Electrical Technology Grade 12

LEARNER’S GUIDE

GRADE 12

ELECTRICAL TECHNOLOGY Trevor Adams Steve Mitchell

CAPS

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Electrical Technology Grade 12 Learner’s Guide

SAMPLE COPY

© Future Managers 2013 All rights reserved. No part of this book may be reproduced in any form, electronic, mechanical, photocopying, or otherwise, without prior permission of the copyright owner. ISBN 978-1-77581-012-4 First published 2013 To copy any part of this publication, you may contact DALRO for information and copyright clearance. Any unauthorised copying could lead to civil liability and/or criminal sanctions.

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Contents Chapter 1: Occupational health and safety Dangerous practices....................................................................................................................................................2 Risk analysis......................................................................................................................................................................3 Human rights in the workplace..........................................................................................................................4 Work ethics........................................................................................................................................................................5 Bleeding...............................................................................................................................................................................8 Chapter 2: Three-phase AC generation Advantages and disadvantages of single-vs. three-phase systems......................................... 12 Single phase generation......................................................................................................................................... 15 Three-phase generation......................................................................................................................................... 16 Three-phase systems: star-vs.-delta (delta-vs.-star)......................................................................... 18 Power in three-phase systems and calculations.................................................................................. 22 Losses.................................................................................................................................................................................. 26 Concept of power factor correction............................................................................................................. 27 kWh meter...................................................................................................................................................................... 30 Measurement of power in three-phase systems.................................................................................. 35 Chapter 3: Three-phase transformers Three-phase transformers................................................................................................................................... 42 How to connect a three-phase transformer in star or delta....................................................... 45 Type of transformers............................................................................................................................................... 46 Transformers (star/star, delta/delta, star/delta, delta/star ......................................................... 51 Calculations (efficiency at 100%)................................................................................................................... 54 Application of transformers............................................................................................................................... 57 Safety and transformers........................................................................................................................................ 58 Chapter 4: Three-phase motors and starters Principle of operation of the three-phase squirrel cage induction motor....................... 66 Connections of motor in star or delta........................................................................................................ 71 Calculations on synchronous speed, slip, power and efficiency............................................. 72 Electrical and mechanical inspections/fault finding....................................................................... 75 DOL starter with overload.................................................................................................................................. 81 Forward/reverse starter with overload...................................................................................................... 86 Automatic star-delta starter with overload............................................................................................. 89 Sequence motor control starter with overload (without timer)............................................. 92 Sequence motor control starter with overload (with timer)..................................................... 94 Chapter 5: RCL circuits Inductive reactance............................................................................................................................................... 106 Power............................................................................................................................................................................... 108 How to make sense of series and parallel circuits.......................................................................... 111 Calculations of series RCL circuits............................................................................................................ 115 Parallel RCL circuits............................................................................................................................................. 118 Calculations of parallel RCL circuits........................................................................................................ 124

Chapter 6: Logic Main components of a PLC............................................................................................................................. 141 Basic operation of PLC....................................................................................................................................... 143 Basic ladder logic instructions...................................................................................................................... 145 Ladder logic diagrams......................................................................................................................................... 146 Logic gates and its ladder equivalents..................................................................................................... 149 Converting Boolean expressions to ladder diagrams.................................................................. 151 Karnaugh maps........................................................................................................................................................ 154 Combination logic networks.......................................................................................................................... 162 Motor starter control: a simple approach............................................................................................. 166 Chapter 7: Amplifiers Characteristics of the ideal operational amplifier.......................................................................... 180 Principle of operation of negative/positive feedback of operational amplifiers....... 183 OP-amp as comparator..................................................................................................184 OP-amp as inverting amplifier...................................................................................................................... 188 OP-amp as a non-inverting amplifier...................................................................................................... 190 OP-amp used as a summing amplifier.................................................................................................... 193 OP-amp used as an integrator....................................................................................................................... 196 OP-amp used as differentiator...................................................................................................................... 198 Bi-stable multivibrator........................................................................................................................................ 201 Mono-stable multivibrator.............................................................................................................................. 203 Astable multivibrator........................................................................................................................................... 205 The Schmitt trigger................................................................................................................................................ 208 Operational amplifier oscillators................................................................................................................. 211 Icon

Description Key word

Did you know?

Take note

Activity

Case study

Chapter 1 Occupational health and safety

A

B OHS Act

A

Medical emergencies

Human rights B Electrical shock

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Electrical Technology

Introduction Safety and first aid are the two most important things that must always be high on the priority list of any person working in any workshop, big or small and on any construction site. Being aware of safety and knowing some basic first aid can be very helpful and can even save a life. We must remember, safety is the responsibility of every single person, not just of the employer or factory owner. Most accidents are caused by the careless acts of humans. Because electricity is very dangerous, care must be taken to be safety conscious all the time when busy in a workshop. In South Africa, we have the Occupational Health and Safety Act (OHS Act) that regulates workplace safety in general in our country. It is the aim of the Act to eliminate or reduce work-related accidents. It also tries to ensure that workers have a safe environment in which to work. According to the Act, an accident is an unplanned, uncontrolled event caused by unsafe acts and conditions. The Act compels the employer to ensure the workplace meets safety standards by appointing people and committees whose task it is to monitor their safety in the workplace. The employers can be punished and fined if machinery or working conditions are found to be unsafe. Hundreds of accidents occur annually on our roads, in mines, in the workplace and on construction sites, not only because of unsafe working conditions but also because people become complacent, act negligently, lose concentration or are too tired or too hasty. To ensure safety in the workplace means that every single person has to be aware of potential dangers at all times. Learners have to acquire the necessary knowledge and skills to enable them to prevent accidents from happening. In general, safety is understood to be the absence of danger or risk. However, accidents occur when workers are too complacent and then take risks because they think nothing will happen to them.

Unsafe actions As mentioned earlier, most accidents are caused by human carelessness. However, accidents also occur when workers are too complacent and then take risks because they think nothing will happen to them. Below are some unsafe actions responsible for most accidents in workshops. • Failure to wear protective clothing and eye wear – when grinding, drilling or working with acid and chemicals. • The unsafe placement of tools. • Horseplay in workshop – running around and playing the fool. • The unsafe use of equipment or incorrect use of equipment. • Trying to make adjustments or working on moving equipment. • Taking up unsafe positions. • Working too quickly.

Dangerous practices

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Dangerous practices most commonly refer to processes or activities that have some form of risk or hazard when performed. In an electrical workshop, many activities or tasks done can be referred to as dangerous practices. If the user of the workshop is not cautious of such situations or practices, they could lead to serious injury or even death. Below are a few such dangerous practices.

Occupational health and safety

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The etching of printed circuit boards (PCBs) by making use of chemicals and acids. The chemicals can be harmful as they could damage your clothes or even be dangerous to your skin. It is therefore important for anybody involved in the etching of PCBs to observe the safety rules, i.e. wear protective clothing, for example, rubber aprons, rubber gloves, respiratory masks and eye protection. Another dangerous practice is the use of power tools (electric hand drill, bench and hand-held grinding machines). Learners are sometime very reluctant to observe all the safety rules regarding the use of power tools, which can lead to serious injury. Drill bits that are not securely tightened, using your hand to hold the piece of material to be drilled, not wearing protective eye wear and protective clothing, not tucking away all loose-hanging clothes and long hair, not inspecting the electrical cords of the tools, can all lead to unnecessary accidents. Incorrect use of hand tools is also a very dangerous practice in a workshop. Tools must be fully functional and must be used for the purpose for which they were made. Under no circumstances should anyone use a file without a handle or with a broken handle. Never use a screwdriver as a lever or a chisel. It is important that you select the correct size screwdriver for a job, failing to use the correct size will result in the tip being damaged, making it less effective. When you use any pliers or cutters, please ensure the handle is insulated to prevent electrical shocks. Always ensure the blade of a blade knife is retracted when not in use to prevent anyone from being cut.

Unsafe conditions Unsafe conditions are also a big contributor to many accidental mishaps in the work-place. Unsafe conditions refer to a hazardous, risky or dangerous environment or surroundings that can lead to accidents. Let us look at a few unsafe conditions. • • • • • • • • •

Inadequate guarding of machines. Bad ventilation. Rough, wet or slippery floors. No personal protective equipment. A disorganised workshop. Overcrowding in a workshop. Badly planned workshop. Loose-hanging clothing and long hair. Insufficient light in workplace.

Risk analysis Risk analysis is the process of defining and analysing the dangers to individuals and businesses posed by potential natural and human-caused adverse events. A risk analysis report can be either quantitative or qualitative. With a quantitative risk analysis, an attempt is made to determine numerically the probabilities of various adverse events and the likely extent of the losses if a particular event were to take place. On the other hand, a qualitative risk analysis does not involve numerical probabilities or predictions of loss. Instead, the qualitative method involves defining the various threats, determining the extent of vulnerabilities and generating countermeasures should something unforeseen occur.

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Electrical Technology It is therefore important for any employer to do a risk assessment in the workplace to make sure that no one gets hurt or becomes ill – that a person returns home safely after work. Risk management is nothing more than a careful examination of what could cause harm to people in your workplace (a qualitative risk analysis). In many instances, straightforward measures can readily control risks, for example, ensuring spillages are cleaned up promptly so people do not slip, or cupboard drawers kept closed to ensure people do not trip. For most, that means simple, cheap and effective measures to ensure that your most valuable asset – your workforce – is protected. The following things must be kept in mind when doing risk management: • involve workers in the process • don’t use it to justify a decision that has already been made • consider good practice in your industry • keep records of any risk management activities undertaken. Risk management is a five step process for controlling exposure to health and safety risks associated with hazards in the workplace. • Identify the hazard. • Decide who might be harmed and how. • Evaluate the risk and decide on precautions . • Record your findings and implement them . • Review your assessment and update if necessary. When thinking about your risk assessment, remember: • a hazard is anything that may cause harm, such as chemicals, electricity, working from ladders, an open drawer, etc; and • the risk is the chance, high or low, that somebody could be harmed by these and other hazards, together with an indication of how serious the harm could be.

Human rights in the workplace Chapter 2 of the South African Constitution contains the Bill of Rights, which contains the human rights which protect all South Africans. Human rights are also called natural rights. It is argued that they belong to people just because they are human beings. People are entitled to them regardless of where they live in the world or of their position in society. It doesn’t matter what a person’s race, sex, age, class, language, beliefs, culture or religion are, or how much money or education a person has, we all have the same human rights. Most people would support human rights that are based on basic values, such as respect for human life and human dignity. Many of the principles of the Human Rights Act are designed to protect you as a worker within the workplace. It is about protecting the human dignity of the workers. Human rights are about ensuring that your human dignity is not infringed and that you, as the worker, are treated with dignity, respect and are not exploited. For example, you have the right to a private and family life. So an employer who discriminates against a gay worker, for example, may be violating that worker’s right to a private life.

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Your employer has the right to monitor communications within the workplace as long as you are aware of the monitoring before it takes place. Monitoring can cover: • e-mails • Internet access • telephone calls • data • images. You have the right to see any information held about you (for example, e-mails or CCTV footage). Your right to a private life means you have the right to some privacy in the workplace. You can’t be monitored everywhere. If your employer doesn’t respect this, he/she will be breaching human rights law. Human dignity at work is also about making sure that working conditions for workers are good, i.e. good ventilation, enough light, enough space to work, an environment that is not harmful to their health or well-being. Other issues that are also covered by human rights are: • Your right to not to be discriminated against because of your race, sex, religion, language, disability, etc. • Your right to earn a living wage. • Your right to work reasonable hours. • Your right to belong to a trade union. • No one may be subjected to slavery or forced labour. • Everyone has the right to fair labour practices.

Work ethics Work ethics include not only how one feels about your job, career or occupation, but also how one performs your job or responsibilities. This involves attitude, behaviour, respect, communication, and interaction; how one gets along with others. Work ethics demonstrate many things about the person and who he/she is and how he/she behaves. Work ethics involve such characteristics as honesty and accountability. Essentially, work ethics break down to what one does or would do in a particular situation. Work ethics are intrinsic; they come from within. Steps towards better work ethics 1. Attendance Attendance and punctuality often have a large impact on individual and team success. Lateness or absenteeism can also greatly impact job performance and retention.

How you can maintain good attendance: • Make work a high priority. • Know your schedule. • Make use of an alarm clock. • Get enough sleep. • Arrange your transportation. • Inform your supervisor of an absence.

2. Character An employer expects employees to work together toward achieving the objectives of the company. The wise employee who is interested in having a good relationship with an employer will try to help the employer achieve success. The

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employer expects employees to develop certain desirable traits that will help them to perform their jobs well so that the company can succeed. Some of these traits include: • loyalty • honesty • trustworthiness • dependability • reliability • initiative • self-discipline • self-responsibility.

3. Teamwork It is vital that employees work as a team. It is important not only to their personal success and advancement, but also to that of their co-workers and to the company. Teamwork involves the following aspects: • Respecting the rights of others. • Being a team worker. • Being co-operative. • Being assertive. • Displaying a positive customer service attitude. • Seeking opportunities for continuous learning. • Demonstrating polite behaviour. • Respecting confidentiality. 4. Appearance The first impression of a person is generally created within three seconds. If you appear untidy and your clothes are creased, you may give the impression that your work is sloppy. If you dress as a professional, the first impression you give will be excellent. 5. Attitude It is very important to demonstrate a positive attitude, appear self-confident, and have realistic expectations for yourself. Developing and maintaining a positive attitude involves setting realistic expectations for ourselves at work. These goals should be challenging, but obtainable. 6. Productivity In order to be a productive employee, a person must follow safety procedures, conserve materials, keep the work area neat and clean and follow directions properly. 7. Organisational skills Employers consider effective time management and organisational skills as good work habits.

Medical emergencies Medical emergencies are injuries, or an illness that is severe and poses a serious risk to a person’s life or long-term health. The treatment of medical emergencies would require the services of a doctor, nurse or any other person with the necessary medical/first aid qualifications. We must also note that not all medical emergencies are life-threatening but only require medical attention to prevent any future medical problems.

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Medical emergencies in the workplace are seldom expected and very rarely properly planned for. Should there be a medical emergency where you work, your level of preparedness could mean the difference between life and death. It is therefore very important that any workshop has a well-trained first aider with a well-stocked first aid box. It is also essential for each workshop to have the telephone numbers of all emergency services close at hand. A number of medical emergencies which could happen in an electrical workshop are listed below.

Burns Burns in a workshop can be caused by steam from hot liquids, contact with flames, contact with hot surfaces (tips of soldering irons), electrical burns caused by someone touching a bare electrical conductor, or chemical burns (caused by chemical spillage on the skin). Managing burn injuries properly is important because they are common, painful and can result in disfiguring and disabling scarring of affected parts or even death in severe cases. For minor burns • Remove the person from the heat source and remove any burned clothing, except clothing imbedded in the skin. • Run cool – not cold – water over the burn or hold a clean, cold compress cloth on it until the pain subsides. Do not use ice. Do not use butter or other types of grease. • Remove jewellery or tight clothing from around burned areas and apply a clean bandage. You can also apply antibiotic cream. • Never break blisters resulting from a burn. • Remember not to remove clothing stuck to burned skin. • If you are helping someone with a serious burn, keep the burned areas elevated to reduce swelling. First aid for electrical burn victims Electrical burns vary in severity, depending on the strength of the current, the duration of the electrical shock as well as the direction of the current through the body. Often these burns are deep. Electrical burn wounds may look small from the outside but could be severe on the inside. The following must be done in case of an electrical burn. • Check that the victim is not in a state of shock (cold, clammy, pale and having a rapid pulse). If the victim is in shock, lay the victim down with feet slightly higher than the head. • Do not apply grease or oil to the burn. • Cover the burn with a dry, sterile dressing. • Do not try and remove clothing stuck to the burn. • Cover the victim with a blanket to maintain body heat. • Call for medical attention. What to do for chemical burns • Dry chemicals should be brushed off the skin by a person wearing rubber gloves. • Remove the person’s clothing and jewellery and rinse chemicals off the skin by placing the person in a shower for 15 to 20 minutes. (Be careful to protect your eyes and the eyes of the injured person.) • Wet chemicals should be flushed off affected areas with cool running water for 20 minutes or longer or until emergency help arrives.

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Electrical Technology • If you or someone else has swallowed a chemical substance or an object that could be harmful (e.g. watch battery), call poison control first and then the medical emergency numbers. It is helpful to know what chemical product has been swallowed. Take it with you to the hospital.

Bleeding

Bleeding is mainly caused by accidents where blood escapes the body when a vein is cut. Excessive bleeding can lead to shock or even death. Any bleeding wound should be treated using medical gloves. Try to use latex gloves when treating someone who is bleeding. Latex gloves should be in every first aid kit. People allergic to latex can use a non-latex, synthetic glove. You can be infected by HIV/ AIDS if infected blood gets into an open wound – even a small one. The following should be done to control bleeding. • Apply direct pressure or use a pressure bandage. • Keep the victim calm. • Keep the bleeding point above the heart level if possible. • When there is severe bleeding, where a major artery has been severed, pressure may be insufficient and a tourniquet may be used. Pressure from tourniquets must be Figure 1.1: Apply direct pressure on external wounds with a sterile relieved periodically to prevent damage to cloth or your hand, maintaining the tissue from lack of oxygen. pressure until bleeding stops

The following should not be done when treating bleeding. • DO NOT apply a tourniquet to control bleeding except as a last resort. • DO NOT peek at a wound to see if the bleeding has stopped. The less a wound is disturbed, the better the control of bleeding. • DO NOT probe a wound or pull out any Figure 1.2: Use a tourniquet only object that might still be inside a wound. as a last resort, if bleeding cannot be stopped and the situation is lifeThis will usually cause more bleeding and threatening harm. • DO NOT remove a dressing if it becomes soaked with blood; just add a new one on top. • DO NOT try to clean a large wound. This may cause heavier bleeding. • DO NOT try to clean a wound after you get the bleeding under control. Get medical help. There are four types of bleeding that can result from wounds: • Arterial bleeding Full of oxygen and bright red in colour, the blood has just come from the heart so it is under pressure and often spurts from a wound in time with the heartbeat. This type of bleeding is the most dangerous because the patient can lose a lot of blood in a short time. • Venous bleeding Containing less oxygen and a darker red, venous blood flows at a lower pressure than arterial blood and will not spurt. It may gush freely if a major vein is torn. • Capillary bleeding Usually minor wounds where blood oozes from the wound. Blood loss is limited but the risk of infection is very high.

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• Internal bleeding: Bleeding may not be visible due to internal injuries. This can be very dangerous and may develop following an injury to the abdomen or the chest.

Wounds at the workplace A worker on a construction site may suffer any of the wounds listed below. The treatment for all wounds is more or less the same, except that bleeding from deep cuts and puncture wounds may be more difficult to stop. Cut Caused by a knife, razor or even the sharp edge of paper. The wound may bleed profusely because cleanly cut blood vessels do not contract easily. Tear wound (laceration) Barbed wire, machinery or the claws of an animal may tear the skin in a ragged way. These wounds tend to bleed less severely because torn blood vessels contract more quickly than those that have been cut cleanly. Puncture wound Nails, needles, garden forks and even teeth may result in serious internal injury. If the wound is deep, the risk of infection is high because dirt and germs may have been carried into it. Graze A graze normally results from a sliding fall. Superficial layers of skin are scraped off, leaving a tender, raw area. These wounds often contain dirt or grit and may easily become infected. Gunshot wound These wounds can result in serious internal injuries. The exit wound is often much larger than the entry wound. Internal organs, tissues and blood vessels may be damaged by the passage of the bullet through the body. In addition to external bleeding, there may also be internal bleeding. Bruise A fall or a blow to the body by a blunt object. The skin is split and the surrounding tissues are bruised. With a bruise, damaged blood vessels leak blood into the tissues although the skin remains unbroken. Listed below are a few simple steps that can be taken in the event of any medical emergency. • Stay calm: The worst thing you can do in any emergency medical situation is to panic. For the sake of the victim and your colleagues, try to remain calm, cool and collected. You will be more effective and efficient this way. • Assess the situation: Quickly assess the scope of the injuries and collect information. If the injured person is conscious, ask him/her to tell you if anything hurts and observe where on the body he/she may be physically injured. Do not move an injured person, especially if he/she is reporting pain, unless there is imminent danger. • Call the emergency services: If a person is badly injured, call the emergency services immediately. If there is any doubt as to whether emergency services are needed, it is better to be safe than sorry. Stay calm and provide your address, location in the building, phone number, name and any information you have gathered about the injuries.

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Electrical Technology • Report the injury to the appropriate authority: Depending on the size of your workplace, you may need to notify management about the situation. • Administer first aid and CPR: If required, CPR or first aid should be performed by a person who is trained. If there is no skilled person, wait for emergency personnel to arrive. Do not administer medical treatment or medications. Be careful not to come in contact with blood, vomit or other bodily fluids.

Activity 1 1. In your own words, what is the general aim of the OHS Act? 2. What is your definition of an accident? 3. Unsafe actions and conditions are the cause of many accidents in workshops. Give THREE examples of each. 4. Why is it important to work in a well-ventilated workshop? 5. Write a short definition for what you understand by ‘dangerous practices’. 6. Why is the incorrect use of tools regarded as a dangerous practice? 7. Name at least four important things that must be taken into consideration when doing risk management. 8. Briefly distinguish the difference between a hazard and a risk. 9. In your own words, briefly explain your understanding of ‘Human rights in the workplace’. 10. Name at least FOUR issues that are also covered by human rights in the workplace. 11. Teamwork can be seen as one aspect that can lead to improvement in work ethics. How is this possible? 12. Name at least three other aspects which play a role in work ethics. 13. Define ‘medical emergencies’ 14. Burns caused by steam, flames or even hot liquids can be very unpleasant and must therefore be treated correctly. Explain the steps to be taken when someone experiences a burn of some kind. 15. When a person experiences bleeding, name at least FOUR things you must not do when you try to help the person who is bleeding. 16. Name the FOUR types of bleeding and briefly explain each one. 17. List and explain the five steps which can be taken in the event of any medical emergency.

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Chapter 2 Three-phase AC generation

A

B Waveforms

Phasor diagrams

A

B Power

Instruments

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Introduction In this chapter, comparisons will be made between single-phase and three-phase generation. Both the waveforms as well as the phasor diagrams will be referred to. However, the focus will be on three-phase systems. Star and delta connections will be shown in diagrammatic as well as schematic format. For calculations, the line and phase relationship between voltage and current will clearly be shown, as well as all aspects of power calculations. Further insights will also be given to matters like losses and efficiency. The measurements required in three-phase systems will be discussed and will include the wattmeter, kWh-meter, power factor meter as well as calculations with respect to these meters.

Principles of three-phase AC generation When mankind first started generating electricity, it was in the form of direct current (DC). While supply and demand circuits were close together, this system worked well, but the moment this electricity had to be fed to circuits further removed, this became problematic as there was no way that DC could be increased or decreased at these faraway points. This led to the development and introduction of single-phase AC systems, which also worked well for a certain period. But with the introduction of single-phase motors, this single-phase AC showed some shortcomings. For example, these motors were not self-starting, meaning something else had to be added to them to ensure they could start rotating with the required torque. With the introduction of three-phase generation came many advantages. A threephase generator means that instead of a single coil rotating though a magnetic field (two wires), three coils now rotate (meaning 6 wires). In the following section, comparisons will be made between the single and the three-phase systems.

Advantages and disadvantages of single-phase compared to three-phase systems Electrical power is generated at the power plant. This means the spinning of a coil in a magnetic field, or spinning a magnetic field in a coil. Something must spin the generator, and in South Africa this is usually done by means of the following: Power plants

Substation

Underground service line Commerical and industrial users

Local distribution 13 800 volts Substation Overhead service line System distribution loop 115 00 volts

Residential users

Underground service line Padmount transformer

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Three-phase AC generation Type of generation

Name of plant

Hydroelectricity

Vanderkloof and Gariep

Nuclear

Koeberg in Cape Town

Pumped storage

Palmiet in the Western Cape, Drakensberg in KwaZuluNatal

Gas turbine

Arcacia, Port Rex, Ankerlig, Gourikwa

Coal (Those in excess of 3 500 MW)

Duhva, Lethabo, Matmba, Tutka, Majuba, Kendal, Matla

Wind

Klipheuwel, Darling

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Take note Electricity can be created by means of a generator or an alternator.

Single-phase is what is used in your house. Generally this is a 240 V 50 Hz supply available at every plug point in the house. Whether it is switching on a kettle, toaster, TV, microwave, radio or lamps, it is all connected to single phase. But that still leaves the problem of single-phase motors. Although additional circuitry can be added to make it self-starting, it is expensive and cheaper alternatives needed to be considered. However, in industry the use of three-phase motors far outweighs that of singlephase motors, as they are self-starting, and have a better efficiency and power factor. And so the ‘three-phase generation’ was born. Three-phase generation is not a magical system. It is simply three coils positioned at 120° with respect to each other. The supplier (Eskom) will generate the power. From there it goes via what we call transmission lines. These are the power lines that carry the power from the supplier to the consumer. The next time you drive down the road and look at the power lines, that is exactly what this section of the work is about. There are numerous advantages to three-phase generation over single-phase generation, not only for the consumer (the people), but also for the provider (Eskom). A lot of these advantages may not make sense immediately, but all these aspects will be addressed in the relevant chapters and they will unfold as we work through each part. The advantages of three phase over single-phase will be split into three categories. 1. The generation process. 2. The transmission and distribution process. 3. The load. 1. The generation process

• For three-phase and single-phase alternators of similar physical sizes, three phase will generate more power. • Three-phase can supply power to single and three-phase loads.

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Electrical Technology • Three phase is cheaper to generate. • Three phase requires less maintenance. • Three phase has two connection options, namely star and delta, each with their own advantages (single-phase only has one option) • In three phase, losses are limited (phase voltages are 57,7% of the line voltages). • The star point can be used as a neutral. • A three-phase supply has three times the power of a single phase. 2. The distribution and transmission process

Power distribution happens over long distances and the idea is to make that happen at higher voltages and lower currents.

• • • • • •

Lower currents means less heat. Less heat means less losses. Lower current means thinner cables can be used and associated cost savings. Pylons required to support thinner cables will require less metal during construction and associated cost savings. A neutral point is available when connected in star. Load distribution and phase balancing becomes possible.

3. The 3-phase load L1 L2 L3 N

Rotation

Generation

Figure 2.1 (a): A three-phase motor

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Three-phase loads 400V

Single-phase loads 230V

Figure 2.1 (b): How three-phase and singlephase loads are connected to a three-phase supply

• Three-phase motors are more efficient. • Three-phase motors have a higher power factor. • Three-phase motors have a much higher starting torque. • Three-phase motors do not need additional starting circuitry. • For three- and single-phase motors of similar physical sizes, three-phase will produce more power. • Three phase can be connected in star or delta. • Three-phase motors are easier and cheaper to manufacture. • Three-phase motors have fewer moving parts, therefore less maintenance. • Three-phase motors do not use centrifugal switches, less arcing, less of a fire hazard.

Three-phase AC generation

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To summarise, most of the above simply means smaller machines are required, less materials are required to manufacture them, they’re cheaper to produce and maintain, shorter manufacturing times, easier to install and longer lasting. Those are more than enough reasons to go with the three-phase systems.

Waveform of single- and three-phase systems Single-phase generation Very basically, a single phase can be generated by rotating a conducting loop of wire through a magnetic field. Field pole

Voltage and current produced

Rotation

Field pole

Figure 2.2: A generator is really a wire moving within a magnetic field

The only problem could be that the wires from the rotor could twist themselves together as the rotor turns, and a split ring commutator is used to prevent this from happening. However, in this section no attention will be paid to that process, as the objective here is three-phase generation. A better way to see that objective is shown by the diagram below.

Rotating phasor

Sinusoidal waveform in the time domain

Figure 2.3: The generation of a single-phase AC supply by rotating a conductor through a magnetic field

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Electrical Technology However, it is a lot easier to explain this generation by using the basic sketches below. B

A

N

B

S

D

A

C

A D

C Figure 2.4: Simplified version of the generated single phase

• • • • •

At position A, the conductor moves parallel (//) with respect to (wrt) the magnetic field (MF) and zero current is induced. At position B, it moves at 90° wrt MF – maximum current induced. At position C, it moves // wrt MF – zero current induced. At position D, it moves at 90° wrt MF but in the opposite direction – max. current induced but flowing in the opposite direction. At position A, it moves // wrt the MF and zero current is induced.

This completes one cycle of the generated AC wave.

Three-phase generation As mentioned before, there is no magic involved in three-phase generation. It is simply three coils placed at 120° with respect to each other. They are usually referred as to the RED, YELLOW and BLUE phases, but occasionally in our descriptions they will simply be called phase 1, 2 and 3. As one can imagine, when the three rotating conducting loops turn in the magnetic field, six wires will twist together. However, this is very easily eliminated, by rotating the magnetic field instead, and keeping the conducting loops stationary. R

N

120˚

Rotating magnet 120˚

120˚ B

Y

Red phase

S Yellow phase

Blue phase Possible 4th wire (Natural)

Figure 2.5: Generation of a three-phase wave by rotating three coils through a magnetic field

It is easier to describe how the red coil’s wave is generated for one rotation, and then fit the yellow and the blue coils, keeping their phase relationship with respect to the red in mind.

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Explanation of how three-phase is generated • At the start, the red phase moves parallel (//) with respect to (wrt) the magnetic field (MF) and zero current is induced. • After rotating 90°, it moves perpendicular with respect to the magnetic field – maximum current induced. • After 180°, it moves parallel with respect to the magnetic field – zero current induced. • After 270°, it moves perpendicular with respect to the magnetic field, but in the opposite direction – max. current induced but flowing in the opposite direction. • After one revolution (cycle), it is moving parallel to the magnetic field again – zero current is induced. This completes one complete cycle for the red phase. The yellow phase follows the red phase exactly, except it lags 120° with respect to the red phase. The blue phase follows the red phase exactly, except it lags 240° with respect to the red phase. R

Y

120

240

Take note When drawing the waves, it is important to indicate exactly at which degrees the phases start.

B

Time

Figure 2.6: The generated three waves with the 120 degrees phase shift indicated between phases

Phasor diagram of single- and three-phase systems Whether talking about single- or three-phase generation, it is always inductive. The reason for this is that conducting loops of wire are used, and this brings a certain amount of inductance into the equation.

Phasor diagram for a single-phase system As the inductive part causes a phase shift between current and voltage, the same principle is used as in RCL circuit, to determine where they will be with respect to each other.

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Electrical Technology COIL Voltage (V) leads current (I)

CIVIL V

90˚

I

Figure 2.7: The phase relationship between current and voltage through a coil (if voltage is the reference)

Phasor diagram for a three-phase system In the three-phase system (if connected in star), the EMFs (voltages) generated in each phase will be 120° out of phase with each other. This means that the voltage in the red, yellow and blue phases would be represented as follows. R

120˚

120˚

120˚ B

Y

Figure 2.8: A phasor diagram showing the three voltages (one per phase)

3-phase systems; Star vs. Delta (Delta vs. Star) Take note Examiners like to ask for either schematic or diagrammatic representations. It is important for all learners to take note of this.

There are three voltages generated, one in each phase. The idea is not to use them as three individual single-phase voltages, but to combine them in specific ways to form a three-phase system. There are two ways of connecting the end point of the three coils (or phases) together. They are known as the star and the delta connections. This can be represented by means of a schematic or diagrammatic representation. Schematic (sketch without indication of components) is making use of lines, while diagrammatic (sketch with components) is almost the way it would be drawn in a circuit diagram.

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In a star-connected system, all the end points of each phase are joined and the beginning points will go to Live 1, Live 2 and Live 3. The beginning points are labeled as a1, b1, c1 and the end points as a2, b2, and c2. The common end points can be used as a neutral. a1

L1

c2

N

a2

L1

a1

N

• a2

c2

b2 L2

b1

c1

b2 b1

c1

L2 L3

L3

Figure 2.9: Diagrammatic versus schematic representations for a star connection

In a delta connection, the end of phase 1 will go to the beginning of phase 2, and the end of phase 2 to the beginning phase 3, and the end of phase 3 to the beginning of phase 1. All the beginning points will go to Live 1, Live 2 and Live 3. L1

c2

a1 a1

L1 a2

c1 L2 L3

c2

b2

b1

L2 L3

c1

b2

b1

a2

Figure 2.10: Diagrammatic versus schematic representations for a delta connection

Only balanced loads Since all alternators, generators and motors use windings of copper wire, each phase would have the resistance of that copper wire, as well as a certain inductance. The reason for the inductance is that the wires of each phase are turned around a core to form a coil.

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Electrical Technology Therefore each phase would have a certain resistance and inductive reactance, and the total ohmic value for each phase is called the impedance of that phase. It is of utmost importance that these rotating coils generate the same voltages and currents in each of the three phases and that is only possible if the impedance of all three-phases is matched. This is referred to as a balanced load. 3-phase transmission line

Z1

Z3

3-phase generator (supply)

Z2

3-phase load

Figure 2.11: A balanced three-phase load showing the impedance of each phase

Introduction to star and delta calculations with reference to basic line/phase values In three-phase systems, reference is made to phase values and line values. A phase value is the voltage across one of the phases and the current flowing through that phase only. Line values refer to the voltage that is measured across any two of the lines and the current flowing between those lines. (e.g. between L1 and L2). Calculations for STAR connections When looking at the star connection, it is clear that the voltage between L1 and L2 is actually across phases 1 and 3. However, this does not mean that one can just add the voltages of phase 1 and phase 3 together to get the line voltage, as the two phases are 120° out of phase with each other. Current on the other hand is the same in the line and the phase. Although there is a logical mathematical deduction of how to get this relationship between line and phase values, we will simply use the formula. L1 Phase 1 VLINE

N Phase 3

Take note VPHASE 1 +VPHASE 3 ≠ VLINE

20

VPHASE Phase 2

L2 L3

Figure 2.12: An indication that the line voltage and phase voltage cannot be the same in a star system


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The relationships between voltages and currents in a star system are as follows. VL = √ 3VPH IL = IPH Example 1: A star-connected alternator generates 240 V per phase. Each phase in this balanced system has an impedance of 16 Ω. Calculate the line voltages and currents. In a star system VL = √ 3VPH = √ 3(240) = 415,69 V IPH = VPH = 240 = 15 A Z 16 In star IL = IPH = 15 A Example 2: In a star system, a 11 kV line voltage and a line current of 20 A are generated by an alternator. Calculate the voltage across each phase as well as the impedance per phase. VPH = VL = 11 000 = 6,35 kV √3 √3 IL = IPH = 20 A ZPH = VPH = 6350 = 317,54 Ω IPH 20 Calculations for DELTA connections When looking at the delta connection, it is clear that the voltage between L1 and L3 is actually only across phase 2 and it stands to reason that they would be equal. The current flowing from L1 would split up between phase 1 and 2, but again there is the phase implication to keep in mind. Again, there is a logical mathematical deduction of how to calculate this relationship between line and phase values but we will simply use the formula. L1 VPHASE

e2

VLINE

as Ph

Phase 1

L2 L3

Phase 3

Figure 2.13: An indication that the phase and line voltages are the same in a delta system

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Electrical Technology The relationships between voltages and currents in a star system are as follows. IL = √ 3IPH VL = VPH Example 1: (The same values are deliberately used to show the different outcomes between star and delta calculations.) A delta-connected alternator generates 240 V per phase. Each phase in this balanced system has an impedance of 16 Ω. Calculate the line voltages and currents. In a delta system VL = VPH = 240 V IPH = VPH = 240 = 15 A Z 16 IL = √ 3 IPH = √ 3(15) = 25,98 A Example 2: In a delta-connected system, a 11 kV line voltage and a line current of 20 A are generated by an alternator. Calculate the voltage across each phase as well as the impedance per phase. VL = VPH = 11 kV IPH = IL = 20 = 11,55 A √3 √3 ZPH =

VPH IPH

= 11 000 = 952,63 Ω 11,55

Power in three-phase systems and calculations Take note Electric power is the rate of energy consumption or energy transfer in an electrical circuit. Electric power is measured by the capacity to do this power transfer and is expressed in kW or MW.

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Fortunately, the calculations that were done in RCL circuits regarding power stay exactly the same. As mentioned before, any alternator, generator or motor consists of coils of wire which basically makes them an RL load. This means the voltage and current in each coil will not be in phase with each other. Electric power is transmitted in overhead lines like those shown in the picture on the right, and also in underground high-voltage cables.


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Power in 3-phase AC circuits Before any calculations are done, there are some basics which need to be spelled out clearly. 1. ANY of the calculations done using these formulae always uses the LINE values of voltage and current. 2. Do not confuse the single-phase formula with the three-phase formula. 3. Always make use of a phasor diagram to get clarity with regards to kW and kVA. One also notices a √ 3 in all the three-phase calculations. This simply comes from the fact that P = IV but in a three-phase star system, the voltage generated in the phase is converted to a line value, which is VL = √ 3 VPH and in delta the current generated must be converted to a line value which is IL = √ 3 IPH. This means either the voltage or the current will have a √ 3 attached to it. Lastly, the phase angle always refers to the phase shift between the voltage and current in the phase and not the line. Active power (also called real or true power) The true power is the power that is effectively being used by the load or the circuit, and this would be the voltage and current values that are exactly in phase with each other. The formula used to calculate this true power is: Pactive (P) =√3 (IL) (VL) cosθ measured in watt (W) or kilowatt (kW) Apparent power The apparent power is the power that is supplied to the circuit. This is the product of the voltage and the current, ignoring the angle between the two. It is measured in VA or kVA, depending on the size of the values. The formula used to calculate this apparent power is: Papp (S) = √3 (IL) (VL) measured in VA or kVA Reactive power Reactive power is the power that is wasted and not used to do work on the load, usually in the form of heat. Preac (Q) = √3 (IL) (VL) sinθ measured in VAr or kVAr

Take note Pythagoras can be used to work out a value, but the other two values must be available before the third one can be calculated. PAPP2 = PTRUE2 + PREAC2 OR PAPP = √ PTRUE2 + PREAC2

3-phase calculations (star and delta) To start three-phase power calculations, the trick is to identify the type of power given. It’s proven that learners who plot the information on a phasor diagram are more successful at calculating the correct answers than those who don’t. Two examples using similar values will be used to try and demonstrate this simple fact.

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Electrical Technology Example 1: An AC star-connected alternator generates 300 kW at a power factor of 0,8 lagging. The phase voltage is 220 V. Calculate the following: a) Line voltages b) Line current c) Apparent power (S) d) Reactive power (Q) . Answers to example 1: a) First determine what is given and plot it on a phasor diagram. kW is always the phasor that is in phase with the supply voltage. Once this is established, it is easy to figure out the rest. Pact = 300 kW θ Pr

VT = 220 V

PApp

a) VL = √ 3 VPH = √ 3 (220) = 381,05 V b) The power given is in kW, therefore it is the active power. That is the power component in phase with voltage. Now select the correct formula. Make sure you use the calculated line voltage, and remember that cos θ = 0,8 and it is not cos 0,8. Pact = √ 3 (IL) (VL) (cos θ)

300 000 Pact IL = ____________ = _______________ = 568,18A √ 3 (VL)(cos θ) √ 3 (381,05)(0,8)

It also helps to fill in values on the phasor diagram as they are calculated.

c) Papp = √3 (IL)(VL) = √3 (568,18)(381,05) = 375 kVA

OR

Pact Papp = _____ cosθ

300 000 = _______ = 375 kVA 0,8

d) Before using the reactive formula, the angle needs to be calculated. Cosθ = 0,8

θ = cos-1 0,8 = 36,87˚

Preac (Q) = √ 3 (IL)(VL)(sin θ) = √ 3(568,18)(381,05)(sin36,87) = 225kVAr

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OR Preac (Q) = √ Papp2 – Pact2 = √ 375 0002 – 300 0002 = 225kVAr (Pythagoras)


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Example 2: (The same example, but the given power is now in kVA.) An AC star-connected alternator generates 300 kVA at a power factor of 0,8 lagging. The phase voltage is 220 V. Calculate the following: a) Line voltages b) Line current c) Active power (P) d) Reactive power (Q) Answers to example 2: First determine what is given and plot it on a phasor diagram. kVA is always the phasor that is at an angle with the supply voltage. Once this is established, it is easy to figure out the rest. Pact VT = 220 V

θ Pr

PApp = 300 kVA

a) VL = √ 3 VPH = √ 3 (220) = 381,05 V b) The power given is in kVA, therefore it is the apparent power. Now select the correct formula. Make sure you use the calculated line voltage, and remember that cosθ = 0,8 and it is not cos 0,8. Papp (S) = √ 3 (IL) (VL) 300 000 = _______________ = 454,55 A √ 3 (381,05)

Papp IL = ____________ √ 3 (VL)

It also helps to fill in values on the phasor diagram as they are calculated.

c) Pact (P) = √3 (IL)(VL) (cos θ) = √3 (454,55)(381,05) (0,8) = 240 kW

OR

Pact = Papp (cosθ) = (300 000)(0,8) = 240 kW

d) Before using the reactive formula, the angle needs to be calculated. Cos θ = 0,8

θ = cos-1 0,8 = 36,87°

Preac (Q) = √ 3 (IL)(VL)(sin θ) = √ 3(454,55)(381,05)(sin 36,87) = 180 kVAr OR Preac = √ Papp2 – Pact2 = √ 300 0002 – 240 0002 = 180 kVAr (Pythagoras)

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Losses Whenever we convert one form of energy into another there are bound to be losses. No machine is perfect, and therefore not 100% efficient as there are always certain losses. Some are considered electrical losses while others are called mechanical losses. The basic losses can be categorised as follows: • copper losses • iron losses or core losses • friction losses • windage losses

Electrical Copper losses

These are basically due to the resistance of the copper wire used in the loops of wire that make up the windings of the alternator/generator. Any current that flows will experience a resistance and create heat.

