D-6 LEARNING GUIDE D-6 ANALYZE ELECTRONIC CIRCUITS

Transcription

1 CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM Level 2 Line D: Apply Circuit Concepts D-6 LEARNING GUIDE D-6 ANALYZE ELECTRONIC CIRCUITS

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3 Foreword The Industry Training Authority (ITA) is pleased to release this major update of learning resources to support the delivery of the BC Electrician Apprenticeship Program. It was made possible by the dedicated efforts of the Electrical Articulation Committee of BC (EAC). The EAC is a working group of electrical instructors from institutions across the province and is one of the key stakeholder groups that supports and strengthens industry training in BC. It was the driving force behind the update of the Electrician Apprenticeship Program Learning Guides, supplying the specialized expertise required to incorporate technological, procedural and industry-driven changes. The EAC plays an important role in the province s post-secondary public institutions. As discipline specialists the committee s members share information and engage in discussions of curriculum matters, particularly those affecting student mobility. ITA would also like to acknowledge the Construction Industry Training Organization (CITO) which provides direction for improving industry training in the construction sector. CITO is responsible for organizing industry and instructor representatives within BC to consult and provide changes related to the BC Construction Electrician Training Program. We are grateful to EAC for their contributions to the ongoing development of BC Construction Electrician Training Program Learning Guides (materials whose ownership and copyright are maintained by the Province of British Columbia through ITA). Industry Training Authority January 2011 Disclaimer The materials in these Learning Guides are for use by students and instructional staff and have been compiled from sources believed to be reliable and to represent best current opinions on these subjects. These manuals are intended to serve as a starting point for good practices and may not specify all minimum legal standards. No warranty, guarantee or representation is made by the British Columbia Electrical Articulation Committee, the British Columbia Industry Training Authority or the Queen s Printer of British Columbia as to the accuracy or sufficiency of the information contained in these publications. These manuals are intended to provide basic guidelines for electrical trade practices. Do not assume, therefore, that all necessary warnings and safety precautionary measures are contained in this module and that other or additional measures may not be required.

4 Acknowledgements and Copyright Copyright 2011 Industry Training Authority All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or digital, without written permission from Industry Training Authority (ITA). Reproducing passages from this publication by photographic, electrostatic, mechanical, or digital means without permission is an infringement of copyright law. The issuing/publishing body is: Crown Publications, Queen s Printer, Ministry of Citizens Services The Industry Training Authority of British Columbia would like to acknowledge the Electrical Articulation Committee and Open School BC, the Ministry of Education, as well as the following individuals and organizations for their contributions in updating the Electrician Apprenticeship Program Learning Guides: Electrical Articulation Committee (EAC) Curriculum Subcommittee Peter Poeschek (Thompson Rivers University) Ken Holland (Camosun College) Alain Lavoie (College of New Caledonia) Don Gillingham (North Island University) Jim Gamble (Okanagan College) John Todrick (University of the Fraser Valley) Ted Simmons (British Columbia Institute of Technology) Members of the Curriculum Subcommittee have assumed roles as writers, reviewers, and subject matter experts throughout the development and revision of materials for the Electrician Apprenticeship Program. Open School BC Open School BC provided project management and design expertise in updating the Electrician Apprenticeship Program print materials: Adrian Hill, Project Manager Eleanor Liddy, Director/Supervisor Dennis Evans, Laurie Lozoway, Production Technician (print layout, graphics) Christine Ramkeesoon, Graphics Media Coordinator Keith Learmonth, Editor Margaret Kernaghan, Graphic Artist Publishing Services, Queen s Printer Sherry Brown, Director of QP Publishing Services Intellectual Property Program Ilona Ugro, Copyright Officer, Ministry of Citizens Services, Province of British Columbia To order copies of any of the Electrician Apprenticeship Program Learning Guide, please contact us: Crown Publications, Queen s Printer PO Box 9452 Stn Prov Govt 563 Superior Street 2nd Flr Victoria, BC V8W 9V7 Phone: Toll Free: Fax: crownpub@gov.bc.ca Website: Version 1 New, August 2011 Corrected, November, 2012

5 LEVEL 2, LEARNING GUIDE D-6: ANALYZE ELECTRONIC CIRCUITS Learning Objectives: Learning Task 1: Describe the operation of single-phase AC rectifier circuits Self-Test Learning Task 2: Describe the operation of filters for rectifier circuits Self-Test Learning Task 3: Determine values for rectified power supplies Self-Test Learning Task 4: Describe the features of the bipolar-junction transistor Self-Test Learning Task 5: Describe the basic applications of the junction transistor Self-Test Learning Task 6: Describe the features of field-effect transistors Self-Test Answer Key CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 5

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7 Learning Objectives D-6 Learning Objectives The learner will be able to describe the application of diodes in rectifiers. The learner will be able to analyze single-phase rectifier circuits. The learner will be able to analyze AC electronic circuits that utilize BJTs. The learner will able to describe operating principles of FETs. The learner will be able to analyze electronic circuits that utilize FETs. Activities Read and study the topics of Learning Guide D-6: Analyze Electronic Circuits. Complete Self-Tests 1 to 6. Check your answers with the Answer Key provided at the end of this Learning Guide. Resources You are encouraged to obtain the following textbook for supplemental learning information: Solid State Fundamentals for Electricians by Gary Rockis; American Technical Publishers Inc. CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 7

