A Sequence of Power Electronics Experiments

Chapter 19 -- A Sequence of Power Electronics Experiments Version 3.01 August, 1997

CHAPTER 19 -- A SEQUENCE OFPOWER ELECTRONICS EXPERIMENTS

19.1 Introduction

19.1.1 Overview

This Chapter lists a series of experiments, with explanatory material, suitable forexploring power electronics in depth. Four sets of experiments are included: a set ofrectifier experiments that includes a laboratory orientation sequence, a set of dc-dcconverter experiments, inverter experiments, and component experiments. An additionalset, directed toward a dc-dc converter design project, is included as well. The Chapterin many ways stands alone as a Laboratory Manual; its inclusion here in the text helpsprovide a complete context for work in power electronics.

19.1.2 Safety information

The experiments discussed here are designed for relatively low power levels ofabout 100 W and below. They therefore can be performed with a minimum investmentin special instrumentation and equipment, and safety concerns are minimized. However,the risks are not negligible, and it is important to make proper preparations and use duecare in performing the work. Safe lab practice is the responsibility of the experimenter.It is not possible to perform completely benign power electronics experiments; damagedcomponents or expensive repairs will be necessary if basic safety rules are not addressed.

It is especially important to be careful when working with spinning motors, andparts which become hot from power dissipation. Even if rugged equipment is chosen,many instruments can be damaged when driven beyond ratings. Please follow the safetyprecautions listed to avoid injury, discomfort, and lost lab time.

• GROUND. Be aware of which connections are grounded, and which are not.The most common cause of equipment damage is unintended shorts to ground.Remember that most oscilloscopes are designed to measure voltage relative toground, not between two arbitrary points.

• RATINGS. Before applying power, check that the voltage, current, and powerlevels you expect to see do not violate any ratings. What is the power expectedin a given resistor or other component? Does the device have polarity, and is itconnected in the proper direction?

• HEAT. Small parts can become hot enough to cause burns with as little as onewatt applied to them. Even large resistors will become hot if five watts or so areapplied.

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P. T. Krein

Chapter 19 of P. T. Krein, Elements of Power Electronics. New York and Oxford: Oxford University Press, 1998. This chapter (c) 1998 Philip T. Krein. All rights reserved. This material may be used for non-profit educational purposes provided that proper credit is given and that the copyright notice is included.


• CAREFUL WORKMANSHIP. Check and recheck all connections beforeapplying power. Plan ahead: consider the effects of a circuit change beforetrying it. Use the right wires and connectors for the job, and keep theexperiment area neat. Avoid manipulating circuits or making changes withpower applied.

• LIVE PARTS. Most semiconductor devices have an electrical connection to thecase. Assume that anything touching the case is part of the circuit and isconnected. Avoid tools and other metallic objects around live circuits. Keepbeverage containers away from the work area.

• Neckties and loose clothing should not be worn when working with motors.Be sure motors are not free to move about or come in contact with circuitry.

The most common cause of trouble in power conversion experiments is impropergrounding practice. Any circuit can have only one reference node, and the circuit isaltered if a probe or tool introduces an external connection. Ratings and polarities arealso common problems. Electrolytic capacitors, for example, often burn or even explodeif connected in reverse. Small resistors do not tolerate excess power levels very long.Neatness is an issue as well. A messy, convoluted circuit is extremely hard to debug.In power electronics, small inductances of wires make a significant difference. A neat,tight package in general operates much better than a layout with long looping wires.