Iron or core losses

The core of a generator/armature is made from soft iron which is a conducting material with desirable magnetic characteristics. Any conductor will have currents induced in it when it is rotated in a magnetic field. The currents that are induced in the generator/armature core are called EDDY CURRENTS. The power dissipated in the form of heat, as a result of the eddy currents, is considered a loss.

Mechanical Take note Why is the efficiency of a generator greater than that of a motor?

Friction losses

The full weight of the rotating parts is carried by bearings. Although these bearings make it easier to rotate and reduce friction to a large extent, they still experience friction, and friction results in unwanted heat.

The generator windings are made of thicker windings and hence have less resistance and hence lower copper losses.

Windage losses

In a motor, the armature produces back EMF which tries to oppose the motion of the motor (there are two opposing voltages), but in a generator there are no opposing voltages.

The alternators are usually designed for maximum efficiency to occur at about 80% of full load. To simplify matters, all the calculations we do will be at 100% efficiency.

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The alternator/generator has fan blades on the rotor that blow air through and over the alternator to keep it cool. The losses due to forced air movement are called windage losses.

Efficiency: (For calculations: η=100%)

The symbol used for efficiency is given by η and the formula associated with calculating these losses is:


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Only concept of power factor correction – no calculations for exam purposes When a load (like a three-phase motor) is connected to a three-phase supply, it is always inductive. The reasons for this have been discussed earlier. This motor will draw a certain amount of current from the supply. The user must pay for the current that is drawn, and the supply cables to the motor need to be of sufficient thickness to carry the current safely. For example, a 20 kVA star-connected motor with a power factor of 0,75 is connected to a supply of 380 V. Firstly, we plot this on a phasor diagram. It is 20 kVA, so it is the one at an angle to the supply voltage. Pact θ Pr

VT = 380 V Cos θ = 0,75

PApp = 20 kVA

The current drawn from the supply will be: Papp = √ 3 (IL) (VL)

20 000 Papp IL = _________ = _______ = 30,39 A √ 3 (VL) √ 3 (380)

(This is the total current drawn from the supply)

This is the line current that is drawn from the supply, and this is what you pay for whether it is used effectively or not. The cables to the motor will also have to be of specified thickness to carry 31 A. However, the horizontal component of current can be determined by: IHORIZONTAL = ILINE × cos θ = (30,39)(0,75) = 22,79 A

Ihor = 22,79 A θ Pr

VT = 380 V Cosθ = 0,75

Itotal = 30,39 A

This means that you pay for 30,39 A, but effectively you are only using 22,79 A. The rest are losses and the major negative implication of this is unwanted heat. The motor runs hotter, the bearings are operating at a higher temperature, maintenance increases and all this time it is completely unnecessary.

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Take note The best power factor is 1, and improving it means changing it from 0,75 to as close to 1 as possible.

A way must be found to reduce the angle between the voltage and the supply current. The 22,79 A will stay exactly the same, but the supply current will become less as the power factor increases. See the diagram below on how supply current decreases with an improved power factor. Cos θ = 1

IT = 22,79 A

Cos θ = 0,9

IT = 25,32 A

Cos θ = 0,8

IT = 28,49 A

Cos θ = 0,75 IT = 30,39 A Figure 2.14: Supply current decreasing as the power factor is increased

The way to achieve this is to connect a capacitor in parallel with the load. By selecting the correct capacitor, one can get very close to unity power factor. It is not possible to have it exactly correct due to varying load conditions. A capacitor would result in a vertical current (plotted in the upwards direction), resulting in the top and bottom verticals cancelling each other out (almost like a parallel RCL circuit at resonance). ICapacitor IT After correction VS Ihor Ivertical

IT Before correction

Figure 2.15: A phasor diagram showing the effect on supply current due to power factor correction

Take note A power factor of 1 is also called unity power factor.

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Figure 2.16: Power factor correction is achieved by connecting a capacitor in parallel with the load


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Advantages of power factor correction For the consumer: • Less current drawn from the supply. • Cost saving from drawing less current. • Less heat generated. • Maintenance reduced due to motor running cooler. • Longer life span of load. For the supplier: • Thinner supply cables required. • Reduced cost as a result of thinner cables. • Suppliers’ machines not under so much stress to produce electricity that is wasted. • Less maintenance. • Longer life span. It can clearly be seen that it is to the advantage of both the supplier and the consumer to make use of power factor correction.

Purpose of the wattmeter Power can be measured by connecting a voltmeter (in parallel) and an ammeter (in series) in a circuit. Record the individual readings and then multiply them together. This might work for low power circuits, but with higher values of voltage and current it becomes dangerous, and the use of potential transformers (PT) and current transformers (CT) becomes necessary. This method also does not take into consideration the power factor and therefore the answer would be in VA or kVA [Apparent power (S)]

Figure 2.17: A voltmeter and an ammeter connected to a load

With normal meters, one would simply use P = I × V (measured in VA or kVA) However, this can be time consuming, and there are far easier and more convenient ways of measuring power. The modern wattmeter enables us to measure power without any difficulty and it measures the active power (P) in W or kW immediately.

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Electrical Technology There are two main types of wattmeters. • moving coil (of which there are a few variations) • digital.

Figure 2.18: A variety of old and modern wattmeters

Purpose of wattmeters • The wattmeter is an instrument for measuring the electric power (or the supply rate of electrical energy) in watts of any given circuit. • It essentially accumulates or averages readings. • Electronic wattmeters are used for direct, small power measurements or for power measurements at frequencies beyond the range of electrodynamometer- type instruments. • It basically samples the voltage and current thousands of times a second. For each sample, the voltage is multiplied by the current at the same instant. • The average taken over at least one cycle is the real power. Digital type It is not that easy to describe the operation of the digital meter, as different types use different technologies. What is important is to compare the advantages of using a digital meter compared to the moving coil type.

Take note All wattmeters are designed to measure the power in a circuit at a specific frequency. If used to measure power in another circuit where the frequency is different, the readings would be inaccurate.

Advantages: • The cost to operate the meter. • Virtually all digital meters can emit a radio signal to forward the power usage for a certain time period to a recording device. This means that a utility provider can drive around the block rather than walking up to each house and recording the power usage every month. • The best electromechanical wattmeter has a power consumption of 24 W. The leading digital meters have a starting power of less than 5 W. • The operating power consumption for electromechanical meters is also more than that of digital meters. • Electromechanical meters have a loss of about 0,7 W compared to the 0,5 W loss of digital meters. • The display could also be considered an advantage because people who are not trained in reading a scaled meter can read a digital meter easily.

kWh meter An electricity meter or energy meter is a device that measures the amount of electrical energy consumed by a residence (household) or business, or an electrically powered device.

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Figure 2.19: A conventional kWh meter and the electronic kWh meter

Electricity meters are typically calibrated in billing units, the most common one being the kilowatt hour [kWh]. Periodic readings of electric meters establish billing cycles and energy used during a certain period of time. http://www.youtube.com/watch?v=AsQwgagD7IU (disc speeds) Line

Stator

Line Potential coil

Permanent magnet (braking)

I–coil L

AL disc

Rotor (disc) V–coil Current coil Load

L O A D

N

Load

Figure 2.20: Main components of the kWh meter, and a simplified version

Basically, it measures power which is a product of current and voltage. Inside the kWh meter one would find a current coil, a voltage coil and a spinning aluminium disc.

The voltage coil

Figure 2.21: The voltage (potential) coil of thin wires inside the kWh meter

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Electrical Technology • • • •

A voltage coil is connected in parallel to the supply (because voltmeters are connected in parallel). It consists of a coil with many turns of fine wire and a laminated core. This coil has a high resistance to prevent as little current as possible flowing through so it does not interfere with the operation of the circuit. The voltage coil is designed to produce an air gap magnetic flux, which is proportional to the applied voltage.

Current coil • • • •

The current coil is connected in series with the live wire (ammeters are connected in series). It consists of a coil of a few thick turns and a laminated core assembly. This coil must have a low resistance (hence thick wire) to not interfere with the operation of the circuit. (The full load current flows through this coil.) The current coil assembly must produce an air gap magnetic flux, which is proportional to the current drawn by the load.

Figure 2.22: The current coil of thick wires inside the kWh meter

Aluminium disc Gear train to dials

Voltage coil Copper shading ring

Magnetic brake Rotating aluminium disc

Current coils Aluminium disc Figure 2.23: The aluminium disc that rotates

• The voltage coil and core assembly and the current coil and core assembly are situated close to each other. • The aluminium disc is located precisely in the air gap between these two assemblies.

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Three-phase AC generation • • •

2

The two magnetic fields of the current and potential coils induce an eddy current in the disc creating a torque, causing it to turn. The more current that is drawn by the user, the stronger the magnetic field, the faster the disc will rotate. The disc is attached to a spindle in which it rotates and in turn it drives a geared mechanism which indicates the units of power used.

Meter register

Display or measuring face Figure 2.24: Meter register

• • •

A part of the meter registers the revolutions of the aluminium disc. This drives a gear train that is attached to the meters (clock mechanism) that are calibrated in kilowatt hours (kWh). This in turn tells us how much electricity we have used over a period of time.

Prepaid electricity meters A prepaid electricity meter is a KWh (Kilowatt Hour) meter, measuring electrical consumption. The main difference is that this KWh meter counts backwards as the electricity is consumed and has a relay (an automatic switch) which disconnects the power when the KWh reading on the meter reaches zero. It further incorporates hardware which has the ability to decipher the pin number entered and convert it to KWh. All prepaid meters in South Africa are STS (Standard Transfer Specification) compliant. This means that they all use the same coding system. In SA this is a 20digit encrypted code, preventing fraudulent vouchers from being generated.

Different types of prepaid meters used in South Africa

Initiated by Eskom The development of STS (Standard Transfer Specification) was initiated by Eskom, since they were purchasing numerous prepayment units for various municipalities from different manufacturing companies, which resulted in their purchasing different vending systems for the operation of these units. The introduction of STS was the only means of ensuring that the prepayment tokens issued by the vending system of one manufacturer could also be used by a prepayment unit of a different manufacturer. As a result, all prepaid units, even though different in looks and manufactured by different companies, function using the same STS encryption technology for the production of vouchers. The Standard Transfer Specification (STS) has become recognised as the only globally accepted standard for prepayment systems. It has become established as a worldwide standard for the transfer of electricity prepayment tokens since its initial introduction in South Africa in 1997. Conclusion The Standard Transfer Specification has found widespread application. Initially created in South Africa, it has subsequently been distributed to many developed and developing countries. To date, over 10-million STS-compliant meters have been installed at 400 utilities in 30 countries around the world. http://www.prepaidelectric.co.za

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Power factor meter when connecting instruments in circuits

Figure 2.25: Different types of power factor meters

The power factor meter simply enables us to determine the power factor in commercial installations and this knowledge is used to try and do something to the load to improve the power factor to as close to unity as possible. As done in previous calculations, we know that power factor is Cosθ = ____ Pact Papp Power factor is simply the ratio between true power (P) and apparent power (S). Power factor meters enable us to measure this reading without much difficulty. The power factor meter has seen major adjustments and developments in the last few years. There are 3 basic types of power factor meters: • moving coil type • fixed digital type • clamp-on type.

The moving coil type (electromechanical) This is a moving coil meter. It has two perpendicular coils on the moving part of the instrument. The coils create electromagnetic fields when current flows in the circuit. The two moving coils are connected in parallel with the circuit load. The first coil will be connected through a resistor and the second coil through an inductor. The current in one coil would be delayed with respect to current in the other coil (or there will be a phase shift between the currents in the two coils). At unity power factor, the currents in both coils are in phase with each other, and the needle would show “1”. The moment there is a phase difference between the two currents it causes the needle to deflect either one way (if it’s leading) or the other way (if it’s lagging) The angle between the currents in the crossed coils is a function of frequency and consequently each power-factor meter is designed for a single frequency and will be in error at all other frequencies.

Digital types There is a fixed type for more permanent use and a clip-on model which is easily moved and used. The main advantage of the clip-on power factor meter is its ability to measure power factor without interrupting the operation or service of equipment. They are in wide demand due to their measuring power factor capacity over a wide current range.

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Measurement of power in three-phase systems The meter of choice to measure power in any system is the wattmeter as it measures true (or active) power in watts (W) or kilowatts (kW). Any wattmeter must have a voltage coil (two wires) and a current coil (two wires), as shown in the sketch below. This is for the moving coil type. (Digital meters use a different method.) The current coil must be connected in series to the live wire, and the voltage coil in parallel between a live and a neutral.

L

Take note A wattmeter takes the power factor into consideration and is measured in W or kW.

L

V–coil N

W

I–coil L

Figure 2.26: Sketch of wattmeter with its four terminals

Figure 2.27: A digital wattmeter

One can unfortunately not rely on measuring voltage with a voltmeter, and current with an ammeter, and then multiply them together to get the power. (Using this method would result in getting the apparent power (S) measured in VA or kVA.) It must be measured with a wattmeter because the wattmeter takes the power factor into consideration and hence measures true (active) power. For the purpose of our sketches, only analogue meters will be shown. How to connect the meter will depend on: • whether it is a star or delta load • if the load is balanced or unbalanced • if the terminals of the load are easily accessible.

Take note The voltmeter and ammeter methods do not take the power factor into consideration and are measured in kVA.

For balanced star and delta loads A single wattmeter can be used to measure the power in one phase. The total power of the system can be determined by PT = P1+ P2 + P3 OR PT = 3 P1. P1 L1

L1

P1

N L2 L3

L2 L3

Figure 2.28: A single wattmeter used to measure the power in a balanced star and a delta load

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Electrical Technology Example 1:

The wattmeter shown above is connected to the balanced star connected circuit in figure 2.28. If the needle deflects to between the 500 and the 1 000 mark, what is the total power of the system? Answer 1: Meter reading is P1 = 750 W Total power PT = 3P1 = 3(750) = 2,25 kW Example 2: The wattmeter is connected to the balanced delta connected circuit in figure 2.28. a) If the needle goes to full scale deflection (fsd), what is the total power of the system? b) Why would this not be recommended? Answer 2: a) Meter reading is P1 = 1 500 W Total power PT = 3P1 = 3(1 500) = 4,5 kW b) If there is a spike in supply voltage or current, it would force the meter past fsd which may result in damage. The star system is relatively easy to connect, as the three lines (L1, L2 and L3) are easily accessible. But with the delta system, the actual coils are placed inside the motor or load and the terminals are not easily available to connect to the wattmeter. For unbalanced star and delta loads: To measure the total power in an unbalanced star or delta load you would have to make use of one wattmeter and repeat the process three times but this leads to inaccuracy. The better, although more expensive option, is to use three separate wattmeters. one for measuring the power in each phase. The total power of the system can then be determined by PT = P1+ P2 + P3 P1

L1

L1

P1 P2

N

P2

L2

L2

L3

L3

P3

P3

36

Figure 2.29: Three wattmeters used to measure the total power in an unbalanced star and delta load


2

Example 3: Three digital wattmeters are used to measure the power in an unbalanced load. These meters have a maximum rating of 2 000 W each (without multipliers). The readings are shown in the photos of the three meters below. What is the total power of the system?

Answer 3: PT = P1+ P2 + P3 = 1 150 + 850 + 1 302 = 3,3kW Two wattmeter method The most common method of measuring power in a three-phase system must be the two wattmeter method, as it has major advantages over the other methods. P1 L1

L O

L2

A D

L3 P2

Figure 2.30: The connections of the two wattmeters connected between three supply lines

The main advantages of the two wattmeter method are: • The total power can be measured in a balanced or unbalanced load. • The total power can be measured in a star or delta system. • Terminal is easily accessible as only the live wires to the system are required to do the connections. • The power factor can be determined. Disadvantages • The availability of two wattmeters at any time. • Cannot determine if the power factor is leading or lagging. The total power in this case is simply calculated by adding the readings of the two meters together. Total power PT = P1+ P2 Calculating the power factor is also possible by using either of the following formulae. Power factor

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Electrical Technology The disadvantage of the second formula is that it needs to be converted to an angle, and then back to cos θ. Example 4: The two wattmeter method is used to determine the power delivered to a threephase load. It is not known if the load is balanced or unbalanced, or star or delta. The reading on wattmeter 1 is recorded as 215 W and the reading on wattmeter 2 is recorded as 189 W. It is connected to a 120 V line voltage. a) Does it matter that we do not know if the load is balanced or not? b) Does it matter that we do not know if it is a star or delta load? c) Calculate the total power delivered to the load. d) Calculate the power factor of the load. e) The line current that is drawn from the supply. Answers 4: a) No, it does not matter. The two wattmeter method measures both without it affecting the readings. b) No, it does not matter. The two wattmeter method measures both star and delta. c) PT = P1+ P2 = 215 + 189 = 404 W d)

OR

e)

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Activity 1 1. What is meant by a motor that is not self-starting? 2. Name the environmentally friendly methods of generation in South Africa. 3. What is the voltage available in the average South African household? 4. What are the advantages of three-phase generation for the consumer? 5. Explain how three-phase power can be generated. 6. Draw a neat wave representation of this generated three-phase. 7. Draw a neat schematic representation of a delta connected system. 8. Draw a neat diagrammatic representation of a star connected system. 9. In a star system, a 380 V line voltage and a line current of 5 A are generated by an alternator. Calculate the current and voltage generated across each phase as well as the impedance per phase. 10. A delta-connected alternator generates 220 V per phase. Each phase in this balanced system has an impedance of 44 Ω. Calculate the line voltages and currents. 11. A star-connected generator produces 1,1 kW per phase. Each phase has an impedance of 44 Ω. Calculate. a) the phase voltage b) the line voltage c) the total power of the system d) the line current if the power factor is unity 12. A three-phase delta-connected supply is rated 50 kW and delivers a line voltage of 415 V at a power factor of 0,85. Assume the efficiency to be 100%. a) Represent the system by means of a neat diagrammatic sketch. b) Calculate the line current and the phase current. 13. The input power of a three-phase delta-connected motor is 60,5 kW with a power factor of 0,91. The line voltage is 415 V. a) Draw a phasor diagram to represent this information. b) Calculate the line current of the motor. c) Calculate the phase current of the motor. 14. An AC star-connected alternator generates 15 kVA at a power factor of 0,87 lagging. The phase voltage is 240 V. Calculate the following: a) line voltages b) line current c) active power (P) d) reactive power (Q) 15. Name three types of losses that occur in generators/alternators. 16. What are the advantages of power factor correction to the consumer. 17. Name three types of meters and state their purpose. 18. Show how a single wattmeter can be used to measure the total power in a three-phase star-connected system if the reading on the meter shows 1 200 W. A sketch and calculations are required. 19. The two wattmeter method is used to determine the power delivered to a three-phase load. It is not known if the load is balanced or unbalanced or star or delta. The reading on wattmeter 1 is recorded as 1 183 W and the reading on wattmeter 2 is recorded as 317 W. It is connected to a 380 V line voltage. a) Calculate the total power delivered to the load. b) Calculate the power factor of the load. c) Calculate the line current that is drawn from the supply.

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Practical Activity 1 To investigate the relationship between the line and phase voltages in a three phase supply. Equipment: • Multimeter with appropriate Voltage scale • A Three Phase supply.. Steps: 1. Switch on the supply.

Take the following readings and record in the table supplied below: 2. Measure the voltage between L1 and L2. 3. Measure the voltage between L1 and L3. 4. Measure the voltage between L2 and L3. 5. . Measure the voltage between L1 and Neutral. 6. Measure the voltage between L2 and Neutral 7. Measure the voltage between L3 and Neutral Connection

Meter readings

Voltage between L1 and L2. Voltage between L1 and L3. Voltage between L2 and L3. Voltage between L1 and Neutral Voltage between L2 and Neutral Voltage between L3 and Neutral Conclusion: 1. What can be observed from the readings between any of the lines? 2. What can be observed from the readings between any of the line and neutral values? 3. What could be the reasons for this?

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Chapter 3 Three-phase transformers

A

Three-phase transformers

B Connection types

A

B Calculations

Construction

A

B Cooling

Protection

3


Introduction In this chapter, specific attention will be given to three-phase transformers. Questions will be asked on why the transformer is not 100% efficient and what constitutes these losses. Basic construction, types of connections and calculations in respect of the turn ratios, voltage, current, power and power factor will be addressed. Important factors covered will be the application of these transformers, cooling methods, safety and protection, focusing specifically on the distribution network used in South Africa.

Three-phase transformers Principle of operation and connections of three-phase transformers Before three-phase transformers are discussed, it is important to know where they fit into the national distribution of power in South Africa. On one side, there is the provider (the generation authority in South Africa is Eskom) and they generate three-phase electricity. On the other side are the users or consumers and many of them use three-phase machines. Getting the generated power from the provider to the consumer is where the three-phase transformer fits in. This process of getting the power from supplier to user is called the transmission and distribution of electricity.

Transmission lines 275/400/765 kV 22 kV Generating station

Generating step up transformer

Substation step down transformer

Subtransmission customer 22 kV

Primary customer 11 kV

Transmission customer 132 kV

Secondary customer 380 V/240 V

Figure 3.1: General format of generation, transmission and distribution to the customer in South Africa

Electricity is transmitted at high voltages (110 kV or above) to reduce the energy lost in long-distance transmission. Power is usually transmitted through overhead power lines. Underground power transmission has a significantly higher cost and greater operational limitations but is sometimes used in urban areas or sensitive locations. A key limitation to the distribution of electric power is that, with minor exceptions, electrical energy cannot be stored and therefore must be generated as needed. A sophisticated control system is required to ensure electricity generation very closely matches the demand. If the demand for power exceeds the supply, generation plants and transmission equipment can shut down which, in the worst cases, can lead to a major regional blackout, as has occurred in SA from time to time. To reduce the risk of such failures, electricity transmission networks are interconnected into regional,

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national or continental networks, thereby providing multiple alternative routes for power to flow should (weather or equipment) failures occur. High-voltage overhead conductors are not covered by insulation. The conductor material is nearly always an aluminium alloy, made into several strands and possibly reinforced with steel strands. Copper used to be used for overhead transmission but aluminium is lighter, yields only marginally reduced performance and costs much less. Overhead conductors are a commodity supplied by several companies worldwide. In order for the electricity to be transmitted safely and efficiently, it must be at a high voltage (pressure) and a low current. If the current is too high, the cable would heat up too much and even melt, and if the voltage was too low, hardly any energy would be carried. The generators in the power stations produce electricity at 22 kV. This voltage is raised by transformers before it is sent out at 275 kV, 400 kV or even 765 kV onto the transmission grid. The electricity is transformed down to 11 kV for local distribution and then further reduced according to the need – for example, 380 V and 240 V for domestic use. The electricity entering your home at 240 volts has had an eventful journey, from the initial high-voltage transmission grid to a lower-voltage distribution network. Travelling over ground and (probably) underground for a great many kilometres, it has been transformed many times on the way. You’ve probably seen some of the equipment which performs these operations in your local area. They are known as substations which can be found in many sizes – small transformers mounted on wooden poles, larger transformers sitting behind high fences and huge arrays of strangely shaped devices on sites occupying several hectares. Eskom is the first utility in the world to successfully operate transmission lines at 765 kV at high altitudes above sea level.

Single-phase transformer Single-phase transformers were studied in Grade 11 in great detail, but it is such an important part in the electrical distribution network that a quick review will be done before moving on to three-phase transformers. Iron core

Primary coil

Secondary coil Magnetic flux

Figure 3.2: A simplified sketch of a single-phase transformer and the actual transformer

Take note Michael Faraday discovered in 1831 that magnets and moving wire had strange effects on each other when they moved closer together. In fact, Faraday found that the mechanical energy used to move a magnet inside a coil of wire could be changed into electrical energy. It was this simple discovery which has led to modern power stations.

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Take note Lenz’s law states that the induced current will always oppose the applied current.

By having less windings on the secondary than on the primary, a voltage can be reduced. Such transformers are called step down transformers and those that cause higher voltages on the secondary are called step up transformers. The application of this concept is extremely useful in the distribution of power on the national grid.

Three-phase transformers Three-phase transformers are actually three separate single-phase transformers which are connected in a certain way. Once they have been connected and put inside a special container, the unit is called a three-phase transformer. It is of vital importance that the three transformers used to form a three-phase transformer are identically matched in all regards. To be identically matched the three transformers must have the same: • turns ratios • voltage and current ratings • power ratings • power factor • efficiency • size.

Figure 3.3: Three single-phase transformers compared to a single three-phase transformer

Three-phase transformers have three coils, which results in six wires on the primary side (2 from each coil) and the same on the secondary. This enables connections in star and delta.

Figure 3.4: Three-phase transformer core

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How to connect a three-phase transformer in star or delta L1

L1

L2

L2

L3

L3

N Figure 3.5: THREE single-phase transformers connected in a star delta configuration

Figure 3.5 shows how the primary side can be connected in the star configuration, while the secondary shows the delta connection. If you follow these steps, it will make the drawing and wiring process a lot easier. It’s like following a recipe in a cookbook. Star – Join all the end points together, and all the beginning points go to L1, L2 and L3. The end points can be connected to the neutral. Delta – Join the end of coil 1 to the beginning of coil 2, the end of coil 2 to the beginning of coil 3, the end of coil 3 to the beginning of coil 1. All the beginning points go to L1, L2 and L3. The delta connection has no neutral. Exactly the same star delta transformer can be drawn by using a SINGLE threephase transformer. This can be drawn schematically or diagrammatically. L1

L1

N L2 L3 L1

L2 L3 L1

N L2

L2

L3

L3

Take note THREE single-phase transformers means three transformers must be connected in star or delta. A SINGLE three-phase transformer has already been wired and placed inside its container ready for use.

Figure 3.6: A schematic (on top) and a diagrammatic (at the bottom) drawing of a SINGLE three-phase transformer

Concept and understanding of losses In the ideal transformer, the input power would match the output power 100%. However, no transformer is 100% efficient. This simply means that the power at the output is always less than the input. For example, if a transformer is rated at 100 kW, but only 97 kW is measured at the output, then that transformer is 97% efficient. There are a number of reasons for this and for the sake of keeping things simple we will focus on the obvious ones.

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Electrical Technology Losses: 1. Copper losses These are losses due to the resistance of the copper wires used for the windings of the primary and secondary coils in the transformer. 2. Iron losses (eddy current losses) These are heat losses that occur in the iron core due to hysteresis of the core materials and induced eddy currents in the laminated plates. 3. Dielectric losses The copper wire used for the windings of the primary and the secondary coils are covered with a thin layer of lacquer to insulate them and prevent shorting between coils. If this lacquer is damaged, or not sufficient, a small leakage current will flow. 4. Stray losses Some of the magnetic field from the primary coil cuts the surrounding metal parts of the transformer casing. This means that some of the magnetic field does not cut the secondary winding to induce a current there and it is considered a loss.

It is of utmost importance to keep losses in transformers as low as possible, as they create heat. This in turn leads to more losses. Most transformers make use of cooling to keep them operating at as low as possible temperature.

Three-phase transformers compared to single phase Single-phase transformers


Has one coil on primary and one on secondary

Has three coils on primary and three on secondary

Has two wires on primary and two on the secondary

Has six wires on primary and six on the secondary

Has only one way to connect the primary and one way for the secondary

Can be connected in two formats (star and delta) on the primary and secondary

Has a third of the power

Has three times the power

Cheaper

More expensive

Weighs less

Weighs much more

Provides less loading compared to three phase

Provides better loading

Less skill required to wire

Need to understand star and delta before wiring

Reduced labour to install. Only live and neutral on the primary

Must first wire in star or delta before connecting to supply

Only one operating voltage on secondary

Two operating voltages if connected in star

Cheaper to replace

If just one of the three is broken, the whole transformer must be replaced

Types of three-phase transformers – basic types of construction http://www.youtube.com/watch?v=FVuEI_G3KNM (2 min 15) Three-phase transformers are basically three single phase transformers mounted on a single core.

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The types of transformers differ in the manner in which the primary and secondary coils are wired around the laminated steel core. According to design, transformers can be classified into two categories: 1. Core-type transformer In a core-type transformer, the windings are given to a major part of the core. This means you see lots of winding and less of the core. The coils used for this transformer are form-wound and are cylindrical in shape. This type of transformer can be used for small sized and large sized transformers. The cylindrical coils will have different layers and each layer will be insulated from the other with the help of materials like lacquer, paper, cloth, to name but a few. The general arrangement of the core-type transformer with respect to the core is shown below.

(a)

(b)

Figure 3.7: Basic sketches of two the transformer types. The core type shown in (a) and shell type shown in (b) Single phase

Three phase

Core type

Shell type

Figure 3.8: The difference between a shell-type and a core-type transformer

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Electrical Technology 2. Shell-type transformer In shell-type transformers, the core surrounds a considerable portion of the windings. This means more of the core is visible compared to the other type. The whole winding consists of discs stacked with insulation spaces between the coils. Such a transformer may have the shape of a simple rectangle.

A strong mechanical support must be given to the cores and coils of the shelltype transformer. A transformer with good support (bracing) will not produce any humming noise while working and will also reduce vibration of the laminated plates.

Usually, these transformers are placed in tightly-fitted sheet-metal tanks filled with special insulating oil.

The general arrangement of the core-type transformer with respect to the core is shown in figure 3.7 and 3.8.

Types of three-phase transformers: Mostly quoted from: http://www.trafoworld.com/en/transformers/different_ types_transformer/#DTH

There are several different types of three-phase transformers available in the world market. They are mainly divided by power and voltage supplied, but also depend on their application: • small distribution transformers • distribution transformers • cast resin transformers • dry-type transformers • large distribution transformers • medium power transformers • large power transformers.

Small distribution transformers

Single-phase transformers, usually made with wound core system and rectangular windings. Power range: 50 to 200 kVA Main use: Distribution in rural areas and countryside Main advantages: Small production costs with the possibility of good automation

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Distribution transformers

Usually three-phase transformers, immersed in liquid oil as dielectric insulation and enclosed in a tank with cooling system. Recently made hermetically sealed for reduced maintenance and better quality. 250 to 2 500 kVA Power range: Main use: Distribution of energy in cities and centres with different houses Main advantages: Great extension of use in different outdoor applications

Cast Resin Transformers CRT

Usually three-phase transformers but instead of being immersed in oil, the high voltage side is cast into a resin which will be its dielectric insulation. Power range: 250 to 4 000 kVA Main use: Underground systems, mines and skyscrapers Main advantages: Fireproof and explosion-proof, particularly adapted for indoor applications

Dry type transformers

Usually three-phase transformers but instead of being immersed in oil, the HV side is dipped into an insulating varnish which will be its dielectric insulation along with open air.

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Electrical Technology Power range: Main use: Main advantages:

250 to 4 000 kVA Underground systems, mines and skyscrapers. Fireproof and explosion-proof, particularly adapted for indoor applications

Large distribution transformers

Three-phase transformers, usually immersed in liquid oil as dielectric insulation and enclosed in a tank with cooling system. Power range: Main use: Main advantages:

2 500 to 20 000 kVA Grid interconnections, industrial applications, special applications such as furnace or railway. Big power with the potential of 35 kV distribution

Medium power transformers

Three-phase transformers, adapted for grid interconnections for short distance transmission lines up to 220 kV. Power range: 250 to 4 000 kVA Main use: Interconnecting grids Main advantages: Big power and high voltage

Large power transformers

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Three-phase transformers, adapted for grid interconnections for long distance transmission lines above 220 kV. Power range: Main use: Main advantages:

250 to 1 000 MVA Interconnecting grids and main power station. Big power and high voltage

Transformers (star-star, delta-delta, star-delta, delta-star) Star-star transformer L1

a1 a1

L1

L2

a2 a2

b1 b1

L2

L3

b2 b2

c1 c1

c2 c2

L3

Figure 3.9: Diagrammatic star-star by means of three single-phase transformers L1

L1

L2

L2

L3

L3

Figure 3.10: Schematic star-star by means of single three-phase transformer

Characteristics of a star-star connection: • Used for small high-voltage applications. • It gives a VL of √3 times the VPH. • Most economical due to minimum insulation required. • Smallest number of turns (windings) required. • Lowest insulation and smallest number of turns per phase. • No phase shift between Vprim and Vsec. • Possibility of a star point forming a neutral on both the primary and the secondary sides.

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Electrical Technology Delta-delta transformer L1

L2

a1 a1

a2 a2

L1

L3

b1 b1

b2 b2

L2

c1 c1

c2 c2

L3

Figure 3.11: Diagrammatic delta-delta by means of three single-phase transformers L1

L1

L2

L2

L3

L3

Characteristics of a delta-delta connection: • Relatively large, low line voltage applications. • No possibility of a neutral on either side. • System that carries large currents for low voltages. • Phase current under balanced conditions only 57,7% of the line current. • Continuous service even if one phase is removed (open VEE system). • No phase displacement between primary and secondary.

Star-delta transformer L1

a1 a1

L1

L2

a2 a2

b1 b1

L2

L3

b2 b2

c1 c1

c2 c2

L3

Figure 3.12: Diagrammatic star-delta by means of three single-phase transformers

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Three-phase transformers L1

L1

L2

L2

L3

L3

3

Characteristics of a star-delta or delta-star connection: • Combines advantages of both star and delta in one transformer. • Used in power distribution systems (where voltage must be stepped up or down). • No neutral displacement under unbalanced loads. • The star side can provide a neutral. • Polarity conscious due to an inherent electrical 30° phase shift between primary and secondary.

Delta-star transformer L1

a1 a1

L1

L2

a2 a2

b1 b1

L2

L3

b2 b2

c1 c1

c2 c2

L3

Figure 3.13: Diagrammatic delta-star by means of three single-phase transformers L1

L1

L2

L2

L3

L3

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Electrical Technology Calculations: (Efficiency at 100%) It is extremely helpful always to make a sketch of the information given. Fill in your calculated values as they are calculated. It helps to keep track of where you are in the process. It may be a three-phase transformer but remember that it is made up of three single-phase transformers placed opposite each other. Therefore ANY calculation that is done using the transformer formulae must be done using phase values only. Example 1: A three-phase delta-star transformer has a winding ratio of 1040:23. a) Calculate the value of the line voltage on the secondary side if the primary side is connected to a line voltage of 11 kV. b) Calculate the line current on the secondary side if the transformer had a rating of 15 kVA. Answer 1: 11kV

1040

:

23

a) On the supply side VL = VPH = 11 kV

The transformer formula may only be used with phase values

VPH

b)

Papparant

Example 2: A 20 kVA star-delta transformer has primary and secondary line voltages of 6 kV and 400 V respectively. Ignore all losses and calculate: a) The secondary line current of the transformer. b) The primary line current of the transformer. c) The windings ratio rounded off to the nearest turn. d) If the power factor is 0,97, calculate the effective (true) power of the transformer on the secondary side.

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Answer 2: a)

Take note Remember that 0,97 is substituted, and not cos 0,97.

Papparant

b) c)

Papparant

(Prim)

d)

Example 3: A 10 kW delta-connected load with a lagging power factor of 0,8 is connected to a delta-star transformer with a windings ratio of 137:3. The transformer is connected to a 11 kV supply. a) Draw a schematic representation of the system. Ignore all losses and calculate: b) The secondary line voltage of the transformer. c) The secondary line current. d) The current flowing through each phase of the load. e) Primary line current. f) The apparent power that the transformer must deliver.

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Electrical Technology Answer 3: a)

Take note Whenever a question refers to supply voltage, it means a line value.

b) On the primary side VL = VPH = 11 000 V (delta) (Remember...all values must be phase)

c) The load is rated as 10 kW and cos θ = 0,8

d) The load is in delta

e) On the secondary star side of the transformer IL = IPH = 17,3 A (star)

Remember, calculation with transformer ratios must be done with PHASE values only.

f) Deliver means on the secondary side

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Construction of transformers This was completed in the section under types of transformers and construction.

Application of transformers 1. Star-star • Used for small high-voltage applications. 2.

Delta-delta • Relatively large, low line voltage applications. • They handle large current at low voltages. • No need for a stable neutral. • Isolation transformers in power converters. • Used in industrial applications if there is no need for a neutral.

3. Delta-star and delta-star • Generator step-up transformers (usually with the generator connected to the delta and the star connected to the secondary) • Substation transformers – with the delta connected to the grid and the star connected to the load substation. • Used in power distribution systems (where voltage must be stepped up or down). • Most common in commercial and industrial delta-star high-voltage transmissions.

Cooling methods A transformer has to work 100% of the time which means current flows all the time. This flow of current causes heat in the transformer and in smaller transformers this heat is mostly manageable and acceptable. In larger transformers, however, this is not the case and alternative methods of cooling need to be added to ensure that the transformer does not overheat and burn out or cause damage elsewhere. The general rule of thumb is that the transformer’s life expectancy is halved for every 8° C increase in operating temperature. The most common methods of cooling are by means of: • circulating air • oil filled self-cooled • oil filled water-cooled. • Air blast type (or air ventilated type) This type is used for transformers that use voltages below 25 000 volts. The transformer is housed in a thin sheet metal box open at both ends through which air is blown from the bottom to the top. This could be natural movement of air or by means of cooling fans. • Oil filled self-cooled type Medium-sized distribution transformers are the oil filled self-cooled type. The windings and core of these transformers are mounted in welded, oil-tight steel tanks provided with a steel cover. The tank is filled with purified, high quality insulating oil. The oil helps to transfer the heat away from the core and the windings to the surrounding case. From there it is radiated out to the surroundings. The movement of the oil inside the tank is a result of convection.

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Take note The oil that is used is a highly refined mineral oil that helps cool the transformer and windings and as a result protects the conductor insulation.

For smaller-sized transformers, the tanks are usually smooth surfaced but for large-size transformers a greater heat radiation area is needed. This is achieved by corrugating the metal case that houses these transformers.

The even larger-sized transformers make use of radiation pipes on the outside of the tank to speed up the cooling process.

Figure 3.14: Shown on the left is a transformer with a smooth tank and on the right one with corrugations. These tanks are filled with special transformer oils.

2. Oil filled water-cooled type The self-cooled type is quite expensive. The water-cooled type offers a more economic alternative. The same method is used where the windings and the core are immersed in the oil. The only difference is that a cooling pipe is mounted near the surface of the oil, through which cold water keeps circulating. This water carries the heat away from the transformer to the outside. This is usually used on transformers in high-voltage transmission lines.

Figure 3.15: An oil filled water-cooled transformer that uses radiator pipes and speeds up cooling of the transformer.

Safety and transformers The electrical substation changes the extremely high voltage carried by the longdistance transmission lines into the lower distribution voltage that serves homes and businesses. In order to perform this task, the substation needs a direct feed from the transmission line into the substation structure. The equipment within the substation is always under high-voltage electrical load. Substation technicians, coop linesmen and maintenance crews are trained to work in high-voltage situations and to recognise and avoid potential hazards in the substation.

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It is important to be extremely careful and take whatever safety precautions one can to ensure one’s safety. It is mostly common sense but people do really silly things. When connecting • A good, permanent low impedance ground connection must be made to the tank by using the ground pad(s) provided. • During installation, the recommended sequence of connections is to make all ground connections first, then the low-voltage connections and finally the high-voltage connections. (The transformer should be removed from service by reversing the above sequence of connections.) • Carefully check the transformer nameplate for its rating and the connections that can be made to it. • Ensure that connecting terminals are attached tightly and that a back-up nut is available on the secondary stud. Mounting and environment • The transformers should be mounted on a flat level concrete pad • The concrete pad should be strong enough to support the full weight of the transformer. • The area should be securely fenced off. This is to keep children, animals and curious people at a distance. • Appropriate ‘High Voltage’ danger signs should be clearly displayed. • The power, voltage and current rating should be displayed on a nameplate.

