Final Exam Preparations - Power Electronics

Tutorial 1 – Power, RMS Values, Distortion Don’t use more than 3 significant digits – 1.23 A, 0. 456 W, 78.9 V P.1.1. The voltage and current for a passive device are periodic with T=200ms (Fig.1) ⎧20V ; 0 < t < 140ms ⎫ v(t ) = ⎨ ⎬; ⎩0V ; 140 < t < 200ms ⎭ ⎧20 A; 0 < t < 60 ms ⎫ ⎪ ⎪ i (t ) = ⎨− 15 A; 60 < t < 100ms ⎬. ⎪12 A; 100 < t < 200ms ⎪ ⎩ ⎭ v 20

t

0

140 ms

i

200 ms

20 12

t

0

60ms

100 ms

200 ms

− 15

Fig.1. Voltage and current profiles Determine (1) the instantaneous power, (2) the average power, (3) the energy absorbed in each period. Solution. 1. Instantaneous power ⎧400W ; 0 < t < 60 ms ⎫ ⎪− 300W ; 60 < t < 100ms ⎪ ⎪ ⎪ p (t ) = ⎨ ⎬ 240 W ; 100 < t < 140 ms ⎪ ⎪ ⎪⎩0W ; 140 < t < 200ms ⎪⎭

(1)

2. The average power

PAV =

400 ⋅ 60 − 300 ⋅ 40 + 240 ⋅ 40 = 108W . (2) 200

3. The energy absorbed in each period

W = PAV T = 108W ⋅ 0.2s = 21.6 J .

(3)

P.1.2. Find the average power absorbed by a 48 Vdc voltage source given that the current into the positive terminal is given in P. 1.1. Solution. 1. The average current

I AV =

20 ⋅ 60 − 15 ⋅ 40 + 12 ⋅100 = 9 A. 200

(4)

2. The average power (positive because absorbed by the source)

PAV = VDC ⋅ I AV = 48V ⋅ 9 = 432W .

(5)

P.1.3. Find the RMS values of the voltage and current waveforms given in P. 1.1. Solution. 1. Voltage RMS

VRMS =

20 2 ⋅140 = 16.7V . 200

(6)

20 2 ⋅ 60 + 15 2 ⋅ 40 + 12 2 ⋅100 = 15.4 A . 200

(7)

2. Current RMS

I RMS =

P.1.4. The voltage and current for a passive device are given by v(t ) = 4 + 7COS (ωt + 45 o ) − 5COS (2ωt + 20 o ); i(t ) = 3 + 2COS (ωt ) − COS (2ωt − 40 o ).

Find (1) the RMS value of voltage, (2) the RMS value of current, (3) the power consumed by the device. Solution. 1. Voltage RMS

VRMS = 42 + 0.5 ⋅ (72 + 52 ) = 7.28V .

2. Current RMS

(8)

I RMS = 32 + 0.5 ⋅ (22 + 12 ) = 3.39 A .

(9)

3. Average power

P = 4 ⋅ 3 + 0.5 ⋅ 7 ⋅ 2 ⋅ COS (45o ) + 0.5 ⋅ 5 ⋅ 1 ⋅ COS (60 o ) = 18.2W . (10)

P.1.5. A sinusoidal voltage source produces a non-linear load current v(t ) = 170COS (314t ); i(t ) = 20COS (314t + 30 o ) + 12COS (628t + 45 o ) + 6COS (1256t + 20 o ).

Find (1) the power consumed by the load, (2) RMS current value, (3) the distortion factor (DF) of the load current, (4) the power factor (PF), (5) the THD of the load current. Solution. 1. Power

P = 0.5 ⋅ 170 ⋅ 20 ⋅ COS (30 o ) = 1470W

(11)

I RMS = 0.5 ⋅ (202 + 122 + 62 ) = 17.0 A .

(12)

2. RMS current value

3. Load current distortion factor

DF =

I1, rms 20 = = 0.832 . Irms 2 ⋅ 17

(13)

4. Power Factor

PF = DF ⋅ COS (30 o ) = 0.832 ⋅ 0.866 = 0.720 .

(14)

5. Load current THD

THD =

1 1 −1 = − 1 = 0.667 . 2 DF 0.8322

(15)

Tutorial 2 – Semiconductor Losses Don’t use more than 3 significant digits – 1.23 A, 0. 456 W, 78.9 V There are 2 basic semiconductor loss mechanisms – conductivity loss and switching loss. Diode reverse recovery caused losses are associated with switching loss but are neglected in the first approximation (in this Tutorial). Switching loss can’t be modelled in elementary PSIM simulation – this requires a detailed consideration of semiconductor switch (power MOSFET / IGBT) dynamic model (capacitances) and gate driver electronic circuit. Conductivity Loss Mechanisms Conductivity losses in power MOSFET and IGBT power stages have essentially different nature and, therefore, are considered on separate. But first consider power diodes. 2.1. Power Diode Conductivity Loss There are 3 power diode models – ideal, offset and offset + slope (Fig.1)

Fig.1. Diode models – ideal (a), offset (b) and offset + slope (c) There are no losses in ideal diode (ideal short + ideal open). The instantaneous losses power for the offset model –

pD (t ) = iD (t )VD ,

(1)

where VD – offset voltage. Average power on a period

PD = I D _ AV VD ,

(2)

I D _ AV – average diode current on the same period. The instantaneous losses power for the offset+slope model –

pD (t ) = iD (t )VD + iD2 (t ) RD , where RD – diode (differential) resistance. Average power on a period

(3)

PD = I D _ AV VD + I D2 _ RMS RD , I D _ RMS – RMS diode current. Problem 2.1. Power diode periodic current is given by (Fig.2)

⎧20 SIN (314t ), 0 < t < 10ms;⎫ iD (t ) = ⎨ ⎬ ⎩0 A, 10 < t < 20ms. ⎭

Fig.2. Diode current Find average diode power dissipation for: (1) ideal diode model; (2) offset model with VD = 0.7V ; (3) offset+slope model with VD = 0.7V , RD = 20mΩ . Solution: (1) for ideal diode, dissipated power is zero. (2) average diode current I D _ AV =

Im

π

=

20 = 6.37 A . 3.14

For the offset diode, dissipated power

PD = I D _ AV VD = 6.37 ⋅ 0.7 = 3.13W .

(4)

(3) RMS diode current I

2 D _ RMS

I D _ RMS

1 I m2 I m2 = = = 100 A2 ; 2 2 4 . I m 20 = = = 10 A. 2 2

For the offset diode, dissipated power

PD = I D _ AVVD + I D2 _ RMS RD = 6.37 ⋅ 0.7 + 100 ⋅ 0.02 = 3.13 + 2 = 5.13W . 2.2. Power MOSFET Conductivity Loss In conductance state, power MOSFET behaves like a resistor and is able to conduct electrical current in both directions. This way, power MOSFET conductivity loss is similar to resistor loss

p(t ) = i 2 (t ) RDS (on) (T ) ,

(5)

where i (t ) - transistor current; RDS (on ) - channel resistance (DS stands for Drain-Source, on – for conductivity state); T - channel (average) temperature. The channel resistance is temperature dependent with a typical graph shown in Fig.3. The dependence is slightly non-linear and may be conservatively approximated by a linear RDS ( on ) (T2 ) = RDS ( on ) (T1 )(1 + α T (T2 − T1 ) ), (6)

α T - positive channel resistance temperature coefficient (PTC).

Normalized Rds(on), mOhm

Rds(on) Temperature Dependence 30 25

Linear approximation

20 15

Accurate

10 5 0 -60 -40 -20

0

20

40

60

80

100 120 140 160 180

T, degrees Celsius

Fig.3. Typical power MOSFET resistance temperature dependence Linear approximation (6) is quite good for analytical calculations. Average dissipated power becomes

2 P(T ) = I RMS RDS (on) (T ) ,

(7)

I RMS - RMS MOSFET current. Modern power MOSFET channel resistance reaches very low values of a few milliohms. MOSFET channel resistance voltage drop is essentially smaller than that of the parasitic anti-parallel intrinsic body diode ~1.0V. Practically, all the reverse current flows through the MOSFET channel that reduces overall power loss. The reverse diodes are conducting only for very small time portions of switching event – of the order of 1 microsecond. Remember that channel resistance is also a function of Gate-Source voltage supplied by Gate Driver (power MOSFET electronic control circuit). The minimum resistance (minimum conductivity losses) is achieved for the Gate-Source voltage close to rated one (typically 13-15V). Problem 2.2. Power MOSFET periodic current is given by (Fig.4)

⎧80COS (3140t ), 0 < t < 0.5ms;⎫ i(t ) = ⎨ ⎬ ⎩0 A, 0.5 < t < 2ms. ⎭

Fig.4. MOSFET current Find average MOSFET power dissipation (conductivity loss) for channel temperatures -20C, 40C, and 100C. For channel resistance temperature dependence use linear approximation red curve in Fig.3. Solution:

From Fig.3 graph, RDS ( on ) (−20C ) = 10 mΩ , RDS ( on ) (40C ) = 15mΩ , RDS ( on ) (100C ) = 20 mΩ . RMS MOSFET current 1 I m2 I m2 = = 800 A2 ; 4 2 8 Im 80 = = = 28.3 A. 2 2 2 ⋅ 1.41

I D2 _ RMS = I D _ RMS

MOSFET conductivity loss for the three temperatures –

P(−20C ) = 800 ⋅ 0.01 = 8W ; P(40C ) = 800 ⋅ 0.015 = 12W ; P(100C ) = 800 ⋅ 0.02 = 16W . 2.3. IGBT Conductivity Loss IGBT conductivity loss is quite similar to that of power diode. However, IGBT forward voltage drop is larger (1.0-1.5V and even more – compare with 0.7-0.8 V of silicon diodes). 3. Switching Loss Switching loss mechanisms for power MOSFET and IGBT are assumed identical and discussed below for MOSFET inductive switching. An example of inductive switching is given in Fig.5.

VDS

i

D

L

+ -

+

G +

-

VGS

-

VDC

VDC

R TS

VDS

I AV 0

a

t

0 i

OFF

OFF

ON

ON

t

TS

b

Fig.5. Inductive switching circuit (a); low resolution switched waveforms When the MOSFET is ON, the supply voltage Vdc is applied to LR-load and the current (exponentially) increases. When the MOSFET is OFF, the current circulates via (freewheeling) diode, LR-load voltage is almost zero and the current (exponentially) decays due to resistor losses. The

average current is defined by the average load voltage and load resistor. For sufficiently large inductance L, the current is almost DC with negligible pulsations. Inductive switching assumes that the RL-load current practically does not change during turn-ON or turn-OFF switching events. We will assume inductive load current I = I AV . 3.1. Turn-ON Switching Loss Turn-ON current and voltage graphs for inductive switching are shown in Fig.6.

V DS I D

1 2

V DC

I

t Δt1ON Δt 2ON

Fig.6. Turn-on switching loss model Current rise time and voltage fall time depend on gate drive circuit that is not shown in Fig.5. While the current in MOSFET increases at turn-ON, the same in diode decreases to zero (the sum of the two equals load current I). Under the assumptions, turn-ON switching loss becomes

Δt + Δt 2ON ⎛ PON ( f , VDC , I ) = f ⎜ IVDC 1ON 2 ⎝

⎞ ⎟, ⎠

(8)

where I - switched current; VDC - DC bus voltage; Δt1ON , Δt 2ON - current rise and voltage fall times (Fig.6), f - device switching frequency. 3.2. Turn-OFF Switching Loss Turn-OFF current and voltage graphs for inductive switching are shown in Fig.7.

V DS

ID

1 2

V DC

I

t Δt1OFF Δt2OFF

Fig.7. Turn-off switching loss model While the current in MOSFET declines to zero at turn-OFF, the same in diode increases to the load current I. Under the assumptions, turn-OFF switching loss becomes

Δt + Δt 2OFF ⎞ ⎛ POFF ( f , VDC , I ) = f ⎜ IVDC 1OFF ⎟, 2 ⎝ ⎠

(9)

where I - switched current; VDC - DC bus voltage; Δt1OFF , Δt2OFF - voltage rise and current fall times (Fig.7); f - switching frequency. Problem 2.3. Calculate power MOSFET switching loss for inductive switching with the following parameters: VDC = 80V ; I = 20 A ; Δt1ON = 0.2µs ; Δt 2ON = 0.3µs ; Δt1OFF = 0.25µs ; Δt 2OFF = 0.4µs ; f = 20 kHz . Solution. Turn-ON loss

Δt + Δt 2ON ⎞ 0.2 + 0.3 ⎛ 4 PON = f ⎜ IVDC 1ON ⋅ 10 −6 = 8W ; ⎟ = 2 ⋅ 10 ⋅ 20 ⋅ 80 ⋅ 2 2 ⎝ ⎠ turn-OFF

Δt + Δt 2OFF ⎞ 0.25 + 0.4 −6 ⎛ 4 POFF = f ⎜ IVDC 1OFF ⋅10 = 10.4W . ⎟ = 2 ⋅10 ⋅ 20 ⋅ 80 ⋅ 2 2 ⎝ ⎠ Overall switching loss becomes

PS = PON + POFF = 8 + 10.4 = 18.4W .