19.1.3 Equipment assumptions

For the purposes of the experiments discussed here, the following minimumequipment is assumed:

1. Oscilloscope (digital preferred), two channels, 60 MHz bandwidth.

2. Function generator, 10 Hz to 2 MHz, sine and square waves.

3. True RMS multimeters.

4. Power supply, 0-24 V, 0-5 A or better.

5. Power supply, +12 V, 0-0.5 A (for control functions).

6. Access to standard ac mains power.

7. FET and SCR control circuits as discussed in Chapter 18, along withtransformers described there.

8. Leads, connectors, various parts, and a breadboard system for circuit prototypes.

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The following additional equipment is very helpful, and is recommended:

1. Magnetic current probe, 0-20 MHz.

2. Digital wattmeter.

3. Power supply, 0-50 V, 0-10 A.

4. Isolation amplifier for oscilloscope channels.

5. Temperature probe for multimeter.

6. Access to three-phase mains.

7. A semiconductor curve tracer.

Be sure to be aware of ground connections or other considerations of the instruments.Many power supplies have ground connections that should be removed before using theequipment with a power converter.

19.1.4 Keeping a laboratory notebook

A proper laboratory notebook is a crucial tool for work in any experimentalenvironment. A notebook used in a research lab, a development area, or on the factoryfloor is probably the most valuable piece of gear in the engineer’s arsenal. The purposeof the notebook is to provide a complete record of practical work. This should help avoidduplication of effort. Human memory is not perfect, but a notebook is a permanentrecord. In industrial practice, the notebook is usually the employer’s property. Manycompanies have specific rules about notebook format and content. The suggestions hereare provided for general guidance, and are not meant to conflict with requirements invarious areas of practice. In general, a notebook should include:

• Diagrams of all circuits used in the lab. The important factors are to be able toreproduce a setup and check for possible errors.

• Procedures and actions. The idea is to provide enough information so that theexperiment could be repeated.

• List of equipment used, especially if unusual items are involved.

• All experiment data. Be sure to include units and scale settings. It is generallygood practice to record data in its most primitive form to avoid errors. Scalingor other calculations can be done later.

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• The date, and the names and signatures of the experimenters.

It is usually permissible to include calculations, observations, and evenspeculations in a notebook, provided these are clearly marked and kept apart fromexperimental data and actual lab work. A bound book with permanent, pre-printed pagenumbers is a good selection for use as a notebook. Loose sheets often get lost or spoiled,and spiral books offer the temptation to tear out unwanted information. Notebook errorsshould be crossed out (not obliterated). A notebook should be kept in ink, consistent withits function as a permanent record. Keeping a complete lab notebook sometimes seemsinconvenient, but in the long run saves a tremendous amount of time and effort.

19.2 Experiment -- Demonstration of Special Equipment.

19.2.1 Introduction

This experiment tests some of the special equipment discussed in Chapter 18.It also provides practice with the general equipment of an electronics lab in preparationfor the full sequence of experimental work. As discussed in Chapter 18, three specialequipment items are proposed to support the full range of power electronics experimentsin a standard laboratory:

• A low-voltage ac mains supply, from a transformer bank, for experiments withrectifiers and other circuits. The suggested arrangement supports both single-phase and three-phase power input. A circuit diagram for the combination wasprovided in Figure 18.4.

• A self-contained SCR control unit. The suggested design contains three siliconcontrolled rectifiers (SCRs) and time delay circuits to produce turn-on controlsignals. The anodes and cathodes are isolated.

• An isolated power FET control box. The FET provides switch action for almostany type of dc-dc converter. A complete pulse-width modulation block createsa PWM switching function, with adjustable frequency and duty ratio. As in theSCR circuit, this means that the terminals of the FET can be connected to anyvoltage level up to the device ratings.

The function of these circuits is to support tests of useful power converters. The internaloperation is important and well worth exploring. The circuits will be considered as "blueboxes" rather than black boxes, meaning that their function is not hidden away.

19.2.2 Demonstration circuits

For this demonstration, both the SCR and the FET will be used in simpleconverter circuits. The SCR set will be used in a controlled full-wave rectifier circuit,

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while the FET unit will be used to form a basic dc-dc converter. These circuits aretypical power electronics applications, not too far removed from simple commercialproducts.

