
12.9.3. Reluctance Motors A salient-pole synchronous motor connected to a voltage source runs at synchronous speed. If its field current is switched off, it contmues running at synchronous speed as a reluctance motor. Thus, a machine designed to operate as a reluctance motor is similar to a salient-pole motor with no field winding on the rotor. Three-phase armature winding produces rotating magnetic field in the air gap. This rotating flux induces a field in the rotor which tends to align itself with the armature field, thus producing a reluctance torque at synchronous speed. r . . Reluctance motors are used for low power drives where constant-speed operation is required and where more than one motor is needed for the Job so that number of Inotors can run in synchronism.
With zero field current, E(= 0 and the phasor diagram for a reluctance m9tor can be drawn from Fig. 12.49 (a) by making E f = 0, this is shown in Fig. 12.50 (a). From this figure, power P is given by
P = Vt . cos '0 . Iq - Vt . sin 'Old It is seen from Fig. 12.50
(a )
Iq =
that
Vtsm'O X . and q
Substitutmg the values of Id and I q , we get power P as .
.
V t 2[ 1-
-
1 )· ..
.. .(12.117 )
P = -2 Xq - Xd sm 28
5
,. \
I
.
~
(0)
Id
Flg. 12.00. Reluctance motor
(b ) (a )
its phasor di agram and
(b)
its torque-angle characteristics.
.r
Electric Drives
[Art. 12.9]
659
Phasor diagram of Fig. 12.50 (a) reveals that
V t . sin 0 =1q Xq
=Xq · l a cos (9 -
o=tan- 1
. . This gives
0).
1a Xq cos 9 VI -la Xq sin e
... (12.118)
Reluctance torque, from Eq. (12.117) is 2
T =1- . -V t e c.os 2
.
1 -1). (sm20
Xq
... (12.119)
Xd
Pull-out torque Tem :,., obtained when 0 = 45° 2
..
= Vt
T
(1:__ 1:_) X
2c.os Xq
e .m
._--. ...(12.120) ----
d .
Variation of reluctance torque with load angle is shoWn in Fig. 12,50 (b). Example 12.34. A 3-phase, 400 V, 50 Hz, 4 pole, star-connected reluctance motor, with negligible armature resistance, has Xd =8 .n and Xq =2 .n. For a load torque of 80 Nm, calculate (,,) the load angle (b) the line current and (c) the input po.lf!.e~ factpr. N eglect rotational losses.
. Solution. (a) Synchronous speed,
OOs
~xW = 4~ P =4 = 501t radls
3 1[1- - -1]. 2
From Eq. (12.117)
P = T . (j) = 80 X 501t = - [400 ~ e
I:
u
. . (b) Per phase voltage, VI
2
s
'V3)
2
8
sm 20
= . - 1 [80 X 501t x 8] = 12 3830
sm 400
80000 x 3
.
=13 = 230.95 V
1d = V t cos 0 = 230.95 x cos 12.382° Xd . 8 .
=28.197 A
1q =V t ~in 0 = 230.95 x ~n 12.383°
=24.761 A
q
:. Armature current, la= ~Ii +1/ =~28.1972 +24.7612 =37.53 A (c) --J3 V t 1a cos a = Te . c.os .; [3 x 400 x 37.53 x cos a= 80 x 501t W 4000 x 1t '. :. Input pf = 'f3 x 400 x 37.53 =0 .4833la~g. .
.
~
12.9.4. Permanent-magnet :Motors A permanent -magnet synchronous motor (PMSM) is similar to a salient-pole synchronous motor without the field winding on the poles. In PMSM, the required field flux is produced by permanent magnet s mounted on the rotor. In these motors, the excitation emf E f cannot be varied. All the equations governing the performance of a salient-pole synchronous motor are also applicable to PMSM with excitation emf Ef taken as constant. The absence offield winding, dc supply t o field winding and two slip rings leads t o reduction in motor losses . For the same frame size, PMSM has higher pull-out t orqu e and more effi ciency as compared to salient-pole motor.
660
Power Electronics
[Art. 12.10]
These mot ors are used in robots and m achine t ools. A PMSM can be fed from rectangular current source or sinusoidal current source. A rectangUlar current-fed motor has concentrated winding on the stator and is used in low-power drives. A sinusoidal current-fed motor has distributed winding on the stator and ' is used in high-power drives.
I..
{
12.10. SOME WORKED EXAMPLES In this article, some typical problems pertaining t o electric drives are solved. Example 12.35. A separately-excited dc motor is fed from 230 V, 50 Hz source through a 1-phase semiconverter. The motor armature resistance is 2 n and its torque constant is 1.2 Nm / A. The thyristors are fired at an angle of 100° and the discontinuous armature current extinguishes at 45° beyond.voltage zero. Calculate the motor speed ifload on the motor is 3 Nm. Solution. Circuit diagram for single-phase semiconverter feeding a separately-excited dc motor is shown in Fig. 12.8 (a) . The waveforms of discontinuous armature current and armature terminal voltage are shown in Fig. 12.51. When thyristor Tll is turned on at a = 100°, armature current ia builds up, reaches some -maximum value and then decays to zero at rot = 225°; 45 ° beyond voltage zero, as specified. For CJJt = 100° to 180°, Tll Dll conduct and from rot = 180° to 225°, FD conducts and so on.
....
When FD stops conducting at rot = 225, motor terminal voltage jU.lllPS fr-am zero to Ea and stays there till T1 2 is triggered at a =1000 beyond rot =1t. In other words; a r inat ure generated voltage Ea stays at constant value for 55° as shown. From the waveform of voltage
VOl
or vt ' mean value of armature terminal voltage is 0
Vt
1 [.
=;
55 x 1t J 180 Ea x 180 + 100 Vmsmrot . a(wt) •
•
]
.
=0.3056Ea + 0.263 x -..f2 x 230 = [0 .3056 Ea + 85.53] V Armature current I
Te 3 a =-Km = -1.2
= 25 A .
v~ s
Vms in wt
Vm
ia
.~
I
FD!
Fig 12.51. Waveforms pertaining to Example 12.35.
f.
Electric Drives
[Art. 12. 0]
661
V t :: Ea + Ia ra = 0.3056 Ea + 85.53 Ea + 2.5 x 2 = 0.3056 Ea + 85.53 80.53 Ea = 0.6944 ~ 115.971 V = Km . (Or
Also or
:. Motor speed,
rom
= 115.971 = 96 042
1.2
ra
.
d/ = 2 1t N s 60
:. Motor speed, N = 922.86 rpm Example 12.36. A separately-excited dc motor, with ra = 4 n, La = 0.04 Hand Km =1 Nm/A, is fed from 230 V, 50 Hz supply via a I-phase semiconverter. For a firing-angle delay of 75°, the motor runs at a speed of 1360 rpm. (a) Derive expression for motor armature current. (b) Sketch the.waveforms of source voltage, load voltage and motor armature current (c) Calculate the average motor torque .
Solution. (a) At firing angle delay of 75°, Tll in Fig. 12.8 (a) is turned on and Tll Dll
begin to conduct. Under this condition, equivalent cir cuit of Fig. 12.52 (a) applies, for which the KVL gives . ra
.
di1
.
·"1 +La' -dt +Ea = Vm sm rot.
The solution of armature current i 1 has the following three components :
(\)
Us
·i 1 = iac ' steady-state ac component + i dc ' steady-state dc component + exponentially decaying component such that i1 = 0 at rot = a = 75°.
(a )
. Vm . "ac = - Z sm rot Impedance
.
Equiva~ent
circuit of Fig. 12.8 .
(a) for example 12.3 6.
= 2It x601360 = 142.42 rad/ sec
rom
Motor emf,
'Ea = Km - rom
ide = -
=1 x 142.42 =142.42 V Ea = _ 142.42 =_35 .6 0 A ra
L
1:
.
~l
= 13.~: ~;~350 sin rot = 24.66 sin (<.ot - 72.35°)
Motor speed,
From Eq (i ),
Fig. 12.52.
Z = 4 + j 21t 50 x 0.04 = 4 + j 12.57 = 13.19 L72.35° n "ac
When
=
Ym sinwt
4
0.04
1
= R = -4- = 0.01, '1 = 100 . A -tl 't + ldc + e 100 t = 24.66 sin (rot - 72.35°) - 35 .60 + A. e=
.
Lac
o
rot = a == 75 or t = 180
75 Xlt X
21t x 50 = 0.004 2 sec,
0 = 24.66 sin (75 - 72. 35) - 35.60 + A e-
...(i) . _
l1 -
100 x 0.0042
0
662
Power Electronics
[Art. 12.10]
= 52.45
A
or
i 1 = 24.66 sin (rot - 72.35°) - 35.60 + 52.45 e-
At rot =
1t,
loOt
...(ii)
when voltage is zero, current is given by i 1 = 24.66 sin (180 - 72.35°) - 35.60 + 52.45 e- 100x
TC
100lt
= 7.194 A
Soon after rot = 7t, freewheeling diode FD begins to conduct; now the equivalent circuit of Fig. 12.52 (b) is applicable, 'for which KVL gives .
di2
rt2
+ L dt + Ea = 0
,
Here time t is measured from the instant FD begins to conduct. Its solution is
E
ra
i2 = - ~ + B . e- t / 't = - 35.6 + B e- 100 t "
~ ' 0 Att=0,i 2 =7.194A,:. 7.194=-35.6+BeorB=42.794
i2
= - 35.6 + 42.794 e- lOOt
.. .(iii)
Expr ession for motor armature current ia (= i 1 -+- i 2 ) is given by Eq. (i ) from rot = a t o 180° when T ll Dll conduct and by Eq. (iii) from rot = 180° to 225° when FD conducts and so on. lOOt
(b) The current i2 would be zero (after rot=7t), whep. i 2 =0 =-35.6 + 42.794eor 180 ' . t = 0.00184 sec or angle = 0.00184 x 2 7t X 50 x ' = 33.12° < a = 75°. For 75 - 33.12° = 41.88°, 1t
no device conducts, therefore, armature terminal voltage jumps to Ea at rot = 213. 12° and stays there for 41.88°. FD conducts for 33.12° from rot = 180° to 213.12°. Waveforms of source voltage, motor terminal voltage and discontinuous armature current are shown in Fig. 12.53. Waveform of voltage vt gives its average value as V t = 1. 7t
[f80 12 x 230 sin e . de + 142.42 x (75 - ~~.~2) x 75
'
1t ] = .1
wI
,
o' FD',:
~~
~= " 75 '
I
j' I
i
Til
,
0 11
,' FD ! "
: I
T12
I ;.
0 17
:j FD!,
C2JI'£N ,
'1 / ,II
271
Til
011
wt
II :'
"FD ;
. , I i ! '~ i I~: 37l'
L/
10 5' 33.12' ' 41. 88'
Fig. 12.53. Waveforms Pertaining to Example 12.36.
.
wI
163.45 V
Electric Drives
[Ar t. 12.10)
Motor emf, Ea
= 142.42 =V t Ia
= ~t -
Jara
Ea
= 163.45 -
= 5.2575 A
142.42
4
ra
:. Motortorque; · Te =Km Ia
663
= 1 X 5.2575 = 5.2575 Nm
Example 12.37. Repeat Example 12.36 with the same magnitude of firing-angle deldy, but with load torque reduced so that motor speed rises to 2100 rpm. . SolutIon. (a) Motor armature e,mf, Ea
Ea ra
27t X 2100 60
=Km rom = 1 X
= 219.91 V
= 21~.91 = 54.9775 ::: 55 A
The expression for the armature current, from the previous example, is given by
= 24.66 sin (rot - ~2.35°) - 55 + A e- lOOt rot = a = 75° or t = 75 x 0.01 = 00042 sec i l = 0 180 . , i1
At
o = 24.66 sin (75 or
A il
72.35°) - 55 + A e- 100 x 0.0042
=81.98 =24.66 sin (rot -
72.35°) - 55 +81.98 e"':' lOO t
...(i)
.
At rot =7t, when the voltage is zero, il
= 24. 66 sin (180 -
72.35°) - 55 + 81.98 e-
100 x 0.01
=-
1.342 A
This shows that i l decays to zero before rot = 7t. The angle « 180°) at which i1 becomes zero can be obtained from Eq. (0, by hit and trial. .
When
rot
= 175°, i l = 24.66 sin (175 -72.35) -
So when rot = 175°, or when t = 1751~g·01
55 + 81.98 e-: 100 x
= 0.00972
175 x 0.01 180 :::
0.0696
sec, the current i l decays to almost
zero value. (b) The waveforms for us' Uo and ia are sketched in Fig. 12.54. It is seen that FD of the I-phase semiconverter d oes not come into play. The current flows from rot = a = 75° to
u:~~,
vo.Vtb!
.
I
I
I
I
I.
/" I "~"" " ""~ o~>.>" Li~/>' L"l.:.i,/ ,"""Ji
- I a-7S': --1 ,
o
",:
Til 011
~ ~ ; ' CC
iI
i :
i
i! 1. 1 : ,' "
TI2 012
A
I,
I
:\
I
:, ;:CC
: I.
il
i' , \'
:1 75~ 100'- i,1- 75'--1- jOO'~~ S'
: ' ,.
Ti l ; . : 011 · :l--a~
ii,, :
!!
:I
:
,
II
V
L:\::I
:
:1
5'
Fig. 12.54. Waveforms pertaining to Example 12.37.
• wt
wt
Power Electronics
[Art. 12.10]
664
wt = 175° . Armature voltage then jumps to Ea and stays there till rot = 255°, •
constant for 255 - 175 = 80° as shown. (c)
1 Mean output voltage, V t = 1t =
Bu t Vt
=E a +"aT ra ,
1
:.
a
[J
~[
.
,
1750
75
C"
.
'
'
i,~.
Ea remains
i ' ~ ·
,
Ea .80X1t]
Vm sm 9 . de + . 180
,j2 x 230 sin 9 d 9 + 219.9i;o80 x x ] = 227.66 V
= ~27.66 -4 219.91 = 1.9375 A ,
Te =Km Ia = 1 X 1.9375 = 1.9375 Nm Example 12.38 : Repeat Example 12.36 with the same firing-angle delay. Now the load . torque is increased so that the motor speed reduces to 840 rpm. • 21t x 840 = 87.96 V Solution. (a) Motor arPlature emf, Ea = Km . rom = 1 X · 60
Ea = 87.96 = 21.99 A ra 4 From Example 12.36, armature current il is given by il
Assum ing
t = 75 ;8~·01
the
= 0.0042
current
=24.66 sin (rot to
be
72.35°) - 21.99 +A . e-
discontinuous
when
rot
1OOt
=ex = 75°,
" .(i)
or
when
sec, current il = O. From Eq. (i) 0= 24.66 sin (75 - 72.35°) - 21.99 + A e- 100 x 0.0042
When rot =1t =180° or t
A
=31 .74
il
=24.66 sin (rot -72.35°) - 21 .99 + 3 1.74 e- 100t
='0.01 sec, then it = 24.66 sin (180 - 72.35°) - 21.99 + 31. 74 e- 1 = 13.186 A
After
rot
=1t, FD begins to conduct. Now when t =0, i2,=13.186 A, Current i2 is given by i2 = - 21.99 + Be-loOt
When t =0, i2 = 13.186 A, :. 3.18 = - 21.99 + B , :. B '= 35.176
...(ii) i2 = - 21.99 + 35.176 e- lOOt Current i2 would decay to zero after sometime t, which an be obtained from Eq (ii) as under: 0= - 21.99 + 35.176 e- 100 t or t = 0.0047 s or 84.6° after voltage zero at Cilt = 1t.
As firing angle Ct. =75° < 84.6°, i2 does not fall t o zero when the next SCR in sequence is
triggered. Thls shows that armature current itl =i2 is continuous. Constant s A and B must be . re-calculated to obtain correct expressions for currents i 1 and i2• 'When eDt = 75°, current
i2
= - 21.99 + 35.176 e- 100x 0.0042 = 1.122 A
Therefo e, when wt =75 , i 1 = 1.122 A or
1.122 =24.66 sin (75 - 72.35) - 21.99 +A e- 100x 0,0042
[Art. 12.10]
Electric Drives
A = 33.443 il = 24 .66 sin (rot 72.35) - 21.99 + 33.443 e- 100 t
or
665
_.
il = 24.66 sin (180 - 72.35) - 21.99 + 33.443 e- 1 =13;812 A =i2 at t =0, i.e. when voltage zero occurs.
"Then rot = 7t, For i 2 , when t = 0,
i2
= 13.812 A
13.812 =- 21.99 + B x 1
or
B = 35.802
or
i2 = - 21..99 + 35.802 e- 100 t (b)
Relevant wavefornls for voltage and armature current are sketched in Fig. 12.55. \)s
OL-__- L_ _ _ _~--------~--------~--------wt FO
Tl l 0 11
'I'
o / // 1-1.1 --i-- i2~
i
1
. wt
I
i wt
Fig. 12.55. Waveforms pertaining to Example 12.38.
.
..
V
(c) Mean voltage applied to armature, Vt
=~ (1 + cos a) 7t
= ..f2 x 230 (1 + cos 75°) = 130.31 V 7t
.. 130.31=87.96+fa x4
: . Annature current, fa = 10.5875 A
ButVt=Ea+fara
Torque,
Te
=K mfa = 1 x 10.5875 = 10.5875 N m
Example 12.39. A single-p hase full converter feeds a separately-excited dc motor having ra = 3 .0, La = 0.06 H and torque constant 1 Nm I A. For a source voltage of 230 V, 50 H~ the m otor runs at a speed of 1400 rpm for a firing-angle delay o{60°. (a) Derive an expression fo r the armature current (b ) d ketch the waveforms of source voltage, load voltage and motor arm ature current t c) Calculate the average m otor torque. Solution . (a) Impedance Z = 132 + (27t X 50 X 0.06)2]1 / 2 = 19.087 n
Z = 19.087 Lt an- 1 18 85 = 19.087 L SO.96°
3
Motor count er emf,
Ea =Km· rom = 1 x
27t x 1400 60
Ea = 146.61 = 48 87 A ra = Ta
3
.
,
L a
= 146.61 V
~ = 50 0.06
666
[Art. 12.10]
Power Electronics
.
l
X 230 . = -{2 19.087 sm (rot -
8096(;) .
-
4887 .
+
A-50 t
e
'
i = 17.04 sin (rot - 80.96°) - 48 .87 + A e- 50 t When
rot = 60°, or t = 1806XO2xlt1tx50 =3~O sec, let the armature current be discontinuous so
that i = O. 50
0= 17.04 sin (60 - 80.96) - 48.87 +Ae- 300 A =64;933 i =.17.04 sin (rot - 80.96) - 48.87 + 64.933 e- 50 t
or
...w
By hit and trial, when rot = 212°, i = 0.01. This shows that armature current decays to zero . a t rot = 212°. But the thyristors in a 1-phase full converter, i.e., T13 T14 in Fig. 12.9' (a), would be triggered at rot = 180 + 60° = 240°. This shows that armature current is discontinuous as depicted in Fig. 12.56. Mter rof = 212°, armature terminal voltage jumps to E and stays there till rot = 240°. Expression for armature current is given by Eq (i) . (b) The waveforms of source voltage, load voltage and discontinuous armature current are sketched in Fig. 12.56.
OL __- L_ _ _ _~--------~--------~------~ wI
wI I'
I
I I :
i
l/N.i .
~1
,/
3R
wI
Fig. 12.56. Waveform pertaining to Example 12.39. (c)
Average value of output voltage,
Vo =
~ I ~: Vms'
":it: 230 B~
.
[cos 60 _ cos 2121
\~~O x 1t 1
+
~= ~= ~ + ~~
:. Armature CUl'rent, Ta =
M otcr torqu e,
rot d (rot ) + Ea
162.357 - 146.61 . 3 = 5.249 A
T = Km 10. = 1 x 5.249 =5.249 Nm
146.6i~028 ~ =162.357 V x
[Art. 12.10]
Electric D ives
667
Example 12.40. In Example 12.39, load torque on the motor is increased and this causes the speed to drop to 600 rpm with the same firing-angle delay. Calculate the load torque and expression for the motor armature current. 21t X 600 . . . - Ea . Solution. Here Ea = 1 x 60 = 62.831 V, r~ = 20.944 A. From Example 12.39, i = 17.04 sin (rot -':80'.96) - 20.944 +A e- 50t
.
Let the armature current be discontinuous, therefore when rot = 60 0 or when t = 0 =17.04 sin (- 20.96) - 20 .944 +Ae-l/6 or A
..
Thyristors T13 T14 are triggered at rot = 240 0 , or at t =
3~ · sec, i = O.
=31.943
3~O = 715 sec.
Armature current at this instant is given by i
= 17.04 sin (240': 80.96) -
This shows that armature current is 1
·
continuo~s.
.20.944 + 31.943 e-
50/ 75
= 1.5515 A
So calculate A again . .
..
When rot = 60 0 , or t = 300 sec, i = 1.5515 A. . .. ..
1.5515 = 17.04 sin (- 20.96) - 20.944 +A e 1I6 A= 33.776
.-
:. Expression for armature current is i = ia = 17.04 sin (rot - 80.96) - 20.944 + 33.776 e- 50t
Since the armature current is continuous, 2 Vm 12 x 2 x 230 V = - - cos a = cos 60 0 o 1t . 1t Vo = 103.52 V = 62.831 + Ia ra
:. Armature current, Ia = Motor torque,
Te
103.52 - 62.831 3 = 13.563 A
=Km Ia = 1 x 13.563 = 13.563 Nm
Exampl e 12.41. An inverter feeds a three-phase induction motor whose parameters are given in Example 12.4. Calculate the source current .and torque at a full-load slip of 0.04 w hen inverter output is 400 V, 50 Hz.
If the inverter output is suddenly reduced to 360 V, 40 Hz ; determ ine, at that moment, the new value of source current and the torque. . .
.,
Solution. Im pedance seen by t h e voltage source at a slip of 0.04,
z=r rl + ; ) +j (XI +x2) =[ + o.O~ 1+ -+ 0. 6
Source current, Synchron oussp ed
-
-
j [1.6
1.6] = 11.073 L 16. 8'
400
11 =12 = 11.073 L 16.8 = 36.124 L - 16.8 0 A
= 1500 rpm, ws =50 1t rad/ sec
668
Power Electronics
[Prob. 12]
Torque, 3 2 0.4 N AB =50n . (36.124) x 0. 04 = 249.23 . m = N r = 1500 (1 - 0.04) = 1440 rpm = 0 B
Rotor speed, New synchronous speed,
Ns1
= 120 4x 45 :: 1350 rpm = 0 D.
ws1
= 45 1t rad/sec.
Te
When the inverter output is suddenly reduced to 45 Hz, the motor speed will remain substantially constant for some time due to drive inertia. .
. New slIp,
51
=
Ns1 - NT Ns1
=
1350 - 1440 1350
=-
1 15
o ~----------~~~~--------.
Impedance, Zl :: [ 0.6 -
= 6.12 L
°i4 x 5 ] +j [ 3.2 x :~ ] 151.96°
Fig. 12.57. Pertaining to Example 12.41.
Source current,
1\ =1'2 = Torque,
6. 12
Te = ~ (12,)2 (Usl
~6~51.93:: 58.823 L r2 5
3
= 45
.1t
-15 1.93° A
(58.823)2 (- 0.4 x 15) = - 440.56 Nm = Be
Since induction machine torque is negative, it is now working as induction generator in regenerative braking mode. The motor operating point A (BA = motoring torque = 249.33 Nm) at once shifts to C so that B C = 440.56 Nm = regenerative braking torque. The induction generator now operates at power factor cos (-151.93) :: 0.8824 or 0.8824 leading. Now the operating point travels from C to D. Before point D is reached, if V/fis further reduced, regenerative braking action can be obtained uptilliow values of operating speeds. PROBLEMS
the concept of electric drive. Illustrate your answer with examples. (b) Give two methods of speed control normally employed for dc motors. Hence, sketch the characteristics of a separately-excited dc motor based on these two methods . Indicate clearly ~ons taID-torque drive and constant-power drive regions. (c) Write down the basic performance equations for it de series motor. Sketch also the ch aracteristics of this motor indicating t he two regions of constant-tor que m03e and constant-power mode. 12.2. (a) Give the general circuit layout for single-phase de drives. Enumerate the various I-phase d c drives used . (b ) Describe single-phase half-wave converter eeding a separately-excited dc motor. Il lustrate your answer with waveforms and appropriate expressions . . (c) The peed of a separately-excited dc motor is controlled through slngle-phase half-wave controlled converter fed from 230-V mains. The motor armature resistance is 0.5 .Q and
12.1.
(a) Give
~
[Prob.12]
Electric Drives
669
motor constant is K = 0.4 V-slrad. For load t orque of 20 Nm at 1500 rpm'and for constant armature current, calculate (i) firing-angle delay of the converter (ii) rms value of thyristor current and (iii) input pf of the motor. [Ans. (e) 45.821°, 30.52 A, 0.6255 lag] 12.3. (a) Describe the working of a single-phase semiconverter fed dc separately-excited motor with relevant waveforms and expressions. State the assumptions made. (b) A single-phase semi converter feeds a separately-excited dc motor. If armature current is ri!,ple free, then shown that input supply power factor is given by
(1 + cos (,()
A
"1tI (2
)
1t-(,(
where (,( = firing-angle delay of semiconverter fceding the armature circuit of the motor. 12.4. A separately-excited dc motor has its armature circuit connected to one semiconverter and field winding to another semiconverter. The supply for both the converters is single-phase, 230 V, 50 Hz. Resistance for the field circuit is 100 n and that for the armature circuit is 0.2 n. Rated load torque is 80 Nm at 1000 rpm. The motor constant is 0.8 V-s/A-rad and magnetic saturation is neglected. For ripple free armature and field currents and with zero degree firing angle for field converter, determine (a) rated armature current (b) firing-angle delay of armature converter at rated load (e) speed regulation at full load (d) input pf of the armature converter and the drive at rated load. . • [Ans. (a) 48.31 A (b) 39.78° (e) 5.571% (d) 0.769 lag, 0.8338 lag] 12.5. A separately-excited dc ,motor is fed from two single-phase semiconverters, one in the arma ture circuit and the other in the field circuit. Field current is constant at 2A. Motor armature resistance is 0.8 n and motor constant is k = 0.5 V-s/A-rad. AC voltage is 230V, 50Hz. For a ripple-free armature current and speed of 1500 rpm, calculate. (a) motor current and torque for a firing angle of 30° and (b) inpu t supply power factor. [Ans. (a) 45.12 A, 45.12 Nm (b) 0.92 la,g] 12.6. (a ) Describe, with appropriate voltage and current waveforms, the working of a single-phase
full-converter fed dc drive. Derive also an expression for its input pf. State the assumptions made. (b) A 200 V, 1000 rpm, lOA separately-excit-ed dc motor is fed from a single-phase full . converter with ac source voltage of 230 V, 50 Hz. Armature circuit resistance is 1 n. Armature current is continuous. Calculate firing angle for (i) rated motor torque at 500 rpm. . [Ans. (b) (i) 59.526° (ii) 115.766°] (ii) half the rated motor torque at (- 500) rpm. 12.7. The speed of a separately-excited dc motor is controlled by two single-phase full converters, one in the armature circuit and the other in the field circuit. Both converters are fed from the same single-phase, 230 V, 50 Hz source. Armature res istance is 0.5 .Q and field circuit resis t nee is 200 n ~ Firing angle for field conv,eTter is zero and motor constant is 0.8 V- s/A-rad . Armat ure and field currents are continuous and ripple free. If annature current is 30 A for a firing angle of 45°, then calculate Ca) m otor speed (6) m otor torq a . (e) inpu t pf of the armature converter and (d ) input pf of the drive. [Ans. (a) 1515.1 r pm (b) 24.845 Nm (e) 0. 6365 lag (d) 0.6672 lag] 12.8. Describe the use of a three-phase semiconverter for the speed control of a de series mo or. I11ust rate your answer with appropriate waveform . Deri Ir e expressioru! for the rms values of sour a and thyristor currents and the average value of SCR current for (a) firi ng angle < 60 0 and ( b) firing angle > 60 0 • 12,9. The speed of a de series motor is controlled by a 3--ph se semiconverter connect ed t o 3-phase, 400V, 50Hz source. The mota consta n i 0. 4 Y-siA. rad. Tot 1 field and armature resisl;ance
670
[Prob. 12]
Power Electronics
is 1 0.. Assuming continuous and ripple free armature current at a firing angle of 400 and speed of 1000 rpm, determine (a) motor current and motor torque (b ) power delivered to motor (c ) reactive power drawn from the supply in VAr. [Ans. (a ) 11.12 A, 49.462 Nm (b ) 5303.46 W (c) 3383.085 VArJ 12.10. (a) Describe how the speed of a separately-excited dc motor is controlled through the use of two 3-phase full converters. Discuss how two-quadrant drive can be obtained from this scheme. Derive expressions for rros values of source and thyristor currents. State the assumptions made. , (b ) ,The speed of a 50 kW, 500V, 120 A, 1500 rpm separately excited dc motor is controlled by a three-phase full converter fed from 400V, 50Hz supply. Motor armature resistance is 0.1 n. Find the range of firing angle required to obtain speeds between 1000 rpm and (-1000) rpm at rated torque. [Ans. (b) 51.35° , 125.46°J 12.11. The speed of a separately-excited dc motor is controlled by means of two 3-phase full con verters, one in the armature circuit and the other in the field circuit and both are fed from 3-phase, 400 V, 50 Hz supply. Resistance of armature and field circuits is 0.2 nand 320 n respectively. The motor constant is 0.5 V-s/A . rad. Field converter has zero degree firing-angle delay. Armature and field currents have negligible ripple. For rated load torque of 60 Nm at 2000 rpm, calculate (a ) rated armature current (b) firing angle delay of the armature converter (c) speed regulation at rated load and (d) input pf of the armature converter an~ the drive at rated load . [Ans. (a) 71.10 A (b) 69.291° (c) 8.045% (d) 0.3376 lag, 0.36011agJ 12.12. A dc motor driven from a 3-phase full converter shown in Fig. 12.17 draws a dc line current of 90 A with negligible ripple. (a) Sketch the line voltageuab taking it zero-crossing and becoming positive at rot = 0 0 • Also, sketch line current iA (for one cycle) and 't hyristor current iT for a firing angle of 30°. Conduction of SeRs must also be indicated. (b) Calculate average and rms values of thyristor current. (c) Compute power factor at ac source. (d) For motor constant of 2.5 V-s/rad ar.d armature circuit resistance of 0.4 n, calculate the motor speed. 12.13. (a) Describe how the speed of a dc series motor can be controlled by means of a dc choPller. (b) A dc series motor, fed from 400V dc source through a chopper, has the following parameters : ra
= 0.05 n, rs = 0.07, n, k = 5 x 10-3 Nm/amp.2
'
The average armature current of 200 A is ripple free . For a chopper duty cycle of 50%, determine , ( i ) input power from the so urce (ii) motor speed an d (ii i) motor torque . [Ans. (b) (0 40 kW (ii) 1680.68 rpm (i i i) 200 Nml
a
12.14. A 230 V de source is connected to separat ly-excited.dc motor th rough a chopper oper ating at 500 H z. The load torque at 1200 rpm is 32.5 Nm . he motor has ra == 0, L(],:;; 2mH and Km = 1.3 V-slrad . Motor a nd chopper losses are n eglected. (a ) Calculate the minimum and maximum values of armature current and the ar mature-cur rent excursion. (b ) Obtain the expI~ssi ons for armature current during on and off periods fa chopper cy.. . le. ' [_09. (a ) 22.634 ,27. 366 A, 4.732 A (b) 22.634 + .3332 t, 27 .366 - 8168 t] 12.15. Repeat Prob 12.14 in case armature circuit of the dc motor has a resistance of 0.3
Q
· EI,ectric Drives
[Prob.12] [Ans. (a) 14.438 A, 18.95 A, 4.512 A (b) 222 .13 (1 - e- 1St) + 14.43 e- 1St; - 544.53 (1 - e-
151)
+ 18.95 e-
671 15t]
12.16. What is regenerative braking? Describe the regenerative braking of a chopper-fed separatEl ly-excited dc motor. Illustrate your answer with circuit diagram and relevant waveforms. Derive expressions for the minimum and maximum braking speeds for obtaining regenerative braking of the dc motor. Show that the. speed range for regenerative braking is (V, + Jara) : Jara'
12.i7. A 220 V, 60 A dc series motor, having combined resistance of armature and field of 0.15 n, is controlled in regenerative braking mode. The dc source voltage is 220 V. Motor constant is 0.05 V-s/A.rad. The average motor armature current is rated and ripple free. For a duty cycle of 50%, determine (a) the power returned to the supply, (b ) minimum and maximum permissible braking speeds and (c) speed during regenerative braking. [Ans. (a) 6.6 kW (b) 28.65 rpm, 728.9 rpm (c) 378.82 rpm] 12.18. (a) Distinguish between two-quadrant and four-quadrant drives. (b) Describe how a four-quadrant drive can be obtained from a chopper-fed separately-excited dc motor. 12.19. (a) What are ac drives? Give the merits and demerits of ac drives with respect to dc drives. (b) From the approximate equivalent circuit ofa 3-phase induction motor, derive therfollowing expressions : Torque at any slip, slip at maximum torque, maximum torque, maximum to~que and slip at which it occurs in case stator resistance is neglected. 12.20. (a) Enumerate the various methods of speed control of a 3-phase induction motor when fed through semiconductor devices. (b) Describe stator-voltage-control t echnique for the speed control of a 3-phase induction motor. 12.21. (a) For fan-type load.( show that rotor current in a 3-phase induction motor is maximum ~hen slip s = 1/3. State the assumptions made. (b) ~ 400 V, 50 Hz, 3-phase SCIM develops full-load torque at 1470 rpm. If supply voltage reduces to 340 V, with load torque remaining constant, calculate the motor speed . Assume speed-torque characteristics of the motor to be linear in the stable region. Neglect stator resist ance. .. [Ans. (b) 1458.5 rpm] 12.22. (a) Induction motor speed control with constant-supply voltage and reduced-supply frequency is r arely used in practice. Justify this statement. (b) Describe stator frequency control for the speed control of 3-phase induction mot or. Derive expressions for motor torque, maximum tor que and the slip at which it occurs. Stat e the various assu mptions made. . . Discuss, why during this method of speed control, an induction motor is said to be working in field-weakening mode.
a
12.23. A 3-phase, 400V, 20 kW, 970 rpm , 50 Hz, delta-conn ect inductionmotoT has rotor leakage impedance of 0.5 + j 2.00 n. Stator leakage impe ance and rotational losses are assumed negligible. If this motor is energised frOlj,l a source of 3-phase, 400 V, 90Hz, t hen compute ( a ) the motor speed at rated torque (b) the slip at which maximum to que occurs and (c) the m aximum torque. [Ans. (a) 1819.8 rpm (b) 0.138 (c) 235.785 Nml 12.24. Explain voltslhertz control for a 3-phase induction motor for its speed control. Enumerat e its advantages.