8 Learning Objectives D-6 BC Trades Modules We want your feedback! Please go the BC Trades Modules website to enter comments about specific section(s) that require correction or modification. All submissions will be reviewed and considered for inclusion in the next revision. SAFETY ADVISORY Be advised that references to the Workers Compensation Board of British Columbia safety regulations contained within these materials do not/may not reflect the most recent Occupational Health and Safety Regulation. The current Standards and Regulation in BC can be obtained at the following website: Please note that it is always the responsibility of any person using these materials to inform him/herself about the Occupational Health and Safety Regulation pertaining to his/her area of work. Industry Training Authority January CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2

9 Learning Task 1: Describe the operation of single-phase AC rectifier circuits A rectifier (sometimes called a converter) is a device that changes alternating current to direct current. In an AC circuit, the current flows in both directions as it alternates. A rectifier (such as a semiconductor diode) allows the current to flow only in one direction. DC is typically thought of as smooth, unidirectional current. The DC output of rectifier circuits is unidirectional, but it pulsates. It requires special devices called filters to make it smoother. Half-wave and full-wave DC The waveforms obtained from AC rectification can be either half-wave or full-wave DC. The way this occurs will be explained later in this Learning Task. When only half of the AC input is converted to unidirectional DC output, it is called halfwave rectification. The resulting waveform is illustrated in Figure 1a. When both alternations of the AC waveform are converted to unidirectional DC output, then the DC is referred to as full-wave rectification. This waveform is illustrated in Figure 1b. Figure 1 Rectifier waveforms Full-wave DC has an advantage over half-wave DC in that both alternations of the AC input can provide power output to a DC load. That is, it has double the power capability. Half-wave rectifier circuits A half-wave rectifier circuit is constructed simply by connecting a diode in series with an AC load. Half-wave rectification occurs because: For half a cycle of AC, the diode is forward biased, and conducts current to the load. For the alternate half cycle (reverse polarity), the diode blocks current flow. CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 9

10 Learning Task 1 D-6 The output DC waveform is as shown in Figure 1a. Figure 2 Half-wave rectifier circuit with waveforms If an oscilloscope is connected to simultaneously display the waveforms of the input and the waveforms across the diode and the load resistor, the shapes are as illustrated in Figure 2. Notice that: For one half-cycle when the diode conducts, there is practically no voltage drop across the diode, but voltage appears across the resistor. For the alternate half-cycle when the diode blocks current, there is no voltage across the load, but there is a voltage drop across the diode. The flat-line portion of the waveforms represents zero volts. For the half-cycle when the diode is forward biased, electrons flow from negative to positive through the circuit components. This is illustrated by the current-direction arrows in Figure 2. The polarity of the load resistor during this half-cycle is also identified. Reversing the line-load connections to the diode also reverses the polarity of the voltage across the load. Refer to the diagram in Figure 3 and compare it to Figure 2. Figure 3 Reversed diode connections 10 CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2

11 Learning Task 1 D-6 Peak values and PIV rating The peak value of the AC input appears across the load for one half-cycle and across the diode for the other half-cycle. Consequently, the PIV (Peak Inverse Voltage) rating for the diode must be greater than the peak value of the AC input. Example 1 If the AC input to the preceding rectifier circuit is 120 V, then the diode must have a PIV rating greater than the peak value of the rms voltage: PIV rating = = 170 V Full-wave rectifier circuits In single-phase AC circuits, full-wave rectification can be achieved by using two different circuit configurations: the biphase and bridge types. Full-wave rectification (biphase type) One method of full-wave rectification uses a centre-tap, (double-wound) transformer and two diodes. This circuit configuration is sometimes called a biphase, full-wave rectifier, and is connected as illustrated in Figure 4. Figure 4 Full-wave rectifier circuit For one half-cycle of AC on the secondary of the transformer, diode D 1 is forward biased and conducts current to the load (as indicated by the solid arrows). During this same time period, diode D 2 is reverse biased and blocks any current flow. For the next half-cycle of AC, D 2 is forward biased and conducts current to the load (as indicated by the broken arrows). During this time period, D 1 is now reverse biased and blocks any current flow. Caution: The two diodes are common-connected on one side of the load. A reversal of either diode results in a short circuit. CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 11