Problem 2.4. Calculate power MOSFET conductivity loss for RDS ( on ) = 15mΩ and the periodic current given by Fig.8.

i, A 70

40 t, ms

0

10

5

Fig.8. MOSFET current Solution. It may be shown by direct computation that for Fig.9 current

i I2 I1

t t1

t2

Fig.9. Linear current segment squared RMS current t

2 I RMS =

I12 + I1 I 2 + I 22 1 2 2 . i ( t ) dt = t 2 − t1 ∫t1 3

(10)

In Fig.9 and formula (10) any combinations of positive and negative I1 , I 2 are valid. By applying (10), Fig.8 squared RMS current 5

1 2 40 2 + 40 ⋅ 70 + 70 2 5 i ( t ) dt = = 1550 A2 . 10 ∫0 3 10 Then MOSFET conductivity loss 2 P = I RMS RDS (on) = 1550 ⋅ 0.015 = 23.3W . 2 I RMS =

Tutorial 3 – Non-Controlled Half-Wave Rectifiers Don’t use more than 3 significant digits – 1.23 A, 0. 456 W, 78.9 V P3.1. For the half-wave rectifier with active load Vrms=120 V, frequency f=60 Hz, load resistor R=5 Ohm. Determine: (a) average load current; (b) average load power; the power factor (PF). Solution. Voltage magnitude Vm = 2VRMS = 170V . Average voltage VO , AVG = Vm / π . Average current

I DC =

Vm 170 = = 10.8 A . πR 5π

Output RMS voltage VO ,RMS

2 Vm2 170 Vm = = 1440W . = = 85V , average load power P = 4R 4 ⋅ 5 2

Resistor and source RMS current I RMS =

Vm 170 = = 17 A . 2R 2 ⋅ 5

Apparent power S = VS ,RMS I RMS = 120 ⋅ 17 = 2040VA. Power factor PF =

1440 ≈ 0.707. 2040

P3.2. For the half-wave rectifier with RL-load R=100 Ohm, L=0.1 H, f =60Hz, and Vm=100 V. Determine: (a) current expression; (b) average current; (c) RMS current; (d) load power; (e) power factor. Hint: from numerical solution, β will equal 3.50 rad or 201 el. degree (Fig.1).

Fig.1. Numerical BETA solution for different Omega*Tau

Solution.

VS

π

β

π

β

π

β

ωt

2π

vO

iO

ωt

2π

VD

ωt

2π

Fig.2. Voltage and current graphs Angular frequency ω = 2πf = 6.28 ⋅ 60 = 377rad / s. Impedance Z = R 2 + (ωL) 2 = 1002 + (377 ⋅ 0.1) 2 = 107Ohm.

⎛ ωL ⎞ −1 ⎛ 37.7 ⎞ Phase angle θ = TAN −1 ⎜ ⎟ = TAN ⎜ ⎟ = 0.361rad . ⎝ R ⎠ ⎝ 100 ⎠

SINθ =

ωL Z

=

37.7 = 0.352. 107

Time constant ωτ =

ωL R

=

Current equation i(ωt ) =

37.7 = 0.377rad . 100

Vm Z

⎡ ⎛ ωt ⎢SIN (ωt − θ ) + SINθ exp⎜ − ωτ ⎝ ⎣

⎞⎤ ⎟⎥, 0 ≤ ωt ≤ β . ⎠⎦

⎛ ωt ⎞ i (ωt ) = 0.936SIN (ωt − 0.361) + 0.331exp⎜ − ⎟, A, 0 ≤ ωt ≤ β . ⎝ 0.377 ⎠ From numerical solution β = 3.50rad .

Fig.3. Current graph Useful integrals: β

∫ SIN (τ − θ )dτ = −COS (τ − θ ) |α α

β

β

⎛ τ⎞

⎛ τ⎞

∫α exp⎜⎝ − T ⎟⎠dτ = −T exp⎜⎝ − T ⎟⎠ |α

β

Average current

IO =

1 2π

3.50

∫ 0

⎡ ωt ⎞⎤ ⎛ ⎢0.936SIN (ωt − 0.361) + 0.331exp⎜ − 0.377 ⎟⎥d (ωt ) = ⎝ ⎠⎦ ⎣

=

1 2π

⎡ ωt ⎞ 3.50 ⎤ ⎛ 3.50 ⎢− 0.936COS (ωt − 0.361) |0 −0.331 ⋅ 0.377 exp⎜ − 0.377 ⎟ |0 ⎥ = ⎝ ⎠ ⎣ ⎦

=

1 [− 0.936(−0.9999 − 0.9355) − 0.125(0.0001 − 1)] = 0.308 A 2π

A simpler way to calculate average current is to use average voltage

Vm (1 − COSβ ) = Vm [1 + COS (β − π )]; 2π 2π V 100 [1 + COS (3.50 − π )] = 0.308 A. = m [1 + COS (β − π )] = 2πR 2π ⋅ 10

VO ,DC = I DC Useful integrals: β

∫ SIN α

2

1 1 (τ − θ )dτ = − SIN [2(τ − θ )] |αβ + (τ − θ ) |αβ 4 2

β

1

1

∫α SIN (τ )SIN (τ − θ )dτ = − 4 SIN [2τ − θ )] |α + 2 COS (θ )τ |α β

β

β

TSIN (τ − θ ) + T 2COS (τ − θ ) ⎛ τ⎞ ⎛ τ⎞ SIN ( τ − θ ) exp − d τ = − exp⎜ − ⎟ |αβ ⎜ ⎟ 2 ∫α 1+ T ⎝ T⎠ ⎝ T⎠ I

2 RMS

=

1 2π

1 = 2π 3.50

∫ 0

3.50

∫ 0

2

⎡ ωt ⎞⎤ ⎛ ⎢0.936SIN (ωt − 0.361) + 0.331exp⎜ − 0.377 ⎟⎥ d (ωt ) = ⎝ ⎠⎦ ⎣

⎡ ωt ⎞ ωt ⎞⎤ ⎛ ⎛ 2 ⎢0.876SIN (ωt − 0.361) + 0.618SIN (ωt − 0.361) exp⎜ − 0.377 ⎟ + 0.110 exp⎜ − 0.189 ⎟⎥d (ωt ) = ⎝ ⎠ ⎝ ⎠⎦ ⎣

1 = (1.389 + 0.000 + 0.020) = 0.224 A2 . 2π

I RMS = 0.224 = 0.474 A. 2 P = I RMS R = 22.4W .

Power supplied by the source β

1 PS = V S (ωt )i (ωt )d (ωt ) = 2π ∫0 = = =

1 2π

3.50

1 2π

3.5

⎡

ωt ⎞⎤

⎛

∫ 100SIN (ωt )⎢⎣0.936SIN (ωt − 0.361) + 0.331exp⎜⎝ − 0.377 ⎟⎠⎥⎦d (ωt ) = 0

⎡

⎛

ωt ⎞⎤

∫ ⎢⎣93.6SIN (ωt )SIN (ωt − 0.361) + 33.1SIN (ωt ) exp⎜⎝ − 0.377 ⎟⎠⎥⎦d (ωt ) = 0

1 (136.8 + 4.1) = 22.4W . 2π

P = PS . Power factor PF =

P P 22.4 = = = 0.67. S VS ,RMS I RMS 70.7 ⋅ 0.474

P3.3. For the half-wave rectifier with RL-load and DC source (Fig.3.1) R=2 Ohm, L=20 mH, Vrms=120 V, f =60Hz, and Vdc=100 V. Determine: (a) current expression; (b) resistor power; (c) the power absorbed by DC source; (d) the power supplied by AC source; (e) power factor. Hint: from numerical solution, β equals 3.37 rad or 193 el. degree.

VS VDC

α

2π

π

β

π

β 2π

π

β 2π

ωt

vO iO VDC

0 α

VD

α

ωt ωt

Fig.2. Voltage and current graphs Solution. Voltage magnitude Vm = 2VRMS = 170V . Angular frequency ω = 2πf = 6.28 ⋅ 60 = 377rad / s. Impedance Z = R 2 + (ωL) 2 = 22 + (377 ⋅ 0.02) 2 = 7.80Ohm.

⎛ ωL ⎞ −1 ⎛ 7.54 ⎞ Phase angle θ = TAN −1 ⎜ ⎟ = TAN ⎜ ⎟ = 1.31rad . ⎝ R ⎠ ⎝ 2 ⎠ Time constant ωτ =

⎛ VDC ⎝ Vm

α = SIN −1 ⎜⎜

ωL R

=

7.54 = 3.77 rad . 2

⎞ ⎟⎟ = 0.630rad . ⎠

Current equation

Vm V V (ωt − α )⎤, ⎡V ⎤⎡ SIN (ωt − θ ) − DC + ⎢ DC − m SIN (α − θ )⎥ ⎢exp− Z R ⎣ R Z ωτ ⎥⎦ ⎦⎣ α ≤ ωt ≤ β . i (ωt ) =

(ωt − 0.63)⎤, ⎡ i (ωt ) = 21.8SIN (ωt − 1.31) − 50 + 63.7 ⎢exp− 3.77 ⎥⎦ ⎣ α ≤ ωt ≤ β . From numerical solution β = 3.37 rad .

Fig.4. Current graph Average current β

1 IO = i(ωt )d (ωt ) 2π α∫ 1 IO = 2π

3.37

⎧

⎡

∫ ⎨⎩21.8SIN (ωt − 1.31) − 50 + 63.7⎢⎣exp−

0.63

(ωt − 0.63)⎤ ⎫d (ωt ) = 3.77

⎥⎦ ⎬ ⎭

⎡ (ωt − 0.63)⎤ 3.37 ⎤ ⎡ 3.37 3.37 ⎢− 21.8COS (ωt − 1.31) |0.63 −50 |0.63 +63.7 ⋅ 3.77 ⎢exp− 3.77 ⎥ |0.63 ⎥ = ⎣ ⎦ ⎣ ⎦

=

1 2π

=

1 (27.1 − 137 + 124) = 2.25 A 2π

A simpler way to calculate average current is to use average voltage

Vm [COSα + COS (β − π )] − VDC (β − α ); 2π 2π V − VDC V V I O = O ,DC = m [COSα + COS (β − π )] − DC (β − α ) = 2.25 A R 2πR 2πR

VO ,DC − VDC =

Power absorbed by the DC source Squared RMS Current

PDC = I OVDC = 225W . Squared RMS Current β

2 I RMS =

1 2 i (ωt )d (ωt ) 2π α∫

I

2 RMS

1 = 2π

1 2π

3.37

=

1 2π

3.37

+ =

2

3.37

⎧ (ωt − 0.63)⎤ ⎫ ⎡ ∫0.63 ⎨⎩21.8SIN (ωt − 1.31) − 50 + 63.7⎢⎣exp− 3.77 ⎥⎦ ⎬⎭ d (ωt ) =

⎡ (ωt − 0.63)⎤ ⎤ ⎡ 2 ⎢475SIN (ωt − 1.31) + 2500 + 4057 ⎢exp− 1.885 ⎥ ⎥d (ωt ) + ⎣ ⎦⎦ 0.63 ⎣

∫

⎧

⎡

∫ ⎨⎩− 2180SIN (ωt − 1.31) + 2777 SIN (ωt − 1.31)⎢⎣exp−

(ωt − 0.63)⎤ − 6377 ⎡exp− (ωt − 0.63)⎤ ⎫d (ωt ) =

0.63

3.77

⎥⎦

⎢⎣

3.77

⎥⎦ ⎬ ⎭

1 (630.9 + 6852 + 5862.2 − 2711 + 1873.1 − 12407.7) = 15.9 A2 . 2π

I RMS = 15.9 ≈ 4 A. 2 P = I RMS R = 32W .

Power supplied by the source β

PS =

1 V S (ωt )i (ωt )d (ωt ) = 2π α∫

1 = 2π

3.37

1 = 2π

3.37

=

⎧

⎡

∫ 169.7SIN (ωt )⎨⎩21.8SIN (ωt − 1.31) − 50 + 63.7⎢⎣exp−

0.63

⎧

(ωt − 0.63)⎤ ⎫d (ωt ) = 3.77

⎥⎦ ⎬ ⎭

⎡

∫ ⎨⎩3699SIN (ωt )SIN (ωt − 0.131) − 8485SIN (ωt ) + 10810 SIN (ωt )⎢⎣exp−

0.63

(ωt − 0.63)⎤ ⎫d (ωt ) = 3.77

⎥⎦ ⎬ ⎭

1 (1945 − 15119 + 14787 ) = 257W . 2π

PS = PR + PDC . Power factor PF =

P P 257 = = = 0.54. S VS ,RMS I RMS 120 ⋅ 4

P3.4. For the half-wave rectifier with RL-load and clamping diode R=2 Ohm, L=25 mH, f =60Hz, and Vm=100 V. Determine: (a) average load voltage; (b) average load current; (c) RMS current; (d) resistor power. Use frequency domain and time domain analysis and compare the results of both.

VS

π

2π

iD

π

2π

iD 1

π

2π

vO i O i20 i10 i20 i10 i20 i10

VD

ωt

ωt ωt ωt

2π

π VD

ωt

V D1

Fig.5. Voltage and current graphs Solution. Average voltage

VO , AVG =

Vm

π

=

100

π

= 31.8V .