Describ e at least t wo inverter cir cuits used for volts/her tz control.
I
672
Power E lectronics
[Prob . 12]
12.25. (a ) Discuss how voltsihertz control for a 3-ph ase induction motor is similar to armature-volt age control of a dc motor. . . (b ) In stator-frequency control of a 3-phase induction motor, explain why (i) ratio Vlfia maintained constant for speeds below base speed (ii) terminal voltage is maintained constant for sl:leeds above base speed. 12.26. A 3-phase, 15kW, 420V, 4-pole, 50Hz, delt a-connected induction motor has the following per-phase parameters referred to stator :
rl :;:: 0.5 n, r2 :;:: 0.4 n, Xl:::: X2 = 1.5 n, Xm = 0
If this motor is operated at 210V, 25Hz with DOL starting, calculate (a ) current and pf at the instant of starting and under maximum torque conditions; compare the results with normal values, (b) starting and maximum torques and compare with normal values. [Ans. (a) 120.05 A, 0.5145 lag, 81.862 A, 0.8112 lag At nonnal : 134.1 A, 0.2873 lag, 90.486 A, 0.763 lag 220.2 Nm, 404.702 Nm At normal: 137.38 Nm, 475.662 Nml 12.27. Describe stator-current-control method for the speed contr!>l of a 3-phase induction motor. Derive expressions for maximum torque, slip at maximum torque' etc. by using approximate equivalent circuit. . Discuss the effect of saturation on the speed-torque characteristics obtained by this method of speed control. . 12.28. A 420 V, 6-pole, 50Hz, 3-phase, star-connected 1M has rl = 0, xi = X2 = 1.2 0, r2 = 0.5 nand Xm =50 n as its per-phase parameters referred to stator. This 1M is fed from (i) constant-volt-. age source of 242.5 V per phase and (ii) constant-current source of 30A. For both types of sources (i) and (ii), calculate (a ) the slip for maximum torque, (b) the starting and maximum torques and (c ) the supply voltage required to sustain the constant current at the rilaxi ~m torque. [Ans. (a ) 0.2108, 0.00976 (b) 136.71 Nm, 12.293 Nm ; 338.676 Nm, 629.47 Nm (c) 1838.1 Vj (b)
12.29. Sketch speed-torque characteristics of a 3-phase induction motor as influenced by different control techniques employed for its speed control by using semiconductor devices. Explain the three regions into which these characteristics can be subdivided. Discuss the various methods empl()yed for obtaining the three regions. 12.30. Discuss, in detail, how Ward-Leonard type of characteristics can be obtained from a 3-phase induction motor.
Explain also how high-speed series-motoring region is obtained in a 3-phase induction motor.
12.31. (a) Describe static rotor-resistance control method for the speed control of a 3-phase induction motor . Derive expressions for inductor current and motor speed in terms of load torque, supply voltage, motor turns ratio, synchronous speed etc. (b) A 3-phase SRI M uses r otor ON-OFF control by means of chopper for its speea control. The effective rotor resistance is increased t o 10 times during off pe riod. If the motor develops 0.4 pu torque at a slip of 0.02 for normal operation, calculate the average torque developed at the same slip for 20%, 50% and 80% dut y cycles of the st a tic rotor chopper. [Hint. (b )At low values of slips,
V2
T = K -s e
r2
[prob. 121
Electric Drives
673
. y2 y2 . Tel = O.4=K- X 0.02 or K - = 20 r2
2
KV
Te2 = 10r2 s =
r2· .
20x 0.02 10 = 0.04 p u
_ Ton ' Tel + T off ' Te2 Net torque, T net T etc.]
[Ans. (b) 0.112 pu, 0.22 pu, 0.328 pu]
- 12.32. A 3-phase, 400V, 6-pole, 50 Hz, star-connected SRIM uses a chopper in its rotor circuit for its speed control. The effective turns ratio from rotor to stator is 0.6. Induc tor current is ripple free . Losses in the rectifier, inductor, chopper and no-load losses of the motor are neglected. Load torque, proportional to speed squared, is 360 Nm at 970 rpm. (a) For a minimum motor speed of 600 rpm, calculate the value of chopper resistance R. For the value of R obtained in part (a) if the speed is to be raised to 800 rpm, calculate (b) inductor current (c) duty cycle of the chopper (d) rectified output voltage (e) efficiency in case per-phase resistances for stator and rotor are 0.015 0 and 0.02 0 respectively. [Ans. (a) 3.4960 (b) 65.93 A (c) 0.7187 (d) 64.823 V (e) 79.285 %]
12.33. (a) In static rotor-resistance control of a 3-phase SRIM, each diode in the rotor circuit conducts for 120°. Assuming ripple free rotor current, derive expressions for rms value of rotor current referred to stator, fundamental component o'f rotor ·c urrent and its value referred to stator. (b ) A 3-phase, 415V, 50 Hz, 1470 rpm, star-connected SRIM has the following per-phase . parameters referred to stator : rl = 0. 120, r2 =0.1 0, xl =x2 =0.4 O,Xm = 0 Effective per-phase turns ratio from rotor to stator = 0.8 Speed of this motor .is controlled by rotor ON-OFF control. For a speed of 1200 rpm, the inductor current is 100 A and chop per resistance is 1.8 O. Calculate (i) the value of chopper duty cycle (ii) efficiency for a power output of 25 kW and for negligible no-load losses (i i i) the input power factor. . [Ans. (b )(i) 0.498 (ii) 67 .84% (iii) 0.822 lag] 12.34 . Describe static Kramer drive for the speed control of a 3-phase SRIM and show that steady stat e torque is not influenced by whether a transformer is used or not. Derive appropriate expressions to obtain speed-torque characteristics of static Kramer drive: 12.35. Speed of a 400V, 6-pole, 50Hz, star-connected SRIM is controlled by static Kramer drive. The effect ive phase turns ratio from rotor to stator is 0.6 and transformer has phase turns ratio from l.v. to h.v. as 0.4. The inductor current is ripple free. Losses in diode rectifier, inductor , inver ter and transformer are negleded. The load torque is proportional to speed squared and its value is 250 Nm at 800 rpm . For a motor operating speed of 700 rpm, calculate (a ) rotor rectified voltage (b) induc tor current (c) delay angle of the inverter (d) efficiency in case inductor resistance is 0.01 0 and per-phase resistances for stator and rotor are 0.015 o and 0.02 0 r espectively (e ) For the firing angle obtained in part (c), the load torque is increased to 350 Nm, find the motor speed. [Ans. (a ) 97.23V (b) 61.84 A (c) 116.743° (d) 98.37% (e) 696.53 rpm)
12.36. (a ) Describe a st tic Kramer drive and show that the slip s at which it operates is given by aT s = - - cos a. a where a a nd aT pertain to per-phase turns ratio for SRIM and transformer respectively. (b ) Speed of a 6-pole SRIM fed fro m 400 V, 50Hz source i s controll ed by sta ic "Kramer driVe . The inverter is directly connected to the supply. If t he motor i requir pd to oper ate a t 800 rpm, determine the firing advan ce angle of the inverter. Volt age · c ro~s the ope n·cir cu·ted
674
(Prob . 12]
Po wer Electronics
s lip rings at stand-still is 600V. Ther e is a voltage drop of 0.7V and 1.5V across each of th e diodes and thyristors respectively. Inductor drop is neglected. In case transformer is to be interposed between supply and the inverter for obtaining a minimum speed of 600 rpm, determine the voltage ratio of the transformer from l.v . to [Ans. (b) 106.97°,0.6] h .v.
12.37. Repeat Problem 12.36(b) in case there is an overlap angle of 12° in the rectifier and 5° in the inverter. . [Ans. 76.305°] 12.38. Explain, with relevant circuit diagrams, both types of static Scherbius drives for obtaining speeds below as well as above synchronous speed. 12.39. (a) Enumerate the various types of synchrouous motors. Derive an expression for power developed in a cylindrical-rotor synchronous motor with negligible armature resistance. (b ) A 415V, 50Hz, 4-pole, star-connected synchronous motor has Xs = 1.5 n. Load torque, propor tional to speed, is 300 Nm at synchronous speed. The speed ofthemotor is lowered by keeping
-7 constant and maintaining 0.8 pf leading by field .control. For the motor operation at 840 rpm, calculate (a) supply voltage (b) the armature current (c) the excitation voltage (d) load angle and (e) the pull-out torque. Neglect rotatiouallosses . [Ans. (a) 232.4 V (b) 44.96 A (c) 276.635 V (d) 10.904° (e) 870.15 Nml
12.40. (a) Derive the expression for power developed in a salient-pole synchronous motor in terms of excitation voltage, load angle etc. Neglect armature resistance. (b) Explain the working and uses of a permanent-magnet synchronous motor. How does the input-current waveforms effect the constructional features of PMSM ?
.I
Chapter 13
Power Factor Improvem ent
......•...•....••..•....•....•.••..•...•.....•........•....•.•••..••......•...........••••....••... In this Chapter • • • •
Effect. of Poor Power-Factor Methods of Reactive Power Compensation Static VAr Compensator (SVC) Some Worked Examples
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• w ••••••••••••••••••••••••••••••• · · · · · ·
There are several techniques available for the improvement of system po\ver factor. Out of all these, the one employing thyristor controlled reactor (TCR) is now the most sought-after method. It is because TCR offers a .continuous and a very fast control of reactiye~power flow in a system for regulating its power factor. Primarily, popularity of this method is due to the use ofthyristors which are fast, reliable, efficient and precise in their operation. In this chapter, power-factor improvement techniques in ac systems, with more emphasis to TCR, are discussed.
13.1. EFFECT OF POOR POWER-FACTOR
.
.
First of all, the effect of poor load power-factor on the system performance would be illustrated through an example. Examp le 13.1. An alternator, with fixed source voltage of 250 V, delivers power to a load. The transmission line reactance is 5 n and load current is 20 A. Calculate the load voltage, voltage regulation, system utilization and the energy consumed for a load power factor of (a) unity and (b) 0.5 lagging. Solution. (a) Unity pf load. The phasor diagram for unity ioad is shown in Fig. 13.1. (b), where VL = load voltage and E = source voltage, 250 V. Fig. 13.1 (a) represents the circuit diagram whereE, VL> load and reactance are shown. 20A
100 V
E = 250 V
90· '--. ilL
+ I
E (\)
250V
J
I =20A
o (3)
(b)
0 (e)
Fig. 13. 1. Perta ining t o Example 13.1 (a) circui t diagram and phas or diagr ams at (b) unity pf and (c) pf = 0.5 laggi ng.
676
Power Electronics
[Art. 13.2]
From Fig. 13.1
(b) ,
Load voltage, V L = 229.13 V . 25 0 - 229.13 Voltage regulatIon = 250 x 100 = 8.35%
or
Load power
=VLI cos e = 229.13 x 20 xl = 4582.6 W
Maximum possible system rating = 250 x 20 x 1 = 5000 W
4582.,6 .
System utilization factor = 5000 x 100 = 91.652% · d per h our Loa d energy d e1were
. 4 .6 um'ts = 4582.6 1000 :::
Assuming Rs. 5 per unit, revenue earned =4.6 x 5 = Rs. 23 per hour. (b) Load power factor = 0.5 lagging. For this power factor, the phasor diagram is as shown in Fig. 13.1 (c). From this figure, OA2+AB2=E2
(0.5 V L)2 + (0.866 V L + 100)2 = 2502 or VL =158.35 V
or
= 250 ;;;8.35 x 100 = 36.66% .
Voltage regulation
Load power =VL I cos System utilization factor
e = 158.35 x. 20 x 0.5 =1583.5 W
1583.5 = 5000 x 100 = 31.67%
· d eIivere. d to 1oad' per h' our = 1583.5 1 58 um'ts E nergy 1000 ':::.
Revenue earned = 1.58 x 5 = Rs. 7.90 per hour It is seen from the above example that for the same source voltage and load (or line) current, a load at a poor pf effects the system performance as under: (i)
Load power is reduced by
45824~;2~:83.5 x 100 = 65.45%.
As a consequence, revenue
earned falls from Rs. 23 per hour to Rs. 7.90 per hour. (ii) Load voltage falls to 158.35 V. As a result; utility devices, like fluorescent tubes, lamps, refrigerators, washing machines etc., designed to operate at 230 V, would operate erratically or may even fail to operate. (iii) System utili zation is reduced from 91. 52% to 31.67%. It means that system infrastructure is n ow utilized up to 31.67% of its installed capacity. Underutilization of the system, associated with ow revenue earnings, cannot be tolerated by an efficient organizatL• . At the same time, consumers are peeved at the poor and unreliable .perlormance of their utility devices. . This example demonstrates that load pf shouid be improved and made as close to unity as is ,economically viable. 13.2. METHODS OF REACTIVE POWER COi'U·ENSATION Industrial loads, which no lal1y operate at poor pf are induction motors, arc and indu ciion furnaces. Fluorescent tubes, fans etc also oper ate at low value of power fact or. All these loade ,
Power Factor Improvement
[Art. 13.2]
677
working at low pf, need large amount ofreactiY~.J2ower which results in.. r.educed voltage level at the load tenninals. A low voltage at the.cOI1stlIner~ tefmina]s is undesirable as it leads to impaired perfonnance of their utility devi~es. · The.various methods of power-factor improvement are as under: (i) Use of capacitor banks . (ii) Use of synchronous condensers (iii) Use of static VAx compensators.
These are now discussed in what follows.
13.2.1. Capacitor Banks "
A bank of capacitors is connected across the load. Since the capacitor takes leading reactive power, overall reactive power taken from source decreases, consequently system power factor improveS. Example 13.2 illustrates how capacitor bank renders the improvement in system power factor..
Example 13.2. A single-phase induction motor, when running from 230 V, 50 Hz supply, gave the following data : . No-load: Half-full load: Full-load : .
2A, pf = 0.3 lag. 5A. pf = 0.5 lag. lOA, pf = 0.7 lag;
Ca l,culate the capacitance required in parallel with the induction motor so that power-factor of the motor-capacitor combination (or the supply power factor) is raised to unity under all the three operating modes listed above. C
Ie 90' lis o ~~----~~------_
B
(b)
(a )
Fig. 13.2. (a) Circuit diagram and its (b) its phasor diagram. Pertaining to Example 13.2.
Solution. A capacitor, connected in parallel with I-phase induction motor, is shown in Fig. 13.2 (a). Its ph asor diagram is drawn in Fig. 13.2 (b ), where OB = motor current 1m lags the source voltages VB by motor operating p f cos e and.OC = capacitor current Ie leads Vs by 900 • For unity power factor o.~he combination, OC must be equ al to AB 1m sin 9 for all the eo operating modes.
=
(a)
At no load,
or B ut
OC = AB I e =1m sin 9 =2 sin [cos- 1 0.3] Va . Ie =X = 27tf C.V. =1.908 A
c
.
.
C = 1.908 X 10
6
27t X 50 X 230
=26.406 )IF
=1.908 A ..
678
[Art. 13.2] (b) At h alf-full load,
Power Electronics
OC =AB
Ie = 5 sin [cos· 1 0.5] = 4.33 A =
or
.
l.V = 21t f C.vs c
6
4.33 X 10 C = 21t X 50 x 230
.. (c)
= 59.925 :~F
At full load, OC =AB
= 10 sin [cos- 1 0.7] =7.1413 A =21t fC.Vs 6 C = 7.1413)~ 10 = 98.834 F
or
Ie
21tx 50 x 230
~
This example illustrates that for keeping the supply power factor u'nity, the value of capacitance across the motor terminals must be varied as the load on the induction motor alters. This is called dynamic VAr compensation or dynamic pf control; that is, reactive power compensation is carried out through switching-in or out of the capacitors so as to achieve a desired pf at all load conditions. A continuous control of the pfwould entail the need of a large number of capacitors of small ratings. The switching-in or out is carried out by means of relays and circuit breakers. But this all is quite cumbersome and expensive. The mechanical switches and relays are sluggish, unreliable, require frequent maintenance and introduce switching transients. However, with the replacement of mechanical switches by thyristors, it has been made possible to continuously (i) regulate the reactive power flow and (ii) control the power factor and voltage profile by rapid switching-in or out of the static capacitors. 13.2.2. Syn ch ronous Condensers A 3-phase synchronous motor, when overexcited, wor ks as a synchronous condenser, or a capacitor. It gives dynamic power-factor correction over a wide range of its excitation. When underexctt ed, it operates at a lagging power factor and therefor e absorbs reactive power from the bus. When over excited, a synchronous motor works at a leading power factor and therefore acts as a generator of reactive power and therefore behaves like a capacitor. A static capacitor bank provides pf control in discrete steps whereas a synchronous condenser fur nishes a continuous control of power-factor improvement and the associated reactive power flow. A synchronous condenser, however, suffers from the following drawbacks.
It has more losses as compared to capacitor bank. (i i) A synchronous condenser can be installed at one place only, whereas capacitor bank can be distributed at many places . A distributed capacitor bank is more effective in contr olling the reactive-power flow and "oltage profile. (iii ) A synchronqus conde nser is slmv in resp onse due to large time constant of its field circuit, whereas a capacitor bank offers faster r esponse. .... 1 3. 2.3 ~ Thyri st or C~ntroll ed R eactors (TCRs) Thyristor controlled r eactor is a m aj or component of ~ atic VAr compens at o . In t his section , ()nIy TCR L des cr i.bed. Static VAr compensator is explruned in t h e next article 13.3. (i)
Static thyristor con trolled re actors are connected in parallel with the load or th e control of r eaGtive pow er fl ow. With increase in the size of industrial connected loads, fast r eactive power . compens ' tion h as become necess ary. F or such loads , t hyristor controlled reactors (TCR ) are now becoming increasingly popular. TCR is also called thyristor controlled inductor (TCl). Fig. 13.3 . (a ) sh ows a linear react or (or inductor ) L connected to ac source Vs th ro ugh two thyr is ors conne ct ed in an t ip arallel. This circuit confi gurat ion is also call e ac voltage controller. Ch£1pt er 9. It can, therefore, be said that in Fig. 13.3 (a), a lin e r in du ctor L is
Power Factoc Improvement
{Art. 13.2]
679
T1
T2
(\) U s
L
=Vm sinwt
(a) "
"
Fig. 13.3. Thyristor controlled reactor (a) circuit diagram and (b) its voltage and current waveforms.
connected to I-phase voltage controller. A reference to Art. 9.3 is therefore of considerable benefit. During positive half cycle of source voltage, T1 is turned on and during the negative half cycle, T2 is tUTIled on. For firing angle ex = 90°, the source current is is continuous as shown in Fig. 13.3 (b) . The circuit behaves as if inductance L is dir ectly connected to as source without .seRs. For ex = 90°, is is a sine wave, its fundamental component ifl is the same as is and is
.
therefore maximum. Af3 a resul~, inductive reactance offered by reactor, XL
V
=i
is minimum f1 when ex =90°. Here Vs = rms value of source voltage and If1 is the rms value of fundaIr..9ntal component of source current, which for ex = 90° is given by If1 =Is (rms value of source current). For firing angle ex >90°, current is is discontinuous, Fig. 13.3. (b) , but its fundamental component ifl again lags Vs by 90°. With ex> 90°, as rms value of fundamental component If1 h as decreased, the inductive reactance offered by reactor (= V/lfl) has become more. If ex is further increased, fundamental component of is wouldpe further reduced and therefore, re actance offered by the reactor wou ld be more" pr o"h o u nce d . For fi ring an gl e , ex = 1800 , is = 0, if1 = 0 and theoretically, the inductive reactance offered by the reactor would be infinite. This shows that with firing angle control from (X = 90° to 180° ; the effective r ea tance of the reactor, as seen by the source, can be regulat d from its actual value XL = 2rr:fL when ex = 90°, to an' nite value when ~ = 180°. As the fun damental component of source current lags the so :Tc e VU .Lage by 90°, the reactor (or inductor) consumes n o power. It draws only the r eactive power. vVhen
C(
= 90°,
Ifl
Vs
=Is = roL '
: . XL
Vs
=Tn s
=rms value of source volt age "Is =rms value of source current Ifl =rIDS value of fun damental compon ent 0
w e e Va
source current .
Power Electronics
[Art. 13.3]
680
Actually, for 0° :S ex :S 90°, there is n o control over the L'1 duct6r L of Fig. 13.3. (a), therefore, ...(13.1a) ...(13.1b)
or
For a:> 90°, the Fourier analysis of inductor current waveform gives the fundamental component Ifl as under: I f1 =
VB
,_. 1tu.u... T
The reactive power drawn at a
[21t - 2a + sin 2a]
At
-Q = VB' Ifl =
a~ 90' , reactive power ( ~
Q (from Eq. (13.4» is zero. -
_...(13.2)
~ =Vs Is =-roL 2
-
2a + sin 2ex]
= 90° or for 0° ~ ex :S 90°, is
Q = VB Ifl For 90° :S ex :S 180°,
VB =X [21t 1t. L
Vs
,.T
1t u.u...
.. .(13.3)
-
.
... (13.4)
[21t - 2a + sm 2a]
:t)
drawn is maximum. When a
~ 180', reactive power
-
.
13.3. STATIC VAr COMPENSATOR (SVC) A static VAr compensator is also called static VAr compensating system (SV8 ) or thyristor controlled compensator (TCC). It consists of a thyristor controlled reactor (TCR) in parallel with a fixed capacitor C. As st atedbefore, TCR (shown in the dotted r ectangle ) is made up of two • 10 Is thyristors Tl, 1'2 connected in antiparallel and a + SVC I series-connected linear inductor L. An SVC, equal rL.--- · to TCR + fixed capacitor C, is shown in dash-dot 10 i .....---+--~-., Ie IL rectangle in Fig. 13.4. Load is connected in parallel ---:---41 I i - with static VAr compensator in Fig. 13.4. i
._----_._.,
I'
r-------
,
Capacitance has a constant value C ; it, ther elore, supplies constant leading-reactive power equal to UlCV;.When both seRs are fired at a = 90 0, L seen by source is minimwn, therefore TCR takes maximum lagging-reactive power equal to V;lroL . The values of-
(\) Vs
i
,I
c
F~~ _
capaCitor
'I 1
i
iI, T2 i
,,
TCR~ _
- _
I
•
T1!
!I
I
i ,,- .I , i I I
: ______J
i
L
o
A D
L._____ __ ._____ .\: j L and C are so selected that m aximum value V; I roL Is 10 5VC is somewhat more than this means that SVC takes lagging-reactive power from the saurre when Fig. 13.4. Static VAr Compensator (SVC). a = 90°. As a result, pf of the combin ation (load plus SVC) gets impaired a little when a = 90°. For firing angle Ct = 90°, TCR current h is maximum. The current Ie th rough capacitor C is less than th e magnitude of IL for c£ = 90 a s stat ed above. For a load p f cos 80 , the ph asPT diagram indicating load current 1o, TCR current IL , capacitor current Ie and source current Is (= phasa! sum of 10' l L' Ie) is as shown in Fig. 13.5 (a) fo r a = 90°. The net reactive current
coev; ;
Power Factor Improvement
[Art. 13.3]
681
OA (= IL - Ie) lags Vs by 90°, th¢ pf of the combination, therefore, deteriorates from cos 90 to cos 9. The reactive power from the supply has increased from Vs 10 sin 60 to Vs Is sin 6. In other words, lagging-reactive power taken from the source has increased by Vs (IL - Ie). Note that fundamental component Ifl ofTCR current IL for 0.= 90° is the same as Iv see Fig. 13.3 (b). In other words, Ifl == IL when a = 90°. For firing angle a = aI' where a l > 90°, the inductive· reactance roL offered by reactor is more, likewise Ifl (fundamental component of h) gets reduced. This is shown in Fig. 13;5(b). Now, net reactive current, leadingVs by 90°, is OB (= Ic- Ifl) and pf gets improved from cos 90 to cos 6 1, The net reactive power taken from the supply is now V Is sin 9 1, less than the load reactive power Vs 10 . sin 60 , Ie Ie . r:J. =90°
p! =cos e togging
For firing angle a = ~, where a2> 0.1' the indu~tive reactance offered is still more and h and therefore Ifl are further reduced. As a result, net reactive current leading Vs is now OC> OB of Fig. 13.5 (b) and power fa etoris sh own to become unity in Fig. 13.5 (c ). The reactive power taken from the supply is now zero. F or firing angle a = a3' where a 3 > Ct2, the inductive reactance is more, IL orlfl is less. As a result , net reactive current OD (= Ie - I fl ), leading Vs by 90°, is more than DC of Fig. 13.5 (c) . Power factor now becomes leading as shown in Fig. 13.5 (d). The react ive power, ~qual to Vs Is sin 63 is now retu~ned to the supply. For Ct = 1800, h or I fl =0 and pf is still further improved (not sh own). . Examination of Fig. 13.4 reveals that fundam ental component of SVC current I is given by
I = Ie + fundam en al component Ifl of TCR current given by Eq. (13.2). As Ie and Ifl oppose ea ch V other (see ph asor diagrams in Fig. 13.5), 1= Xs -Ifl c
From Eq. (1 3.2),
Vs . Vs 1= - - --
Xc
7t
CJ)
L
(21t - 2 a + sin 2 a)
682
Power Electronics
(Art. 13.3]
or
I
1 (21t = Vs [ we - .'1twL
2 ex + sin 2a)
1... for '2 < ex < 1t
1t
... (13.5)
If SVC net current I from Eq. (13. 5) turns out to be positive, SVC delivers reactive power to load to improve (i) system pf and also (ii) the load voltage profile. In case net current I is negative, SVC absorbs reactive power from source and impairs the system pf as wen as the voltage profile. It is seen from above that by appropriately controlling the firing angle of TCR from 90 0 to 180 , the reactive power taken from the supply can be regulated continuously and power-factor of the combined static VAr compensator and the load can be improved. In other words, source p f gets improved from lagging to leading as the firing angle ofTCR is progressively altered from 90 ~ t o 180°. Since the circuit response is fast, dynamic stability of the system voltage also im proves. 0
Fig. 13.3 (b) shows that inductor current waveform for ex > 90° is not a sine wave. Fourier an alysis of this waveform shows it to consist of fundamental component as given by Eq. (13.2) and higher harmonics of order n = 3,5,7,9, 11 etc. In 3-phase circuits, the TCI (or TCR) is connected in delta so that triplen harmonics are confined to this closed delta and do not enter the ac system. The fixed capacitor C in parallel with TCR delivers VAr to the system and at the same time, filters out high-frequency harmonics . Fifth and seventh order harmonics are also . filtered out by suitably designed series-tuned filters which, in addition, contribute some leading VAr to the system; these series tuned filters are appropriately connected in parallel with SVC. Examp le 13.3. A reactive load is connected to i-phase 230 V, 50 Hz source. Load current is observed to vary between two extreme limits of (4 - jO) A and (6 - j 10) A. It is required that supply pf be maintained at unity by using a fixed capacitor and TCR of linear characteristics in parallel with the load. Determine the required values ofcapacitor and indu.ctor. What should be the fi rin u angle of TCR at the two extreme limits of load current? Solution . When load current is (6 - j 10) A, the phasor diagram is as shown in Fig. 13.6 (a). Now capacitor current Ic must be equal to 10 A so as to cancel 90° lagging current h = 10 A in order that source pris unity.
Vs Ic = 10 = X
= 21trC.Vs
c.
10 X 10 6 C = 21t X 50 x 230
= 138.396 IlF
For load current of (6 - j 0) A, TCR should not take any current , therefore its firing angle should be 180 0 • When load current is (4 - )0) A, t he current Ie in the fixed capacitor must be cancelled by in u ctive current.
Vs 90° 4.ll,
(a )
6A
(b )
Fig. B .S. Phasor diagram pe taining to Exampl e (a ) 13. 3 and (b) 13.4.
[A rt. 13.3]
Power Factor Improvement
683
As Ie = 10 A, inductive current IL has to be 10 A.
Vs Vs IL=XL =27tx50xL =10A 230 x 103
L = 27t X 50 x 10 = 73 .211 mHo
or
For load current (4 - jO), a = ~Oo :. Firing angle of TCR for the two extreme limits is 90 0 and 180 0. Examp le 13.4. In Example 13.3, if load current attains a value of (6 - j4) A, find the firing angle of the TCR.
Solution. Load current of (6 - j4) A is shown in Fig. 13.6 (b). Here Ie = 10 A, load current = 4A lagging V s by 90 0 • Therefore, fundamental component of TCI (or TCR) current 1/1 must be 10 - 4 = 6 A. In Example 13.3, XL
. From Eq. (13 .2),
Vs
230
=1L = 10
= 23
n.
Vs
1/1 = - X ,[27t - 2 a + sin 2 a] 7t.
L
230 .
or
6 = 7tx23 [27t - 2a + sin 2a]
or
2a- sin 2a = 4.39823
By hit and trial,
a
= 108.65 0
Example 13.5. A TCI is fed from 230 V, 50 Hz and has an inductance of 10 mHo Calculate" the effecti ve indu ctance seen by the source for firing angles of 90 0 , 120 0 , 150 0 , 1700 , 175~ and 180 0 • . . Solution. From Eq. (13.1b) , effective load inductance as seen by the source,
Vs
L ff = - e W Ifl
From Eq. (13.2),
Vs
1/1 = -X [27t - 2a + sin 2a] 7t
L eff =
. when
(J.
= 90
0
,
L
Vs 7t. X L [
co .T
1 ] 27t - 2a + sin 2a
= [27t -
7t L 2a + sin 2a]
. 7t L L eff = [27t - 7t + SIn . 1800] = L = 10 mH
L
eff
=
7t X 1 0 x 10- 3 = 25. 575 mR 47t 0 [27t - 3 + sin 240 ]
7t x 10 x 10- 3
L eff = (360
L ert' ~
0 -
300) l~ O + sin 300" 7t x 10
x 10- 3
(3 60 - 340) I~O + sin 340
= 173 .40 mB
= 4.45 9 H 0
684
Power Electronics
[Art. 13.4]
L
.:. . eff- [
.' . For a = 180
0
,
.
.'
7txlOx lO7t
3 .
(360 - 350) 180 + sin 350
=35.51H
0
L eff =00 .