12 Learning Task 1 D-6 Peak values and PIV rating (biphase) Note that in determining load polarity, the anode (+) side of each diode connects to the negative terminal of the load. In essence, the two diodes alternately switch the load from line to neutral for each half-cycle of AC input. The maximum value of the AC waveform appearing across the load is one-half of the full secondary (L 1 L 2 ) voltage of the transformer. In the half-cycle when diode D 1 is conducting (neglecting the forward voltage drop of D 1 ): The load is connected across the (L 1 N) voltage. At this same instant of polarity, diode D 2 (which is blocking) is connected directly across the (L 1 L 2 ) voltage. For the next half-cycle of AC, diode D 2 is conducting from (L 2 N): The load is connected across the (L 2 N) voltage. Diode D 1 now blocks the full (L 1 L 2 ) voltage. Therefore, the PIV rating of each diode must be greater than the maximum value of the full secondary voltage (L 1 L 2 ) of the transformer. In other words, the PIV rating for each diode must be more than twice the peak voltage appearing across the load. Example 2 If the secondary line-to-neutral voltage of the transformer is 120 V, then the peak value of voltage appearing across the load is: = 170 V However, since each diode must block the peak value of the line-to-line voltage, each diode must have a PIV rating of: = 340 V Full-wave rectification (bridge type) Full-wave rectification can also be achieved by connecting four diodes in a bridge configuration, as shown in Figure 5.,, Figure 5 Full-wave bridge rectifier 12 CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2

13 Learning Task 1 D-6 The bridge rectifier is constructed so that for each half-cycle of AC input, there are two forwardbiased diodes connected in series with the load. The other two diodes are connected so that they are reverse biased at the same instant. Figure 6 shows how each pair of diodes conducts load current for each half-cycle of AC input to a bridge rectifier. To remember how to connect diodes in a bridge-rectifier circuit, it may help to visualize the two current paths through the diode pairs as shown. Figure 6 Conducting pairs of diodes Another way to illustrate a bridge rectifier circuit is shown in Figure 7. Although this circuit appears different, the connections are really the same as those shown in Figure 5. Compare the current paths illustrated with those in the preceding diagrams.,, Figure 7 Alternative schematic for a bridge rectifier Notice again that the cathode ( ) side of the rectifier diodes is the positive terminal with respect to the load, and that the anode (+) side of the diodes is the negative terminal with respect to the load. This is true for all rectifier circuits. CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 13

14 Learning Task 1 D-6 Semiconductor manufacturers build bridge rectifiers in compact phenolic packages with four terminals for circuit connections. For easy identification, these terminals are sometimes colourcoded. The two terminals marked yellow are for the AC input. One terminal marked red is the positive load terminal, and one terminal marked black is the negative load terminal. Study Figures 5 and 7 carefully. Caution! When you are constructing a bridge rectifier using four single diodes, it is important that you make the proper connections. If any one of the diodes is reverseconnected, a short circuit will result. Before connecting the load or the AC supply, test for a short circuit using an ohmmeter connected across the input terminals of the bridge (Figure 8). In whichever direction the leads are connected, the ohmmeter should read infinity. This confirms that none of the diodes are incorrectly connected. Peak values and PIV rating (bridge) In a bridge-rectifier circuit, each diode must have a PIV rating greater than the peak value of AC input. As before, if the AC input is 120 V, then the peak voltage across the load is 120 V = 170 V. Each diode must have a PIV rating greater than 170 V. Figure 8 Ohmmeter test for bridge rectifier Ripple frequency All rectifiers convert an AC input to a pulsating DC output. The amount that the DC output fluctuates above and below its average value is referred to as ripple. The number of output pulses of DC is known as ripple frequency. In half-wave rectifiers, the frequency of the ripple is the same as the frequency of the AC input. For each cycle of AC input, a DC output of one pulse per cycle is produced. A 60 Hz AC input produces 60 pulses of DC output, or a ripple frequency of 60 Hz. 14 CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2

15 Learning Task 1 D-6 Figure 9 Ripple frequency In full-wave rectifiers, the ripple frequency is twice that of the AC input, since there are two pulses per cycle of DC output. A 60 Hz AC input will produce 120 pulses of DC output, or a ripple frequency of 120 Hz (Figure 9). Now complete Self-Test 1 and check your answers. CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 15

16 Learning Task 1 D-6 Self-Test 1 1. Which one of the diodes in the bridge rectifier circuit shown in Figure 1 is incorrectly connected? Figure 1 Bridge rectifier for Questions 1, 2 and 3 2. If the incorrect diode in Figure 1 is changed so that the bridge rectifier is correctly connected and the AC input is 240 V, 50 Hz, what is the: a. peak voltage across the DC load? b. ripple frequency? c. PIV rating required for each of the diodes? 3. If one of the diodes in the preceding bridge rectifier develops an open, what is the: a. peak voltage across the load? b. ripple frequency? 4. A diode tester is constructed using two indicating lamps, as shown in Figure 2. The diode to be tested is connected at points 1 and 2. Determine whether the test diode is open, shorted or okay under the following conditions: a. Both lamps A and B are on. b. Neither lamp is on. 16 CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2

17 Learning Task 1 D-6 c. Only lamp B is on. d. Only lamp A is on. Figure 2 Diode-testing circuit for Question 4 5. A bridge rectifier has its four terminals colour-coded as follows: one is red, one is black and two are yellow. What do these colours mean? 6. In the rectifier circuit shown in Figure 3, terminal A with respect to B is (positive or negative). CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 17

18 Learning Task 1 D-6 Figure 3 Circuit for Questions 6 and 7 7. If the line-to-neutral voltage of the transformer in Figure 3 is 24 V, then the calculated PIV rating for each of the diodes should be V. Go to the Answer Key at the end of the Learning Guide to check your answers. 18 CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2