Average current I DC =

Vm 100 = = 15.9 A. πR 2π

Frequency domain solution. Voltage harmonic magnitudes

Vm = 50V ; 2 2V V2 = 2 m = 21.2V ; (2 − 1)π 2V V4 = 2 m = 4.24V ; (4 − 1)π 2V V6 = 2 m = 1.82V ... (6 − 1)π V1 =

Angular frequency ω = 2πf = 6.28 ⋅ 60 = 377rad / s. Harmonic impedances

Z1 = 2 2 + (377 ⋅ 0.025) 2 = 9.63Ohm; Z 2 = 2 2 + (2 ⋅ 377 ⋅ 0.025) 2 = 18.96Ohm; Z 4 = 2 2 + (4 ⋅ 377 ⋅ 0.025) 2 = 37.75Ohm; Z 6 = 2 2 + (6 ⋅ 377 ⋅ 0.025) 2 = 56.58Ohm... Current harmonics

I1 = V1 / Z1 = 5.19 A; I 2 = V2 / Z 2 = 1.12 A; I 4 = V4 / Z 4 = 0.11 A; I 6 = V6 / Z 6 = 0.03 A... RMS current

I RMS ≈ I O2 , AVG +

I12 I 22 I 42 I 62 + + + = 16.3 A. 2 2 2 2

2 P = I RMS R = 532W .

Time domain solution. Current expression

Vm ⎡V ⎤ ⎛ ωt ⎞ SIN (ωt − θ ) + ⎢ m SIN (θ ) + i10 ⎥ exp⎜ − ⎟; 0 ≤ ωt ≤ π ; Z ⎝ ωτ ⎠ ⎣Z ⎦ ⎡ (ωt − π ) ⎤ i (ωt ) = i20 exp⎢− ; π ≤ ωt ≤ 2π . ωτ ⎥⎦ ⎣ i (ωt ) =

⎛ π ⎞ 1 + exp⎜ − ⎟ ⎝ ωτ ⎠ i10 = ⎛ π ⎞ ⎛ π exp⎜ ⎟ − exp⎜ − ⎝ ωτ ⎠ ⎝ ωτ

Vm SIN (θ ) = 10.7; ⎞ Z ⎟ ⎠

⎛ π ⎞ exp⎜ ⎟ +1 ωτ ⎠ ⎝ i20 = ⎛ π ⎞ ⎛ π exp⎜ ⎟ − exp⎜ − ⎝ ωτ ⎠ ⎝ ωτ

Vm SIN (θ ) = 20.9. ⎞ Z ⎟ ⎠

⎛ ωt ⎞ i (ωt ) = 10.4 SIN (ωt − 1.36) + 20.9 exp⎜ − ⎟; 0 ≤ ωt ≤ π ; ⎝ 4.71 ⎠ ⎡ (ωt − π ) ⎤ i (ωt ) = 20.9 exp⎢− ; π ≤ ωt ≤ 2π . 4.71 ⎥⎦ ⎣

Fig.6. Current graph

I

2 RMS

1 + 2π

2

π

1 ⎡ ⎛ ωt ⎞ ⎤ = 10.4 SIN (ωt − 1.36) + 20.9 exp⎜ − ⎟⎥ d (ωt ) + ⎢ ∫ 2π 0 ⎣ ⎝ 4.71 ⎠⎦ 2π

2

π

⎧ 1 ⎡ (ωt − π ) ⎤ ⎫ 2 ∫π ⎨⎩20.9 exp⎢⎣− 4.71 ⎥⎦ ⎬⎭ d (ωt ) = 2π ∫0 107.7 SIN (ωt − 1.36)d (ωt ) +

π

π

+

1 2 ⎛ ωt ⎞ ⎛ ωt ⎞ 433.2 SIN (ωt − 1.36) exp⎜ − 435.4 exp⎜ − ⎟d (ωt ) + ⎟d (ωt ) = ∫ ∫ 2π 0 2π 0 ⎝ 4.71 ⎠ ⎝ 2.36 ⎠

=

1 (169.2 + 0.0 + 1511) = 267.4 A2 . 2π

I RMS = 267.4 = 16.35 A. 2 P = I RMS R = 534 .8W .

Frequency and time domain results are in good agreement.

Tutorial 4 – Controlled Half-Wave Rectifiers Don’t use more than 3 significant digits – 1.23 A, 0. 456 W, 78.9 V

Useful integrals: β

∫ SIN (τ − θ )dτ = −COS (τ − θ ) |α α

β

β

⎛ τ⎞

⎛ τ⎞

∫ exp⎜⎝ − T ⎟⎠dτ = −T exp⎜⎝ − T ⎟⎠ |α α β

∫ SIN α β

2

β

1 1 (τ − θ )dτ = − SIN [2(τ − θ )] |αβ + (τ − θ ) |αβ 4 2 1

1

∫ SIN (τ )SIN (τ − θ )dτ = − 4 SIN [2τ − θ )] |α + 2 COS (θ )τ |α α β

β

β

TSIN (τ − θ ) + T 2COS (τ − θ ) ⎛ τ⎞ ⎛ τ⎞ SIN ( τ − θ ) exp − d τ = − exp⎜ − ⎟ |αβ ⎜ ⎟ 2 ∫α 1+ T ⎝ T⎠ ⎝ T⎠

Lecture 3 (cont.) P3.5. For the half-wave rectifier with capacitive filter Vrms=120 V, frequency f=60 Hz, load resistor R=500 Ohm, filter capacitance C=100 uF. Determine: (a) output voltage expression; (b) peak-to-peak output voltage; (c) capacitor current expression; (d) the peak diode current. Hint: from numerical solution, α equals 0.843 rad or 48 el. degree (Fig.1).

Fig.1. Numerical solution for ALPHA Solution. Voltage magnitude Vm = 2VRMS = 170V . Angular frequency ω = 2πf = 6.28 ⋅ 60 = 377rad / s.

ωRC = 377 ⋅ 500 ⋅ 100 ⋅ 10 −6 = 18.9rad . θ=

⎛ 1 ⎞ −1 ⎛ 1 ⎞ + TAN −1 ⎜ ⎟ = 1.57 + TAN ⎜ ⎟ = 1.62rad . 2 ⎝ ωRC ⎠ ⎝ 18.9 ⎠

π

Vm SIN (θ ) = 169.5V . From numerical solution of equation

⎛ 2π + α − θ ⎞ SIN (θ ) exp⎜ − ⎟ = SIN (α ) ωRC ⎠ ⎝

α equals 0.843 rad.

VO (ωt ) = 170SIN (ωt ), 0.843 ≤ ωt ≤ 1.62; ⎛ ωt − 1.62 ⎞ VO (ωt ) = 169.5SIN (θ ) exp⎜ − ⎟, 1.62 < ωt ≤ 0.843 + 2π . 18.9 ⎠ ⎝ Peak-to-peak output voltage

ΔVO = Vm [1 − SIN (α )] = 170[1 − SIN (0.843)] = 43V . Capacitor current

iC (ωt ) = ωCVm COS (ωt ) = 6.4COS (ωt ), 0.843 ≤ ωt ≤ 1.62; ⎛ ωt − θ ⎞ ⎛ ωt − 1.62 ⎞ iC (ωt ) = ωCVm COS (θ ) exp⎜ − ⎟ == −0.339 exp⎜ − ⎟, 1.62 < ωt ≤ 0.843 + 2π . 18.9 ⎠ ⎝ ωRC ⎠ ⎝ Peak diode current ⎡ SIN (α ) ⎤ ⎡ SIN (0.843) ⎤ iDPEAK = iD (α ) = Vm ⎢ + ωCCOS (α )⎥ = 170⎢ + 377 ⋅ 10 −4 COS (0.843)⎥ = 0.25 + 4.26 = 4.51 A. 500 ⎣ R ⎦ ⎣ ⎦

Lecture 4 P4.1. For controlled half-wave rectifier with active load (Fig.1.1) R=100 Ohm, f =60Hz, and Vrms=120 V. Select delay (control, or firing) angle α to produce average output voltage Vdc=40 V. Next, determine (1) active power; (2) power factor. Solution. From

VDC =

Vm [1 + COS (α )] 2π

firing angle

⎡

⎛ VDC ⎞ ⎤ ⎡ ⎛ 40 ⎞ ⎤ ⎟⎟ − 1⎥ = COS −1 ⎢2π ⎜ ⎟ − 1⎥ = 1.07 rad (61O ). ⎣ ⎝ 2 ⋅ 120 ⎠ ⎦ ⎝ Vm ⎠ ⎦

α = COS −1 ⎢2π ⎜⎜ ⎣

RMS voltage

VRMS =

Vm α SIN (2α ) 2 ⋅ 120 1.07 SIN (2 ⋅ 1.07) 1− + = 1− + = 75.6V . 2 π 2π 2 π 2π

RMS current

I RMS =

VRMS 75.6 = = 0.756 A. R 100

(1) Active power 2 75.6 2 VRMS P= = = 57.1W . R 100

(2) Power Factor

PF =

P 57.1 = = 0.63. S 120 ⋅ 0.756

P4.2. For controlled half-wave rectifier with RL-load R=20 Ohm, L=0.04 H, f =60Hz, and Vrms=120 V. The delay (control) angle α = 45 el. degrees. Determine: (1) output current expression; (2) average current; (3) load power; (4) the power factor. Hint: from numerical solution, β equals 3.79 rad or 217 el. degrees. Solution. Voltage magnitude Vm = 2VRMS = 170V . Angular frequency ω = 2πf = 6.28 ⋅ 60 = 377rad / s. Impedance Z = R 2 + (ωL) 2 = 20 2 + (377 ⋅ 0.04) 2 = 25.0Ohm. Normalized time constant ωτ =

ωL R

=

15.08 = 0.754rad . 20

⎛ ωL ⎞ −1 Phase angle θ = TAN −1 ⎜ ⎟ = TAN (0.754) = 0.646rad . ⎝ R ⎠

α = 45O = 0.785rad . (1) Output current expression

Vm Z

⎡ ⎛ ωt − α ⎞⎤ ⎛ ωt − 0.785 ⎞ ⎢ SIN (ωt − θ ) + SIN (θ − α ) exp⎜ − ωτ ⎟⎥ = 6.78SIN (ωt − 0.646) − 0.941exp⎜ − 0.754 ⎟, ⎝ ⎠⎦ ⎝ ⎠ ⎣ α ≤ ωt ≤ β . i (ωt ) =

From numerical solution β equals 3.79 rad.

Fig.2. Load current (2) Average current by direct integration

1 IO = 2π

3.79

⎧ (ωt − 0.785)⎤ ⎫d (ωt ) = ⎡ ⎨6.78 SIN (ωt − 0.646) − 0.941⎢exp− ⎬ 0.754 ⎥⎦ ⎭ ⎣ 0.785 ⎩

∫

⎡ (ωt − 0.785)⎤ 3.79 ⎤ ⎡ 3.79 ⎢− 6.78COS (ωt − 0.646) |0.785 −0.941 ⋅ 0.754⎢exp− 0.754 ⎥ |0.785 ⎥ = ⎣ ⎦ ⎣ ⎦

=

1 2π

=

1 (13.48 − 0.696) = 2.04 A 2π

There is a shortcut based on average output voltage (be advised by the lecture material). RMS current square 2 I RMS =

1 = 2π =

1 2π

3.79

2

⎧ (ωt − 0.785)⎤ ⎫ d (ωt ) = ⎡ ⎨6.78 SIN (ωt − 0.646) − 0.941⎢exp− ⎬ ∫ 0.754 ⎥⎦ ⎭ ⎣ 0.785 ⎩

3.79

⎡ (ωt − 0.785)⎤ + 0.886⎡exp− (ωt − 0.785)⎤ ⎤d (ωt ) = ⎡ 2 ⎢45.97 SIN (ωt − 0.646) − 12.76 SIN (ωt − 0.646) ⎢exp− ⎢⎣ 0.754 ⎥⎦ 0.754 ⎥⎦ ⎥⎦ ⎣ 0.785 ⎣

∫

1 (72.1 − 5.52 + 0.334) = 10.65 A2 . 2π

I RMS = 10.65 = 3.26 A. (3) Load power 2 P = I RMS R = 10.65 ⋅ 20 = 213W .

Power supplied by the source

β

PS = 1 = 2π

1 V S (ωt )i (ωt )d (ωt ) = 2π α∫ 3.79

⎧ (ωt − 0.785)⎤ ⎫d (ωt ) = ⎡ 169.7 SIN (ωt )⎨6.78 SIN (ωt − 0.646) − 0.941⎢exp− ⎬ 0.754 ⎥⎦ ⎭ ⎣ ⎩ 0.785

∫

3.79

⎧ (ωt − 0.785)⎤ ⎫d (ωt ) = ⎡ ⎨1151SIN (ωt ) SIN (ωt − 0.646) − 159.7 SIN (ωt ) ⎢exp− ⎬ 0.754 ⎥⎦ ⎭ ⎣ 0.785 ⎩

=

1 2π

=

1 (1434.6 − 96.9) = 213W . 2π

∫

PS = P. (4) Power factor

PF =

P P 213 = = = 0.54. S VS ,RMS I RMS 120 ⋅ 3.26

P4.3. For controlled half-wave rectifier with RL-load and DC source R=2 Ohm, L=20 mH, Vrms=120 V, f =60Hz, and Vdc=100 V. The delay (control) angle α = 45 el .degrees. Determine: (1) output current expression; (2) the power absorbed by the DC source; (3) resistor power; (4) power supplied by the source. Hint: from numerical solution, β equals 3.37 rad or 193 el. degree. Solution. Voltage magnitude Vm = 2VRMS = 170V . Angular frequency ω = 2πf = 6.28 ⋅ 60 = 377rad / s. Impedance Z = R 2 + (ωL) 2 = 22 + (377 ⋅ 0.02) 2 = 7.80Ohm.

⎛ ωL ⎞ −1 ⎛ 7.54 ⎞ Phase angle θ = TAN −1 ⎜ ⎟ = TAN ⎜ ⎟ = 1.31rad . ⎝ R ⎠ ⎝ 2 ⎠ Time constant ωτ =

⎛ VDC ⎝ Vm

α MIN = SIN −1 ⎜⎜

α = 0.785 > α MIN .

ωL R

=

7.54 = 3.77 rad . 2

⎞ ⎟⎟ = 0.630rad . ⎠

(1) Current equation

Vm V V (ωt − α )⎤, ⎡V ⎤⎡ SIN (ωt − θ ) − DC + ⎢ DC − m SIN (α − θ )⎥ ⎢exp− Z R ⎣ R Z ωτ ⎥⎦ ⎦⎣ α ≤ ωt ≤ β . i (ωt ) =

(ωt − 0.785)⎤, ⎡ i (ωt ) = 21.8SIN (ωt − 1.31) − 50 + 60.9 ⎢exp− ⎥⎦ 3.77 ⎣ α ≤ ωt ≤ β . From numerical solution β = 3.37 rad .

Fig.3. Load current Average current

1 IO = 2π

3.37

⎧ (ωt − 0.785)⎤ ⎫d (ωt ) = ⎡ ⎨21.8SIN (ωt − 1.31) − 50 + 60.9 ⎢exp− ⎥⎦ ⎬ 3.77 ⎣ ⎭ 0.785 ⎩

∫

⎡ (ωt − 0.63)⎤ 3.37 ⎤ ⎡ 3.37 3.37 ⎢− 21.8COS (ωt − 1.31) |0.785 −50 |0.785 +60.9 ⋅ 3.77 ⎢exp− 3.77 ⎥ |0.785 ⎥ = ⎣ ⎦ ⎣ ⎦

=

1 2π

=

1 (28.9 − 129 + 113.8) = 2.19 A 2π

(2) Power absorbed by the DC source

PDC = I OVDC = 219W .