This example demonstrates that effective inductance seen by the source rises as the firing angle of both the thyristors is increased from 90° to 180 0 • Example 13.6. A single-phase inductive load of 100 k W + j 50 k VAr is supplied from 11 kV, 50 Hz source. The static VAr compensator (SVC) has fixed capacitors of rating 100 kVAr whereas TCl can draw a maximum of 100 kVAr. For raising the system power factor to unity, find the firing-angle delay of TCland its effective inductance. .
'
x: =
100 kVAr
11000 2 50 x 100,000
= 3.8515 H
Solution~ Maximum value of Tel reactive power,
. ~
:. Inductance of Tel,
L = 27t/ Q = 27t X
Q ==
~.
I t is seen'from Fig. 13.5 that for unity pf operation, the reactive power of fixed capacitors ,.; reactive power ofTeI + reactive power ofload.
V;
100,000 = - X [27t - 2a + sin 2al + 50,000 7t.
V2
But
X:
L
,;:~::,.. )'-
= 100 kVAr = 100,000 VAr
100000= 100,000 [27t - 2a + sin 2al + 50,000 , 7t or
2a - sin 2 a
= 1.5 x 1t
By hit and trial, a = 113.83 Effective value of Tel inductance at this firing-angle delay is given by (from Example 13.5)
, 7t x 3.891 5
L eff = [27t _ 2 x 113.83 _ sin (2 x 113.83)] = 7.7039 H
0
13.4. SOME WORlillD EXAMPLES
In this article, some typical examples on some topics in power electronics are presented.
Example 13.7. Define the following parameters pertaining to inverters,'
harmonic factor of nth harmonic, total harmonic distortion.
H ow do these param eters help in evaluating the quality of inverters.
Solution. The output voltage obtained from inverters is not sine wave. It consists of
fundament al compon ent-plus certain harmonics. Lower the harm onic content in the output
voltage wave, bett er is the quality of an inverter. More important performance parameters of
inverters are harmonic factor for any (nth) harmonic and total h armonic distortion (THD).
These are explained below. .
1. Harmonic factor of nth harmonic (Pn)' It is defined as the ratio of rms value of nth
harmonic voltage component, to the rms value of th e fun dament al voltage component.
Pn =
~on
I
01 I
... (13.6)
..
I
,!
Power Factor Imp rovemen t
[Art. 13.4]
685
where Von = rms value of the nth harmonic component of output voltage and VOl = rms value of fundamental component of output voltage. For example, harmonic factor of 5th harmonic is . V05
Ps=
VOl
where Vos
= rms value of 5th harmonic component of output voltage.
Harmonic factor (HF) is a measure of the contribution of any individual harmonic to the inverter output voltage. Higher the value ofHF for anyone harmonic component, greater is the contribution of that particular harmonic component. 2. Total harmonic distortion (THD). It is defined as the ratio of rms value of all the harmonic components, to the rms value of fundamental component. . . T H D = Voh
...(13 .7a)
VOl
where Voh
001
= ~V;r -
... (13.S)
= rms value of all harmon~c components present in the inverter output voltage and
Vor
= rms value of inverter output voltage, including fundamental
plus all the har
monies.
= ~V~I" -
THD
001 =["( Yor )2,_1]112
...(13.7b)
I VOl
VOl
THD is a measure of the waveform distortion. Lower the '{alue of THD, closer is the waveform to sine wave.
Example 13.8. A single-pha'se full -bridge inverter, using transistors and diodes, is feeding a load ofR = 30. with input dc voltage of 60V Calculate (a) rms value of (i) output voltage and (ii) fundamental-component of output voltage (b) output power (c) fundamental-frequency output power (d) average and peak currents of each transistor (a) harmonic factor for third harmonic and (g) THD. Solution. Here Vs (a)
=60 V,
R
=3 0.
Rms value of output voltage, VOl" = ~V;
X
~ =Vs = 60 V
1t
Fr om Eq. (S.26), rms value of fundamental component of output voltage, 4 x 60
-
.I
VOl
= T2 x 'It = 54.02 V
~r -60 = 1200 W P 0=-= 2
(b) Output power,
R
3
V~l 54.02 =If = 3 = 972. 72 W 2
(c)
F undament al-frequency output power, PO l
P eak current of each transistor == 60 3
every cycle of 2'1t radians .
(d )
= 20 A. Each
transistor conducts for 'It radi n s for
686
Power Ele.ctronics
[Art. 13.4]
:. Average current of each transistor . (e)
=.20 x ~ = 10 A 2n
Peak reverse blocking voltage of each transistor
=V, =60 V.
4VII 4Vs V 03 = :3 . ...f2 x n' VOl = T2 x n
(f) Here
4Vs
...f2xn
1
.
P3 = 3 . 0\f2 x n x 4Vs = 3" = 0.3333 or 33.33% (g) From Eq. (13.8), rms value of all harmonic voltages is
Voh ="';602 - 54.022 = 26.112 V From Eq. (13 .7a),
Voh 26.112 .
THD = VOl = 54.02 = 0.4834 or 48.34%.
Example 13.9. A single-phase full bridge inverter, employing transistors, is fed from 220 V dc and output frequency is 50 Hz. Load is RLC with R = 60, L =30 mH and C = 180 pF. (a) Calculate THD of the output voltage (b) Obtain an expression for load current in Fourier series. Also, compute (c) THD of the load current (d) load power and average dc source current. Considering only the fundamental component of load current, calculate (e) conduction time of each transistor and diode and (f) peak and rms current of each transistor. Solution. Here Vs = 220 V, f = 50 Hz, R = 60, L = 30 mH and C = 180 ~F.
.
-3 . 106 XL == 21t X 50 x 30 x 10 = 9.425 n and Xc = 2n x 50 x 180 = 17.6840.
~
Zn [ 6' + ( 9.425 n _ 17;84 (a)
Rms value of output voltage, Vor = Vs
J'
r
~
~ 17;84]
anHn tan- {ii.425 n
=220 V
4x 220
=.198.071 V Rms value of fundamental component of output voltage, VOl = :ln2 ~~ xn Rms value of all harmonic voltages, Voh:::: -..JV~r - ~l = "';2202 - 197.071 2
=95.751 V
Voh 95.751 THD = VOl = 198.071 = 0.4834 or 48.34% (b)
Fundamental component ofload current,
101
Vol = Zl
198.071 = 19403 A [62 + (9.425 _ 17.684)2]112 .
1· = 01
and
1 03 -
4 x 220 x Z - 3n x T2 [
V03 _ 3
and
(j>5=tan-l [
1
28 9A
2 ]1/2
6' + ( 9.425 x 3 - 17 ; 84 9.425 x 3 - 17.684 ] 3 =75 0 6
J
=.4
Power Factor Improvement .'
.
[Art. 13.4]
- Vos _ 4 x 220 x Z - 51t X ~ [
I
1
05 -
2
9.425 x 5 - 17.: 1
Il>s=tan- [ Similarly,
]"2 =0.9A
6' + [ 9.425 x 5 _ 17.;84 )
5
and
687
84
6
1=82.16°
= 0.444 A and (1)7 = 84.6° Therefore, load current expression in Fourier series is io (t) = 19.403 sin (wt + 54°) + 2.849 sin (3rot - 75°) + 0.9 sin (5wt - 82.16°) + 0.444 sin (7rot - 84.6°) + ... 107
2
2
2
(c) Peak load cu rrent, 1m = [19.403 2 + 2.849 + 0.9 + 0.444 ]112
.
Rms value of harmomc load current, loh =
[I; -1~1 ]1/2.
= 19.637 A
2
f'
=[ 19.637'; 19.403' =2.1372 A ..
THD
= ~:: == 2·J13;!~~ == 0.155773 or 15.5773o/~ · = T2 1m = 19.637 = 138855 A ~ ' .
(d) Rms load current,
I
Load power,
Po = l~r x R
Or
= 13,8855 2 X 6 = 1156.843 W Po
11568<13 220 - == 5.2584 A
Average value of source current = V = s
(e) Expression for fundamental component of load current, i 01 == 19.~0? sin(wt + 54°). It
shows that current leads the voltage by 54°. :. Conduction time of each transistor = (180 - 54) x Conduction time of each diode (f) Pe8.k transistor current,
=
if-
7.0 x 10- 3 =
1~0 x 2 1t ~ 50 = 7.0 ms 1~0 -
7 x 10- 3 = 3 ms
1p = 101m = 19.403 A
Since each transistor conducts for 126° for every 360° of output cycle, rms value of transistor current, from Example 8.6 is
IT! =
~'t [126
0
x ;;;0 - sin;52
t'
=
0.461 3510 1",
= 0.46135 x 19.403 = 8.952 A
E xample 13.10. A 3-phase 180°·m ode bridge inverter has star-connected load of R = 4 n and L = 20 mHo The inverter is fe d from 220 V dc an d its output freq uency is 50 Hz. Obtain Fourier series expressions fOT line voltage vab (t ), phase voltage va (t) and line current ia (t) . Also, calculate (aj rms values of phase a nd line voltage s (b) rm s value of fu n damen tal component of phase and line voltages (c) TE D for voltages (dJ loa d po US T a nd a verage source current and (e) a verag e value of thy ristor curre n t.
688
[Art. 13.4]
Power Electronics -.
Solution. Here Vs = 220 V, R = 4 n, L = 20 mH, wL = 21t X 50 x 20 x 10- 3 = 6.283 n From Eq. (8.44), Fourier series expression for line voltageYab is vab
(t) =
4:s [cos ~ sin,(rot + 30) + ~ cos i sin
kcos
51t sin (5oot + 150°) 6
+ ~ cos 76 sin (7~ + 210°) + 1\ 11 1t
. =
(3rot + 90°) +
· 1t
6
sin (11 rot +'330°) ]
4 x 220 [-{3 . 1 -{3 . . 1-{3 1t 2 8m (rot + 30°) + 0 - 5"' 2" sm (5rot + 150°) --;;. 2 cos (7rot + 210°) .
.
1 ' -{3 ] + ~ x 2 sin (1100t + 330°)
= 242.585 sin (rot + 30°) - 48 .517 sin (5rot + 150°) - 34.655 sin (7oot + 210°) + 22.053 sin (11oot + 330°) From Eq. (8.47), phase voltage va(t) in Fourier series is va
Here
(t) = 2:s [ sin rot +
i
sin 500t +~ sin 7oot) + IIi sin Hoot + '" ]
2Vs = 2 x 220 = 140.06 V 1t 1t va (t) = 140.06 sin rot + 28.011 sin 5rot + 20.008 sin 700t + 12.732 sin 11rot + ... Zn = [R 2 + (nOOL)2] 112 = [42 + (6.283 n)2] 112 and «I>n
Here
I DIm
I
=tan- 1 [
6.2:3 n ]
j_ tan- 6.283 = 18.804 L - 57.52° 4 = 2 28.011 j_ tan- 6.2834 x 5 =0.8845 L _ 82.7440 (4 + (6 .283 5)2]112 _ = ' 140.06 [4 2 + 6.283 2]1 / 2
1
_ .
1
05m
X
Similarly
107m = 0.455 L - 84.803° and 10 .lIm = 0.184 L - 86.68° ia (t) = 18.804 sin (rot - 57.52°) + 0.8845 sin (rot - 82.744°) + 0.455 sin (7rot - 84.803°) + 0.184 sin (1100t - 86.68°) (a) From Eq. (8.51), rms value of phasor voltage, .
V
p
= -{2 V = -{2 x 220 = 103 71 V 3 a 3 . .
From Eq. (8.50), rms value of line voltage, VL = -{3Vp = (b)
- f2 -. '13 x 220 = 179.63 V
From E q. (8.52), fund amental component of phase voltage is
V
1
p
= -{2 . Va = 'J'2 x 220 =99.035 V 1t
1t
Fundamental component of line voltage, VLl =~ . Vp1 =-{3 x 99.035 = 171. 534 V (c )
Rms value of all h armonic volt ages, V ah
= ['1 -
1'HD =
111] 112 = [179.632 - 171 .5342]112 = 53 .32 V
~:~ = 1~i:~~4 = 0.31 08421 or 31.08421%
Power Factor Improvement (d)
[Art. 13.4]
689
Rms value of phase, or line, current .
2
2
lor == [ 18.804 + 0.8845:+ 0.455 + 0.184
1'2
2
] '
== 18.831 A
== 3 x I~r . R == 3 x 18.831 x 4 == 4255.3 W
2
Load power
. . 42553
. Average value of source current, Is == 220' = 19.342 A (e) For calculating the average thyristor current, Fig. 13.7(a) is 8.21 (a) for a 3-phase bridge inverter. For average source current Is,
drawn for step-I from Fig. current in SCR1 == current in SCR5 == 111/2, as these two SCRs 1 and 5 are in parallel and current Is is equally shared. For SCR1, current iTl is shown 1/2 for n/3 in Fig. 13.7 (b) for step I which is of n/3 radians duration. Reference to Fig. 8.21 shows that in =Is for step II and iTl == 1/2 for step III ; current iTl is sketched according-Iy in Fig. 13.7 (b). Periodicity of iT 1 is 2n radians. Is
"
15 /2 5
3
Is
VS
Is 6
4
Is
'2
T . Is
il l
2
Is Is
Is
a
2
~[
b
'IT
1[.
3
27r L:!lj
-.l-K3 -!-..!3 -l
wt
3
c
(a)
(b)
Fig. 13.7. Pertaining to three-phase bridge inverter; Example 13.10.
From Fig. 13.7 (b), average value of thyristor current is given by ITA
PROBLEMS
=2~ [( ~ J~+I.. ~+ ( ~ H]=; ~ 19.:42
=
64473A
. . . ."
13.1. What is thyristor controlled inductor? Explain, how the inductance seen by the source can . be altered in TCl. Illustrate your answer with appropriate waveforms and phasor dia grams . Hence, for any firing-angl e delay, derive an expression for th e reactive power h an dled by TCI. 13.2. What is SVC ? Describe how SVC regulates the reactive po wer flow and improves the system power factor. Illustrate your answer with relevant phasor diagrams . . • Derive an expression for the resultant current drawn 'by an SVC and comment on the fl ow of reactive power. 1 ~ 3. (a) Derive an expression for the effective indu ct ance in case of TCI. ( b ) A TCI, fed from 6.6 kV, 50 Hz source , has an inductance of 1H. Calcu lat e reactive power h andled by TCl for firin g angles of 90°, 120°, 1500, 175° and 180° , [Ans. (b) 138.656 kVAr, 54.215 kVAr, 7 ,996 kV,Ar, 0.039 kVAr, zero]
690
Power Elec tronics
[Prob.13)
13.4. A sl ngle-phase load of 20 k'W+ j 12 k VAI is fed from 400 V, 50 Hz source. TCl in SVC has an ind uctance of 0.8 H. For a firing- angle delay of 120°, it is found that the system operates at un ity pf Find the value o(capacitance of fixed capacitor and kVAr delivered by it. [Ans. 243.68452 jlF, 12248.92 VArl 13.6. Which device will be the optimum choice for the following applications? Give reasons for your choice. (a) Single-stage, 50 kVA, self-commutated ae to de converter with variable 'o utput voltage. (b) Single-stage, 50 kVA, line-commutated ae to de converter with variable output voltage. (e) 200 VA self-commutated de to ae inverter having 100 kHz as device switching frequency. (d) 5 kVA self-commutated dc to ac inverter having 5 kHz as device switching frequency. Ce) 2 MVA self-commutated dc to ac inverter having 400 Hz as device switching frequency. (/) 300 kVA self-commutated dc to ac inverter with 100 kHz as the device switching frequen cy. [Ans. (a) GTO (b) SCR (e) PMOSFET (d) BJT (e) GTO (/) S1TH. 13.6. Show that for a 3-phase full converter, if the firing-angle delay is a, the fundamental com ponent of the input phase current lags the respective phase voltage by angle a. [Hint. Here it is to be shown that displacement angle of fundamental current is - a so that displacement factor = cos a.l 13.7. "If the operation of inversion is not required from a line-commutated ac to dcconverter, a semiconverter possesses better performance characteristics than a full converter." Justify. this statement. [Hint. Read Art. 6.5.1 and 6.5.2 and compare the various performance parameters for any one firing-delay angle a. For example, for a = 45°, I-phase full converter: rectification 11 = 63.66%,p{= 0.6365 lag, reactive power = Volo etc. I-phase semiconverter : rectification 11 = 80.58%, p{ = 0.8892 lag, reactive power = 0.4142 Volo etc.] 13.S. A single-phase full-bridge inverter is fed from 230 V dc, its output frequency is 100 Hz and load is RLC with R == 6 n, L = 20 mH and C = 100 f.LF. (a) Calculate THD of the output voltage. (b) Obtain an expression for load current in Fourier series. Also, calculate (e) THD of the load current (d) load power and a verage source current.
Considering only the fund a mental component ofloadcurrent, calculate (e) conduction time of
each thyristor and diode a n d ({) peak and rms current of each thyristor.
[Ans. (a ) 48.346% (b) 30.124 sin (rot + 29.205°) + 2.095 sin (3wt - 79.506)
+ 0.691 sin (5rot - 84.256°) + 0.3444 sin (7rot - 88°). (e) 7.428% (d)
2737.5 W, 11.9022 A (e) 4.189 ms ; 0.811 ms (/) 30.124 A; 15.861 Al
18.9. A 3-phase 120°-mode bridge inverter feeds a star-connected load of R == 5 n. DC source voltage is 230 V and output frequency is 50 Hz. Obtain Fourier series expression for line voltage vab (t), ph ase vol tage va (t) and line current ia (t). Also, calculate (a) rms value of phase and line voltages (b) rms value of fundamental component of phase and line voltages (e) THD for voltages (d)load power and average source current and (e) average and rms val ue of thyristor cu rrent. LAns. Vab (t) == 21'9 .634 sin (rot + 60 + 43.93 sin (5rot + 300°) + 3 1.376 sin (7 wt + 60°) 0
)
+ 19.!;l67 sin (11<.Ot +- 30a O); va (t)
== 126.802 sin (rot + 30°) - 25.36 sin (5rut
+ 150°)
- 18.115 sin (7wt T 210°) + 11.53 sin (l1wt + 330°) ; ia (t) == 25.3604 sin (tOt + 30°) - 5.072 sin (5rut + 150°) - 3.623' sin (7wt + 2 10°) + 2.306 sin ( l l wt + 33 0°). (a )
93 .897 V, 162.635 V (b) 89.66 5 V, 155.304 V (c) 8 1.0864% (d)
10312.33 vii, 44.836 A (e) 7.667 A, 13.28 AJ
Appendix: A
Fourier An lysis ;' . I I 'I~II"
I~ .' J"'I 'I"
••• • ' . I I' I " I ' I I I I " " ' I •• ' . ' ••• • ••••• , •••••••• , ••
,J •• J •• , ••• •••• a
.,'
'''_
In power electronics, voltage and current waveforms are usually non-sinusoidal. These waveforms are, however, periodic in nature. A periodic function is one which repeats itself after regular intervals of time. A function f(t) is said to be periodic function of time if f(t)=f(t+T) ... CA-]) where T =period, or periodic time of one cycle of f(t). In power electronics, let f(t) be the output voltage function of a power converter. Therefore, Eq. CA.l) is written as vo(t) = Vo (t + 1') ...(A.l) If f is the frequency of output voltage in hertz, then f 00
Eq.
(Al)
=~ and angular frequency ro is given by
21t =21t{= T
... (A.2)
can now be written as Vo (oot) =Vo (oot ± 2 1t)
A periodic function v0
...(A3)
(t) with period T can be expanded into fourier series as under :
Vo (t)
=ao + al cos rot + a2 cos rot + ... + an cos nCJx + b l sin rot + b2 sin 2rot + ... + b" sin nrot ... (A.4 )
,,= 1,2,3 wher.e a o is the average value, or de component, of the periodic wave over a complete cycle. In Eq. (A.4), coefficients ao,a" and b" can ~e determined from the following expr essions: a o = average value, or de value, of Vo (t) over a period. x
lr =T1(1' L Vo (t) dt = -2 J_ o 1t 0
VO
... (A.S)
(oot), d (rot)
, an = amplitude of cos n rot component of Vo (t) = T2
fo
vI) (t).
cos nffit . dt =.! 1t
b" = amplitude of sin n oot component of vI) (t)
= T2
fo
v() (t)
sin nwt . dt) =.!
n
flt
Vo (rot).
cos n rot. d (rot )
r
VI) (rot) . sin nrot . d (wt)
0
Sine and cosine terms of Eq. (A,4 ) can be combined into one term as under: . Von (t ) = a" cos n rot + b" sin nwt a b = + [ ;Jan2 +" b~2 cos n rot + ",f an2 +" b112 sin nrot ]
~a! b~
. .. .(A.6)
0
...CA.7) .,
Power Electronics
692
Let us defme 9n as shown in Fig. AI. :. 9n
~ tan- 1 [ ~: ]
Von (t) = en [sin en cos nrot + cos en sin nrot] = Crr §in (nrot + en)
..
an
90° Va
(t)
=a o + L Cn sin (nrot + en)
... (A-B)
n'" 1,2,3.
where
Cn
,
= ....fa! + b!
and en
= tan- 1 [
bn
Fig. A-I. Pertaining to angle en·
J
a b:
...(A-9)
Here C n and en are respectively, the amplitude and the delay angle of the nth harmonic component of output voltage Va (t). A.I Half-wave or Mirror Symmetry. A waveform possesses half-wave symmetry if Vo (t) = - Va (t + T/2) } ...(AID) { or va (rot) = - Va (rot + 7t ) Eq. (A10) implies that for every positive value of va (t) in a period, there is a corresponding negative value of Va (t), of the same magnitude at a distance ; or
7t.
Waveforms shown in Fig.
A2 possess half-wave symmetry. The distance {, or 7t, can be taken anywhere, positive value' of va (t) must be equal to negative value of va (t) for half-wave symmetry. Also, a wave possesses half-wave symmetry if negative half-cycle is the mirror image of the positive half-cycle displaced horizontally by a distance T 12 or 1t. Since areas of positive and negative half cycles are identical foy waveforms having half-wave symmetry, average value of waveform aa = O. Also, a waveform with half-wave symmetry contains only odd harmonics (i .e. n = 1, 3, 5 ...). r--T/2~
:'--1/2 or
IT
J:'\ ' --' ; --' 1 1 Lv 3Jwt i
,
t
i
r-T/2
1I
L..---J........
j
7l--i
~1
t
(C)~
wt'
,
(a)
or
wt'
(b)
Fig. A.2. Pertaining to half-wave symmetry.
Summarizing, a waveform with half-wave symmetry has the following attributes: (i) aa = 0 (ii) From Eq. (A -6), for n = 1, 3, 5, ......
-2f
an = T
va (t) cos nrot d t
o
(i ii) From E q. (A7), for n
bn, = T2 uo(t ) = n
r..
1t
I..
= 1, 3, 5
1t
. cos n wt . d (CDt)
...(All )
0
= 1, 3, 5, ...
va (t) sin nrot dt
o
1
r Va (rot) =- J_
=1. f1t Va (CDt) sin T!.cot . d (rot ) 'It
Cn sin (n rot + 8T)
...(A12)
0
... (A13)
693
Appe dix
A.2. Odd and Even functions. It is seen from above that a waveform having half-wave
symmetry is free from even harmonics and its average value is zero. When a waveform is odd or even, some more terms from Fourier series vanish. A. 2.1. Odd function. A function is said to be an odd function if va (t)
= - Vo (- t)
Some of the odd functione are shown in Fig. A.3. For a point P, there is a corresponding point P at an equal distance from the origin t = 0, or rot =0. Note that the magnitude of vo(t) at P =- magnitude of va (t) at P. Vo (t)
t
t
wt
(a)
(c)
Fig. A.3 . Waveforms for odd functions
Ifthe positive cycle is folded about the horizontal reference axis and then again folded about the zero vertical axis, positive cycle will overlap the negative half cycle. This shows t hat odd function has (a) symmetry about time origin and (b) symmetry about the zero -vertical axis. Thus, odd functions (i) possess half-wave symmetry (ii) are . symmetry about horizontal reference line, or t-axis, or rot-axis, (iii) are symmetrical about zero v ertical axis (iv) have aa = 0 and an = (v) contain sine terms only and (vi)
r
°
sin n rot . dt = ~ Vo (rot) sin nrot d (rot) o It 0 A.2.2 . E v en Function. A function is said to be an even function if va bn
= T4
fl2 va (t)
... (A.14) (t) =
va (- t).
Some even f1.mctions are shown in Fig. A.4. It is observed that even function exhibits mirror symmetry about the vertical axis at origin t = 0, or rot = 0, and also with respect to all vertical axis at a distance nT12, or mt from the origin. Thus, even functions (i) possess mirror symmetry about vertical. axis only (ii) have bn = (i i i) Contains cosine t erms only and (iv) an = T4
° fl2o
va (t)
r
cos nrot = ~
Va (rot).
... (A.15)
cos nrot. d (rot)
0
It
Va ( t ) .
r- T
T
-"2 , ,
0
2 ,, ,
,
t
.~ . L ,
0
7I
(e)
(a)
( b)
Fig. A.4. Waveforms having even symmetry.
I
--I---- 17 ~
wt
694
Power Electronics
A.2.3. Quarter-wave Symmetry. A combination of half-wave symmetry and odd symmetry gives a function odd-quarter-wave symmetry. Therefore, an odd-quarter wave should have ' .
.
v, (t) = - v, ( t + ;
Jand v, (t) = - v. (- t)
The waveforms shown in Fig. A.3 possess odd quarter-wave symmetry. For these waves; ao = 0, an = 0 and b n = T8
f/4 o .
Va (t)
sin nrot . dt = 1.
(2
Va (rot).
sin nrot. d (rot)
... (A.16)
1t 0
An even-quarter-wave has Va (t)
=- Va (t + T12) and va (t) =va (-t)
Waveforms shown in Fig. A.4 possess even-quarter-wave symmetry. For these waves, bn =o. 8 (1'/4
an = T .L
and
o
va
t
4 1(tI2 (t) cos nrot dt = Va (rot) cos nrot d (Olt) 1t
A.a. Applications. Now some waveforms encountered in power electronic circuits are resolved into Fourier series. A.a.l. Square wave. The square-wave voltage shown in Fig. A.5 (a) is obtained as the output voltage wave from a single-phs.se full-bridge inverter. It is a pure ac voltage wave.
(ii)v, (t
I + ;) =- v•.
=T /4) =V. and v. (t =
T/2
TIl.
T 2'lf
t
o ~-~----+~--~----~--~
'1(/2
1(
371 12
wt
- Vs - ' - - - - - - - .....
Fig. A.5 (a) Square-wave voltage .
Symmetry conditions :- (i) An areas of positive and negative half cycles are equal, aa = 0 .
... (A.17)
0
As v(t) =- vJ t + ; ) the wave has half-wave
symmetry,-therefore only add harmonics are present in the Jourier series expansion.
=~ J= V• .~d v, (t =- T/4) =- V•. As v. ~I) =- v. (- tl, it is odd function. It is observed fr~m (u) and (m) above, that square wave of Fig. A-5 (a) possesses odd-quarter wave (iii) vJ t
symmetry. Thus, aa = 0 and an = 0 , 8 (1'/4 b n = T J0 VB'
•
,
SIn mot • dt
r /2 Va sm • n rot . d (rot)
8
or
.
b n = -2 J_ 1t 0
.
4 Vs =-I-cos nrot n7t
For
4 Vi IaltI2 = n1t
l
l - cos 1t.n ]
n7t
n = 1, 3, 5,.. . cos 2 = 0 4V b11'' -- nn" Va (t)
=aa + L an cos nwt + b", sin nCDt n = 1,3,5
2
.
Appendix
695
4V
=L __s sin noot 00
n :: 1,3,rf.
... (A.1B)
1t
4 Vs [ sm . rot + 3" 1. 1. 1. =---;:t sm 3 oot + '5 sm 5rot + '7 sm 7 00t -:- .•. ] =V OI +V03+ V 05+ V 07+'"
Rm s value of nth harmonic component ofuo(t) square wave 4 Vs 1 Von = --r::::'"2 - = - [0.900 Vs]l "J~ • n1t n Rms value of fundamental component,
VOl
= 0.900 Vs
Rms value of third-harmonic component, T/03
VOl
Fig. A.5. (b) Summation of fundamental vOl' third harmonic V03, V05 etc gives square wave.
=0.300 Vs and so on.
The sum of the ordinates of the various harmonic components of Fourier series i.e. + V03 + V05 + V 07 + .. , gives the actual square wave. In Fig. A.5(b), only vOl and V03 are shown. Note that
or
wI
Von _ 1:. VOl - n 1
Von =-. n VOl'
A .3.2 . Q uasi -square Wave. This type of waveform shown in Fig. A-6 is obtained as the output voltage from a single-phase full-bridge iuverter with single-pulse modulation. This wavefo rm is also specified by the conditions of square wave.
Fig. A.6.Quasi·square wave.
Therefore, a o = a and an = O. 4 r /2 bn = - J'_ va (oot) . sin nrot . d (rot) 1t
0
4
Jlt/2
=-
1t
(
!-d 2
4 Vs lV 2 )Vs.sinnwt.d(c.at)= - - I -cosnrot I--d n.1t 2
4 Vs [n1t . -n"Jt sm . =-cos - cos nd + sm nd + cos n7t ] n1t. 2 _ 2 2
For all values of odd n, cos n1t is zero. 2 4 V, . n7t , b = - - . sm-sm nd n n1t 2
"" 4 V s • n7t . d . ~ va () t =.£... --;- . sm 2 sm n . sm nwt /l.=
1.
3,f .
... (A. 19)
Power Electronics
696
1]
n
l 'sm 3 dsm ' 3 rot + '5 l' 5 dsm ' 5cot -7'" v0 () t = 4Vs['d' sm 8m rot - 3' sm
or
4 V,
. n 7t . sm nd,V01
4 Vs
.
,
=;/2. 1t sm a
= "'12 , n1t' sm 2
Here
Van
"
Von 1 sin nd . n7t [ 1 sin nd . -V = - -'-d-' slnor Von = . d' sm- . VOl 2 2 01 n SIn n sm
mt] .
A-3.3. Single-phase half-wave rectified voltage. Rectified output voltage wave obtained from single-phase half-wave diode rectifier is shown in Fig, A. 7. This waveform is seen to have no symmetry. Therefore, coefficient ao' an and btl would have to be determined. :, a o = de component of output voltage, Vo (t)
fit Vo (rot), d (cot)
1 ="2 7t
0
Vm
ix[f
=
Uo(t)
V.
=--..!!!.
Vm sin rot d Crot) + 0
0] °
0 .. ,(i)
rfit
n
w.
2.
Fig. A. 7. Single-phase half~wave rectified . voltage wave.
1t
an = 1. 7t 1
=-
vo(rot) . cos nrot. d (cot)
0
. .
Vm sin cot. cos nrot. d (cot)
7t o . . Now sin (cot + ncot) = sin rot cos ncot + cos cot . sin nrot and sin (rot - nrot) = sin rot cos n{J)t - cos cot . sin nrot Adding, we get sin rot cos nc.ot = -~ [sin (1 + n) cot + sin (1
V . . an = 2;
rI IJ
=V m [ 27t
. =Vm [ 21.
.
sin (1 + n) cot + sin (1- n) cot) . d (wt)
1 - cos (1 + n) 7t + 1 -- cos (1- n) 7t ] 1+n 1- n
1 - n2
27t
1 _ n2
.
1
.
' .= -V m [ l+-l(= -Vmlr-1+coSn1t] . l t ] , as cos Tl7t = ( 7t 1 - n - 1t 1 _ n2 .
.
.
For n == 1, al is in determinate. F or n odd, an .
: . .
1 - n + 1 + n _ cos 1t cos n1t - n cos 1t cos n7t + cos 7t cos n7t + ncos 1t cos n7t 1 - n2 1 _ n2
. = Vm [ ~_ + 2 cos n 1t '.
~ n) rot]
~
.
= 0 and for n even, an =: .
f Vm sin cot cos V2m f sin 2wt d
a 1 = 1. 7t =
.
0
7t
(ot .
2 m
1t (n. - 1)
d (cot) .
.
0
2V
(cot) = O.
1)"
1
697
Appendix
r
b n = 1.
0
1t
Vm sin rot. sin nrot . dewt)
From trigonometry, sin rot sin Tirot bn
· V = 2;
~lcos (1 -
= Vm
[
\
1-n
1
r 0
1t
+ n) rot]
+ it) rot} d(rot)
n) rot - cos (1
1+n
The above expression is indeterminate for n . 1 b =-
n) rot - cos (1
sin (1 - n) rot ~ sin (1 + n) rot
.
~
= ~ [cos (1 -
.\It] =0 o
.
= 1 for b l .