19 Learning Task 2: Describe the operation of filters for rectifier circuits For many electronic circuit applications, the pulsating DC output of single-phase rectifiers is often not satisfactory. It is usually necessary to provide a filter for these power supplies to smooth out the output DC. Capacitors and inductors are used to construct filter circuits. Capacitor filter The simplest way to reduce the ripple of a rectified output is to use a capacitor filter. This consists of connecting a capacitor in parallel with the load. Figure 1 Capacitor filter with half-wave rectifier As you learned in earlier studies, the capacitor responds to any changes in voltage. Figure 1 shows a filter circuit with a half-wave rectifier: As the load voltage rises from zero, the capacitor quickly charges. As the rectifier voltage drops, the capacitor slowly discharges through the load to maintain a more constant voltage level. During each half-cycle, the capacitor charges to the peak value of the rectified output voltage. As the rectified voltage drops toward zero, the capacitor tends to prevent this change by discharging through the load. The average level at which the output voltage is maintained (while the diode is not conducting) depends mainly on the RC time constant, which is determined by the amount of load resistance and the size of capacitance. If the discharge rate of the capacitor is long compared to the time between the rectifier output pulses, then, instead of falling to zero, the load voltage is maintained at a much higher value. Since the RC time constant determines the rate at which the capacitor discharges, a low value of load resistance results in a more rapid discharge, and the average level of voltage drops. See Figure 2. CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 19

20 Learning Task 2 D-6 Figure 2 Partially filtered DC load voltage Figure 3 illustrates how the amount of load current affects the ripple. Notice that, as the amplitude of ripple increases, the average DC voltage decreases. The change in voltage level with load is known as voltage regulation. Capacitor filters are not normally used where large load currents and good voltage regulation are required. Figure 3 Output voltage waveforms Polarity Normally, electrolytic capacitors are used as filters because of their large capacity. Since they are polarized, it is extremely important to properly connect these capacitors in the circuit. You must connect the positive lead of the capacitor to the positive terminal of the load, and so on. Ratings The capacitor charges to the peak value of voltage. Therefore, the DC working-voltage rating of the capacitor must be greater than the peak AC input. This protects against dielectric failure. You will not need to calculate the proper size of capacitance for filtering action. However, note that too large a value of capacitance results in high inrush currents during the charging periods. (Circuit designers must consider this factor.) 20 CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2

21 Learning Task 2 D-6 When using a capacitor sized for maximum filtering action, the average DC voltage across the load is equal to the peak AC voltage input to the rectifier circuit. During discharge of the filter capacitor, the rectifier diode must block the alternate half-cycle of AC from the load. As shown in Figure 4a, the capacitor polarity and the line polarity are connected in series, aiding during this period. As a result, the diode must be capable of blocking this total combined voltage (Figure 4b). In other words, the PIV rating of the diode must be more than twice the peak value of AC input. Figure 4 Voltage effects of maximum filtering Using the values listed in Figure 4, you can determine the following: V DC working voltage of the capacitor = E = 120 max V = 170 Average DC voltage (with maximum filtering) = E max = 170 V PIV rating of the diode = 2 E max = 340 V Caution!: Since electrolytic capacitors can maintain a charge for a long time, a voltage may be present after the circuit is switched off. Before working on the circuit, always discharge capacitors through a bleeder resistor. Inductance filters L filters An inductor, or choke, connected in series with a rectifier load can also create a filtering action. These filters are also called choke filters or L filters. As the load current rises, the inductor opposes the increase in current. As the load current drops, the inductor opposes the decrease in current. The output waveform of a half-wave rectifier using a choke filter is illustrated in Figure 5. CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 21

22 Learning Task 2 D-6 - Figure 5 Simple choke filter LC filters Inductors are commonly used with capacitors as LC filters. See Figure 6. Make a choke input filter by connecting an inductor in the input line to a capacitor filter. Make a capacitor input filter by connecting a capacitor across the input side of a choke filter. Figure 6 LC types of filters For the choke input filter, the addition of the inductor reduces the voltage level to which the capacitor will charge. As a result, the voltage level across the load is closer to the rms value of input AC. For the capacitor input filter, the capacitor discharge to the load through the choke also results in the load voltage being closer to the rms value of AC input. Compared to filtering with the capacitor only, the added effect of inductance improves the voltage regulation with changing loads. The typical waveform from an LC filter is shown in Figure 7. Figure 7 Typical waveform of LC filter 22 CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2

23 Learning Task 2 D-6 Pi filter To remove ripple to an even higher degree, a pi filter is commonly used. The construction of a pi filter is shown in Figure 8. As you can see, it is simply a combination of the two types of LC filters described above. It gets its name because it resembles the Greek letter π, pronounced pi. + + Rectifier output Load 1 2 Figure 8 Pi filter The action of the pi filter is as follows: Capacitor C 1 charges to the peaks of the rectified output pulses. Capacitor C 2 charges to a lesser value through the inductor L. C 2 provides filtering at the lesser voltage level. C 1 discharges through L to provide filtering action to the load. The resulting load voltage is at a value close to the rms value of the AC input. There are several arrangements of filter circuits: capacitor only, inductor only, and several inductor-capacitor or even resistor-capacitor combinations. All give slightly different voltage characteristics under circuit-loading conditions. Some are best suited for light loads, others for heavier loads. Filter design considerations involve: Allowable ripple Allowable regulation Output voltage Rectifier peak current limits Load current Now do Self-Test 2 and check your answers. CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 23