I

2 RMS

1 = 2π

3.37

2

⎧ (ωt − 0.785)⎤ ⎫ d (ωt ) = ⎡ ⎨21.8SIN (ωt − 1.31) − 50 + 60.9⎢exp− ∫ ⎥⎦ ⎬ 3.77 ⎣ ⎭ 0.785 ⎩

3.37

=

1 2π

⎡ (ωt − 0.785)⎤ ⎤ ⎡ 2 ⎢475SIN (ωt − 1.31) + 2500 + 3709 ⎢exp− 1.885 ⎥ ⎥d (ωt ) + ⎣ ⎦⎦ 0.785 ⎣

1 2π

3.37

+ =

1 (605.1 + 6449.6 + 5216.4 − 2891.6 + 2098.5 − 11382.3) = 15.2 A2 . 2π

∫

⎧ (ωt − 0.785)⎤ − 6090⎡exp− (ωt − 0.785)⎤ ⎫d (ωt ) = ⎡ ⎨− 2180SIN (ωt − 1.31) + 2655SIN (ωt − 1.31) ⎢exp− ⎥⎦ ⎢⎣ ⎥⎦ ⎬ 3.77 3.77 ⎣ ⎭ 0.785 ⎩

∫

(3) Resistor power

I RMS = 15.2 = 3.9 A. 2 PR = I RMS R = 30.4W .

(4) Power supplied by the source β

PS = 1 = 2π

1 V S (ωt )i (ωt )d (ωt ) = 2π α∫ 3.37

⎧ (ωt − 0.785)⎤ ⎫d (ωt ) = ⎡ 169.7 SIN (ωt )⎨21.8SIN (ωt − 1.31) − 50 + 60.9⎢exp− ⎥⎦ ⎬ 3.77 ⎣ ⎩ ⎭ 0.785

∫

3.37

⎧ (ωt − 0.785)⎤ ⎫d (ωt ) = ⎡ ⎨3699 SIN (ωt ) SIN (ωt − 0.131) − 8485SIN (ωt ) + 10335 SIN (ωt ) ⎢exp− ⎥⎦ ⎬ 3.77 ⎣ ⎭ 0.785 ⎩

=

1 2π

=

1 (2160 − 14274 + 13680) = 249.4W . 2π

∫

PS = PR + PDC .

Lecture 5 P5.1. For the full-wave rectifier with active load Vrms=120 V, frequency f=60 Hz, load resistor R=5 Ohm. Determine: (1) average load current; (2) average load power; (3) the power factor (PF). Solution. Voltage magnitude Vm = 2VRMS = 170V . (1) Average load current

VO ,DC =

I RMS =

2Vm

π

, I DC =

2Vm 2 ⋅ 170 = = 21.6 A. πR 5π

VRMS 120 = = 24 A. R 5

(2) Average load power 2 VRMS P= = 2880W . R

(3) The power factor

PF =

P P 2880 = = = 1. S VS ,RMS I RMS 120 ⋅ 24

P5.2. For the full-wave rectifier with RL-load R=10 Ohm, L=10 mH, f =60Hz, and Vm=100 V. Determine: (1) the average load current; (2) estimate peak-to-peak load current variation based on the first AC term in Fourier series; (3) load power; (4) the power factor; (5) the diodes average current; (6) the diodes RMS current. Frequency domain solution. (1) Average voltage

VO ,DC =

2Vm

π

=

2 ⋅ 100

π

= 63.7V .

(1) Average load current

I DC =

VO ,DC 63.7 = = 6.37 A. R 10

Amplitudes of first voltage harmonics

V2 =

4Vm 4 ⋅ 100 = = 42.4V ; 2 π (2 − 1) π (4 − 1)

V4 =

4Vm 4 ⋅ 100 = = 8.48V . 2 π (4 − 1) π (16 − 1)

Harmonic impedances

Z 2 = 10 2 + (2 ⋅ 377 ⋅ 0.01) 2 = 12.5Ohm; Z 4 = 10 2 + (4 ⋅ 377 ⋅ 0.01) 2 = 18.1Ohm. Current harmonics

I 2 = V2 / Z 2 = 3.39 A; I 4 = V4 / Z 4 = 0.47 A. (2) As the second harmonic is dominating, it can be used to estimate current peak-to-peak variation –

Δi ≈ 2I 2 = 2 ⋅ 3.39 = 6.78A. RMS current 2 I RMS ≈ I DC +

I 22 I 42 + = 6.81A. 2 2

(3) Load power 2 P = I RMS R = 464W .


PF =

P P 464 = = = 0.964. S VS ,RMS I RMS (100 / 2 )6.81

(5) Diodes’ average current

I D , AVG =

I DC 6.37 = = 3.19 A. 2 2

(6) Diodes’ RMS current

I D,RMS =

I RMS 6.81 == = 4.82 A. 2 2

Time domain solution. Impedance Z = R 2 + (ωL) 2 = 10 2 + (377 ⋅ 0.01) 2 = 10.7Ohm . Time constant ωτ =

ωL R

=

3.77 = 0.377rad . 10


Current expression

Vm ⎡V ⎤ ⎛ ωt SIN (ωt − θ ) + ⎢ m SIN (θ ) + i0 ⎥ exp⎜ − Z ⎝ ωτ ⎣Z ⎦ i (π ) = i0 . i (ωt ) =

⎞ ⎟; 0 ≤ ωt ≤ π ; ⎠

π ⎞ ⎛ π ⎞ ⎛ 1 + exp⎜ − 1 + exp⎜ − ⎟ ⎟ ⎝ ωτ ⎠ Vm SIN (θ ) = ⎝ 0.377 ⎠ 100 0.353 = 3.30 A. i0 = π ⎞ 10.7 ⎛ π ⎞ Z ⎛ 1 − exp⎜ − 1 − exp⎜ − ⎟ ⎟ ⎝ ωτ ⎠ ⎝ 0.377 ⎠ ⎛ ωt ⎞ i (ωt ) = 9.36 SIN (ωt − 0.361) + 6.60 exp⎜ − ⎟; 0 ≤ ωt ≤ π . ⎝ 0.377 ⎠

Fig.4. Load current Squared RMS current

I

2 RMS

1

2

π

π

1 ⎡ ωt ⎞⎤ 1 ⎛ 2 = ∫ ⎢9.36 SIN (ωt − 0.361) + 6.60 exp⎜ − ⎟⎥ d (ωt ) == ∫ 87.6SIN (ωt − 0.361)d (ωt ) + π 0⎣ 0 . 377 π ⎝ ⎠⎦ 0 π

π

ωt ⎞ 1 ωt ⎞ ⎛ ⎛ + ∫ 123.6SIN (ωt − 0.361) exp⎜ − ⎟d (ωt ) + ∫ 43.6 exp⎜ − ⎟d (ωt ) = π 0 π 0 ⎝ 0.377 ⎠ ⎝ 0.189 ⎠ 1 = (862.5 + 0.0 + 2367.7) = 1028 A2 . π

I RMS = 46.4 = 6.81A. 2 P = I RMS R = 464W .


PS = =

1

1

π

π

∫V

S

(ωt )i (ωt )d (ωt ) =

0

π

⎡

⎛

ωt ⎞⎤

100SIN (ωt ) ⎢9.36 SIN (ωt − 0.361) + 6.60 exp⎜ − ⎟⎥d (ωt ) = π∫ ⎝ 0.377 ⎠⎦ ⎣ 0

π

1 ⎡ ωt ⎞⎤ ⎛ 936SIN (ωt ) SIN (ωt − 0.361) + 660SIN (ωt )⎜ − ⎟⎥d (ωt ) = ⎢ ∫ π 0⎣ ⎝ 0.377 ⎠⎦ 1 = (1373 + 82) = 464W . 2π =

PS = P . Compare with the above frequency domain results.

Tutorial 5 – Non-Controlled Full-Wave Rectifiers Don’t use more than 3 significant digits – 1.23 A, 0. 456 W, 78.9 V


∫ SIN (τ − θ )dτ = −COS (τ − θ ) |α α

β

β

⎛ τ⎞

⎛ τ⎞

∫ exp⎜⎝ − T ⎟⎠dτ = −T exp⎜⎝ − T ⎟⎠ |α α β

∫ SIN α β

2

β


1


β

β


Lecture 5 P5.1. For the full-wave rectifier with active load Vrms=120 V, frequency f=60 Hz, load resistor R=5 Ohm. Determine: (1) average load current; (2) average load power; (3) the power factor (PF). Solution. Voltage magnitude Vm = 2VRMS = 170V . (1) Average load current

VO ,DC =

I RMS =

2Vm

π

, I DC =

2Vm 2 ⋅ 170 = = 21.6 A. πR 5π

VRMS 120 = = 24 A. R 5

(2) Average load power

P=

2 VRMS = 2880W . R


PF =

P P 2880 = = = 1. S VS ,RMS I RMS 120 ⋅ 24

P5.2. For the full-wave rectifier with RL-load R=10 Ohm, L=10 mH, f =60Hz, and Vm=100 V. Determine: (1) the average load current; (2) estimate peak-to-peak load current variation based on the first AC term in Fourier series; (3) load power; (4) the power factor; (5) the diodes average current; (6) the diodes RMS current. Frequency domain solution. Average voltage

VO ,DC =

2Vm

π

=

2 ⋅ 100

π

= 63.7V .

(1) Average load current

I DC =

VO ,DC 63.7 = = 6.37 A. R 10

Amplitudes of first voltage harmonics

V2 =

4Vm 4 ⋅ 100 = = 42.4V ; 2 π (2 − 1) π (4 − 1)

V4 =

4Vm 4 ⋅ 100 = = 8.48V . 2 π (4 − 1) π (16 − 1)

Harmonic impedances

Z 2 = 10 2 + (2 ⋅ 377 ⋅ 0.01) 2 = 12.5Ohm; Z 4 = 10 2 + (4 ⋅ 377 ⋅ 0.01) 2 = 18.1Ohm. Current harmonics

I 2 = V2 / Z 2 = 3.39 A; I 4 = V4 / Z 4 = 0.47 A. (2) As the second harmonic is dominating, it can be used to estimate current peak-to-peak variation –

Δi ≈ 2I 2 = 2 ⋅ 3.39 = 6.78A. RMS current

I RMS ≈ I

2 DC

I 22 I 42 + + = 6.81A. 2 2

(3) Load power 2 P = I RMS R = 464W .


PF =

P P 464 = = = 0.964. S VS ,RMS I RMS (100 / 2 )6.81

(5) Diodes’ average current

I D , AVG =

I DC 6.37 = = 3.19 A. 2 2

(6) Diodes’ RMS current

I D,RMS =

I RMS 6.81 == = 4.82 A. 2 2

Time domain solution. Impedance Z = R 2 + (ωL) 2 = 10 2 + (377 ⋅ 0.01) 2 = 10.7Ohm . Time constant ωτ =

ωL R

=

3.77 = 0.377rad . 10

⎛ ωL ⎞ −1 Phase angle θ = TAN −1 ⎜ ⎟ = TAN (0.377) = 0.361rad . ⎝ R ⎠ Current expression

Vm ⎡V ⎤ ⎛ ωt SIN (ωt − θ ) + ⎢ m SIN (θ ) + i0 ⎥ exp⎜ − Z ⎝ ωτ ⎣Z ⎦ i (π ) = i0 . i (ωt ) =

⎞ ⎟; 0 ≤ ωt ≤ π ; ⎠

π ⎞ ⎛ π ⎞ ⎛ 1 + exp⎜ − 1 + exp⎜ − ⎟ ⎟ ωτ ⎠ Vm 0.377 ⎠ 100 ⎝ ⎝ i0 = SIN (θ ) = 0.353 = 3.30 A. π ⎞ 10.7 ⎛ π ⎞ Z ⎛ 1 − exp⎜ − 1 − exp⎜ − ⎟ ⎟ ⎝ ωτ ⎠ ⎝ 0.377 ⎠ ⎛ ωt ⎞ i (ωt ) = 9.36 SIN (ωt − 0.361) + 6.60 exp⎜ − ⎟; 0 ≤ ωt ≤ π . ⎝ 0.377 ⎠

Fig.4. Load current Squared RMS current

I

π

π

π

ωt ⎞ 1 ωt ⎞ ⎛ ⎛ 123.6SIN (ωt − 0.361) exp⎜ − ⎟d (ωt ) + ∫ 43.6 exp⎜ − ⎟d (ωt ) = ∫ π 0 π 0 ⎝ 0.377 ⎠ ⎝ 0.189 ⎠ 1 = (862.5 + 0.0 + 2367.7) = 1028 A2 . +

1

2

π

1 ⎡ ωt ⎞⎤ 1 ⎛ 2 = ∫ ⎢9.36 SIN (ωt − 0.361) + 6.60 exp⎜ − ⎟⎥ d (ωt ) == ∫ 87.6SIN (ωt − 0.361)d (ωt ) + π 0⎣ π 0 ⎝ 0.377 ⎠⎦

2 RMS

π

I RMS = 46.4 = 6.81A. 2 P = I RMS R = 464W .


PS = =

1

1

π

π

∫V

S

(ωt )i (ωt )d (ωt ) =

0

π

⎡

⎛

ωt ⎞⎤

100SIN (ωt ) ⎢9.36 SIN (ωt − 0.361) + 6.60 exp⎜ − ⎟⎥d (ωt ) = π∫ ⎝ 0.377 ⎠⎦ ⎣ 0

π

1 ⎡ ωt ⎞⎤ ⎛ 936SIN (ωt ) SIN (ωt − 0.361) + 660SIN (ωt )⎜ − ⎟⎥d (ωt ) = ⎢ ∫ π 0⎣ ⎝ 0.377 ⎠⎦ 1 = (1373 + 82) = 464W . 2π =


P5.3. For the full-wave rectifier with RL-load and DC source R=2 Ohm, L=10 mH, Vrms=120 V, f =60Hz, and Vdc=80 V. Determine: (1) the resistor power; (2) the power absorbed by the DC source. Hint: the rectifier operates in CCM. Frequency and time domain solutions possible - compare both.