Protection http://itee.uq.edu.au/~elec4302/Document%20Folder/Lecture%20PDF%20files/ Transformer%20Protection%206%20slides%20per%20page.pdf There are a number of protection devices associated with transformers. Their main function is to prevent the transformer from overheating due to the high currents that flow through them. Excessive high currents can damage or burn the insulation between the windings and cause a short circuit or start a fire. Some of the protection devices are listed below with a brief explanation of their purpose. A transformer does not use all these listed options as protection. The application of the transformer and load conditions will determine whether one or more are used. 1. Electrical • Fuses • Over-current protection • Earth-fault protection • Differential protection • Over-exitation protection 2. Mechanical • Accumulated gases (Bucholz relay) • Pressure relays 3. Thermal • Hot spot temperature protection • Heating due to over exitation

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Electrical protection Fuses (also called power fuses) Have been used for many years to provide transformer fault protection. They are used in transformers up to 10 MVA and provide reasonable protection at low cost. If the current exceeds the rating of the fuse, it breaks the circuit and disconnects the supply from the load. Over current These are used in transformers up to a rating of 50 MVA. Between 10 MVA and 50 MVA, they are sufficient to serve as the main protection, but for transformers with a rating of more than 50 MVA they should only serve as back-up protection. These devices have higher sensitivity and fault clearing times when compared to fuses. They do not have the same maintenance and cost advantages of power fuses. Earth-fault relay This relay detects either the current in the neutral or the sum of line current (which will be the residual sum of the line currents). Depending on location of the current transformer (CT) to detect these currents, it will in all likelihood only see earth faults.

Take note When a slight fault occurs in the transformer, the small bubbles of gas which pass upwards towards the oil conservator tank are trapped in the Bucholz relay housing causing its oil level to fall. As a result, the upper float drops and activates the external alarm switch. If gas continues to be generated, then the second float operates the second switch that is normally used to isolate (trip) the transformer.

Differential protection These are the most widely accepted relays for transformer protection. They compare the current values flowing into and out of the transformer and allow for a 15% mismatch in values before they activate, disconnecting the supply from the transformer. Their fast operating speeds vary from 10 ms to 40 ms. Over exitation These are activated when an increase in the system’s voltage or a decrease in frequency occurs. Transformers can withstand an increase in voltage with an associated increase in frequency (up to a point) but not the other way around.

Mechanical protection Accumulated gases (Bucholz relay) This system provides very sensitive protection for oil-filled transformers. However, they are only suitable for transformers fitted with an oil-filled conservator. The conservator will capture gas as it rises through the oil. The accumulated gases are caused as a by-product of the insulating oil decomposition, which in turn is created by the excessive heating of the transformer. The Bucholz relay provides sensitive protection for oil-filled transformers. It provides very good transformer protection for internal faults. Conservator

Bucholz relay

Transformer tank

Wiring to trip and alarm circuits

Figure 3.16: A sketch that shows the position of the Bucholz relay as well as what it really looks like

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Pressure relays The pressure relay relies on the transformer’s internal pressure rise that occurs due to a fault. A fault would cause more current to flow, which in turn will heat up the transformer oil. This leads to rising bubbles which will cause internal pressure in the tank. The relay accommodates for pressure changes due to outside temperatures.

Thermal protection Hot spot temperature In any transformer design, there is a place in the winding that the manufacturer believes to be the hottest spot in that transformer. A sensor is placed there and an instantaneous alarm or trip setting is set for that hot spot area. Heating due to overexitation Windings temperature and hot spot temperature are the main areas of concern but both transformer oils (and water temperature if it is an oil filled water-cooled type) are usually measured. If it exceeds a pre-set value, it sets off an alarm or initiates circuit breakers to disconnect the supply from the load. Temperature measurement is done using a Bourdon tube, which actually detects the associated rise in pressure as a result of the high currents. Pointer Cross section no internal pressure Bourdon tube Cross section with internal pressure Bourdon tube

Pinch

Pivot

Mechanical linkage

Bourdon tube

Inlet pressure

Figure 3.17: A sketch of Bourdon’s tube, and the meter as we know it

Take note The strain of the material of the tube is magnified by forming the tube into a C shape or even a helix, such that the entire tube tends to straighten out or uncoil, elastically, as it is pressurised. Eugene Bourdon patented his gauge in France in 1849 and it was widely adopted because of its superior sensitivity, linearity and accuracy.

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Activity 1 1. What is the purpose of transformers? 2. What is the basic principle of operation of a transformer? 3. Explain the operation of the transformer. 4. Why can a transformer not work on DC? 5. What is meant by a laminated iron core and what is its main function? 6. What do we know about the power rating of transformers? 7. Show by means of neat sketches the difference between a SINGLE three phase transformer and THREE single-phase transformers (Diagrammatic). 8. Name FOUR types of losses that occur in transformers. 9. Name three advantages of three-phase transformers over single-phase transformers. 10. Name two core types commonly used to make transformers. 11. Name the characteristics of a star-star connection. 12. A farmer uses a three-phase delta-star transformer with a windings ratio of 150:3 to reduce a line voltage from 11 kV. He uses both single- and three-phase motors on his farm. a) Represent the system by means of a neat schematic sketch. Use ONE three-phase motor and ONE single-phase motor to represent the load on your sketch. b) If the three-phase motor is star connected, determine the voltage across each phase of the motor. c) This three-phase motor has a rating of 12 kW at a power factor of 0,8 lagging. d) Calculate the line current supplied at the secondary of the transformer. e) Calculate the kVA rating of the transformer (assume efficiency is 100%). The transformer is positioned at the back of a shed on the farm. One day the transformer suddenly starts to get very hot. f) Name TWO reasons why it may be getting hot. g) Name TWO ways of reducing this temperature. 13. A balanced three-phase star-connected load draws 10 A from a star delta transformer with a 100% efficiency. The transformer has a transformer ratio of 15:1. The primary winding is connected to a supply voltage of 11 kV/50 Hz. The power factor of the load is 0,9. a) Draw a fully labelled diagram of the above circuit. b) Calculate the secondary line voltage. c) Calculate the total kVA rating of the transformer. d) Write down the disadvantage of a low power factor on a load. e) Why will a transformer NEVER have 100% efficiency? Give FOUR reasons. f) State any TWO methods of how overheating in transformers can be prevented. 14. Name some electrical protection employed on power transformers.

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Practical Activity 1 To investigate the relationship between the voltages and currents in a three phase Star-Delta Transformer. Equipment: • A clamp on ammeter • Multimeter with appropriate Voltage scale • A Star Delta connected transformer using three single phase transformers of the same rating. Steps: 1. Connect the three transformers in a star delta configuration as shown below. L1

L2

L3

L1

L2

L3

Star-Delta Three Phase Transformer Take the following readings and record in the table supplied below: 2. Measure the line and phase voltages on the primary side. 3. Measure the line and phase voltages on the secondary side. 4. Measure the line and phase currents on the primary side. (Use the clamp on ammeter.) 5. Measure the line and phase currents on the secondary side. (Use the clamp on ammeter.) Connection

Meter readings

1. STAR PRIMARY

VLINE

1. STAR PRIMARY

VPHASE

2. DELTA SECONDARY VLINE 2. DELTA SECONDARY VPHASE 3. STAR PRIMARY

ILINE

3. STAR PRIMARY

IPHASE

4. DELTA SECONDARY ILINE 4. DELTA SECONDARY IPHASE

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Electrical Technology Conclusion: 1. What can be observed from the line and phase voltages on the primary side? 2. What can be observed from the line and phase voltages on the secondary side? 3. What can be observed from the line and phase currents on the primary side? 4. What can be observed from the line and phase currents on the secondary side?

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Chapter 4 Three-phase motors and starters

A

B Three-phase

Applications

A

B Calculations

Testing

A

B Starting methods

Protection

4


Introduction In this chapter, attention will be given to three-phase motors and how these motors can be safely connected to the three-phase supply. Matters discussed will be the advantages of a three-phase motor compared to a single-phase motor. Basic construction will be covered as well as practical applications. Calculations will include the line and phase voltages, power, power factor, synchronous speed, slip and efficiency. Specific attention will be given to losses that occur in motors. This refers to electrical (fault finding) as well as mechanical losses. Once the theory is covered, the students will then learn how to design starter circuits safely (control and main circuits) and connect motors safely to the supply for specific functions.

Principle of operation of the three phase squirrel cage induction motor

Figure 4.1: Three-phase squirrel cage induction motor

An electric motor is used to convert electrical energy into mechanical energy, usually by the rotating shaft driving or powering a load. In this chapter, the focus will be on how the motor is put together, the purpose of the various parts, applications for the motor and practical application for students by physically wiring a motor to a supply, with all the control and safety features required that go with it.

Construction Any motor consists of a lot of parts. These are: • The rotor itself. As the name indicates, this is the part that rotates. • An axle (or shaft) on which the rotor turns. • The metal frame that houses all the other parts (called the stator, as it is stationary). • End plates that hold the bearings and also seal off the rotor. • The windings of wire in special slots placed on the metal frame. • The cooling fan. • The protection cover for the cooling fin. • The terminal box that holds the six end wires of the three coils.

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Three-phase motors and starters

4

This is basically summarised into three main parts. 1. The rotor 2. The stator 3. The stator windings Lifting eye Grease plug

Field stator windings Grease plug

Cooling fan

Shielded roller bearing Shaft Drain plug Cast iron frame and end bracket

Take note Another kind of threephase motor that is used is called the ‘wounded rotor’ because the rotor has windings of copper wire wound around the rotor core. It is also known as a slip ring motor.

Cooling fan cover Laminated steel rotor Figure 4.2: Cutaway view of a motor showing most of the parts

Each of the words in the name of the ‘three phase squirrel cage induction motor’ spells out exactly what it is. Three phase

The motor requires a three phase supply in order to create a rotating stator field.

Squirrel cage Refers to the type of rotor. A squirrel cage rotor has metal rods (or strips) that are short circuited at the ends by means of a metal ring.

Induction Current is induced in the metal strips of the rotor.

Motor A motor is used to convert electrical energy into mechanical energy. The induced currents cause the rotor and the shaft to turn.

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The stator windings and the rotating stator field Take note The generator principle using the right-hand rule. Grab the coil with the right hand with the thumb pointing in the direction of the current flow and the encircling fingers will show the direction of the magnetic field around that conductor.

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Although the process of creating a rotating stator field is in fact far more complex that it sounds here, we have kept it as simple as possible and have made use of sketches to get the basic concept across. The stator windings are the three coils of wire that are placed in special slots on the inside of the frame. Their starting points are placed 120° apart and remember that a coil goes around a core a few times. To keep the explanation simple, only one loop of each phase is indicated and it shows where the beginning point enters on one side and the end point exits on the other side. We will show, by using 30° intervals, how the resultant field shifts and results in a rotating stator field.

Three-phase motors and starters R1

R1

Y2

B2

B1

Y1 R2

Figure 4.3: Magnetic field at A

R1 B2

B1

4

Y1 R2 Magnetic field at B

B2

B1

Y1 R2

Magnetic field at C

At point A indicated on the waveform • Phase 1 will have maximum current flowing through it (in a positive direction). • Phases 2 and 3 will have equal currents but in the negative direction to phase 1’s current. The resultants of these three magnetic fields have been plotted on the sketch shown in figure 4.3 on the left. At point B indicated on the waveform • Phase 1 will have less than maximum current flowing through it but in the positive direction. • Phase 2 has no current flowing through it and hence no magnetic field. • Phase 3 will have more current flowing through it but in the negative direction to phase 1’s current. The resultants of these two magnetic fields have been plotted in figure 4.3 in the middle. The 30° shift is already clearly visible. At point C indicated on the waveform • Phase 1 will have even less current flowing through it but still in the positive direction. • Phase 2 will now have a current also in the positive direction. Therefore the magnetic fields of phase one and two will be in the same direction. • Phase 3 will have maximum current flowing through it, but in the negative direction. This results in a further 30° shift of the resultant magnetic field as shown figure 4.3 on the right. Basically, we cannot go through this whole process when we want to explain how a motor works, but the bottom line is that when a three-phase supply is connected to the stator windings, it results in a rotating stator field.

Did you know? http://www.you tube.com/watch?v= TxUWXgqEvXo Excellent video on YouTube to show the rotating stator field.

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Electrical Technology To change direction of rotation This concept can also be used to show how the direction of the rotating stator field can be reversed, thereby changing the rotational direction of the rotor. Ph1 (L1)

Ph3 (L3)

Ph2 (L2)

Ph3

1.

Take note To swop the direction of rotation of a three-phase motor, any TWO lives can be swopped.

• • •

Ph1

Ph1

Ph2 Ph3

2.

Ph2

3.

In sketch 1, the first phase is at maximum and has the strongest magnetic field. In sketch 2, the second phase will be at maximum, although 120° later. It will then have the strongest magnetic field. In sketch 3, the third phase will be at maximum but 120° after the second phase.

This shows a clockwise rotation of the stator magnetic field. If any TWO lines are swopped, it will change the direction of rotation of the stator field and, as a result, the rotor as well. We have swopped line 2 and line 3 around. The black dot now rotates anticlockwise. Ph1 (L1)

Ph2 (L2)

Ph3 (L3) 1.

Ph1

Ph1

Ph2

Ph3 Ph2

2.

Ph3

3.

Operation of the three-phase squirrel cage induction motor • When the three-phase supply is connected to the motor, a rotating stator field (RSF) is generated automatically. • This rotating stator field cuts the metal rods of the squirrel cage rotor (which is standing dead still at this moment), inducing a large current in it. • These induced currents in the rotor create their own rotor magnetic field. • The rotating stator field and the rotor magnetic fields react with each other. • A force is exerted between the two fields (it is called torque) and the rotor starts turning in the same direction as the rotating stator field. • As the rotor speed increases, less current is induced in the metal rods of the rotor. (The reason is that the relative speed between the rotating stator field and the rotor speed is reduced as the rotor speed increases.) • The current induced in the rotor when starting is typically 10x higher than at full speed. In an ideal world where there are no losses, the rotor speed should catch up to the speed of the rotating stator field. But it doesn’t, due to the following factors: • friction of the bearings • eddy currents (induced in the laminated plates) • induction resistance losses • windage losses due to the wind resistance of the fan.

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Advantages of the three-phase induction motor over single-phase motors The squirrel cage motor is one of the most popular motors around. Some factors for this are their simple construction and robustness. They have many advantages compared to most other motors. • Self-starting (they do not need additional circuitry like some other motors). • Robust. • Simple construction. • Low maintenance due to less moving parts that cause wear and tear. • Reduced fire risk as there are no moving parts that cause sparking. • Reduced running cost due to less maintenance. • High starting torque. • Wide range of applications. • Easy to change direction of rotation. • Can be connected in star or delta. • Higher efficiency. • Higher power factor. • Wide range of power ratings.

Connections of the motor in star or delta One of the advantages of the three-phase motor is that it can be connected to run in star or in delta. The 6 ends of the THREE coils of the stator winding (two per phase) are brought out to the terminal box and attached to 6 bolts. They are always connected to the same terminals so that when the connections for star or delta are made they are always the same. • Star To make the star connection, the bottom three terminals are joined together. • Delta To make the delta connection, the vertically opposite terminals are joined together.

Figure 4.4: Terminal boxes of electrical motors showing a star connection on the left and a delta connection on the right

Applications This motor has many many applications in everyday life. • lathes • pedestal drilling machines • motors that operate the lifts in skyscrapers • bench grinders • fans for ventilation shafts in mines • conveyor belts in factories • motor assembly plants • water pumps on farms • commercial hoists • cranes.

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Electrical Technology Calculations on synchronous speed, slip, power and efficiency Before any calculations can be done, there has to be a clear understanding of what the words mean. Synchronous speed This is the speed at which the rotating stator field rotates. Slip

The slip is an indication of how much the rotor speed differs from the speed of the rotating stator field. The larger the load, the more the slip.

Full-load slip varies from less than 1% in high torque motors and varies between 5% to 6% in lesser torque motors. Motor size kW

0.5

5

15

50

250

Typical slip (%)

5

3

2,5

1,7

0,8

Figure 4.5: A table showing typical slip for various power rating motors

Synchronous speed Take note When the motor starts rotating, the slip is 100%. This is due to the fact that the rotating stator speed is instantaneous when the supply is connected but the rotor is standing still. The slip is reduced when the rotor starts to turn.

The synchronous speed of an AC motor is the speed of the rotating magnetic field created by the stator windings due to the supply. The synchronous speed ns is given by: 60 × f ns = _____ p n f p

synchronous speed in revolutions per minute (RPM) n is always an integer fraction of the supply frequency frequency in hertz (Hz) the number of magnetic pole pairs per phase

Example 1: A small 3-phase 50Hz motor has six magnetic poles (organised as three opposing pairs 120° apart, each powered by one phase of the supply current). This means 2 poles per phase (meaning one pair of poles per phase). Calculate the synchronous speed. f = 50 p=1

(This means pole pairs per phase)

Synchronous speed ns = 60 × f = 60 × 50 = 3 000 RPM p 1 A table that shows typical synchronous speeds for various pole pairs is given below. It is not realistic to change the frequency as that is what the service provider (Eskom) delivers to the national grid and we cannot change that easily. However, the speed of the motor can be controlled by using different numbers of pole pairs.

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Three-phase motors and starters No. of poles/phase (Not pole pairs)

Speed at 50 Hz

Speed at 60 Hz

2 (1 pair)

3 000

3 600

4 (2 pairs)

1 500

1 800

6 (3 pairs)

1 000

1 200

8 (4 pairs)

750

900

10 (5 pairs)

600

720

12 (6 pairs)

500

600

16 (8 pairs)

375

450

20 (10 pairs)

300

360

4

Take note The frequency of distribution in South Africa is 50 Hz while in America it is 60 Hz.

Figure 4.6: The table shows synchronous speeds for different numbers of poles per phase used

Example 2: Calculate the total number of poles in a three-phase motor if the supply frequency is 50 Hz and the synchronous speed of the motor is rated as 500 RPM. Answer 2: f = 50 Hz ns = 500 RPM Pole pairs per phase p = 60 × f = 60 × 50 = 6 500 ns This means 6 × 2 = 12 poles per phase and 12 × 3 = 36 total poles in total Example 3: Calculate the synchronous speed of a 50 Hz three-phase motor with 12 poles in total. f = 50Hz Poles per phase = 12 = 4 poles per phase = 2 pairs 3 Synchronous speed ns = 60 × f = 60 × 50 = 1 500 RPM p 2

Calculations on slip The slip is an indication of how much the rotor speed differs from the speed of the rotating stator field. The larger the load, the more the slip. It can be calculated using the following formula. ns – nr Slip = _______ ns S ns nr

slip as a percentage synchronous speed in RPM rotor speed in RPM

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Electrical Technology Example 1: Calculate the slip of a motor with a synchronous speed of 3 000 RPM and an associated rotor speed of 2 800 RPM.

Take note The answer on the calculator reads 0,066666 but this must be converted to a percentage which equates to 6,67%.

ns – nr Slip = _______ ns

3 000 – 2 800 = ___________ = 6,67% 3 000

Example 2: Calculate the speed of the rotor if the synchronous speed is 3 000 and the slip is 5% Answer 2: The formula must be mathematically manipulated, and this gives Rotor speed (nr) = ns (1-S) = 3 000 (1 – 0,05) = 2 850 RPM OR if we have lost 5%, it means that the rotor is turning at 95% of the synchronous speed Rotor speed (nr) = ns × 95% = 3 000 × 95 = 2 850 RPM 100

Calculations on efficiency

Take note All motors should carry a nameplate and the power rating given on this nameplate is usually the full load power ability of the motor.

No motor is 100% efficient as there are losses involved. These losses can be summarised as: • Copper losses Due to the resistance of the copper wires in the rotor and the resistance of the metal rods of the stator. • Iron losses Due to the hysteresis of the core materials as well as the eddy currents induced in the rotor. • Mechanical losses Due to friction of the bearings and the wind resistance of the cooling fin. These losses can be considered a power loss and must be factored in when calculations are done to determine the active power (P) the motor can deliver at the shaft.

Figure 4.7: Nameplate of an induction motor. It shows all the required information regarding phases, type of connection, power, voltage, current and speed.

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Example 1: A three-phase 415 V induction motor draws a current of 29 A at a power factor of 0,87 and the motor is 83% efficient. Determine the maximum power the motor can deliver at the shaft. Pact (P) = √3 (IL)(VL)(cos θ)ŋ = √3 (29)(415)(0,87)(0.83) = 15,05 kW This means that 17% of the power is gone due to losses. Example 2: Determine the efficiency of a rated 20 kW motor if the power losses are calculated at 3 kW.

Take note The answer on the calculator reads 0,85 but this must be converted to a percentage which equates to 85%.

20 000 – 3 000 Pin – losses = ____________ = 85% Efficiency (ŋ) = __________ Pin 20 000 The nameplate of a motor usually gives the shaft output power measured in watts (or kW). Sometimes a customer wants to install a motor but the motor must meet the efficiency demands according to certain standards. The table below shows these standards for motors constructed according to NEMA (an American electric motor corporation) Power (kW)

Minimum nominal efficiency(%)

1–4

78,8

5–9

84,0

10 – 19

85,5

20 – 49

88,5

50 – 99

90,2

100 – 124

91,7

> 125

92,4

Figure 4.8: Electrical motors constructed according to NEMA must meet the efficiencies above

Electrical and mechanical inspections/faultfinding After the manufacture of any motor, it needs to be tested comprehensively to ensure that it works as intended. There cannot be any faults which occurred during the manufacturing process. When the customer buys the motor, it needs to be 100% operational. When a company installs a motor, they have a detailed maintenance schedule that is followed to ensure an optimal life span for the motor. (Try and get the motor to last as long as possible.) There are certain tests that must be performed during these scheduled checks. Sometimes a motor malfunctions, burns out, gets wet or full of dust. Such motors will be repaired where possible. After repairs, the motor will have to be tested again to ensure everything is in proper working order before reinstallation. Knowledge of these tests enables one to do faultfinding and/or troubleshooting if a problem occurs.

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Electrical Technology Electrical inspections The following are categorised as electrical inspections that should be carried out before commissioning an electric motor: • Continuity/resistance test. • Insulation resistance between coils (windings). • Insulation resistance between coils and earth. • Check for obvious open/exposed wires. These would be leads or cables joining the motor to the supply. Any bare wire carrying electricity is just bad news. • Check the electrical connections in the terminal box. It is important that the nuts where the supply leads/cables are joined to the terminal box are tight. If not, this may lead to sparking, unwanted heat and unstable supply voltages and current required for optimal performance.

Take note Although 1 MΩ is generally accepted in motor testing, some companies who work with precision motors demand 0,5 MΩ or even lower.

Figure 4.9: The terminal box for a three-phase motor showing the nuts that must be tight for proper electrical connection

• Check for a proper earth connection. After installation the motor should be earthed to the frame on which it is mounted.

Mechanical inspections The following are categorised as mechanical inspections that should be carried out before commissioning an electric motor: • Check the shaft for movement. As bearings wear out, they start causing a certain ‘play’ in the shaft. If the play becomes excessive, it will cause unwanted heat and the rotor could start scraping against the inside of the stator (this test is not possible if the motors are really big). • Check the bearings for smooth rotation. When the rotor is turned by hand, it should turn smoothly (usually without the load connected). Any ‘rough’ feeling should lead to the bearings being replaced. • Check the housing for cracks. • Check that the motor is bolted to the frame. Check that the bolts securing the motor to the frame/platform are tight. Powerful motors will actually rip themselves off the frame if not secured. • Check the end plates are fastened properly. Sometimes the slight vibrations as the motor turns cause these bolts to loosen themselves. If not tightened, the rotor might shift slightly and scrape against the stator’s inside.

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Three-phase motors and starters • •

4

Check cooling fan blades. They should be intact and turn with the rotor. Any part broken off the fan can cause the motor to be unbalanced and vibrate excessively (almost like a car with an unbalanced wheel). Check the protection cover over the fan. The cover should be firmly in place and the fan should not scrape against the protection cover.

Figure 4.10: An example of a broken fan and the cover in place on the motor

Motor testing and troubleshooting

Figure 4.11: Meggers are used to test the continuity and insulation of electric motors

Take note Electrical measurements MUST be done with a Megger or insulation tester and NOT a multimeter. Meggers/ insulation testers perform tests at a much higher voltage than the supply to the motor and the logic is that if it’s good at higher value, it will definitely be acceptable at the operating voltage.

To recap, the three electrical tests, the reasons for testing as well as the settings on the Megger are set out below. Type of test

Reason for test

Meter settings

Continuity/ resistance test

To ensure that the resistance of stator windings is realistic (this will depend on the size of the motor) and the three coil readings, which should be within 10% of each other.

Set to the lowest ohm scale

Insulation resistance between coils (windings)

To ensure that there is no electrical connection between any of the three coils. Ideally, it should be infinity but any value above 1 MΩ will be acceptable.

Set to the MΩ scale

Insulation resistance between coils and earth

To ensure that there is no electrical connection between any of the three coils and earth. Ideally, it should be infinity but any value above 1 MΩ will be acceptable.

Set to the MΩ scale

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Figure 4.12: The settings of the Megger to show the Ω scales and the MΩ scales

The three electrical tests will be carried out on the terminal box of the motor. This process will be split into three steps.

Figure 4.13: Terminal box where testing must take place

Strep 1 Take the measurements between each of the six terminals and earth, and record them in a table. Step 2

Interpret the readings and put them onto a sketch.

Step 3

Write a comprehensive report on the condition of the motor by referring to each of the electrical tests individually.

Step 1: Recording of readings Remember to use a Megger or an insulation tester when performing this task. Multimeters operate either off a 1,5 V or 9 V battery, so using them to test 220 V/380 V motors is completely useless. The recording form will help to ensure that testing was done between all the terminals in a logical manner. Missing just one recording renders the whole test invalid. Take note The British method for labelling terminal boxes is to use the letters A, B and C.

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The German method for labelling the terminal box will be used in this section of the work (U, V and W). All readings greater than 1 MΩ can simply be recorded as infinity, as this is interpreted as an ohm value that is large enough to have no effect on the outcome of the results.


4

Recording template for 3-phase motor testing U2 – V2

V2 – U1

W2 – E

U2 – W2

V2 – W1

U1 – W1

U2 – U1

V2 – V1

U1 – V1

U2 – W1

V2 – E

U1 – E

U2 – V1

W2 – U1

W1 – V1

U2 – E

W2 – W1

W1 – E

V2 – W2

W2 – V1

V1 – E

Step 2: The sketch Use all the recorded values and fill them in on the sketch of the terminal box shown below. Doing this will enable one to form a better idea (picture) of where the readings fit in and it makes the interpretation for the report a lot easier. To make this process of transferring values from the table to the sketch much easier and to eliminate silly mistakes, it helps to do them in this order. • Fill in the values of the coils first (draw them as coils). • All short circuits (if any). • All low ohm values. • All high ohm values. • Use common sense. For example, 20 Ω plus 900 kΩ is going to show a Megger reading of 900 kΩ. The 20 Ω is so small compared to 900 kΩ it will not even feature. But we must keep this in mind when doing the sketch.

Earth

Step 3: The report Using the sketch to write the report makes things a lot easier. 1. Continuity of the coils Winding

Reading

Accept or not

Motivate

Winding U Winding V Winding W

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Electrical Technology 2. Insulation between windings: Winding

Reading

Accept or not

Motivate

Reading

Accept or not

Motivate

Winding U – V Winding U – W Winding V – W

3. Insulation to earth: Winding Winding U – E Winding V – E Winding W – E

Example 1: Step 1: Recording of readings A motor was tested and the following readings were recorded. Recording template for 3-phase motor testing U2 – V2



V2 – U1



W2 – E

82 Ω

U2 – W2



V2 – W1

85,3 Ω

U1 – W1



U2 – U1

1,5 kΩ

V2 – V1

82 Ω

U1 – V1



U2 – W1



V2 – E

85,3 Ω

U1 – E



U2 – V1



W2 – U1



W1 – V1

167,3 Ω

U2 – E



W2 – W1

82 Ω

W1 – E

0Ω

V2 – W2

3,3 Ω

W2 – V1

85,3 Ω

V1 – E

167,3 Ω

Step 2: The sketch Coil values Short circuits Low ohm High ohm

U = 1,5 kΩ V = 82 Ω W = 82 Ω W1 – E = 0 Ω 3,3 Ω, 82 Ω, 85,3 Ω, 167,3 Ω nothing in this particular example

Earth

V1 80


4

Step 3: The report Using the sketch to write the report makes things a lot easier. 1. Continuity of the coils Winding

Reading

Accept or not

Motivate

Winding U

1,5 kΩ

No

Not realistic and not within 10% of V and W

Winding V

82 Ω

Yes

Realistic and within 10%

Winding W

82 Ω

Yes

Realistic and within 10%

2 Insulation between windings Winding

Reading

Accept or not

Motivate

Winding U – V



Yes

Reading above 1 MΩ

Winding U – W



Yes

Reading above 1 MΩ

Winding V – W

3,3 Ω

No

Reading below 1MΩ

3. Insulation to earth Winding

Reading

Accept or not

Motivate

Winding U – E



Yes

Reading above 1 MΩ

Winding V – E

85,3 Ω

No

Reading below 1MΩ

Winding W -E

0Ω

No

Short circuit

Direct-on-line starter with overload We have learnt about three phase generation, three-phase transformers and threephase motors. This section will focus on how to connect the motor to the supply in such a way that all safety precautions are adhered to and all safety mechanism are included.

Function of components on diagrams (Most of the components shown in photos are from specially designed modules to speed up the wiring process for learners under training. They might not look like this in industry situations.) Stop button

This is a normally closed (N/C) push button. Its purpose is to disconnect the supply from the motor the moment it is pressed. It is red in colour (like a red traffic light means stop). In this photo the top button is red, so terminals 1 and 3 will form the N/C contact. The bottom button is green, so terminals 2 and 4 will form the N/O contact.

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Electrical Technology Start button

This is a normally open (N/O) push button. Its purpose is to connect the supply to the motor the moment it is pressed. It is green in colour (like a green traffic light means go). (Both the START and the STOP buttons are spring loaded. This means they return to their normal states if the pressure is released.)

Overloads

Various types making use of different principles are available. However, we will focus on its purpose. Its purpose is to disconnect the supply from the motor the moment the current exceeds the value that the overload is specifically designed for or pre-set to.

Main contactor

To connect the power physically from the supply to the motor terminals. This is designed to carry much larger currents than a normal on/off switch. It also carries auxiliary contacts that will be used in the design of the control circuit.

Holding in contacts Without this contact wired in place, the operator would have to stand with his/ her hand on the start button all the time the motor needs to run. This enables the operator to press the start and then release it. The holding in contact, which is in parallel to the start button, now keeps the main contactor activated or latched. Zero volt coil This is exactly the same as the holding in contact, but it serves a further purpose. In the event of a power failure, the system will switch off, but when the power is restored again (which could be at 3 o’clock in the morning) the system will not start by itself. It will have to be manually restarted by the operator. This connection also provides a small degree of protection against low power supply voltage and loss of a phase. However, since contactor coils will hold the circuit closed with as little as 80% of normal voltage applied to the coil, this is not a primary means of protecting motors from low-voltage operation.

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Principle of operation

Figure 4.14: A direct on line starter connected to a motor control board

A direct on line (DOL) or across the line starter applies the full line voltage to the motor terminals. This is the simplest type of motor starter. A DOL motor starter also contains protection devices and, in some cases, condition monitoring. Any circuit design is usually done in two parts. • Control circuit This part normally operates off a single-phase supply (any line and neutral). It is at a lower voltage and this contains all the parts that control and protect. It NEVER contains any loads/lights, etc. • Main circuit It contains the overloads, main contactor and load. And nothing else. Explanation for the circuit diagram shown in Figure 4.15 and 4.16 • When the start button is pressed, current can flow from the live through all the closed contacts points. That is the overload (O/L), the stop button (N/C), the start (pressed and closed), the main contactor (MC) to neutral. • This activates the MC 1 and closes the holding in MC 1 (N/O). • This holding in MC 1 is now in parallel with the start button. • The operator removes his/her hand from the start, and current bypasses the start via the holding in MC 1 and keeps the MC 1 itself activated. • At the same time, L1, L2 and L3 (in the main circuit) are connected through the overload (N/C), though the three contacts of MC 1 and the supply is safely connected to the motor. When the operator presses the stop button, it opens the circuit, current flow is disrupted, the MC 1 deactivates, disconnecting L1, L2 and L3 from the motor. Wiring diagram The wiring (control as well as main) for the DOL is shown below.

Hold in Start

Figure 4.15: Control wiring diagram for the DOL starter

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Take note Please make sure that the isolator switch is off and the key removed before starting any wiring. This is to ensure that the operator does not work on the system with live wires causing possible shock.

Take note This is a great activity and will help the learner to understand much better what he/she is doing. If the motor does not start, the learner will have the knowledge and the background to deal with the problem.

84

Figure 4.16: Main wiring for the DOL starter

Wiring on a panel This circuit can now be built on the wiring boards. It is strongly recommended that the control circuit be built first and tested, before the main circuit is built. This way it can be established that the control circuit works 100%. If it is not done, there is a real possibility that a faulty (or wrongly wired) control circuit connects the supply to the motor causing some damage to it and endangering the operator.

Figure 4.17: The isolator switch must be off and the key removed before any wiring is done

Troubleshooting/Faultfinding Should there be a problem with the circuit, especially the control circuit, switch OFF the supply and remove the key (NB: it must be switched off). A multimeter can now be used to check for the following. In fact, if time allows, this is a great and interesting activity.


4

Use the multimeter to check (set to the lowest ohm scale) that: • The stop switch Is actually N/C, and whether it changes state if it is pressed. • The start button Is actually N/O, and whether it changes state if it is pressed. • Main contactor Measure the resistance value of the coil itself (always between A1 and A2). It is usually a value between 180 Ω and 600 Ω, depending on the type of contactor. If not, it is problematic. • The overload The terminals should be closed • MC The 3 contacts connecting the three lives to the motor should be open, but when the contactor is pressed in manually it should close. Contacts are sometimes dirty and cause problems. Start and stop The control circuit can now be built exactly according to the design. Once completed, the supply can be switched on. Press the start button The MC 1 should activate and stay activated once the start is released. If the stop is pressed, the MC should deactivate. Switch the supply OFF again. The main circuit can then be wired, again strictly according to the diagram. Press the start button The motor should rotate. Switch off the supply when done. One can change the direction of rotation and check the result by simply swopping any two lives around manually. This can be done either at the supply or at the motor.

Practical: Connecting the DOL starter to a load The following would be typical instructions for a practical activity for a DOL motor starter circuit. This falls under applied theory, as learners would first have to acquire the theoretical knowledge on all the parts before they could go on to the design and build stage. 1. Purpose: To connect a three-phase squirrel cage induction motor to the supply by means of a direct-on-line starting connection. (All interlocks, overloads and safety mechanisms must be shown.)

If the start button is pressed the motor must run and continue to run (even after the start button is released). If the stop button is pressed, the motor must stop. If a power failure occurs, the motor should only be started manually once power is restored.

2. Equipment: 1 × start button 1 × stop button 1 × main contactor with auxiliary contacts 1 × overload 1 × three-phase squirrel cage motor Connecting wires

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Electrical Technology 3. Control circuit diagram: Draw a neat control circuit diagram. All connecting points must be clearly shown. 4. Main circuit diagram: Draw a neat main circuit diagram to show how the main wiring should be connected. All connecting points must be clearly shown. 5. 6. 7. 8.

Connect the control circuit according to your diagram and call the educator when you are sure your circuit is correct.

Phase 1_________ Phase 2__________ Phase 3___________

Connect the main circuit diagram according to your diagram and call the educator when you are sure of your circuit is correct. Explain how you would change the direction of rotation of the motor. Make use of a Megger and measure the resistance value of the 3 coils of the motor: (The supply must be off)

Forward and reverse starter with overload A forward reverse circuit is designed to change the direction of rotation of a motor without having to rewire anything in the circuit. This saves down time (time taken to rewire the circuit), as the company would lose money as a result of machinery standing still. The circuit must have two starting buttons, the first one to make the motor run forward and the second one to reverse the rotation of the same motor. The circuit must have one stop button only. (All interlocks, overloads and safety mechanisms must be shown.)

Function of components on diagrams 1 × stop button Its purpose is to disconnect the supply from the motor the moment it is pressed. 2 × start buttons The first start is to connect the supply to the motor so that it runs in one direction. If the other start is pressed, the motor must reverse the direction of rotation. 1 × overload Its purpose is to disconnect the supply from the motor the moment the current exceeds the value that the overload is specifically designed for or pre-set to. 2 × main contactors The first contactor is to connect the power from the supply to the motor terminals. The second contactor is to swop any two of the live wires around, so enabling a reverse in direction. Holding in contacts The holding in contacts MC 1 (N/O) will keep in MC 1 (it is in parallel to the start forward button), and MC 2 (N/O) will keep in MC 2. This keeps the main contactors activated.

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Zero volt coil: MC 1 (N/O) and MC 2 (N/O) In the event of a power failure, the system will switch off, but when the power is restored again the system will not start by itself. It will have to be manually restarted by the operator. These are MC 1 and 2 (N/O). Holding out contacts: MC 1 (N/C) and MC 2 (N/C) The purpose of the holding out contact is to keep the reverse contactor from activating while the motor is running in the forward direction. The motor must first be stopped before the reverse can be pressed. If the reverse button is now pressed (activating the reverse contactor), the other holding out contact should keep the forward contactor from activating.

Principle of operation The wiring (control as well as main) for the forward reverse is shown below.

MC2 (N/C)

FWD

REV

Figure 4.18: Control wiring diagram for the forward reverse

Figure 4.19: Main wiring for the forward reverse (MC2 swops L2 and L3)

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Electrical Technology • When the 1st button is pressed, current can flow from the live through the overload (O/L), the stop button (N/C), the start (pressed and closed), MC 2 holding out (N/C), through MC 1 to neutral. This activates MC 1, which activates multiple contacts. • It closes the holding in MC 1 (in parallel to the start forward button), opens the holding out (MC1) situated just above MC2 (which now prevents anyone from activating MC2, even if the other start button is pressed), and closes the three contacts of MC 1 that connect L1, L2 and L3 to the motor and it runs in one direction. • The system must be stopped first. Now the reverse button can be pressed. • When the reverse button is pressed, current can flow from the live through the overload (O/L), the stop button (N/C), the 2nd start (pressed and closed), MC 1 holding out (N/C), through MC 2 to neutral. • This activates MC 2, which activates multiple contacts. • It closes the holding in MC 2 (in parallel to the start reverse button), opens the holding out (MC 2) situated just above MC 1 (which now prevents anyone from activating MC 1, even if the other start button is pressed), and closes the three contacts of MC 2 that switch L2 and L3 around and the motor runs in the opposite direction.

Wiring on a panel: (start and stop) This circuit can now be built on the wiring boards. It is again strongly recommended that the control circuit be built first and tested, before the main circuit is built. Take note The isolator switch must be off and key removed before any wiring is done.

Take note A sequential check is a number of checks done in a particular order to ensure that all design requirements have been adhered to.

88

Once everything has been checked in the control circuit, the supply can be switched on and the start button (forward) pressed. A sequential check should be carried out to ensure that all the design requirements were adhered to. The sequential check for the forward reverse: • Press start 1 MC 1 should activate and stay activated once the start 1 is released. If stop is pressed, MC 1 should deactivate. MC 2 should activate and stay activated once the start 2 • Press start 2 is released. If stop is pressed, MC 2 should deactivate. • Press start 1 If start 2 (reverse) is now pressed, nothing should happen. Press stop to deactivate MC 1 • Press start 2 If start 1 (forward) is now pressed, nothing should happen. Press stop to deactivate MC 2. • All possibilities have been tested for and if OK then the control circuit is 100% functional.


4

ONLY NOW can the main circuit be wired. Remember, only neutrals should be connected to the bottom of the MCs and nowhere else. If checked and all looks right, reconnect the supply. • • • •

Press start 1 (forward) Press stop Press start 2 (reverse) Press stop

The motor will rotate in one direction. The motor will rotate in the opposite direction. Motor should stop.

Automatic star-delta starter with overload (Not manual) The automatic star-delta motor starter is designed to start the motor in a star connection to reduce starting currents and voltages and once turning almost at full speed to change the terminal connections to delta. This is done automatically, so after the start button is pressed the motor runs in star immediately and after a preset time on a timer the change to delta happens automatically. The circuit must have one start and one stop button. (All interlocks, overloads and safety mechanisms must be shown.)

Function of components on diagrams 1 ×stop button Its purpose is to disconnect the supply from the motor the moment it is pressed. 1 × start button Its purpose is to connect the supply to the motor. 1 × overload Its purpose is to disconnect the supply from the motor the moment the current exceeds the value that the overload is specifically designed for or pre-set to. 3 × main contactors The first contactor is to connect L1, L2 and L3 from the supply to the top three terminals of the motor. The second contactor (star) is to connect the 3 bottom terminals of the terminal box together. The third contactor (delta) is to connect the three vertically opposite terminals of the terminal box together. Holding in contacts: (MC 1 (N/O) The holding in contacts MC 1 (N/O) will keep MC 1 activated after the start button is released (it is in parallel to the start forward button). Zero volt coil: MC 1 (N/O) In the event of a power failure, the system will switch off, but when the power is restored again the system will not start by itself. It will have to be manually restarted by the operator.