There are two major ways to form a full-wave rectifier. One is the rectifier

Figure l9.1 -- Basic diode and SCR full-wave rectifier circuits

bridge circuit, which converts a single ac voltage source into a full-wave rectifiedwaveform. The second involves a center-tapped transformer as a “two-phase” ac source.This second form requires a more expensive transformer, but needs only two rectifiers.Both are shown in Figure 19.1. As shown in the Figure, SCRs can be substituted for thediodes in a full-wave rectifier. The SCRs are operated half a cycle apart, with anadjustable phase angle delay.

19.2.3 Procedure

Part 1 -- Rectifiers and the SCR box1. Connect the transformer set to single-phase input to provide both positive and

negative outputs. Connect two 1N4004 diodes for full-wave output (anodes tophase A and B output, cathodes in common to load). See Figure 19.2.

2. Connect a load of approximately 50Ω from the common cathode to ground.What resistor power rating is needed?

3. Observe the resistor voltage waveform. Observe diode current and comment.Measure the resistor RMS voltage, RMS current, power, and average voltage.What is the relevance of each?

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4. Remove the diodes, and substitute SCRs labelled A and B from the SCR unit.

Figure l9.2 -- Two-phase diode rectifier test circuit.

Set up the SCR box for two-phase operation. Use a 25Ω load in this case.What should the power rating be?

5. Turn the phase delay control to the lowest setting. Double check all connections,then turn the power on.

6. Observe the voltage waveform across the resistor. Notice how the waveformchanges with changes in the control setting. Again measure RMS voltage, RMScurrent, power, and average voltage, and consider the relevance of each.

Part 2 -- Dc-dc conversion and the FET boxThe FET unit will be used in a simple circuit which converts a dc voltage to a

lower level with minimal power loss. In essence, the output is presented with a rapidswitching of the input. The average output level (the dc portion) is lower than the inputsince the switch cannot be on more than 100% of the time.1. Set up the FET control box as shown in Figure 19.3.

2. Observe the drain-to-source voltage. Turn the unit on. Adjust the output for

Figure l9.3 -- Simple dc-dc converter circuit.

50% duty ratio and about 50 kHz.3. Connect a voltage source of 20 to 30 V to the input. Observe the output

waveform and average voltage. Adjust the duty ratio and notice the change.Adjust the frequency and notice the change. Explain the results.

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Part 3 -- Power semiconductor devices and the curve tracerPower semiconductors are important to the field of power electronics. The curve

tracer is a helpful measuring instrument to characterize these devices. Recall thecharacteristic curves for diodes, bipolar transistors (BJTs), and FETs. Examples are givenbelow. The diode shows exponential V-I behavior. Transistors show a family of curves,dependent on the base or gate input operation. For switching operation, only a smallportion of these curves is relevant. (Which portion would it be?)

The curve tracer is an instrument intended to measure these device curvesdirectly. The BJT, for example, is characterized by a certain collector-emitter voltage,collector current, and base current. Recall that the BJT is basically a current-controlledcurrent source. To measure BJT curves, the curve tracer applies a specified range ofcollector-emitter voltage and collector current values. Several discrete base current valuesare then applied in steps, and the results are displayed on an oscilloscope. The FET isa voltage-controlled current source. The drain and source terminals function very muchlike those of the BJT collector and emitter. The gate must have a voltage applied relativeto the source, rather than a current. In the procedure below, four classes of devices willbe tested on the curve tracer.

1. A rectifier diode (1N4004) is to be tested on the curve tracer. Since the defining

Figure l9.4 -- Voltage-current characteristic curves of major devices.

characteristics are forward voltage and current, the device should be connectedas if its anode were a collector and cathode were an emitter.

2. With the curve tracer socket off, set up the unit for limits of about 1 A and 1 W.The 1N4004 can handle up to 400 V in its off state.

3. Check your connections. Apply power to the socket and observe the diodewaveform. Notice that the typical model of a constant 0.7 V drop has limited

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accuracy. A better model would also include a resistor to model the upwardslope of the curve. What resistance and forward voltage provide the mostaccurate model for your device?