I DC =

2Vm VDC 2 ⋅ 2 ⋅ 120 80 − = − = 14.0 A . πR R 2π 2

(2) PDC = I DCVDC = 14.0 ⋅ 80 = 1120W . Frequency domain solution. Voltage Fourier components

VO ,DC =

2 2VRMS

π

=

2 ⋅ 2 ⋅ 120

π

= 108V ;

V2 =

4Vm 4 ⋅ 2 ⋅ 120 = = 72.0V ; 2 π (2 − 1) π (4 − 1)

V4 =

4Vm 4 ⋅ 2 ⋅ 120 = = 14.4V . 2 π (4 − 1) π (16 − 1)

Harmonic impedances

Z 0 = R = 2Ohm; Z 2 = 2 2 + (2 ⋅ 377 ⋅ 0.01) 2 = 7.8Ohm; Z 4 = 2 2 + (4 ⋅ 377 ⋅ 0.01) 2 = 15.2Ohm. Current components

I DC = VO ,DC / Z 0 = 14.0 A; I 2 = V2 / Z 2 = 9.23 A; I 4 = V4 / Z 4 = 0.90 A. RMS current 2 I RMS ≈ I DC +

I 22 I 42 + = 15.5 A. 2 2

2 (1) P = I RMS R = 478W .

Time domain solution. Voltage magnitude Vm = 2VRMS = 170V . Angular frequency ω = 2πf = 6.28 ⋅ 60 = 377rad / s. Impedance Z = R 2 + (ωL) 2 = 22 + (377 ⋅ 0.01) 2 = 4.27Ohm. Time constant ωτ =

ωL R

=

3.77 = 1.89rad . 2

⎛ ωL ⎞ −1 Phase angle θ = TAN −1 ⎜ ⎟ = TAN (1.89 ) = 1.08rad . ⎝ R ⎠ Current equation

i (ωt ) =

Vm V V ⎡V ⎤ ⎛ ωt ⎞ SIN (ωt − θ ) − DC + ⎢ m SIN (θ ) + DC + i0 ⎥ exp⎜ − ⎟; 0 ≤ ωt ≤ π . Z R ⎣Z R ⎝ ωτ ⎠ ⎦

⎛ π ⎞ ⎛ π ⎞ 1 + exp⎜ − 1 + exp⎜ − ⎟ ⎟ ⎝ ωτ ⎠ Vm SIN (θ ) − VDC = ⎝ 1.89 ⎠ 170 0.883 − 80 = 11.5 A. i0 = R 2 ⎛ π ⎞ Z ⎛ π ⎞ 4.27 1 − exp⎜ − 1 − exp⎜ − ⎟ ⎟ ⎝ ωτ ⎠ ⎝ 1.89 ⎠

⎛ ωt ⎞ i (ωt ) = 39.8SIN (ωt − 1.08) − 40 + 86.6 exp⎜ − ⎟, 0 ≤ ωt ≤ π . ⎝ 1.89 ⎠

Average current π

1 ⎡ ⎛ ωt ⎞⎤ I O = ∫ ⎢39.8SIN (ωt − 1.08) + 86.6 exp⎜ − ⎟⎥d (ωt ) − 40 = π 0⎣ ⎝ 1.89 ⎠⎦ 1⎡ ⎛ ωt ⎞ π ⎤ − 39.8COS (ωt − 1.31) |π0 +86.6 ⋅ 1.89 exp⎜ − ⎟ |0 ⎥ − 40 = ⎢ π⎣ ⎝ 1.89 ⎠ ⎦ 1 = (37.3 + 132.4) − 40 = 14 A. 2π =

Power absorbed by the DC source (2) PDC = I OVDC = 1120W .

I

2

π

1 ⎡ ⎛ ωt ⎞⎤ = ∫ ⎢39.8SIN (ωt − 1.08) − 40 + 86.6 exp⎜ − ⎟⎥ d (ωt ) = π 0⎣ ⎝ 1.89 ⎠⎦

2 RMS

π

=

1 ⎡ ⎛ ωt ⎞⎤ 1584 SIN 2 (ωt − 1.08) + 1600 + 7400⎜ − ⎟⎥d (ωt ) + ⎢ ∫ π 0⎣ ⎝ 0.943 ⎠⎦ π

1 ⎧ ⎛ ωt ⎞ ⎛ ωt ⎞⎫ ⎟ − 6928⎜ − ⎟⎬d (ωt ) = ⎨− 3184 SIN (ωt − 1.08) + 6894SIN (ωt − 1.31)⎜ − ∫ π 0⎩ ⎝ 1.89 ⎠ ⎝ 1.89 ⎠⎭ 1 = (2484 + 5027 + 6819 − 2982 + 0.0 − 10595) = 240 A2 . +

π

I RMS = 239.7 = 15.5 A. 2 (1) PR = I RMS R = 479W .


PS = =

1

1

π

π

π

∫V

S

(ωt )i (ωt )d (ωt ) =

0

⎡

⎛

ωt ⎞⎤

169.7 SIN (ωt ) ⎢39.8SIN (ωt − 1.08) − 40 + 86.6 exp⎜ − ⎟⎥d (ωt ) = π∫ ⎝ 1.89 ⎠⎦ ⎣ 0.

π

1 ⎡ ⎛ ωt ⎞⎤ 6754 SIN (ωt ) SIN (ωt − 1.08) − 6788SIN (ωt ) + 14696 SIN (ωt )⎜ − ⎟⎥d (ωt ) = ⎢ ∫ π 0⎣ ⎝ 1.89 ⎠⎦ 1 = (4968 − 13577 + 13638) = 1600W . =

π

PR + PDC = 479 + 1121 = 1600W = PS . Compare with the above frequency domain results.

P5.4. For the full-wave rectifier with capacitive filter Vrms=120 V, frequency f=60 Hz, load resistor R=500 Ohm, filter capacitance C=100 uF. Determine: (1) output voltage expression; (2) peak-to-peak output voltage; (3) capacitor current expression; (4) the peak diode current. Hint: from numerical solution, α equals 1.06 rad or 61 el. degree. Voltage magnitude Vm = 2VRMS = 170V . Angular frequency ω = 2πf = 6.28 ⋅ 60 = 377rad / s.

ωRC = 377 ⋅ 500 ⋅ 100 ⋅ 10 −6 = 18.9rad .

θ=

⎛ 1 ⎞ −1 ⎛ 1 ⎞ + TAN −1 ⎜ ⎟ = 1.57 + TAN ⎜ ⎟ = 1.62rad . 2 ⎝ ωRC ⎠ ⎝ 18.9 ⎠

π

Vm SIN (θ ) = 169.5V . ⎛ π + α −θ ⎞ From numerical solution of equation SIN (θ ) exp⎜ − ⎟ = SIN (α ) α equals 1.06 rad. ωRC ⎠ ⎝ VO (ωt ) = 170SIN (ωt ), 1.06 ≤ ωt ≤ 1.62;

(1)

⎛ ωt − 1.62 ⎞ VO (ωt ) = 169.5SIN (θ ) exp⎜ − ⎟, 1.62 < ωt ≤ 1.06 + π . 18.9 ⎠ ⎝

(2) Peak-to-peak output voltage ripple

ΔVO = Vm [1 − SIN (α )] = 170[1 − SIN (1.06)] = 21.7V . (3) Capacitor current expression iC (ωt ) = ωCVmCOS (ωt ) = 6.4COS (ωt ), 1.06 ≤ ωt ≤ 1.62; ⎛ ωt − θ ⎞ ⎛ ωt − 1.62 ⎞ iC (ωt ) = ωCVmCOS (θ ) exp⎜ − ⎟ == −0.339 exp⎜ − ⎟, 1.62 < ωt ≤ 1.06 + π . 18.9 ⎠ ⎝ ωRC ⎠ ⎝

(4) Peak diode current ⎡ SIN (α ) ⎤ ⎡ SIN (1.06) ⎤ iDPEAK = iD (α ) = Vm ⎢ + ωC ⋅ COS (α )⎥ = 170⎢ + 377 ⋅ 10 −4 COS (1.06)⎥ = 0.30 + 3.13 = 3.43 A. R 500 ⎣ ⎦ ⎣ ⎦

Tutorial 6 – Controlled Full-Wave Rectifiers Don’t use more than 3 significant digits – 1.23 A, 0. 456 W, 78.9 V


∫ SIN (τ − θ )dτ = −COS (τ − θ ) |α α

β

β

⎛ τ⎞

⎛ τ⎞

∫ exp⎜⎝ − T ⎟⎠dτ = −T exp⎜⎝ − T ⎟⎠ |α α β

∫ SIN α β

2

β


1


β

β


Lecture 6 P6.1. For controlled full-wave rectifier with active load R=20 Ohm, f =60Hz, and Vrms=120 V. Delay (control, or firing) angle α =40 el. degrees. Determine (a) active power; (b) power factor. Voltage magnitude Vm = 2VRMS = 170V . Average voltage VO , AVG = Average current I DC =

Vm

π

(1 + COSα ) = 170 40O = 95.4V . π

VO , AVG 95.4 = = 4.77 A. R 20

Angle in radians α = 40

π 180

Output RMS voltage I RMS =

= 0.698rad . Vm α SIN (2α ) 170 0.698 SIN (2 ⋅ 0.698) 1− + = 1− + = 5.80 A. π 2π π 2π 2R 2 ⋅ 20

2 Average load power P = I RMS R = 5.80 2 ⋅ 20 = 673W .

Apparent power S = VS ,RMS I RMS = 120 ⋅ 5.80 = 696VA. Power factor PF =

672 = 0.967. 696

P6.2. For controlled full-wave rectifier with RL-load R=10 Ohm, L=20 mH, f =60Hz, and Vrms=120 V. The delay angle α = 60 el. degrees. Determine: (a) average current; (b) load power. Hint: check that the rectifier operates in DCM. From numerical solution, β equals 3.78 rad or 216 el. degree. Voltage magnitude Vm = 2VRMS = 170V . Angular frequency ω = 2πf = 6.28 ⋅ 60 = 377rad / s. Impedance Z = R 2 + (ωL) 2 = 10 2 + (377 ⋅ 0.02) 2 = 12.5Ohm. Normalized time constant ωτ =

ωL R

=

377 ⋅ 0.02 = 0.754rad . 10

⎛ ωL ⎞ −1 Phase angle θ = TAN −1 ⎜ ⎟ = TAN (0.754) = 0.646rad . R ⎝ ⎠

α = 60O = 1.047rad . Vm Z

⎡ ⎛ ωt − α ⎞⎤ ⎛ ωt − 1.047 ⎞ ⎢ SIN (ωt − θ ) + SIN (θ − α ) exp⎜ − ωτ ⎟⎥ = 13.6SIN (ωt − 0.646) − 5.29 exp⎜ − 0.754 ⎟, ⎝ ⎠⎦ ⎝ ⎠ ⎣ α ≤ ωt ≤ β . i (ωt ) =

From numerical solution β equals 3.78 rad. As π + α = 4.19 > β = 3.78 it is DCM indeed.

Average load current

IO =

3.78

⎧ (ωt − 1.047)⎤ ⎫d (ωt ) = ⎡ ⎨13.6 SIN (ωt − 0.646) − 5.29 ⎢exp− ⎬ π 1.047 ⎩ 0.754 ⎥⎦ ⎭ ⎣ 1

∫

(ωt − 1.047)⎤ |3.78 ⎤ = 1⎡ ⎡ 78 − 13.6COS (ωt − 0.646) |13..047 −5.29 ⋅ 0.754⎢exp− 1.047 ⎥ ⎢ π⎣ 0.754 ⎥⎦ ⎣ ⎦ 1 = (26.02 − 3.88) = 7.05 A =

π

RMS current square I

2 RMS

3.78

2

⎧ (ωt − 1.047)⎤ ⎫ d (ωt ) = ⎡ = ⎨13.6SIN (ωt − 0.646) − 5.29 ⎢exp− ⎬ ∫ π 1.047 ⎩ 0.754 ⎥⎦ ⎭ ⎣ 1

3.78

⎡ (ωt − 1.047)⎤ + 28⎡exp− (ωt − 1.047)⎤ ⎤d (ωt ) = ⎡ 185SIN 2 (ωt − 0.646) − 144SIN (ωt − 0.646) ⎢exp− ⎢ ⎢⎣ π 1.047 ⎣ 0.754 ⎥⎦ 0.377 ⎥⎦ ⎥⎦ ⎣ 1 = (284.6 − 76.1 + 10.5) = 69.7 A2 . =

1

∫

π

I RMS = 69.7 = 8.35 A.

2 P = I RMS R = 69.7 ⋅ 10 = 697W .


PS = =

1

π

β

∫α V

S

(ωt )i (ωt )d (ωt ) =

3.78

⎧ (ωt − 1.047)⎤ ⎫d (ωt ) = ⎡ 169.7 SIN (ωt )⎨13.6SIN (ωt − 0.646) − 5.29 ⎢exp− ⎬ π 1.047 0.754 ⎥⎦ ⎭ ⎣ ⎩ 1

∫

3.78

⎧ (ωt − 1.047)⎤ ⎫d (ωt ) = ⎡ ⎨2308SIN (ωt ) SIN (ωt − 0.646) − 898SIN (ωt ) ⎢exp− ⎬ π 1.047 ⎩ 0.754 ⎥⎦ ⎭ ⎣ 1 = (2740 − 550.4) = 697W . =

1

∫

π

PS = P. Power factor PF =

P P 697 = = = 0.70. S VS ,RMS I RMS 120 ⋅ 8.35

P6.3. For controlled full-wave rectifier with RL-load R=10 Ohm, L=100 mH, f =60Hz, and Vrms=120 V. The delay angle α = 60 el. degrees. Determine: (a) average current; (b) load power. Hint: check that the rectifier operates in CCM. Frequency and time domain solution possible – it is 1 Voltage magnitude Vm = 2VRMS = 170V . Angular frequency ω = 2πf = 6.28 ⋅ 60 = 377rad / s. Frequency domain Fourier analysis.