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Take note Alternatively, MC 1 (N/O) may be left out (if there are not enough contacts on MC 1 available), but then the timer must be fed from the bottom of the start button.

Holding out contacts: MC 2 (N/C) and MC 3 (N/C) The purpose of the holding out contacts is to ensure that, during the split-second switchover from star to delta both MC 2 and MC 3 cannot be on at the same time. If it were not for these two holding out contacts, it would cause a short circuit between L1, L2 and L3 which is definitely not acceptable. Holding out contact: MC 1 (N/O) The only purpose for this contact is to ensure that, in the event of MC 1 being faulty, no power can be connected to the rest of the contactors and timers.

Timer: on delay This timer has two contacts, one N/O and the other N/C. When the timer is activated, these two contacts will keep their normal state, and after a pre-set time selected on the timer will change state simultaneously. The function of the on delay is to use the timer (N/C) contact to bring in the star contactor at the same time as MC 1, and after a pre-set time to disconnect automatically the star and bring in the delta contactor via timer (N/O).

Figure 4.20: The on delay timer and a close-up of the available time range on this particular timer

Principle of operation The wiring (control as well as main) for the star-delta starter is shown below.

HOLD OUT

Figure 4.21: Control wiring diagram for the star-delta starter

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Figure 4.22: Main wiring for the star-delta starter

When the start button is pressed, current can flow from the live through the overload (O/L), the stop button (N/C), the start (pressed and closed), through the first three branches. Branch 1 Branch 2 Branch 3

MC 1, activating MC 1 (N/O), and MC (N/O) hold out The timer, which activates The timer (N/C), MC 3 (N/C) and MC 2 itself, activating MC 2

At the same time, the three contacts of MC 1 (that connect L1, L2 and L3 to the motor), as well as the three contacts of MC 2 (that create the star point on the bottom three terminals) close and the motor runs in star. When the pre-set time on the timer expires, the contacts belonging to the timer change state, and MC 2 deactivates while MC 3 activates. The connections have now been changed from star to delta and the motor continues to run in delta until the stop button is pressed.

Wiring on a panel: (start and stop) This circuit can now be built on the wiring boards. It is again strongly recommended that the control circuit be built first and tested. It would also help to use some masking tape and to label the three main contactors as MC 1, MC 2 and MC 3. This will prevent a lot of confusion during the wiring process. Once everything has been checked in the control circuit, the supply can be switched on and start button pressed. MC 1, MC 2 and the timer must all activate simultaneously. After a certain time, MC 2 must deactivate and MC 3 activate.

Take note The isolator switch must be off and key removed before any wiring is done.

ONLY NOW can the main circuit be wired. Remember, only neutrals should be connected to the bottom of the MCs, and nowhere else. If checked and all looks right, reconnect the supply.

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Electrical Technology Press the start button and the motor will start rotating. After a certain time, as the change-over happens, one should hear a change in the motor, especially if under load conditions. If all is well, the stop button can be pressed to stop the motor.

U1

U2

V1

V2

W1

W2

Figure 4.23: How six 240 V lamps can be wired to simulate the phases of a three phase load

Sequence motor control starter with overload (without timer): Practical 1. Purpose: To design a circuit with the given equipment so that when start 1 is pressed a red light must be switched on first before a three-phase motor can be switched on. In other words, if the red light is not switched on, then the motor must not switch on either, even if start button 2 is pressed.

Your circuit must have two starting buttons and one stop button. If the stop button is pressed, then both the light and the motor must switch off immediately.

All interlocks, overloads and safety mechanisms must be shown.

2. Equipment: 2 start/stop stations 2 main contactors with auxiliary contacts 1 light station 1 three-phase motor Connecting wires

Take note An alternative to using a motor is possible. It makes use of two 220 V lamps per phase connected in series. When the motor runs in star, there is effectively 220 V across the two lamps. They will burn dimly. After the change-over, there is now 380 V across the two lamps and they will burn brightly. This is an effective way to see the change as the circuit switches from the star to the delta connection. Two lamps MUST be used in series. If only one is used, then in the delta connection there would be 380 V across the 220 V lamps and it would blow the lamps.

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3. Explain the following with regards to this design • Function of components on diagrams • Principle of operation 4. Control circuit diagram: Draw a neat control circuit diagram. All connecting points must be clearly shown. 5. 6.

Main circuit diagram: Draw a neat main diagram to show how the main wiring should be connected. All connecting points must be clearly shown.

7.

Building the main circuit: Wire the circuit exactly according to the design. Once complete and all checks have been carried out, connect the supply and press the start button. If all works according to the criteria, then press the stop button.

Building the control circuit: Wire the circuit exactly according to the design. Once complete and if all checks have been carried out, connect the supply and press the start button. If all works according to the criteria, then press the stop button.

STOP 1

HOLD OUT

O/L1

MC1 (LIGHT)

O/L2

MC2 (MOTOR)

Figure 4.24: Control circuit for the manual sequence start

O/L1

O/L2

Figure 4.25 Main circuit for the manual sequence start

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Electrical Technology Sequence motor control starter with overload (with timer): Practical 1. Purpose: To design a circuit with the given equipment so that when the start button is pressed the first three-phase motor is activated. After 10 seconds, a second threephase motor should automatically be activated and both motors should run.

Your circuit must have one start button and one stop button.

All interlocks, overloads and safety mechanisms must be shown.

2. Equipment: 1 start/stop station 2 main contactors with auxiliary contacts 1 on-delay 2 three-phase motors Connecting wires 3. Explain the following with regards to this design: • Function of components on diagrams • Principle of operation 4. Control circuit diagram: Draw a neat control circuit diagram. All connecting points must be clearly shown. 5. Main circuit diagram: Draw a neat main diagram to show how the main wiring should be connected. All connecting points must be clearly shown. 6. Building the control circuit: Wire the circuit exactly according to the design. Once complete and after all checks have been carried out, connect the supply and press the start button. If all works according to the criteria, then press the stop button. 7.

94

Building the main circuit Wire the circuit exactly according to the design. Once complete and after all checks have been carried out, connect the supply and press the start button. If all works according to the criteria, then press the stop button.


4

HOLD OUT

O/L1

O/L2

(MOTOR 1)

(MOTOR 2)

Figure 4.26: Control circuit for the automatic sequence start

O/L1

O/L2

Figure 4.27: Main circuit for the automatic sequence start

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Activity 1 1. What is the function of the electric motor? 2. Name the three main parts of the electric motor. 3. How can rotational direction of the motor be changed? 4. Explain the operation of the squirrel cage induction motor. 5. Name some advantages of three-phase motors over single-phase motors. 6. Name two connection methods for the three-phase motor and show by means of two neat sketches how that can be achieved. 7. Name some applications in industry for motors. 8. Calculate the following for a 50 Hz 415 V 3-phase motor with a total of 18 poles: a) The synchronous speed in RPM. b) The percentage slip if the shaft speed is 955 RPM. 9. Calculate the total number of poles in a three-phase motor if the supply frequency is 50 Hz and the synchronous speed of the motor is rated as 600 RPM. 10. Calculate the synchronous speed of a 50 Hz three phase motor with 4 poles per phase. Also calculate the slip if the shaft speed is 1 450 RPM. 11. Calculate the shaft speed if a 60 Hz three-phase motor with a total of 36 poles is connected to a 400 V supply. The slip is 4%. 12. A 90 kW three-phase electrical motor has 12 poles in total and has a rated lagging power factor of 0,87. It is connected to a 400 V/50 Hz supply. Calculate: a) The apparent power (S). b) The line current. c) The phase angle. d) The synchronous speed of the motor. e) The slip if the shaft speed of the rotor is 1 400 RPM. 13. A three-phase delta-connected squirrel cage motor has an input power of 58,8 kW and a power factor of 0,87. The line voltage is 400 V. a) Calculate the phase current of the motor. b) What is meant by a squirrel cage rotor? c) The motor will not start when the start button is pressed. Name three possible causes why the motor will not start. 14. After a motor has been installed, basic electrical and mechanical inspections should be carried out on it, before the motor is started. a) Name TWO basic electrical inspections. b) Name TWO mechanical inspections. c) With reference to motor testing, what would the minimum acceptable insulation value be for this insulation test? 15. A three phase delta connected motor is operating off a 415 V/50Hz system is developing 20kW and the power factor of 0,82. Calculate: a) The line current. b) Current flowing in each phase. c) The reactive power (Q). d) The efficiency if the power on the shaft is 15kW.

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Practical Activity 1 Motor control cirucits (automatic switch off) 1. Purpose To design a circuit with the given equipment so that light 1 is switched on. After 10 seconds the light stops burning automatically. Your circuit must have one starting button and one stop button. All interlocks, overloads and safety mechanisms must be shown. Use a multimeter and make sure you know how the off-delay works. 2. Equipment • 1 start/stop station • 1 1 main contactors with auxiliary contacts • 1 1 off-delay • 1 1 light station • 1 connecting wires 3. Control circuit diagram Draw a neat control circuit diagram. All connecting points must be clearly shown. 4. Main circuit diagram Draw a neat main diagram to show how the main wiring should be connected. All connecting points must be clearly shown. 5. 6.

Connect the control circuit according to your diagram, and call your teacher when you are sure your circuit is correct. • correct 1st time: 10 points • correct 2nd time: 5 points • thereafter: 3 points Connect the main circuit diagram according to your diagram, and call your teacher when you are sure of your circuit is correct. • correct 1st time: 10 points • correct 2nd time: 5 points • thereafter: 3 points

97

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Practical Activity 2 3 Phase motor testing Simulation 1 Equipment: • Three phase motor • Megger or insulation tester An electrical 3 phase induction motor were tested with an insulation/continuity tester, and the following readings were obtained. U2 – V2



V2 – U1



W2 – E



U2 – W2



V2 – W1



U1 – W1



U2 – U1

82kΩ

V2 – V1

55,8Ω

U1 – V1



U2 – W1



V2 – E



U1 – E

0Ω

U2 – V1



W2 – U1



W1 – V1



U2 – E

82kΩ

W2 – W1

55,8Ω

W1 – E



V2 – W2



W2 – V1



V1 – E



1.1 Interpret the readings and draw a possible electrical circuit (internal connections) of the faulty/simulated motor. 3 Labels 3 Nodes U2

V2

W2

Earth

V1

W1

U1

1.2 Write a brief report about the electrical condition of the motor with reference to: 1.2.1 Continuity test: Winding Winding U Winding V Winding W

98

Reading

Accept or not

Motivate


4

1.2.2 Insulation between Windings: Winding

Reading

Accept or not

Motivate

Winding U and V Winding U and W Winding V and W

1.2.3 Insulation to Earth: Winding

Reading

Accept or not

Motivate


1.2.4 What is the minimum acceptable value for the insulation test according to regulation?

3 Phase motor testing Simulation 2 Equipment: • Three phase motor • Megger or insulation tester An electrical 3 phase induction motor were tested with an insulation/continuity tester, and the following readings were obtained. U2 – V2



V2 – U1



W2 – E



U2 – W2



V2 – W1

42.3Ω

U1 – W1



U2 – U1

198kΩ

V2 – V1

39Ω

U1 – V1



U2 – W1



V2 – E



U1 – E



U2 – V1



W2 – U1



W1 – V1

81,3Ω

U2 – E



W2 – W1

39Ω

W1 – E



V2 – W2

3,3Ω

W2 – V1

42.3Ω

V1 – E



1.1 Interpret the readings and draw a possible electrical circuit (internal connections) of the faulty/simulated motor. Labels 4 Nodes 2

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U2

V2

W2

Earth

V1

W1

U1

1.2 Write a brief report about the electrical condition of the motor with reference to: 1.2.1 Continuity test: Winding

Reading

Accept or not

Motivate



Reading

Accept or not

Motivate



Reading

Accept or not

Motivate


100



4

3 Phase motor testing Simulation 3 Equipment: Three phase motor Megger or insulation tester An electrical 3 phase induction motor were tested with an insulation/continuity tester, and the following readings were obtained. U2 – V2



V2 – U1



W2 – E

82Ω

U2 – W2



V2 – W1

85,3Ω

U1 – W1



U2 – U1

1,5kΩ

V2 – V1

82Ω

U1 – V1



U2 – W1



V2 – E

85,3Ω

U1 – E



U2 – V1



W2 – U1



W1 – V1

167,3Ω

U2 – E



W2 – W1

82Ω

W1 – E

0Ω

V2 – W2

3,3Ω

W2 – V1

85,3Ω

V1 – E

167,3Ω

1.1 Interpret the readings and draw a possible electrical circuit (internal connections) of the faulty/simulated motor. Labels 4 Nodes 3 1.2 Write a brief report about the electrical condition of the motor with reference to: 1.2.1 Continuity test: Winding

Reading

Accept or not

Motivate



Reading

Accept or not

Motivate


101

4



Reading

Accept or not

Motivate



Practical Activity 3 Automatic Star-Delta Investigation To investigate the start current versus the running current in a star delta starter under starting and then running conditions: The automatic star-delta motor starter is designed to start the motor in a star connection to reduce starting currents and voltages, and once turning almost at full speed to change the terminal connections to delta. This is done automatically, so after the start button is pressed the motor runs in start immediately, and after a preset time on a timer the change to delta happens automatically. The circuit must have one and one top button. (All interlocks, overloads and safety mechanisms must be shown). Equipment: • Three phase motor • One start and one stop button • 3 × main contactors • 1 × On –Delay with N/O and N/C contacts • Connecting wires • Clamp on ammeter • Three phase supply

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Wiring diagram 1. Build the control circuit for the Star-Delta starter as is shown below.

2. Build the Main circuit for the Star-Delta starter as shown below.

3. Use the clamp on meter and connect across any one of the lines coming from the supply to the motor. 4. Start the motor and record the current when the STAR contactor is in. 5. Record the current when the DELTA contactor is activated. 6. What do you observe?

103

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104


Chapter 5 RCL circuits

A

B Series circuits

Parallel circuits

A

B Calculations

A

Resonant frequency

Phasor diagrams B Q-factor

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Introduction The effect of alternating current on R, L and C components in series and parallel circuits. In this chapter, attention will be given to resistors, coils (inductors) and capacitors that are connected in series and parallel and the difference between the two. Calculations will be done in detail and this will include inductive reactance, capacitive reactance, impedance, power, phase angle, power factor, resonance and Q-factor. This chapter will lead the learner through the steps of how to convert these calculated values to phasor diagrams and how the wave representations would look. The effect of AC on a pure resistor, coil or capacitor was looked at in detail in the Grade 11 book. The series RL, RC and RCL circuits were also completed, but as an introduction this will be treated as revision before the transition to parallel circuits is made.

Inductive reactance The value of a coil is always measured in Henries. However, the inductive reactance of the coil is the equivalent ohm value of a coil at a certain frequency. This fact is the same for a series or a parallel circuit. The only factors that have an influence on the inductive reactance are the frequency and the actual value of the coil (in henries). The formula for calculating this inductive reactance is given by: XL = 2 πfL (and is measured in ohms) Take note In physics, and electronics, the henry (symbol H) is the SI derived unit of inductance. It is named after Joseph Henry (1797–1878), the American scientist who discovered electromagnetic induction independently of, and at about the same time as, Michael Faraday (1791–1867) in England.

Therefore a coil with an inductance of 36 mH at 5 kHz would have an inductive reactance of: XL = 2 πfL = 2 π(5 000)(36 mH) = 1,13 kΩ

Figure 5.1: Typical inductors used in electronic/electrical circuits

Capacitive reactance The same can be said for a capacitor. The value of a capacitor is always measured in Farads although most capacitors are measured in µF, nF of pF. However, the capacitive reactance of the capacitor is the equivalent ohm value of a capacitor at a certain frequency. This fact is the same for a series or a parallel circuit. The only

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factors that have an influence on the capacitive reactance are the frequency and the actual value of the capacitor. The formula for calculating this capacitive reactance is given by: 1 XC = _____ 2 πfC

(and is measured in ohms)

Therefore a capacitor with a capacitance of 9 nF at 5 kHz would have a capacitive reactance of: 1 1 XC = _____ = _____________ = 3,54 kΩ 2 πfC 2 π(5 000)(9 nF)

Take note The farad (symbol: F) is the SI derived unit of capacitance. The unit is named after the English physicist Michael Faraday. Figure 5.2: The sizes of capacitors vary from really small to really large. A standard 30 cm ruler is used to illustrate the physical difference in size of the capacitors

Remember…… Kilo (k) = 103 Mega (M) = 106 Milli (m) = 10-3 Micro (µ) = 10-6 Nano (n) = 10-9 Pico (p) = 10-12

Impedance For RCL circuits, there would be a combination of resistors, capacitors and inductors (coils). Impedance is the total ‘resistance’ a RCL circuit would offer to the flow of current in a circuit. The value of the impedance is always measured in ohms. This fact is the same for a series or a parallel circuit. The factors that have an influence on the impedance are the frequency, the values of the resistor, coils and the capacitor. The formula for calculating this impedance is given by: Z = √ R2 + (XL – Xc)2 (and is measured in ohms) Therefore a resistor of 2 kΩ, a coil of 36 mH and a capacitor of 9 nF at 5 kHz would have an impedance of:

Take note Substitutions must be done in the same order as the formula. It does not matter that XC is bigger than and subtracted from XL. Since the bracket is squared it makes the answer a positive irrespective of the sign.

Z = √ R2 + (XL – Xc)2 = √ (2 000)2 + (1 130 – 3 540)2 = 3,13 kΩ (The values used are taken from above calculations)

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Power Take note The watt (symbol: W) is a derived unit of power in the International System of Units (SI), named after the Scottish engineer James Watt (1736–1819).

Electric power is the rate of energy consumption or energy transfer in an electrical circuit. Electric power is measured by the ability to transfer this power and is commonly expressed in milliwatt (mW) in electronic circuits to megawatts (MW) when talking about the national distribution of electricity in South Africa. The basic formulae for calculating power is shown below: Electric power calculation

2 P = V · I or P = I 2 · R or P = V R

P is the electric power in watts (W). V is the voltage in volts (V). I is the current in amps (A). R is the resistance in ohms (Ω). These formulae are true only if the current and the voltage are exactly in phase with each other. However, inductors (coils) and capacitors do cause a phase shift between voltage and current and therefore the formulae above cannot be used with calculation of power in an RCL circuit.

Power in AC circuits Apparent power The apparent power is the power that is supplied to the circuit. This is the product of the voltage and the current, ignoring the angle between the two. It is measured in VA or kVA, depending on the size of the values. The formula used to calculate this apparent power is: Papp = I V

measured in (VA or kVA)

True power (also called real or active power) The true power is the power that is effectively being used by the load or the circuit, and this would be the voltage and current values that are exactly in phase with each other. The formula used to calculate this true power is: Papp = I V cosθ

measured in watt (W) or kilowatt (kW)

Reactive power Reactive power is the power that is wasted and not used to do work on the load. Usually it is in the form of heat. Preac = I V sinθ

measured in VAr or kVAr

The relationship between the three powers can be given by the power triangle shown below. Apparent power (VA) Reactive power VAR

Real power (W)

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RCL circuits Phase angle and power factor The ratio of true power to apparent power is called the power factor and is a number always between 0 and 1. (That means a maximum of ONE.) It is actually pure trigonometry where cos θ = Adjacent Hypotenuse

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Take note True and active means exactly the same thing.

but in this case the true power is the adjacent and the apparent power is the hypotenuse, therefore Ptrue cos θ = Papparent PAPP PREAC PTRUE

Since it is a right angle triangle, Pythagoras also plays a part and it does help with some calculations. Alternatively, the following formula could be used to calculate the reactive power. (PAPP)2 = (PTRUE)2 + (PREAC)2 OR (PAPP)2 = √ Ptrue2 + Preac2 Remember that power factor is always written as 0,98 and NEVER as 980 × 10-3. Phasor and wave representation A vector is an arrow that is used to represent a mechanical object that has direction and speed, like a car traveling in a certain direction at 60 km/h. 60

Take note Power factor is always written as zero comma E.g.: 0,98.

Figure 5.3: A vector diagram showing speed and direction for the car

With electrical quantities like current, voltage, power and ohms there is a size aspect but not really a direction aspect. Instead, reference is made to the phase relationship between these values. The phase relationship between voltage values can be observed on an oscilloscope. The arrows that represent these electrical values show the size and phase relationship. It is called a ‘Phasor Diagram’. There are very definite differences between the phasor diagrams for resistive, inductive and capacitive components, and these differences between waves and phasor diagram have been summarised on the next page.

Take note It is important to remember that the direction of rotation must be shown all phasor diagrams. What makes it easier is that it is always anticlockwise.

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Electrical Technology The resistive circuit VR I

Take note Note that the current and voltage are always in phase with each other.

I

VR

Waves and phasor representation for a resistor-resistor (RR) circuit. Across a resistor, the two waves are always in phase with each other, which means the phase angle between them is 0°. This means a unity power factor of ‘1’. Take note Note that the current always lags the voltage. In other words, current is always below voltage.

The inductive circuit VL I

How the observed waveforms would be drawn. VL

I

Waves and phasor representation for a resistor-coil (RL) circuit. Across an inductor, the two waves are always 90° out of phase with each other and current will always lag the voltage. This is referred to as a lagging power factor. The capacitive circuit I V I Take note Note that the current always leads the voltage. The current is always above the voltage.

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VC

Waves and phasor representation for a resistor-capacitor (RC) circuit.

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Across a capacitor, the two waves are always 90° out of phase with each other and current will always lead the voltage. This is referred to as a leading power factor. How to make sense of series and parallel circuits (CIVIL) For any series circuit, the current flowing in the circuit is the same everywhere and therefore the current would be used as a reference. For any parallel circuit, the voltage across all the branches would be the same and therefore the voltage would be used as a reference. It is also known that a resistor does not cause a phase shift between voltage and current whether it is series or parallel, and VR is always in phase with current (I). The power factor is always referred to as leading or lagging. The question asked to establish this is, “Where is supply current with respect to supply voltage?” The following aid can be used to help establish what fits where on a phasor diagram, as it gets a little tricky when the move from series to the parallel circuits is made. The big advantage is that this works for series and parallel circuits. So use will be made of something called CIVIL. CAPACITOR (C) Current (I) leads voltage (V)

CIVIL COIL (L) Voltage (V) leads current (I)

CIVIL How to draw phasor diagrams To draw a phasor diagram, three basic steps are followed: this will make the drawing of the phasor diagrams a lot easier. 1. What stays constant? (In series it would be current, and in parallel it would be voltage.) 2. VR is always in phase with current. 3. Where is VT with respect to VR. Resonance with its characteristic curves Resonance is a special frequency in an RCL circuit where certain characteristics are only true at that frequency. The phasor diagrams for voltage values and ohm values, as well as the waveforms on the scope, clearly show these characteristics. VL

XL

VT

Z

I

R

VR

VC

I

XC

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Figure 5.4: Waves on the oscilloscope in phase with each other at resonant frequency VT VR

Figure 5.5: How the drawing of the two waves in phase with each other will look at resonant frequency

Characteristics for the series RCL circuit at resonance: • XL = XC • R = Z • Z is min. • I is max. • Cos θ = 1 • θ = 0° • VL = VC • VR = VT • VR exactly in phase with VT Q-factor The Q-factor of a resonant RCL circuit that is connected to an AC supply is the ratio of the energy stored to the energy wasted. Normally we refer to ‘Q’ as the Q-factor or the amplification factor. At resonance the value of the resistor is small when compared to the values of the inductive or capacitive reactances. Therefore the voltage drops across the coil (VL = I.XL) and the capacitor (VC = I.XC) will be larger than the supply voltage, and how many times VL or VC is greater than VT is referred to as the Q-factor of the circuit. The quality factor (Q) of an inductor can also be the ratio of its inductive reactance to its resistance at a given frequency and is a measure of its efficiency. The higher the Q factor of the inductor, the closer it approaches the behaviour of an ideal, lossless, inductor.

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The following formulae can be used to calculate Q-factors. Q-factor (using ohm values)

=

Q-factor using voltage values

=

XL XC OR R R VL

VT

OR

VC

VT

Take note Q-Factor can only be calculated at resonant frequency.

When drawing a phasor diagram for a series RCL circuit there are THREE possiblites: • below resonant frequency • at resonant frequency • above resonant frequency. For the first scenario, where the frequency falls below resonance, the condition is that XL < XC. and therefore VL < VC (this circuit would be more capacitive). VL

(VC – VL)

VR I

θ

VT VC

The phasor diagram representing all the voltage values in a series RCL circuit at a low frequency is shown above.

Figure 5.6: Waveforms on an oscilloscope where VR leads V T (VR starts left centre)

In this case, supply current leads the supply voltage and the power factor is leading overall.

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Electrical Technology At resonant frequency For the second scenario, the frequency is at resonance and only at this frequency will XL = XC and VL = VC. VL VT

I

VR VC

The phasor diagram representing all the voltage values in a series RCL circuit at resonance.

Figure 5.7: Waveforms on an oscilloscope where V T and VR are in phase with each other

In this case, the supply current and the supply voltage are exactly in phase with each other and this is the unity power factor (unity means 1). Above resonant frequency For the third scenario, where the frequency is above resonance the condition is that XL > XC and therefore VL > VC (this would be more inductive). VL VT (VL – VC)

θ

I VR

VC

The phasor diagram representing all the voltage values in a series RCL circuit at a high frequency.

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Figure 5.8: Waveforms on an oscilloscope where V T leads VR

In this case, supply current leads the supply voltage and the power factor is leading overall.

Calculations for series RCL circuits A number of examples will be discussed to combine all these aspects in series RCL circuits. These examples will be used as revision to expand the way one thinks about RCL circuits. On completion, the move to parallel RCL circuits will be made. Question 1: A series RL circuit consists of a resistor of 470 Ω and a coil of 47 mH that is connected to a 0,6 V 2 kHz supply. R

L

Figure 5.9: A series RL circuit diagram

Calculate the following: a) inductive reactance b) impedance c) the current drawn from the supply d) the power factor e) the voltage drop across the coil. f) What will happen to the current flowing in the circuit if the frequency is increased? Answers for question 1: a) XL = 2 π(2 000)(47 mH) = 590 Ω b)

Z = √ R2 + Z2 = √ (470)2 + (590)2 = 754,81 Ω

c) IT =

VT Z

=

0,6 = 794,91 µA 754,81

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Electrical Technology d)

Cos θ = R = 470 = 0,62 lagging Z 754,81

e) VL = I.XL = (794,91 µA)(590) = 469 mV (or 0,47 V) f)

f

XL

Z

I

Question 2: With reference to the phasor diagram below, answer the following questions. XL = 51,23 Ω

R = 20 Ω I (XL – XC)

Z XC = 73,66 Ω

Figure 5.10: A phasor diagram showing ohm values

a) Draw the labelled circuit diagram that is represented by the above phasor diagram. (4) b) Is the circuit more capacitive or more inductive? Motivate your answer. (2) (3) c) Calculate the total circuit impedance. d) Calculate the power factor of the circuit. (3) e) Name THREE things that can be changed to ensure resonant frequency. (3) f) Give FOUR characteristics of the circuit in resonance. (4) g) Calculate the supply voltage if the current at resonance in the circuit is 5 mA. (3) Answers for question 2: a)

Figure 5.11: A series RCL circuit diagram

b) Capacitive, because XC > XL c) Z = √ R2 + (XL – Xc)2 = √ (20)2 + (51,23 – 73,66)2 = 30,05 Ω

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20 d) cos θ = R = = 0,67 leading Z 30,05 Z e) • Change the capacitor, • Coil, or • Change supply frequency f) • XL = XC • R=Z • Z is min. • I is max. • Cos θ = 1 • θ = 0° • VL = VC • VR = V T g) VT = I × Z = (5 mA)(20) = 100 mV (at fr R = Z = 20 Ω) Question 3: Refer to the circuit diagram shown below and calculate the: a) total current in the circuit b) resistance of the 60 watt 110 volt lamp c) impedance of the circuit d) inductance of the coil that must be connected in series with the 60 watt 110 volt incandescent lamp so that it can operate from a 220 volt 50 Hz AC (assume the resistance of the coil is negligible).

(3) (3) (3) (6)

[15]

Answers for Question 3: a) I = P = 60 = 0,55 A (545,45 mA) V 110 b) RLAMP

=

VLAMP = 110 = 201,67 Ω 0,55 ILAMP

VT 220 c) Z = ___ = ____ = 403,34 Ω I 0,55 d) XL = √ Z2 – R2 = √ 403,342 – 201,672 = 349,3 Ω L = XL = 349,3 = 1,11 H 2 πf 2 π(50)

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Parallel RCL circuits Take note Kirchoff ’s current law for a series circuit states that: The current is the same everywhere. Kirchoff ’s voltage law for a parallel circuit states that: The voltage across all the branches is equal.

Apart from the difference in circuit construction between the series and the parallel circuits, there are many other changes. However, if the basics are kept simple, then by using common sense one can navigate one’s way through parallel circuits without too much drama. A main factor that will help is the question of ‘reference’. By reference is meant ‘what stays constant’ in a particular circuit. Kirchoff ’s voltage and current laws fortunately help to make this easier in a big way. Until now, all the circuits used were series circuits and that is why the current was always the reference phasor. However, with the parallel circuit voltage becomes the reference phasor. A parallel RC circuit In the parallel RC circuit, the voltage across the components would be equal (VT = VR = VC), but the supply current (IT) would split up between the two branches. This will give IR and IC . In the parallel RC circuit shown below, the value of the resistor is 2 kΩ and the capacitor is 10 nF. The supply is a 6V/4 kHz AC supply.

Take note An important fact to remember is that the parallel circuit cannot be demonstrated by using an oscilloscope, as the oscilloscope cannot measure current. (Kirchoff ’s current law for parallel circuits – the current entering a point is the same as the current leaving a point.)

Figure 5.12: A parallel RC circuit diagram

As the two component values cannot be added (since they are in ohm and farads) the capacitive reactance would have to be calculated first. XC =

1 = 2 πfC

1 = 3,98 kΩ 2 π(4 000)(10 nF)

The supply current splits up and flows through the two branches before joining together again. VT = VR = VC = 6 V IR = IC =

VR R VC XC

=

6 = 3 mA 2 000

=

6 = 1,51 mA 3 980

It must be kept in mind that the capacitor causes a phase shift between voltage and current.

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RCL circuits To plot these current values on a phasor diagram, extreme care must be taken. It is important that the current and voltage values are filled in, keeping in mind their phase relationship with respect to each other. Make use of CIVIL. CAPACITOR (C) Current (I) leads voltage (V)

CIVIL

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Take note When drawing a phasor diagram, remember the following three facts: • VT is reference • IR always in phase with VT • Plot IC or IL with respect to VT.

IT

IC = 1,51 mA θ

IR = 3 mA

VT = 6 V

As seen on the phasor diagram, it is not possible just to add IR and IC together algebraically as they are out of phase with each other. But since they form a right angle triangle, Pythagoras can be used to calculate the supply current (IT). IT = √ IR2 + IC2 = √(3 mA)2 + (1,51 mA)2 = 3,36 mA Only at this stage can the impedance of the circuit be calculated, as the supply voltage and the supply current are available. Z =

VT 6 = = 1,79 kΩ IT 3,36mA

From the phasor diagram, the power factor can also be determined. Cos θ = IR IT

=

3 mA 3,36 mA

= 0,89 leading

Remember…. Where is IT with respect to VT? It is leading since IT is above VT . Hence the phase angle will be: θ = cos-1 0,89 = 26,77° leading

A parallel RL circuit In the parallel RL circuit, the voltage across the components would be equal (VT = VR = VL), but the supply current (IT) would split up between the two branches. This will give IR and IL . In the parallel RL circuit shown overleaf, the resistor is 2 kΩ and the coil (or inductor) is 200 mH. The supply is a 6 V/4 kHz AC supply.

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Figure 5.13: A parallel RL circuit diagram

The inductive reactance would have to be calculated first. XL = 2 πfL = 2 π(4 000)(200 mH) = 5,03 kΩ VT = VR = VL = 6 V (Voltage across all branches are equal) The supply current splits up and flows through the two branches before joining together again. IR = IL =

VR R

=

6 = 3 mA 2 000

VL = 6 = 1,19 mA XL 5 030

Remember, the inductor (coil) causes a phase shift between voltage and current. To plot these current values on a phasor diagram, extreme care must be taken. It is important that the current and voltage values are filled, in keeping in mind their phase relationship to each other. Take note When drawing a phasor diagram, remember the following three facts: • VT is reference • IR always in phase with VT • Plot IC or IL with respect to VT.

Make use of CIVIL. COIL (L) Voltage (V) leads current (I)

CIVIL IR = 3 mA θ IL = 1,19 mA

VT = 6 V

IT

Once again, it is not possible just to add IR and IL together algebraically as they are out of phase with each other. But since they form a right angle triangle, Pythagoras can be used to calculate the supply current (IT). IT = √ IR2 + IL2 = √(3 mA)2 + (1,19 mA)2 = 3,23 mA

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Only at this stage can the impedance of the circuit be calculated, as the supply voltage and the supply current are available. Z =

VT 6 = = 1,86 kΩ IT 3,23 mA

From the phasor diagram, the power factor can also be determined. Cos θ = IR IT

=

3 mA 3,23 mA

= 0,93 lagging

Remember Where is IT with respect to VT? It is lagging since IT is below VT . Hence the phase angle will be: θ = cos-1 0,93 = 21,75° lagging

A parallel RCL circuit In the parallel RCL circuit, the voltage across the components would be equal (VT = VR = VC = VL), but the supply current (IT) would split up between the three branches. This will give IR , IC and IL . In the parallel RCL circuit shown below, the resistor is 3 kΩ, the coil (or inductor) is 100 mH and the capacitor is 18 nF. The supply is a 4 V/5 kHz AC supply.

Figure 5.14: A parallel RCL circuit diagram

The inductive and capacitive reactances are calculated first. XL = 2 πfL = 2 π(5 000)(100 mH) = 3,14 kΩ XC =

1 = 2 πfC

1 = 1,77 kΩ 2 π(5 000)(18 nF)

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Electrical Technology The supply current splits up and flows through the three branches before joining together again. VT = VR = VL = VC = 4 V IR =

VR 4 = 1,33 mA = R 3 000

IL =

VL 4 = 1,27 mA = XL 3 140

IC =

VC 4 = 2,26 mA = XC 1 770

IC = 2,26 mA IT (IC – IL)

θ

VT = 4 V IR = 1,33 mA

IL = 1,27 mA

Since the currents are all out of phase with each other, Pythagoras will be used to determine the total current (supply current). IT = √ IR2 + (IC – IL)2 = √ (1,33 mA)2 + (2,26 mA – 1,27 mA)2 = 1,66 mA Once again, the impedance of the circuit can only be calculated by using the supply voltage and the supply current. Z =

VT 4 = = 2,41 kΩ IT 1,66 mA

From the phasor diagram, the power factor can also be determined. Cos θ =

IR IT

=

1,33 mA 1,66 mA

= 0,801 leading (because IT is above VT)

Hence the phase angle will be: θ = cos-1 0,801 = 36,75° leading The phasor diagrams for series and parallel circuits may be confusing, as in series the ohm value of the coil (XL) is at the top, while for the parallel circuit the current value for the capacitor (IC) is at the top. The only thing that stays constant is CIVIL, and a lot of mistakes can be eliminated by using CIVIL.

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Resonant frequency for the parallel circuit is identical to that of the series circuit, as the only things that determine resonant frequency are the value of the capacitor and the value of the coil. fr =

1 1 = = 3,75 kHz 2 π √ LC 2 π √ (100 mH)(18 nF)

Once again, at this frequency a number of interesting facts emerge. Firstly, this will be proved by means of calculations, followed by a summary of the facts. XL = 2 πfL = 2 π(3 750)(100 mH) = 2,36 kΩ XC =

1 2 πfC

=

1 = 2,36 kΩ 2 π(37 500)(18 nF)

VT = VR = VL = VC = 4 V IR =

VR R

IL =

VL 4 = 1,69 mA = XL 2 360

IC =

VC XC

=

=

4 = 1,33 mA 3 000

4 = 1,69 mA 2 360

IC = 1,69 mA

IT VT = 4 V IR = 1,33 mA

IL = 1,69 mA

IT = √ IR2 + (IC – IL)2 = √ (1,33 mA)2 + (1,69 mA – 1,69 mA)2 = 1,33 mA Z =

VT 4 = = 3,01 kΩ IT 1,33 mA

From the phasor diagram, the power factor can also be determined. Cos θ =

IR IT

=

1,33 mA 1,33 mA

= 1 (unity power factor)

Hence the phase angle will be: θ = cos-1 1 = 0°

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Electrical Technology Summary of facts at resonant frequency for a parallel circuit Characteristics of the series RCL circuit at resonance: • XL = XC • Z is max. • I is min. • Cos θ = 1 • θ = 0° • IL = IC • IR = IT • IT exactly in phase with VT

Q-factor The Q-factor of a parallel resonant RCL circuit that is connected to an AC supply is the ratio of the energy stored to the energy wasted. Normally we refer to ‘Q’ as the Q-factor or the amplification factor. At resonance the value of the supply current is small when compared to the values of the current through the inductor or capacitor. How many times IL or IC is greater than IT is referred to as the Q-factor of the circuit. The following formulae can be used to calculate Q-factors. (But remember that they can only be used at resonant frequency.) Q-factor (using current values) =

IL I = C IT IT

Q-factor for the example used is Q =

IL = 1,69 mA = 1,27 IT 1,33 mA

A curve that represents the current in the parallel circuit versus frequency will look as follows. At resonance, the current would be a minimum and as frequency increases or decreases from resonance so the current increases.

Calculations for parallel circuits A number of examples will be discussed to combine all these aspects in parallel RCL circuits. These examples will be used as revision to expand the way one thinks about RCL circuits. Question 1: A parallel RC circuit consists of a resistor of 10 Ω and a capacitor of 450 µF is connected to a 240 V 50 Hz supply. Calculate the following: a) capacitive reactance b) current through the resistor and the capacitor. c) the current drawn from the supply

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d) the power factor. e) Represent the current values by means of a neat phasor diagram. f) What will happen to the current flowing in the circuit if the frequency is decreased? Answers to Question 1: a) XC =

1 1 = 2 πfC 2 π(50)(450 μF)

= 7,07 Ω

b) VT = VR = VC = 240 V IR = IC =

VR R VC XC

240 10

=

= 240 7,07

= 24 A = 33,93 A

c) IT = √ IR2 + IC2 = √ 242 + 33,932 = 41,56 A d) Cos θ = e)

IR IT

=

IC = 33,93 mA

24 41,56

= 0,58 leading

IT = 41,56 A

θ IR = 24 A

f) f ↓

XC ↑

IC ↓

VT = 240 V

IT ↓

Question 2 The phasor diagram for an RLC circuit is shown. Interpret the information and answer the questions that follow.

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Electrical Technology a) Draw the circuit that is represented by the above phasor diagram. b) Is the circuit more inductive or capacitive? Motivate your answer. c) Calculate the total current in the circuit. d) Calculate the power factor in the circuit. e) What will happen to the phase angle if the frequency should decrease? Explain your answer. Answers to question 2: a)

A parallel RCL circuit diagram

b) Capacitive because IL > IC which means that XL < XC c) IT = √ I2R + (IL – IC)2 IT = √ 102 + (10,61 – 4,72)2 IT = 11,61 A

d) Cos θ = IR ___ IT =

10 11,61

= 0,86 lagging e) θ will increase because Xc ∞ 1 if F decreases then Xc will increase which in F turn will let IC decrease. Therefore the difference between IL and IC will become larger which means that IT will tend more towards IL on the phasor diagram. Question 3: A parallel RLC network has the following components connected across a 230 V/ 50 Hz alternating current supply:

resistor = 12 Ω

capacitor = 70 µF

inductor = 100 mH

Determine the: a) inductive reactance of the coil b) capacitive reactance of the capacitor c) current flowing through each of the branches d) total current flowing in the network e) phase angle between the supply voltage and the current f) impedance of the network. g) Draw a fully labelled phasor diagram that represents all the current and the voltage values.