Vm
2
V sin rot d (rot) = m 2 1
=a o + L
Vo (t)
b n sin nwt +
Vm .
2 Vm
L
1t
n =2,4,6
= - + - s m r o t - -·1t
an
cos nrot
n = 2,4,6
n=l
Vm
L
2
1 n -1
(A.20 a)
- cos nrot 2
. V m· V m . 2Vm 2Vm 2Vm =-1t + - 2 sm rot - -3 COS 2rot - - - COS 4rot - - - COS 6mt 1t 151t 351t
...(A.20 b)
A.3.4. Single-phase Full-w ave Rectified voltage. Rectified voltage obtained from a single·phase full-wave diode rectifier is shown in Fig. A.8. Such waveforms are encountered in chapter 3. Waveform of Fig. A·8 may be treated as the sum of two waveforms shown in Fig. A.9. Waveform V02 is shown to lag VOl waveform by 1t radians but Vo = VOl + V02'
"o]CVV=VV~~ 71
3lT
2rr
47r
Srr
wt'
Fig. A.B. Single-phase two-pulse rectified voltage wave.
Fig. A.9. Sum of VOl (t) and
Voltage
VOl (t),
VOl
V02 (t)
gives va
(t ) ofFig~ A.S.
identical with Va (t) and given by Eq. (A.20 a), is expressed as
Vm + -2 Vm' sin rot - 2 Vm " - ~
(t) = -
1t
1t
1 cos nrot n-l n = 2, 4, 6 --2-
.. ~ (A. 2 1 a)
·Waveform v 02 lags waveform VOl by 11: radian s. Therefore, Fou rier series express' on for v 02 can be obtained from Eq. (A.21 a) by putting (rot -1t) for rot .
698
Power Electronics
v02
(t)
Vm
Vm
=- + -2 sin (OOt -7t) 7t
= Vm
_
7t
2 Vm
~-
1
- - £..J -- 2 - 7t n-246 n -: 1
i
I
cos n (rot -1t)
,
V m sin rot _ 2 V m cos mot 2 7t n2 - l II. = 2, 4, 6
.:.(A.21b)
By adding Eqs. (A.21 a) and (A.21 b), we get
~
2 Vm 4 Vm va (t) = - - - - 7t
or
2 Vm
- Va
(t)
= -7t- -
L.J
cos n rot
-2.,-----
=2, 4, 6 n - 1 4 Vm 4 Vm ' 4 V m ~ cos 2rot - 151t cos 4 rot - 351t cos 6 rot 7t
... (A.22 a)
II.
...(A.22 b)
..
Appendix: B
Laplace Transform.s
.1 1 1 • •••• I ' I . I . IJ I. I . I ' I IS I • • I ••
•• ••• • • • • • • • • ' • • • • 11
II ' ~ III
•• !
l l. I I I. ~t1.i ,
1. Impulse function: li(t)
1. 1
2. Unit step function: U(t)
1 2. ,s
3. t
3.
1 s2
4.
1 s+a
I
4.
5. sin rot
(0
5.
6. cos rot
s
7.
8. e- at . sin rot
9. .
11.
10.
1- e- tiT
1 + a)2
(s + a)2 + 0)2 (s
s+a + a)2 + (02
(8
1 + a)(s + b)
bt
11.
1 s (1 + s1')
12.
(0
-.
12. 1 - cos wt I
(8
(0
8.
9. e at . cos rot -at -e10. e b-a
s2 + (02
6. s2 + (02
7. t . e- at
~J';"~ ~ ~.".JI
Laplace transform, F(s) : s-domain
Time domain, fit): time domain
e- at
••
s (s2 + (02)
,• •
, ~ .~
Appendix: C
Objective Type Questions
In this appendix, multiple choice questions pertaining to Chapters 2 to 13 are given. The questions are se framed that four alternatives are provided, out of which one correct answer is to be tick-marked. . POWER SEMICONDUCTOR DIODES AND TRANSISTORS 1. Fora diode, reverse recovery time is defined as the time between the instant diode
current becomes zero and the instant reverse recovery current decays to (a) zero (b) 10% of reverse peak currentIRM (c) 25% of IRM (d) 15% of IRM
.
2. In a diode, the cut-in voltage and forward~voltage drop are respectively · (a) 0.7 V, 0.7 V (b) 0.7 V, 1 V (c) 0.7 V, 0.6 V (d) 1 V, 0.7 V 3. The softness factor for soft-recovery and fast-recovery diodes are respectively (a) 1, > 1 (b) < 1, 1 (c) 1, 1 (d) 1, < 1 4. Reverse recovery current in a diode depends upon (a) forward
field current
(c) temperature
(b)
(d)
storage charge PIV
5. In a BJT,
(a) ~
(c)
a a+l
=-
a=-.lL ~...;.1
a
a-I
(b) ~= ~
(d)
- .l!..±..l ~
a-
6. A power MOSFET has three terminals called
collector, emitter and base . (b) drain, source and base drain) source and gate (d )· collector, emitter and gate
. 7. AB compared to power MOSFET, a BJT has
(a) lower switching losses but higher conduction loss (b) higher s"Witching losses and higher conduction loss (c) higher switching losses but lower conduction loss (d) lower switching losses and lower conduction loss 8. Choose the correct statement:. (a ) MOSFET has positive temperature coefficient (TC) whe eas BJT has negative TC (b) Both ~ IOSFET andBJT have positive TC (c) Both MOSFET and BJT h ave negative TC (d) MOSFET h as ne~ativ e TC whereas BJT has positive TC (a) (c)
...
701
Appendix
9. Choose the correct statement : (a) Both MOSFET and BJT are voltage controlled devices (CDs) (b) Both MOSFET and BJT are current CDs (c ) MOSFET is a voltage CD whereas BJT is a current CD (d) MOSFET is a: current CD whereas BJT is a voltage CD 10. Secondary breakdown occurs in (a ) MOSFET but not in BJT
(b)bbth MOSFET andBJT
(c) BJT but not in MOSFET
, (d) none of these
11. At present, the state of the art devices are available as under:
MOSFET BJT (a) 1200 V,800 A 50d V, 140 A (b) 500 V, 140 A 1200 V, 800 A
, (c) 800 V, 1000 A 1000 V, 1200 A
(d) 200V, 140 A 1500 V, 800 A 12. An IGBT has three terminals called (a ) collector, emitter and base (b) drain,source and base (c) drain, source and gate . ' (d) collector, emitter and gate 13. An MCT has three terminals called (a) anode, cathode and gate (b) collector, emitter and gate (c ) drain, source and base (d ) drain, source and gate 14. For the switching waveform shown in Fig. C.1 for a power transistor, the peak instantaneous power loss is Vs I s 4
Vs Is ' , 6
Vs Is 3
(c) - -
(b) - -
(a) - -
(d)
V . Is 8
-~-
1 5. In Fig. C.1, if V~ = 100 V, Is = 10 A, then peak instantflneous power loss in watts is (a) 500 (b) 166.67 ' (c ) 333.33 (d) 250 16. For the swit ch ing waveform shown in Fig. C.1 for a power transistor, the average val u e of switch-on power loss at a switching frequ ency f is if = 1/ T) . 'V s ' s V s Is (a ) - 4 - ' ton' f (b) -6- ' ton·f
Vs Is
( c) - 3-' ton
.f
'
(d)
t
Fig. C.l
Vs' Is - 8 - . ton .f.
17. In Fig. C.l, ifVs = 100 V , Is = l OA, ton = 1.2 ~s and f = 10 kH z, then average value of svvitch-on power loss is ( a ) 1 W (b) 4 W (c) 2 W (d) 3W
702
Power Electronics
18. Match the devices on the left hand side with the circuit symbols on right h and side and give the correct answer from the codes given below: . ..: - (A) BJT
(E) MOSFET
(C)
IGBT
(D)
MCT
(1)
)A ~lc
Code., D A B C D C (a) (b) 4 2 3 1 4 1 (c) (d) 2 4 1 4 3 1 19. Power-electronic equipment has very high efficiency, because (a) the devices always operate in active region (b) the devices never operate in active region (c) the devices traverse active region at high speed and stay at the two states, on and off .• (d) cooling is very efficient 20. In the conduction mechanism of Schottky diode (a) only electrons can participate (b ) only holes can participate (c) both holes and electr on s participate (d) n one of the above. A 2 3
B 3 2
21. Common emitter current gain hFE of a BJT is (a) d ependent on collector current Ie (b ) dependent on collect or -emitter voltage, VeE (c ) dependent on base-emitter voltage, VEE (d) always constant 22. The semiconductor device which is suitable for induction h ardening in radio fre quen cy ran ge is
703
Appendix
MeT (b) BJT (c ) IGBT (d) MOSFET 23. High-frequency operation of a circuit is limited by (a) On-state loss in the device (b) off-state loss in the device (c) switching losses in the device (d) all of the above. 24. Read the following statements carefully: 1. PMOSFET is a majority carrier device 2. IGBT is a bipolar device 3. BJT is a majority carrier device 4. MCT is unipolar device
From above, the correct statements are
(a)
(b)2,4
(a) 1, 3 (c) 1,4
1,2 25. A bipolar junction transistor (BJT) is used as a power control switch by biasing it in the cut-off region (off state) or in the saturation region (ON state). In the ON state, for the BJT (a) both the base-emitter and base-collector junctions are reverse biased (b) th e base-emitter junction is reverse biased, and the base-collector j~nction is forward biased (c) the base-emitter junction is forward-biased, and the ' base-collector junction is reverse biased (d) both the base-emitter and base-collector junctions are forw ard-biased. (d)
ANSWERS 1. (c ) 7. (c)
2. (b )
3. (d)
4. (a)
5. (b )
6. (c)
8. (a)
9. (c)
10. (c)
13. (a) 19. (c )
14. (a) 20. (a)
15. (d)
16. (b)
11. (b) 17. (c )
12. (d ) 18. . (a)
21. (a)
22. (d)
23. (c)
24. (d)
25. (d)
DIODE CIRCUITS AND RECTIFIERS 1. In Fig. C.2, capacitor C is charged to Vo = 50 V with upper plate positive. Switch S
is closed at t = O. Current through the circuit at t = 0 and final voltage across Care r espectively, (a ) 15 A, 200 V (b) 20 A, 200 V (c) 25 A, 250 V (d) 15 A, 150 V R=10n
+
D "l
200V
Va
Fig. C.2
ei;
v,
Fig. C.3
2 . In Fig. C .2, suppose capacitor C is charged to 50 V with lower plate positive. Switch S is cl osed at t = O. Current through 'the diode at t = 0 an d fi n al voltage across C are r especti vely (0 ) 25 A, 25 0 V ( b) 25 A, 200 V (c ) 20 A, 200 V (d) 15 A, 150 V
Power Electronics
704
3. In the circuit of Fig. C.3, switch S is closed at t = 0 with iL (0) =() and steady~state, Vc equals
Vc
(0) =O. In
(b) 100 V (d) - 100 V
(a) 200 V (c) zero
4. In Fig. C.4 ; VI, V 2 and V3 are zero centre PMMC voltmeters. The circuit is initially relaxed. Switch S is closed at t =O. In steady state, readings of voltmeters VI, V 2 and V3 are respectively (a) 100 V, 100 V, - 100 V (b) 100 V, 0, 200 V (c) ...:. 100 V, O! 200 V (d) 100 V, 0, 100 V
Fig. C.4 .
Fig. C.5
5. In Fig. C.5, initial voltage across capacitor is Vo= 50 V with the polarity as shown. Switch S is closed at t =O. In steady state, Uc and UD are respectively given by (a)400 V,-200 V (b) 350 V, -150 V (d) 450 V, - 250 V .
(c) 200 V, Q
6. In Fig. C.5, if C =811P and L = 0.2 niH, the peak current handled by diode is (a) 40 A
(b) 50 A
(c ) 10 A
(d) 30A
7. In Fig. C.6, initial voltage across capacitor is Vo =50 V with the polarity as shown. Switch S is closed at t = O. In steady~state, Vc and UD are respectively given by (a) 400 V, ~ 200 V (b) 350, - 150 V (c) 200 V, 0 (d) 450 V, - 250 V 8. In Fig. C.6, if C =8 I1F and L = 0.2 mH, the peak current through diode is given by (a) 40A (c) lOA
(b) 50A (d) 30A
i
R
Fig. C.B
Fig. C.7
9. In Fig. C. 7, capacitor C is initially charged with voltage Vo with upper plate positive. Switch S is closed at t == O. At t =0+, Vc and i are given by
Vo . Vo
(a) 0,
R
(b) ·- V o,
v:O
(c ) - Va' - R
(d ) V a.
If
Vo
If
Appendix
70
10. In. Fig. C.B, · capacitor C is initially charged with voltage Vo. Switch S is closed at t = O. In stead! state, Vc and VD are respectively given by (a) V o, - V o (b ) 0, - V o (c) - V o, 0 (d) V o, V o
L
+ 11. Each diode in Fig. C.g can be described by a cut-in voltage and zero resistance. If the cut-in voltage of Fig. C.S diode D1 is 0 ~ 2 V and of diode D2 is 0.6 V, the magnitude of current II through D1 is .... .. rnA and magnitude of current through D2 is .. .. .. rnA.
o
10 kn
. I, IOOV
L= O.5H
D1
Fig. C.9
fig. C.10
12. In Fig. C.lO, ideal PMMC ammeter M will read (a) zero (b) 1.414A (c) 0.707A (d) 1A 13. In Fig. C.lO, if ammeter M is an ideal MI ammeter, then it will read (a) 0.707 A (b) 1.414 A (c) 1.225 A (d) 1 A 14. In Fig. C.U, ideal moving iron voltmeters M1 and M2 will respectively read (a) 141.4 V, 141.4 V (b) 0, 141.4 V (c) 0, 200 V (d) 141.4 V, 0
D ru 10sin314t
Fig. C.11
15. In Fig. C.12, an ideal moving iron voltmeter ·Mwi1l read (a) 7.07 V (b) 12.25 V (c) 14.14 V (d) 20.0 V 1 . In .Fig. C.13, zer o-centre a..l1.d ideal PMMC voltmeters Ml and M2 will read (a) - 10 V, 10 V (b) - 0, 10 V (c) - 10 V, 7.07 V (d) 10 V, 7.07 V
C
100,uF
Fig. C.12
Fig. C.13
706
Power Electronics
17. In Fig. C.14, PIV required for the diode is (a) 300 V (b) 100 V (c) 200 V (d) 400 V 18. A single-phase one-pulse diode rectifier is feeding an RL load with freewheeling diode across the load. For conduction angle~, the main diode and freewheeling diode would, respectively, conduct for (a)1t , 1t-~
(b)1t,~-1t
(c) ~, 1t
(d) ~ -1t, 1t
l00V ·
Fig. C.14
19. A single-phase full-bridge diode rectifier delivers a load current of 10 A, which is ripple free. Average and rms values of diode currents are respectively (a) 10 A, 7.07 A (b) 5 A, 10 A (c) 5 A, 7.07 A (d) 7 .07 A, 5 A 20. A single-phase full-bridge diode rectifier delivers a constant load current of 10 A. Average and rms values of source current are respectively (a) 5 A, 10 A (b) 10 A, 10 A (c) 5 A, 5 A (d) 0, 10 A 21. A voltage v = 4 sin rot is applied to the terminals A and B of the circuit shown in Fig. C.15. The diodes are assumed to be ideal. The impedance offered by the circuit across the terminals A and B in kilo-ohms is (a) 5 (b) 20 (c) 10 . (d) none of these .. A
1kn
D2
~D
10 k,(l
~4V B
Fig. C.15
iFig. C.16
22. The peak current through the resistor of circuit of Fig. C.16, assuming the diodes to be ideal, is (a ) 12 rnA (b) 4 rnA (c) 16 mA (d) 8 rnA
23. In Fig. C.17, VI and V2 are zero-centre PMMC voltmeters. When a sinusoidal signal is applied, V2 reads + 20 V. The reading of the voltmeter VI is .... .. volts.
D
I'
R
Fig. C.17
F ig. C.l8
24. F or a symmetrical squa e w ave of 800 V peak to peak and for .deal diode, t he voltmet er in Fig. C.l8 will read (a ) 200 V (b) 400 V (c ) 800 V (d) zero
Appendix
707
' 25. The circuit in Fig. C.19 shows zener-regulated dc power supply. The zener-diode is ideal. The minimum value of RL down to which the output voltage remains constap.t is (a) 27 n . (b) 45.0 (e) 15.0 (d ) 24n
won
27SL
+ . 24 V
15 V
RL
Fig. C.19
Fig. C.20
26. In the circuit of Fig. C.20, the 5V zener diode requires a minimum current of 10 rnA. For obtaining a regulated output of 5 V, the maximum permissible load current h is ... ... rnA and the minimum power rating of zener diode ,is .. .. .. W. 27. Fig. C.21 shows an electronic voltage regulator. The zener diode may be assumed to require a minimum current of 25 rnA for satisfactory operation. The value of R required for satisfactory operation is .... .. ohms. D1
R j) . \
20V
JOV
D2
lOOn.
E
Fig. C.2!
Fig. C.22
28. In the circuit of Fig. C.22; the diode states 'at the extremely large negative value of the input voltage vi' are (a) Dl off, D2 off (b) Dl on, D2 off (c) D1 off, D2 on (d) D1 on, D2 on . 29. A s ingle-phase half-wave diode rectifier, connected to V m sin wt, feeds a load R = 4.5 n . The diode has forward resist ance of 5 n and the r em aining parameters are s ame as those of an ideal diode. The de component of the source current is Vm Vm Vm . . 2 Vm (a) 501t (b) 50 1t T2 (e) 100 1t 'l2 (d) 501t
.''i
30. In a single-phase diode bridge rectifier, the source voltage is 200 sin wt with w = 21t X 50 r ad/ sec and th e load is R = 50 n. The power dissipated in th e load resistor R is ' (a) 3200
1t
W
(b)
400 W 1t
:
(e)
400 W
(d )
806:W
31 . The cen tre-tap full-wave single phas e rectifi er circui t us es two diodes. The t r an~- . former turns ratio from primary t o each secondary is 2. In case transformer input . voltage is 200 V at 50 Hz , then rms voltage acr oss each diode is (a ) 565.6 V (6) 282.8 V (c) 70. 7 V (d) 141.4 V
708
Power Eledronics
32. A single-phase two-pulse diode bridge has input supply of 200 sin rot with load R ~ 50 .0. Rms voltage across each diode is (a) 100
V
(b)
141.4 V
(c)
200 V
(d)
200 V 1t
33. The average current rating of a semiconductor. diode will be maximum .for (a) full-wave rectified ac (b) half-wave rectified ac (C) pure ac (d) pure de 34; The selection of rectifier diode depends mostly on (a) forward voltage (b) reverse voltage (e) fault current (d) average load current 35. A single-phase bridge rectifier (a) can operate without an isolating transformer ·· (b) cannot operate without an isolating transformer (c) cannot operate with isolating transformer (d) none of these .36. The rms value of ~alf-wave r ectified symmetric~l square-wave current of 2 A is (a) {2 A
(b) 1 A
(c)
* A
(d)-{3 A
37. In a single-phase diode rectifier fed from Vm sin (2 rtft), the lowest ripple frequency . and PIV are respectively 2. {, Vm for full-wave circuit. 1. f 12, V m for half-wave circuit 4.. . {, Vm for half-wave circuit. 3. 2{, 2 Vm for M-2 circuit. From these, the correct statements are (d) 2,3,4 (a) 1,2,3 (b) 2,4 (e) 3,4 38. Current-ripple factor for a rectifier circuit is defined as (0)
-V(~: J-1
(b)
~i.' -1
(b ) -V(::,
_~ \J T -1
. ·~(l )2
(c)
~-1
(d)
~ :0 -1
10 lo/' O/' where lor = rms value of output current and 10 = de value of output current . 39. A single-phase diode bridge rectifier has load R and filter capacitor across it. If one of the diodes is defective, then 1. the de load voltage would be lower than its expected value 2. ripple frequency would be lower than the expected value
3; the capacitor surge current would increase manifold.
Of the", statements,
(a) 1 and 2 are correct ( b ) 1 and 3 are correct (c) 2 an d 3 are correct (d ) 1,2 and 3 are correct. 40. Total ha rmonic distortion (THD) is defi ned as (a )
J -1
(e)
~r--(1-;~·-"'J---1
I
(d ) _ S lsI
wher e Is = r ms value of supply phase current in cludi ng fundamental and harmonies an d l s I = rm s value of fun damental component of supply current.
Ap pendix
709
41. Reactive power requirement of a rectifier system depends upon 1. displacement factor 2. input power factor 3. current distortion factor 4. crest factor
From these, the correct statements are
(d) 1,2,3 (a) 2 alone (b) 1,3,4 (c) 2,3,4 42. In a I-phase full-wave rectifier with R load and parallelcapacitor filter C, with the increase of wCR, the average value of output voltage (a) increases (b) decreases . (c) remains unaltered (d) i ncreases, reaches maximum value and then decreases. 43. Capacitor fIlter is ideal for currents which are (a) small (b) medium (c) iarge (d) very large . 44. An inductor filter at the output of a rectifier results in ripple which (a) increases with load resistance R (b) decreases with R (c) remains unaltered with increase of R (d) remains "maltered with decrease of R. 45. A capacitor filter "at the output of a rectifier results in ripple which (a) increases with load resistance R (b) decreases with R (e) remains unaltered with increase of R (d) remains unaltered with decrease of R. 46. The functionofa filter in a rectifier is to (a) limit the total current in the rectifier (b) limit the. peak voltage of the rectifier (c) limit the de current (d) reduce the ripple voltage in the output. 47. The disadvantage of half-wave diode rectifier circuit is that the (a) diode must have high PlV rating (b) diode must have high power rating (e) output volt age is difficult to filter (d) diode must have h igh current rating. 48. The fun ction of centre-tapping on the secondary in a full-wave rectifier is to (a) step-up the Voltage
( ~ step-down the voltage
(e) isolate the load form ground (d) cause the diodes to conduct alternately. 49. A 3-phase M-6 (or six-phase half-wave) diode rectifier i fed from 400/1000 V, delta-star transformer, PN of each diode is (a) 1414 V (b) 707 V (e) 1732 V (d ) 866 V 50. delta-star t ransformer , with out put line voltage of 1000 sin 100 1tt Yolts, feeds 3-ph ase diode rectifier cir cuits. Their PlV is 2 . 1155 V for M-B rectifier 1. 000 V for 3-pulse rectifier
710
Power Electronics
4. 1000 V fo r B-6 rectifier 3.141 4 for 3-pulse rectifier 5. 1414 for B-6 rectifier.
From these, the correct statements are
(d) 2,3,5 (a) 1, 2, 5 (b) 1, 2, 4 . (c) 2, 3, 4 :H. A 3-phasehalf-wave diode rectifier feeds a load of R = 100 n. For an input supply of 400 V,50 Hz, the power delivered to load is . (a) 753.73 W (b) 974.23 W (c) 376.98 W (d) 487.26 W . 52. Read the following statements: 1. If load resistance R is low, ripple factor for L- filter is high 2. For C-filter, ripple factor gets reduced if R is increased 3. L-filter is suited to low-current loads ' .. 4. for C-filter, better-waveform in obtained if R is increased.
From these, the correct statement are
(d) 2,3 (a) 1,4 (b) 2, 4 (c) 1,3
53. In rectifier circuits, if supply frequency is increased, then 1 .. ripple voltage increases withC-filter 2. ripple voltage decreases with L-filter 3. currer: t ripple decreases with L-fiJter 4. output voltage increases with C~filter .
From these, the correct statements are
(a ) 1,2,3,4 (b) 2,4 (c) 3,4
.
t .
.. •
. ,",...
(d) 1,4
54. A single-phase diode bridge rectifier is feeding a parallel combination of resistance load ar..d a capacitor of high value. The nature of the input current drawn by the rectifier from the ac source will be (a) a square wave of 180 0 duration (b) a square wave of 120 0 duration (c) peaky, with peaks at both positive and negative half .cycle of the input voltage (d) peaky, with peaks only at positive half cycle of the input voltage .
55. The current through the zener diode in Fig. C.23 is (a) 33 rnA (b) 3.3 rnA (c) 2 rnA (d) 0 rnA
[GATE, 2004]
R
.' I l
2.2kD. Y
+ RL
~.~
Fi g. C.23
R
B
3.5 V
N
Fig. C.24
56. The circuit in Fig. C.24 showns a 3-phase half-wave rectifier. The source is a symmetrical 3-phase four wire system. The line to line voltage of the source is 100 V. Th e supply fr equency is 400 Hz. The ripple frequency at the output is (a) 400 Hz (b) 800 Hz (c) 1200 H z (d ) 2400 Hz. [GATE, 2004]
-------. 111
Appendix
57. Assuming that the diodes are ideal in Fig. C.25, the current in diode Dl is (a) 8 rnA (b) 5 rnA (c) 0 rnA (d) - 3 rnA [GATE, 2004]
, lk.C1.
lk.C1.
lmA (DC)
Fig. C.25
Fig. C.26
' 58. Assume thatDl and D2 in Fig. C.26 are ideal diodes. The value of the current I is , (a) 0 rnA. (b) 0.5 rnA (c) 1 rnA (d) 2 rnA
ANSWERS 1. (a) , 7. (d)
13.
(c)
19. (c) 25.
(b)
30.
(c)
2. (b) 8. (b)
3. (a)
4. (c)
9. (b) 15. (b)
10. (a)
20. (d) 21. (c) 26. 40 rnA and 0.05 W 31. (d) 32. (a) (c) 37. 38. (a)
22. (d)
14. (b)
16. (a) ,
n,
11. 10 mA, 0 . 17. (d). ' "
23., - 20V 28; (b)
12. (d)
18. (b) 24. (a)
33. (d)
34. (b )
29. (a) 35. (a )
3 9. (a)
40. (b)
41. (d)
44. (a)
45. (b)
46. (d )
47. (c )
49. (a)
50. (b)
52. (b)
53. (c)
55. (c)
56. (c)
51. (a) , 57. (b)
36. (a) 42. (a)
43. (a)
48. (d) 54. (c )
THYRISTORS
6. (d)
5. (b)
27. 80
58. (a )
.
,i. The number of p-n junctions in a thyristor is (a) 1 (b) 2 (c) 3 (d) 4
, 2. When a thyristor is forward biased, the number of block ed p-'TI. j unctions is
(a) 1 (b) 2 (c) 3 (d ) 4 3. When a thyristor is reverse biased, the number of blockQ
, 4. In a thyrist or , anode current 1'3 made up of
(a) electronic only (b) electrons or holes (c ) electrons and h oles (d) holes only. 5. A thyristor , when triggered, will ch arge from forward-bl ocking state to con duction state if its anode to ca thod voltage is equal to (a) peak r epetitive off-state forward volt age (b ) peak working off-state forward voltage (c ) peak working off-state reverse voltage (d) peak n on-re etitive off-state forw ard volt age
712
Power Electronics '. '.~.·i.,. ..
0
':I •
.•
6. Commutation or turn-off of a thyristor requires t hat 1. anode current is reduced below holding current 2. anode voltage is reduced to zero 3. anode current is allowed to reverse 4. anode voltage gets r eversed 5. reverse voltage is applied to it.
From these, the correct statements are
(a) all (b) 1, 3, 4 (c) 1, 3, 4, 5 (d) 1, 2, 4 7. In a thyristor, the ratio of holding current to latching current is (a) 0.4 (b) 1.0 (c) 2.5 (d) 4.00 8. When a thyristor gets turned on, the. gate drive (a) should not be removed as it will turn-off the SCR (b)
mayor may not be removed
(c) should be removed
should be removed to avoid increased losses and higher junction temperature. 9. For normal SCRs, turn-on time is (a) less then turn-off time, tq (b) more then tq (c) equal to tq (d) about halfoftq. (d)
10. The forward voltage drop during SCR on-state is 1.5 V. This voltage drop (a) remains constant and is independent ofload current (b ) increase slightly with load current (c) decreases slightly with 'load current (d) varieus linearly with load current. 11. On-state voltage drop a cross a phase-controlled thyristor is as under: 1. 0.5 V to 1.0 V for 250 V devices 2. 1.5 V for 1 kV devices 3. 1.5 V for 6 kV devices 4. 3 V for 6 kV devices 5. 1.5 V for 250 devices
From these, the correct statements are
(a) 1, 2, 3, 5 (b) 1,2,4 (c) 2,3, 4, 5 (d) 2, 4, 5 12. The average current r ating of a thyristor, as supplied by t h e manufacturers, cor responds to (a) resistive current (b) indu ctive current (c) capa citive current (d) none of th e above 13. For reliable gate triggering of thyristors, it is advisable to employ (a ) s ligh t over triggering (b) very soft triggering (c ) v ery hard triggerin g (d) n one of th e above 14. It is easier to m anufact ur e (a) fast SCR of high PN (c) fast SCR of low losses
(b ) (d)
fast SCR ofl ow PlV none of the abcve.
Appendix
713
15. The most efficient gate-triggering signal for SCR is (a) a steady de level (b) a short duration pulse (c) a high-frequency pulse train (d) a low-frequency pulse train. 16. A driver circuit is required between the controller and the power circuit mainly for (a) isolation (b) voltage level change (c) polarity change (d) necessary drive power. 17. During forward blocking state, a thyristor is associated with (a) large current, low voltage (b) low current, large voltage (c) medium current, large voltage (d) low current, medium voltage 18. Once SGR starts conducting a forward current, its gate loses control over (a) anode circuit voltage only· (b) anode circuit current only (c) anode circuit voltage and current •.C~~ ' ,.". (d) anode circuit voltage, current and time. 19. In a thyristor (a) latching current h is associated with turn-off process and holding current IH . with turn-on process· (b) both I L and I H are associated with turn-off process (c) IH is associated with turn-off process and h with turn-on process · (d) both hand IH are associated with turn-on process. 20. The SCR ratings, dildt inAI/lsec and duldt in VI)lse~, may vary, respectively, be tween (a) 20 to 500, 10 to 100 (b) both 20 to 500 (c) both 10 to 100 (d) 50 to 300, 20 to 500. 21. A thyristor can be termed as (a) DC switch (b) AC switch (c) either (a) or (b) (d) square-wave switch. 22. Turn-on time of an SCR can be reduced by using a (a) . rectangular pulse of high amplitude and narrow width (b) rectangular pulse of low amplitude and wide width (c) triangular pulse (d) trapezoidal pulse. 23. Turn-on time of an SCR in series with RL circuit can be r educed by (a) increasing circuit resistance R . (b) decreasing R (c) increasing circuit inductance L (d) decre asing L. 24. For an SCR wit h turn-on time of 5 microsecond, an ideal trigger pulse sh ould have (a) short rise time with pulse width =3 /lsec (b) long rise tim e with pulse width = 6 )lsec (c) short rise time with pulse width = 6 /lsec (d) long rise tim e with pulse width = 3 )lsec. 25. A forward voltage can be applied t o an SCR aft er its (a) anode cur r ent reduces t o zero ( b) gate r ecovery time (c) r everse r ecovery tim e (d ) anode voltage r educes to zer o.
Power Electronics
714
26. Turn-off time of an SCR is measured fr om the instant (el) anode current becomes zero (b) anode voltage becomes zero· (c) anode voltage and anode current become zero at the same time (d) gate current becomes zero. 27. Turn-on time for an SCR is 10 Ilsec. If an inductance is inserted in the anode circuit, then the turn-on time will be (a) 10 Ilsec (b) less than 10 Ilsec (c) more than 10 ~sec (d) about 10 ~sec. 28. In an SCR, anode current flows over a narrow region near the gate during (a) delay time td (b) rise time tr and spread time tp (c) td and tp (d) td and t r . 29. Gate characteristic of a thyristor (a) is a straight line passing through origin (b) is of the type Vg = a + b Ig (c) is a curve between Vg and Ig (d) has a spread between two curves ofVg - Ig. 30. The average on-state current for an SCR is 20 A for a conduction angle of 1200 : Its average on-state current for 60 0 conduction angle would be (a) 20 A (b) 10 A (c) less than20 A (d) 40 A 31. The average on-state current for an SCR is 20 A for a resistive load. If an inductance of 5 mH is included in the load, then average on-state current would be (a) more than 20 A (b) less than 20 A (c) 15 A (d) 20 A
32. Specification sheet for an SCR gives its maximum rms on-state current as 35 A This r ms rating for a conduction angle of 1200 would be (a) more than 35 A (b) less than.3 5 A (c) 35 A (d) 52.5 A 33. Surge current rating of an SCR specifies the maximum (a) repetitive current with sine wave (b) non-repetitive current with rectangular wave (c) non-repetitive current with sine wave (d) r epetitive current with rectangular wave.