VO (ωt ) = VO,DC +

∞

∑V COS (nωt + θ n

n

).

n=2, 4, 6

VO,DC =

2VmCOS (α )

π

=

2 ⋅ 170 ⋅ COS (60 O )

π

= 54.0V .

2Vm ⎛ COS [α (n + 1)] COS [α (n − 1)] ⎞ − ⎜ ⎟; π ⎝ n +1 n −1 ⎠ 2V ⎛ SIN [α (n + 1)] SIN [α (n − 1)] ⎞ bn = m ⎜ − ⎟; π ⎝ n +1 n −1 ⎠ an =

Vn = an2 + bn2 .

a2 = −90.0; b2 = −93.5; V2 = 129.8V . a4 = 46.8; b4 = −18.7; V4 = 50.4V . a6 = −3.19; b6 = 32.0; V6 = 32.2V . Harmonic impedances

Z 2 = 10 2 + (2 ⋅ 377 ⋅ 0.1) 2 = 76.0Ohm; Z 4 = 10 2 + (4 ⋅ 377 ⋅ 0.1) 2 = 151.1Ohm; Z 6 = 10 2 + (6 ⋅ 377 ⋅ 0.1) 2 = 226.4Ohm. Average current and harmonics

I O , AVG = VO , AVG / R = 5.40 A; I 2 = V2 / Z 2 = 1.71 A; I 4 = V4 / Z 4 = 0.33 A; I 6 = V6 / Z 6 = 0.14 A... RMS current

I RMS ≈ I O2 , AVG +

I 22 I 32 I 62 + + = 5.54 A. 2 2 2

2 P = I RMS R = 307W .

Time domain solution. Impedance Z = Z1 = R 2 + (ωL) 2 = 10 2 + (377 ⋅ 0.1) 2 = 39Ohm . Normalized time constant ωτ =

ωL R

=

377 ⋅ 0.1 = 3.77 rad . 10

⎛ ωL ⎞ −1 Phase angle θ = TAN −1 ⎜ ⎟ = TAN (3.77 ) = 1.31rad . α = 1.047rad . R ⎝ ⎠ Current expression

Vm ⎡V ⎤ ⎛ ωt − α ⎞ SIN (ωt − θ ) + ⎢ m SIN (θ − α ) + iα ⎥ exp⎜ − ⎟; α ≤ ωt ≤ π + α ; Z ωτ ⎠ ⎝ ⎣Z ⎦ ⎛ π ⎞ 1 + exp⎜ − ⎟ ⎝ ωτ ⎠ Vm SIN (θ − α ). iα = ⎛ π ⎞ Z 1 − exp⎜ − ⎟ ⎝ ωτ ⎠ i (ωt ) =

⎡ (ωt − 1.047) ⎤ i (ωt ) = 4.35 SIN (ωt − 1.31) + 4.02 exp⎢− ⎥⎦; 1.047 ≤ ωt ≤ 4.189. 3.77 ⎣

Squared RMS current

I

2 RMS

1

1

2

4.189

4.189

⎡ 1 ⎡ (ωt − 1.047) ⎤ ⎤ = 4.35 SIN (ωt − 1.31) + 4.02 exp⎢− d (ωt ) = 18.9SIN 2 (ωt − 1.31)d (ωt ) + ⎢ ⎥ ∫ ∫ ⎥ π 1.047 ⎣ 3.77 π 1.047 ⎣ ⎦⎦ 4.189

4.189

1 ⎡ (ωt − 1.047) ⎤ ⎡ (ωt − 1.047) ⎤ + 35.0SIN (ωt − 1.31) exp⎢− d (ωt ) + 16.2 exp⎢− ∫ ∫ ⎥ ⎥⎦d (ωt ) = π 1.047 3.77 π 1.047 1.89 ⎣ ⎦ ⎣ 1 = (29.7 + 42.0 + 24.7) = 30.7 A2 .

π

I RMS = 30.7 = 5.54 A. 2 P = I RMS R = 307W .


PS = =

1

1

π +α

π

∫V

S

(ωt )i (ωt )d (ωt ) =

α

4.189

⎧ ⎡ (ωt − 1.047) ⎤ ⎫ 170SIN (ωt )⎨4.35 SIN (ωt − 1.31) + 4.02 exp⎢− ⎥⎦ ⎬d (ωt ) = π 1.047 3.77 ⎣ ⎩ ⎭

∫

π

1 ⎡ ⎡ (ωt − 1.047) ⎤ ⎤ 740SIN (ωt ) SIN (ωt − 1.31) + 683SIN (ωt ) ⎢− ⎢ ∫ ⎥⎦ ⎥d (ωt ) = π 0⎣ 3.77 ⎣ ⎦ 1 = (297.4 + 667.4) = 307W . 2π =


P6.4. For controlled full-wave rectifier with RL-load and DC source (Fig.3.1) R=5 Ohm, Vrms=240 V, f =60Hz, and Vdc=100 V. Inductance is large enough to cause CCM. Determine: (a) the delay (control) angle α such that power absorbed by the DC source is Pdc=1000 W; (b) estimate from the first AC term the inductance that limits peak-to-peak current variation to 2 A. Voltage magnitude Vm = 2VRMS = 339V . The DC current component

IO =

PDC 1000 = = 10 A . VDC 100

DC voltage component

VO = VDC + I O R = 100 + 10 ⋅ 5 = 150V . ⎛ VOπ ⎞ V 150 ⋅ π ⎞ O ⎟⎟ = COS −1 ⎛⎜ ⎟ = 46 = 0.803rad . 2 V 2 ⋅ 339 ⎝ ⎠ ⎝ m⎠

α = COS −1 ⎜⎜

The first AC term is 2nd harmonic.

2Vm ⎛ COS (3α ) ⎞ ⎜ COSα − ⎟ = 203V ; π ⎝ 3 ⎠ 2V ⎛ SIN (3α ) ⎞ b2 = m ⎜ SINα − ⎟ = 107V ; π ⎝ 3 ⎠ a2 =

V2 m = a22 + b22 = 230V . The required load 2nd harmonic impedance

Z2 =

V2 m 230V = = 230Ohm. I 2m 1A

Required inductance

Z 22 − R 2 2302 − 52 L= = = 0.31H . 2ω 2 ⋅ 377 Time domain analysis. Voltage magnitude Vm = 2VRMS = 339V . Impedance Z = R 2 + (ωL) 2 = 52 + (377 ⋅ 0.31) 2 = 117Ohm. Normalized time constant ωτ =

ωL R

=

377 ⋅ 0.31 = 23.4rad . 5


α = 46 O = 0.803rad . Current expression

i (ωt ) =

Vm V V ⎡V ⎤ ⎛ ωt − α ⎞ SIN (ωt − θ ) − DC + ⎢ m SIN (θ − α ) + DC + iα ⎥ exp⎜ − ⎟; α ≤ ωt ≤ π + α . Z R ⎣Z R ωτ ⎠ ⎝ ⎦

⎛ π ⎞ 1 + exp⎜ − ⎟ V ωτ ⎠ Vm ⎝ iα = SIN (θ − α ) − DC . R ⎛ π ⎞ Z 1 − exp⎜ − ⎟ ⎝ ωτ ⎠

⎡ (ωt − 0.803) ⎤ i (ωt ) = 2.90 SIN (ωt − 1.53) − 20 + 30.6 exp⎢− ⎥⎦; 0.803 ≤ ωt ≤ 3.945. 23.4 ⎣

∞

THDI =

∑I

2 n , RMS

n=2

I1,RMS

=

2 I RMS − I12,RMS

I1,RMS

=

44.1 − 9.27 2 / 2 = 0.162 = 16.2%. 9.27 / 2

Tutorial 7 – Non-Controlled Three-Phase Rectifiers Don’t use more than 3 significant digits – 1.23 A, 0. 456 W, 78.9 V


∫ SIN (τ − θ )dτ = −COS (τ − θ ) |α α

β

β

⎛ τ⎞

⎛ τ⎞

∫ exp⎜⎝ − T ⎟⎠dτ = −T exp⎜⎝ − T ⎟⎠ |α α β

∫ SIN α β

2

β


1


β

β


Lecture 7 P7.1. For a three-phase non-controlled full-wave rectifier with RL-load line-to-line voltage Vrms,L=480 V, R=25 Ohm, L=50 mH, f =60Hz. Determine (a) output DC voltage; (b) DC component and first AC (6x) harmonic magnitude of the load current; (c) average diode current; (d) RMS diode current; (e) RMS source current; (f) active power and apparent power from the source; (g) power factor.

i

+

+

D1

D3

vL

D5 -

vO

+

VBN

VAN ~

VCN ~

~

D6

D4

vR

D2

-

N

-

Fig.1. A three-phase non-controlled full-wave rectifier with RL-load

VAN

VS

VBN

i

VCN

π 6

π

2π

π

ωt

5π 6

7π 9π 6 6

3π 6

5π 6

7π 9π 6 6

3π 6

5π 6

7π 9π 6 6

3π 6

5π 6

7π 9π 6 6

3π 6

5π 6

7π 9π 6 6

3π 6

5π 6

7π 9π 6 6

3π 6

5π 6

7π 9π 6 6

3π 6

5π 6

iD1 π 6

iD2

VMAX VMIN

3π 6

π 6

π /6

5π / 6

3π / 2

π /6

5π / 6

3π / 2

ωt ωt

iD3 π 6

iD4 π 6

iD5 π

VO

6

iD6

i

π 6

π 6

3π 6

5π 6

7π 9π 6 6

ωt

iA π 6

Fig.2. Voltage and current graphs

ωt ωt ωt ωt ωt ωt ωt ωt

Solution. (a) Output DC voltage

VO , AVG =

3VmL

=

π

3 2VRMSL

=

π

3 2 ⋅ 480

π

= 648V .

(b) DC current component

I DC =

VO , AVG 648 = = 25.9 A. R 25

The first AC (6x) voltage harmonic

Vn =

6VmL 6 2VRMSL = , n = 6,12,18... 2 π n − 1 π n2 − 1

V6 =

6 2VRMSL 6 2 ⋅ 480 = = 37.0V . π 62 − 1 π 62 − 1

(

)

(

)

(

)

(

)

Load impedance for this harmonic

Z6 = R 2 + (6ωL) 2 = 252 + (377 ⋅ 0.05) 2 = 116Ohm . The first AC (6x) harmonic of the load current

I6 =

V6 37.0 = = 0.32 A. Z 6 116

L 50m 1 =3 = 6ms is larger than the output voltage period = 2.8ms , R 25 6 ⋅ 60 therefore, the load current can be considered pure DC. Three time constants 3

(c) Average diode current

I D ,DC =

I DC 25.9 = = 8.63 A; 3 3

(d) RMS diode current

I D,RMS = 3I D ,DC =

I DC 25.9 = = 15.0 A. 3 3

(e) The RMS source (phase) current

I A,RMS ≈

2 2 I DC = 25.9 = 21.1A. 3 3

(f) Active and apparent power

P = I 2 RMS R = 25.92 ⋅ 25 = 16.8kW ; S = 3VRMS I A,RMS = 3VRMSL I A,RMS = 3 ⋅ 480 ⋅ 21.2 = 17.6kVA. (g) Power Factor

PF =

P 16.8 = = 0.955. S 17.6

Tutorial 8 – Controlled Three-Phase Rectifiers Single-Phase AC-AC Controllers Don’t use more than 3 significant digits – 1.23 A, 0. 456 W, 78.9 V


∫ SIN (τ − θ )dτ = −COS (τ − θ ) |α α

β

β

⎛ τ⎞

⎛ τ⎞

∫ exp⎜⎝ − T ⎟⎠dτ = −T exp⎜⎝ − T ⎟⎠ |α α β

∫ SIN α

2

β


β

1

1


β

β

TSIN (τ − θ ) + T 2COS (τ − θ ) ⎛ τ⎞ ⎛ τ⎞ exp⎜ − ⎟ |αβ 2 ∫α SIN (τ − θ ) exp⎜⎝ − T ⎟⎠dτ = − 1+ T ⎝ T⎠

P8.1. For a three-phase controlled full-wave rectifier with active load R=100 Ohm, f =50Hz, and lineto-line RMS voltage Vrms,L=415 V. For the delay angles α = 45 and 90 el. degrees determine: (a) average load voltage; (b) load power. Solution. For 0 < α < 60 o

VO , AVG =

3VmL

π

COS (α ) =

3 2VRMSL

π

COS (α ).

For α = 45 o

VO , AVG =

3 2VRMSL

π

COS 45O =

For 60 o < α < 120o

3 2 ⋅ 415 2 = 396V . π 2

VO , AVG =

3VmL ⎡ π ⎞⎤ 3 2VRMSL ⎛ 1 + COS ⎜ α + ⎟⎥ = ⎢ π ⎣ 3 ⎠⎦ π ⎝

⎡ π ⎞⎤ ⎛ ⎢1 + COS ⎜ α + 3 ⎟⎥. ⎝ ⎠⎦ ⎣

For α = 90 o

VO , AVG =

3 2VRMSL ⎡ π ⎞⎤ 3 2 ⋅ 415 ⎡ ⎛ ⎛ π π ⎞⎤ 1 + COS ⎜ α + ⎟⎥ = 1 + COS ⎜ + ⎟⎥ = 75.5V . ⎢ ⎢ π 3 ⎠⎦ π ⎝ ⎝ 2 3 ⎠⎦ ⎣ ⎣

For 0 < α < 60 o 2 3Vm2 2VRMSL P= 2π + 3 3COS (2α ) = 2π + 3 3COS (2α ) . 4πR 4πR

[

]

[

]

For α = 45 o

P=

2 ⋅ 4152 2π + 3 3COS 2 ⋅ 45O = 1722W . 4π ⋅ 100

[

)]

(

For 60 o < α < 120o

P=

2 3Vm2 [4π − 6α − 3SIN (2α − π / 3)] = 2VRMSL [4π − 6α − 3SIN (2α − π / 3)]. 4πR 4πR

For α = 90 o

2 ⋅ 4152 [4π − 6 ⋅1.571 − 3SIN (π − π / 3)] = 149W . P= 4π ⋅ 100

P8.2. For a three-phase controlled full-wave rectifier with RL-load R=10 Ohm, f =50Hz, line-to-line RMS voltage Vrms,L=240 V and smoothing inductance is sufficiently large. For the delay angles α = 45 and 90 el. degrees, determine: (a) average load voltage; (b) load power; (c) power factor; (d) RMS current of SCR. Solution. For 0 < α < 90 o

VO , AVG =

P=

3VmL

π

COS (α ) =

3 2VRMSL

π

COS (α );

2 27Vm2 18VRMSL 2 ( ) COS α = COS 2 (α ); 2 2 π R π R

PF =

3COS (α )

π

;

3 3Vm 3 2VRMSL COS (α ) = COS (α ); πR πR I 6VRMSL = O ,RMS = COS (α ). πR 3

I O ,DC ≈ I O ,RMS = I SCR ,RMS

For α = 45 o

VO , AVG =

3 2VRMSL

π

COS 45O =

3 2 ⋅ 240 2 = 229V . π 2

2 18 ⋅ 2402 1 18VRMSL 2 O P= COS 45 = = 5250W . π 2R π 210 2

( )

PF =

3COS (α )

π

I SCR ,RMS =

=

( )= 3

3COS 45O

2

2π

π

= 0.675.