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Answers to question 3: a) XL = (2 πfL) = 2 π(50)(100 × 10-3) = 31,42 Ω b) XC =

1 1 = = 45,47 Ω 2 πfC 2 π(50)(70 × 10-6)

c) I = VT = L XL IR =

VT R

IC =

VT = XC

=

230 = 7,32 A 31,42 230 12

= 19,17 A

230 = 5,06 A 45,47

d) IT = √ IR2 + (IL – IC)2 = √ 19,172 + (7,32 – 6,06))2 = 19,3 A e) Cos θ = f) Z = g)

VT IT

IR

IT

=

19,17 = 0,993 lagging 19,3

= 230 = 11,92 Ω 19,3

IC = 5,06 A

IR = 19,17

VT = 230 V

IT = 19,3 A

IL = 7,32 A

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Electrical Technology Practical: Simulate an RLC circuit and display waveform on oscilloscope To investigate the relationship between the voltage across the resistor, the voltage across the capacitor, the voltage across the coil and the supply voltage in a series rcl – Alternating Current Circuit. Apparatus 1. 1,0 kΩ resistor (R1) 2. 0,1 micro farad capacitor (C1) 3. 26 mH coil 3. Matrix board with hook-up wires 4. Audio signal generator (50 Hz 10 kHz) 5. Dual trace oscilloscope Method 1. Draw a series circuit with the resistor, coil and capacitor connected to the signal generator. Show on your sketch how the oscilloscope is connected to the circuit to comply with the following. • Channel 1 (or Y1) of the oscilloscope must be connected across the resistor to measure VR, and • Channel 2 (or Y2) across the signal generator to measure VT . The triggering of the scope should be across the resistor. (NB Ensure that all the earth connections are connected on one side of the resistor.) 2. Build the circuit exactly according to the circuit diagram. (NB Check that all the earth connections are connected on one side of the resistor. In other words, all black croc clips should be together.) 3. Follow these steps. • Set the signal generator to 500 Hz. • Go to channel 2 and set the supply voltage to 0,6 V. • Go to CH 1 and make sure that the wave starts left centre. (The trigger must be CH 1.) • Make the necessary adjustments on the oscilloscope so that one complete cycle is displayed over 10 cm on the screen. • Set the V/div setting so that the wave is as high (big) as possible. • Select the ALT or CHOP mode so that both waves are displayed simultaneously. 4. Draw the oscillograms as displayed on the screen. All labels must be clearly indicated. (Remember that the scales of CH 1 and CH 2 on the oscilloscope may possibly be different.)

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Take the following readings: a) VR = V/div × no. blocks × 0,707 = __________________________________

b) VT = V/div × no. blocks × 0,707 = __________________________________

c) The phase angle = _____________________________________________ (State which wave is leading.) VR

d) The supply current =

e) What does it mean if the triggering is set across a certain component?

R

= _____________________________________

5. Convert the oscillograms in Step 4 to phasors. (They must be drawn to scale.) Follow these steps: • Start the phasor diagram in the middle of the page. • Plot the reference phasor. • Draw in VR. • Draw in VT with respect to VR. • Graphically subtract VR from VT (This will determine the voltage drop which represents the difference between VL and VC.) • At this stage, one should have good understanding of RL and RC circuits. Estimate the sizes of VL and VC and indicate them on your phasor diagram. • All labels (including the scale and phase angle) must clearly be indicated. 6. Change the frequency to 10 kHz and repeat points 3 to 5 above. 7. Without doing ANY CALCULATIONS, adjust the frequency to resonance.

Explain why you selected this particular frequency. Draw the waveforms. (Remember that the scales of CH 1 and CH 2 on the oscilloscope may possibly be different.)

Take the following readings: a) VR = V/div × no. blocks × 0,707 = _________________________________

b) VT = V/div × no. blocks × 0,707 = _________________________________

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Electrical Technology c) The phase angle = ______________________________________________ (State which wave is leading.)

d) The supply current = VR = _____________________________________ R

e) What does it mean if the triggering is set across a certain component?

8. Draw a neat phasor diagram, not to scale, but representing all the voltage drops in the circuit as accurately as possible. Observations: Answer the questions below by comparing the phasor diagrams. Establish how a change in frequency will influence the relationship between VR and VT . 1. 2.

What is the phase relationship between VR and VT at 500 Hz? (Refer to angle, as well as leading or lagging.) What is the phase relationship between VR and VT at 10 kHz? (Refer to angle, as well as leading or lagging.)

3. Did the change in frequency influence the phase relationship (angle) between the two phasors VR and VT? 4. What effect does the change in frequency have on the magnitude of VT , VR, VL and VC? 5. What influence does the change in frequency have on the current flowing in the circuit and thus on the impedance of the circuit? Draw a neat curve that represents frequency versus current. 6. What is the reason for all the changes that occurred as a result of the change in frequency (increase in frequency)?

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Activity 1 1.

The sketch below depicts the reactance of an inductor and a capacitor and the resistance of a resistor versus frequency. Interpret the information and answer the questions that follow.

a) Draw TWO approximate impedance triangles of an RLC series circuit next to each other, with the frequency at point A and C respectively. b) The circuit represented by the above graph consists of a 3kΩ resistor, a 50µF capacitor and a 0,1H coil. Calculate the frequency at point B. 2. A series circuit with a 250µF capacitor and a 120mH coil and a resistor of 15Ω is connected across a 240V/50Hz supply. Calculate: a) The total impedance in the circuit. b) The voltage drops across the coil. c) The phase angle. d) Is the circuit more inductive or capacitive? Motivate your answer. e) Must the frequency increase or decrease for the circuit to resonate? Motivate your answer. 3. Name THREE measurements we can take with the scope.

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Electrical Technology 4.

Use the oscillograms below to answer the questions that follow.

V2 V1

Determine the following if both V/div = 0,5V and the T/div = 10µs

a) Frequency b) V1 (RMS) c) Would the circuit be inductive or capacitive if V1 is the voltage across a supply and V2 is the resistor voltage? Explain your answer. 5.

A 15V/10kHz supply is connected to the following components connected in series: R = 2kΩ, L = 60mH and C = 8nF.

a) Draw a neat diagrammatic circuit diagram for the circuit. b) Calculate the inductive and capacitive reactances and then the impedance of the circuit. c) Represent all the calculated ohm values on a neat phasor diagram. (not to scale but in good proportion). d) What will happen to capacitance if the frequency is decreased? Give a reason for your answer. e) Which factors will determine the resonant frequency. 6. VL = 8 V IT = 2 mA VR = 3 V

f = 7 kHz

VC = 12 V

Use the phasor diagram to answer the following questions: a) Calculate the supply voltage. b) Is the power factor leading or lagging? Explain your answer. c) Calculate the reactive value of the coil. d) Calculate the actual value of the coil in mH.

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e) Name 5 characteristics of a series RCL circuit if it is set to resonant frequency. f)

Draw a neat curve that shows the relationship between current and frequency in a series RCL circuit

7.

A parallel circuit consisting of a capacitor of 18 nF is connected in parallel to a resistor of 2 kΩ. The circuit is connected to a 5 V/8 kHz. Calculate the following: a) The capacitive reactance. b) The branch currents. c) Draw a neat phasor diagram to represent all the current values.

8. An alternating current circuit of an inductor of 200 mH, a capacitor of 300 μF and a resistor of 15 Ω is connected in parallel to a 120 V/50 Hz supply. Answer the following questions. a) Draw the circuit with labels. b) Calculate the capacitive and inductive reactance respectively. c) Calculate the current through each component d) Calculate the total current drawn from the supply. e) If the total current through the circuit is 5,63 A, calculate the phase angle for the circuit. f) Is the circuit more inductive or capacitive? Motivate your answer. 9.

A parallel RCL circuit consists of a resistor of 4 kΩ, a coil of 36 mH and a capacitor of 9,1 nF. The circuit is connected to a 12 V AC supply voltage of which the frequency can be changed. All the required calculations were done and recorded in the table shown below. Freq

XL (Ω)

XC(Ω)

IR(mA)

IL(mA)

IC(mA)

IT(mA)

Cos θ

θ

4 000

904,78

4 370

3

13,26

2,75

10,93

0,27

74,1°

6 000

1 360

2 910

3

8,82

4,12

5,58

0,53

57,5°

8 793

1 990

1 990

3

6,03

6,03

3

1

0°

11 000

2 490

1 590

3

4,82

7,55

4,06

0,74

42,4°

13 000

2 940

1 350

3

4,08

8,89

5,67

0,53

58,1°

Use the information provided in the table to answer the following questions: What observation can be made regarding the following as frequency is increased? a) Inductive reactance (XL) b) Capacitive reactance (XC) c) Resistance d) Capacitor current (IC) e) Resistor current (IR) f) Supply current (IT) g) Power factor h) Phase angle

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Electrical Technology i) What is the relationship between XL and IL? j) What is the relationship between XC and IC? k) What is the relationship between f and IR? l) What is significant about the circuit at 8 793 Hz? m) What are the characteristics of the circuit (as seen on the table) at this frequency? n) Draw a graph to represent IT vs frequency and Z vs frequency. o) What can we deduce from the graphs drawn above?

Practical Activity 1 Aim This module can be done in class without the use of any equipment, and it will test if the learners has mastered all the basic concepts before they do real experiments using real equipment. To investigate the relationship between the voltage across the resistor, the voltage across the capacitor and the supply voltage in a series rc-alternating current circuit. Apparatus 1. 1,0 kohm resistor (R1) 2. 0,1 micro Farad capacitor (C1) 3. Matrix Board with hookup wires (intermediate-board) 4. Audio signal generator (50Hz 10kHz) 5. Dual trace oscilloscope Method 1. Draw a series circuit by making use of the resistor and capacitor that is connected to a signal generator. Indicate how the oscilloscope is connected to the circuit. The one input (Y2) of the oscilloscope is connected across the signal generator and the other input (Y1) across the resistor. The triggering must be set across Y1 (the resistor). Ensure that all the earth connections are connected on one side of the resistor. 2. Assume that this circuit was built correctly. 3. The necessary adjustments were done on the oscilloscope so that one complete cycle of the reference signal (Vr) is displayed over 10 cm on the screen. The V/div setting were adjusted so that the waves are as high (big) as possible.

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T/div = 0,2ms V/div = 0,2V

4. These were the oscillograms (waves) observed on the oscilloscope. Interpret the information and do the necessary calculations for the following:

V = (V/div)(No blocks)(0,707)

Vs = ________________________________________________________

Vr = ________________________________________________________

f = __________________________________________________________

Phase angle = _________________________________________________

The supply current = Vr/R = _______________________________________ 5. Convert the oscillograms to phasors by drawing the phasor diagram to scale (1:10) on the same page. Graphically subtract Vr from Vs to determine the voltage drop across the capacitor. Measure this voltage. All labels (including the scales and phase angle) must clearly be indicated.

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Electrical Technology 6. The frequency is now changed.

T/div = 10μs V/div = 0,2V These were the oscillograms (waves) observed on the oscilloscope at the new frequency. Interpret the information and do the necessary calculations for the following:

V = (V/div)(No blocks)(0,707)

Vs = ________________________________________________________

Vr = ________________________________________________________

f = _________________________==_______________________________

Phase angle = _________________________________________________

The supply current = Vr/R = _______________________________________ 7. Convert the oscillograms to phasors by drawing the phasor diagram to scale (1:10) on the same page. Graphically subtract Vr from Vs to determine the voltage drop across the capacitor. Measure this voltage (NB – This must be done at both frequencies). All labels (including the scales and phase angle) must clearly be indicated. 8. Observation Answer the questions below by comparing the phasor diagrams. Establish how a change in frequency will influence the relationship between Vr and Vs: 1. What is the phase relationship between Vr and Vs at frequency 1. (refer to angle as well as leading or lagging) 2. What is the phase relationship between Vr and Vs at frequency 2. (refer to angle as well as leading or lagging)

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3. Did a change in frequency influence the phase relationship (angle) between Vr and Vs ? 4. What effect does the change in frequency have on the magnitude of Vs, Vr and Vc? 5. What influence does the change in frequency have on the current flowing in the circuit, and hence the impedance of the circuit? (NO calculations allowed to answer this question).

6. What is the reason for all the changes that occurred due to the change in frequency?

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138


Chapter 6 Logic

A

B PLCs

Ladder diagrams

A

B Boolean algebra

Logic circuits

6


Introduction This chapter is only about an introduction into programmable logic control devices (PLCs). Aspects that will be covered in this chapter range from the advantages of PLCs, introduction to ladder logic, conversion from hard-wired schematics to ladder logic, to writing programs for various motor controlled circuits as real time application of the PLC. A PLC can be defined as a specialised computer system that can be used to monitor different inputs, perform a certain function based on the conditions of such inputs (i.e. make decisions e.g. logic, sequencing, timing, counting and arithmetic and data handling) and then control various types of machines or processes. PLCs can be used in a multitude of industries: • manufacturing/machining • food/beverage • metal • power • mining • petrochemical/chemical. PLCs can vary in size, from very small to very large. The smaller units can have up to 128 Input/Outputs ( I/Os) and a memory capacity of about 2 KB, and the larger units can have up to 8192 I/Os and a 750 KB memory capacity.

Short history of PLCs The first programmable logic controller was developed by an American car manufacturer (General Motors) which started the process in 1968. Their primary goal was to eliminate the high cost of inflexible, relay controlled systems. In 1969, the first PLC was installed for industrial use. These hardware CPU controllers used logic instructions, had 1KB of memory and 128 I/O points. In the early seventies, PLCs were used to send and receive varying voltages allowing them to enter the analogue world. Because of the software programs and standardised communication that was developed in the 80s, PLCs became smaller and could even be programmed using a personal computer instead of a hand-held unit or a dedicated programming terminal. All of this had a great impact on the cost of PLCs, making them more affordable to many industries. As time went on, international standards were developed to communicate with PLCs. We now have computers that can be programmed in functional blocks, instruction lists, ladder logic, C (a programming language) and structured text at the same time.

Common brands of PLCs American: Texas Instruments and Allen Bradley European: Siemens, Klockner Moëller and Téléméchanique Japanese: Toshiba and Mitsubishi

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Main components of a common PLC

Processor

Output module

From sensors Push buttons, contacts, limit switches, etc.

Input module

Power supply

To output Solenoids, contactors, alarms, etc.

Programming device

Figure 6.1: Main components of a PLC

Power supply The power supply provides the voltages needed to run all the PLC components. Most PLC controllers work either from 24 V DC or 240 V AC. Some PLC controllers have electrical supply as a separate module, while small and medium series already contain the supply modules.

Input/output module Input modules convert signals from discrete or analogue input field devices to logic levels acceptable to the PLC’s processor. Inputs monitor field devices, such as switches and sensors. Inputs to the devices are connected to the processor by means of an opto-isolator (photo diodes and transistors), i.e. the input is connected to the processor by means of an optical connection. This is to prevent voltage spikes from damaging the processor of an optical connection and also to reduce the effect of electrical noise. Output modules convert signals from the processor to levels capable of driving the connected discrete or analogue output field devices. Outputs control other devices, such as motors, pumps, solenoid valves and lights. Also on the output side the output devices are connected to the processor by means of an opto-isolator (photo diodes and transistors), i.e. the output is connected to the processor by means of an optical connection. This is to prevent voltage spikes from damaging the processor and also to reduce the effect of electrical noise.

Processor This is the ‘brain’ of the PLC. The processor is the unit that is responsible for the processing of all arithmetic operations, logic operators, storage of information, computer interface, local area network, etc. It provides the intelligence to govern and direct all activities of the PLC.

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Electrical Technology The main function of the processor is therefore to analyse data coming from input sensors through input modules, make decisions based on the user’s defined control program and return signals back through output modules to the output devices.

Programming device The programming device is used to enter the necessary program that will determine the sequence of events in the memory of the processor. The program is developed in the programming device and then transferred to the memory unit of the PLC.

Advantages of PLC over hard-wired logic control Hard-wired logic control used to be considered as the first step towards automation. The use of contactors and relays, in conjunction with timers and counters, served their purpose, but the system had some drawbacks, e.g. size, complexity of the wiring, as well as time taken to implement changes. All of this led to the development of the PLC. The advantages of the PLC compared to hard-wired logic control are: • Reduced space: PLCs are solid-state devices (no moving parts) and therefore are very compact and small compared to hard-wired logic. • Use less energy: A PLC uses on average about 1/10th the power consumed by an equivalent relay logic controller. • Less maintenance: PLCs have error diagnostic units, making it easy to troubleshoot. The modular system makes replacement of modules easy. • Economical: Cost of the PLC is recovered over a very short period. It is a onetime investment. • More reliable: It uses static devices, hence no wear and tear on moving parts. Greater ON TIME (where the system works without any interruptions) compared to hard-wired logic systems. • Flexibility: It is easy to change a program. The PLC can carry out more complex functions. • Faster response time: It can process thousands of items per second.

Comparison between relay logic and the PLC

142

Relay

PLC

Large complicated system that takes up lots of space

Very compact solid-state device that is relatively small

Uses much more energy than PLC system

Uses about 1/10th the amount of energy of relay system

Limited mechanical life and more maintenance

A solid-state device with a very long life and less maintenance

More expensive than PLC system – the more the relays, the greater the cost

Costs much less compared to relay logic system (one-time investment)

Mechanical breakdowns can occur at any time – many moving parts

More reliable because of no moving parts

Difficult to update or modify an existing program

Very flexible because it is easy to change or modify a program

Response time slower than that of the PLC

Fast response time – can process thousands of items per second

Logic

6

Basic operation of a PLC PLCs are devices that are programmable, i.e. they need to be programmed to perform a specific function. Programs for PLCs can range from the more common ladder logic to the less common functional blocks. A program is a set of instructions written in a specific programming language for a PLC to perform a specific function. The execution of each instruction requires three basic steps. The three steps are: 1. Checking the inputs – (INPUT): The PLC reads the inputs via the input interface. 2. Execution of instructions – (PROCESS): The PLC will now look at the first instruction in the program and execute it. 3. Updating the outputs – (OUTPUT): After the execution of the first instruction, all outputs will now be updated (changed) accordingly.

STEP 1 Checking inputs

STEP 2 Executing instructions

STEP 3 Update outputs

Figure 6.2: PLC scan cycle

After the completion of the third step, the PLC will start back at step 1 and repeat the process all over again for every instruction. This process of sequentially checking the inputs, executing the program in memory, and updating the outputs is known as scanning.

Introduction to ladder logic As was mentioned earlier, ladder logic is a very common and popular programming language used to write programs for PLCs. The PLC makes use of simple instructions to write a ladder logic program. These instructions are computer codes that make the inputs and outputs do what you want, in order to get the desired result you want. All PLCs have two basic types of instructions. • contacts • coils. Contacts: refer to the information supplied by the input field devices. Different contacts will monitor the status of different input field devices. Depending on the information from the field device, the contactor then tells the PLC what to do. Input field devices refer to: push buttons, limit switches, various sensors, etc.

Coils: refer to what each particular output field device is supposed to do in the system. As with contacts, each coil also monitors the status of a certain output field device; however, unlike contacts that monitor the status of the input field devices and then tell the PLC what to do, the coils tell the field devices what to do. Output field devices refer to: indicating lights, alarm horns, motor starters, etc.

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Electrical Technology Example Let us see what will happen if we connect an on/off switch and a single-phase motor to a PLC. If the switch is supposed to turn the motor on, the PLC would then have a contact that examines the input field device (the on/off switch) and a coil that references the motor. As soon as the switch is turned on, the contact will energise, and this is then relayed to the PLC which in turn relays this information to the coil instruction by energising it. This will tell the coil that it needs to tell the referenced output device (the motor) to start running. Output module (coil)

Input module (contacts)

Motor contactor

On/off switch

PLC

Figure 6.3: Motor and switch connected to PLC

Input and output connections to the PLC Input field devices Input module (contacts)

Output module (coils)

Start button

Output field devices

Lamp

Stop button Overload contact Motor contactor Various input sensors. e.g. limit switch, pressure switch, temp switch, etc.

Common supply for outputs

Figure 6.4: Inputs and outputs connected to PLC

From the above diagram, we can clearly see that all input and output field devices are outside the PLC and the contacts and coils are inside the PLC. It is important that, when writing a program, all inputs and outputs are labelled to make it easier to write the program. Another important point to remember is that the PLC will only react on the input from the input field devices, process the information and let the output field device react accordingly.

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Basic ladder logic instructions Symbol

Definition Normally open contact (Input) The normally open (NO) is true when the input or output status bit controlling the contact is 1 Normally closed contact (Input) The normally closed (NC) is true when the input or output status bit controlling the contact is 0.

or

Coil (Output)

Ladder diagrams consist of two vertical lines representing the power rails. Circuits are connected as horizontal lines between these two vertical lines. A ladder logic program consists of a number of units called ‘rungs’. A rung can be defined as number of symbols (instructions) connected together. The number of symbols (instructions) per rung will vary from PLC to PLC. There is no limit to the number of rungs in a program.

Rung 1

Rung 2

Figure 6.5: Rungs in ladder logic

Rungs are labelled sequentially and are also executed in the same sequence. Each rung must contain a number of inputs and one output. Rungs are executed according to the following rules in any one rung. 1. If the rung only contains series connections between input and output, then the rung is executed from left to right. 1

2

3

Figure 6.6: Rung with only series connections

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Electrical Technology 2. If the rung contains a vertical connection (OR function), the OR function is executed first. 1

3

4

2

Figure 6.7: Rung with vertical connections (OR function)

3. If there is a branch, it is executed in order from the top line to the bottom line. 1

2

3

4

5

Figure 6.8: Rung with a branch

Ladder logic diagrams Inputs and outputs

Take note Fig. 6.9 (a) & (b) represent the ladder diagrams. The switch and the lamp are not part of this ladder diagram. The switch and lamp are outside the PLC and the ladder diagram is the program inside the PLC. Please refer to figure 6.4 on page 138.

As was mentioned earlier, each rung is connected between two power rails (live and neutral), so for current to flow through the circuit and activate the output field device, there must be a continuous link from live to neutral. If the link is broken, the output field device will stop operating. Let us consider the following circuits. Figure 6.9 (a) shows the connection of a lamp through an open contact. When the switch is closed, the contact (X0) is closed and the coil (Q0) will be activated and therefore switch the lamp on. Figure 6.9 (b) shows the lamp connected through a closed contact keeping the lamp on. The lamp will only switch off when the switch is activated, thus opening the contact (X1) and de-energising the coil (Q1). L

N XO

Q0

Inside PLC

Figure 6.9 (a): NO contact to switch one output

L

N X1

Q1

Inside PLC

Figure 6.9 (b): NC contact to switch one output

From figures 6.9 (a) & (b), it is evident that X0 & X1 represent the inputs (contacts) and Q0 & Q1 represent the outputs (coils). Remember, ladder diagrams can have more than one input to activate one output. On the next page is an example of a ladder diagram with two input contacts in series, activating one output coil that in turn could switch a lamp or motor on. In this example, the lamp/motor will only be activated (switched on) when both the input contacts X0 & X1 are closed, therefore energising the coil (Q1) allowing the lamp/motor to be switched on.

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Logic L

6

N X0

X1

Q0

Figure 6.10: Two NO contacts to switch one output

Timers It is important to note that the way timers work differs from PLC to PLC but that the basic operational principles are the same. A timer can therefore be seen as an instruction that waits a set amount of time before doing something. Timer instructions are output instructions used to time intervals for which their rung conditions are true (On delay timer, T-ON), or false (Off delay timer, T-OF). These are software timers that must be programmed to perform a certain function. A second important input to a timer is the enable/reset input. When this input is activated, it will reset the timer back to zero. Once again, the operation of the enable/reset depends on the type of PLC. You will have to study/familiarise yourself with the operation of the PLC you will be working with. On delay timers: This type of timer delay turns the output field device on for a certain amount of clock pulses. In other words, after the input field sensor turns on, it will wait X seconds before switching the output field device on (a motor). This is one of the most commonly used timers. Off delay timer: This timer, on the other hand, will delay turning the output field device off after a certain number of clock pulses. The output field device (a motor) would be on, and eventually switch off after X seconds. Let us look at an example: With reference to figure 6.11, when contact X0 is closed, the timer (T1) starts running and after 5 seconds (indicated by k5) the timer contact (T1) closes and switches the coil (Q0) to the motor on. In this example, when contact X1 is closed, it resets the timer back to zero. Please also note that figure 6.11 is a ladder diagram (inside the PLC) and the switches and motor is not shown in this figure. They are external devices to the PLC. X0

T1/k5

T1

Q0

X1

Reset T1

Figure 6.11: Ladder diagram for a timer circuit

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Electrical Technology Counters Once again, it is important to note that the way counters work differs from PLC to PLC but that the basic operational principles are the same. Counters are instructions to do only one thing and that is count. We basically get two types of counters, one that counts up (i.e. 1,2,3,4, etc.), the up-counter and the other that counts down (i.e. 4,3,2,1, etc.), the down-counter. Counters do not physically exist. They are simulated counters and they can be programmed to count pulses. These pulses could be the number of bottles or boxes passing a specific point and being picked up by a sensor. Counters will also have a set/reset input to set the counter to its starting count or to reset the counter back to zero. Let us look at an example: Figure 6.12 represents a ladder diagram for a counter that is set to 9 and counting down. Each time contact X0 is closed, the counter decreases by 1 (counting down). So if the contact X0 is repeatedly opened and closed 9 times, at the ninth count the counter contact (C1) closes and the coil (Q0) that activates a motor will be energised. When contact X1 is closed, it sets the counter back to its initial state (9). L

N X0

C1/k9

C1

Q0

X1

Set/Reset C1

Figure 6.12: Ladder diagram for a counter circuit

When we use an up-counter on the other hand, every time the input contact X0 is closed, the counter increases (counts up) by one. At the 9th pulse, the counter contact (C1) will close and the coil (Q0) will then be activated, which in turn will switch on the motor. When contact X1 is closed, it resets the counter back to its initial state (0). We could, for example, have a system in a factory where 9 items need to be placed in a box. The counter will count the 9 items, and after the 9th count the full box is then removed and an empty box is put in its place to be filled again. After the 9th count, the counter will also get a reset pulse to set it back to zero.

Data storage (markers, internal relays or flags) These are simple registers in the memory used to store information or data. Information or data is only stored temporarily in these registers. They are also sometimes used for retentive/accumulative data storage. If, for example, the output

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of one rung must be used as an input to another rung (figure 6.13), then we will use a marker/internal relay/flag as a temporary storage device. The input to the second rung depends on the condition of the marker/ internal relay/flag. The marker is represented by coil (M) and the marker contact represented by contact (M). In other words, if marker coil (M) is energised then the marker contact (M) will be closed. However, if the marker is not energised, the marker contact will be open. X0

X1

M

X3

X2

M

Q

X4

Figure 6.13: Use of the marker

Latches Latching is about using an ‘imaginary’ contact that is linked to your output device (coil) and uses such imaginary contact within your program, allowing you to create a latch or some kind of memory element. Let us look at an example. In figure 6.14 below, the contact labelled Q0 is an imaginary contact that is tagged (linked) to the output coil (Q0). In other words, the condition of the output coil will determine the condition of the contact that is tagged (linked) to it. When contact X0 is closed, the coil (Q0) will be energised and the imaginary contact called Q0 will also be energised, resulting in the coil (Q0) staying energised, even though the input contact (X0) is opened again. As with a DOL motor starter, where the normally open switch of the main contactor acts as the holding in contact, a latch keeps the main contactor activated once the start button is released. X0

Q0

Q0

Figure 6.14: A latch circuit

Logic gates and their ladder equivalents The following section of work deals with simply converting the basic logic gates to their equivalent ladder diagrams. By using these simple logic gate equivalents you will be able to implement more complicated gate combinations in ladder logic.

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Electrical Technology When working with Boolean expressions on the other hand, it is important that the Boolean expression be in its simplest form, resulting in the simplest ladder diagrams. Simplification of Boolean expressions can either be done by using the basic Boolean laws or making use of Karnaugh maps. Both methods are relatively easy if you know what it is all about.

Logic gate equivalents Logic gates

150

Truth table

Ladder diagram

Logic

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Converting Boolean expressions to ladder diagrams Let us now look at a few examples of how to convert a simple Boolean expression to a ladder diagram. The table on page 150 already shows us how the five basic logic functions are converted from Boolean expressions to ladder diagrams. The following points are very useful to remember when implementing ladder diagrams from Boolean expressions: • The OR function is represented by parallel contacts. • The AND function is represented by series contacts. • The NOT function is represented by a normally closed contact.

Boolean expressions in sum-of-product (SOP) notation Boolean expressions in the SOP notation are simply an expression consisting of a number of product terms (AND terms), separated by a sum sign (OR sign) e.g. F = A.B + C.D. SOP expressions lend themselves to implementation as a set of AND gates (products) feeding into an OR gate (sum). The following examples show how a Boolean expression in the SOP notation can be converted to a ladder diagram. • F = A.B + C: this expression consists of one AND term (A.B) OR-ed by a single term C. A

B

F

C

Figure 6.15: F = A.B + C

• F = A.B + C.D: this expression consists of two AND terms (A.B) as well as (C.D) separated by an OR sign. A

B

C

D

F

Figure 6.16: F = A.B + C.D

• F = A.B + A.B: this expression consists of two AND terms (A.B) as well as (A.B) separated by an OR sign. If you examine this equation more closely, you will see that it represents the expression for the exclusive OR gate. A

B

A

B

F

Figure 6.17: F = A.B. + A.B

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Electrical Technology • F = A.B.C + A.B.C + A.B.C: this expression consists of three AND terms (A.B.C, A.B.C as well as A.B.C) separated by an OR sign. This expression is, however, not in its simplest form and must still be simplified by making use of Boolean algebra or Karnaugh maps. The simplified expression will look like this: F = B (A + C). A

B

F

C

Figure 6.18: F = A.B.C + A.B.C + A.B.C (non-simplified) F = B (A +C) (simplified)

Boolean expressions in product-of-sum (POS) notation Boolean expressions in the POS notation are simply expressions consisting of a number of sum terms (OR terms), separated by a product sign (AND sign) e.g. F = (A+B).(C+D). POS expressions lend themselves to implementation as sets of OR gates (sums) feeding into a single AND gate (product). The following examples show how a Boolean expression in the POS notation can be converted to a ladder diagram. • F = (A+B).C : this expression consists of one sum term (A+B) AND-ed with a single term C. A

C

F

B

Figure 6.19: F = (A+B). C

• F = (A+B).(C+D): this expression consists of two sum terms (A+B) as well as (C+D) separated by an AND sign. A

C

B

D

Figure 6.20: F (A+B).(C+D)

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F

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• F = (A+B+C).(A+B+C): this expression consists of two sum terms (A+B+C) as well as (A+B+C) separated by an AND sign. A

A

B

B

C

C

F

Figure 6.21: F (A+B+C).(A+B+C)

Boolean laws Boolean laws are logic laws that will help us to mathematically simplify long and complex Boolean expressions and allow us to make our designs simpler. Rules

Explanation of rule

Rule 1

A . 0 = 0 } A variable ANDed with 0 is always equal to 0.

Rule 2

A . 1 = A } A variable ANDed with 1 is always equal to the variable.

Rule 3

A . A = A } A variable ORed with 1 is always equal to 1.

Rule 4

A . A = 0 } A variable ANDed with its complement is always equal to 0.

Rule 5

A + 0 = A } A variable ORed with 0 is always equal to the variable.

Rule 6

A + 1 = 1 } A variable ORed with 1 is always equal to 1.

Rule 7

A + A = A } A variable ORed with itself is always equal to the variable.

Rule 8

A + A = 1 } A variable ORed with its complement is always equal to 1.

Rule 9

A = A } A double inverted function = the function

Rule 10 Commutative Law

A + B = B + A } The order in which two variables are ORed makes no difference.

Take note Boolean laws are used to simplify complicated Boolean expressions.

A.B = B.A } The order in which two variables are ANDed makes no difference. Rule 11 Associative Law

A + (B + C) = (A + B) +C } Free association (grouping) of any two terms A (B.C) = (A.B) C } Free association (grouping) of any two terms

Rule 12 Distributive Law

A (B + C) = A.B + A.C } Simplification by the removal of the brackets

Rule 13 De Morgan’s Law

1 A + B = A . B } The compliment of the sums equals the product of the compliments. Or (When you break the line you change the sign) 2 A . B = A + B } The compliments of the product equals the sum of the compliments. Or (When you break the line you change the sign)

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Simplification by using Karnaugh maps Just as with Boolean algebra, Karnaugh maps are use to simplify Boolean expressions or truth tables. It is just a more graphical form of simplification. A Karnaugh map is simply a big block with a number of smaller cells where each cell represents a specific row in the corresponding truth table, or a term in the Boolean expression. A Karnaugh map can be filled in or populated from a Boolean expression or a truth table. Karnaugh maps comes in different sizes depending on the number of cells. If the Boolean expression or truth table has 2 variables, then the Karnaugh map will have 22 = 4 cells, if there is 3 variables, then the Karnaugh map will have 23= 8 cells and if there is 4 variables then the the Karnaugh map will have 24 = 16 cells. You can get Karnaugh maps with more cells, but we will stop at the 4 variable Karnaugh map. It is important to note that Karnaugh maps are also labelled in a specific way and it is important that you remember it. Have a close look at the Karnaugh maps below and learn how to draw and label them. 2 Variable map: A.B

3 Variable map: A.B.C

OR

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4 Variable map: A.B.C.D

Figure 6.22: Different size Karnaugh maps

Populating (filling in) of the Karnaugh map Now that you can draw and label the Karnaugh maps, let us fill the information from the Boolean expression or truth table into the map. The steps are the same for all sizes of maps, but we will do an example of a 4 variable Karnaugh map. It must be remembered that each term of the Boolean expression is represented by a ‘1’ in the Karnaugh map and all the other cells of the map will have a ‘0’ in it. With reference to the truth tables, the output of each row is plotted in the respective position in the Karnaugh map. Simplify the Boolean expression and or truth tables by making use of Karnaugh maps. Example: F = A.B.C.D + A.B.C.D + A.B.C.D + A.B.C.D + A.B.C.D + A.B.C.D Remember that each term of the Boolean expression has a corresponding place in the Karnaugh map as indicated in figure 6.22 above e.g. Term A.B.C.D is represented by position where A=1, B = 1, C= 1 and D=1 in the Karnaugh map i.e. position (1111) With reference to the truth table below each row is represented by a specific position in the Karnaugh map eg row 1111 is represented by the position where A=1, B=1 , C=1 and D=1 in the Karnaugh map ie position (1111). A 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

B 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1

C 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

D 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

OUT 0 0 1 0 1 0 1 0 0 0 0 0 1 0 1 1

Figure 6.23: Four variable truth table

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Electrical Technology Let us now refer back to our example as set out above. Step 1 Draw and label the Karnaugh map.

Step 2 Plot the information from the Boolean expression/ truth table into the map. (Note for our example the information for the Boolean expression and truth table is exactly the same, so the Karnaugh map will also be the same.)

Step 3 Group the ‘1s’ together in specified group sizes. The aim of grouping the 1s together is to get the biggest possible group of 1s together, in order to give us the simplest output term. These groups must consist of only 1s that are adjacent to one another. This means the 1s must only be above, below and alongside each other. A group cannot consist of 1s diagonally across from each other. All groups should be marked with a loop/circle as will be indicated below. Groups can overlap ie. a 1 can be shared by two or more groups, provided it makes the other groups bigger. Let us look at some examples to show what grouping is all about before we continue with this example.

Take note The bigger the group the simpler the term.

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Possible group sizes are as follow: The smallest group consist of only 1 logic 1 The next group size is 2 1s The next group size is 4 1s The next group size is 8 1s And the biggest possible group size will consist of 16 1s.

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Examples of Karnaugh map groupings

Two groups of 1 only

Two groupls of 2 1’s

Two groups of 2 1’s only

One group of 4 1’s connected by sides of Karnaugh map and 1 group of 2

One group of 4 1’s and one group of 2 1’s

One group of 4 1’s and one group of 2 1’s

One group of 8 1’s

Two groups of 4 1’s and one group of 2 1’s connect top and bottom sides of Karnaugh map

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One group of 8 1’s (connected by the sides of the Karnaugh map and one group of 4 1’s

Two groups of 4 1’s and two groups of 2 1’s

Figure 6.24: Different examples of grouping of Karnaugh maps

Step 4 Extracting the simplified Boolean expression from the Karnaugh map. Each group in the Karnaugh map represents a term, the bigger the group the simpler the term. All terms are separated by an OR (+) sign. The following example will explain the process more clearly. Example 1 Simplify the following Boolean expression. F = A.B + A.B + A.B

To get the simplified expression for each group, examine each group individually. Remember each group represents a simplified term. This Boolean expression will have two simplified terms. Let us examine group 1 first. Compare the condition of A and B in both cells making up group 1. If the condition of a variable changes from one cell to the next within the same group (when a variable is complimented) then that variable is discarded/eliminated, and if the condition of a variable remains the same within the group then that variable remains as part of the simplified answer. In group 1, A is changing from a 0 to a 1 and can therefore be eliminated. Variable B remains constant at 0 and will be part of term 1. Our simplified term for group 1 = B

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Now let us examine group 2. Variable A remains constant at 1 in both cells, but variable B changes from a 0 to a 1. Our simplified term for group 2= A Our simplified Boolean expression will therefore be F = B + A This whole process sounds very complicated, but with some practice it will become quite easy. The following basic rules will help you do Karnaugh map simplifications. • • • • •

Determine the size of the Karnaugh map. Label the Karnaugh map correctly. Record the 1s and 0s in the Karnaugh map. Group the adjacent 1s in correct size groups Simplifying by eliminating variables that is changing within the group (variables that is complimented within a group) • Write the simplified answer in sum-of-products notation. Let us do a few more examples to consolidate Karnaugh map simplification. Example 2 Simplify the following Boolean expression by making use of Karnaugh maps. F = A.B.C + A.B.C + A.B.C + A.B.C

Group 1 = B.C (A changed from 0 → 1) Group 2 = A.B (C changed from 0 → 1) Group 3 = A.C (B changed from 0 → 1) ∴ Simplified equation F = B.C + A. B + A.C Example 3 Simplify the following Boolean by making us of Karnaugh maps. F = A.B.C.D + A.B.C.D + A.B.C.D + A.B.C.D + A.B.C.D + A.B.C.D

Group 1 = B.D (A & C changed from 0 → 1) Group 2 = A.B.D (C changed from 0 → 1) ∴ Simplified equation F = B.D + A.B.D

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Boolean algebra & Karnaugh map simplification As was mentioned earlier, any Boolean expression must be in its simplest form before it can be converted to ladder diagrams. Not simplifying the equation can lead to a very complicated and difficult ladder diagram. Below, we show how simplification of Boolean expressions can make the ladder diagrams less complicated. Example 1 F = A.B.C + A.B.C + A.B.C: this expression is not in its simplest form and can therefore be simplified by making use of Boolean algebra or Karnaugh maps. Boolean algebra simplification

Karnaugh map simplification

F = A.B.C + A.B.C + A.B.C = A.B (C+C) + A.B.C = A.B (1) + A.B.C = A.B + A.B.C = B (A + A.C) = B (A + C) F = A.B + B.C

F = A.B.C + A.B.C + A.B.C

F = A.B + B.C OR

OR F = A.B.C + A.B.C + A.B.C = (A.B.C + A.B.C) + (A.B.C) + (A.B.C) = A.B( C + C) + B.C (A +A) = A.B (1) + B.C (1) = A.B + B.C F = A.B + B.C

Non-simplified Boolean expression: F = A.B.C + A.B.C + A.B.C Non-simplified ladder diagram A

B

C

A

B

C

A

B

C

F

Figure 6.25: Non-simplified ladder diagram

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Simplified Boolean expression: F = A.B + B.C = B(A + C) Simplified ladder diagram A

B

F

C

Figure 6.26: Simplified ladder diagram

Example 2 F = A.B.C +A.B.C +A.B.C +A.B.C: this expression must now be simplified by means of Boolean algebra or Karnaugh maps.

Boolean algebra simplification

Karnaugh map simplification

F = A.B.C + A.B.C + A.B.C + A.B.C = A.C (B+B) +A.C (B+B) = A.C + A.C = C (A+A) F=C

F = A.B.C + A.B.C + A.B.C +A.B.C

F=C OR

F=C

Non-simplified Boolean expression: F = A.B.C + A.B.C + A.B.C + A.B.C Non-simplified ladder diagram A

B

C

A

B

C

A

B

C

A

B

C

F

Figure 6.27: Non-simplified ladder diagram

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Electrical Technology Simplified Boolean expression: F = C Simplified ladder diagram: C

F


Combination logic networks Combination logic is used for building circuits where we want to obtain certain outputs, given certain input conditions. Combination logic circuits have no feedback, therefore the output is dependent at all times on the combination of its inputs. Combination logic circuits can be made up by making use of the basic logic gates (AND, OR, NOT, NAND, NOR, etc.) These combination logic circuits can be very simple or very complicated. These combination logic networks can be implemented by making use of either sum-of-product or the product-of-sum methods. Very often, logic designs are initiated by the designers describing in words how a system should work. This verbal information can then be translated into a truth table, from which a person can then develop a Boolean expression (SOP or POS). The Boolean expression is then simplified and implemented as a simplified combination of logic gates. The simplified expression can also be implemented as a simple ladder diagram. To see how this works, let us look at a few examples. Example 1 Design a three bit logic counter that must produce a logic 1 at the output only when it counts an odd number. Implement this design by making use of ladder logic. Step 1: Compile a complete truth table for all the input and output bits from the information given to you. Truth table A

B

C

0

0

0

0

0

0

1

1

0

1

0

0

0

1

1

1

1

0

0

0

1

0

1

1

1

1

0

0

1

1

1

1

Inputs

Output F

Output

Figure 6.29: Truth table for a three bit counter

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Step 2: From the truth table, now develop the Boolean expression. This can be in SOP or POS notation. The notation that is chosen will depend on which notation will give us the simplest design. For this example we will work with the SOP notation. Boolean expressions (SOP) F = A.B.C + A.B.C + A.B.C + A.B.C This expression is not simplified and can therefore be simplified by either using Boolean algebra or Karnaugh maps. Simplification with Boolean algebra F = A.B.C + A.B.C + A.B.C + A.B.C = B.C (A + A) + BC (A + A) = B.C (1) + BC (1) = C (B + B) =C OR

F=C Step 3: From the simplified Boolean expression we can now implement the logic gate combination. These logic gate circuits are in its simplest form and cannot be simplified any further.