24 Learning Task 2 D-6 Self-Test 2 1. Why are electrolytic capacitors generally used as filters? 2. What is the main disadvantage of using only a capacitor for filtering? 3. What is a pi filter? Use the half-wave rectifier circuit shown in Figure 1 to answer Questions 4, 5, 6 and 7. Figure 1 Half-wave rectifier circuit for Questions 4 to 7 4. Capacitor C is an electrolytic type. Is terminal 1 the red or black lead of the capacitor? 5. What is the minimum DCWV rating required for the filter capacitor? 24 CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2

25 Learning Task 2 D-6 6. If maximum filtering is provided, what is the average value of voltage across the load resistor? 7. What is the minimum PIV rating required for the diode? Go to the Answer Key at the end of the Learning Guide to check your answers. CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 25

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27 Learning Task 3: Determine values for rectified power supplies You have learned about different significant values of AC voltage. They are the instantaneous voltage value, E inst ; the effective or root-mean-square value, E rms ; and the average value, E av. These values are calculated from the peak or maximum value as follows: E E E inst rms av = E sin θ max Emax = Emax = 2 2 = Emax = E max π Typically, voltage calculations in rectifier circuits involve solving for the: Peak value of AC, to determine the PIV rating of diodes Average value of the rectified waveform, to determine what a DC meter would indicate Effective (rms) value, to determine the power developed in the circuit Average values in rectifier circuits Since there is an equal quantity of positive and negative instantaneous values in a sine wave, the average value of the wave is zero. Since the waveform of a rectifier circuit is unidirectional, a net average value can be calculated. Note that this average value of a rectified waveform is used ONLY for calculating the readings on DC meters in the rectifier s load circuit. In half-wave rectifier circuits, because current is conducted to the load for only one-half of the time (compared to full-wave rectifier circuits), the average value of voltage (or current) is only half of what it is in a full-wave circuit with the same peak voltage input. Figure 1 Average values in rectifier circuits As illustrated in Figure 1, the average value of rectified DC is approximately: 64% of the peak AC input in full-wave circuits 32% of the peak AC input in half-wave circuits Remember, average values are only used to calculate DC meter readings. CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 27

28 Learning Task 3 D-6 Effective (rms) values in rectifier circuits DC meters indicate average values of voltage and current. However, for power calculations in rectified DC circuits, you must use the effective or rms value of sine-wave voltages and currents. Figure 2 Effective or rms values in full-wave rectifiers Full-wave circuits As far as the load is concerned, a full-wave rectifier simply changes the direction of the AC. All of the instantaneous values are still the same as in the AC. Therefore, the value of voltage (or current) used for power calculations is still the same rms or effective value that you have used in AC circuits. In full-wave rectifier circuits, the rms value of the rectified waveform is 70.7% of the peak or maximum value as shown in Figure 2. In your calculations, you will neglect any power losses in the diodes. Half-wave circuits In half-wave rectifier circuits, current flows for only one half-cycle of AC input. For half a cycle the current is zero. Therefore, when the instantaneous values for one half-cycle are averaged with zero for the alternate half-cycle, the rms value is not the same as for an equivalent full-wave circuit. It calculates as 50% of the peak or maximum value (Figure 2). This value is not very significant. You will be shown an easier way to calculate power in rectifier circuits. Power in rectifier circuits The peak or maximum values of sinusoidal waves are necessary to calculate average and effective values. Therefore, it is sometimes handy to use peak values to also solve for power in rectifier circuits. Full-wave circuits The power formula for full-wave rectifier circuits is derived as follows: P= E I Emax Imax = 2 2 E = rms max rms I 2 max 28 CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2

29 Learning Task 3 D-6 Half-wave circuits In half-wave rectifier circuits, current only flows for half a cycle. Therefore, power is only produced for half a cycle as well. With input values as for a full-wave circuit, the same power is produced for half a cycle, but then zero power is produced for the next half-cycle. As a result, the net power produced over the same time period is one-half of what it would be in an equivalent full-wave circuit. Therefore, the power formula for half-wave rectifier circuits is: E P = max I 4 max For half-wave rectifier circuits, note that you don t have half the voltage for developing power, but rather the same voltage for half the time. Sample calculations You will often be required to solve problems that involve determining the reading on different meters and calculating the power of resistance loads. Example 1 For the rectifier circuit shown in Figure 3, solve for the: Reading on the AC ammeter (A 1 ) Reading on the DC ammeter (A 2 ) Reading on the DC voltmeter (V) Power developed by the load resistor Neglect any diode losses. Solution: The AC input to the rectifier circuit is 120 V rms. Calculate the instantaneous maximum or peak value as follows: E max Erms = V = = 170 V CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 29