6 ⋅ VRMSL 6 ⋅ 240 2 COS 45O = = 13.2 A. πR π ⋅ 10 2

( )

For α = 90 o , average load voltage, load power, power factor, and RMS current of SCR all equal zero.

P8.3. For the single-phase AC controller with active load R=15 Ohm, f =60Hz, and Vrms=120 V. Determine: (a) the firing angle α required to deliver 500W to the load current; (b) voltage source RMS current; (c) the RMS and average currents of SCRs; (d) the power factor; (e) the total harmonic distortion (THD) of the source current. Solution.

VO2,RMS P= . R The required output squared RMS voltage

VO ,RMS = PR = 500 ⋅ 15 = 86.6V . Vm α SIN (2α ) 1− + ; π 2π 2 2 VRMS α SIN (2α ) α SIN (2α ) 7500 α SIN (2α ) − −1+ 2 = − −1+ = − − 0.479 = 0. π 2π VO ,RMS π 2π 1202 π 2π

VO ,RMS =

From numerical solution of the above equation α = 1.54 rad = 88.1O. Source RMS current

I O ,RMS =

VO ,RMS 86.6 = = 5.77 A. R 15

SCR RMS current

I SCR ,RMS =

I O , AVG =

I O ,RMS 2

=

5.77 = 4.08 A. 2

VO , AVG 2VRMS (1 + COSα ). = R πR

SCR average current

I SCR , AVG =

I O , AVG 2VRMS (1 + COSα ) = 2 ⋅ 120 1 + COS 88.1O = 1.86 A. = 2 2πR 2π ⋅ 15

(

Power factor

PF =

P VO ,RMS α SIN (2α ) = = 1− + ; S VRMS π 2π

PF =

VO ,RMS 86.6 = = 0.72. VRMS 120

Fundamental harmonic calculation

V1O = a1COS (ωt ) + b1SIN (ωt ) = V1m SIN (ωt − ϕ1 ); 2VRMS [COS (2α ) − 1] = −54.0V ; 2π 2VRMS [SIN (2α ) + 2(π − α )] = 88.2V . b1 = 2π a1 =

V1m = 54.0 2 + 88.2 2 = 103.4V ; I1,RMS =

THDI =

V1m 103.4 = = 4.89 A. 2R 215 2 I RMS − I12,RMS

I1,RMS

5.77 2 − 4.89 2 = = 0.63 = 63%. 4.89

)

P8.4. For the single-phase AC controller with RL-load R=20 Ohm, L=50 mH, f =60Hz, Vrms=120 V, and the the firing angle α =90 el.deg. Determine: (a) the load current expression for the first half-period; (b) the load RMS current; (c) the SCR RMS current; (d) the SCR average current; (e) the output (load) power; (f) the power factor (PF) Hint: from numerical solution, β equals 3.83 rad or 220 el. degree. Solution. Voltage magnitude Vm = 2VRMS = 170V . Angular frequency ω = 2πf = 6.28 ⋅ 60 = 377rad / s. Impedance Z = R 2 + (ωL) 2 = 20 2 + (377 ⋅ 0.05) 2 = 27.5Ohm . Normalized time constant ωτ =

ωL R

=

377 ⋅ 0.05 = 0.942rad . 20


α = 90 O = 1.57 rad . Vm Z

⎡ ⎛ ωt − α ⎞⎤ ⎡ (ωt − 1.57 ) ⎤ ⎢ SIN (ωt − θ ) + SIN (θ − α ) exp⎜ − ωτ ⎟⎥ = 6.18 SIN (ωt − 0.756) − 4.49 exp⎢− 0.942 ⎥, ⎝ ⎠⎦ ⎣ ⎦ ⎣ α ≤ ωt ≤ β . i (ωt ) =

From numerical solution β equals 3.83 rad.

RMS current square

I

2

3.83

⎧ (ωt − 1.57 )⎤ ⎫ d (ωt ) = ⎡ = ∫ ⎨6.18 SIN (ωt − 0.756) − 4.49 ⎢exp− ⎬ π 1.57 ⎩ 0.942 ⎥⎦ ⎭ ⎣ 1

2 RMS

3.83

⎡ (ωt − 1.57 )⎤ + 20.2⎡exp− (ωt − 1.57 )⎤ ⎤d (ωt ) = ⎡ 38.2SIN 2 (ωt − 0.756) − 55.5SIN (ωt − 0.756) ⎢exp− ⎢ ⎢⎣ π 1.57 ⎣ 0.942 ⎥⎦ 0.471 ⎥⎦ ⎥⎦ ⎣ 1 = (53.9 − 40.3 + 9.4) = 7.34 A2 . 1

=

∫

π

I RMS = 7.34 = 2.71A. SCR RMS current

I SCR ,RMS =

I RMS 2.71 = = 1.92 A. 2 2

SCR average current

I SCR , AVG

1 = 2π

3.83

⎧

⎡

∫ ⎨⎩6.18SIN (ωt − 0.756) − 4.49 ⎢⎣exp−

(ωt − 1.57 )⎤ ⎫d (ωt ) = 0.942

1.57

⎥⎦ ⎬ ⎭

⎡ (ωt − 1.57 )⎤ 3.83 ⎤ ⎡ 3.83 ⎢− 6.18COS (ωt − 0.756) |1.57 −4.49 ⋅ 0.942⎢exp− 0.942 ⎥ |1.57 ⎥ = ⎣ ⎦ ⎣ ⎦

=

1 2π

=

1 (10.4 − 3.85) = 1.04 A. 2π

2 Load power P = I RMS R = 7.34 ⋅ 20 = 147W .


PS = = = =

1

π 1

π 1

π

1

π

β

∫α V

S

(ωt )i (ωt )d (ωt ) =

3.83

⎧

⎡

∫ 169.7SIN (ωt )⎨⎩6.18SIN (ωt − 0.756) − 4.49 ⎢⎣exp−

1.57

3.83

⎧

⎡

(ωt − 1.57 )⎤ ⎫d (ωt ) = 0.942

∫ ⎨⎩10.49 SIN (ωt )SIN (ωt − 0.756) − 762SIN (ωt )⎢⎣exp−

1.57

(888.6 − 437.8) = 147W .

PS = P. Power factor PF =

P P 147 = = = 0.45. S VS ,RMS I RMS 120 ⋅ 2.71

⎥⎦ ⎬ ⎭

(ωt − 1.57 )⎤ ⎫d (ωt ) = 0.942

⎥⎦ ⎬ ⎭

Tutorial 9 – DC-DC Converters – Buck Converter Don’t use more than 3 significant digits – 1.23 A, 0. 456 W, 78.9 V

P9.1. For buck converter Vs=50V, D=0.4, L=400 uH, C=100uF, f =20kHz, and R=20 Ohm. Assuming ideal component, determine: (a) the output voltage; (b) the maximum and minimum inductor current; (c) the output voltage peak-to-peak ripple. Solution. Suppose CCM. Then output voltage

VO = DVS = 50 ⋅ 0.4 = 20V . Maximum and minimum inductor currents

⎛ 1 1− D ⎞ 1 − 0.4 ⎛ 1 ⎞ ⎟⎟ = 20⎜ + I MAX = VO ⎜⎜ + ⎟ = 1.75 A; ⎝ 20 2 ⋅ 0.0004 ⋅ 20000 ⎠ ⎝ R 2 Lf ⎠ ⎛ 1 1− D ⎞ 1 − 0.4 ⎛ 1 ⎞ ⎟⎟ = 20⎜ − I MIN = VO ⎜⎜ − ⎟ = 0.25 A. ⎝ 20 2 ⋅ 0.0004 ⋅ 20000 ⎠ ⎝ R 2 Lf ⎠ For instance, average inductor current equal to load current is 1A, peak-to-peak inductor current – 1.5A Minimum inductor current is positive that confirms CCM. Output voltage ripple

ΔVO 1− D 1 − 0.4 = = = 0.0047 = 0.47%. 2 VO 8LCf 8 ⋅ 0.0004 ⋅ 0.0001 ⋅ (20000) 2

P9.2. Design a buck converter to produce an output voltage of Vo=18V for load resistor R=10 Ohm with voltage ripple (ΔVO / VO )=0.5%, Vs=48V, switching frequency f=40kHz. Take inductance 1.25 times minimal inductance required for CCM. Solution. Determine: (a) duty cycle D; (b) required inductance; (c) inductor minimum, maximum and RMS currents; (d) filter capacitance; (e) peak and RMS capacitor currents. Duty cycle

D=

VO 18 = = 0.375 . VS 48

Minimum inductance

LMIN =

(1 − D) R (1 − 0.375)10 = = 78uH . (microHenry) 2f 2 ⋅ 40000

With 25% margin

L = 1.25 ⋅ LMIN = 1.25 ⋅ 78 ≈ 100uH . Average inductor (load) current and current ripple magnitude

IL = IR = ΔiL =

VO 18 = = 1.8 A . R 10

VO 18 (1 − D)T = (1 − 0.375) ⋅ (40000) −1 = 2.80 A L 0.0001

ΔiL 2.8 = 1.8 + = 3.2 A; 2 2 Δi 2 .8 = I L − L = 1.8 − = 0.4 A. 2 2

I MAX = I L + I MIN

RMS inductor current

I L ,RMS =

2 2 I MAX + I MAX I MIN + I MIN 3.2 2 + 3.2 ⋅ 0.4 + 0.4 2 = = 1.97 A. 3 3

Capacitor is selected for (ΔVO / VO )=0.5%

C=

1− D 0.375 = ≈ 0.0001 = 0.1mF . 2 8(ΔVO / VO )Lf 8(0.005) ⋅ 0.0001 ⋅ 40000 2

Peak and RMS capacitor currents

I C ,PEAK = I C ,RMS =

ΔiL 2.8 = = 1.4 A; 2 2 1.4 2 I C2 ,PEAK = = 0.81 A. 3 3

Tutorial 10 – DC-DC Converters – Boost Converter Don’t use more than 3 significant digits – 1.23 A, 0. 456 W, 78.9 V

P10.1. Design boost converter to have Vo=30V, Vs=12V, load resistance R=50 Ohm, f =25kHz and 1% voltage ripple for continuous conduction mode (CCM). Note: select inductance with a 25% margin. Solution. The duty cycle (duty ratio)

VS ; 1− D V 12 D =1− S = 1− = 0.6. VO 30

VO =

Minimum inductance

LMIN

D(1 − D) 2 R 0.6(1 − 0.6) 2 50 = = = 96uH . 2f 2 ⋅ 25000

With 25% margin, minimum inductance becomes L=120uH. Average inductor current

IL =

VS 12 = = 1.5 A. 2 (1 − D) R (1 − 0.6) 2 50

Inductor current ripple magnitude

ΔiL =

VS 12 ⋅ 06 DT = = 2.4 A L 0.00012 ⋅ 25000

Minimum and maximum inductor currents

ΔiL 2.4 = 1.5 + = 2.7 A; 2 2 Δi 2.4 = I L − L = 1.5 − = 0.3 A. 2 2

I MAX = I L + I MIN

Minimum required capacitance

C=

D 0.6 = = 0.000048 = 48uF . R(ΔVO / VO ) f 50(0.01)25000

P10.2. Boost converter has the following parameters: Vs=20V, D=0.6, L=0.1mH, R=50 Ohm, C=0.1mF, f=15kHz. Verify that inductor current is discontinuous (DCM). Determine: (a) the output voltage Vo; (b) the maximum inductor current. Solution. Suppose that inductor current is continuous. Then minimum current

⎡ 1 1 I MIN = VS D ⎢ − 2 ⎣ (1 − D) R 2 Lf

⎤ ⎡ ⎤ 1 1 ⎥ = 20 ⋅ 0.6⎢ (1 − 0.6) 2 50 − 2 ⋅ 0.0001 ⋅ 15000 ⎥ = −1.5 A. ⎦ ⎣ ⎦

Negative minimum current means that the boost converter operates in DCM. Accurate output voltage

VO =

VS 2

2 ⎛ ⎜1 + 1 + 2 D RT ⎜ L ⎝

2 ⎞ 20 ⎛ ⎞ ⎟= ⎜1 + 1 + 2 ⋅ 0.6 ⋅ 50 ⎟ = 60V . ⎟ 2 ⎜ 0.0001 ⋅ 15000 ⎟⎠ ⎠ ⎝

Approximate output voltage

⎛1 ⎛1 ⎞ RT ⎞ 50 ⎟ = 20⎜ + 0.6 ⎟ = 59V . VO ≈ VS ⎜⎜ + D ⎟ ⎜2 ⎟ 2 2 L 2 ⋅ 0 . 0001 ⋅ 15000 ⎝ ⎠ ⎝ ⎠ Compare with boost converter output voltage for CCM –

VO =

VS 20 = = 50V . 1 − D 1 − 0.6

Maximum inductor current

I MAX =

VS DT 20 ⋅ 0.6 = = 8 A. L 0.0001 ⋅ 15000

??? P10.3. Buck-boost converter has the following parameters: Vs=24V, D=0.4, L=0.02mH, R=5 Ohm, C=0.08mF, f=100kHz.