Figure 6.30: Simplified logic gate combination

Step 4: The final stage is the implementation of the ladder diagram from the simplified logic circuit. Always remember: • The OR function is represented by parallel contacts • The AND function is represented by series contacts • The NOT function is represented by a normally closed contact.

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Electrical Technology Ladder diagram

Figure 6.31: Ladder diagram for the counter

Example 2 In the control room of a factory, three motors are monitored. If two or more of the motors develop a fault, a buzzer must go off in the control room, notifying the person in the control room. Design a logic circuit that will be able to do this function, making use of ladder logic. Step 1: Compile a complete truth table for all the input and output bits from the information given to you. Truth table

Take note 0 = no fault 1 = fault

Figure 6.32: Truth table for motor monitoring alarm

Step 2: From the truth table, develop the Boolean expression. This can be in SOP or POS notation. The notation that is chosen will depend on which notation will give us the simplest design. For this example we will work with the SOP notation. The Boolean expression to satisfy our logic design is: F = A.B.C + A.B.C + A.B.C + A.B.C From this expression, it is clear that the expression is not in its simplest form. The expression must first be simplified before we can implement the logic circuit or ladder diagram.

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Simplification of Boolean expression F = A.B.C + A.B.C + A.B.C + A.B.C = B.C (A +A) + A.B.C + A.B.C = B.C+ A.B.C + A.B.C = C (B + AB) +A.B.C = B.C + A.C + A.B.C = B.C + A (C + B.C) F = B.C + A.C + A.B (simplified expression) OR

F = B.C + A.C + A.B Step 3: From the simplified Boolean expression, we can now implement the logic gate combination. These logic gate circuits are in their simplest form and cannot be simplified any further. Logic gate combination From the simplified Boolean expression, we can now implement the combination logic gate. This logic gate circuit is in its simplest form and cannot be simplified any further. A B C

F

Figure 6.33: Simplified logic gate combination

Step 4: Ladder diagram The final stage is the implementation of the ladder diagram from the simplified logic circuit. Always remember: The OR function is represented by parallel contacts. The AND function is represented by series contacts. The NOT function is represented by a normally closed contact. B

C

A

C

A

B

F


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Motor starter control: a simple approach Example 1 Instructions: With the aid of a personal computer (PC), design and write a simple ladder logic program that will be able to control a three-phase motor (direct–on–line motor starter controller). Your design must have one start button, one stop button as well as an overload relay. Once you have completed your program, download it to the PLC and then run the program. (First simulate the program on your PC before it is downloaded.) STEP 1 Switch the motor on without any latching

X0

Q0

L Start MC 1

X0 – Start Q0 – Output to main contactor of motor

N Step 2 Introduction of the latch

X0 – Start Q0 – Output to main contactor of motor Q0 – Holding in for main coil of motor

Step 3 Introduction of a stop button to stop the motor.

X0 – Start X1 – Stop Q0 – Output to main coil of motor Q0 – Holding in for main coil of motor

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Step 4 Introduction of overload protection to the circuit.

X2 – Overload X0 – Start X1 – Stop Q0 – Output to main contactor of motor Q0 – Holding in for main contactor of motor

From the above example, it is pretty clear that if you break down the question into smaller chunks, it is much easier to solve such questions. Example 2 ( Forward and reverse starter) • With the aid of a personal computer (PC), design and write a simple ladder logic program that will be able to control a three phase motor. The design must be so that when the first start button (Forward) is pressed, the motor must run in one direction and when the second start button (Reverse) button is pressed, the motor must run in the opposite direction. The design must be so that the rotational direction cannot be change while the motor is running. If the stop button is pressed, the motor must stop. The design must include an overload protection. • Once you have completed your program, download it to the PLC and then run the program. (First simulate the program on your PC before it is down loaded) Let us start with the wiring diagram of the control circuit first and then do the ladder logic program.

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Electrical Technology Wiring diagram (Control circuit)

Figure 6.35: Wiring diagram of a forward/reverse starter

Ladder program for a forward/reverse starter

Figure 6.36: Ladder program of a Forward/Reverse starter

Example 3 (Star/Delta starter) • With the aid of a personal computer (PC), design and write a simple ladder logic program that will be able to control a three phase motor. The design must be so that when the start button is pressed, the motor must start as a star connected motor and after about 5 seconds it must switch the motor connections to delta. The design must have interlocking between the star and delta contactors. If the stop button is pressed, the motor must stop. The design must include an overload protection. Note: A star-delta starter is used to reduce the starting current of a motor at start and reduce unnecessary tripping.

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• Once you have completed your program, download it to the PLC and then run the program. (First simulate the program on your PC before it is down loaded) • Let us start with the wiring diagram of the control circuit first and then do the ladder logic program. Wiring diagram (Control circuit)

Figure 6.37: Wiring diagram of a Star/Delta starter

Ladder program of Star/Delta

Figure 6.38: Ladder program of a Star/Delta starter

Example 4 (Sequence starter without a timer) With the aid of a personal computer (PC), design and write a simple ladder logic program that will be able to control a three phase motor and a lamp. The design must be so that the light must first be on before the motor can be started. If the stop button is pressed, both the lamp and the motor must stop. The design must include an overload.

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Electrical Technology Once you have completed your program, download it to the PLC and then run the program (First simulate the program on your PC before it is down loaded). Let us start with the wiring diagram of the control circuit first and then do the ladder logic program. Wiring diagram (Control circuit)

Figure 6.39: Wiring diagram of a sequence starter without a timer

Ladder program for sequence starter without a timer

Figure 6.40: Ladder program of a sequence starter without a timer

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Example 5 (Sequence start with a timer) • With the aid of a personal computer (PC), design and write a simple ladder logic program that will be able to control a three phase motor and a lamp. The design must be so that the light must first be on for 10 seconds before the motor can be started. If the stop button is pressed, both the lamp and the motor must stop. The design must include an overload. • Once you have completed your program, download it to the PLC and then run the program. (First simulate the program on your PC before it is down loaded) Let us start with the wiring diagram of the control circuit first and then do the ladder logic program. Wiring diagram (Control circuit)

Figure 6.41: Wiring diagram of a sequence starter with a timer

Ladder program for sequence starter with a timer

Figure 6.42: Ladder program of a sequence starter with a timer

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Activity 1 1. Write down a simple definition of a PLC. 2. Explain what the abbreviation PLC stands for. 3. Many industries make use of PLCs, name two industries using PLCs. 4. What was the main reason for the introduction of PLCs in the early 1960s? 5. Write down the main components of a simple PLC. 6. With the aid of a simple block diagram, explain the main components of a simple PLC. 7. What are the main advantages of using the PLC over hard-wired logic relays? 8. With the aid of a simple block diagram, explain how a PLC instruction is executed. 9. Write down in your own words the definition for a PLC logic program. 10. Name the two basic types of instructions of a ladder logic program and also explain the function of each instruction. 11. Draw the ladder logic symbols for the following: • A normally closed contact • A coil 12. Name THREE programming methods used in programmable logic controllers. 13. The following operands are used in the programming of programmable logic controllers. Write down TWO examples of each. • Inputs • Outputs 14. Why is it important to label all the components in a ladder diagram? 15. Draw the logic symbol, truth table as well as ladder diagram for a NAND gate. 16. What is the difference between the sum-of-products notation and the product–of–sums notation? 17. Draw the ladder diagram for each of the following Boolean expressions. • Q = A.B + A.C • Q = (A + B) +C • Q = A.B.C + A.B.C • Q = B ( A + C) 18. By making use of Boolean algebra, simplify the following Boolean expressions and then implement the ladder diagram for the simplified expression. • Q = A.B.C + A.B + C • Q = (A + B) (A + C) • Q = M.N.O + M.N.O • Q = X.Y.Z + X.Y.Z + X.Y.Z + X.Y

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19. From the truth table below, determine the simplified SOP Boolean expression, and also draw the ladder diagram for this expression. Inputs

Output

W

X

Y

(Q)

0

0

0

0

0

0

1

0

0

1

0

1

0

1

1

0

1

0

0

1

1

0

1

1

1

1

0

0

1

1

1

0

20. Write a ladder logic program that will perform the function as per the diagram below.

21. What logic gate does the switching diagram below represent? Also compile a truth table for this gate.

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Electrical Technology 22. Write down the Boolean expression that is represented by the ladder diagrams below.

23. Write a ladder logic program that will turn on a 3-phase motor and after five seconds switch on a green lamp, indicating the motor is running. Your design must have one stop button, one start button and an overload. 24.

A fruit packing plant has a conveyer belt with three position sensing devices. Each sensing device produces an output of 1 when a box of fruit is sensed in that position. The boxes of fruit must be loaded onto another conveyer belt only when two or more of the sensing devices are producing signals of 1.

You are required to design a logic system that will perform the above mentioned task. Also write a ladder logic program that will perform this function. Your design must include the following: • Truth table • Boolean expression • Karnaugh maps • The gate network • Ladder program

25.

Explain the following terms with reference to PLC’s • Economical • Reduced maintenance • Ladder logic

26.

Give examples and uses for the following terms and devices with referenced to programming PLC’s eg. Output (y) can be a light or buzzer to alert the operator about a faulty condition in a plant. • Markers/internal relays or flags • Timers • Counters

27. What is the purpose of the following devices of a PLC. • Central Processing Unit 28. PLC’s are very versatile in their applications, give some applications and explain the impact they have on production.

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Practical Activity 1 Part1 With the aid of a simple sketch, show how you would connect the PLC available to you to a motor control board. Please ensure that you label all inputs and outputs correctly. It must be noted that all switches, contactors, motors and overloads are external to the PLC and must be drawn as such on the sketch. You must indicate at least six inputs and two outputs. Part 2 Sequence motor starter controller (without a timer) Form of activity: PLC programming Material and equipment: • PLC unit • Connecting wires • Motor control board with the following components: – 1 × Stop button – 2 × Start buttons – 1 × Overload relay – 2 × Contactors – 1 × Three-phase motor – 1 × Lamp Instructions: • With the aid of a personal computer (PC), design and write a simple ladder logic program that will be able to control a three-phase motor and a lamp. The design must be such that the light must be on before the motor can be started. If the stop button is pressed, both the lamp and the motor must stop. The design must include an overload. • Once you have completed your program, download it to the PLC and then run the program (first simulate the program on your PC before it is downloaded). Part 3 Sequence motor starter controller (with a timer) Form of activity: PLC programming Material and equipment: • PLC unit • Connecting wires • Motor control board with the following components: – 1 × Stop button – 2 × Start buttons – 2 × Overload relays – 2 × Contactors – 1 × Three-phase motor – 1 × Lamp – 1 × Timer unit (On-delay)

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Electrical Technology Instructions: • With the aid of a personal computer (PC), design and write a simple ladder logic program that will be able to control a three phase motor and a lamp. The design must be such that the light must be on for 10 seconds before the motor can be started. If the stop button is pressed, both the lamp and the motor must stop. The design must include an overload. • Once you have completed your program, download it to the PLC and then run the program (first simulate the program on your PC before it is downloaded). Part 4 Forward and reverse motor starter controller Form of activity: PLC programming Material and equipment: • PLC unit • Connecting wires • Motor control board with the following components: – 1 × Stop button – 2 × Start buttons – 2 × Overload relays – 2 × Contactors – 1 × Three-phase motor Instructions: • With the aid of a personal computer (PC), design and write a simple ladder logic program that will be able to control a three-phase motor. The design must be such that when the first start button (forward) is pressed, the motor must run in one direction and when the second start button (reverse) button is pressed, the motor must run in the opposite direction. The design must be such that the rotational direction cannot be changed while the motor is running. If the stop button is pressed, the motor must stop. The design must include an overload protection. • Once you have completed your program, download it to the PLC and then run the program (first simulate the program on your PC before it is downloaded). Part 5 Automatic star-delta motor starter controller Form of activity: PLC programming Material and equipment: • PLC unit • Connecting wires • Motor control board with the following components: – 1 × Stop button – 2 × Start buttons – 2 × Overload relays – 3 × Contactors – 1 × Three-phase motor – 1 × Timer unit (On-delay)

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Logic Instructions: • With the aid of a personal computer (PC), design and write a simple ladder logic program that will be able to control a three-phase motor. The design must be such that when the start button is pressed, the motor must start as a star-connected motor and after about 5 seconds it must switch the motor connections to delta. The design must have interlocking between the star and delta contactors. If the stop button is pressed, the motor must stop. The design must include an overload protection. • Once you have completed your program, download it to the PLC and then run the program (first simulate the program on your PC before it is downloaded).

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Take note A star-delta starter is used to reduce the starting current of a motor at start and reduces unnecessary tripping.

Practical Activity 2 Part 1 Form of activity: PLC programming Material and equipment: • PLC unit • Desktop /Laptop computer with PLC software Instructions: • With the aid of a personal computer (PC), design and write a simple ladder logic program that will allow you to activate a guillotine only if two normally open (N/O) switches are pressed at the same time (simultaneously) The guillotine must de-activate (stop) when any one of the two switches is released. • First simulate the program on your PC before it is down loaded to the PLC Part 2 Form of activity: PLC programming Material and equipment: • PLC unit • Desktop /Laptop computer with PLC software Instructions: • With referenced to the above design now add a timer that must first allow 10 seconds to elapse before the guillotine will activate after both switches have been pressed. • First simulate the program on your PC before it is down loaded to the PLC Part 3 Form of activity: PLC programming Material and equipment: • PLC unit • Desktop /Laptop computer with PLC software Instructions: • Extend the above program by adding a buzzer that must also go on during the 10 second period before the guillotine is activated. • First simulate the program on your PC before it is down loaded to the PLC

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Electrical Technology Part 4 Form of activity: PLC programming Material and equipment: • PLC unit • Desktop /Laptop computer with PLC software Instructions: • Add to your program a counter that will be able to count the number of times the guillotine is activated for the day. You must be able to reset the counter at the end of the day. • First simulate the program on your PC before it is down loaded to the PLC Part 5 Form of activity: PLC programming Material and equipment: • PLC unit • Desktop /Laptop computer with PLC software Instructions: • With the aid of a personal computer (PC), design and write a simple ladder logic program that will allow you switch on your water pump that will supply water to your three pop up sprinklers. The design must have an over load to protect the motor as well as a timer that must keep the system on for 30 minutes before it must be switched off automatically. • First simulate the program on your PC before it is down loaded to the PLC.

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Chapter 7 Amplifiers

A Feedback

Applications of operational amplifiers

A

B

B Oscillators

Multivibrators

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Introduction In Grade 11, you started working with the transistor used as an amplifier. In this chapter we will once again look at amplifiers, but the focus will be on operational amplifiers commonly known as op-amps. We will focus on the different applications of op-amps, the different circuit constructions, input output waveforms as well as the construction of these circuits on a breadboard. The concept of positive and negative feedback will also be looked at, as well the use of op–amps in different oscillator circuits. Op-amps are integrated circuits (ICs) with a high voltage gain connected to a DC voltage source. They have two inputs and usually a single output. An op-amp produces an output that is amplified hundreds of thousands of times. Op–amps are one of the basic building blocks of analogue electronic circuits. They are linear devices and can therefore be used in a variety of different circuit applications. Characteristics of a circuit using an op-amp are set by the external components. Op-amps are cheap to manufacture and are very versatile devices. The design of opamps is such that they try to be as close as possible to an ideal amplifier.

Characteristics of the ideal operational amplifier As indicated previously, operational amplifiers are high gain amplifiers that can be used in a number of applications, depending on the feedback and the components connected to them. Amplifiers are rarely ideal but the op-amp comes very close to an ideal amplifier. Below are some open loop characteristics of an ideal amplifier. • Infinite open loop gain ( AV = ∞ ). • Infinite bandwidth (amplifies both AC and DC signals without any loss in gain). • Infinite input impedance (Zin = ∞). Zero current flows from V+ to V–. • Zero output impedance (Zout = 0). • Infinite common mode rejection ration (CMRR= ∞). • Very high stability.

The differential amplifier The principle of operation of the op-amp is based on the simple differential amplifier as indicated in Figure 7.1

Figure 7.1: Differential amplifier

The circuit is constructed by using two identical transistors, biased at the same operating point with the emitters connected through RE to the negative supply. The value of both collector resistors is also identical. The circuit makes use of a dual power supply (+VCC and –VEE). This circuit has two inputs (V1& V2) and one output (Vout).

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The voltage that appears at the output, Vout of the amplifier is the difference between the two input signals as the two base inputs are out of phase with each other. So, as the forward bias of transistor TR1 is increased, the forward bias of transistor TR2 is reduced and vice versa. Then if the two transistors are perfectly matched, the current flowing through the common emitter resistor, RE will remain constant. This differential amplifier, as the name indicates, will only amplify the difference between the two input signals. If V1 and V2 are equal to each other, then Vout will be zero.

Symbol for an op-amp Figure 7.2 shows the basic symbol used for the op-amp. From the symbol, we can see that the op-amp has two inputs, one output and a dual power supply. The input connected to the –ve input is referred to as the inverting input and the input connected to the +ve input is referred to the non-inverting input. This figure also indicates the dual power supply. Normally, the supply inputs are not indicated on the symbol, only the inputs and output.

Figure 7.2: Op-amp

Figure 7.3 shows the layout of the Dual-in-line IC package of the most common opamp package, the LM 741.

LM741

Figure 7.3: Dual in-line op-amp IC package

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Electrical Technology Outputs of ideal amplifiers with different inputs In the examples below, we will look at what the outputs would look like for a combination of different input signals. V1 is connected to the inverting input and V2 is connected to the non-inverting input. Example 1 Input signal is applied to V1 (inverting input) and V2 (non-inverting INPUT: input) is grounded. OUTPUT: Output signal is amplified but 1800 out of phase with input signal.

Example 2 Input signal is applied to V2 (non-inverting input) and V1 (inverting INPUT: input) is grounded. OUTPUT: Output signal is amplified and in phase with input signal.

Example 3 INPUT: The same input signal is applied to both V1 and V2. OUTPUT: Output signal equals zero.

Example 4 INPUT: OUTPUT:

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Input signals V1 and V2 have the same amplitude but are 1800 out of phase. Output signal will be amplified to about double the input signal, but with the same phase relationship as input V2.

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Principle of operation of negative/positive feedback of operational amplifiers With reference to op-amps, feedback is the process of feeding a small portion of the output signal back to the input. The portion that is fed back to the input could either be in phase with the input signal (the fed-back signal is added to input signal) or 1800 out of phase with the input signal (the fed-back signal is subtracted from the input signal). A circuit can either make use of positive feedback or negative feedback, depending on the application. Amplifier circuits most commonly make use of negative feedback and oscillator circuits will make use of positive feedback. We will first look at negative feedback and then look at positive feedback later when we deal with oscillators.

Did you know? Robert Widlar was responsible for the development of the integrated circuit operational amplifier.

Negative feedback: This occurs when a portion of the output signal is fed back to the input but it is 1800 out of phase with the input signal (the fed-back signal is subtracted from the input signal). A negative feedback amplifier is a system consisting of three elements (see Figure 7.4): an amplifier with gain Av , a feedback network β and a summing circuit acting as a subtractor (the circle in the figure). It is important to note that the output (gain) of the circuit with negative feedback will always be less than the gain of the circuit without any feedback.

Figure 7.4: Negative feedback

Advantages of negative feedback: • The increase in bandwidth of the amplifier, i.e. the frequency response of the amplifier, is greatly improved. The circuit will have a constant response over a wider frequency range. • Increased stability of the amplifier. The amplifier can maintain a constant value in circuit gain. • Reduces distortion and noise. • Improved input and output impedances. • Allows us to design for a specific gain. Disadvantages of negative feedback: • The only disadvantage of negative feedback is that the gain/output of the circuit is reduced.

Different applications of op-amps As was said earlier, op-amps are devices that are very versatile and easy to use. By changing just a few components externally connected to the op-amp, it can be used in a wide variety of applications, ranging from amplifiers to oscillators. We will now look at a few of the basic applications of op-amps in simple circuits.

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Op-amp as comparator

Take note Comparators compare the input voltage with a reference voltage and always give a square wave as output.

A voltage comparator is an electronic circuit that compares the input voltage Vin to that of a reference voltage Vreference . When your input voltage is applied to the noninverting input of the circuit, then we have a non-inverting voltage comparator and when the input signal is applied to the inverting input, then we have an inverting voltage comparator, as indicated in figures 7.5 (a) and 7.7 (a) respectively. It is important to note that the comparator is an example where no feedback is used. In the non-inverting voltage comparator (Fig 7.5 (a)), if Vin is smaller than Vreference, then the output will be switched to about –Vsupply and when Vin becomes greater than Vreference, the output will switch to +Vsupply . Applications of comparators Comparators are very versatile and can be used in a number of applications. The circuit configuration will vary depending on the application. Let us look at some applications. Zero level detector: This is a comparator with its reference level set at zero. It is used for the detecting the zero crossing of AC signals. The output voltage will switched between +Vcc and –Vcc depending on the configuration of the detector circuit Threshold detector: This is when the input voltage is compared with a threshold voltage that is normally generated with a voltage divider network connected to noninverting input or inverting input. Null detectors: A null detector is a device that functions to identify when a given value is zero. Comparators can be a type of amplifier distinctively for null comparison measurements. Comparators are also commonly used in devices such as analog –to- digital converters, Relaxing oscillators, bar graph drivers of graphic equalisers.

Rin Rreference

Vin Vout

Vreference

Input voltage Vin

+Vsupply

Output voltage Vout

Figure 7.5 (a): Non-inverting voltage comparator –Vsupply

Figure 7.5 (b): Waveforms for the non-inverting voltage comparator

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Example With reference to figure 7.5 (a) on page 184 (Non- inverting comparator), what would the output be with the following input conditions? Vref = 2 V Vin = 3 V peak AC When Vin is smaller than Vref , then the output will be –Vsupply and when Vin is greater than Vref , the output will be +Vsupply . Vin = 3 Vpeak

Input voltage more than 2 V

Vreference = 2 V

+Vsupply

Input voltage less than 2 V

Sqaure wave output –Vsupply

Figure 7.6: Non-inverting comparator output

In the inverting voltage comparator (Fig 7.7 (a)), if Vin is less than Vreference, then the output will be switched to about +Vsupply and when Vin becomes greater then Vreference, the output will switch to -Vsupply . Vin

Rin Vout

Vreference

Figure 7.7 (a): Inverting voltage comparator

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Take note The input voltage can take any form or shape and the output voltage will always swing between +Vsupply and –Vsupply (a square wave).

Rreference Input voltage Vin

+Vsupply

Output voltage Vout –Vsupply

Figure 7.7 (b): Waveforms for the inverting voltage comparator

Example With reference to figure 7.7 (a) on page 185 (inverting comparator), what would the output be with the following input conditions? Vref = 5 V Vin = 7 V peak AC When Vin is less than Vref , then the output will be +Vsupply and when Vin is greater than Vref , the output will be -Vsupply . Vin = 7 Vpeak

Input voltage more than 5 V

Vreference = 5 V

+Vsupply

Input voltage less than 5 V

Sqaure wave output –Vsupply

Figure 7.8: Inverting comparator output

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Breadboard construction of inverting comparator

Figure 7.9: Inverting comparator

Waveform representation on oscilloscope and measurements (inverting comparator) Oscilloscope settings: Channel 1 = 5 V/Div Time/Div = 0,2 ms Rin = 10 kΩ & R1 = 10 kΩ

Input voltage

Output voltage

Figure 7.10 (a): Vin is less than the reference voltage of 5 V, therefore Voutput = + Vsupply = 5 V × 3 Div = +15 V

Input voltage

Output voltage

Figure 7.10 (b): Vin equals the reference voltage of 5 V, therefore Voutput = 0 V

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Input voltage

Output voltage

Figure 7.10 (c): Vin is more than the reference voltage of 5 V, therefore Voutput = – Vsupply = – (5 V × 3 Div) = –15 V

Op-amp as an inverting amplifier With reference to the inverting op-amp circuit, a portion of the output is fed back to the input through the feedback resistor Rf but 180° out of phase with the input signal at the point at which they join. The output of this type of amplifier is always amplified, but 180° out of phase with the input. The amplification factor (voltage gain) is determined by the ratio of Rin and Rf . The formula for calculating the Rf voltage gain is: Vout = – ___ AV = ___ Vin Rin Take note V R out Av = ___ = __f Vin Rin Rf Vout = – ___ × Vin Rin

and therefore the formula to calculate the output voltage is given by: R Vout= – __f × Vin Rin The negative sign is an indication that the output signal is 180° out of phase with the input signal. It is important to note that the input voltage is always applied to the inverting input of the op-amp through Rin. Applications of inverting op amps Inverting amplifier can be used in audio applications where the signal needs to be inverted. Other applications are filters, oscillators, controllers etc.

Input wave

Figure 7.11 (a): Inverting amplifier

Output wave – amplified and 180° out of phase with the input wave

Figure 7.11 (b): Waveforms for the inverting op-amp

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Example Calculate the output voltage of the amplifier shown in figure 7.11 (a) if the following information is given to you. Rf= 100 kΩ Rin = 10 kΩ Vin = 5 V Take note The negative sign in front of the 50 V indicates an 180° phase shift between input and output.

R Vout = – [ ___f × Vin] Rin 100 × 103 = – [________ 10 × 103 × (5)] = – [10 × 5] = – 50 V Breadboard construction of an inverting op-amp

Figure 7.12: Inverting op-amp

Waveform representation on an oscilloscope and measurements Oscilloscope settings: Channel 1 = 1 V/Div

Channel 2 = 5 V/Div

Time/Div = 0,1 ms

1 Frequency = 1 = T 0,1 × 10-3 × no Div 1 = = 1 000 Hz or 1 kHz 0,1 × 10-3 × 10 Rf =10 kΩ Rin = 10 kΩ Channel 1 Input signal = 1 V × 1 Div = 1 V peak

Figure 7.13 (a): Input signal

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Channel 2 Output signal = 5 V × 2,5 Div = 12,5 V peak Output signal = 180° out of phase with input signal Figure 7.13 (b): Output signal

Combination of channel 1 & 2 Vout Gain = ____ Vin = 12,5 V 1V = 12,5 Figure 7.13 (c): Input and output signals combined

Op-amp as a non-inverting amplifier In the non-inverting op-amp circuit, the input signal is now applied to the noninverting input of the op-amp. Feedback of this circuit is through Rf back to the inverting input, with resistor Rin used to couple the inverting input to ground. The voltage gain for this circuit can be calculated as follows. R Vout = 1 + ___f , Av = ___ Vin Rin and therefore the formula to calculate the output voltage is given by: Rf Vout = [ 1 + ___ ] × Vin Rin It is important to note that the input and output waves are in phase, therefore there is no negative sign in front of the formula. Take note V 1+R out Av = ___ = _____f Vin Rin Rf Vuit = [1 + ___ ] × Vin Rin

Application of the non-inverting op amp The non-inverting op amp can be used in various audio applications where a phase shift is not required like in audio amplifiers. Other applications are filters, oscillators, controllers, etc. + –

Figure 7.14 (a): Non-inverting amplifier

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Figure 7.14 (b) shows another way that the non-inverting op-amp is sometimes drawn. Always take note that the input voltage is still connected to non-inverting input.

Figure 7.14 (b): Non-inverting amplifier (different circuit layout)

Input wave

Output wave – amplified and in phase with the input wave

Figure 7.14 (c): Waveforms for the non-inverting op-amp

Example Calculate the gain of the amplifier shown in figures 7.14 (a) & (b) if the following information is given to you. Rf = 50 kΩ Rin = 10 kΩ Rf Vout Av = ___ = 1 + ___ Vin Rin 50 × 10_3 = 1 + _____ 10 × 103 =1+5 =6

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Breadboard construction of a non-inverting op-amp

Figure 7.15: Non-inverting op-amp


Channel 2 = 2 V/Div

Time/Div = 0,1 ms 1 Frequency = 1 = T 0,1 × 10-3 × no Div 1 = = 1 000 Hz or 1 kHz 0,1 × 10-3 × 10 Rin = 10 kΩ Rf = 10 kΩ

Channel 1 Input signal = 1 V × 1 Div = 1 V peak


Channel 2 Output signal = 2 V × 2 Div = 4 V peak Output is in phase with the input

Figure 7.16 (b): Output signal

Combination of channel 1 & 2 Vout Gain = ____ Vin

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Figure 7.16 (c): Input and output signal combined

= 4V 1V = 4

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Op-amp used as a summing amplifier The summing amplifier is an inverting amplifier with more than one input added to the inverting input. The summing amplifier is used to add a number of analogue signals (e.g. inputs from different musical instruments) or different voltages together. This circuit is commonly used as an audio mixer. Application of summing amplifier Just as with comparators the summing amplifier also has some meaningful applications. If the input resistances of a summing amplifier are connected to potentiometers the individual input signals can be mixed together by varying amounts. For example, measuring temperature, you could add a negative offset voltage to make the display read “0” at the freezing point or produce an audio mixer for adding or mixing together individual waveforms (sounds) from different source channels (vocals, instruments, etc) before sending them combined to an audio amplifier. Another useful application of a Summing Amplifier is as a weighted sum digital-to-analogue converter. This application will produce an output which is the weighted sum of the digital inputs. Input V1

Input V2

Input V3

Output wave = the algebraic sum of the input waves but 180° out of phase with the input waves

Figure 7.17 (a): Summing amplifier

Figure 7.17 (b): Waveforms for the summing op-amp

The formula to calculate the output voltage can be written as follows.

If Rf and the input resistors have the same value, they will cancel each other out and the output voltage can then be calculated as follows: Vout = – [V1 + V2 +V3]

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Electrical Technology A combination of different voltages that have opposite polarities can be applied to the inputs of a summing op amp. The following table shows the output of a summer with different input conditions. Input voltages V1

V2

V3

Algebraic sum of output voltages

+1

+1

+1

-3

+1

-1

-1

+1

+2

-1

-1

0

-3

-1

+3

+1

+1

+2

-1

-2

Example With reference to the summing amplifier in figure 7.17, calculate the output voltage if the following information is given to you. R2 = 15 kΩ R3 = 20 kΩ R1 = 10 kΩ V1 = 2 V V2 = –10 V Rf = 47 kΩ V3 = 5 V

Breadboard construction of a summing op-amp

Figure 7.18: Summing op-amp

Waveform representation on an oscilloscope and measurements Oscilloscope settings: Channel 1 = 1 V/Div Time/Div

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Channel 2 = 1 V/Div

= 0,1 ms

1 Frequency = 1 = T 0,1 × 10-3 × no Div 1 = = 1 000 Hz or 1 kHz 0,1 × 10-3 × 10 NB: R1 = R2 = R3 = Rf = 10 kΩ The gain for this circuit = 1

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Channel 1 Input voltage Vin = 1 V × 1 Div = 1 V peak NB!!! V1 = V2 = 1 Vpeak

Figure 7.19 (a): Input voltage

Channel 2 Output voltage Vout = 1 V × 1 Div = 1 Vpeak

Figure 7.19 (b): Output voltage

Output showing V1 only in relation to the input. We also see the output is 180° out of phase with input. Vout = 1 V × 1 Div = 1 Vpeak

Figure 7.19 (c): Output showing V1 only in relation to the input

Output voltage showing V1 + V2 in relation to the input i.e. Vout = 1 V × 2 Div = 2 Vpeak

Figure 7.19 (d): Output voltage showing V1 + V2 in relation to the input

Output voltage showing V1 + V2 +V3 in relation to the input i.e. Vout = 1 V × 3 Div = 3 Vpeak

Figure 7.19 (e): Output voltage showing V1 + V2 +V3 in relation to the input

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Electrical Technology Op-amp used as an integrator With reference to figure 7.20 (a), it can be seen that the circuit resembles that of an inverting op-amp amplifier except that the feedback element is a capacitor and not a resistor. The use of a capacitor as a feedback element impacts the circuit in a special way. As its name implies, the op-amp integrator is an operational amplifier circuit that performs the mathematical operation of integration, that is finding the area under a specific curve between two points. As mentioned earlier integrators performs the mathematical operation of integration with respect to time; that is, its output voltage is proportional to the input voltage over time. Let us briefly consider the basic operation of the integrator. When the input voltage is applied to the inverting input, the capacitor initially has very little resistance and acts like a short circuit. As the feedback capacitor begins to charge, its resistance also starts to increase, increasing the output voltage until the capacitor is fully charged. Now the capacitor acts as an open circuit that blocks DC. The rate at which the output voltage increases is determined by the value of the resistor and the capacitor, the ‘RC time constant’. By changing this RC time constant value, either by changing the value of the capacitor (C), or the resistor (R), the time in which it takes the output voltage to reach saturation can also be changed. By applying a square wave input to the integrator, it could be changed to a sawtooth waveform at the output. Application of the integrator One application for this device would be to keep a “running total” of radiation exposure, or dosage, if the input voltage was a proportional signal supplied by an electronic radiation detector. Nuclear radiation can be just as damaging at low intensities for long periods of time as it is at high intensities for short periods of time. An integrator circuit would take both the intensity (input voltage magnitude) and time into account, generating an output voltage representing total radiation dosage. Another application would be to integrate a signal representing water flow, producing a signal representing total quantity of water that has passed by the flowmeter. This application of an integrator is sometimes called a totalizer in the industrial instrumentation trade. Differentiator is essentially low pass filters where the reactance of the capacitor is high at high frequencies therefore only allowing low frequencies through. It can also be used as a ramp generator used to deflect an electron in a CRT to draw a horizontal line across face of the CRT.

Figure 7.20 (a): Op-amp integrator circuit

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Figure 7.20 (b): Waveforms for the op-amp as integrator

Bread board construction of an integrator

Figure 7.21: Integrator

Waveform representation on an oscilloscope and measurements Oscilloscope settings: Channel 1 = 0,5 V/Div

Channel 2 = 5 V/Div

Time/Div = 50 µs

1 Frequency = 1 = T 50 × 10-6 × no Div 1 = = 4 000 Hz or 4 kHz 50 × 10-6 × 5

C1 = 1 nF and Rin = 10 kΩ

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Channel 1 Input signal = 0,5 V × ±3 Div (depending on scope calibration) = 1,5 V peak


Channel 2 Output signal Vout = 5 V × ±3 Div (depending on scope calibration) = 15 V peak


The output showing channels 1 and 2 combined

Figure 7.22 (c): Combination of input and output

Op-amp used as differentiator Looking at figure 7.23 (a), we can once again see the circuit resembles the inverting op-amp circuit, with the input resistor being replaced with a capacitor. This circuit performs the mathematical function known as differentiation, i.e. it will produce an output voltage that is directly proportional to the rate-of-change of the input voltage with respect to time. The input capacitor blocks DC and only allows AC through. The reactance of the capacitor will vary from low for high frequencies to high for low frequencies. With the high input frequencies, the gain of the op-amp will also be high, resulting in a higher output voltage. The reverse will happen with lower frequencies, i.e. the gain will be less therefore a lower output voltage. By applying a sawtooth to the input of the differentiator, a square wave output will result.

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Application of the Differentiator As said previously, the differentiator op amp is one type of op amp where the magnitude of the output is determined by the rate at which the voltage is applied to its input changes. The faster the input voltage changes, the greater the output One such rate-of-change signal application might be for monitoring (or controlling) the rate of temperature change in a furnace, where too high or too low of a temperature rise rate could be detrimental. The DC voltage produced by the differentiator circuit could be used to drive a comparator, which would signal an alarm or activate a control if the rate of change exceeded a pre-set level. Differentiator is essentially high pass filters where the reactance of the capacitor is high at low frequencies therefore only allowing high frequencies through. Differentiators can be used as square wave generators when a rectangular wave is fed to the input. Differentiators are an important part of electronic analogue computers and analogue Proportional-intergral-derivative controller (PID Controllers). A PID controller calculates an ‘error’value as the difference between a measured process variable and a desired set point eg automatic ship steering and head positioning of a disk drive. PID controllers are also used in industry to regulate temperature, flow rate, pressure and many other variables for which a measurements

Figure 7.23 (a): Op-amp differentiator circuit

Figure 7.23 (b): Waveforms for the op-amp as a differentiator

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Electrical Technology Breadboard construction of a differentiator R1

Figure 7.24: The differentiator

Waveform representation on an oscilloscope and measurements Oscilloscope settings: Channel 1 = 0,5 V/Div

Channel 2 = 5 V/Div

Time/Div = 0,1 ms 1 Frequency = 1 = T 0,1 × 10-3 × no Div 1 = = 200 Hz 0,5 × 10-3 × 5

C = 100 nF & Rf = 10 kΩ

Channel 1 Input signal = 0,5 V × 3 Div = 1,5 V peak


Channel 2 Output signal Vout = 5 V × 3 Div = 15 V peak


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The output showing channels 1 and 2 combined

Figure 7.25 (c): Input and output signal combined

Op-amps used as multivibrators Bi-stable multivibrator The bi-stable multivibrator circuit is a timing circuit with two stable output states. The output will either be +Vcc or –Vcc, depending on the condition of the input trigger pulse. If the input trigger pulse is a positive pulse, the output will go to –Vcc. It will remain in this state until a negative trigger pulse is received. When this happens, the output will now change to +Vcc. The feedback resistors R1 & R2 are responsible for keeping the output in the one stable state (+Vcc or –Vcc) until they are triggered by another input pulse. Application of the Bi-stable multivibrator A typical application of a bi-stable multivibrator is as a contact bounce eliminator. Once the set input is pressed, no other interferences will have an impact on the output, because the output will be latched. The bi-stable multivibrator can also be used a relaxation oscillator with a continuous square wave output.

Figure 7.26 (a): Op-amp bi-stable multivibrator circuit

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Trigger pulse

Vout

Figure 7.26 (b): Waveforms for the op-amp as bi-stable multivibrator

Breadboard construction of a bi-stable multivibrator

Figure 7.27: Bi-stable multivibrator

Waveform representation on an oscilloscope and measurements Oscilloscope settings: Channel 1 = 5 V/Div Time/Div = 0,2 ms

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C = 10 nF & Rf = 10 k ohm

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Channel 1 Output with negative trigger pulse Output voltage = 5 V × 3 Div = +15 V

Figure 7.28 (a): Output switches to +15 V

Channel 1 Output with positive trigger pulse Output voltage = 5 V × 3 Div = –15 V

Figure 7.28 (b): Output switches to –15 V

Mono-stable multivibrator The mono-stable multivibrator is a timing circuit that changes state once triggered but returns to its original state after a certain time delay (t1). It got its name from the fact that only one of its output states is stable. It is also known as a ‘one-shot’ multivibrator. A negative trigger pulse at the input forces the output of the op-amp to a ‘high’. This charges up C2 , which keeps the non-inverting input of the op amp temporarily higher than the inverting input, maintaining the output high for a certain period of time (t1). Eventually C2 discharges to ground and the op-amp output swings back to a ‘low’. The duration of the pulse is defined by R2 and C2. Applications of the mono stable multivibrator The pulse detector is classified as a monostable multivibrator because it has only one stable state. By stable, I mean a state of output where the device is able to latch or hold to forever, without external prodding. Another application for a retriggerable one-shot is that of a single mechanical contact debouncer. That is the output will remain high despite “bouncing” of the input signal from a mechanical switch.