34. Th e dil dt r ating of an SCR is specified for its (a) decaying anode current (b) decaying gate current (c) rising gate curr ent (d ) rising anode current. 35. For an SCR , dvld t protection is achieved through the use of (a) RL in series with SCR (b) RC across SCR (c) L in series with SCR (d) RC in series with SCR. 36. For an SCR, d il dt protection is achieved through the us e of (a ) R in series with SCR . (b) RL in series with SCR (c) L in series with SCR (d) L across SCR. 37. The fUnction of snubber circuit connected across an SCR is to (a) suppress d v l dt
715
Append'x
(b) increase dv l dt (c) decrease dvldt (d) keep transient overvoltage at a constant value. 38. The object of connecting. resistance and capacitance acr oss gate circuit is to protect the SCR gate against (a) overvoltages . (b) dv l dt (c) noise signals (d) overcurrents.
39. During the turn-off process in a thyristor, the current flow does not stop at the instant current reaches zero but continues to flow to a peak value in the reverse direction. This is due to (a) commutation failure (b) hole-storage effect . (c) presence of reverse voltage across the thyristor (d) protective inductance in series with the thyristor. 40. The maximum dUdt in a thyristor circuit is l. directly proportional to maximum value of supply voltage Vm 2. inversely proportional to Vm 3. inversely proportional to circuit inductance L 4. directly proportional to L
Fromabove, the correct statements are
(a) 1,3.· (b) 2,4 (c) 2,3 (d) 1,4 41. Thermal resistance. in SCRs has the units of (a) °C / W and heat sinks (HS) are made from aluminium (b) W1°C and HS are m a de from steel (c) °C/W and HS are made from copper (d) °C/W and HS are made from copper alloy.
42. A power semiconductor device of thermal resistance 0.6°C/W has its heat sink at 90°C . In case the junction drop is l.5 V for a load current of 30 A dc, the junction temperature would be (a) 63°C . (b) 107°C (c) 117°C (d) 127°C 43. A thyristor converter of 415 V, 100 A is operating at rated load. Details of thyristor used are as follows : . Thermal resistance: junction to case = O.OPC/W case to sink = 0.08°C / W sink to at mosphere = 0.09°C/ W . . . For an ambient te mperature of 35°C, the junction temperature of 100% load would be (a) 48.5°C (b) 54.5°C (c) 60°C (d) 62°C 44. A thyristor can withstand a m aximum junction temperature of 120°C with an ambient temper ature of 75°C. If this SCR has thermal resistance from junction to ambient as l.5°C/W, the maximum internal power dissipation allowed is (a) 30
W
(6) 60 W
(c)
80 W
(d ) 50 W
45. Practical way of obtaining static voltage equalization in series- connected SCRs is by the use of (a) . one resistor across th e string (6) r esistors of different va.lues a cross each SCR
716
Power Electronics
resistors of the same value across each SCR resistor in series with the string. 46. For series connected SCRs, dynamic equalizing circuit consi~ts of (a ) resistor R and capacitor e in series but with a diode D acrosse (b ) series R and D circuit but with e across R (c ) series Rand e circuit but with D across R (d) series e and D circuit but with R across e. 47. For dynamic equalizing circuit used for series connected SCRs, the choice of e is based on (a ) reverse recovery characteristic
. (b) turn-on characteristics
(c) turn-off characteristics (d) rise-time characteristics. (c )
(d) one
48. In an UJT, with VBB as the voltage across two base terminals, the emitter potential at peak point is given by (a) 11VBB (c) 11 V BB + Vn
(b) 11VD (d) 11VD + V BB .
49. An UJT exhibits negative resistance region (a) before the peak point (b) between peak and valley points (c) after the valley point (d) both (a) and (c). - 50. In an UJT, maximum value of charging resistance is associated with (a) peak point (b) valley point (c) any point between peak and valley points (d) after the valley point. 5!. vVhen an UJT is used for triggering an SCR, the waveshape of the voltage obtained from UJ T circuit is a (a) sine wave (b) saw-tooth wave (c) trapezoidal wave (d) square wave. 52. For an UJT employed for the triggering of an SCR, stand-off ratio 11 = 0.64 and de source voltage VBB is 20 V. The UJT would trigger when the emitter voltage is (a) 12.8 V . (b ) 13.5 V (c) 10 V (d) 5 V, . 53. UJTs are used for oscillators for the existence of (a) peak-point potential .· ( b ) valley-point potential (c ) positive resistance part of VA charact~iisttcs
Cd ) n egative resistanc:3 part of VA charact eristics.
54. An UJT i s employed to fabricate a relaxation oscillator . When energised, it fails to oscillate . This may be due to 1. high base-terminal voltage VBB 2. too large a capacitor 3. low value ·of chargin ", resistor 4. large interbase r esistance.
From these, the correct statements are
(a ) 1, 3 ( b) 1, 2,3 (c) all (d) 2, 4
71 7
Appendix
55. A power semiconductor may undergo damage due to (a) high dildt (b) low dildt . (e) high dv/dt (d) low dv/dt
56. A triace is equivalent to (a ) two diodes in antiparallel . (e) two thyristors in parallel
(b) (d)
,'"
one thyristor and one diode in parallel
two thyristors in antiparallel
57. For a triac and SCR, (a) both are unidirectional devices (b) triac requires more current for turn-on than SCR at a particular voltage (e) a triac has less time forturn-offthan SCR
Cd) both are available with comparable voltage and current ratings.
58. Consider the following statements : 1. The triac is a five layer device 2 . The triac may be considered to consist of two parallel sections PI nI P2 n2 and P2 nIpI n4
3. An additional latera r egion serves as the control,gate 4 . . The triac is a double ended SCR.
From above, the correct statements are
(a) all (b) 1,2,3 (e) 1 only (d) 1,4 59. In a conventional reverse blocking thyristor (a) external layers are lightly doped and internal layers are heavily doped (b) external layers are heavily doped and internal layers are dightly doped
. (e) the p-Iayers are heavily doped and the n-Iayers are lightly dope d
(d ) the p-layers are lightly doped and the n-Iayers are heavily doped. 60 . If the amplitude of the gate pulse to thyristor is increased, then (a) both delay time and rise time would increase .. (b ) the delay time would increase but the rise time would decrease (c) the delay time would decrease but the rise time would increase (d) the delay time would decrease while the rise time remains unaffected. 61. Use of a reverse conducting thyristor in place" of antiparallel combination ofthyristor and feedback diode in an inverter . (a) effectively minimises the peak commut ating current (b) decreases the operating frequency of operation (e) minimises effects oflead inductances on the commutation performance (d) causes deterioration in the commutation performance 62. 7. iacs are most suit able wh en the supply voltage is (a) de (b) low-frequencyae (e) high-frequencyae (d) full-wave rectified ae 63. Which one of the following statements is correct? A riac is a (a) 2 t erminal switch (b) 2 ter minal bilateral switch (e) 3 terminal unilateral swit ch (d) 3 ter minal bidirectional switch
718
Power Electrenics
64. In the circuit of relaxation oscillator shown in Fig. C. 27, wh a t will be the change in voltage waveform
R across capacitor, if the voltage Vl is doubled.
(a) The amplitude as well as the frequency of the
waveform will get doubled.
(b) The amplitude will get doubled, but the fre
c . quency will reduce to half its value.
(c) The amplitude will get. doubled, but the fre Fig. C.2? . quency will remain unchanged . .. (d) The amplitude will remain unchanged, but the frequency will get doubled? 65. Which of the following characteristic of a silicon p- n junction diode makes it suitable for use ·as an ideal diode? 1. it has very low saturation current 2. It has a high vahle of forward cut-in voltage 3. It can withstand large reverse voltage 4. When compared with germanium diodes, silicon diodes show a lower degree of temperature dependence under reverse bias conditions.
Select the correct answer using the codes given below:
Codes:
(a) 1 and 2 (b) 1,2,3 and 4 (c) 2,3 and 4 (d) 1 and 3 66. Which one of the following stateIr..ents regarding the two-transistor model of the p-n-p-n four-layer device is correct? . (a) It explains only the turn-on portion ofthe device characteristics. (b) It explains only the turn-off portion of the device characteristics (c) It explains only the negative-region portion of device characteristics (d) It explains all the regions of the device characteristics. 67. Thyristors can be turned off by 1. reducing the current below the holding current 2. applying a negative voltage to the anode of the device 3. reducing thE:) gate current.
Of these statements,
(a) 1 and 3 are correct (b) 2 and 3 are correct (c) 1, 2 and 3 are correct (d) 2 and 3 are correct. 68. The turn-on time for an SCR is 30 ~s. The pulse train at the gate has a frequen<;:y of 2.5 kHz with a mark/space ratio of 0.1. This SCR will (a) turn-on (b) not turn-on _ . (c) turn-on if pulse-frequency is increased (d) turn-on if pulse-frequency is decreased [Hint: Here 8 = 1/ 11J 69. A thyristor is t riggered by a pulse train of 5 kHz. The duty r atio is 0.4.· If the allowable average power is 100 W, the maximum all owable gat e-drive power is (a) 100 {i w (b) 50 W (c) 150 W (d) 250 W 70. During turn-on process of a thyristor, maximum power losses occur during (a) tp ( b) tr (c) td (d) equal in all
.i t
Appendix
719
71. A pulse transformer is used in a driver circuit (a) to prevent a de triggering (b) to shape the trigger signal (e) to generate high-frequency pulses (d) to provide isolation 72. Current unbalance in the parallel~connected SCRs i~ due to the non-uniformity in the (b) reverse characteristics (a) forward characteristics (d) dvldt withstand capability. (c) dildt withstand capability 73. In synchronized UJT triggering of an SCR, voltage Vc across capacitor reaches UJT thresh-hold voltage thrice in each half cycle so that there are _three firing pulses during each half cycle. The firing angle of the SCR can be controlled (a) once in each half cycle (b) thrice in each half cycle (c) twice in each half cycle (d) four times in each half cycle. 74. The function of connecting a zener diode in an UJT circuit, used for the triggering of SCRs, is to . (a) expedite the gene~ation of triggering pulses (b) delay the generation of triggering pulses (e) provide a constant voltage to UJT to prevent erratic firing· (d) prqvide a variable voltage to UJT as the source voltage changes. 75. A metal oxide varistor (MOV) is used for protecting (a) gate circuit against overcurrents (b) gate circuit against overvoltages (c) anode circuit against overcurrents . (d) anode circuit against overvoltages. 7H. The functions of connecting a resistor in series with gate-cathode circuit and a zener-diode across gate-cathode circuit are, respectively, to protect the gate circuit of a thysistor from (a) overvoltages,overcurrents (b) overcurrents, overvoltages (e) overcurrents, noise signals (d) noise signals, overvoltages. 77. In a GTO, anode current begins to fall when gate current (a) is negative peak at time t = 0 (b) is negative peak at t = storage period ts (c) just begins to become negative at t = 0 (d) is n egative peak at t = (ts + fall time). 78. For a pulse transformer, the material used for its core and the possible turn-ratio fr om primary to secondary are .~ e spectively, \ (a) ferrite; 20 : 1 · (b) laminated iron; 1: 1 (e) ferrite; 1 : 1 (d) powdered iron; 1 : l. 79. Protection against d ild t stress in a device is necessary because (a) it interferes with control electronics (b) it introdu ces voltage s rges on supply lines (c) it destr oys t he device (d) n one of the above are valid.
720
Power Electronics
80. Overcurrent protection of thyristors is provided by (a) use of saturable di / dt coils (b) use of circuit breaker and a fuse · (c) use of snubber circuit (d) liberal heat sinking. 81. The capacitance of a reverse biased junction of a thyristor is 20 picofaracd. The charging current of this thyristor is 4 rnA. The limiting value' of du / dt in V /JlSis (a) 50 (b) 100 (c) 200 (d) 500 82. SCR can be turned on by · 1. applying anode voltage at a sufficiently fast rate 2. applying sufficiently large anode voltage 3. increasing SCR temperature to a sufficiently large value 4. applying adequately large gate current.
From these, the correct statement are
(a) all (b) 4 only (c) 2,4 (d) 1,2,4 83. During forward blocking of two series-connected SCRs, a thyristor with 1. high leakage impedance shares lower voltage 2. high leakage impedance shares higher voltage 3. low leakage impedance shares higher voltage 4. low leakage impedance shares lower voltage.
From these, the correct statements 'are
(a) 1,3 (b) 1,4 (c) 2,3 (d) 2,4 84. Thyristor A has rated gate current of 2A arid thyristor B a rated gate current of 100 rnA .
1.. A is a GTO and B is a conventional SCR
2. B is a GTO and A is a conventional SCR 3. A may operate as a transistor 4. B may operate as a thyristor.
From the above the correct statements are
(a) 1,4 (b) 1,3 (c) 2,3 (d) 2, 4 85. A resistor connected across the gate-cathode terminals of a thyristor increases its 1. dul dtrating holding current 3. noise immunity 4. turn-off time.
From these, the correct statements are
(a) all (b) 2,3 (c) 1,2,3 (d) 2,3,4 86. To generate gate-t riggering signals IC 555 is often used in (a) monostable mode (b) bistable mode (c) astable mode (d) none of the above. 87. A GTO with anode fing er s has (a) no reverse blocking capability (b) r educed tail current (c) high turn-off time (d) r edu ced turn-off gain 88. A gold-doped GTO h as (a) high revers e blocking capability (b) increased tail current (c ) low on-state voltage drop (d) low turn-off tim e
2:
721
Appendix
From these, the correct statements are (d) 1,2,4 (a) 1,4 (b) 1,3,4 (c) all 89. The correct sequence of the given devices in the decreasing order of their speed of operation is (a) power BJT, PMOSFET, IGBT, SCR (b) IGBT, PMOSFET, powerBJ'T, SCR (c) SCR, PBJT, IGBT, PMOSFET (d) PMOSFET, IGBT, PBJT, SCR 90. Consider the following statements: . l. . BJT has lower power losses than PMOSFET ' 2. PMOSFET has lower power losses than IGBT 3. SeRs have lower power losses than PMOSFET and IGBT.
Which of these statements are correct?
(d) 1 and 3 (a) 1,2 and 3 (b) 1 and 2 (c) 2 and 3
9t. Consider the following statements about SITH: l. It is a p+nn+ diode 2. It is normally-on device 3. It has low reverse blocking capability
4. It has p + gate electrodes buried in n layer 5. Its turn-off gain varies from 4 to 6.
From these , the correct statements are
(a) I, 2, 4 (b) 1,2,3,5 (c) 2,3,4, 5
(d)
1,3,4,5
92. Consider the following statements about GTO and SITH. 1. . Anode-shorting in GTO reduces its reverse-voltage blocking capability to zero 2. Anode-shorting in SITH reduces its reverse-voltage blocking capability to zero 3. Anode-fingers in GTO increase its turn-off gain
Both GTO and SITH have n + type fingers in the anode.
From above, the correct statements are
(a) all (b) 2,3,4 (c) 1,2,3 (d) 2,4 93. Cosine triggering control is used to linearise the relation between (a) Ec and Cl. (b) CI. and Vo (c) Vo and 10 (d ) Ec an d Vo 4.
where Vo = outpu t voltage of converter, 10 = out
put current , Ec = control voltage to triggering
scheme and CI. = trigger angle.
94. The triggering circuit of a thyristor is shown in Fig. C.28. The thyristor requires a gate cu_ rent of 10 rnA, for guaranteed t urn-on . The value of R required for the thyri stor to turn on reliably under all con ditions of Vb variation is (a) 10,000 n (b) 1600 n (c) 1200 n
Fig. C.28 (d ) 800
n [GATE,2004]
;
i.
722
P ower Electronics
95. An electronic switch S is r equired t o block voltages of either polarity during its OFF state as shown in Fig. C.29 (a). The switch is required to conduct in only one direction during its ON state as shown in Fig, C.29 (b) , 1 5 l' ~o----
1
5
l'
0--0-0---0
2::;:
~
(a)
( b) ,
, Fig. C.29
Which of the following are valid realization of the switch S ? '
(P)
(R)
(Q)
(a) Only P
(b)PandQ
(c)
(8)
P and R
(d) Rand S
[GATE, 2005)
".
ANSWERS 1. (c)
2.
7.
8.
(a)
13. (c) 19. (c)
14.
(a)
3. (6)
(d)
9.
"
(a)
, 20. (b)
/-
....
4. (c)
5. (b)
6. (c)
10. (b)
11. (6)
12. Ca)
15. (c)
16.
(d)
17. (6)
18. (c)
21.
(a)
22.
(a)
23.
(d)
24. (c)
(a)
25. (6)
26.
(a)
27. (c)
28.
(d)
29.
(d)
30. (c)
31.
(a)
32. (c)
33. (c)
34.
(d)
35. (b)
36. (c)
37.
(a)
38. (c)
39. (b)
40.
(a)
41.
(a)
42. (c)
43.
(d;
44.
(a) ,
45. (c)
'46. (c)
47.
(a)
48. (c)
49.
(6)
50.
(a)
51. (6)
52. (6) ,
53.
(d)
54.
(a)
55.
(a)
56.
(d)
57. (6) ,
58.
59. (6)
60.
(d)
61.
(c)
62. (b)
63.
(d)
64. (6)
65.
(d)
6 6.
(a)
69.
(d)
70. (b)
71.
(d)
72.
(a)
76. (6)
77. (b)
78. (c)
82.
(a)
83.
(d)
84. (6)
88. (c)
89.
(d)
90.
94.
95. (c)
67. (6)
68.
73.
(a)
74. (c)
75., (d)
79. (c)
80. (6)
81.
85. (c)
86. (c)
87. (c)
91. (a)
92. (6)
93.
(a) ,
(c)
(d) ,
(a)
(d)
(d)
.1
'
~
723
Appendix
THYRISTOR COMMUTATION TECHNIQUES
1. For the circuit shown in Fig. C.30, the conduction time for thyristor in microseconds is (a) 0.393 (b) 2.546 (c) 25.133 (d) 8.0
2. For the circuit in Fig. C.30, the capacitor voltage after .SCR gets ·self-commutated is (a) 200 V (b) 400 V (c) 300 V
+ 200 V
Fig. C.30 (d)
100 V
3. For the circuit shmvn in Fig. C.30, the voltage across thyristor, after it is self-com mutated is (a) zero (b) -1.5V (c)- 200 V (d) - 400 V 4. In the circuit of Fig. C;30, the peak thyristor current is (a) 100 A (b) 50 A (c) 400 A .. (d) 800 A !>. In the circuit of Fig. C.31, the maximum value of current thrQugh thyristors T1 and TA can respectively be Vs Vs ·
(a)
. (b)
_rr;-;T
R' Ii + Vs .... CI L
i
D
+ Vs ..JCIL, Vs ..JCIL
(c) V s ..JelL ,
v
+
.
R
i
TA
(d) ; , Vs ..JCIL
Fig. C.31
6. For the circuit shown in Fig. C.31, the peak value of resonant current is twice the load current. In case Vs = 200 V, the magnitude of voltage across the main thyristor, when it gets turned off, is equal to . (ci) 86.6 V (b) 100V . (c) 173.2 V (d ) 200 V 7. For the circuit in Fig. C.31, the peak value of current through auxiliary SCR is twice that through the main SCR. In case Vs = 100 V, C = 10 iJ.F an d constant load current = 40 A, the circuit turn-off time for main SCR, in microseconds is . (a) 12.5 (b) 21. 65 (c) 25 · (d ) .10 8. Read the following statements with regard to Fig. C.32, where cap acitor C is charged to voltage Vs with polarity as sh own : 1. In order to turn-offT1, turn on T2. 2. In order to turn-off T2, turn on T l. 3. At the t ime of turn-on of SCR, initial thyrist or
cur rent is Vs [
;1;2 ] ;1;2} +
T2
4. At the time ofturn -on of SCR, initial thyristor .curr ent is V s [
+
Frain above, the correct statements are 2,4 (b) 1,3 (c) 1, 4
(a)
Fig. C.32 (d)
2,3
Power Electronics
724
9. In the circuit shown in Fig. C.32, Rl =50 n, R2 = 100 Q and Vs = 100 V. The possible . peak values of current through SCRs 'i'l''''and T2 are respectively ., (a) 2A,5A (b) 1A,2A (c) 4A,2A (d) 4A,5A 10. In the circuit shown in Fig. C.33, Vs =200 V, C =4 IlF,L = 16 IlH and R =20 n. The peak value of current through T1 and D can respectively be + (a) 110 A, 100 A (b) 10 A, 110 A (c) 110 A, 10 A (d) 100A, 110 A
11. In the circuit of Fig. C.33 and for the parameters given in Prob. 10, the circuit turn-off time for main and auxiliary SCRs in microseconds are respectively (a) 8, 12.566 (b) 40, 1.2566, (c) 80, 12.566 (d) 80,25.132 ' ,
R
D 0 - - - ' - - - - ' - - - - --
--'
Fig. C.33
12. In the circuit configuration shown in Fig. C.34, the T1 circuit turn-off time for main thyristor is 34.657Ils. The , +. value of capacitor C required, in this circuit, is ' c (a) 5 IlF (b) 3.4661lF 200V (c) 1.7331lF (d) 10 IlF TA I 13. Match List I with List II and select the correct answer by using the codes given below the lists : Fig. C.34 List I List II Type of commutation ' Power Circuit Diagram
.
:" '.
+
(1) A Self-commutation
+ B. Complementary commutation
(2)
Vs
(3)
Vs
. (4 )
Vs
-'
C. Impulse commutation
D. Resonant-pulse commutation
'• •<: ..
• ......
1011
725
App ndix
Codes: A
B
c
4 1
1 4
2 3
(a) (c)
D 3 2
14. A series circuit consists of R
(b) (d )
A 1 4
B
c
4 .1
2 3
D 3 2
=2.4.0, L =25 jJ.H, C and a
thyristor. For obtaining self-commutation in the circuit, the value of C should be equal to (a) 50 J.1F (b) 30 jJ.F (c) 20jJ.F (d) 10 jJ.F 15. Match the type of commutation in List I with their alternative names in List II and tick ·the correct answer from the codes given below : List I List II
. Type ofcommutation . Alternative title
. A. Class A . 1. Voltage commutation B. Class B 2. Parallel-capacitor commutation C. Class C 3. Complementary-impulse commutation 4. Self-commutation D. Class D 5. Natural commutation 6. Current commutation
Codes:
D C A B A B C D (a) (b) 6 4 4 6 5 1 3 1 4 2 (c) 4 6 (d) 4 6 2 3 16. Match the type of commutation in List I with those in List II and give the correct answer by using the codes given below the lists : List I List II
Type ofcommutation Alternative title
A. Load commutation 1. Voltage commutation B. Impulse commutation 2. Natural commutation C. Line commutation 3. Resonant commutation 4. Parallel-capacitor commutation D. Resonant-pulse commutation 5. Current commutation
Codes :
A B D A C D B C (a) 5 1 (b) 5 2 2 3 3 4 (c) 3 4 (d) 4 2 5 3 2 1 · 17. In a commut ation circuit employed to turn-off an SCR, satisfact ory t urn-off is ob ained when (a ) circu it turn-off time < device turn-off time ( b) circuit turn-off time > device turn-off time (c ) circuit time constant > device turn-off time (d ) circuit time constant < device tum-off time
ANSWERS 1. (c)
2. (b)
7 . (b)
8 . (b)
1 $, (d)
14. (d)
3. (c) 9-. (d) 15. (a )
4. (a ) 1D. (a)
5. (d)
6. (e)
11. (e)
12. Cal
I S. (e )
17. (b)
Power Electronics
726 PHASE CONTROLLED RECTIFIERS
1. A sirrgl-;ph ase half-wave controlled rectifier has 400 sin 314 t as the input voltage
and R -as the load . For a firing angle of 60 0 for the SCR, the average output voltage IS
. (c) 240/1t 300/1t (d) 20011t. 2. A single-phase one-pulse controlled circuit has resistance and counter emf load and , 400 sin 314 't as the source voltage. For a load counter emf of 200 V, the range of firing angle control is . (a) 30 0 to 150 0 (b) 30 0 toI8.0° . (c) 60 0 to 120 0 (d) 60 0 to 180 0 3. In a single-phase half-wave circuit with RL load, and a freewheeling diode across the load, exti:p.ction angle ~ is more than ' 1t. For a firing angle Ct, the SCR and freewheeling diode would conduct, respectively, for (a) 1t - Ct, ~ (b) ~ - Ct, 1t - Ct (c) 1t - Ct, ~ -1t (d) 1t - Ct, 1t - ~. 4. In a single-phase one-pulse circuit with RL load and a freewheeling diode, extinction angle ' ~ is less than 1t. For a firing angle Ct, the SCR and freewheeling diode would, respectively, conduct for (a) ~ - Ct, 00 (b) 1t - Ct, 1t - ~ , (c)Ct, ~ - Ct . (d) ~ - Ct, Ct. 5. A single-phase full-wave mid-point thyristor converter uses a 230/200 V transformer with center tap on the secondary side. The P.I.v. per thyristor is . (a) 100 V (b) 141.4 V (c) 200 V (d) 282.8 V. 6. A single-phase two-pulse bridge converter has an average output voltage and power output of 500 V and 10 kW respectively. The SCRs used in the two-pulse bridge converter are now re-employed to form a single-phase two-pulse mid-point converter. This new controlled converter would give, respectively, an average output voltage imd power output of (a) 500 V, 10 kW .)fI (b) 250 V, 5 kW (c) 250 V, 10 kW (d) 500 V, 5 kW. 7. In a single-phase full converter bridge, the average output voltage is given by (a) 400/1t
1 (a)-
r+
ex
.
r
1t
1 (c)-
(b)
ex
Vmcos9d9
+ (1t/ 2)
.
Vmcos9 · d9
1
rx+
1t
(b)-L 1t 0
Vmcos9·d9
r
1 7t/2 ) + ex (d)-l V m cos9 · d9
1t ex - (1t/ 2) 1t (7t/ 2) - ex 8 . In a single-phase semiconverter, the average output voltage is given by
1 (a)-
f1t
1t ex
.
Vmc os9 · d9
1 J1t/ 2
(c)- ·
V m cos 9 ·d9
r
1 7t/2 )
(b)-J.
1t
.1 (d .) -
(1t/ 2) -
fTt
Vmcos9·d9 ex
Vmcos9 · d9
1t ex -(1t / 2 ) 1 t ex-(1t/ 2) 9: In a single-phase full converter, for continuous conduction, each pair of SCRs conduct for (a)n:-:-Ct
(b)1t
(c) Ct
(d) 1t+a.
10. In a sin gle-phase full converter, for discontinuous load current and extinction angle ~ ' > 7t, each S CR condu cts for (a) a (b ) ~ - a (c) ~ (d ) Ct + ~
727
Appendix
11. In a single-phase semi-converter, for continuous conduction, each SCR conducts for (a) ex (b) 1t (c) ex + 1t (d) 1t - a 12. In a single-phase semiconverter, for discontinuous conduction and extinction angle ~ > 1t, each SCR conducts for (a) 1t-a (b) ~-ex (c) a (d) ~. 13. In a single-phase semiconverter, for discontinuous conduction and extinction angle ~ < 1t, each SCR conducts for (a) 1t-ex (b) ~-ex (c) a (d) I~ 14. In a single-phase semiconverter, for continuous conduction, freewheeling diode con~ . ducts for (a) . ex (b) 1t - ex (c) 1t (d) 1t + a . 15. In a single-phase semiconverter, with discontiilUous conduction and extinction angle ~ > 1t, freewheeling diode conducts for (a) ex (b) ~ -1t (cf 1t + a (d) ~ . 16. In a single-phase semiconverter, with discontinuous conduction and extinction angle ~ < 1t, freewheeling diode conducts for (a ) . a (b) 1t - ~ (c) ~ - 1t . (d) zero degree. 17. In a single-phase full converter, if a and ~ are firing and extinction angles respec tively, then the load current is (a) discontinuous if (~ - ex) < 1t (b) discontinuous if (~ - a) > 1t (c) discontinuous if (13 - ex) =1t (d) continuous if (13 - a) < 1t. 18. In a single-phase semiconverter with resistive load and for a firing angle a, each SCR and freewheeling diode conduct, respectively, for (a) ex, 0° (b) 1t - ex, ex (c) 1t + ex, ex (d) 1t - a, 0°. 19. In controlled r ectifiers, the nature of load current, i.e. whether load current is continuous or disco~tinuous (a) does not depend on type ofload and firing angle delay (b) depends both on the type ofload and firing angle delay (c) depends only on the type ofload (d) depends .only on the firing angle delay. 20. In a single-phase full converter with resistive load and for a firing angle delay a, the load current is . 1. zero at a , 1t + ex, 27t + a ... 2. remains zero for duration a. Vm . 3' }fsmexatex, 1t+ex, 21t+a ... 4. remains zero for duration 1t - a Vm .
5. R sm a at ex only. From these, the correct statements are (b) 2, 3,5 (c ) 2,3 (d) 4,5 21. In a sin gle-phase full converter, if outpu t voltage h as peak and average values of 325 V and 133 V respectively, then t h e firing angle is (a) 40° (b) 140° (c) 50° (d) 130° (a) 1,2,3
22. In a single-phase semlconverter, if output voltage has pea..\ and average valu es of 325 and 133 V respectively, t h e firin g angle is (a) 40° ( b ) 73.40° (c) 80° (d) 140°
728
Power Electronics
23. A single SCR is inserted in between voltage source 200 sin 314 t and a load R = 10 n. If the gate trigger voltage lags the ae supply voltage by 120°, then average load current is 15 15 5 5 (a) - - A (b) - A (e) --A {d)-A 7t
7t
7t
7t
24. The average value of de voltage of a single-phase semiconverter, under continuous conduction, is
-12 . Vs (1 - cos a)
-12 . Vs (1 + cos a) (a) - - - " - - - - - -
(b)-~---
1t
1t
(e)
-12 (1 ;7tcos a)
-12. Vs (1 (d)
cos a)
21t
25. A freewheeling diode is placed across the de load 1. to prevent reversal ofload voltage 2. to transfer the load current away from the source 3. to transfer the load current away from conducting thyristor.
The correct statements are
(a) 1,3 (b) 2,3 (e) 1,2 (d) 1,2,3 26. A freewheeling diode across inductive load will provide (a) quick turn-on (b) slow turn-off (c) reduced utilization factor (d) improved power factor 27. When a flywheeling diode is connected across a full converter supplying ripple-free current at contron ed output voltage 1. de voltage increases at higher value of a 2. converter pf is improved 3. SCR heating is reduced.
From these, the correct statements are
(a) 2 only (b) 1,2 (c) 2,3 (d) 1,2,3 28. Consid~r the following statements: Phase controlled converters 1. do not provide smooth variation of output voltage 2. inject harmonics into the power system 3. draw non-unity pf current for finite triggering. angle.
Which of these statements are correct?
(a) 1,2 and 3 (b) 1 and 2 (e ) 2 and 3
(d)
-
1 and 3
29. The purpose of commutating diode in a thyristor controlled ac t o de converter is to (a) reduce the current of its associated SCR t o zero so that commutation can take place (b) share the load current of its associated SCR (c ) conduct the load current when its associated SCR is turn ed off (d) m aintain voltage across load at const ant value. 30. Con sider the following statements : The overlap angle of a phase-controlled converter will incr ease 1. as the firing angle increases 2. as the frequency of supply incre as es 3. as th e supply voltage decreases
Appendix
729
Of these statements, (b) 2 and 3 are correct (a) 1,2 and 3 are correct (d) 1 and 2 are correct (c) 1 and 3 are correct 31. A single-phase two pulse converter feeds RL load with sufficient smoothing so that the conduction is continuous. If the resistance of the load circuit is increased (a) the ripple content of the load current will remain the same (b) the ripple content of the load current will increase (e) the ripple content of the load current will decrease (d) there is a possibility of discoutinuous conduction. 32. Neglecting drops across SCRs and the circuit resistance except that in the load, the regulation of a converter, 6Vd, is given by IdB where R is an equivalent o~tput resistance. It is proportional to (a) input line voltage (b) trigger angle ex (c) back emfin the de circuit (d) number of commutations per second 33. Reactive loading of supply lines by a converter is directly dependent on (a) displacement angle only (b) displa'cement angle and distortion factor (e) back emfin the load circuit (d) circuit configuration . 34. A single-phase flill converter operates as an inverter, when (a) 0 0 ~ ex ~ 90° (b) 90° ~ ex ~ 180° (c) it supplies to a back-emfload (d) 90° ~ ex ~ 1800 and there is a suitable de source in the load cir cuit. 35. An inductance is inserted in the load circuit of SCR. With this 1. the turn-on time of SCR is increased 2. de output voltage is reduced for the same firing angle 3. conduction continues even after reversal of phase of input voltage 4. ex by-pass diode is connected in such circuits.