4. A BJT is to be tested on the curve tracer. Let us assume that the device is ageneric, unknown part. To be conservative, limit current to 0.1 A, power to 0.5W, and voltage to 20 V. Connect the part with the socket off.

5. Double check connections. Allow base current of 0.1 mA per step to be applied.Turn on the socket. Adjust as necessary to observe the curves. Keep powerlevels below 0.5 W in any case if a small device is being tested. Which portionsof the curves are useful for switching action? How much base current will beneeded if the device is to be used to switch 0.5 A?

6. A power FET is to be tested on the curve tracer. This power part should belimited to 5 A, 1 W, and 50 V. Connect the part with power off. The drainterminal should be connected as the collector, source should be connected asemitter, and gate should be connected as base.

7. Double check your connection. We want to vary gate-source voltage fromapproximately zero to about ten volts.

8. When ready, apply power to the socket and observe the curves which result.Which portions are relevant for switch action? As with the diode, a resistanceis often used to model the slope of the curves. What value of resistance providesa good model when the gate-source voltage is 10 V?

9. The SCR can be thought of as a diode with an extra gate terminal. Whensufficient gate current flows, the device operates as a diode. Without it, thedevice is off. For the SCR here, limits of 10 A, 2 W, and 100 V are appropriate.To test it, associate the anode with a BJT collector, the cathode with an emitter,and the gate with a base.

10. Double check connections. The gate current can be made as high as 0.2 Awithout damage, but try values around 50 mA to best see the device curves.

11. Consider what curves are expected (e.g. diode when on), then apply power. Arethe results what you anticipated? Is there a family of curves, as for thetransistors?

19.2.4 Study questions

In each experiment, a set of questions to guide further study and reports will beprovided.

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1. In dc output converters, what is the relevance of RMS output voltage? Whatabout average voltage? (Hint: consider the effects on loads which are purelyresistive, and on loads which consist of a large inductor in series with a resistor.)

2. For the simple dc-dc converter tested in Part 1 of the above procedures, explainhow to predict the average value of output voltage from the input voltage, theswitching frequency, and the switch duty ratio (the fraction of the time duringwhich the switch is on).

3. Which portion of the curve tracer device characteristics would you expect to usein a switching application? What would the curve tracer display if a current-controlled ideal switch were to be measured?

19.3 Experiment -- Basic Rectifier Circuits

19.3.1 IntroductionThe objective of this experiment is to provide additional practice with power

electronics measurement techniques. These will be studied in the context of simplecontrolled and uncontrolled rectifier circuits. Properties of simple R-C, R-L, and R-L-Ccircuits are studied in introductory circuit analysis courses. Properties of “D-C” circuits(diode-capacitor circuits) are less well known. Similarly, D-L, D-L-C, and various D-R-xcircuits are not widely studied. The behavior of these circuits provides a practical lookat power electronic converters, both from the standpoint of energy conversion applicationsand from the standpoint of laboratory measurements. This is the focus of the BasicRectifier Circuits experiment. The operation of the SCR will be examined briefly as well.An R-C timing circuit for adjusting the phase delay of an SCR will be constructed andtested.

19.3.2 Basic theory

Consider the simple half-wave rectifier shown in Figure 19.5. The state of thediode depends on the input voltage polarity -- and also on the load. With no informationabout the load, it isnot possibleto predict either the load current or voltage. With aresistive load, the load voltage is zero whenever the diode is off. One simple way to findthe circuit action (even though we already know what the circuit does), is to take atrialmethodapproach as in Chapter 2. Since an off diode cannot block forward voltage, thediode will be on if and only if the input voltage is positive. The current is given inFigure 19.6. All currents and voltages in a diode circuit must be consistent with therestrictions imposed by the diodes. Even a complicated diode circuit combination can beunderstood quickly with the trial method.