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Figure 7.29 (a): Op-amp mono-stable multivibrator circuit

Trigger pulse

Vout

Figure 7.29 (b): Waveforms for the op-amp as mono-stable multivibrator

Breadboard connection of mono-stable multivibrator

Figure 7.30: Mono-stable multivibrator

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Amplifiers Waveform representation on an oscilloscope and measurements Oscilloscope settings Channel 1 = 5 V/Div Channel 2 = 2 V/Div R3 = R2 = 1 kΩ & C1 = C2 = 0,1 µF Supply

Supply Output (duration of output will depend on R2 & C2

Output (with no trigger pulse)

Figure 7.31(a): Only the supply voltage, with no trigger pulse

Figure 7.31(b): Supply voltage and output signal after the trigger pulse

Astable multivibrator An astable multivibrator, or free-running multivibrator, is a timing circuit that generates square waves without any external trigger. It has no stable state but has only two quasi stable (half stable) states between which it keeps on oscillating of its own accord without any external trigger. Depending upon initial conditions, the op amp’s output will switch to either the positive or negative rail voltage (+Vccor –Vcc). When this happens, the capacitor will either charge or discharge through the resistor Rf , its voltage slowly rising or falling. As soon as the voltage at the op amp’s inverting terminal reaches that of the non-inverting terminal (the op amp’s output voltage divided by R1 and R2), the output will drive to the opposing rail and this process will be repeated with the capacitor discharging if it had previously charged and vice versa. Thus, the astable multivibrator creates a square wave with no inputs. Applications of the astable multivibrator Astable multivibrators are used in amateur radio equipment to receive and transmit radio signals. One radio application could be a code practice audio-frequency oscillator while it can also be used for a simple electronic keyer for keying a transmitter. Astable multivibrators are also used in morse code generators, timers, and systems that require a square wave, including television broadcasts and analog circuits. Regenerative switching circuits such as Astable Multivibrators are also used as a type of relaxation oscillator as they produce a constant square wave output waveform as well as their simplicity, reliability and ease of construction. Another common application is the simple LED flashing lights of toys or even used as dummy alarm.

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Figure 7.32 (a): Op-amp astable multivibrator circuit

Waveform at the inverting input

Waveform at the noninverting input

Waveform at the output

Figure 7.32 (b): Waveforms for the op-amp as an astable multivibrator

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Breadboard connections of an astable multivibrator

Figure 7.33: An astable multivibrator


Channel 2 = 5 V/Div

Time/Div = 1 ms

1 Frequency = 1 = T 0,1 × 10-3 × no Div 1 = = 2 000 Hz or 2kHz 0,1 × 10-3 × 5

C1 = 100 nF, R1 = 10 kΩ, R2 = 22 kΩ & Rf = 22 kΩ Channel 1 Waveform at the inverting input Vin = 5 V × 2 Div = 10 V

Figure 7.34 (a): Waveform at the inverting input

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Channel 2 Waveform at the output Vout = 5 V × 3 Div = 15 V

Figure 7.34 (b): Waveform at the output

Combination of channels 1 & 2

Figure 7.34 (c): Combination of channels 1 & 2

The Schmitt trigger

Did you know? The Schmitt trigger was first invented by Otto H. Schmitt in 1934. The name he gave it was ‘thermionic trigger’.

The Schmitt trigger is a type of comparator with two different trigger voltage levels. Whenever the input voltage goes over the upper trigger level, the output of the comparator is HIGH (positive supply voltage, if it is a non-inverting Schmitt trigger) or LOW (negative supply voltage, if it is an inverting Schmitt trigger). The output will remain in this state, as long as the input voltage is above the second trigger level, the lower trigger level. When the input voltage goes below this level, the output of the Schmitt trigger will switch. The Schmitt trigger can be seen as a device that will change a slow varying input signal into a square wave output signal. When the non-inverting input (+) is higher than the inverting input (–), the comparator output switches to the POSITIVE voltage supply but when the noninverting input (+) is lower than the inverting input (–), the output switches to the NEGATIVE voltage supply. Characteristics of a Schmitt trigger • Voltage sensitive switch • Have two fixed trigger values (voltages) • Supplies a digital output • Output frequency is the same as input frequency • Used as a wave shaper. Applications of the Schmitt trigger Schmitt triggers are typically used in the close loop configurations to implement function generators.

208

A Schmitt trigger is a bi-stable multivibrator and it can be used to implement another type of multivibrator, the relaxation oscillator with a continuous square wave output. The Schmitt trigger changes any input wave into a square wave output. It can therefore be used to ‘clean’ noisy signals. The Schmitt trigger is often used when interfacing an analogue signal to digital circuitry. The Schmitt trigger can be used to measure the frequency of an AC signal by first converting it to square waves, before being counted by digital counter.

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Figure 7.35 (a): Circuit for an inverting Schmitt trigger

Figure 7.35 (b): Output waveforms for a Schmitt trigger

Breadboard connections of a Schmitt trigger

Figure 7.36: Schmitt trigger

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Electrical Technology Waveform representation on an oscilloscope and measurements Oscilloscope settings: Channel 1 = 0,5 V/Div (DC coupled) Channel 2 = 5 V/Div (DC coupled) Time/Div = 20 µs R1 = 10 kΩ

Rf = 100 kΩ

Upper trigger level = UTL Lower trigger level = LTL Output signal

Vin = 0,5 V × 1Div = 0,5 V (sine wave) Vout = +5 V × 3Div = +15 V The input signal has not yet reached the UTL, therefore the output will either be +15 V or –15 V Input signal

Figure 7.37 (a): Input and output signals Input (sine wave)

Vin = 0,5 V × 2,5Div = 1,25 V (sine wave) Vout = +5 V × 2,5Div = 12,5 V (square wave) When Vin > UTL then Vout = –15 V and when Vin < LTL then Vout = +15 V Output (square wave)

Figure 7.37 (b): Square wave output and sine wave input combined

Because of component tolerances and the calibration of the scope the output is not always a pure square wave. Input (triangular wave)

Even with a triangular input wave the output is still a square wave Output (square wave)

Figure 7.37 (c): Square wave output with a triangular input

210

Because of component tolerances and the calibration of the scope the output is not always a pure square wave.

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Operational amplifier oscillators Positive feedback Earlier in this chapter, when we dealt with amplifiers we discussed negative feedback as a means of stabilising the output of the amplifier. When dealing with oscillators, we make use of positive feedback however. Positive feedback is where a portion of the output wave is fed back to the input, where the feedback signal is in phase with the input signal. The fed-back signal is added to the input signal. The result of this is a more unstable amplified signal that will lead to oscillations. A system with positive feedback consists of three elements (see Figure 7.38): an amplifier with gain Av , a feedback network β and a summing circuit acting as an adder (the circle in the figure). Fig. 7.38 represents the block diagram showing how positive feedback is accomplished.

Figure 7.38: Positive feedback

Hartley/Colpitts oscillators Oscillators can be divided mainly into two types, the one using a combination of inductors and capacitors in its feedback circuit and the other one using a combination of resistors and capacitors in its feedback circuit. The Hartley and Colpitts oscillators are oscillators which make use of LC networks in their feedback circuits, forming resonating tank circuits. The Hartley and Colpitts oscillators are commonly used where frequencies of 100 kHz and higher are required. The Hartley oscillator will have two inductors and one capacitor in its feedback tank circuit, whereas the Colpitts oscillator has two capacitors and one coil in its feedback tank circuit. Figure 7.39 is the circuit for a Hartley oscillator. In this circuit, the op-amp is used as an inverting operational amplifier, where the gain of the circuit is determined by Rf and R1. The resonating tank formed by L1, L2 and C1 is responsible for the positive feedback as well as the 1800 phase shift of the feedback signal to ensure the feedback and output signals are in phase with each other. The frequency at which this circuit resonates is referred to as the resonating frequency and can be calculated by: r =

1 where Lt = L1 + L2 2 π√LC

Application of the Hartley oscillator Hartley oscillator can be used over a wide range of frequencies, making it a very popular oscillator. The Hartley oscillator is widely used as local oscillators in radio receivers. Hartley oscillator is not suitable for low frequency work because at low frequency, the value of inductance required becomes large. Hartley oscillator can also be suitably used for generating RF signals.

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Figure 7.39: The Hartley oscillator

Example Calculate the resonating frequency of a Hartley oscillator consisting of two coils of 45 mH each and a capacitor of 0,47 µF.

r =

1 2 π√LC

1 2 π√ (45 × 10-3 )×(0,47 × 10-6 ) = 1 094,37 Hz = 1,094 kHz

=

Breadboard connections of the hartley oscillator

Figure 7.40: The Hartley oscillator

Waveform representation on an oscilloscope and measurements Oscilloscope settings: Channel 1 = 5 V/Div Time/Div = 1 ms

1 Frequency = 1 = T 1 × 10-3 × no Div 1 = = 200 Hz 1 × 10-3 × 5

212

R1 = 10 kΩ

Rf = 10 kΩ

C1= 0,47 µF

L1 = L2 = 45 mH

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Because of the sensitive nature of the oscillator, change R1 to a 10 kΩ pot when you build the circuit to enable you to adjust R1 to a value to start the oscillations. Channel 1 Output waveform Vout = 5 V × 2 Div = 10 Vpeak

Figure 7.41: Oscillator output

RC phase shift oscillator As mentioned earlier, the RC phase shift oscillator makes use of resistors and capacitors in its feedback circuit to create an oscillator. This circuit is mainly used for generating frequencies in the lower range, i.e. frequencies less than 10 kHz. The circuit also makes use of an inverting op-amp with three RC networks responsible for creating positive feedback as well as the 180° phase shift to make input and feedback in phase. Each of the RC networks generates a 60° phase shift, so at point 1 there will be a 60° phase shift, at point 2 a 120° phase shift and at point three there will be a 180° phase shift. The phase relationship between point 3 and Vout will be the same. The phase shift is only done to one particular frequency and that is the frequency of oscillations which can be calculated by: =

1 2 π√6RC

Application of the RC Phase shift oscillator RC Oscillators are stable and provide a well-shaped sine wave output with the frequency being proportional to 1/RC and therefore, a wider frequency range is possible when using a variable capacitor. However, RC Oscillators are restricted to frequency applications because of their bandwidth limitations to produce the desired phase shift at high frequencies. Phase shift oscillators are often used at audio frequency as audio oscillators. Other applications of this type of oscillator are: electronic organs and other musical instruments, voice synthesis, GPS units and other equipment that emits a beep.

Figure 7.42: The RC phase shift oscillator

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Electrical Technology Example Calculate the oscillating frequency for an RC oscillator making use of three RC networks. We assume all resistor values are the same and all capacitor values are the same. The value of the resistor is 10 kΩ and that of the capacitor 250 pF.



=

1 2 π√6RC

1 2 π√ 6 (10 × 103)×(250 × 10-12) = 41,09 Hz

=

Breadboard connections of RC phase shift oscillator

Figure 7.43: RC phase shift oscillator


Channel 2 = 5 V/Div

Time/Div = 20 µs

1 Frequency = 1 = T 20 × 10-6 × no Div 1 = 20 × 10-6 × 6

= 8 333 Hz or 8,33 kHz

C1 = C2 = C3 = 0,01 µF, Rf = 50 kΩ pot & R1 = R2 = R3 = 1,2 KΩ Channel 2 Output waveform Vout = 5 V × 2,5 Div = 12,5 V

Figure 7.44 (a): Channel 2

214

Amplifiers Channel 2

7

Channel 1

Channel 1 & Channel 2 Output at point 1 (Channel 2) compared with final output as seen on figure 7.42 (60° phase shift between output and point 1)

Figure 7.44 (b): 60° phase shift Channel 1

Channel 2


Figure 7.44 (c): 120° phase shift Channel 2

Channel 1


Figure 7.44 (d): 180° phase shift

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Activity 1 1. 2. 3. 4. 5.

Write down at least FIVE characteristics of an ideal op-amp. What is the working principle of the differential amplifier? Why are op-amps so popular in the electronics field? The ideal op-amp has an infinite bandwidth. What does this mean? Is the op-amp a voltage or current amplification device? Motivate your answer. 6. Draw a neat labelled symbol for an op-amp. 7. With reference to the following diagrams, what will the output of the following op-amp circuits be with the following input conditions?

a)

b)

8. Briefly explain the term ‘negative feedback’ and why is it employed in op-amp circuits. 9. What is the main difference between a non- inverting op amp and an inverting op-amp? 10. With the aid of a simple sketch, explain the concept negative feedback. 11. With reference to the figure below, what will the output waveforms be if this input is applied to the input of the following op-amps? Draw both the input and output waveforms in your book. a) Non-inverting op-amp b) Integrator

12. Calculate the gain of a non-inverting op-amp if Rin = 10 kΩ and Rf = 100 kΩ. 13. Make a net labelled circuit diagram of a summing amplifier, also calculate the output voltage with the following input voltages. V1 = 3 V peak, V2= 1 V peak and V3= 5 V peak. Remember that all resistors have exactly the same value (10 kΩ).

216

Amplifiers

7

14. What type of output will a differentiator have if the input is a rectangular wave? 15. Identify the circuit below, and also calculate the output voltage with the following data given Rf = 100 kΩ¸Rin = 10 kΩ and Vin = –10 V.

16. What is the difference between an astable and a bi-stable multivibrator? 17. Make a neat labelled drawing of a bi-stable multivibrator. 18. Briefly explain the operation of the Schmitt trigger. 19. What will the output waveform of a Schmitt trigger look like if it receives an input signal as indicated below? Draw both waveforms to show their relationship to one another.

Upper trigger level Input signal Lower trigger level

20. Briefly explain what positive feedback is and in what kind of circuits is it used extensively. 21. What is the difference between a Hartley oscillator and a RC phase shift oscillator? 22. Calculate the resonant frequency of a Hartley oscillator with a LC network consisting of two coils of 45 mH each and a capacitor of 0,47 µF. 23. With reference to the non-inverting op-amp, what must the feedback resistor be for the amplifier to have a gain of 10 and 100 respectively? 24. Describe the TWO requirements for oscillation to occur in all oscillators. 25. Operational amplifiers are commonly used in complex circuits (between stages) to link the stages. State, with a reason, the application (function) of the operational amplifier when utilised between stages. 26. Comparators are user to compare two signals. State, with a reason, what the output will be if the non – inverting input is grounded and a signal is applied to the inverting input. Describe one practical application of the comparator.

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Electrical Technology 27. Give two applications of Integrators. 28. Negative feedback decreases the gain of an amplifier circuit. Is negative feedback in or out of phase with the input of the circuit? Motivate your answer. 29. With reference to the diagram below, answer the following questions: • Identify the circuit below. • If Rf is increased, how will this influence the gain of the circuit? • Explain the function of Rin. Rf

Iin

Vi

218

Rin

If

_

+

Vout

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Practical Activity 1 Inverting comparator Form of activity: Investigation Material and equipment: • Bread board • 0.5 mm solid connecting wire • 2 × 10kΩ resistors • 1 × 741 Op amp IC • Multimeter • Variable dual power supply (+-15V ) • Variable DC supply • Fix 5V supply • Dual trace oscilloscope Pin connections of the 741 op amp IC

NC

Input Offset

+Vcc Supply

Inverting Input

Output

Non-inverting Input

Input Offset

-Vcc Supply

Instructions: • Connect the circuit of the inverting comparator as indicated in the diagram below on the bread board. Vin

Rin

_ Vout & Channel 1 of oscilloscope

Vreference

• Connect the oscilloscope to the output of the op amp and ground. • Adjust Vreference to 5V DC

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7

Electrical Technology • Adjust the variable DC supply (Vin) to the voltages as indicated in the table below and observe the output on the oscilloscope. Record the output voltages in the table below. Vin

Vref

4V

5V

4.5V

5V

5V

5V

5.5V

5V

6V

5V

Vout

• From the information in the table above, briefly explain the operation of this circuit. • Will the shape of the input voltage have any influence on the output?

Practical Activity 2 Inverting Op-amp Form of activity: Investigation Material and equipment: • Bread board • 0.5 mm solid connecting wire • 2 × 10kΩ resistors • 1 × 741 Op amp IC • Multimeter • Variable dual power supply (+-15V ) • Variable DC supply • Dual trace oscilloscope • Function generator Pin connections of the 741 op amp IC

Input Offset Inverting Input Non-inverting Input -Vcc Supply

220

NC +Vcc Supply Output Input Offset

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Instructions: • Connect the circuit of the inverting op amp as indicated in the diagram below on the bread board. Rf

Rin Vin (from function generator) Oscilloscope channel 1

• • • • • • • • •

– Vout Oscilloscope channel 2

+

Connect channel 1& 2 of the oscilloscope as indicated in the circuit above. Set the function generator to give a sine wave output. Adjust the function generator to 1000Hz (1kHz) at a voltage of 1V peak and connect as indicated on the circuit above Switch on the power to the circuit and observe the input and output waveforms as displayed on the oscilloscope. Make neat labelled drawings of the waveforms in your book. Calculate the amplitude and frequency of the output waveform. Now replace Rf with a 100kΩ resistor and observe the output waveform. By increasing Rf to 100kΩ, how did this influence the circuit? What is the phase relationship between the input and output waveforms in this activity?

Practical Activity 3 Non-Inverting Op-amp Form of activity: Investigation Material and equipment: • Bread board • 0.5 mm solid connecting wire • 2 × 10kΩ resistors • 1 × 741 Op amp IC • Multimeter • Variable dual power supply (+-15V ) • Variable DC supply • Dual trace oscilloscope • Function generator Pin connections of the 741 op amp IC Input Offset Inverting Input Non-inverting Input -Vcc Supply


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Electrical Technology Instructions: • Connect the circuit of the non-inverting op amp as indicated in the diagram below on the bread board. Rf

Rin –

Vin from function generator & scope channel 1

• • • • • • • • • •

Vout Scope channel 2

Connect channel1 & 2 of the oscilloscope as indicated in the circuit above. Set the function generator to give a sine wave output. Adjust the function generator to 1000Hz (1kHz) at a voltage of 1V peak and connect as indicated on the circuit above Switch on the power to the circuit and observe the input and output waveforms as displayed on the oscilloscope. Make neat labelled drawings of the waveforms in your book. Calculate the amplitude and frequency of the output waveform. Now replace Rf with a 100kΩ resistor and observe the output waveform. By increasing Rf to 100kΩ, how did this influence the circuit? What is the phase relationship between the input and output waveforms in this activity? From your observations of the input and output waveforms, calculate the gain of the amplifier.

Practical Activity 4 Summing amplifier Form of activity: Investigation Material and equipment: • Bread board • 0.5 mm solid connecting wire • 4 × 10kΩ resistors • 1 × 741 Op amp IC • Multimeter • Variable dual power supply (+-15V ) • Variable DC supply • Dual trace oscilloscope • Function generator

222

+

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Pin connections of the 741 op amp IC

NC

Input Offset

+Vcc Supply

Inverting Input

Output

Non-inverting Input

Input Offset

-Vcc Supply

Instructions: • Connect the circuit of the summing op amp as indicated in the diagram below on the bread board. V1

Vin from signal generator & scope channel 1

V2 V3

R1

Rf

R2 R3

–

+

• • • • • • • • • • •


Connect channel1 & 2 of the oscilloscope as indicated in the circuit above. Set the function generator to give a sine wave output. Adjust the function generator to 1000Hz (1kHz) at a voltage of 1V peak and connect to input V1 as indicated on the circuit above Switch on the power to the circuit and observe the input and output waveforms as displayed on the oscilloscope. Make neat labelled drawings of the waveforms in your book. What can you conclude from the above waveforms? Switch the circuit off and now connect the signal generator to all three inputs simultaneously (all together). Observe the input and output waveforms as displayed on the oscilloscope. Make neat labelled drawings of the waveforms in your book. What can you conclude from the second set of waveforms? What is the phase relationship between the input and output waveforms in this activity? From your observations, what is the main function of this operational amplifier circuit?

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Practical Activity 5 Op – amp Integrator Form of activity: Investigation Material and equipment: • Bread board • 0.5 mm solid connecting wire • 100nF, 10nF Capacitor • 10kΩ, 100kΩ Resistor • 1 × 741 Op amp IC • Multimeter • Variable dual power supply (+-15V ) • Variable DC supply • Dual trace oscilloscope • Function generator Pin connections of the 741 op amp IC

NC

Input Offset

+Vcc Supply

Inverting Input

Output

Non-inverting Input

Input Offset

-Vcc Supply

Instructions: • Connect the circuit of the op amp Integrator as indicated in the diagram below on the bread board. C

Rin Vin from function generator & scope channel 1

• • • •

224

–

+


Connect channel1 & 2 of the oscilloscope as indicated in the circuit above. Set the function generator to give a square wave output. Adjust the function generator to 2000Hz (2kHz) at a voltage of 1.5V peak and connect to input V1 as indicated on the circuit above Switch on the power to the circuit and observe the input and output waveforms as displayed on the oscilloscope.

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• Make neat labelled drawings of the waveforms in your book. • Calculate the amplitude and frequency of this waveform. • Switch the circuit off and now replace the 1nF capacitor with a 10nF capacitor. • Switch the power back on and observe how this influenced the output of the circuit? • Switch the circuit off and now replace the 10kΩ resistor with a100kΩ resistor also put back the 1nF capacitor. • Switch the power back on and observe how this influenced the output of the circuit? • Calculate the time constants for each of the three combinations of resistor and capacitor values.

Practical Activity 6 Op –amp Differentiator Form of activity: Investigation Material and equipment: • Bread board • 0.5 mm solid connecting wire • 1nF Capacitor • 10kΩ Resistor • 1 × 741 Op amp IC • Multimeter • Variable dual power supply (+-15V ) • Variable DC supply • Dual trace oscilloscope • Function generator Pin connections of the 741 op amp IC



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Electrical Technology Instructions: • Connect the circuit of the op amp differentiator as indicated in the diagram below on the bread board. Rf

C Vin from function generator & scope channel 1

–

+


• Connect channel 1 & 2 of the oscilloscope as indicated in the circuit above. • Set the function generator to give a saw tooth wave output. • Adjust the function generator to 1000Hz (1kHz) at a voltage of 1V peak and connect to input Vin as indicated on the circuit above • Switch on the power to the circuit and observe the input and output waveforms as displayed on the oscilloscope. • Make neat labelled drawings of the waveforms in your book. • Calculate the amplitude and frequency of this waveform. • Switch the circuit off and increased the input frequency to 4000Hz (4kHz) • Switch the power back on and observe how this influenced the output of the circuit. • Calculate the time constants for this circuit.

Practical Activity 7 Op –amp Bi-stable multivibrator Form of activity: Investigation Material and equipment: • Bread board • 0.5 mm solid connecting wire • 0.1µF Capacitor • 2 × 10kΩ .(R1 &R2) 1kΩ (R3) Resistors • 1 × 741 Op amp IC • Multimeter • Variable dual power supply (+-15V ) • Variable DC supply • Dual trace oscilloscope

226

Amplifiers

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Pin connections of the 741 op amp IC

NC

Input Offset

+Vcc Supply

Inverting Input

Output

Non-inverting Input

Input Offset

-Vcc Supply

Instructions: • Connect the circuit of the bi-stable multi-vibrator as indicated in the diagram below on the bread board. +Vcc

C –

Trigger

+ R3

–Vcc

R1 Vout Channel 2 of Oscilloscope R2

• • • • •

Switch on the power to the circuit and observe the output waveform as displayed on the oscilloscope. Apply a positive trigger pulse to the trigger output and observe the output. (Ensure the positive trigger pulse is large enough to trigger the circuit) Make neat labelled drawings of the waveforms in your book. Now apply a large enough negative trigger pulse to the trigger input and observe the output waveform displayed on the oscilloscope. Make neat labelled drawings of the waveforms in your book.

Practical Activity 8 Op –amp Mono-stable multivibrator Form of activity: Investigation Material and equipment: • Bread board • 0.5 mm solid connecting wire • 0.1µF (C1), 0.01µF (C2) Capacitor • 22kΩ (R1), 10kΩ & 100kΩ (R2), Resistors • 1 × 741 Op amp IC • Multimeter

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Electrical Technology • Variable dual power supply (+-15V ) • Variable DC supply • Dual trace oscilloscope Pin connections of the 741 op amp IC

NC

Input Offset

+Vcc Supply

Inverting Input

Output

Non-inverting Input

Input Offset

-Vcc Supply

NB: The tolerances of components and calibration of the instruments is crucial to the correct operation of this experiment. Instructions: • Connect the circuit of the mono stable multi-vibrator as indicated in the diagram below on the bread board. +Vcc

C1 Trigger

– +

C2

–Vcc

R3

R2

Vout Channel 1 of Oscilloscope

+Vref

• • • • • •

228

Switch on the power to the circuit and observe the output waveform as displayed on the oscilloscope. Apply a large enough trigger voltage to the trigger input and observe what happens at the output. Make neat labelled drawings of the waveforms in your book. Switch the power to the circuit off and change the value of R2 to100kΩ. Switch the power back on and observe the output waveform as displayed on the oscilloscope. What influence did the increase in resistor value had on the working of the circuit? Motivate your answer.

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Practical Activity 9 Op –amp Astable multivibrator Form of activity: Investigation Material and equipment: • Bread board • 0.5mm solid connecting wire • 100nF Capacitor • 10kΩ (R1), 22kΩ(R2) & 22kΩ, 47kΩ(Rf), Resistors • 1 × 741 Op amp IC • Multimeter • Variable dual power supply (+-15V ) • Variable DC supply • Dual trace oscilloscope Pin connections of the 741 op amp IC

NC

Input Offset

+Vcc Supply

Inverting Input

Output

Non-inverting Input

Input Offset

-Vcc Supply

NB: The tolerances of components and calibration of the instruments is crucial for the correct operation of this experiment. Instructions: • Connect the circuit of the astable multi-vibrator as indicated in the diagram below on the bread board. Rf

+Vcc

C – +

–Vcc

R1 Vout & Channel 1 of Oscilloscope R2

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7

Electrical Technology • Switch on the power to the circuit and observe the output waveform as displayed on the oscilloscope. • Make neat labelled drawings of the waveforms in your book. • Calculate the amplitude and frequency of this wave. • Switch the power off and replace Rf with a 47kΩ resistor. • Switch the power back on and observe the output waveform displayed on the oscilloscope. • Make neat labelled drawings of the waveforms in your book. • Now calculate the amplitude and frequency of this wave. • What type of waveform do you observe on the output? • How does the astable multivibrator differ from the bi-stable multivibrator?

Practical Activity 10 Op –amp Schmitt trigger Form of activity: Investigation Material and equipment: • Bread board • 0.5 mm solid connecting wire • 100nF Capacitor • 10kΩ (R1), 100kΩ(Rf), Resistors • 1 × 741 Op amp IC • Multimeter • Variable dual power supply (+-15V ) • Variable DC supply • Dual trace oscilloscope • Function generator Pin connections of the 741 op amp IC



NB: The tolerances of components and calibration of the instruments is crucial for the correct operation of this experiment.

230

Amplifiers

7

Instructions: • Connect the circuit of the Schmitt trigger as indicated in the diagram below on the bread board. +Vcc

Vin from the signal generator Channel 1 of the oscilloscope

–

Vout & Channel 2 of Oscilloscope

+

Rf –Vcc

R1

• Set the function generator to give a sine wave output. • Adjust the function generator to 500Hz and amplitude at zero and connect to input Vin as indicated on the circuit above • Switch on the power to the circuit and observe the output waveform as displayed on the oscilloscope. • Slowly increase the amplitude of the signal generator to its maximum • Observe the output waveform as displayed on the oscilloscope. • Make a neat drawing of this waveform in your book. • Switch the power off and set the function generator on a rectangular wave, keep the amplitude at maximum. • Switch power back on and observe the output waveform as displayed on the oscilloscope. • What do you observe about the two waveforms? • Calculate the frequency and amplitude of both waveforms.

Practical Activity 11 Op –amp Hartley oscillator Form of activity: Investigation Material and equipment: • Bread board • 0.5 mm solid connecting wire • 0.47µF Capacitor • 10kΩ (R1), 10kΩ Pot(Rf), Resistors • L1=L2 =45mH Inductors • 1 × 741 Op amp IC • Multimeter • Variable dual power supply (+-15V ) • Variable DC supply • Dual trace oscilloscope

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7

Electrical Technology Pin connections of the 741 op amp IC

NC

Input Offset

+Vcc Supply

Inverting Input

Output

Non-inverting Input

Input Offset

-Vcc Supply

NB: The tolerances of components and calibration of the instruments is crucial for the correct operation of this experiment. Because of the sensitive nature of the oscillator change R1 to a 10kΩ pot when you build the circuit to enable you to adjust R1 to a value to start the oscillations Instructions: • Connect the circuit of the Hartley oscillator as indicated in the diagram below on the bread board. Rf

R1 – +


R2 L2

L1

C

• • • •

232

Switch on the power to the circuit and observe the output waveform as displayed on the oscilloscope. If no output is displayed, set the 100kΩ pot to minimum and slowly adjust until the oscillations starts. Neatly draw the waveform in your book. Calculate the amplitude as well as frequency of the displayed waveform.

Amplifiers

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Practical Activity 12 The RC Phase shift oscillator Form of activity: Investigation Material and equipment: • Bread board • 0.5 mm solid connecting wire • 3 × 1.2kΩ resistors • 3 × 0.01μF Capacitor • 1 × 47kΩ Variable resistor • 1 × 741 Op amp IC • Multimeter • Variable dual power supply (+-15V ) • Dual trace oscilloscope Pin connections of the 741 op amp IC

NC

Input Offset

+Vcc Supply

Inverting Input

Output

Non-inverting Input

Input Offset

-Vcc Supply

Instructions: • Connect the circuit of the RC phase shift oscillator as indicated in the diagram below on the bread board. Feedback Rf

C1

C2

C3

+V –


A + R1

R2

R3

–V Ov

• • • • •

Connect channel1 of the oscilloscope across the output of the circuit above. Switch on the power to the circuit and observe the output waveform as displayed on the oscilloscope. Make neat labelled drawings of the output waveforms in your book. Calculate the amplitude and frequency of the output waveform. What is the phase relationship between the output and the signal across R2?

233


Glossary of formulae – Chapter 2 Single-phase

3 phase STAR

DELTA

POWER

Two wattmeter method

234

Annexure

Glossary of formulae – Chapter 3

235


Glossary of formulae – Chapter 4

236

Annexure

Glossary of formulae – Chapter 5 RCL series and parallel XL = 2πfL XC =

1 2πfC 1 fr = 2π √LC Series IT=IR=IC=IL Z = √ R2 + (XL – XC)2 VL = I.XL VC = I.XC VT = I.Z VT = √ VR2 + (VL – VC)2 IT = VT Z Cos θ = R Z Cos θ =

VR VT

Parallel VT = VR = VC = VL IR =

VR R

IC = VC XC IL = VL XL IT = √ Ir2 + (IL – IC)2 IT = VT IT Cos θ =

IR IT

237


Glossary of formulae – Chapter 7 Gain Av =

Inverting operational amplifiers

Gain Av =

Non-inverting operational amplifiers

Hartley oscillator

RC phase shift oscillator

Annexure – Preferred SI Prefixes

238

Factor

Factor in words

SI prefix

SI symbol

1 000 000 000 000 000 000

or 1018

trillion

exa-

E

1 000 000 000 000 000

or 1015

billiard

peta-

P

1 000 000 000 000

or 1012

billion

tera-

T

1 000 000 000

or 109

milliard

giga-

G

1 000 000

or 106

million

mega-

M

1 000

or 103

thousand

kilo-

k

0,001

or 10-3

thousandth

milli-

m

0,000 001

or 10-6

millionth

micro-

μ

0,000 000 001

or 10-9

millardth

nano-

n

0,000 000 000 001

or 10-12

billionth

pico-

p

0,000 000 000 000 001

or 10-15

billiardth

femto-

f

0,000 000 000 000 000 001

or 10-18

trillionth

atto-

a

Annexure

Bibliography Three phase generation http://www.science.smith.edu/~jcardell/Courses/EGR220/ElecPwr_HSW.html http://en.wikipedia.org/wiki/Three-phase_electric_power http://www.google.co.za/imgres?imgurl=http://www.rfcafe.com/references/ electrical/Electricity%2 http://www.google.co.za/imgres?start=183&um=1&hl=en&biw=1280&bih=685&tb m=isch&tbnid https://www.google.co.za/search?hl=en&q=single-phase+generation+and+supply& aq=2K&aqi http://2.bp.blogspot.com/_U3umjVzCXTA/S913w4wb9FI/AAAAAAAAAE8/ dapdc3qeaQ8/s1600/Transformer-1ph-assembly-3ph.jpg http://3.bp.blogspot.com/-NfCBQuHN-eQ/TejLIekXFJI/AAAAAAAAAHw/ ZXYW4uE2yA0/s1600/Reversible+Star+Delta+Motor+Control+CKT+2.jpg http://www.electricneutron.com/wp-content/uploads/2011/02/rotation-changecopy.jpg http://2.bp.blogspot.com/-wGGvTEX9L_k/Tc1igx6thQI/AAAAAAAAAFA/ MG4wCe7yh8c/s1600/Star+Delta+Relay+Control.jpg http://answers.yahoo.com/question/index?qid=20081207135123AA3xtX0 http://www.powerelectricalblog.com/2007/03/generator-losses-copperhysteresiseddy.html http://www.powerelectricalblog.com/2007/03/generator-losses-copperhysteresiseddy.html Measuring instruments http://en.wikipedia.org/wiki/Electricity_meter http://hackedgadgets.com/2009/10/15/inside-an-electric-power-usage-meter/ http://en.wikipedia.org/wiki/Power_factor http://www.answers.com/topic/power-factor-meter http://wiki.answers.com/Q/How_the_power_factor_meter_work&src=ansTT http://www.answers.com/topic/power-factor-meter#ixzz1z0UD59SK http://wiki.answers.com/Q/What_is_the_advantage_of_a_digital_wattmeter_over_ analog_wattmeter#ixzz1z0xrcgOZ http://wiki.answers.com/Q/What_is_the_advantage_of_a_digital_watt... http://cr4.globalspec.com/thread/58398 Three Phase Transformers http://en.wikipedia.org/wiki/Electric_power_transmission http://www.eskom.co.za/content/TD_0003TxDxelecRev4~1.pdf http://www.google.co.za/imgres?imgurl=http://www.fullmarks.org.za/Members/ wllhea002/ http://answers.yahoo.com/question/index?qid=20080209233311AAYnlcv http://en.wikipedia.org/wiki/Transformer_types http://www.brucelectric.com/three_phase.html http://www.trafoworld.com/en/transformers/different_types_transformer/#DTH http://itee.uq.edu.au/~elec4302/Document%20Folder/Lecture%20PDF%20files/ Transformer%20Protection%206%20slides%20per%20page.pdf

239

Electrical Technology Three phase motors http://www.youtube.com/watch?v=TXUWXgqEvXo http://www.engineeringtoolbox.com/synchronous-motor-frequency-speed-d_649. html http://www.engineeringtoolbox.com/electrical-motor-slip-d_652.html http://www.engineeringtoolbox.com/electrical-motor-efficiency-d_655.html http://www.google.co.za/search?q=motor+terminal+box+star&um=1&hl=en&biw= 1280&bih=685&tbm=isch&ei=UuHuT7OyE4LRhAf48pWBDQ&start=40&sa=N RCL circuits http://en.wikipedia.org/wiki/File:Electronic_component_inductors.jpg http://en.wikipedia.org/wiki/Henry_(unit) http://en.wikipedia.org/wiki/Farad http://en.wikipedia.org/wiki/File:Capacitors_Various.jpg http://en.wikipedia.org/wiki/Watt http://en.wikipedia.org/wiki/Electric_power http://en.wikipedia.org/wiki/File:Power_Triangle_01.svg http://whatis.techtarget.com/definition/electric-power http://www.rapidtables.com/electric/electric_power.htm Logics and Amplifiers http://en.wikipedia.org/wiki/Negative_feedback http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6071j-introduction-to-electronics-signals-and-measurement-spring-2006/lecturenotes/24_op_amps3.pdf HYPERLINK "http://www.coolcircuits.com" www.coolcircuits.com http://www.technologystudent.com/elec1/opamp3.htm HYPERLINK "http://www.elecfree.com" www.elecfree.com http://www.bristolwatch.com/ele/vc.htm http://zebu.uoregon.edu/~rayfrey/431/notes10.pdf http://www.wisc-online.com/objects/ViewObject.aspx?ID=SSE7906 http://www.wisc-online.com/objects/ViewObject.aspx?ID=SSE3003 http://www.ecircuitcenter.com/Circuits/opint/opint.htm http://www.calvin.edu/~pribeiro/courses/engr332/Handouts/oscillators.pdf http://www.radio-electronics.com/info/circuits/opamp_schmitt_trigger/op_amp_ schmitt_trigger.php HYPERLINK "http://www.101science.com" www.101science.com http://ptuece.loremate.com/pulse-and-digital-switch-circuits/node/5 http://www.engineersgarage.com/tutorials/understanding-op-amp-3 HYPERLINK "http://www.kpsec.freeuk.com" www.kpsec.freeuk.com HYPERLINK "http://www.howstuffworks.com" www.howstuffworks.com http://www.learnabout-electronics.org HYPERLINK "http://www.williamson-lab.com" www.williamson-lab.com http://mycircuits9.blogspot.com HYPERLINK "http://www.aaroncake.com" www.aaroncake.com HYPERLINK "http://www.daviddarling.info/ensyclopedia" www.daviddarling.info/ ensyclopedia HYPERLINK "http://www.doctronics.co.uk" www.doctronics.co.uk http://hyperphysics.phy-astro.gsu.edu HYPERLINK "http://www.ecelab.com/circuit" www.ecelab.com/circuit HYPERLINK "http://www.practicalphysics.org" www.practicalphysics.org http://www.play-hookey.com http://www.sci.brooklyn.cuny.edu/~goetz/projects/logic/test.html HYPERLINK "http://www.circuitdb.com" www.circuitdb.com http://www.silver-fox.ca/index.html http://www.electronics-tutorials.ws/opamp/opamp_4.html

240

Annexure http://www.allaboutcircuits.com/vol_3/chpt_8/11.html Digital Electronics (Third Edition), Roger L Tokheim, McGraw-Hill publishing Company, 1990 Electronics Fundamentals N2 N3, C Nel, McGraw-Hill publishing Company, 1981 Huges Electrical Technology (sixth edition), Edward Huges, Longman Scientific & Technical,1987 Introductory Digital systems for Engineering, Mahomed Rafi Bera, Juta & Co, Ltd, 1990 Electronics for today and tomorrow, Tom Duncan,John Murray publishers, 1985 First steps in Electronics (book 3), E Glasspoole, Shades Technical Publications, 1993 Electrical Technology, Jan Randewijk & Raimund Swart, Guidelines, 2007

Glossary of terms AC

Alternating Current is the movement of current in a circuit, in one direction during the positive half cycle and in the opposite direction in the negative half cycle.

Acceptor impurity

Used in the manufacture of semiconductors and is an impurity element containing three electrons (Tri-valent) in its outer shell.

Accumulator

Also termed a battery.

Active

The active (true) power is the power that is effectively being used by the load or the circuit, and this would be the voltage and current values that are exactly in phase with each other.

Active component

A component capable of current or voltage gain, i.e. transistor.

Active power

See Real Power.

Active transducer

A device capable of producing voltage or current activated by another source without an external source of electrical energy being supplied.

Admittance

The reciprocal of impedance and is the ratio of current to impedance, quantified as Y and has a unit of Siemens.

Algorithm

A set of mathematical “rules” applied to an input. Generally used to describe a section of computer code which performs a specific function

Alpha ( “ ) gain

The ratio of collector current to emitter current in a bi-polar transistor and is also referred to as hfb.

Alpha cut-off

The frequency at which alpha drops to 0,707 of its low frequency value and is also referred to as f ” or f hfb.

Alternating Current (AC)

A current of which the polarity alternates from positive to negative over time. The rate of such “alternations” is measured in cycles per second – more commonly known as Hertz (Hz)

Alternator

An electric generator designed to produce alternating current. Usually consists of rotating parts which create the changing magnetic field to produce the alternating current.

Ambient temperature:

The surrounding temperature of an area.

Amp / Ampère

The basic unit of current flow

Ampacity:

The current, in amperes, that a conductor can carry continuously under the conditions of use without exceeding its temperature rating.

241


242

Ampere hour (Amp hour, Ah)

A measurement of the capacity of a storage medium (a single cell or a battery). A cell which can supply 1 Amp for 1 hour before it is discharged to a specified minimum level is said to have a capacity of 1 Amp hour

Ampere

A type of electric current produced by one volt applied across a resistance of one ohm. It is also equal to the flow of one coulomb per second. Named after French physicist Andre M. Ampère 1836.

Amplification

A method for increasing the amplitude (or loudness) of electrical signals

Amplifier

A device so designed to provide for a gain in current and/or voltage and it may also in some instances be used to match impedances.

Amplifier

An electronic device which generates a high power signal based on the information supplied by a lower powered signal. A perfect amplifier would add or subtract nothing from the original except additional power – these have not been invented yet. A device designed to provide for a gain in current and/or voltage and may also in some instances also be used to match impedances

Amplitude

The loudness of sound waves and electrical signals. Amplitude is measured in decibels (dB) or volts

Amplitude modulation (AM)

A system of modulating; a carrier in which the amplitude of the carrier is changed in sympathy with the modulating signal

Analog

A measuring or display methodology which uses continuously varying physical parameters. In contrast, digital represents information in discrete binary form using only zeros and ones.