30 Learning Task 3 D-6 Figure 3 Full-wave rectifier circuit for Example 1 The DC voltmeter (V) registers the average value of the rectified DC voltage, which you calculate as follows: Eav = Emax = 170 V = 108 V The DC ammeter (A 2 ) registers the average value of the rectified current, use Ohm s law, to calculate it as follows: I av Eav = R 108 V = 12 Ω = 9 A The AC ammeter (A 1 ) indicates an rms current which you calculate as follows: I rms Erms = R 120 V = 12 Ω = 10 A 30 CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2

31 Learning Task 3 D-6 Calculate the power in this circuit in one of the following ways: or E P = max I 2 max 170 V A = 2 = 1200 W P= E I rms rms = 120 V 10 A = 1200 W The next example uses the same value of AC input connected to a 12 Ω resistor, so that you can compare calculated values with the preceding circuit. Example 2 For the rectifier circuit shown in Figure 4, determine the: Reading on the DC voltmeter Power developed by the load resistor Figure 4 Half-wave rectifier circuit for Example 2 Solution: Since the instantaneous maximum (or peak) value of AC input is the same as before, you can calculate the average value as follows: Eav = Emax = 170 V = 54 V Therefore, the DC voltmeter indicates approximately 54 V, or half of what it was in the equivalent full-wave circuit. CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 31

32 Learning Task 3 D-6 Note that if a DC ammeter were placed into the circuit, it would also show a reading of one-half of the full-wave rectifier circuit value: I av Eav = R 54 V = 12 Ω = 45. A Remember, in half-wave rectifier circuits, for one half-cycle the power (I 2 R) developed by the load is the same as in AC or full-wave DC rectifier circuits. But, for the next half-cycle, the power is zero. As a result, the net power is half the value of an equivalent AC or full-wave DC circuit. Since the peak values are the same as in the previous example, you can calculate the power developed in the load as follows: Emax Imax P = V A = 4 = 600 W If there is no diode and AC is applied to the load, then the power developed is 1200 W. However, with the diode inserted, the power is 600 W. For half-wave rectifier circuits, it is sometimes easier to calculate the AC power and then simply divide the result by two. The power in this half-wave rectifier circuit is 600 W. Therefore, you can solve for the value of the effective current, I rms, that produces this power in the 12 Ω resistor by transposing the following formula: 2 P= I R rms I rms = = P R 600 W 12 Ω = 707. A This is the value that an AC ammeter would read if connected into this circuit. If you compare this value of 7.07 A to the instantaneous maximum (or peak) value of A, you can see that it is approximately 50%. 32 CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2

33 Learning Task 3 D-6 Form factor Form factor is a ratio for full- and half-wave rectifier circuits. It compares the AC input to the average DC output. It applies to resistive loads only. Since the value of the form factor varies for different types of rectifier circuits, it is of little practical significance. However, the form factor is a convenient way of converting AC values to average DC values in rectifier-circuit calculations. E Form factor = E rms av For single-phase, full-wave rectifier circuits: E Form factor = E = 111. max max For single-phase, half-wave rectifier circuits: E Form factor = E = 222. max max Example 3 What value of AC input is required to give an average value of 12 V output in a single-phase, bridge rectifier? Solution: Factor for a full-wave rectifier is 1.11, you can solve for the AC voltage value by transposing the formula: E E E rms av rms = 111. = 111. E av = V = V That is, to obtain an average DC output of 12 V on a resistive load in a bridge-rectifier circuit, you would need an AC input of approximately 13 V. Example 4 The AC input to a half-wave rectifier circuit is 53 V. What is the average DC voltage obtained? Since the form factor for a half-wave rectifier is 2.22, you can solve for the value of average DC by transposing the formula: CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 33

34 Learning Task 3 D-6 E E rms E av av = 222. Erms = V = 222. = V That is, an AC input of 53 V in a half-wave rectifier would result in an average output to a resistive load of approximately 24 V DC. Now do Self-Test 3 and check your answers. 34 CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2

35 Learning Task 3 D-6 Self-Test 3 1. The AC input to the bridge-rectifier circuit in Figure 1 is 110 V as indicated by the voltmeter. Neglecting diode losses, calculate the: a. ohmic value of the load resistor to limit the circuit power to 1 kw b. reading on the DC voltmeter (V 1 ) c. reading on the DC ammeter (A 1 ) Figure 1 Circuit for Questions 1 and 2 2. For the preceding question, if diode D 1 burned open, calculate the: a. reading on the DC meters b. power developed by the load resistor 3. For the circuit in Figure 2, if the DC ammeter indicates 10 A and the load resistor is 10 Ω, calculate the: a. reading on the DC voltmeter b. reading on the AC voltmeter c. power developed by the resistor d. minimum PIV rating for each of the diodes CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 35

36 Learning Task 3 D-6 Figure 2 Circuit for Question 3 4. For the circuit in Figure 3, if the instantaneous maximum (peak) value of the AC input is 150 V and the wattmeter registers 1060 watts, calculate the: a. reading on the AC voltmeter b. reading on the DC voltmeter c. reading on the DC ammeter d. ohmic value of the load resistor e. minimum PIV rating required for each of the diodes Figure 3 Circuit for Question 4 36 CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2