From above, the correct statements are
(a) 1,2,3,4 (b ) 1,3,4 (e) 2,3,4 (d) 1,2,3 36. Overlap in a phase-controlled converter, under continuous conduction, does not depend on (a) frequency (b) applied voltage (c) load current (d) load inductance. 37. Com mutation overlap in the ph ase-con tr olled ac to dc converters is due to (a) load inductance (b) harmonic content ofload current . (c) switch ing operation in the converter (d) sour ce in ductance 3S. A sin gle-phase fun converter would los'e its controllability ifit is feeding a load having (a) resistance and a current s ource in it (b) par allel combination of a r esistance and capacit ance (c) resist anc.e and indu ctance in series (d ) none of the above 39. Modern ac to de convertel"S employ GTOs instead of SeRs in order to have (a ) low reactive volt-amper e flow ( b ) r eliable commuta tion (c) low switching loss (d ) smaller h eat sink.
Power Elertronks
730
40. Each diode of a 3-phase half-wave diode rectifier con ducts for (a) 60° (b) 120° (c) 180° (d) 90°. 41. Each diode of a 3-phase, 6-pulse bridge diode rectifier conducts for (a) 60 (b) 120 (c) 180° (d) 90°. 42. In a 3-phase half-wave diode rectifier, if per phase input voltage is 200 V, then the average output voltage is (a) 233.91 V (b) 116.95 V (c) 202.56 V (d) 101.28 V. 43. In a a-phase half-wave diode rectifier, the ratio of average output voltage to per phase maximum ac voltage is (a) 0.955 (b) 0.827 (c) 1.654 (d) 1.169. 44. For a 3-phase, six-pulse diode rectifier, the average output voltage in terms of maximum value of linev61tage V m is 0
(a)
0
(b) 3VT1Il .
3f2"V 7t
ml
. 7t
45 . . In a 3-phase half-wave rectifier, dc output voltage is 230 V. The peak inverse voltage . across each diode is . (a) 481.7 V ' ; . (b) 460 V (c) 345 V (d) 230 V. · 46. Ina 3-phase full-wave diode rectifier, the peak inverse voltage in te~ms of average output voltage is (a) 1.571 (b) 0.955 (c) 1.047 (d) 2.094.
47. In a 3-phase half-wave diode rectifier, if V mp is the maximum value of per phase voltage, then each diode is subjected to a peak inver se voltage of (a) V~p . (b) {3vmp . (c) 2Vmp (d) 3Vmp ' . 48. In a 3-phase full-wave diode rectifier, if Vml is the maximum value of line voltage, then each diode is subjected to a peak inverse voltage of (a) V ml
.
(b) {3 V ml
(c) 2Vml
(d) 3Vml
49. In a 3-phase full-wave diode rectifier, if Vis the per phase input voltage, then average output voltage is given by (a) 0.955 V (b) 1.35 V (c) 2.34 V . (d) 3 V. 50. A converter which can operate in both 3-pulse and 6-pulse modes is a
. (a ) I -phase full converter (b) 3-phase half-wave converter
(c ) 3-phase sem 'converter (d) 3-phase full converter. 5!. In a 3-phase semi-converter, for firing angle less than or equal to 60°, each thyrist or and diode conduct, respectively, for . 0 0 0 (a ) 60 , 60 (b) 90 ,30° (c ) 120 , 120° (d) 180° 180 0 • 52. In a 3-ph ase semiconverter, for firing angle less than or equal t o 60°, freewheeling diode conduct s for J
(a) 30°
(b) 60°
(c) 900
(d)
zero degree 53. In a 3-phase semiconverter, for a . firing angle equal to 90° and for contim. ous conducticn, each SCR and diode con duct , respectively, for (a ) 30°, 60° (b) 600 ,30° 0 (c ) 90°, 90 (d) 30 0 , 30°.
Appendix
731
54. In a 3-phase semi converter, for a firing angle equaL to , 90° and for continuous conduction, freewheeling diode conducts for ' -- ',' (a) 30 0 (b) 60 0 (c) 90 0 . (d) zero degree, 55. In a 3-phase semiconverter, for firing angle equal to 1200 and extinction angle equal to 1100; each SCR and diode conduct, respectively, for
0 0 00
(a) 30 0,60 0 (b) 60 0,6_ (c) 90 0,30 0 (d) 110 ,30 , 56. In a 3-phase semiconverter, for firing angle equal to 1200 and extinction angle equal to 1100, freewheeling diode conducts for 0 (a) 100 (b) 30° , (c) 50 . (d) 110 0 , . 57. In a 3-phase semiconverter, for firing angle equal to 1200 and extinction angle equal to 1000, none of the bridge elements conduct for (a) 100 (b) 20 0 (c) 30 0 (d) 60 0 , 58. A 3-phasesemiconverter can work as (a) converter for ex = 00 to 180° (b) converter for ex = 0 0 to 90 0 0 (c) inverter for ex = 90 0 to 1800 (d) -i nverter for ex = 0° to 90 , 59. In a 3-phase semicoilverter, the three SCRs are triggered at an interval of (a)' 60c;~ (b) 90° (c) 120 0 '.. (d) 1800 , 60. In a 3-phase full converter, the six SCRs are fired at an 'inter;~l of 0 (a) 30 0 (b) 60 0 (d) 120°, . (c) 90 . 6 1. In a 3-phase full converter, three SCRs pertaining to one group are fired at an interval of (a ) 30 0 (b) 60 0 (c) 90 0 (d) 120 0 , 62. For "a single-phase two-pulse phase-controlled rectifier, with a freewheeling diode across RL load, . (a) the instantaneous output voltage Vo is always positive (b) Vo may be positive or zero (c) Vo may be positive, zero or negative ' (d) Vo is always zero or negative, 63. The frequency of the ripple in the output voltage of a 3-phase semiconverter depends
upon (a) firing angle and load resistance (bY firing angle and load inductance (c) th e supply fre quency (d) firing angle and the supply frequency. 64. In a single-phase full convert er, if load current is I and ripple free, then average and rms values of thYTistor current are 1 I ' 1 I II I (a) '2 I, T2 (b ) '3 If T3 (c) "4 I, '2 (d) I'T2' "
65. In a 3-ph ase full converter, if load curr ent is I and ripple fre e, then average and rms values of thyristor cur rent are 1 I II 1 1 I (a) 2 1, 12 (b) '3 I, T3 (c) "4 I, 2" (d) I, Ts' 66. The effect of source ind1.:ctance on th e performance of single-phase and three-ph ase full converters is to (a ) redu ce the ripples in the load current (b) m ake discontinuous current as continuous
Power Electronics
732 (c) reduce the output voltage (d) increase the load voltage.
67. In a single-phase full converter, the number of SCRs conducting during overlap is (a) 1 (b) 2 (c) 3 (d) 4. 68. In a single-phase full converter, the output voltage during overlap is equal to (a) zero · . (b) source voltage (c) source voltage minus the inductance drop (d) inductance drop. 69. In a 3-phase full converter, the output voltage during overlap is equal to (a) zero (b) source voltage (c) source voltage minus the inductance drop (d) average value of the conducting-phase voltages. 70. The totaLnumber of SCRs conducting simultaneously in 3-phase full converter with ov~rlap considered has the sequence of (a) 3,3,2,2 (b) 3,3,3,2 (c) 3,2,3,2 (d) 2,2,2,3. 71. A 3-phase full converter has an average output voltage of 200 V for zero degree firing angle and for resistive load. For a firing angle of 90°, the output voltage would be (a) zero (b) 50 V (c) 100 V (d) 26.8 V. 72. A four quadrant operation requires (a) two full converters in series (b) two full converters connected back to back (c) two full converters connected in parallel (d) two semiconverters connected back to back. 73. In
a 3-phase full
converter, the output voltage pulsates at a frequency equal to (0) supply frequency,{ (b) 2 f (c) 3f (d) 6f. 74. In circulating-current type of dual converter, the nature of voltage across reactor is (a) alternating (b) pulsating (c) direct (d) triangular. 75. For the same ac input volt age, the peak inver se voltage in ac to dc converter systems is highest in (a) single-phase full wave M-2 converter (b) single-phase full converter (c) 3-phase bridge converter (d) 3-phase M-3 converter.
76 • .:rhe three-phase ac to dc converter which requires neutral point connection is
(a) 3-phase semiconverter (b) 3-phase full converter (c) 3-phase half-wave converter (d ) 3-phase full con verter with diodes.
77. A 3-pulse converter fee ds RLE loa d. The source h as a definite induct ance causin g ov erlap. The thyristors are i deai. It h as an overlap angle ~ of 20" at the min imum firing angle 'a'. The current remains const ant in tne com plet e range of firing angle. The r ange of firing angle for the converter woul d be (a) 0 < a < 180° (b) 20 < ex < 180° (c ) 20 < ex < 16 0° (d ) 0° < ct < 160
'733
Appendix
78. The range of firing angle for a 3-phase, 3-pulse conv~rt~r f~e.ding a resistive load is (a) 0° to 180° (b) 0° to 150° (c) 30° to 150E) ..,.: (~1) 30° to 180° . . .... 79. A 3-pulse converter has a freewheeling diode across it~: load. The operating range of the converter is (a) 0° ~ a ~ 150° (b) 60° ~ a ~ 120° (c) 30° < a. ~ 150° (d) 180° ~ a ~ 360° 80. If the commutation angle of a diode rectifier (due to source inductance) is Il, then inductive voltage regulation is (d) 1- cos Il (c) 1 _ cos Il (a) 1 + ~os p. (b) 1 +~ ,
.
2
2
.·
2
81. In a 3-phase bridge rectifier fed from star-connected secondary winding of a trans former, let the voltage to the neutral of A-phase (phase sequence A, B, C) be Vm sin rot. At the instant,when voltage of A-phase is maximum, the output voltage at the rectifier terminals would be (a) V m l{2
(c) 1.5 Vm
(b) Vm
(d)
.J3. Vm
82. Three-phase voltages Va, Vb, Vc are applied to 3-phase M-3 diode rectifier. When Va is passing through zero and becoming positive, the load voltage Vo would be (Vmp = maximum value of phase voltage) (a) Vb = 0.5 V mp (b)v c = 0.866 V mp (c) vb = 0.866Ymp ': '~ (Yi,) Vc = 0.5 V mp 83. Line voltages Vab, Vac, Vbc, Vba etc. are ' applied to 3-phase, six-pulse diode bridge rectifier. When Vab is zero and becoming positive, the output voltage Vo would be (Vml = maxim urn value of line voltage), (a) Vcb=Vml
.
.J3
(b) vbc=TVml
fa
(c) vcb=TVml
.J3
(d) Vac=TVml
84. Line voltages Vab, Vac, Vbc, Vba etc are applied to 3-phase semi converter. For firing angle equal to 15°, when Vab is zero and becoming negative, the load voltage would be (Vml = maximum value of line voltage)
rs
(a) Vac=TVml
(b) vbc=0.866Vml
(c) vca=0.866Vml
(d) vac=-0.866Vml
85. When firing angle a of a single-phase full converter feedi~g constant dc current into a load is 30°, the displacement factor of the rectitier is
1
(d) fa (a) 1 (b) 0.5 (c) 13 2
86. A 3-phase fully-controlled converter is feeding power into a dc load at a constant current of 150 A. The rms current through each thyristor of the converter is (a) 50 A (b) 100 A (c) 100 . "2/ 3 A (d ) 50 . .J3 A "87. In a 3-phase full converter, the ratio of average voltage to maximum line voltage is (a) 0.9549 cos a (b) 0.9549 sin a (C) 0.4775 cos ex (d ) 0.9549 (1 + cos et) In phase-controlled converters feeding RL load, th e ripple content of load current is decided by (0) load resistance alone (b) load inductance alone " (c) both R and L (d) neith er R nor L 89. The function of centr e-tapping on th e secondary in a full-w ave rectifier is to (0) step-up the voltage (b ) step-down the voltage (c ) isolate the load from groun d (d ) cause the diodes to conduct alternately.
734
Power Electronics 90. The peak inverse voltage rating of a diode in M-2 rectifier is 'X' time larger than that of a bridge rectifier yielding the same dc output voltage, where the value of 'X'
is (d) .2.0 (c) {2 (b) 1.0 (a) 0.5 91. PIV rating of a diode in M-2 rectifier is 'X' times that of a B-2. rectifier yielding the same de output voltage, where the value of 'X' is (a) 0.5 (b) 1.0 (c) {2 (d) 2.0 92. PIV rating of a diode in M-3 rectifier is 'X' times that of a 3-phase full converter yielding the .same dc output voltage, where the value of 'X' is (a) 0.5 (b) 1.0 (c) {2 (d) 2.0 93. For the same ac input voltage, PIV rating of a diode in M-3 rectifier is 'X' times that of a 3-ph ase B-6 rectifier, where the value of 'X' is (a ) 0.5 (b) 1.0 . (c) {3 (d) 2.0 94. A six-pulse thyristor rectifier bridge is connected to a balanced 50 Hz three-phase ac source . Assuming that the dc output current of the rectifier is constant, the lowest frequency harmonic component of the ac source line current is (a) 100 Hz (b) 150 Hz (c) 150 Hz . (d) 300 Hz 95. The output of a single-phase full-wave rectifier contains (a) dc plus even harmonics (b) dc plus odd harmonics (c) dc plus both odd and even harmonics (d) dc and no harmonics . 96. A single-phase half-wave converter feeds RLE load from 220 V mains. The value of E may lie between 0 and {2 Vs. The PIV stress on the SCR is (a) 220 ..-f2 V (b) 220/{2V . (c ) 2 x 22Q x vl2v (d) 2x 220 V 97. A 3-pulse converter supplies from 'an ideal transformer, a load current 10, which is tipple fr ee, to RLFD load. The trigger angle is such that the freewheeling diode FD conducts each time for the same duration as each SCR. The rms values of load current, SCR current and FD current, respectively, are (a) 1 : 1/..J6 : 1/..J6 . (b) 1 : 1/{3 : 1/{3 (c) 1 : 1/..J6 : 1/{2 (d) 1 : 116 : 1/2 98. A 2-pulseconverter supplies RLE load with R = 5Q, L = 20 H, E = 160V from a 220 V, 50 Hz supply. Load dr aws an average current 10 of 7.614 A If the value of E is . ch anged to 155 V, the new value of 10 will be (a) 7.914A (b) 8.614 A (c) 7.214A (d) 8.414A 99. In ter-group reactor in a dual converter is needed (a) to absorb the instanta."leou.....inequalities between outpu t voltages ofthe converters (b) to absorb the r egener ated energy from motor and load inertia (c ) to eliminate interconverter cir culating current (d) t o avoid abrupt transfer of current from converter to another. . 100. Athyris oris ed, thr ee phase, fully controlled converter feeds a de load that dr WE a constant current . Then th e input ae line current of the converter h as (a) an rms value equal t o the de load current ( b ) an average valu e equal to the de load curr ent (c) a peak value equ al to the de load current
'\
735
Ap pendix (d) a
fundamental frequency component, whose rms value is equal to the dc load current. [GATE, 2000]
101. A fully controlled natural commutated 3-phase bridge rectifier is operating with a firing angle a = 30°. The peak to peak voltage ripple expressed as a ratio of the peak output dc voltage at the output of the converter bridge is (a)
0.5
.
(b)
-{3 2"
(c)
( 1- 2 -{3
J
.
(d)
-{3 3 - 1 [GATE; 2003]
102. A rectifier type ac voltmeter consists of a series resistance R s , an ideal full-wave rectifier bridge and a PMMC instrument as shown in Fig. C.35. The internal resis tance of the instrument is 100 n and a
full-scale deflection is produced by a de
PMMC current of 1 rnA. The value of Rs required
/ milio mmeter to obtain full scale deflection with an ac
. voltage of 100 V (rms) applied to the
input terminals is (b) 89.93 n (a) 63.56 n (d) 141.3 kn
(c) 89.93 kn Fig. C.35
[GATE, 2003]
103. A phase-controlled half-controlled single-phase converter is shown in Fig. C.36. The control angle a = 30°. The output de voltage waveshape will be as shown in
Fig. B
t
•
t ./
Fig. C.36 (a) Fig. A (c ) Fig. C
(b) (d)
Fig. B Fig. D
[GATE', 2003}
104. Analysis of voltage waveform of a single-phase bridge converter shows that it con tains x% of 6th harm onic. The 6th harmonic content of the voltage waveform of a . . 3-phase bridge convert er would be (a) less than x%due to an incr ease in the number of pulses (b) equal t o x %, the s ame as that of the single-phase converter . (c) greater than x% du e t o changes in t he input and output voltages of the converter (d) difficult to predict as the analysis of converters is n ot governed by any generalised theory
Power Elect:r'oni~s
736
105. Fig. C. 37 shows the voltage across a power semiconductor device and the current through the device during a switching transition. Is the transition a turn ON transition or a turn OFF transition? What is the energy )oss . . ,' ' during the transition? v.l VI (a) Turn ON, 2" (tl + t 2) (b) Turn OFF, VI (tl + t 2) (c) Turn ON, VI (tl + t 2) VI (d) Turn OFF, 2 (tl + t 2)
v
[GATE, 2005}
106. Consider a phase controlled converter t shown in Fig. C.38. The thyristor is fired at an angle a in every positive half cycle of , Fig. C.37 the input voltage. If the peak value of the
instantaneous output voltage equals 230 V,
. the firing angle a is close to
(a) 45° (b) 135° (c) 90° (d) 83 .6° [GATE, 2005] 230V (rms ) (\) 107. A 3-phase full converter, fed from 3-phase, 400 V, 50Hz 50 Hz source, delivers power to load R. Each SCR is triggered sequentially. If the peak value of the instantaneous output voltage is 400 V, the firing Fig. C.38 angle of 3-phase full converter would be (a) 30° (b) 45° (c) 60° 108. A 3-phase semicon.verter, fed from 3-phase, 400 V, 50 Hz source, delivers power to load such that load current is continuous. The triggering angle for each SCR is such that FD conducts for 60°. Under these conditions, the firing angle. of each SCR is (a) 90° (b) 120° (c) 150° (d) 160° 109. A 3-phase full converter delivers power to a load R = 50 .0. The source voltage is 400 V, 50 Hz. For a firing-angle delay of 45°, the power delivered to load R is (a) 3200 W (b) 2918 W (c) 4800 W (d) 5846.4 W 110. When a line commutated converter operates in the inverter mode 1. it draws both real and reactive power from the ac supply 2. it delivers both r eal and r eactive power to the ac supply 3. it delivers real power to the ac supply 4. it draws reactive power from the ac supply.
From these, the correct statements are
(a) 1 only (b ) 2 only (c) 3 only _ (d ) 3,4 111. In a 3-phase semiconverter, fre quen cy of the ripple in th e output voltage wave may be 1. 3 times the supply frequency f for fir ing angle ct < 60~ 2. 3 f for a> 60° 3. 6f for a < 60° 4 . 6f for a > 60° From above, the correct statem ents are
. (a ) 1,3 (b) , 4 (c ) 2,3 Cd) 2,4
Appendix ~
737
. .. .
112. For the same' ac volt'age arid load impedance, read the following statements about rectifiers : 1. The average load current in a full-wave rectifier is twice that in a half-wave .' rectifier. 2. The average lo ad current in afull-wave rectifier is 1t times that in a half-wave rectifier. 3. Half-wave rectifier will have bigger sized transformer compared to full-wave rectifier . . 4. Half-wave rectifier will have a smaller . tansformer compared to a full-wave rectifier.
From these, the correct statements are
(a) 1,4 (b) 2, ,: (c) 1,3 (d) 2,3 113. For the triangular waveform shown in Fig. C. 39, the rms value of the voltage is equal to .
-------- -------------------
T
3T
T
"2
-------;-----------
t
"2 Fig. C.39
(a) ~ 1 / 6
(b) "113
1 (c ) 3
(d) "2/3
(GATE, 2005)
114. A three-phase diode bridge rectifier is fed from a 400 V, 50 Hz, three-phase ac source. If the load is purely resistive, the peak instantaneous output voltage is equal to (a) 400 V (b) 40012v (c) 400 . "2/3
(d)
~V
(GATE, 2005)
ll ij . For the current waveform shown in Fig. C. 40, the rms value of the current is equal to
I
T
-
I
f-2-+~t-~
3T
"2
2T
•
Fig. C.40 (a)
2
-J3
(b) 3 -{3
(c) 2 ,, 5
. (d)
2 ..f6
738
Power Electronics
116. M atch List-I with List II and select the correct a swer by using the codes given below: List I List II
Single phase full converter Output voltage waveform
with 50 Hz supply
A. RC (parallel) load
B. Continuous conduction mode
G. Resistance load
tlJa D. Inverter mode
4.
10
Codes: A (a)
(c)
j
'1
4 4
B 3 3
C
D
1
2
.
t (ms) . .
B
A 3 3
(b)
·30
20
4
C 2
40
D 1
2
(d) 1 4 1 117. Match List I with List II and select the correct answer by using the codes given below (Vm = maximum value of source voltage) : List I List II
Single-phase converters Output voltage and PIV .
V
A. B-2 1. ~ (1 + cos ex), PIV = V m 1t 2Vm B . Half-vvave 2. - - cos ex, PIV = 2 Vm 1t Vm C. S emicon verte r 3. 21t (1 + cos u ), PN = Vm
2
2 Vm 4. - - ' 1t
D . lYl-2 Cod es : A (a ) 4 2, (c)
2
D ,...
(b)
1
4
(d)
B 3
C
r)
.:.>
A 4 3
C03
ex, PIV=Vm
B 3 4
c
D
1 1
2
2
-
739
Appendix
118. Match List I with List II a.nd select the correct answer by using the codes given below (Vmp = maximum value of per-phase supply voltage) : List I 3-phase converters
List II .Output voltage 3Vmp . 1. cos a 1t (1 ) V 2·. 3.--f3 21t mp + cos a
A. M-3 B . semiconverter C. B-6
3·--f3V . 3. ~ mpcosa
D. M-6
4.
3--f3 -n V
cos a
mp
Codes
D D A C B (a) 4 2 4 (b) 1 1 3 2 (c) 1 1 2 4 (d) 3 3 2 4 119. Match List-I with List-II and select the correct answer by using the codes given below the lists: List I List II 50-Hz system measurements,' Waveforms
A 3
B
C
A. Voltage across an R-C. (parallel) load connected through a full-wave bridge
L . I~.~£::C
'~ .. I
: I I I
:
.
I I I
.
t(ms)
I
B. Instantaneous power consumed by a resistor.
2.
C. Output voltage of a positive clamped circuit.
3.
: : : I
D. Instantaneous power
consumed by an R -L cir cuit.
n
I
4.J\l L f\J
t (ms)
..........~ t( ms )'>
......,
Codes: A
(a)
1
(c )
120. Match List the lists :
B 2 1
C 3 2
D
4 4
(b) (d )
A 1 1
B
c
3 3
4
D 2
2
4
wiuh List II an sel ect the correct answer from the codes given below
740
Power Ele tronies
List It
Output voltage waveforms
List I Controlled rectifiers with 50 Hz supply
A. I-phase full converter with source inductance
l.VO~ 102030
Vo
2.
B. 3-phase full converter
C. 3-phase semiconverter
~ ~ ':: : ; : , : I
D. 3-phase half-wave converter.
(a)
(c)
3 4.
C
4 3
2 2
t (ms)
3.
Vo
B
I
:
10
Codes A
•
t (ms)
4.
D 1
(b)
1
(d)
20
30
~ :
I
.
t (ms)
I
~..
A 3 4
B 4 3
C
D
1 1·
2
t(ms)
2
121. A single-phase diode bridge rectifier is feeding a parallel combination of R load and a capacitor of high value. The nature of input current drawn by the rectifier from ac source is (a) a square wave of 180 0 duration (b) a square wave of 120 0 duration (c) peaky, with peaks at both positiveand negative half cycles of the input voltage (d) peaky, with peaks only at positive half cycle ofthe input voltage. 122. Match List I with List II and select the correct answer using the codes given below the lists: List I (Devices)
List II ,Pl'Operties)
A. Triac
1.
Good ~~ behaviour even at low gate currents
B. Reverse conducting thyristor 2. ' Normally provided ,vith a small continuous (ReT) negative gate pulse during off state (' Gate turn-off thyristor (G.T.O.) 3. N egat ive g ate cu rrent for reverse '-' . con duction D. Amplifying gate thyrist or 4 . No gat e pulse for r everse conduction Codes: A (a) 4 (c)
3
B
C
D
3
1 1
2 2
4
(b) (d)
A 3 4
B 4
C' 2
3
2
D 1 1
741
Appendix
ANSWERS
1. (b)
2. (a)
3. (c)
4. (a)
5. (d)
6. (b)
7. (c)
8. (c)
9. (b)
10. (b)
11. (d)
12. (a)
13. (b)
14. (a)
15. (b)
16. (d)
17. (a)
18. (d)
19. (b)
20. (c)
21. (c)
22. (b)
23. (d)
24. (a)
25. (d)
26. (d)
27. (c)
28. (a)
29. (c)
30. (b)
.31. (b)
32. (d)
33. (d)
34. (d)
35. (b)
36. (d)
37. (d)
38. (a)
39. (b)
40. (b)
41. (b)
42. (a)
43. (b)
44. (b)
45. (a)
46. (c)
47. (b)
48. (a) .
49. (a)
50. (c)
51. (c)
52. (d)
53. (c)
54. (a) .
55. (b)
56. (c)
57. (b)
58. (a)
59. (c)
60. (b)
61. (d)
62. (b)
63. (d)
64. (a)
65. (b)
66. (c)
67. (d)
68. (a)
69. (d)
70. (c)
71. (d)
72. (b)
73. (d)
74. (a)
75. (a)
76. (c)
77. (d)
78. (b)
79. (a)
. 80. (d)
81. (c)
82. (b)
83. (c)
84. (a)
85. (d)
86. (d)
87. (a)
88. (c)
89. (d)
90. (b)
91. (d)
92. (d)·
93. (b)
94. (b)
95. (a)
96. (c)
97. (c)'~~ .
98. (b)
99. (a)
100. (c)
101. (a)
102. (c). • .
103. (b)
104. (b)
105. (a)
106. (b)
107. (d)
108. (b)
109. (a) .
110. (d)
111. (c)
112. (a)
113. (a)
114. (b)
115. (d)
116. (c)
117. (b)
118. (b)
119. (d)
120. (a)
121. (d)
122. (b)
CHOPPERS 1. In dc choppers, if Ton is the on-period and f is the chopping frequency, then output voltage in terms of input voltage Vs is given by (a) Vs . Tonlf (b) Vs . flTon . (c) V/f· Ton (d) Vs . f· Ton
2. In dc choppers, the waveforms for input and output voltages are respectively (a ) discontinuous, continuous (b) both contin uous (c) both discontinuous (d) continuous , discontinu ous. 3. A chopper can be used on (a) pulse-width modulation only (b) frequency modulation only _ (c) amplitude modulation only (d) both PWM and TIl! 4. In PWM method of controlling the average output voltage in chopper, 1. on-t ime Ton is varied and chopping frequency f is k ept constant 2. Ton is k ept constant and f is varied 3. both Ton and off-time Tof{ are varied and f is k ept constant 4. T off i:,; varied and T is kept constant. F . om above, the correct st atements are (d ) 3, 4 (a ) 1, 3 (b) 1,3, 4 (c) 2, 3, 4
a
Powe r Electronics
742
5. In FM method of controlling the aver ge output voltage ina chopper, L on-time Ton is kept constant and chopping period T is varied 2. turn-off time Tott is kept constant and T is varied 3. Tanis kept constant and Tott is varied 4. Tott is kept constant and Ton is varied
From these the correct statements are
(a) 1,3,4 (b) 2,3,4 (e) 1,2,3,4 (d) 1,2,3 6. A step-down chopper is operated in the continuous conduction mode in steady state with a constant duty ratio D. If Va is the magnitude of the dc output voltage and if Vsis the magnitude of the dc input voltage, the ratioVolVs is given by (a ) D
(b) 1-D
(e)l
~D
(d) 1
~D (GATE, 2002)
7. A step-down chopper is operated in the discontinuous conduction mode in steady state with a constant duty ratio a. If tx = extinction time, Vs =de source voltage, T = chopping period and E = constant dc load voltage, then the magnitude of the average output voltage is given by tr tx 1- IE (a) aVs (b) exVs + 1- E . T T
t
I' (e) exVs + 1+~ IE (d) aVs + T_1 E T , tr 8. For type-A chopper, Vs is the source voltage, R is the load resistance and ex is the duty cycle. The average output voltage and current for this chopper are respectively (a) aVs, a · (VsIR) (b) (1- a) V s' (1- a) Vs I R (e) V/a, VslcxR (d) V/(l- a), V/(l - a)R. _9. A chopper has Vs as the source voltage, R as the ]oad resistance and a as the duty cycle. For this chopper, rms value of output voltage is (a) aVs (b) fa· Vs . (e) V/fa (d) ~. Vs' 10. For a chopper, Vs is the source volt age, R is the load resistance and a is the duty cycle. Rms and average values of thyristor currents for this chopper are . Vs _C- Vs _C- Vs _C- Vs (a) a ·
Ii' -va . Ii
_rVs
(e) -:va
(b) -va .
Ii' -va . Ii
_~ Vs
Vs
Ii' a Ii
(d) -v1 - a .
R' (1 ~ ex) V/R.
11. In dc ch oppers, per unit ripple is maximum when duty cycle a is (a ) 0.2
_
(b) 0.5
(d) 0.9. 12. In the circuit shown in Fig. C. 41, L = 5 IlH and C = 20 IlF. C i initially charged to 200 V. Mter the switch S is closed at t = 0, th e maximum value of current and the tim e at which it r eaches this value
ar e, resp ectively,
(a) 400 A, 15.707 )ls (6) 50 A, 30 IlS (e) 100 A, 62.828 ~s (d ) 400 A, 31 .414 I1-s. (c ) 0.7
I
Fig. C.4l
Appendix
743
13. A voltage commutated chopper has t h e following parameters: . V s = 200 V, Load circuit parameter : 1 n, 2 mH, 50 V Commutation circuit parameters, L = 25 IlH, C = 50 IlF Ton = 500 Ils, T =~WOO Ils
J
For a constant load current of 100 A, the effective on period and peak current through the main thyristor are respectively (a) 1000 Ils, 200A . (b) 700 IlS, 382.8 A (c) 700 Ils, 282.8 A (d) 1000 Ils, 382.8 A. . 14. For the voltage-commutated chopper of Frob. 13, the turn-off times for main and auxiliary thyristors are, respectively, . ( a ) 120 Ils, 60 Ils . (b) 100 Ils, 0.51ls (c) 120 Ils, 55 I l s ( d ) 100 Ils, 55.54Ils. 15. In the current-commutated chopper shown in Fig. C. 42, thyristor T1 is conducting a Dl load current 10 , When thyristor TA is turned on, with capacitor polarity as shown, the 10 10 capacitor current ic would flow through. + T1 (a) diode Dl because it provides an easy
D2 path.
L TA (b) · thyristor .T1 because it is already con 0 Vs C A ducting DI (c) diode D1 because thyristor T1 ·is
. L unidirectional device and therefore cur
rent ic cannot flow from cathode to anode (d) SCR T1 because diode D1 is reverse Fig. C.42 biased by voltage drop across Tl. 16. A load commutated chopper, fed from 200 V dc source, has a constant load current of 50 A. For a duty cycle of 0.4 and a chopping frequency of 2 kHz, the value of commut ating capacitor and the turn-off time for one thyristor pair are r espectively (a) 25 IlF, 100 IlS (b) 50IlF, 50lls (c ) 25 IlF, 25 Ils (d) 50 IlF, 251ls . . 17. A dc battery is charged from a constant dc source of 200 V through a chopper. The dc battery is to be charged from its internal emf of 90 to 120 V. The battery has internal resistance of 1 n. For a constant charging current of 10 A, the range of duty cycle is (a) 0.45 to 0.6 (b) 0.5 to 0.65 (c) 0.4 to 0.55 (d) 0.5 to 0.6 18. For type-A ch opper; Vs , R , 10 and a are respectively the dc source v oltage, load r esist ance, constant loadcurrent and duty cycle . F or tills chopper, av er age an d rms valu es of fr eewheeling diode currents are (a) a 10 , ..fCi . 10 (b ) (1- a)Io, ..)1 ~ a .10 (c) a · VsIR, ..fCi. VsIR (d) (1- a) 10 , W; . 10 , 19. A step-up chopper h as Vs as the source voltage and a as the duty cycle. The outpu t voltage for this chopper is given by (a) V s (1 + a) (b) / (1 - a) (c) Vs (1- a ) (d) V / (l + a.) .