Determine: (a) the output voltage Vo; (b) inductor maximum, minimum, average, and RMS currents; (c) the output voltage ripple (percent). Solution. Output voltage

VO = −

D 0.4 VS = − 24 = −16V . 1− D 1 − 0.4

Average inductor current

IL =

DVS 0.4 ⋅ 24 = = 5.33 A. 2 (1 − D) R (1 − 0.4) 2 5

Peak-to-peak inductor current

ΔiL =

VS 0.4 ⋅ 24 DT = = 4.8 A. L 0.00002 ⋅ 100000

Minimum and maximum inductor currents

ΔiL 4.8 = 5.33 + = 7.33 A; 2 2 Δi 4.8 = I L − L = 5.33 − = 2.93 A. 2 2

I MAX = I L + I MIN

Positive minimum current confirms CCM operation. Output voltage ripple

ΔVO D 0.4 = = = 0.01 = 1%. VO RCf 5 ⋅ 0.00008 ⋅ 100000

Assignment 1 – Power Electronics

Sanzhar Askaruly

Assignment 1_AH Lecture 1 P.1.1. (10 points) The current and voltage of a passive device are periodic with T=100ms (Fig.1).

i, A 70

40 20 0 30

t, ms

v,V

30

80

100

t, ms

0

60

100

− 20 Fig.1. Determine (1) the average power, (2) the energy absorbed in each period. Solution. Instantaneous power: "1650W; 0 < t < 30ms & $ $ $600W; 30 < t < 60ms $ p(t) = # ' $−400W; 60 < t < 80ms$ $%0W; 80 < t < 100ms $(

1. The average power

PAV =

1650 ⋅ 30 + 600 ⋅ 30 − 400 ⋅ 20 = 595W 100

2. The energy absorbed in each period W = PAV T = 595W ⋅ 0.1s = 59.5J


Sanzhar Askaruly

P.1.2. (10 points) Find the average power absorbed by a 12 Vdc voltage source for the current into the positive terminal given in P. 1.1. Solution. The average current

I AV =

55⋅ 30 + 20 ⋅ 50 = 26.5A . 100

The average power (positive because absorbed by the source) PAV = VDC ⋅ I AV = 12V ⋅ 26.5A = 318W .

P.1.3. (10 points) Find the RMS values of the current and voltage waveforms given in P. 1.1. Solution. Voltage RMS

VRMS

30 2 ⋅ 60 + 20 2 ⋅ 40 = = 26.5V . 100

Current RMS

I RMS =

552 ⋅ 30 + 20 2 ⋅ 50 = 33.3A . 100

P.1.4. (10 points) The voltage and current for a passive device are given by v(t) = 5 + 8COS(ω t) − 6COS(2ω t − 20 o ); i(t) = 4 + 3COS(ω t + 45o ) − 2COS(2ω t + 70 o ).

Find (1) the RMS values of voltage and current, (2) the power consumed by the device. Solution. 1a. Voltage RMS

VRMS = 52 + 0.5⋅ (82 + 6 2 ) = 8.66V .

1b. Current RMS

I RMS = 4 2 + 0.5⋅ (32 + 2 2 ) = 4.74A .


Sanzhar Askaruly

2. Average power

P = 5⋅ 4 + 0.5⋅8⋅ 3⋅ COS(−45o ) + 0.5⋅ 6 ⋅ 2 ⋅ COS(−90 o ) = 28.5W

P.1.5. (10 points) A sinusoidal voltage source produces a non-linear load current

v(t ) = 100COS (314t ); i(t ) = 15COS (314t + 60 o ) + 10COS (628t + 45 o ) + 6COS (1256t + 70 o ). Find (1) the power consumed by the load, (2) the distortion factor (DF) of the load current, (3) the power factor (PF), (4) the THD of the load current. Solution. 1. Power

P = 0.5⋅100 ⋅15⋅ COS(60 o ) = 375W

(11)

RMS current value

I RMS = 0.5⋅ (152 +10 2 + 6 2 ) = 13.4A .

(12)

2. Load current distortion factor DF =

I1, rms 15 = = 0.792 . Irms 2 ⋅13.4

(13)

3. Power Factor

PF = DF ⋅ COS(60 o ) = 0.792 ⋅ 0.5 = 0.396 .

(14)

4. Load current THD THD =

1 1 −1 = −1 = 0.771 . 2 DF 0.792 2

(15)


Sanzhar Askaruly

Lecture 3 Useful integrals: β

∫α SIN (τ − θ )dτ = −COS (τ − θ ) |α

β

β

⎛ τ ⎞

⎛ τ ⎞

∫ exp⎜⎝ − T ⎟⎠dτ = −T exp⎜⎝ − T ⎟⎠ |α α β

∫α SIN

2

β


β

1

1


β

β

TSIN (τ − θ ) + T 2COS (τ − θ ) ⎛ τ ⎞ ⎛ τ ⎞ exp⎜ − ⎟ |αβ 2 ∫α SIN (τ − θ ) exp⎜⎝ − T ⎟⎠dτ = − 1+ T ⎝ T ⎠

Make analytical calculations for Lecture 3 problems. For problems 3.2-3.3, compare analytical results with those obtained from PSIM simulations. Include in your report some PSIM screenshots and data (measurements) to substantiate your results. P3.1. (10 points) For the half-wave rectifier with active load Vrms=210 V, frequency f=50 Hz, load resistor R=10 Ohm. Find: (1) average load current; (2) average load power; (3) the power factor (PF). Solution. Voltage magnitude Vm = 2VRMS = 297V . Average voltage VO , AVG = Vm / π . Average current I DC =

Vm 297 = = 9.45A . π R 10π

Output RMS voltage VO,RMS =

V 2 2972 Vm = 2205W . = 149V , average load power P = m = 4R 4 ⋅10 2

Resistor and source RMS current I RMS =

Vm 297 = = 14.9A . 2R 2 ⋅10

Apparent power S = VS,RMS I RMS = 210 ⋅14.9 = 3129VA.


Power factor PF =

Sanzhar Askaruly

2205 ≈ 0.705. 3129

P3.2. (20 points) For the half-wave rectifier with RL-load R=50 Ohm, L=0.2 H, f =50Hz, and Vrms=100 V. Find: (1) current expression; (2) average current; (3) RMS current; (4) load power; (5) power factor. Hint: β from numerical solution can be found from the graph (Fig.2).

Fig.2. Numerical solution for BETA Solution. VS

π

β

π

β

2π

π

β

2π

2π

ωt

vO

iO

VD

ωt

ωt

Fig.3. Voltage and current graphs Angular frequency ω = 2π f = 6.28⋅ 50 = 314rad / s. Impedance Z = R 2 + (ω L)2 = 50 2 + (314 ⋅ 0.2)2 = 80.3Ohm .


Sanzhar Askaruly

"ωL % −1 " 62.8 % Phase angle θ = TAN −1 $ ' = TAN $ ' = 0.898rad. # R & # 50 &

SINθ =

ω L 62.8 = = 0.782. Z 80.3

Time constant ωτ =

ω L 62.8 = = 1.26rad R 50

1) Current equation i(ωt ) =

Vm Z

⎡ ⎛ ωt ⎢SIN (ωt − θ ) + SINθ exp⎜ − ωτ ⎝ ⎣

⎞⎤ ⎟⎥, 0 ≤ ωt ≤ β . ⎠⎦

" ωt % i(ω t) = 1.76SIN(ω t − 0.898) +1.37exp $ − ', A, 0 ≤ ω t ≤ β. # 1.26 & At

ωτ = 1.26rad. from Figure 2, approximately β ~ 4.0rad .

A simpler way to calculate 2) Average current is to use average voltage: Vm V (1− COSβ ) = m "#1+ COS ( β − π )$%; 2π 2π V 141 " = m "#1+ COS ( β − π )$% = #1+ COS ( 4.0 − π )$% = 0.742A. 2π R 2π ⋅ 50

VO,DC = I DC


∫α SIN

2


β

1

1


β

β

TSIN (τ − θ ) + T 2COS (τ − θ ) ⎛ τ ⎞ ⎛ τ ⎞ SIN ( τ − θ ) exp − d τ = − exp⎜ − ⎟ |αβ ⎜ ⎟ 2 ∫α 1+ T ⎝ T ⎠ ⎝ T ⎠ 3) RMS current is found from: 4.0

2

( " ω t %+ ∫ 1.76 2 *)SIN(ωt − 0.898) + SIN(0.898)exp $#− 1.26 '&-, d(ωt) = 0 4.0 ( " ωt % " 1 ω t %+ 2 = 3.1*SIN 2 (ω t − 0.898) + 2SIN(0.898)SIN(ω t − 0.898)exp $ − ' + SIN(0.898) exp $ −2 '-d(ω t) = ∫ # 1.26 & # 0.63 &, 2π 0 ) 2 I RMS =

1 2π

= 1.09A 2 .


Sanzhar Askaruly

I RMS = 1.09 = 1.044A.

(4) Load power 2 P = I RMS R = 54.5W.

Power supplied by the source 5) Power factor PF =

P P 54.5 = = = 0.522. S VS,RMS I RMS 100 ⋅1.044

Practical Part According to PSsim software simulation, the following results were obtained. The graphs is shown below: 𝐼!" = 1.2 𝐴 𝐼!"# = 1.38 𝐴 𝑃 = 118 𝑊 𝑃𝐹 = 0.854

Fig.5 Vsource, Vload and I simulation. P3.3. (20 points) For the half-wave rectifier with RL-load and clamping diode R=5 Ohm, L=60 mH, f=50Hz, and Vm=170 V. Determine: (1) average load voltage; (2) average load current; (3) RMS current; (4) resistor power. Hint: use frequency domain (DC + harmonics 1, 2, 4, 6) and time domain analysis and compare the results of both.


Sanzhar Askaruly

VS

π

2π

vO i O i20 i10

iD

π

2π

iD 1

π

2π

i20 i10

ωt

ωt ωt

i20 i10

VD

ωt 2π

π VD

ωt

V D1

Fig.5 Voltage and current graphs Solution. 1) Average load voltage: VO,AVG =

Vm 170 = = 54.1V. π π

2) Average current: I DC =

Vm 170 = = 10.8A. π R 5π

Frequency domain solution. Voltage harmonic magnitudes

Vm = 85V; 2 2V V2 = 2 m = 36.1V; (2 −1)π 2V V4 = 2 m = 7.22V; (4 −1)π 2V V6 = 2 m = 3.09V... (6 −1)π V1 =

Angular frequency ω = 2π f = 6.28⋅ 50 = 314rad / s. Harmonic impedances


Sanzhar Askaruly

Z1 = 52 + (314 ⋅ 0.06)2 = 19.49Ohm; Z 2 = 52 + (2 ⋅ 314 ⋅ 0.06)2 = 38.01Ohm; Z 4 = 52 + (4 ⋅ 314 ⋅ 0.06)2 = 75.53Ohm; Z 6 = 52 + (6 ⋅ 314 ⋅ 0.06)2 = 113.15Ohm... Current harmonics

I1 = V1 / Z1 = 4.36A; I 2 = V2 / Z 2 = 0.950A; I 4 = V4 / Z 4 = 0.0956A; I 6 = V6 / Z 6 = 0.0273A... 3) RMS current: 2 I RMS ≈ IO,AVG +

I12 I 22 I 42 I 62 + + + = 11.3A. 2 2 2 2

4) Resistor power: 2 P = I RMS R = 638W.

Time domain solution. Current expression

Vm ⎡V ⎤ ⎛ ωt ⎞ SIN (ωt − θ ) + ⎢ m SIN (θ ) + i10 ⎥ exp⎜ − ⎟; 0 ≤ ωt ≤ π ; Z ⎝ ωτ ⎠ ⎣ Z ⎦ ⎡ (ωt − π ) ⎤ i (ωt ) = i20 exp⎢− ; π ≤ ωt ≤ 2π . ωτ ⎥⎦ ⎣ i (ωt ) =

" π % 1+ exp $ − ' Vm 67.183 # ωτ & i10 = SIN(θ ) = = 3.45; "π % " π % Z Z exp $ ' − exp $ − ' # ωτ & # ωτ & "π % exp $ ' +1 Vm 154.58 # ωτ & i20 = SIN(θ ) = = 7.93. "π % " π % Z Z exp $ ' − exp $ − ' # ωτ & # ωτ & 𝜃 = tan!! (𝑤𝐿/ 𝑅) = 0.54 𝑟𝑎𝑑𝑖𝑎𝑛𝑠; 𝜏 = 𝑍=

𝐿 = 0.012; 𝑤 = 2𝜋𝑓 = 314; 𝑤𝜏 = 3.768; 𝑅

𝑅! + 𝑤𝐿

!

= 19.5


Sanzhar Askaruly

" ωt % i(ω t) = 19.5SIN(ω t − 0.54) +13.47exp $ − '; 0 ≤ ω t ≤ π ; # 3.768 & ) (ω t − π ) , i(ω t) = 7.93exp +− ; π ≤ ω t ≤ 2π . * 3.768 .-

I

2 RMS

+

1 2π

1 = 2π 2π

2

π

( " ω t %+ ∫ *)19.5SIN(ωt − 0.54) +13.47exp $#− 3.768 '&-, d(ωt) + 0

/

( (ω t − π ) +2 2 -,3d(ω t) = 127.24A . 3.768 4

∫ 017.93exp*)− π

3) RMS current: I RMS = 127.24 = 11.28A.

4) Resistor power: 2 P = I RMS R = 636W.

Frequency and time domain results are matched well. Practical Part According to PSsim software simulation, the following results were obtained. The graphs is shown below: 1) Average load voltage: 64.9 V 2) Average current: 7.146 A 3) RMS current: 8.522 A 4) Resistor power: 801 W

Fig.6 Vsource, Vload and I simulation.

Final Exam Preparations - Power Electronics

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