37 Learning Task 3 D-6 5. In the circuit in Figure 4, if the peak value of AC input is 150 V, and the wattmeter registers 1060 watts, calculate the: a. reading on the AC voltmeter b. reading on the DC voltmeter c. ohmic value of the load resistor Figure 4 Circuit for Question 5 Go to the Answer Key at the end of the Learning Guide to check your answers. CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 37

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39 Learning Task 4: Describe the features of the bipolar-junction transistor The transistor was one of the first solid-state devices developed and it replaced the vacuum tube. The term transistor was derived from the words transfer and resistor. A variety of transistors have been developed over the years. The term bipolar means that it uses both electrons and holes as current carriers. The PN junction diode is composed of two layers of semiconductor material (one P-type and one N-type). The bipolar-junction transistor (abbreviated BJT) is made up of three layers of semiconductor material. Construction of bipolar-junction transistors In a simple block diagram, the transistor can be visualized as being constructed like a semiconductor sandwich: If two blocks of N-type semiconductor material are separated with a very thin layer of P-type semiconductor material, then an NPN transistor is created. If the two outer blocks are made of P-type semiconductor, and if the separating layer is N-type semiconductor, then a PNP transistor is formed. See Figure 1. Figure 1 Simple transistor construction For simple applications like ohmmeter testing, the two types of transistors can be looked upon as back-to-back diodes (Figure 2). This helps to identify polarity for forward and reverse measurements. CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 39

40 Learning Task 4 D-6 Figure 2 Junction diode analogy for transistors The transistor may be either germanium or silicon. Most transistors use silicon semiconductor doped to produce N- or P-type material. This withstands higher current, temperature and voltage than germanium. Transistor symbols and lead designation The transistor has three leads, each connected to one of the semiconductor regions: The lead to the separating (centre) layer is called the base. One outer lead is called the collector. The other outer lead is called the emitter. Figures 3 and 4 show the symbols for the NPN and PNP transistor leads. Notice that each lead has a unique symbol. Even though the collector and emitter are constructed of the same type of semiconductor material, the emitter (which is doped to a slightly different level) is identified with an arrow. Figure 3 NPN transistor Figure 4 PNP transistor The emitter arrow always points from P-type to N-type material (just like a diode). This identifies the transistor symbol as being either an NPN or PNP type of transistor. The only basic difference between using NPN and PNP styles of transistors is in the polarity of the connections. The circle may be omitted from these symbols; both forms are used in these Learning Guides. 40 CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2

41 Learning Task 4 D-6 Common transistor-case styles Based on their current/voltage ratings, transistors are packaged in different case styles and sizes. Transistors are identified by a standard numbered code such as 2N3719; you can locate these codes in a specification manual. This manual lists the ratings and also identifies the leads of a transistor. Some typical case styles and lead identifications are illustrated in Figure 5. Figure 5 Common transistor case styles Signal transistors Signal transistors are used in applications where power levels are in the milliwatt range. Therefore, their case size can be relatively small and still have effective heat dissipation. Power transistors Larger power transistors may be used in applications with power up to 100 W or more, so they must be designed to dissipate more heat energy. Quite often, transistors are equipped with heat sinks to help reduce their operating temperature. Now do Self-Test 4 and check your answers. CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 41

42 Learning Task 4 D-6 Self-Test 4 1. What is meant by the abbreviation BJT? 2. For each of the two transistor symbols shown in Figure 1 and Figure 2 below: a. Identify the transistor type (NPN or PNP). b. Label the leads. Figure 1 Figure 2 3. Complete the diagrams in Figures 3 and 4 showing how transistor symbols could be visualized as back-to-back diodes. Figure 3 Figure 4 4. What is the distinction between the terms signal and power as applied to transistors? 42 CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2

43 Learning Task 4 D-6 5. As shown in Figures 5 and 6, if an ohmmeter is used to test a good transistor, you would expect the ohmmeter in position A to indicate a (high or low) value of resistance, and the ohmmeter in position B to indicate a (high or low) value of resistance. Figure 5 Figure 6 6. To identify voltage and current ratings for a given transistor, you should look in a. 7. The two semiconductor elements most commonly used to construct transistors are and. Go to the Answer Key at the end of the Learning Guide to check your answers. CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 43

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45 Learning Task 5: Describe the basic applications of the junction transistor In electronics, a simple definition of an amplifier is a device that controls a large current with a very small current. The transistor does exactly this: it doesn t generate current; it merely regulates it. The actual source of current is a DC power supply such as a battery. Although a transistor is often used to amplify AC signals such as audio-frequency signals, it is essentially a DC device. Comparing a transistor with a relay To simplify, you can compare a transistor with an electromagnetic relay: When a small control current is passed through the coil of the relay, a much larger load current can be conducted by its contacts. Refer to the relay circuit in Figure 1. When the coil circuit is energized by the push-button, the relay contacts allow a larger load current to flow through the lamp. Visualize the two current paths from the batteries (V 1 and V 2 ). Figure 1 Simple relay circuit Notice that one lead of the coil circuit is common with one lead of the load circuit. This is not usually done in electrical power circuits, but is typical of transistor circuits. Figure 1 shows: Control circuit lead (b) Load circuit lead (c) Common circuit lead (e) Basically, a transistor circuit operates in the same manner as this simple relay circuit. The transistor has, of course, more advantages. It has no contacts to pit, no coil to burn out and no moving parts to wear down; it is completely solid state. CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 45