Analogue to Digital Converter (ADC)

A device that converts the infinite range of an analogue signal into discrete “steps”. Normally, a good audio ADC will use sufficient “steps” to resolve the smallest musical detail. For CD, this is a 16 bit converter, having 65,536 discrete levels covering the most negative signal level to the most positive

AND-gate

An AND-gate is a logic gate that will have an output when and only when all the inputs are at logic 1.

Anode

The positive electrode of different electronic components i.e. diode. The electrode devoid of any electrons, thus possessing a positive charge.

Apparent

The apparent power is the power that is supplied to the circuit. This is the product of the voltage and the current, ignoring the angle between the two. It is measured in VA or kVA, depending on the size of the values.

Appliance

Utilization equipment, generally other than industrial, normally built in standardised sizes or types, that is installed or connected as a unit to perform one or more functions such as clothes washing, air-conditioning, food mixing, deep frying, etc.

Armstrong oscillator

An oscillator circuit which employs a transformer to feed part of the output of the circuit back to the input.

Attenuation

The decrease of a signal’s amplitude level over any distance during transmission or through purpose designed attenuators. Attenuation measures signal loss in decibels (dB)

Axil/shaft

The centre part of a motor around which the rotor turns. The shaft is usually mounted on bearings to reduce friction.

Back diode

Also referred to as a tunnel diode with a low peak tunnel current and is used as a low-voltage rectifier biased in the reverse direction.

Balanced loads

A three-phase load in which all three branches have exactly the same impedance, and each branch has the same voltage across it.

Annexure Bandwidth

That range of frequencies that lie between the upper half-power and lower halfpower (3 dB ) points on the frequency response of a system. The measure of a range of frequencies containing an upper and lower limit. Also refers to the speed of an internet or digital connection. The higher the bandwidth, the more spectrum is being used and thus more data is transferred per second.

Bare

A conductor having no covering or electrical insulation whatsoever.

Battery

A group of two or more cells connected together to provide electrical current. Sometimes also used to describe a single cell which converts chemical energy to electrical current. A bank of individual cells connected together to provide the required voltage

Baud

A unit of speed in data transmission equal to one bit per second. Baud rate equals the amount of bits per second.

Bearings

A closed, circular, metal container with ball bearings that turn. This is to reduce friction and associated heat.

Beta

The ratio of collector current to base current in a bi-polar transistor and is also referred to as hfe.

Bi – polar transistor

Is a PNP or an NPN transistor consisting of two PN-junctions and which operates using positive or negative charge carriers for the flow of current.

Bias

Voltages and/or currents required by electronic components to ensure proper operation and could be a DC bias, forward or reverse, or an AC bias.

Binary

The basic counting system used in computer logic. Two values are available – 0 and 1. A zero is normally represented by a 0 Volt signal, and a one by a voltage of approximately 5 Volts. These levels are dependent upon the type of logic used.

Binary code

A coding scheme that communicates information by using a series of “1s” and “Os” that are represented, respectively, by the digital “ON” and “OFF” states.

Binary number

A number system to the base 2

Bipolar-transistor

A transistor in which current is carried through the semiconductor both by holes and electrons

Bit

A unit of the binary code that consists of either a single “1” or “O.” (Commonly 5V or 0V respectively.)

Bit stream

The bit rate, or flow of information, between a sender and receiver in digital communication. Also called Digital Bit Stream.

Boolean algebra

A system of formal logic used for minimising complex digital systems capacitor – a component used in electronic circuits, exhibiting the property of capacitance conductor – a material through which an electric current can flow relatively easily conventional current – electric current, regarded as flowing from positive to negative

Bus

A pathway that connects devices, enabling them to communicate. May be digital or analogue, including power and earth (ground).

Byte

A unit of the binary code that consists of eight bits. One byte is required to code an alphabetic or numeric character, using an eight-bit character set code.

Cable

A type of linear transmission medium. Some of the common types of cables include: hook up wire, coaxial (shielded) cables, lamp and mains cable, figure-8 (zip) cable and fibre optics

Capacitance

The property of a capacitor to store an electrical charge and is measured in Farad.

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244

Capacitive reactance

The total ohms value of a capacitor in an AC circuit. This ohm value opposes the flow of current when connected to an AC supply.

Capacitor

A pair of parallel “plates” separated by an insulator (the dielectric). Stores an electric charge, and tends to pass higher frequencies more readily than low frequencies. Does not pass direct current, and acts as an insulator. Electrically it is the opposite to an inductor. Basic unit of measurement is the Farad, but is typically measured in micro-farads (µF = 1 x 10-6F) or nano-farads (nF – 1 x 109 F). A device that stores electrical charge usually by means conducting plates or foil separated by a thin insulating layer of dielectric material. The effectiveness of the device, or its capacitance, is measured in Farads.

Capacity

The rating of the current – delivering capability of a battery and is measured in ampere – hours (Ah).

Cascade

A term used to describe a number of systems connected in series.

Cathode

The negative electrode of different electronic components i.e. diode.

Cell:

A single device which converts chemical energy into electrical current. Sometimes also referred to as a battery. Cell: one section of a battery. The common carbon or alkaline cells used in battery operated equipment, for example.

Centrifugal switch

An effect by which a spring-loaded switch (N/C) that is placed on the shaft of an electric motor opens as the shaft reaches a certain speed.

Choke

A component capable of blocking an AC quantity above a specified frequency and is also referred to as an inductor or a coil.

Circuit breaker

A device designed to open and close a circuit by non-automatic means and to open the circuit automatically on a pre-determined overcurrent without damage to itself when properly applied within its rating.

Circuit extensions

To extend or add-on to an existing circuit to provide an additional power source.

Clamper

An electronic circuit used to re-establish a DC-level on an AC waveform.

Class-A amplifier

An amplifier in which the collector current will flow for 360Eof the input waveform.

Class-B amplifier

An amplifier in which the collector current will flow for 180º of the input waveform.

Class-C amplifier

An amplifier in which the collector current will flow for less than 180º of the input waveform.

CMMR

Common Mode Rejection Ratio and is the ratio of amplifier gain for differentially applied signals to its gain for signals applied in common to the two inputs of a system and is normally associated with operational amplifiers.

CMOS

Complementary Metal-oxide Semiconductor that uses field-effect transistors. A family of digital logic devices. Some CMOS devices can operate with power supplies from 3 Volts to 15 Volts – others are limited to the traditional logic 5 Volt power supply.

Coaxial cable

A metallic cable constructed in such a way that the inner conductor is shielded from EMR (electromagnetic radiation) interference by the outer conductor. Coaxial cable is less susceptible to more transmission impairments than twisted pair cable, and it has a much greater bandwidth; thus coaxial cable is used by most analogue and digital systems for the transmission of low level signals.

CODEC

COder/DECoder – the component of any digital subsystem which performs analogue to digital and digital to analogue conversions.

Annexure Colour code

Used to identify resistors and some capacitors, as well as wires in telephony. For telephone cables, the basic colour code for the first group of pairs is Blue, Orange, Green, Brown, Slate (grey), with white “Mates”. The Mate is the most positive lead, and is the Tip connection.

Colpitts oscillator

An oscillator circuit which employs a phase-shifting network consisting of an inductor and two capacitors to feed part of the output of the circuit back to the input.

Condenser

Also termed a capacitor.

Conductance

The reciprocal of resistance which is the ratio of current to voltage and quantified as G and has a unit of Siemens.

Continuous load

A load where the maximum current is expected to continue for 3 hours or more.

Controller

A device or group of devices that serves to govern, in some predetermined manner, the electric power delivered to the apparatus to which it is connected.

Covered

A conductor encased within material of composition and thickness that is NOT recognized by this Code as electrical insulation.

CRO

Cathode-Ray-Oscilloscope which is used for the graphical study of waveforms.

Crossover

A filter network which separates frequencies into “bands” which match the capabilities of the loudspeaker drivers within an enclosure.

Crosstalk

A noise impairment when a signal from one pair of wires affects adjacent wires or one channel affects the adjacent channel.

CRT

Cathode-Ray-Tube is a display device used in a CRO.

Current

The flow of electricity commonly measured in amperes.

Cut-off

The zero-current condition or point of a transistor or similar device.

Cut-off frequency

Normally defined as the frequency where the output from a filter has fallen by 3dB from the maximum level obtainable through the filter.

Damp

The process of making an oscillation come to rest or reduce to zero.

Darlington connection

Two bi-polar transistors connected in such a way with the emitter of the first transistor feeding the base of the second transistor, and the two collectors of the transistors connected together.

dB – Decibel – (0.1 Bel)

Defined (more or less) as the smallest variation of volume detectable by ear (under laboratory conditions). This is measured on a logarithmic scale, so a change of 3dB from 1 Watt is equivalent to 0.5 Watt or 2 Watts. A change of 10dB from 1 Watt is equivalent to 100mW or 10 Watts. In electronics, 0dBm is a reference value corresponding to 1mW at 600 Ohms – this equates to approximately 775mV. The threshold of sound is 0dB SPL, and typical sounds can reach 140dB SPL or more. Any prolonged sound above 90dB SPL will cause hearing damage. A logarithmic measure of the ratio of two quantities. Abbreviated dB. For electrical power, 1 dB = 10 x log10 P1/P2. For electric voltage or current, 1 dB = 20 x log10 E1/E2.

DC

Direct Current is the movement of current in a circuit in one direction only, usually at a constant rate.

De-coupling

A filter process which eliminates unwanted signals passing between stages of a system and normally uses inductors and capacitors for such action.

Demodulation

The recovery of a modulating signal from a modulated carrier

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246

Depletion zone

Also referred to as the depletion layer and is that portion near the junction of a PN- junction which is void of charge carriers.

Device

A unit of an electrical system that is intended to carry but not utilize electric energy.

DIAC

A two terminal semi-conductor device or component that will conduct at a specified breakdown voltage in forward or reverse bias. A bi-directional break over a diode; often used for triggering a triac.

Differential

An amplifier constructed in such a way that an output will be provided which is the difference between the two input signals.

Diffused junction

A PN-junction formed by diffusion when charge carriers move from a region of high density to a region of low density.

Digital electronics

The branch of electronics concerned with the processing of digital systems, usually in binary.

Digital/Analogue conversion

A method used to recreate an analogue signal that has been coded into binary data and transmitted as a digital signal.

Digital/Analogue Converter (DAC)

A device used to generate a replica of the original analogue signal that has been coded into binary data and transmitted as a digital signal.

DIL

A dual – in – line package normally associated with integrated circuits.

Diode

An electronic semiconductor device that predominantly allows current to flow in only one direction. A component, either semiconductor or thermionic, that permits current to flow through it in one direction only

Direct coupling

A signal path between two amplifier stages by means of a conductor, resistor or a battery.

Direct Current (DC)

A current flow which is steady with time, and flows in one direction only. Conventional current flows from positive to negative. Electron flow is from negative to positive.

Distortion (1)

Any modification to a signal which results in the generation of frequencies which were not present in the original.

Distortion (2)

Of phase, any modification of the phase relationship between two or more signals which causes the observed waveform to differ from the original.

Distribution equipment

A device designed to provide electricity to multiple connections.

DMM

Digital-Multi-Meter capable of measuring current, voltage and resistance.

Donor Impurity

Used in the manufacture of semiconductors and is an impurity element containing five electrons (Penta-valent) in its outer shell.

Doping

Process whereby impurities (Tri-or Penta-valent) are added to pure or intrinsic semi-conductor material and then obtaining an impure or extrinsic material (Nor P- type).

DVM

Digital Volt Meter.

Dynamic Resistance

This is the ratio of a change in voltage to a change in current as opposed to static resistance.

Earth

A term used to indicate an electrical ground.

Earth (1)

Also known as ground – commonly used to describe the chassis and other materials that provide a return path for power supplies and signals within any electronic device.

Annexure Earth (2)

Also known as ground – a protective connection from wall outlet to equipment chassis to conduct fault currents away from human contact.

Eddy currents

A current that moves against the main current in a particular pattern. (Usually the current induced in each of the laminated plates of the core)

Electric Resistance Heating:

A type of heating system that generates heat by passing current through a conductor, causing it to heat up. These systems usually use baseboard heaters, often with individual controls. They are inefficient and are best used as a backup to more efficient options, such as solar heating or a heat pump.

Electromagnetic Interference (EMI)

An unwanted (possibly interfering) signal emitted by any electronic apparatus. The emission of EMI is heavily regulated in most countries.

Electromagnetic Radiation (EMR)

A transmission medium that includes radio waves and light waves.

Electronic

The use of active electronic components (integrated circuits, transistors, valves etc) which require a power supply to function. Such “active” components will always be used in conjunction with passive components.

EMF

Electromotive force measured in volt and is that value measured when the source carries no load.

Enclosure

The case or housing of an apparatus, fence, or walls that prevent persons from accidentally contacting energizing parts, or to protect the equipment from physical damage.

Energizing

Electrically connected to a source of potential difference.

Energy

Focuses on domestic, industrial, generation and transmission services. The capacity for, or the ability to do, mechanical work. Electrical energy is measured in kilowatt-hours for billing purposes.

Energy-saving devices

Devices utilized within a dwelling designed to more efficiently make use of energy sources while providing heating, cooling, and light.

Enhancement-mode FET

An insulated-gate field-effect transistor that is non-conductive at a zero-gate source voltage and switches or turns on with forward gate bias.

Entrepreneur(ship)

A person who invents or identifies a useful idea and then sets up a business using that idea fuse; a protective device that opens a circuit on overcurrent and is used for protection of systems

Equipment

A general term including materials, fittings, devices, appliances, fixtures, apparatus, and the like used as a part of, or in connection with, an electrical installation.

Eskom

Electricity Supply Commission in South Africa

Exclusive OR-gate

An exclusive OR gate is a logic gate that will have an output when and only when one of the inputs are at logic 1.

Extrinsic semi – conductor ( N- or Ptype )

A semi – conductor material which has been doped and the conducting properties have been changed.

Fall-time

The time required for a quantity to fall from 90 % to 10 % of its peak or maximum value.

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Farad

The base unit of capacitance – equal to the capacitance of a capacitor having an equal and opposite charge of 1 coulomb on each plate and a potential difference of 1 volt between the plates (Abbreviation – F). The Farad is a very large value, and is more commonly referred to as the pico-Farad (pF, 1 x 10-12 Farad), nanoFarad (nF, 1 x 10-9 Farad), micro-Farad (uF, 1 x 10-6Farad), and (less common) milli-Farad (mF, 1 x 10-3 Farad).

Fault

A short circuit in an electrical system.

FET

Field-effect Transistor and is a semiconductor amplifying or switching device in which the current through a single-polarity channel is controlled by the voltage applied to a reverse-biased gate electrode and its high input impedance is its main characteristic.

Filter

A circuit which is frequency dependent. The “pass band” is the range of frequencies allowed through, and the “stop band” is that range of frequencies which are blocked.

Filtering

A process used to remove or accentuate specific frequencies or frequency ranges of a signal.

Fitting

An accessory such as a locknut, bushing, or other part of a wiring system that is intended primarily to perform a mechanical rather than an electrical function.

Fluorescent lamps

Fluorescent lamps produce light by passing electricity through a gas, causing it to glow. The gas produces ultraviolet light; a phosphor coating on the inside of the lamp absorbs the ultraviolet light and produces visible light. Fluorescent lamps produce much less heat than incandescent lamps and are more energy efficient. Linear fluorescent lamps are used in long narrow fixtures designed for such lamps. Compact fluorescent light bulbs have been designed to replace incandescent light bulbs in table lamps, floodlights, and other fixtures.

Frequency

In alternating current, the rate at which the current changes direction. One complete cycle is a unit of 1 Hertz, named after the Physicist who researched AC (Alternating Current). The standard frequency in SA is 50 Hz.

Frequency Modulation (FM)

A modulation technique that records changes in an information signal by modifying the frequency of the carrier signal according to changes in the amplitude of the information signal.

Fuse

A protective device that opens a circuit upon overcurrent and is used for protection of systems.

Generator

A rotating machine which converts mechanical energy into electrical energy. In the automotive industry traditional terminology uses generator to refer to only those machines designed to produce dc current through brushes and a commutator (as opposed to alternator).

Grid

In an electrical system, a term used to refer to the electrical utility distribution network.

Ground (Wire)

A conducting connection, whether intentional or accidental, between an electrical circuit or equipment and the earth, or to some conducting body that serves in place of the earth.

Grounded

Connected to earth or to some conducting body that serves in place of the earth.

Hall-effect

The development of a voltage across a metal or a semiconductor material placed in a magnetic field.

Hartley oscillator

An oscillator circuit which employs a tapped inductor feed part of the output of the circuit back to the input.

Annexure Heater

A heat source (gas or electric) used to adjust the temperature inside a dwelling from a cold to a warm condition.

Heavy Duty

A lightning impulse classifying current category for distribution class arresters defined by ANSI/IEEE C62.11. A heavy duty rated arrester has a 10,000 amperage impulse value crest (refer to normal duty).

Henry

The basic unit of inductance in which an induced electromotive force of one volt is produced when the current is varied at the rate of one ampere per second (Abbreviation – H).

Hertz (Hz):

The unit of frequency (not just electricity, but also, for example, sound waves. Hertz represents the number of cycles of an electrical signal measured in one second.

High-pass.

A filter which allows high frequencies to pass while blocking low frequencies.

Hole

A vacancy in the conduction band of a semiconductor material and regarded as a positive charge carrier.

Horsepower

A unit of power equal to 746 watts.

I/P–O/P

Refers to input / output of a device and / or system.

IC

Integrated circuit.

IEC

International Electro technical Commission who is responsible for world – wide standards.

Impedance

The total ohms value of a RL, RC or RCL circuit. This ohm value opposes the flow of current when connected to an AC supply

Impedance

Impedance: A load applied to an amplifier (or other source) which is not a pure resistance. This is to say that its loading characteristics are frequency dependent. Impedance consists of some value of resistance in conjunction with capacitance and/or inductance. The equivalent circuits can vary from two components to hundreds. The vector sum or combination of resistance and reactance quantised as Z and measured in ohms. The ratio of the voltage applied to a circuit to the current flowing in the circuit; similar to resistance, but applicable to alternating currents and voltages The total effects of a circuit that oppose the flow of an AC current consisting of inductance, capacitance, and resistance. It can be quantified in the units of ohms.

Impulse

A current surge.

Incandescent light bulbs

Incandescent light bulbs produce light by passing electricity through a thin filament, which becomes hot and glows brightly. Incandescent light bulbs are less energy-efficient than fluorescent lamps, because much of the electrical energy is converted to heat instead of light. The heat produced by these bulbs not only wastes energy, but can also make a building’s air conditioning system work harder and consume more energy.

Inductive reactance

The total ohms value of a coil in an AC circuit. This ohm value opposes the flow of current when connected to an AC supply.

Inductor

A coil of wire which exhibits a resistance to any change of amplitude or direction of current flow through itself. Inductance is inherent in any conductor, but is “concentrated” by winding into a coil. An inductor tends to pass low frequencies more readily than high frequencies. Electrically it is the opposite of a capacitor. Basic unit of measurement is the Henry (H), in crossover networks it will typically be measured in milli-henrys (mH = 1 x 10-3H) and for RF micro-henrys (µH) are common.

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Infrared cameras

Energy contractors use infrared cameras to look at the heat leaking into or out of your house. The infrared camera “sees” the heat and can show “hot spots” where a lot of heat is being lost. This helps to identify the places where your home’s energy efficiency can be improved.

Isolating switch

A switch that is turned and simultaneously disconnects the live and the neutral (in case of single phase) or all 3 live wires (in the case of three phase)

Insulated

A conductor encased within material of composition and thickness that is recognized by this Code as electrical insulation.

Insulation

A material having a high resistance to the flow of electric current; insulation over underground conductor is made of either EPR or XLPE material.

Insulator

A material that prevents the passage of electricity, heat or sound. The plastic coating on wires is an insulator, preventing the wires from coming into electrical contact with each other. Insulators are extensively used in electronics. Most good electrical insulators are also good thermal insulators. Any material which does not allow electrons to flow through it.

Integrated Circuit (IC)

A collection of active and passive devices (e.g. transistors and resistors) mounted on a single slice of silicon and packaged as a single component. Examples include operational amplifiers, Central Processing Units (CPUs), random access memory (RAM), etc. An electronic system, or part of a system, produced on a silicon chip using microelectronic techniques

Intrinsic semiconductor

A pure semi-conductor material that has not been doped.

Inverter

An electrical device which is designed to convert direct current into alternating current. This was originally done with rotating machines which produced true sine wave ac output. More recently this conversion has been performed more economically and efficiently using solid state electronics. However, except for the most expensive models, these devices usually do not produce perfect sine wave output. This sometimes can result in electromagnetic interference with other sensitive electronic devices.

Ion

An Ion is a positively or negatively charged atom or molecule.

Joule

A unit of work or energy equal to one watt for one second. One kilowatt hour equals 3,600,000 Joules. Named after James P. Joule, an English physicist 1889.

Joule’s law

Defines the relationship between current in a wire and the thermal energy produced. In 1841an English physicist James P. Joule experimentally showed that W = I2 x R x t where I is the current in the wire in amperes, R is the resistance of the wire in Ohms, t is the length of time that the current flows in seconds, and W is the energy produced in Joules.

Kilovolt

A Unit of electrical potential equal to 1,000 volts. Abbreviated kV or KV.

Kilowatt (kW)

Real power delivered to a load (W x 1,000 VA).

Kilowatt-hour

A unit of energy or work equal to one kilowatt for one hour. Abbreviated as kwh or KWH. This is the normal quantity used for metering and billing electricity customers. The price for a kwh varies from approximately 4 cents to 15 cents. At a 100% conversion efficiency, one kwh is equivalent to about 4 fluid ounces of gasoline, 3/16 pound LP, 3 cubic feet natural gas, or 1/4 pound coal.

Kirchhoff ’s current law ( KCL )

Kirchhoff ’s current law states that the algebraic sum of the current and/or currents entering a point is equal to the algebraic sum of the current and/or currents leaving that point.

Annexure Kirchhoff ’s voltage law ( KVL )

Kirchhoff ’s voltage law states that the algebraic sum of the voltage drops in a closed network is equal to the algebraic sum of the applied voltage and/or voltages.

Kirchhoff ’s current law (KCL)

The algebraic sum of the current and/or currents entering a point is equal to the algebraic sum of the current and/or currents leaving that point

Kirchhoff ’s voltage law (KVL)

The algebraic sum of the voltage drops in a closed network are equal to the algebraic sum of the applied voltage and/or voltages

Lagging

Occurring later in time.

Lagging Load

Inductive type load.

Laminated Core

Several layers of thin metal plate bolted together.

Laser

Light Amplification by Stimulated Emission of Radiation. Originally, lasers were either gas or precious stone (e.g. ruby), but are now made using semiconductors. Laser light is coherent, meaning that the emitted light waves are in phase, which gives the light a strange appearance since our eyes were never designed to observe coherent light.

Lattice structure

A pattern of positions on a regular grid of lines and is descriptive of the atomic structure of intrinsic or extrinsic semi-conducting materials.

LDR

See Photo Resistor

Leading

Occurring earlier in time.

Leading load

Capacitive load.

LED

Light Emitting Diode and is a semiconductor that emits light when activated by a suitable voltage source.

Limit switch

A switch that is operated by some part or motion of a power-driven machine or equipment to alter the electric circuit associated with the machine or equipment.

Line

The voltage measured between any line and neutral

Lissajous figure

The pattern obtained on the screen of an oscilloscope when two sine waves of related frequencies are applied to the vertical and horizontal deflection systems respectively and this figure may be in the form of a line, an ellipse, a figure eight or a more complex series of loops.

Load

The device or component that receives the output from a signal source.

Load

The load of a transformer is the power, in kVA or volt-amperes, supplied by the transformer.

Load Factor

Represents how efficiently the electrical system capacity is being used. The higher the load factor the higher the efficiency.

Load Switching

Transferring the load from one source to another.

Loadbreak

The ability of a switching device to disconnect a load current without damage.

Logic

The basic principles and applications of truth tables, interconnections of on/ off circuit elements and other factors involved in mathematical computation in a computer; also used as a general term for various types of gates, flip-flops, and other on/off circuits used to perform problem-solving functions in a digital computer

Low voltage

A wiring system that provides power to some electronic devices operating on a voltage level much lower than the standard 110 volts. Such devices might be doorbells and thermostats.

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Low-pass.

A filter which allows low frequencies to pass while blocking high frequencies.

LVDT

Linear Variable Differential Transformer classified as a transducer and is a positioning-measuring device utilising an AC excitation.

Majority carriers

Classified as electrons in an N-type extrinsic semi-conductor material and holes in a P-type semiconductor extrinsic material.

Megger

A measuring instrument that can measure low ohm values or high ohm values at a high operating voltage.

Metal Enclosed

Surrounded by a metal case of housing, usually grounded.

Modulation

Variation of the frequency, phase or magnitude of a high frequency waveform in accordance with a waveform of lower frequency

MOS

Metal Oxide Semiconductor and is a technology used for the production of transistors and integrated circuits using Field-effect transistors and having the gates insulated from the channel by an oxide of silicon.

MOSFET

Metal Oxide Semiconductor Field-effect Transistor.

Motors

Electronic device used to move, switch, or adjust one or more of the systems within a dwelling.

Multimeter

A general-purpose measuring instrument, usually able to measure resistance, current and voltage

NAND-gate

A NAND-gate is a logic gate that will have an output when one or all the inputs are at logic 0.

Negative feedback

Feedback applied to a system in such a way that it tends to reduce the input signal that results in the feedback

Neutral

The junction point of the legs in a Wye circuit. (Star)

NOR-gate

A NOR-gate is a logic gate that will have an output when and only when all the inputs are at logic 0.

Norton’s theorem

Norton’s theorem states that a complex network consisting of impedances and voltage may be replaced by a constant current generator with a parallel impedance.

NOT-gate

A NOT-gate is a logic gate where the output is always the opposite of the input.

Ohm

The unit of measure for resistance.

Ohm’s law

Ohm’s law states that the current is directly proportional to the applied voltage and indirectly proportional to the resistance of that network. ( Only applicable to direct current networks). The current is directly proportional to the applied voltage and inversely proportional to the resistance of that network, when the temperature is kept constant (only applicable to direct current networks)

Operating point

A combination of current and voltage at which a transistor is biased when supplied by a signal source.

Operational amplifier

A highly stable, gain, DC amplifier, usually produced as a single integrated circuit opto-electronics – electronic systems or devices that involve the use of light.

Oscillator

An electronic system that produces a regular periodic output.

Oscilloscope

An electronic measurement tool which allows one to view a waveform. The vertical axis shows amplitude and the horizontal axis shows time. An instrument for displaying electrical waveforms on a cathode ray tube or LCD or LED screen.

Annexure Outlet

A point on the wiring system at which current is taken to supply utilization equipment.

Overload

A protection device that will disconnect the supply from the load, usually when current exceeds a predetermined value.

Overload

Operation of equipment in excess of normal, full-load rating, or of a conductor in excess of rated ampacity that, when it persists for a sufficient length of time, would cause damage or dangerous overheating. A fault, such as a short circuit or ground fault, is not an overload.

Overvoltage

A voltage above the normal rated voltage or the maximum operating voltage of a device or circuit. A direct test overvoltage is a voltage above the peak of the line alternating voltage.

Parameter

The measurable characteristics or variables of a circuit, system or component.

Passive

Passive: Containing no devices which require a power supply. Passive devices include resistors, capacitors and inductors.

Passive component

A component not capable of amplification or switching ie. resistors, inductors and capacitors.

PCB

Printed Circuit Board. A fibreglass substrate board laminated / plated with copper used to manufacture circuit boards for electronic devices. Multilayered PCB contains multiple layers of Copper. Double-sided PCB has copper tracks on both sides and uses THP (Through-Hole Plating) to connect the different layers.

Peak Demand

The maximum integrated demand during a time period.

Phase

The time relationship of one waveform with respect to another of the same frequency expressed in degrees or radians with 360º representing one full cycle. Phase is measured in degrees of rotation. Classification of an AC circuit usually single-phase, two wire or three wire; twophase, three wire or four wire; or three-phase, three wire or four wire.

Phase

The voltage measured between any two lines.

Phasor diagram

The representation of voltage values in a RC, RL, RCL circuit, taking into account the size of the voltages and the phase relationship between them.

Photo-resistive

A component that will change resistance when subjected to a source of light.

Photo-voltaic

A component that will generate a voltage when subjected to a source of light.

Photo-Resistor – also known as an LDR (light-dependent resistor)

Also known as an LDR (light-dependent resistor), a resistor whose value depends upon the amount of light falling on it

Piezo-electric effect

Takes place in a quartz crystal when subjected to a mechanical pressure will generate a voltage or will undergo dimensional changes when subjected to an ac voltage.

PIN diode

A semi-conductor formed by a P-type extrinsic material, an Intrinsic material and an N-type extrinsic material.

Pinch-off voltage

The reverse gate-source voltage that will reduce channel current of an FET to a near-zero level.

PIV

The maximum Peak-Inverse-Voltage that a diode may be subjected to and is also termed the Peak-Reverse-Voltage (PRV).

Point-contact

An early process for diode and transistor manufacturing.

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Positive feedback

Feedback applied to a system in such a way that the feedback tends to increase the input signal causing the feedback

Pot.

Abbreviation for a potentiometer, which is a three-terminal adjustable resistor.

Potential difference

An electromotive force of voltage measured when a load is inserted.

Potentiometer

A potentiometer which is a three-terminal adjustable resistor.

Power

The rate at which work is performed or that energy is transferred. Electric power is commonly measured in watts or kilowatts. A power of 746 watts is equivalent to 1 horsepower.

Power amp

An amplifier that is designed to drive loudspeakers or other relatively low impedance loads. Usually combines voltage and current amplification. May be integrated with the preamp (see below).

Power factor

The ratio of the true power compared to the apparent power.

Power outage

An interruption of power.

Power outlet

An enclosed assembly that may include receptacles, circuit breakers, fuse holders, fused switches, buses, and watt-hour meter mounting means; intended to supply and control power to mobile homes, recreational vehicles, park trailers, or boats; or to serve as a means for distributing power required to operate mobile or temporarily installed equipment.

Power supply

Source of electrical energy

Pre-amp

Multiple meanings, but in hi-fi generally refers to a separate section of circuitry that includes source switching, volume and balance controls (as well as tone controls in many cases). to raise the level from tape decks, turntables, CDplayers and other music sources to a level suited to the power amplifier.

Primary voltage rating

Designates the input circuit voltage for which the primary winding is designed.

Programmable Logic Controller (PLC)

A control device, normally used in industrial control applications, that employs the hardware architecture of a computer and a relay ladder diagram language

Protective Device

A particular type of equipment used in electric power systems to detect abnormal conditions and to initiate appropriate corrective action

Pylons

A series of very tall metal structures used to hold or carry the electric cables high above the ground.

The Q-factor

This is referred to as the amplification factor of a resonant RCL circuit that is connected to an AC supply and it is the ratio of the energy stored to the energy wasted.

Quasi

To some degree or in some manner, resembling. For example, a quasi complementary-symmetry output stage in an amplifier is not in fact complementary-symmetry, but appears to be, and acts in a similar manner.

Quiescent

At rest with no signal applied as applied to electronic circuitry and is also known as the Q – point. Being still or at rest, in an inactive state. The quiescent current in an amplifier is that current drawn when the amplifier is “at rest” – i.e. not amplifying a signal, but supplied with power.

Rating

The rating of an arrester – either duty cycle or MCOV rating.

Reactance

The value of opposition offered by an inductor and/or a capacitor to an ac quantity in which the voltage and current is not in phase and is a ratio of voltage to current.

Annexure Reactive

Reactive power is the power that is wasted and not used to do work on the load.

Reactive Power

The mathematical product of voltage and current consumed by reactive loads. Examples of reactive loads include capacitors and inductors. These types of loads when connected to an ac voltage source will draw current, but since the current is 90o out of phase with the applied voltage they actually consume no real power in the ideal sense.

Reactor

A device for introducing inductive reactance into a circuit for motor starting, operating transformers in parallel, and controlling current.

Real Power

The rate at which work is performed or that energy is transferred. Electric power is commonly measured in watts or kilowatts. The term real power is often used in place of the term power alone to differentiate from reactive power. Also called active power.

Rectification

The process of converting an alternating current to a unidirectional current relay – an electromechanical device in which an electric current closes a switch resistance – the property of a material that resists the flow of electrical current

Resin

A transparent liquid that solidifies after application and provides electrical insulation

Resistance

The value of the opposition offered to the flow of an AC or DC in-phase current and is the ratio of voltage to current.

Resistor

An electrical device which impedes (resists) current flow regardless of frequency. Basic unit of measurement is the Ohm. Any device of material that limits the flow of current when voltage is applied.

Resonance

This is a special frequency in a RCL or LC circuit where certain characteristics are only true at that frequency.

Resonance

A condition which will exist in a network containing an inductor and a capacitor when the two reactances equal one another. The natural frequency at which a physical body will oscillate. An example is when you blow gently across the top of a bottle, the enclosed air resonates at a frequency determined by the internal volume. Also refers to the natural resonance of loudspeaker drivers, cabinets and ports, or the frequency where an inductance and capacitance have the same impedance (this causes maximum impedance with a parallel circuit, and minimum impedance for series circuits).

Rheostat

A two-terminal high current carrying variable resistor.

Rise – time

The time required for a quantity to rise from 10% to 90% of its peak or maximum value.

RMS

The root of the mean of the square value of an alternating quantity which is used to determine the dc equivalent of an AC quantity in terms of power – delivering capabilities. Applies to voltage and current, but is commonly (although incorrectly) applied to power. Defined as an alternating voltage (or current) which has exactly the same energy content (power) as the same value of direct current.

Saturation

The point at which an increment in one quantity will no longer lead to an increase in the other quantity and is commonly applied to the base-current vs. collector- current of an amplifier.

SCR

Silicon controlled rectifier which is a three-terminal device that will only conduct when the gate is triggered at a specific value provided that the device is forward biased.

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Semiconductor

Silicon (or various other materials) that are specially treated so as to form diodes, transistors, MOSFETs, light emitting diodes (LEDs), etc. The basis of all modern electronics.

Servo motor

Servomotors are geared DC motors with a positional feedback control that allows the rotor to be positioned accurately.

Sinusoidal

A changing quantity in the shape of a sine wave irrespective of its phase.

Slip

The slip is an indication of how much the rotor speed differs from the speed of the rotating stator field. The larger the load, the more the slip.

Squirrel cage

Refers to the type of rotor. A squirrel cage rotor has metal rods (or strips) that are short circuited at the ends by means of a metal ring.

SSB

Single Sideband Suppressed Carrier Modulation. Modulation which involves the suppression of the carrier wave in order to limit bandwidth and to improve the efficiency of the transmitter circuit. Upper Side Band (USB) and Lower Side Band (LSB) refer to the portion of the sideband being used for the purpose of propagation of the RF signal.

Stepper motor

A stepper motor (or step motor) is a brushless DC electric motor that can divide a full rotation into a large number of steps. The motor’s position can be controlled precisely without any feedback mechanism (an open-loop controller), as long as the motor is carefully sized to the application. Stepper motors are used in Computer Hard Drives, actuators, robotics, DVD and CD players and the manufacturing industry.

Strain gauge

A resistive element that is capable of changing its resistance when subjected to mechanical stress.

Superposition theorem

The superposition theorem states that all current magnitudes and directions may be determined by considering each supply on its own.

Susceptance

The negative reciprocal of reactance quantified as $ and measured in Siemens.

Switch Limit

A switch that is operated by some part or motion of a power-driven machine or equipment to alter the electric circuit associated with the machine or equipment.

Switchboard

A large single panel, frame, or assembly of panels on which are mounted, on the face or back, or both, switches, overcurrent and other protective devices, buses, and usually instruments. Switchboards are generally accessible from the rear as well as from the front and are not intended to be installed on cabinets.

Switches

Circuit interruption devices used to control the flow of electricity to lights, appliances, and outlets.

Synchro

A device used to duplicate an angular position irrespective of the distance.

Synchronous speed

This is the speed at which the rotating stator field rotates.

Tap

A connection brought out of the winding at some point between its extremities, usually to permit changing the voltage or current ratio.

Thermal Coefficient (1)

Coefficient of expansion, describes the amount by which a material expands when heated. Commonly expressed as a percentage per degree Celcius so the exact size at various temperatures may be calculated. Knowledge of the expansion characteristics of different materials is important in high power semiconductor manufacture, since differing expansion rates may cause device failure due to temperature cycling fractures.

Annexure Thermal Coefficient (2)

Coefficient off resistance, describes the change in resistance at various temperatures. Most metals have a positive temperature coefficient of resistance, which means that the resistance increases with increasing temperature. Carbon and some alloys have a negative temperature coefficient of resistance, so as temperature is increased, resistance decreases.

Thermal Résistance

The resistance of various materials to the passage of heat energy. Most electrical conductors are also thermal conductors, with the higher electrical conductivity materials usually having higher thermal conductivity. Important in the design of high power electronics, heat sinks, semiconductor casings, etc.

Thermal Runaway

A condition that may occur in a semiconductor transistor when an increase in current leads to an increase in temperature which leads to a decrease in resistance which leads to an increase in current ending in a spiral of saturation or self – destruction of the transistor.

Thermostat

A low voltage electronic switching device that monitors temperatures inside the home and turns on and off the heating or cooling system in the home.

Thevenin’s theorem

Thevenin’s theorem states that a complex network consisting of impedances and voltage sources may be replaced by a constant voltage source with a series impedance.

Thyristor

The family of semiconductor switching devices and include SCR’s, diac’s, triac’s quadrac’s. A component similar to a semiconductor diode but having in addition a gate connection by which the component, normally non- conducting, can be triggered into conduction

Timer: On delay

This timer has a number of contacts, some N/O and others N/C. When the timer is activated, these contacts will keep their normal state and, after a preset time set on the timer, will change state simultaneously.

Torque

A force that causes something to turn around a central point.

Transducer

A device that is capable of converting one form of energy into another form of energy ie. mechanical to electrical, electrical to mechanical, etc.

Transformer

A static electrical device which by electromagnetic induction transfers electrical energy from one circuit to another circuit usually with changed values of voltage and current in the process.

Transformers

A piece of electrical equipment that changes one voltage value to a lower or higher

Transient

A short pulse or oscillation of current and / or voltage instead of a steady-state condition. A high amplitude, short duration pulse superimposed on the normal voltage.

Transistor

Derived from TRANSfer resISTOR and is an active semiconductor device having three terminals namely a base, a collector and an emitter.

Transmission lines

A process of moving or sending energy or power from one place to another.

TRIAC

A three-terminal semiconductor device similar to an SCR but which is capable of conducting in both directions when the gate is triggered.

Trimmer / Trimpot

A small adjustable capacitor or resistor used for alignment purposes.

Tunnel diode

A semiconductor device that will exhibit a negative resistance between the values of 0,2 volts and 0,4 volts when forward biased and is also termed an Esaki diode.

Turn ratio

The ratio of the number of turns in the high voltage winding to that in the low voltage winding.

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Uninterruptible Power Supply

A device that provides a constant regulated voltage output in spite of interruptions of the normal power supply. It includes filtering circuits and is usually used to feed computers or related equipment which would otherwise shutdown on brief power interruptions. Abbreviated UPS.

Varactor / Varicap diode

A semi – conductor diode which operates like a capacitor since its capacitance will decrease with an increase in reverse bias and is also known as a varicap, voltage- variable capacitance diode.

Variable Capacitance diode

See Varicap

Ventilated

Provided with a means to permit circulation of air sufficient to remove an excess of heat, fumes, or vapours.

Volt

The electrical potential difference or pressure across a one ohm resistance carrying a current of one ampere. Named after Italian physicist Count Alessandro Volta 1745-1827. Volt: The basic unit of “electromotive force”. One Volt applied to a resistance of one Ohm will force a current of one Ampere to flow (Abbreviation – V).

Voltage regulation

A process to maintain the terminal voltage within required limits despite variations in input voltage or load

Volt-ampère

A unit of apparent power equal to the mathematical product of a circuit voltage and amperes. Here, apparent power is in contrast to real power. On ac systems the voltage and current will not be in phase if reactive power is being transmitted. Usually abbreviated VA.

VOM

Volt-Ohm meter or Volt-Ohm millimetre.

Watt

A unit of power equal to the rate of work represented by a current of one ampere under a pressure of one volt. Named after the Scottish engineer James Watt, 1819.

Wavelength

The length of one cycle of an AC signal. Determined by Wavelength = c / f where “c” is velocity and “f ” is frequency. The wavelength of a 345Hz audio signal in air is one metre.

Wiring

A distribution network of wire that conducts electricity to receptacles, switches and appliances throughout a building/home to provide electricity where needed.

Xenon

A gas commonly used in flash tubes, HID (High Intensity Discharge) automotive headlamps and cinema projection lamps, and having an intense white light output with a colour temperature close to that of daylight.

Zener diode

A diode used in reverse biased mode as a voltage regulator device. See Voltage Regulator.

0005

Electrical Technology Grade 12

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