Power Electronics
744
20. A dc chopper is fed from 100 V dc. Its load voltage consists of rectangular pulses of duration ~ msec in an over all cycle time of 3 msec. The average output voltage and ripple factor for this chopper are respectively (a) 25 V, 1 (b) 50 V, 1 (c) 33.33 V, 12 (d) 33.33 V, 1 21. When a series LC circuit is connected to a dc supply of V volts through a thyristor, then the peak current through thyristor is (a) V· -vLC (c) V· -vCIL
,
.
(b) VI-VCL (d) V ·-vLIC
22. For the arrangement shown in Fig. C. 43, the circuit is initially in steady state with thyristor T off. After thyristor Tis turned on, the peak thyristor current would be
c
T
20 0V
0.0; H R=100 .n
Fig. C,43 (b) 22 A (d) 42 A.
(a) 2 A (c) 40 A
23. In type-A ch opper, s our ce voltage is 100 V dc, on-period = 100 ~s, off-period = 150 Ils and load RLE consists of R = 2 n, L = 5 mH, E = 10 V. For continuous condu ction, ,a verage output voltage and average output current for this chopper are respectively: (a) 40 V, 15 A (b) 66.66 V, 28.33 A,
(c)60V,25 A (d) 40 V, 20 A.
24. Refer to the circuit in Fig. C. 44. The maximum current in the main S CR M can be (a) 200A (b) 170.7A (c) 141.4 A (d) 70.7 A.
+
M
A
l
200 V
b~-----~______~~~__~T F ig. C .44
745
Appendix
25. Refer to the circuit in Fig. C. 44 . The m aximum tLlrn-off time of the main SCR M to ensure its proper commutation, in JiS, is (a) 2n (c) 6n
'(b) 4n (d) 87t
26. Match List I with List II and give the correct answer by using the codes given belpw :
List I Chopper conr~urations
List II Output Voltage Waveforms
A. Voltage:"commutated chopper
1.
B. Load-commutated chopper
2.
~ I :0:1_.
~:======-T--------~------~.~!-----· --------~·
Vs~------------~
O~----------------~--~----~ ~---------T----------~
C. Current-commutated chopper
[
I [ T--J
D. Ideal dc chopper
Codes:
A
B
(a)
2
4
(c)
4
2
C
,
D
A
B
C
D
..L
3
(b)
2
4
3
1
1
3
(d)
2
1
4
3
27. Match List I with List II and give the correct answer by using h e odes given be10w the lists.
Power Electronics
746
List 11
(Circuit Configurations)
.List 1 (Types of choppers)
A. Type-A chopper
r-- -, I I : -I
1.
I I IL ___ I
~ .
+
B . .Type-B chopper
C.
Type~C
2.
chopper
3.
D. 'f.;pe-D chopper
4.
vs
LOA 0
.. 1 '.10
L 0 · A
0
Codes:
A
B
3 1
1 3
C
A
D
B 3 1
C 4 2
D
(b ) z 2 4 1 4 2 4 (d) 3 . . 28. A step-up chopper is fed from a 220 V dcsource to deliver a load vOltage of 660 V. · If the non-conduction time of the thyristor is 100 ~s, the required pulse width would be . . (a )
(c) ·
(a) 100 ~s (c) 220 ~s
(b) 200~s
660 ~ 29. A chopper, where voltage as well as current remain ;;'egative, is known as (a) type-A . . (b) type-B (c) type-C (d) type-D 30. A chopper, in which current remains positive butvoltage may bepositive or negative, . . is known as (a) type-A . (b ) type-B (c) type-C (d) type-D 3 1. The freewheeling diode is subjected to double the s o :;:' e voltage in the following ch opper configurations : 1. Voltage commutated chopper 2. Current commu tat ed chopper (d)
747
Ap pendix
3. Load commutated chopper 4. J one's chopper
From these, the correct statements are
. (d) 2, 3,4 (a) 1,3,4 ( b) 1,2,3 (c ) 1,2,4 , 32. The effective on period in a voltage commutated chbpper (a) increases with load current 10 as well as with the commutating capacitance C . (b) decreases with 10 as well as C (c) decreases with 10 but increases with C (d) increases with 10 but decreases with C 33. A step-down chopper operates from a dc voltage source Vs and feeds a dc motor armature with a back emf Eb. From .oscilloscope traces, it is found that the current .increases for time t r , falls to zero over time tf and remains zero for time to, every chopping cycle. Than the average dc voltage across the freewheeling diode is
in
.(a) Vs.ir/(tr+tf+to)
. tf)/(tr + t f + to) . (c) (Vs . tr + Eb . to)/(tr + t f + to)
(d)[Vs . tr + Eb (tf+ to)]I(tr + t f + to)
.:
(b) (Vs . tr + Eb
34. Fig. C. 45 shows a step-down chopper switched at 1 kHz with a duty ratio D ::: 0.5. The peak.. peak ripple in the load current is close to (a) 10 A (b ) 0.5 A (c) 0 .125 A (d) 0.25 A
[GATE, 2005]
[GATE, 2000]
so
~5.
A dc chopper . is fed from constant voltage mains. The duty ratio ex of the chopper is progressively increased while the chopper feeds RL load. The per unit current ripple Fig. C.45 would . (a) increase progressively (b) decrease progressively (c) decrease to a minimum value at ex = 0.5 and then increase (d) increase to a maximum value at ex = 0.5 and then decrease . 36. In a two-quadrant dc to dc chopper , th e load voltage is varied from positive maximum to negative maximu m by varying the time-ratio of the chopper from (a) zero to unity (b) unity to zero (c) zer o to 0.5 . (d) 0.5 to zero 37. A chopper cir cuit is oper ating on TRC principle at a fre quency of 2 kHz 0 a 22 0 V dc supply. If the load voltage is 170 V, then the con duction period of thyristor in each cycle is (a) 3. 86 ms (b) 7.72 ms (c) 0.772 ms (d ) 0.386 ms
38. A chopper is employed to charge a battery as shown in Fig. C. 46. The ch arging current is 5 A. The duty ra tio is 0.2 . The chopper outut voltage is Iso shown iIi Fig. C. 46 . The pea to peak ripple current in th e ch arging current is
Power Electronics
748
- - 60V SA L=20 mH
Ch o ppe r
12 V
I l----- 1ms ------1
t
W200/J S
Fig. C.46 (a)
0.48 A
(c) 2.4 A
(b) 1.2 A (d) 1 A
[GATE, 2003]
, 39. Fig. C. 47 shows a chopper operating from a 100 V dc input. The duty ratio of the main switch S is 0.8. The load is sufficiently inductive so that load cur
rent is ripple free. The average current
through the diode D under steady state
is
(a) 1.6 A
(c) 8.0 A
(b ) 6.4 A (d) 10.0 A
[GATE, 2005]
40. · Fig. C. 48 shows a chopper. The
device 81 is the main switching
device. 82 is the auxiliary com
m utation device. 81 is rated for
400 V, 60 A. 82 is rated for 400 V, . 30 A. The load current is 20 A. The · main device operates with a duty ratio of 0.5 . The peak current
through S 1 is
(a) lOA (c) 30A
+ 100 V
10n
D
Fig. C.47
20A
200 V
200 /-lH
FD
(b)20A (d) 40A
Fig. C,4S
{GATE, 2004]
4J... For eliminating fifth harmonic from the output voltage wave of a dc chopper, the ripple factor should be (a) 1 ( b) 2 (c) 3 (d) 4 42. In a chopper, for eliminating third harmonic from the output volt age wave, the duty cycle should be equal to (ci) 1/5 (b) 114 (c) 1/3 (d) 1/2 43. A chopper circuit , fee d from an input voltage of 20 V dc, delivers a load power of 16 vvattB. For a ch opper efficiency of 0.8, the input current is (a) 0 .64 A
( b) 0. 8 A
(c) 1 A
(d ) 1.25 A
749
Appendix
44. Type-A chopper, fed from 200 Vdc, is connected to loadR = 5Q. This chopper ope~ates with on and off periods of2 ms and 3 ms respectively. The peak value ofload current and ripple factor are (a) 16 A, 1.225 (b) 25.3 A, 1.225 (c) 24 A, 0.8165 (d) 40 A, 1.225 ANSWERS 1. (d)
2. (d)
3. (d)
4. (b)
5. (c )
6. (a)
7. (b)
8. (a)
9. (b)
10. (c)
11. (b)
12. (a)
13. (b)
14. (d)
15. (d)
16. (a)
17. (b)
18. (b)
19. (b)
20. (c)
21. (c)
22. (b)
23. (a)
24. (b)
25 . (d)
26. (a)
27. (d)
28. (b)
29. (b )
30. (d)
31. (a)
32. (c)
33. (c)
34. (c)
35. (d)
36. (b j"
37. (d )
38. (a )
39. (a )
40. (d)
41. (b)
42. (c)
43. (c )
44. (d )
INVERTERS
1\ If, for a single-phase half-bridge inverter, the amplitude of output voltage is Vs and the output power is P, then their corresponding values for a single-phase full-bridge inverter are (a) V s ' P (b) 2Vs , P (c) 2 V s ' 2P (d) 2 V s ' 4P. 2. In voltage source inverters (a) load voltage waveform vo depends on load impedance Z, whereas load current waveform io does not depend on Z (b) Both vo and io depend on Z (c) vo does not depend on Z whereas io depends on Z (d) both vo and io do not depend upon Z. 3 . A single-phase full bridge inverter can operate in load-commutation mode in case
load consists of (a) RL (c) RLC overdamped
(b) RLC underdamped ' (d) RLC critically damped. 4. A sin gle-p hase bridge inverter delivers pow er to a ser ies connected RLC load with R = 2 n, wL = 8 n. For this inver t er-load combin ation, load commutation is possible in case the magnitude of lIwC in ohms is (a) 10 (b) 8 (c) 6 (d) zero. 5. In the half-b ridge inverter of Fig, C. 49, m ain thyr ist or Tl is conducting a load current. With polarity of the capacit or voltage as shown, when auxiliary t hyristor TAl is turned on, capacitor current ic 1. would fl ow through Dl, because capacitor voltage V c forward biases diode D1 2. cann ot fl ow through Dl , bec ause volt age drop across Tl r evers e biases Dl 3. cannot flO'N through Tl , becau se thyristo is unidir ection al device
4, wo uld How through T l, such that load current minus i:-: flows fr om anode t o cathode.
750
Power Electronics
Vs
01
lc
j+
T1
i
Vs
L
C
Vs
T
-
TAl
"2
~
T LO AD
a
b
c
Fig. C.50
Fig. C.49
From these, the correct statements are (a) 1,
3
(b)
2, 4
(e)
3 only
(d)
4 only
6. For a 3-phase bridge inverter in 180° conduction mode, Fig. C. 50, the sequence of SCR conduction in the first two steps, beginning with the initiation of thyristor 1, is (a) 6, 1, 2 and 2, 3, 1 (b) 2,3, 1 and 3,4,5 (e) 3,4, 5 and 5, 6,1 (d) 5, 6, 1 and 6, 1, 2.· 7. For a 3-phase bridge inverter in 120° conduction mode, Fig. C. 50, the sequence of SCR conduction in the first two steps, beginning with the initiation of thyristor 1, is, (a) 6, 1 and 1, 2 (b) 1,2 and 2, 3 (e) 1, 6 and 5, 6 (d) 1, 3 and 3, 4. 8. In single-pulse modulation of PWM inverters, third h ar m onic can be eliminated if pulse width is equal to (a) 30 0 (b) 60° (e) 1200 (d) .150 0 9. In single-pulse modulation of PWM inverters, fifth harmonic can be eliminated if pulse w idth is equal to (a) 30° (b) 72° (e) 36° (d) 108 0 • 10. In singl e-ptlse modulation of PWM inverters, the pulse width is 120 0 • For an input voltage of 220 V dc, the r.m.s. value of output voltage is (a) 179.63 V (b) 254.04 V (e) 127.02 V (d) 185.04 V. 11. In single-pulse modulation used in PWM inverters, Vs is the input dc voltage. For· eliminatin g third harmonic, the magnitude of rms value of fundament al component of output voltage and pulse width are respectively (a ) 2~ V 120° 1t S'
(b)
-Y6 V 60° 1t S'
(c) 2:: V s'
(d )
~ Vs ' 120°.
60°
12. In multiple-pulse modulation used in PWM inverters, t he amplitudes of reference square wave and t riangular carrier wave are r espectively 1 V and 2 V. For generating 5 pulse per h alf cycle, t he pulse width sh ould be (a) 36° (b) 24° (c) 18° (d) 12°. 13. In m ultiple-pulse modulation used in PWM inverters, the amplitud e and frequency for t riangular carrier and square reference signals are respectively 4 V, 6 k Hz and 1 if 1 kHz. The number of pulses per h alf eye e and puls e width are respectively (a ) 6,90° (b) 3} 45° (c 4,60° (d 3, 40°.
75 1
Ap pendi..
14. In sinusoidal-pulse modulation used in PWM inverters, amplitude and frequency for triangular carrier and sinusoidal reference signals are respectively 5 V, 1 kHz and 1 V, 50 Hz. If zeros of the triangular carrier and reference sinusoid coincide, then the modulation index and order of significant harmonics are respectively (a) 0.2, 9 and 11 (b) 0.4,9 and 11 (c) 0.2, 17 and 19 (d) 0.2, 19 and 21. 15. Which of the following statement/statements is/are correct in connection with in verters: (a) 'lSI and CSI both require feedback diodes (b) Only CSI requires feedback diodes (c) GTOs can be used in CSI (d) Only VSI requires feedback diodes . . 16. In a CSI, if frequency of output voltage is {Hz, then frequency of voltage input to CSI 'is (a) { (b) 2{ (c) {/2 (d) 3f. 17. In sinusoidal-pulse modulation used in PWM inverters, amplitude and frequency of triangular carrier and sinusoidal reference signals are respectively 5 V, 1 kHz and 1 V, 50 Hz. If peak of the triangular carrier coincides with the zero of the reference sinusoid, then the modulation index and order of significant harmonics are (a) 0.2, 9 and 11 (b) 0.4,9 and 11 (c) 0.2, 17 and 19 (d) 0.2, 19 and 21. 18. In sinusoidal PWM, there are 'm' cycles of the triangular carrier wave in the half cycle of reference sinusoidal signal. If zero of the reference sinusoid coincides with zero/peak of the triangular carrier wave, then number of pulses generated in each half cycle are respeJ:!tively' (a) (m - l) / m (c) m / m
(b) (m - l)/(m - 1) (d) m /(m - 1).
19. In an inverter with fundamental output frequency of 50 Hz, if third harmonic is eliminated, then frequencies of other components in the output voltage wave, in Hz, wo uld be (a) 250, 350,450, high frequencies (b) 50,250,350, 450 (c) 50, 250, 350, 550 (d) 50, '100,200, 250. 20. A single-ph ase CSI has capacitor C as the load. For a constant source curr ent, the voltage a cr oss the capacitor is (a) sq are wave (b) triangular wave (c ) st ep function (d) pulsed wave.. 2 1. A single-phase f 11 bridge VSI has inductor L as the load. For a constant source voltage, the current th r ough the inductor is (a) square wave (b) triangular wave (c) sine wave (d ) pulsed wave. 22. A VSI will have better performance if its (a) load in uctance is small and source inductance is large (6) both load inductance and source inductan ce are sma 1 (c) bo th load induct n ce and source induct ance ar e lar ge (d) load indu ctCl.ncL is large and source induct ance is sm all.
752
Power Electronics
23. A series capacitor commutated inverter can operate satisfactorily if
L~ > : '
(a)
(b) LIe
~( ~
J
(e)
LIe < (
~J
(d) irrespective ofthe values of R, Land C
24. Simplest method of eliminating third harmonic from the output voltage waveform of a single-phase bridge inverter is to use (a) inverters in series (b) single-pulse modulation (c) stepped-wave inverters . (d) multiple-pulse modulation. . 25. A PWM switching scheme is used in single-phase inverters to (a) reduce the total harmonic distortion with modest filtering (b) minimise the load on the dc side (c) increase the life ofthe batteries (d) reduce low-order harmonics and increase high-order harmonics. 26. In three-phase 180°-mode bridge inverter, the lowest order harmonic in the line to neutral output voltage (fundamental frequency output = 50 Hz) is (a) 100 Hz . (b) 150 Hz (c) 200 Hz (d) 250 Hz 27. A single-phase inverter has square wave output voltage. What is the percentage of the fifth harmonic component in relation to the fundamental component? (a) 40% (b) 30% (c) 20% (d) 10% 28. A time-mar gin for series inverter ensures (a) low power loss (b) safety ofthe device (c) improved power factor (d) absence of harmonics
29 . A single-ph ase full-bridge VSI operating in square-wave mode supplies a purely inductive load. If the inverter time period is T, then the time duration for which each of the feedback diodes conduct in a cycle is (a) T (b) TI2 (c) TI4 (d) TI8
30. Consider the following statements: 1. Inherent short-circuit operation 2. Regeneration capability 3. Need for inverter grade thyristors 4. Voltage spikes across the load
Which of these features are associated with CSI ?
(a) 1, 2 and 3 (b) 2, 3 and 4 (c) 1, 3 and 4 (d) 1,2 and 4 31 . The output voltages el and e2 of two ftiU-b idge inver ters are added using output transformers . In order to eliminate the fifth harmonic from the output voltage, the phase an gle between el and e2 should be (a )
11:
"3 rad 11:
(c)- rad
11:
(b)
"4 rad
(d)
"611: rad
5 32. Full br idge inv erter is shown in Fig. C. 51. The m aximum rm s output voltage VOl at fundamen t al fr equency is (a ) 24 V (b ) 21. 61 V (c) 43.22 V (d ) 48 V
24 \1
D3
T1
24 V
D2
Fig. C. 51
•y
Appendix
753
33. In a I-phase bridge inverter, the m aximum value of fundamental component ofload current is 1. For a load which is highly inductive in natur~, the maximum value of nth harmonic component of load curren t would be ..~ . I I I
n
(a)
(b)
--r-
n"ln
(c) 2
n
(d) I
34. In a I-phase bridge inverter, the maximum value of fundamental component ofload current is I. For a load which is highly capacitive in nature, the maximum value of nth harmonic component of load current would be I I I (a) -
n
(b) - - , n"ln
(c) 2
n
(d) I
35. In a I-phase bridge inverter, the maximum value of fundamental component ofload current is I. For a load which is highly resistive in nature, the maximum value of nth harmonic component of load current would be (a)
l n
Cb) ~ I_ n {iL
(c)
.£2 n
Cd) I
36. Output voltage of a single-phase bridge inverter, fed from a fixed dc source, is varied by (a) varying the switching frequency (b)puls.e~width modulation (c) pulse amplitude modulation Cd) all of the above.
37. A 3-phase VSI supplies a purely inductive three-phase load. Upon Fourier analysis, the output voltage waveform is found to have an h-th orde~ harmonic of magnitude Ct.h times that of the fundamental component (Ct.h < 1). The load current would then have an h-th order harmonic of magnitude (a) zero (b) Ct.h times the fundam ental frequenr:y component Cc) h . Ct.h times the fundamental frequency component Cd) Ct.h1h times the furidamental frequency component. [GATE, 2000J 38. A 3-phase VSI supplies a purely capacitive 3-phase load. Upon Fourier analysis, the . output voltage waveform is found to have an h-th Drder harmonic of magnitude Ct.h . times that of the fundamental component (Ct.h < 1). The load curr·e nt would then have an h-th order harmonic of magnitude
.. (a ) zero
(b ) Cl.h times the fund ament al frequency component (c) h • Ct.h times th e fundam ental frequency component
Cd ) Cl.h1h times the fundamental frequency component.
39. A 3-phase VSI supplies a purely resistive three-phase load . Upon F ourier analysis, the output voltage waveform is found to have an h-th order harm ~ ic of magnitude Cl.h times that of the fundamental component (Cl.h < 1). The load current would ~h en have an h-th order harmonic of magnitude (a) zero (b) Ct.h times the f1.mdamental fre qu ency compon ent (c) h . Cl.h times the fundamental frequency component
Cd ) Ct.h1h times the fun damental frequency component.
Power Electronics
754
40. What is the rms value of the voltage waveform shown in Fig. C. 52 ? (a)
200 V
(b) 100 V
7t
7t
(c) 200 V
100 V
[GATE, 2002]
41. The output voltage waveform of a 3-phase square-wave in verter contains (a) only even harmonics (b) both odd and even harmonics (c) only odd harmonics · (d) only triplan harmonics
1)
+ 100 v
11
(d)
I
I 'IT
I I -IOOV ------- . . ~ TC 2n
3
e
I
T
J
4n
5n
3
3
"-
2rr
Fig. C.5-2
42. Fig. C. 53 (a) shows an inverter circuit with a de source voltage Vs. The semiconductor switches of the inverter are operated in such a manner that the pole voltages V10 and V20 are as shown in Fig. C. 53 (b). What is the rms value of the pole to pole voltage v 12 ?
vfI
IT
Vs 1),
..
e
2rr
h J2'O
'our I
Vs
j
o (a)
e'
(b) Fig. C.53 (c)
Vs'
~
\
43. An inverter has a periodic output volt age with the output waveform as shown
in Fig. C. 54. When con duction angle
Cl = 120°, t he rms funda ment al com
ponent of the output voltage is (a) 0.78 V (b) 1.10 V (c) 0.9 V (d ) 1.27 V
[GATE, 2003]
Vs
n
(d) [GATE, 2002] [Hint. Here vI2 = vIO - v2J
,..-----,---1
44. With reference to the f'utput voltage waveform given i converter will be free fr m 5th h annonic when
(a ) a = 72° (b Ci. =36 0 (c ) a =120 0
21l
- 1_____ "----_ _
~
I
Fi g. C.54
Fig. C. 54, the output of the '
(d)
a. = 150
0
[GATE , 2003]
Appendix
755
. 45. The output voltage waveform of a 3·phase square·wave inverter contains no third harmonics in 1. line voltages in 1800 mode 2. phase voltages in 120 0 mode 3. line voltages in 1200 mode 4. phase voltages in 180 0 mode From these, the correct statements are (d) 1,2,3,4 (a) 1,3,4 (b) 1,2,3 (c) 2,3,4 46. McMurray commutation is superior to parallel capacitor commutation in respect of (a) number of components . (b) overvoltage spike at the output (c) instantaneous reduction in SCR current (d) trig:ger circuit 47. A single-phase bridge inverter can be desimed by having thyristors without forced commutation circuitry if the load it is handling is (a) series combination of resistance and a large inductance (b) series combination of resistance and a large capacitance (c) series combination of resistance, capacitance and inductance with resonant fre .quency of the circuit being lower than the invert~r switching frequency (d) series combination of resistance, inductance and capacitance with resonant fre quency of the circuit being higher than the inverter switching frequency . . 48. The power delivered to a star-connected load of R .n per phase, from a 3-phase bridge inverter fed from fixed dc source, is 10 kW for 180 0 mode. For 1200 mode, the power delivered to load would be (a) 10 kW (b) 5 kW (c) 6.667 kW (d) 7.5 kW 49. Match List I with List II and select the correct answer using the codes given below the lists: List I List II A. Freewheeling diode 1. Voltage spikes in the output voltage B. Feedback diode 2. Peaks in the inverter current 3. Inductive loads of phase-controlled converters C. Curren t source inverter D. Voltage source inverter 4. Inductive loads of dc to ac inverters Codes AB c D C D A B (a) 4 2 3 1 (6) 1 2 3 4 (c) (d) 3 1 4 2 1 4 2 3 50. Contr ol of frequency and control of voltage in 3-phase inverters operating in 1200 mode or 1800 mode of conduction is
.(a ) possible only through inverter control circuit
(b ) possible through the control circuit of inverter and converter simultaneously (c) possible through inverter con trol for frequency and through convert er control for voltage (d) possible through converter control only 51. In a series r eson ant inverter (a) t he load current h as squ are waveform (b) trigger frequen cy is higher than damped resonant frequency (c) change of frequency does not alter trans er red power (d) output voltag . . depends upon damping factor of the 'load
756
Power Electronics
52. In McMurray commutation circuit, the circuit turn-off time is (a) dependent on load cur ent and independent of operating frequency (b) dependent on load current and also on load power factor (c) independent of load current and dependent on operating frequency (d) independent of load current and dependent on recovery period 53. In a 3-phase bridge inverter, the line to line voltage waveform is 1. square wave for 1800 mode 2. square wave for 1200 mode 0 3. stepped wave for 180 mode 4. stepped wave for 1200 mode
FroIl1 these, the correct statements are
(a) 1,3 (b) 2,3 (c) 1,4 (d) 2,4
54. In a 3-phase 180 0 mode bridge inverter, feeding a star-connected load with open neutral, the third harmonic component will be present in (a) voltage of each inverter phase with re~pect to the mid-point of the dc source (b) line to line output voltage (c) line currents (d) none ·ofthe above 55. A single"phase bridge inverter shown 12 V in Fig. C. 55. has an ideal trans~ormer
with primary turns equal to 10. For
obtaining a fundamental frequency
output voltage of 240 V, the number
of secondary turns in transformer
Fig. C.55 should be equal to (take 1t = 3)
(a) 12 (b) 150 (d) 150/...[2 (c) 150;f2 56. The operating freqpency of a self-oscillating inverter using a saturable core is dependent upon (a) battery voltage only (b) battery voltage and saturation flux density (c ) battery voltage, saturation flux density an d number of turns on primary winding (d) load circuit power factor
57. Match List I -#fth List II and select the correct answer using the codes given below · the lists.
List II
List I A. Phase-controlled rectifier feeding RL load with perfect smoothing B. Single-pulse, converter feeding RL load
C. A constant dc voltage fed dc to ac inverter feeding RL load D. A const an t de current-fed dc to ac inver t er feeding RL load Codes A BC D (a ) 2 3 1 (b) 4 (c) 1 (d ) 3 4 2·
1. Depends on the values of R and L
of the load 2. Depends on firing angle
3. Constant and independent of Rand L of the load 4. Depends on firing angle and also impedance angle of the load.
A 1
2
B 4 4
C
D
3 3
2 1
757
Appendix.
ANSWERS
1. (d)
2. (e)
3. (b)
4. (a)
5. (b)
S. (d)
7. (a)
S. (e)
9. (b)
10. (a)
11. (d)
12. (e)
13. (b)
14. (e)
15. (d)
16. (b)
17. (d)
IS.
19. (e)
. 20. (b)
21. (b)
22. (b)
23. (a)
24. (b)
. 25. (d) 31. (e) .
26. (d)
27. (e)
2S. (b)
29. (e)
30. (d)
32. (e)
33. (e)
34. (d)
35. (a)
36. (b)
37. (d)
3S. (e)
39. (b)
40. (d)
41. (e)
42. (b)
. 43. (a)
44. (a)
45. (d)
46. (b)
47. (d)
48. (d)
49. (b)
50. (e)
51. (d)
52. (a)
53. (e)
54. (a)
55.
56. (d)
57. (d)
(e) .
(a)
AC VOLTAGE CONTROLLERS
1. A single-phase voltage controller feeds an induction motor (A) and a heater (B) Ca) In both the loads, fundamental and harmonics are useful (b) In A only fundamental and in B only harmonics are useful (e) In A only fundamental and in B harmonics as well as fundamental are useful (d) In A only harmonics and in B only fundamental are useful.
2. A load resistance of 10 n is fed through a I-phase voltage controller from a voltage · source of 200 sin 314 t. For a firing angle delay of 90°, the power delivered to load · in kW, is Ca) 0.5 (b) 0.75 (e) 1 (d) 2 . . ·3. · A single-phase voltage controller is employed for controlling the power flow from 260 V, 50 Hz source intoa load consisting of R = 5 Q and (OL = 12 Q . The value of.maxi~ mum rms load current and the firing angle are respectively . (a) 20
A, 0°
(b)
1~~~1 A, 0°
(e)
20 A, 90°
(d)
1~~gl A, 90°.
4. A load, consisting of R = 10 nand roL = 10 n, is being fed from 230 V, 50 Hz source through a I-phase voltage cor..troller. For a firing angle deLay of 30 ~, the rIDS value of load current would be (a)23A
(b)*A
(e)
23
> :;r2A
23
Cd) < ~A.
5. In a single-phase volt age controller with RL load, ac output power ca be controlled if Ca) firing angle Ci > ~ (load phase angle) and conduction angle y =1t (b) Ci>~andY<1t (e) Ci < ~ and y = 1t (d) c( < ruld y > 1t.
6. A single-phase voltage contr oller fe e s power to a resistance of 10 n. The so rce voltage is 200 V rms . F or a firing angle of 90°, the rms value of thyristor current in ampe es is (d) 5. (6) 15 ( a ) 20 (e) 10
Power Electronics
·758
7. A single-phase voltage controller is connected to a load of resistance 10 n. and a '
supply of 200 sin 314t volts. For a firing angle of 90°, the average thyristor current
in amperes is
(a) 10 (c) 5-fiI Tt
(b) 10/Tt (d)
5:.f2.
8. A single-phase voltage controller, using two SCRs in antiparallel, is found to be
operating as a controned rectifier. This is because
(a) load is R and pulse gating used , (b) load is R and high-frequency carrier gating is used (c) load is RL and pulse gating is used (d) load is RL and continuous gating is used. 9. A single-phase ac voltage controller (or regulator) feCi from 50 Hz system supplies a
load having resistance and inductance of2.0 nand 6.36 mH respectively. The control '
range of firing angle for thisregulator is '
'(a) 0° < CJ. < 180° (b) 45° -< a< 180° ,
(c) 90° < CJ. < 180° (d) 0° < CJ. < 45°. 10. Two identical SCRs are connected back to back in series with a load. If each SCR is fired at 90°, a PMMC voltmeter across the load would read
is
.
(a) peak
,
1
voltage
(b)- x Tt
(c) zero
(d)
peak voltage
'21 x peak voltage
11. Two identical SCRs, connected back to back, feed a load R . If each SCR is fired at 90°, a MI voltmeter across the load would read
1
'
'2 x peak voltage
(a)
1
(c ) - x 1t
peak voltage '
,
(b)
peak voltage
(d)
zero
12. A purely inductive load is controlled by a single-phase ac voltage controller using back to back connected SCRs. If firing angle of each SCR is 75°, the current through two SCRs will flow for (a) 285° and 0° ' (b) 210° and 0° (c) 105° and 105° (d) 180° and 180° 13. A purely inductive load is controlled by a sin gle-phase ac voltage controller using , ba ck to back connected SCRs. If firing angle of each SCR is 100°, the curren t through' tw o SCRs will flow for ' (a) lEWo and 180° (b) 160 0 and 1600 (c ) 100° and 100° (d) 160° and 0 14. A single-phase voltage contr oller, using one SCR in antiparallel with a diode, feeds a load R and the supply is 230 V, 50 Hz. For a firing angle of 90 0 for the SCR, a PMMC voltmeter connected across the load woul d read (a) zero (b) - 51.8 V (c ) 51.8 V (d ) - 36.63 V 15. A single-ph ase voltage controller, using back to back connected an SCR and a diode, fe ds a load R from 200 V, 50 Hz source. F or a firing angle of 90° for the SCR, a M1 voltmet er across the load would read (a ) 230 V (b ) 173.2 V ' (c) - 173.2 V (d) 51.8 V
759
Appendix
16. A single-phase half wave ac voltage controller feeds a-Ioad R . For a firing angle of 180°, a PMMC voltmeter across the load would read
~ x peak voltage
(a)
(c) -
1 21t x peak voltage
(b) -
~ x peak voltage
(d) zero
17. A single-phase half-wave ac voltage controller feeds a load 180°, a MI voltmeter across the load would read 1
·
2 x peak voltage
(a)
1
(c) - 21t x peak voltage
1
·
(b)
-;t x peak voltage
(d)
peak voltage
E..
For a firing angle of
.