46 Learning Task 5 D-6 The circuit in Figure 2 shows simple NPN transistor in semiconductor block form in. Compare the transistor to the preceding relay circuit. Notice that the: Base (b) is the control circuit lead. Collector (c) is the load circuit lead. Emitter (e) is the common circuit lead. Figure 2 Simple transistor circuit Control circuit open With the push-button in the control circuit open, no load current is allowed to flow because the thin layer of P-type material (the base) prevents conduction of any load current through the N-type semiconductor materials. Notice that if the P-type layer were removed, then load current would flow through the N-type regions of semiconductor to complete the load circuit. The base layer of P-type semiconductor blocks conduction through the load circuit, since a P-type semiconductor must be made positive before it will conduct. Control circuit closed If the push-button is pressed, the emitter base (PN junction) is forward biased, and emitter-tobase (control circuit) current now flows. This is where the magic of the transistor occurs! Once electrons have entered the P-type material of the base, they are attracted toward the more positive potential appearing at the collector. (Notice that the battery V 2 in the load circuit is much larger than the battery V 1 in the control circuit.) For each single electron attracted towardthe positive battery terminal connected to the base, typically, one hundred times more electrons are allowed to flow toward the more positive battery point at the collector. When a small emitter-to-base current is established, a much larger emitter-to-collector current can occur. The emitter lead (common) carries both the collector (load) and base (control) circuit currents. 46 CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2

47 Learning Task 5 D-6 For ALL transistor circuits: The base-emitter junction is forward biased (in this case, + to P-type material). The collector is reverse biased (in this case, + to N-type material). As in the relay, once control current is established, load current can flow. However, the transistor is much more versatile than a simple relay. The transistor amplifier Now consider the same transistor circuit with a variable resistor in place of the push-button. The circuit in Figure 3 uses the proper symbol for the NPN transistor. Figure 3 Simple NPN transistor amplifier If the base resistance is set to minimum value, a large base current is permitted to flow. This allows a much larger current to pass through to the collector region and the load circuit. If the base resistance is increased, then less base current flows. This, in turn, results in less collector current being conducted through the semiconductor layers. This is the real advantage of the transistor over the relay. The magnitude of the base current actually controls the amount of collector current that can flow. That is, the amount of base current causes the amount of resistance between the collector and emitter leads to vary. A transistor amplifier circuit can be compared to a water tap. If the tap (base) is opened more, a larger flow of water (collector current) is allowed. If the tap is closed, then less water flows. Like a water tap, base current can control the collector current of a transistor from fully on to fully off. Basic transistor terms and abbreviations An intense study of transistor circuits would require many new terms and abbreviations. However, to identify simple transistor-circuit action, you will need a working knowledge of just a few basic terms. CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 47

48 Learning Task 5 D-6 Knowing the lead designations (collector, base, emitter) of the transistor symbol makes it easy to identify many of the terms and abbreviations for transistor circuits. Figure 4 illustrates the use of common transistor-circuit abbreviations. Symbol Q I b I c I e Rb R c V Rb V Rc V be V ce V cb V cc Usual meaning A transistor Base current Collector current Emitter current Base circuit resistance (control) Collector circuit resistance (load) Voltage drop across base circuit resistance Voltage drop across collector circuit resistance Voltage drop across base-emitter leads Voltage drop across collector-emitter leads Voltage drop across collector-base leads Voltage potential of source or supply Figure 4 Transistor-circuit abbreviations 48 CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2

49 Learning Task 5 D-6 Current relationships Since all of the electrons that enter the emitter must come out of either the base or the collector, the following current relationship is true for all transistor circuits: I e = I b + I c Beta The term beta, the Greek letter β, is important for transistors. Beta defines the transistor s current gain or amplification factor, and is determined by the following equation: β = I Ic b The beta of a transistor expresses the magnitude of collector current controlled by a specific value of base current. For example, if 1 ma of base current results in 100 ma of collector current, then the beta (or current gain) of the transistor is 100. If the value of collector current and the beta of a transistor are known, then you can determine the value of base current. The beta of a transistor is governed by its design. This ratio could be as low as 10, or as high as 500. You can identify the beta of any transistor by using a specifications manual. Sometimes β is listed as h FE. For most simple circuit designs and calculations, a nominal value of β = 100 is used. Alpha Alpha, the Greek letter α, is another term sometimes used in transistor circuits to express current gain. It is determined by the following equation: α = I Ic e Since the emitter current is always greater than the collector current, alpha is always less than 1. Remember that an NPN transistor consists of P-type material sandwiched between two pieces of N-type material. Figure 5 Construction of emitter and collector regions CONSTRUCTION ELECTRICIAN APPRENTICESHIP PROGRAM: LEVEL 2 49