18. A single-phase voltage controller has input voltage of 240V, 50Hz and a load R :;:: 5 n. For three cycles on and two cyCles off, a PMMC voltmeter across the load would read (a) 160 V (b) 80 V (c) zero (d) 195.96 V 19. · A single-phase voltage controller has input voltage of 240 V, 50 Hz and a load R :;:: 5 n. For three cycles on and two cycles off, a MI voltmeter across the load would . read (a) 144 V (b) 151.79 V (c) 185.9 V (d) 96 V 20. A single-phase voltage controller has input voltage of 240V, 50Hz and a load of R :;:: 6 n. For 3 cycles on and two cycles off, the load would consume a power of (a) 2880 W (b) 5760 W (c) 3456 W (d) 11520 W ·21. In a single-phase voltage controller feeding RL load, when 1. firing angle a. < <\> (load phase angle), load voltage Vo is sinusoidal 2. a. > <\>, Va is non-sinusoidal 3. a. < <\>, Va is non-sinusoidal 4. a. :;:: <\>, Vo is sin usoidal
From these, the correct statements are
(a) 2, 3,4 (b) 1, 3, 4 (c) 1,2,4 (d) 1,4 22. In a single-phase voltage controller with RL load, (X is the firing angle, is the load phase angle and ~ is the extinction angle. For this voltage controller, output power can be controll ed if a_ <\> and l. (p - a.) :;:: 1t 2. (~- a) < 1t 3. P> 1t 4. P< 1t 5. p:;:: 1t
From these, the correct s tatements are
(a) 2,3 (b) 2, 5 (c) 2, 4 (d) 1, 5
760
Powe Electronics
23. A single-phase voltage controller, using a triac, controls t he ac output power to the · . resistive load R = 10 n. The input voltage is 230 ..f2 sin rot and firing angle of the triac is 45°. The peak power dissipation in the load is (a) 3968 W (b) 5290 W (c) 7935 W (d) 10580 W [GATE,2004] 24. A single-phase voltage controller, using a triac, controls the ac output power to the resistive load R = 10 n. The input voltage is 230..f2 sin wt and firing angle . of the triac is 120°. The peak power dissipation in the load is . (a) 3968 W (b) 5290 W (c) 7935 W (d) 10580 W 25. Output voltage wave from a converter is distorted and is found to have fundamental and third harmonic components. The distorted voltage wave differs from fundamen tal voltage as under : 1. Peak value of distorted wave (DW) > Peak valUe of fundamental wave (FW) 2. Rms va:lue of DW > Rms value of FW 3. Average value of DW > Average value of FW
. From these, the correct statements are
(a) 1,2 (b) 2,3 (c) 2 only (d) 1, 2, 3 26. Output voltage wave from a converter is distorted and is found to have fundamental and second harmonic components. The distorted voltage wave differs from fun damental voltage as u nder: 1. Peak value of distorted wave (DW) > Peak value of fund amental wave (FW) 2. Rms valu e of DW > Rms value of FW 3. Average value of DW > Average value of FW 4. Average value of DW over h alf cycle> Average value of FW over half cycle 5. Average value of DW over half cycle = Average value of FW over half cycle From above, th e correct state~ents are (a) 1,2,5 (b) 1,2 (c) 2, 3, 5 . (d) 1, 2, 4 27. When a single-phase ac voltage controller supplies power to an inductive load, control is lost if : (a) a < ~ - 7t (b) a = ~ (c) a>~-
wher e ex = firing angle, ~ = extinction angle with triggering of one of the SeRs an d
28. Integral cycle control (a) is very fast in action . . (b) does not introduce sub-harmonics in th e supply lines which are difficult to filter (c) cannot be used oninductive loads (d ) can be advised only for loads with high time constants and limited r ange control. 29. Match Lis t-I with List-II and select the correct answer using t he codes given below the lists :.
761
Appendix
List II
(Output voltage waveforms )
List I (Power circuit diagrams) Vo
(1) L
(A)
0
A
wt
0
( B)
~v,
Cl ;O~
Vo
o '"
wt
(3)
fe)
( D)
.
(2)
wt
0
~v,
Vo
;O~
(4) 'If'
27T
3'77
wt
Codes:
A
B 3 2
C
D 4
A C B D 1 (b) 2 4 1 3 (c) . (d) 4 1 2 4. 3 1 3fr. Match List I with List II and select the correct answer using the codes given below the lists: (a)
2 3
List I List II (Firing angle = a, load phase (Load Current Wa veforms) angle = <1>. Nature of load on single phase voltage controllers)
A. R load
1.
~~o~ ."
I
I
I
I
io B. RL load, Ct. >
2.
i
2"
~
L wt
I
0
wt
Power Electronics
762
C.RLload,a<¢
3.
D. L load, a > ~ Codes: A (a) 3 (c) 1
B 1
3
4.
C 2 2
D 4 4
..
i.o~
i ' Ie
I
o~w L
.
..
n~~2~~-----w-t-~
!~I
:1·
ch ~~ ,P~/2 31T/2~ A 3 3
(b) (d)
TT
B 1 2
C
D
4
2
1
4
ANSWERS 1. (e)
2. (e)
3. (a)
4. (b)
5. (b)
6. (c)
7. (b)
8. (c)
9. (b)
10. (c)
11. (a)
12. (d)
13. (b)
14. (b)
15. (b)
~6.
(b)
17. (a)
18. (c)
. 19. (c)
20. (b)
21. (c)
22. (a)
23. (d)
24. (c)
25. (c)
26 . (a)
27. (a)
28. (d)
29. (b)
30. (a)
CYCLOCONVERTERS 1. A cycloconverter is a frequency converter from 1. higher to lower frequency with one-stage conversion 2. higher to lower frequency with two-stage conversion 3. lower to higher frequency with one-stage conversion 4. ac at one frequency to dc and then dc to ac at a different frequency From these, the correct statements are (a) 2,4 (b) 1 only (c) 2,3 (d) 1,3 2. The cycloconverters (CCs) require natural or for ced commutation as under: (a ) natural commutation in both step-up and step-down CCs (b) forced commut b on in both step-up and step-down CCs (c) forced commutation in step-up CCs (d ) forced c mmut ation in step-down ces. 3. Consider t he following stat ements regarding cycloconverters : 1. In I-phase to I-phase c e , firing angle may be varied 2. In 3-ph ase to I -phase c e, firing angle may be kept constant 3. In I -phase to I-ph ase ee, firing angle may be kept constant 4. In 3-phase to I -phas e e e , firing anlge may be varied 5. In 3-phase to i-phas e e c, firin g angle must be varied. Fr om th ese, th e correct statem ents are (a) 2,4, 5
(b) 1, 3,5
(c) 2,3, 5
(d) 2, 3 4
wt
763
Appendix
4. Three-phase to three-phase cycloconverters employing 18 SCRs and 36 SCRshave the same voltage and current ratings for their component thyristors. The ratio of VA rating of 36-SCR device to that of 18-StR device is (a) 1/2 (b) 1 (c) 2
(d) 4.
5. Three-phase to 3-phase cycloconverters employing 18 SCRs and 36 SCRs have the same voltage and current ratings for their component thyristors. The ratio of power output from 36-SCR converter to that outpatted by 18-SCR converter is
(a) 4 (b) 2 (c) 1 (d) 1/2. 6. The number of thyristors required for single-phase to single-phase cycloconverter of the mid-point type and for three phase to three~phase 3-pulse type cycloconverter are respectively (a) 4,6 (b) 8, 18 (c) 4, 18 (d) 4,36. 7. A 3-phase to single':'phase conversion device employs a 6-pulse bridge cycloconverter. For an input voltage of 200 V per phase, the fundamental rms value of output voltage is (a) 60017t V
(c) -3001n V
(b) 300~/n V (d) 600~/n V.
8. A three-phase to single-phase cycloconverterconsists of positive and negative grQUp of converters. In this device on e of the two component converters would operate as a 1. rectifier if the output voltage Vo and output current 10 have the same polarity 2. inverter if Vo and 10 have the same polarity 3. rectifier if Vo and 10 are of opposite polarity 4. inverter if Vo and 10 are of opposite polarity. From above , the correct statements are
- (a) 1,4 (b) 2j 3
(c) 3,4 (d) 1,2 9. A 3-phase to 3-phase cycloconverter requires 1. 18 SCRs for 3-pulse device 2.18 SCRs for 6-pulse device 3. 36 SCRs for 3-pulse device _ 4. 36 SCRs for 6-pulse device.
From these, the correct statements are
(a) 1, 3 (b) 2,3 (c) 2, 4 (d) 1,4
10. Which of the following statem ents are correct for cycloconverters ? 1. Step-down cycloconverter (cc) works on n atural commutation 2. Step-up cc requires forc ed commutation 3. Load commutated cc works on line commutation 4. Load com mut at ed cc requires a generated emf in the load circuit. Fr om above! the correct statements are (a) 1,2 (b) 1,2, 4 (c) 2, 3,4 (d ) 1,2, 3
Power Electronics
764
11. Match List I with List II and select the correct answer using the codes given below the lists:
List I (Power electronic controller)
List II
(Applications)
A. Con trolled rectifier
1. Aircraft supplies
B. .Chopper
2. Electric car
C. ·.Cycloconverter
3. Induction heating
. D. Inverter
C.odes:
A (a) 4 (c) 4
4. Rolling mill drive.
B 2 2
C 3 1
D · 1 3
(b)
(d)
A 2 4
B .4 1 .
C 1 2
D
3 3
;
12. Match List I with List II and select the correct answer using the codes given below the lists:
List II
.(Applications)
·List I (Power electronic controller) A. Controlled rectifier
1. High-power ac drive
B. Voltage con troller
2 . Solar cells
C. Cycloconverter
3. . Ceiling fan drive ·
D. Inverter
4 . Magnet power supply.
Codes: A
(a) (c)
.
4 3
B 3
C 1
4
1
D 2 2
(b) (d)
A 4 4
B 3 1
C 2 3
D 1 2
.13. Match List I with List II and select the correct answer using the codes given below the lists : . . List I (Power electronic controller)
.
List II
(Applications)
A. Inverter
1. Fork-lift truck
B. Controlled rectifier
2. illumination control
C. Voltage controller
3. Uninterruptible power supply
D. Chopper
4. Hydrogen production.
Codes : A (a) 4 (c) 3
B 3 4
C 2 1
D 1
(b )
A 3
2
(d)
3
B 4 2
C 2 4
D
1
1
765
Appendix
14. Match List I with List II and select the correct answer using the codes given below the lists:
List I
(Types of cycloconverters)
A. - I-phase to I-phase with continuous
conduction
B. I-phase to I-phase with
discontinuous conduction
C. Step-up device D. 3-phase to I-phase device List II
(Output Voltage Waveforms)
- 1. ,
,
wt
2. wt
wt
..
~
wt
Codes: A
B
3
4
C 2
D 1 2
B C D 4 3 2 1 (c) 3 1 4 (d ) 3 4 1 2 15. In a 3-phase t o l-phase cycloconverter employin g 3-pulse positive and negative group conv erters, if the input voltage is 200 V per phase, the fundamental rms value of output voltage would be (a)
(a) 600 1t
V
(b )
30C..J3 V
A
(b)
(c ) 30013 V 1t
(d )
300 V 1t
166
Power Elertronic§
16. In a single-phase cycloconverter arrange ment shown in Fig . . C.56, the positive direction of currents ipl, ip2; inl, in2 is as indicated. The turns ,ratio from primary to each secondary is unity. The source current is is given by (a)'i s =ipl +
ip2
(b) i s = ip2 -
ipl -
+
i nl
+
in2
(c) is = i pl - ip2 -
+ in2 inl + in2
+ ip2 -
inl - in2
(d) is = i pl
i nl
Fig. C.56
ANSWERS 1; (d)
2. (c)
3. (b)
4. (c)
5. (a)
6. (c)
7. (d)
8. (a)
9. (d)
10. (b)
11. (c)
12. (a)
13. (b)
14. (d)
15. (c)
16. (c)
.. SOME APPLICATIONS 1. SMPSs are superior to linear power supplies in respect of (a) size and efficiency (c) regulation and noise
(b) (d)
efficiency and regulation noise and cost.
2. Consider the following statements: Switched mode power supplies are preferred over the continuous types, because these arc 1. suitable for use in both ac and dc 2. more efficient 4. suitable for high power circuits 3. suitable for low power circuits Of these statements, the correct is (a) 1 and 2 (b) 1 and 3 (c) 2 and 3 (d) 2 and 4 3. Bulk power transmission over long HVDC lines are preferred. on account of (a) low cost ofINDC terminals (b) no hannonic problems (c) minimum line power losses (d) simple protection · 4. Resonant mode power supplies in comparison to square mode ones · (a) have sm aller component count (b) have negligible power loss (c) do n ot cause overvoltages (d ) slower in control action 5. Resonant converters ar~ basically used to (a) gener ate large peak voltages (b) reduce t he switching losses · (c) eliminate harmonics (d) convert a square wave int o a sine wave. 6 . HVD C transmission is preferred to EHV-AC b cause (a) HVDC terminal equipm ent are expensive (b) VAr compensation is not required for I-IVDC systems (c) system stability can be imp rove d Cd) h armom problem is avoided.
767
Appendix
7. In a 3-phase converter used in HYDC transmission, the three anodes conduct se quentially. Due to overlap caused by the circuit inductance&, ·two anodes conduct .simultaneously during the overlap period. The output voltage waveform during this period is the . . (a) voltage of the 1st anode, because the 2nd anode has not completely taken over (b) mean of the tWQ anode voltages, as they conduc~ together (c) voltage of the 2nd anode, because the voltage of this anode is greater than that of the 1st . (d) sum of the 1st and the 2nd anode ~oltages, because both the anode are conduct ing. 8. Which one of the following statements in respect of HVDC transmission line is not correct? (a) The power transmission capability of bipolar line is almost the same as that of single circuit ac line. . (b) HVDC link line can operate between two ac systems whose frequencies need not be equal. (c) There is no distance limitation for HVDC transmission by UG cable (d ) Corona loss is much higher in HVDC transmissienlIne.
ANSWERS 1. (a)
2. (c)
7. (b)
8. (d)
3. (c)
4. (b)
5. (b)
6. (b)
ELECTRIC DJUVES 1. A separately-excited dc motor is required to be controlled from a 3-phase source for operation in the first quadrant only. The most preferred converter would be (a) fully-controlled converter (b) fully-controlled converter with freewheeling diode (c) half-controlled converter (d) sequential control of two series connected fully-controlled converters.
2. A sep arately~excited dc m otor, when fed from I-phase full converter with firi g angle (x , r uns at a speed of N rpm. When this motor is fed from I-phase semiconverter but with the same firing angle as for full-converter, the motor speed is foun d to be 2N rpm . Th e value of firing angle is . (a) 70.528° (b ) 75.572° (c) 70° (d) 69.88° 3. A separately-excited dc motor , when fed from I-phase full converter "'lith fir ing an gle 60° r uns at 1000 rpm. If t his motor is connected to I-phase semiconverter with the sam e firing angle of 60 the m otor would n ow run at (a) 200 0 rpm {b) 1500 rpm (c) 1450 rpm (d) 1000 rpm 0
,
4. A si.ngle-phase half-wave converter wit!-t freewheeling diode, drives a sep aratel "-ex cit ed dc m otor at 90 0 rpm with firing angle 60°. vvnen t his motor is fe d fro m I-phase full con verter with a = 60°, the m otor speed wou ld be (a) 600 rpm (b ) 900 rpm (c ) 1200 rp m (d ) 1800 rpm
768
Power Electronics
5. A single-phase half-wave converter with freewheeling diode, drives a separately-ex cited de motor at 900 rpm with firing angle 60°. When this motor is fed from I-phase semiconverter with ex = 60", the motor speed would be (a) 1800 rpm (b) 1200 rpm (c) 900 rpm (d) 1500 rPm 6. A separately-excited dc motor, when fed from I-phase full converter, runs at a speed of 1200 rpm. ,L oad current remains continuous. If one of thef our seRs gets open circuited, the' motor speed will reduce to (a) 900 rpm (b) 800 rpm (c) 600 rpm (d) 400 rpm 7. A I-phase semiconverter delivers power to a separately-excited dc motor. Armature current is ripple free at 20 A. For a firing angle delay of 45°, the average and rms values of freewheeling diode current would respectively be (a) lOA, 5A (b) 5A, lOA (c) l5A, 17.32 A (d) 17.32 A, l5A 8. A single-phase half-controlled rectifier is driving a separately-excited dc motor. The de motor has back emf constant of 0.5 V/rpm. The armature current is 5 A without any ripple. The armature resistance is 2.0. The converter is working from a 230 V, I-phase ac source with a firing angle of 30°. Under this operating condition, the speed of the motor will be (d) ·356 rpm . (a) 339 rpm (b) 359 rpm (c) 366 rpm .. I[GATE, 20041 . 9. A 3-phase semiconverter feeds the armature of a separately-excited de motor, sup plying a non-zero torque. For steady-state operation, the motor armature current is found to drop to zero at certain instances of time. At such instances, the voltage assumes a value that is (a) equal to the instantaneous value of the acphase voltage (b) equal to the instantaneous value of the motor back emf (c) arbitrary (d) zero [GATE, 20001
10. A 3-phase bridge inverter is used for controlling the speed of a squirrel cage induction
-motor. If frequency of supply voltage is decreased with
1. constant supply voltage VI, starting torque Test decreases 2. constant VI , Test increases 3. constant VI/fl, Test increases 4. constant VI/fl, maximum torque Te .~ must remain constant
· 5. constant V I/fl, T est decreases
. 6. constant VI/fl and at very low frequencies, T e .m may decrease.
From these, the correct statements are (a) 1, 4, 5 (b) 2,3,4,6 (c) 1,3,6 (d j 2,3,6 . 11. For an ac volt~e controller fed induction motor drive negotiating a load whose t orque requirement does n ot vary with speed (a ) the fu ndam ental component of current drawn fr om the supply decr eases as speed is reduced (b) th e fundamental component of curren t drawn from the supply increases as speed is r edu ced (c ) the fundamental component of current drawn from th e supply is independent of speed (d ) none 0 the above
769
Appendix
12. A delta-connected induction motor being fed by a 3-phase ac to dc inverter and operated in constant V If control mode r equires during starting a (a) star-delta starter · (b ) DOL starter (c ) auto-transformer starter (d) n one ofthe above 13. An ac induction motor is used for speed control application. It is driven from an inverter with a constant Vlf control. The motor name-plate details are as follows: V : 415V Ph: 3 f: 50 Hz N : 2850 rpm The motor is run with the inverter output frequency set at 40 Hz and with half the rated slip. The running speed of the motor is (a) 2400 rpm (b) 2280 rpm (c) 2340 rpm (d) 2790 rpm [GATE, 2003] 14. A variable speed drive rated for 1500 rpm, 40 Nm is reversing under no load. Figure C. 57 shows the reversing torque and the speed during the transient. The moment of inertia of the drive is ( a ) 0.048 kgm2 (b) 0.064 kgm2 (c) 0.096 kgm2 (d) 0.128 kgm2 [GATE, 2004] 15. Slip-power control schemes provide a range of speed control of a 3-phase induction motor. The range is (a ) 0 toNs (c) 0 to 2Ns
I
1_ 20N~
:
Speed
!
t l .t--- O. 5s ---+oj, 500 rpm : .-'---'- :
'.
t - 1500 r pm
(b) - Ns toNs (d) - 2Ns to 2Ns ·
Fig. C.57
where Ns is the synchronous speed.
ANSWERS 1. (c)
2. (a)
3. (b)
4. (c)
5. (a)
6. (c)
7. (b)
8. (c)
9. (b)
10. (d)
11. (c)
12. (b)
13. (c)
14. (a )
15. (c)
• •
•
•
POWER FACTOR IMPROVEMENT ' ....
:..
',' 1
.
·
...
.
.
1. The most acc- at e and versatile method of achie ving r eactive power compensation , is by usin g (a) swit ch ed capacitors (b) fixed cap acitor with controlled r eactor (c) saturable r eactor with capacitor bank (d) saturable reactor with controlled react or
2. Thyris t or controlled compensa tor is usu ally designed to oper ate at (a ) a slightly lagging pf (b ) a sligh tly leadin g p f (c ) u ity pf (d) zero pf eading
Power Electronics
770
3. Read the fonowing statements regarding thyrist or controlled react or : 1. It takes maximum reactive power at firing angle a.::;: 90° 2. It delivers maximum reactive power at a = 180° 3. It de ivers no reactive power at.f.!,';" 0° 4. It takes maximum reactive power at a. = 45°
Fr om these, the correct statements are
(a) 1 only (b) 2,4 (c) 1,4 (d) 1,3 4. The use of static VAI compensator in an ac system 1. improves t he supply power factor
.2. reduces the source current
3. improves the load voltage profile 4. reduces the load reactive power
From these, th e correct statements are
(d) 1,2,4 (a) 1,3 (b) 1,2,3 (c) 1,2,3,4 5. The effective inductance offered by a thyristor controlled inductor is 20 mHo The effective load inductance, seen by the source, would be 1. 20 mH for firing angle a::;: 60° 2. 20 mH for a. = 90° 3. less than 20 mH for a. = 120° 4 . more than 20 mH.for a. = 120° 5. less than 20 mR for a = 60°
From these, the correct statements are
(d) 1,2,4 (a) 1,2,3 (b) 2,4,5 (c) 1, 2, 4,5
ANSWERS L (b )
2. (a )
3. (c)
4. (b)
5. (d)
Appen'd ix : D ·
References
1. SCR Manual, 5th Edition, N.Y., General Electric Company, 1972. . 2. F. Csaki et. aI., 'Power Electronics', Budapest: Akademiai Kiado, 1975. 3. F.E. Gentry et. aI., 'Semiconductor Controlled Rectifiers', Prentice-HaH ofIndia New Delhi, 1964. . .
'/
"
,
. "
.
.
4. B.K. Bose, 'Power Electronics and AC Drives', Prentice-Hall, Englewood Cliffs, New Jersey 07632, 1986.
5~~P.C. Sen, 'Thyristorised DC Drive', New York: Wiley Intel-science, 1981.
6. J .M.D. Murphy and F.G. Turnbull 'Power Electronic Control of AC Motors', Per gamon Press, Oxford, 1988.. 7. B.K. Bose, 'Evaiuation of Modern Power Semiconductor Devices and Future Trends of Converters', IEEE Trans. Industry Applications, vol. 28. No. 2.pp. 403 - 413, March/April, 1992. 8. NED MOHAN et. al. , 'Power Electronics', J6~ Wiley andSons, 1989. 9. M.H. Rashid, 'Power Electronics', Prentice-Hall of India, New' Delhi. 1993. 10. M. Chilikan, 'Electric Drive', Mir Publishers, Moscow, 1970. 11. G.K. Dubey and C.R. Kasarabada, "Power Electronics and Drives", IETE Book Series, Vol. 1, TM HILL P.C. Ltd., New Delhi - 1993. 12. B.K. Bose, "Energy, environment and advances in Power Electronics", IEEE Tram~. on P.E. Vol 15, No.4, July 2000.
INDEX
AC drives, 620
AC link cnopper,34 7
AC voltage conti-oilers, 504
. harmonics, 513
integral cycle control, 508
phase ·control, 504 .
power fador, 515
sequence control, 523
single-phase, 511, 517
multistage, 527
two-stage, 523
sinusoidal, 527
types, 505
Angle, conduction, 251
extinction, 251
firing, 250
Angle of overlap, 313
Anode, 121
Applications,
GTO,193
!GBT, 35
PMOSFET, 30
ASCR,188
Basic performance equations,
dc motors, 585
induction motors, 621, 622
synchronous motors, 653, 654
Breakover voltage, 123
By-pass diode, 253
.\
Cathode, 121
Choppers,
control strategies, 349
current limit control, 351
time ratio control, 349
extinction time , 367 .
forced commu tion, 377
Fourier analysis, 358
load commut ation, 378
multiphase, 399
princi ple of operation, 348
steady-state analysis, 361
ste ady-state ripple, 355
step-up, 351
types, 355
Chopper drives , 61 0
four-quadran , 619
motoring c ntrol, ~ 11
regenerative braking control,
616
two-quadrant, 618
Commutating diode, 253
Commutation angle, 313
Commutation techniques, 228
commutation,
class-A, 228
class-B, 230
class-C, 233
class-D, 235
class-E, 237
class-F, 238
load,228
Cosine firing scheme, 217
Crest factor, 77
Current-commutated chopper, 388
design considerations, 392
commutation interval, 393
Current distortion factor, 76
Current ratings,
average on-state, 144
rms, 146
surge, 148
12t, 148
Current ripple factor, 78
Current source inverters, 468
single-phase, 468
ASCI,474
Cycloconverters, 532
load commutated, 548
output voltage, 543
principle of o~ration, 533
single-phase to single-phase,
step-down, 534
step-up, 533
three-phase half-wave
3- to 1-phase, 538
3- to 3-phase, 541
DC drives, 584
single-phase, 587
dual converter, 597
full converte , 592
half-wave converter, 588
semiconverter, 590 ·
three-phase, 597
dual converter, 610
full converter, 602
half-wave converter, 598
semiconvertar, 599
Der ating fact or, 172
Diac 185
Diode ircuits, ~2
773
Diode rectifier,
single-phase half-wave, 58
single-phase full-wave, 6'7
mid-point, 67
bridge, 69
Discontinuous load current,
single-phase, 275
full converter, 275
semiconverter, .276
3-phasesemiconverter, 300
dildt, protection, 153
rating, 149
Displacement factor, 76
Dual converters, 324
ideal, 325
with circulating current, 327
waveforms, 328
without circulating current, 326
dvldt, protection, 153
rating, 143
turn-on, 154
triggering, 125
Dynamic equalizing circuit, 175
Dynamic pf control, 676
Dynamic resistance, 179
Dynamic VArcompensation, 678
Effect of source inductance, 313
1~phase full converter, 314
3-phase full converter, 317
Eiectric drive, concept, 583
Electronic crowbar protection, 160
Equalizing circuit
. - dynamic, 175
static, 173
External overvoltages, 158
Extinction an gle, 251
Feedback diodes, 416
Filters, 106
capacitor type, 107
inductor type, 111
L-C type, 11 3
Finger voltage, 144
Firing angle, 249, 250
Firing circ uits, 195
R and RC cirl:Uits. 196, 198
UJ T triggering t::ircuit, 203
ramp-and- edestal, 206
Power Electronics
774 synchronized, 205
Form factor, 78
Fo:ward blocking mode. 123
Four-quadrant chopper, 359
Four-quadrant converter, 324
Freewheeling diodes, 55
Fully-controlled converter, 265
Full-wave controlled converters,
single-phase. 263
full converter, 265
mid-point converter. 263
semi converter, 268
Gate characteristics, 132
Gate pulse amplifiers, 157
Gate trigger circuit, 133
Gate triggering, 124
Gating circuits,
I-phase converters, 156
pulse-train. 158
GTO, 188
switching performance, 191
Half-controlled converter. 265
. Hal(wave thyristor circuit.
R load. 249
RL load , 251
freewheeling diode, 253
RLE load , 256
Harmonic factor, 76
High-voltag ~ dc transmission, 559
Holding current, 125
IGBT,31
Induction motOl' drives. 621
speed control, 624
[ 2t rating, 148
Intelligent module, 7
Intrinsic stand-off r atio, 201
Inverters , 414
single-phase.
full-bridge. 416
half-bridge. 415
har monic r eduction. 464
current source, 468
force-com mutated . 430
good, 497
parallel. 492
ser ies. 4 2
three-p hase, 442
voltage control, 45 2
Junction capacitance, 126
Junction temperature, 166
Latching current, 125 Light-activated SCR, 184 Line commutated inverter, 267 Line commutation, 238 Load commutated, chopper, 39.6
cycloconverter, 548
inverter, 419
McMurray-bedford inverter, 437
Methods of turning-on SCRs, 123
Modified McMurray inverter, 431
half-bridge, 431
full-bridge, 436 .
MOS-controlled thyristor, 37
Multiphase choppers, 399
Non-circulating current dual
converter, 326
Overcurrent protection. 159
Overlap angle. 313
Overvoltage protection, 157
Parallel inverters. 492
Parallel operation of SCR.". 178
Peak point, 202
P erformance parameters, 75
diode rectifiers,
single-phase. 82
three-phase, 104
controlle two-pulse
1-ph s _ B- 2, 278
l -p se semiconverter, 280
3-phase full converters, 309
p-n jun ction. 9
barrier potential , 11
depleti on . 10
Power diodes.
basic structure. 11
characteristics, 12
types, 14
Power-electronic
modules , 7
systems , 3
types of converters, 6
. Power electronics .
concept. 1
applications, 1. 2
Power-factor
improvement, 675
effect of poor pf. 675
Power MOSFET, 26
Power semiconductor device!!. 4
Power transistors, 15
Principle of phase control, 249
Protection of seRs; 153
Pulse transformers, 208
Pulse-width modulation.
chopper, 349
inverter, 454
PUT,182
PWM inverters. 454
modulation,
single-pulse, 455
multiple-pulse, 456
sinusoidal-pulse, 459
Ramp triggering, 205
Ramp and pedestal triggering , 206
ReT,188
Reactive power compensation, 676
. capacitor banks, 677
. synchronous eondensers. 678
Realization of PWM
full-bridge inverter. 461
half-bridge inverter. 462
Rectification ratio, 78
Relaxation oscillator. 203
Resonant converters. 572
zes , 572
zvs, 578
Reverse blocking mode . 122
Reverse recovery time . 131
Rise time. 129
SCRs, .
p rallel operation. 178
series operat" on. 172
SCS, 183
Single-phase fu Il-w 3ve
bridge can erters. 265
fu ll conve rters . 265
semiconverters, 268
Si ngle-phase half-wave circuit .
.R load. 249
Index RL load, 251
RLE load, 256
RL load andFD, 253
I
I
y,,
I
I
Single-phase inverters,
steady-state analysis, 417
Fourier analysis, 424
PWM,
multiple-pulse, 456
single-pulse, 455
sinusoidal-pulse, 459
reduction of harmonics, 464
voltage control, 452
::iingle-t'hase semi converters,
asymmetrical, 284
symmetrical, 283
SIT, 36
SITH,193
Slip-power recovery schemes, 645
Smart power, 7
Snubber circuits, 154
Solid state relays, 570
Static circuit breakers, 568
Static Kramer drive, 645
Static rotor~resistance control, 639
Static Scherbius drive, 651
Static switches, 564
dc switches, 567
design of, 567
I-phase ac, 46 5
Static VAr compensators, 680
Step-up
choppers, 351
cycioconverters, 533
Surge current, 148
SUS, 183
Switched-mode power supply, 551
SynChronous-motor drives, 652
cylindrical-rotor motors, 653
permanent magne. motors, 6Q9_
rductance motors, 658
salient-pole motors, 656
Temper- ture,
case, junction, 166
Therma l, resistance, 165
equivalent circuit, 166
Three-phase bridge inve rters, 442
180 0 mode, 442 .
120" mode, 447
Fo urier analysis, 446, 450
Three-phase diode rectifiers,
775 bridge, 94, 97
Triac, 185.
full~wave, 97
firing circuit, 211
half-wave, 85
. Types ofcommutation,
, M-6, 89
complementary, 233
M\lltiphase, 92
external pulse, 237
twelve-pulse, 101
forced, 239
TUF; 79
impulse, 235
Three-phase thyristor converters,
line, 238
-full-converters, 294
load,228
M-3,286
resonant-pulse, 230
Multiphase controlled, 308
self, 229
semiconverters, 298
Thyristor-controlled,
compensators, 680
inductors, 678
reactors, 678
Uncontrolled converter, 265
Thyristors,
Unijunction transistor, 200
characteristics, improved, 163
Uninterruptible power supplies,
commutation techniques, 228
557
dynamic characteristics, 127 .
UJ~(L<-ctltator triggering, 2~ _ .
turn-ofT, 130
turn-on, 128
firing circuits, 195
gate characteristics, 132
heating, cooling, mounting, 165
Valley point, 202
mounting techniques, 168
Vernier winding, 527
other members, 182
Voltage clamping device, 158
parallel operation, 178
Voltage-commutated chopper, 379
protection, 153
design considerations, 382
crowbar, 160
gate, 161
Voltage control,
overcurrent, 159
I-phase inverters, 452
overvoltage, 157
Voltage controller, ac, 505
pulse triggering, 134
Voltage reduction factor, 545
ratings, 142
Voltage ripple factor, 78
current, 144
Voltage source inverters, 415
surge current, 148
singlt~-phase, 415
voltage, 142
Fourier analysis, 424
series operation, 172
operating principle, 415
static characteristics, 122
. voltage control, 452
switching characteri tics, 127
three-phase, 442
turn-ofT, 130
PWM,454 ,
turn-on, 128
terminal characteristics, 121 ·
triggering,
forward voltage, 124, 141
du /dt, 125,141
Zener diode, 73
gate, 124, 141
Zero-current switching resonant,
light, 126, 142
converters, 572
temperature , 125, 141 .
L-type, 572
turn-on methods , 123
M-type, 575
two- tran"s istor model, 139
Total harmo nic dist ortion (THD) , 76 Zero-voltage switching resonant
converters, 578
Transformer tap cha nge rs, 523