Now, look at the inductive load of Figure 19.7. Assume that the inductor islarge. If current is initially flowing in the inductor, the diode is on. Inductor voltage VLwill be positive or negative, depending on the input voltage and the inductor current.

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Figure l9.5 -- Basic half-waverectifier diode circuit.

Since there is current flow, the diode will stayon for some time regardless of the Vin value.If V L = L(di/dt) is negative, the inductor

current will fall, possibly even to zero. The diode must stay on until the current reacheszero.

Capacitive loads act much differently. Consider the capacitor of Figure 19.8,

Figure l9.7 -- Half-wave circuit with inductive load.

Figure l9.8 -- Half-wave circuit with capacitive load.

fully charged and supplying energy to the resistor. As long as Vin < Vout, the diode willbe off. When Vin becomes larger than Vout, the diode must turn on. But then a largeforward current C(dv/dt) will flow until Vin becomes less than Vout. Brief, large currentspikes are a feature of Diode-Capacitor circuits. Such currents are typical of the simpledc power supplies in most cheap electronic equipment.

Diodes offer no control means, and a switch capable of blocking forward voltageis needed to add a control function. The SCR is one provides this function. Nevertheless,it fulfills the necessary forward-conducting bidirectional blocking characteristic. The SCR

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is called alatchingdevice, because the on-state

Figure l9.6 -- Half-wave inputcurrent, resistive load.

is self-sustaining. Once a turn-on command isapplied, the SCR will remain on until theforward current is removed. This means thatthe device is exactly equivalent to a diode onceon.

Consider the half-wave resistive

Figure l9.9 -- Basic SCR half-wave circuitwith resistive load.

circuit of Figure 19.9. Turn-on of theSCR can be delayed to alter thewaveshape. The turn-on delay istraditionally measured in degrees relativeto a full diode waveform. The action ofthis control is not hard to determine for aknown load, since the input waveform is

simply being switched on and off.

19.3.3 Measurement issuesIn power electronics, measurement interpretation is an important ingredient of

obtaining data and coming to conclusions about circuit behavior. Like any electronicslaboratory, we will use voltage waveforms from an oscilloscope, and voltage and currentmagnitudes obtained from meters. Some important additional measurements includecurrent waveforms and phase angles. Many of the waveforms are not sinusoidal, so theactual wave shape is important information.

For average and RMS measurements, a typical laboratory multimeter such as theFluke model 8010A provides a useful basis for discussion. It displays the average valueof the input whenever it is set for dc. When set to ac, this meter directs the input througha capacitor, and computes the RMS value of the ac portion which passes through. Someof the important meter specifications are given in Table 19.1. Many power converterwaveforms contain both dc and ac. The “full RMS” value might be useful in this case.With most multimeters, even "true RMS" units, such as the 8010A, the RMS values forac and dc portions must be measured separately and combined to give the correct result.

Power itself is important. There are growing numbers of manufacturers offeringdigital power meters appropriate for complicated waveforms. One type, which will beused here as an example, is represented by the Valhalla Scientific model 2101 voltage-current-power meter. This instrument samples the incoming voltage and currentwaveforms, and computes RMS values as well as the average power. The current ismeasured by sensing the voltage across a fixed 0.01Ω resistor. A meter configured inthis way need not distinguish between dc and ac waveforms, and can provide accuratepower readings over a wide frequency range. A functional symbol is shown in Figure

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19.10. The voltage is sensed across the input. The output connection forces the current

Table I -- Summary Specifications of the Fluke 8010A Multimeter

Quantity Dc voltage Ac voltage Dc current Ac current Resistance

Measurement

range

1 mV - ±1kV average

1 mV - 750V RMS

1 µA - 10 Aaverage

1 µA - 10 ARMS

1 Ω - 20MΩ

Validfrequencie

s

0 Hz(rejects acabove 45

Hz)

45 Hz - 50kHz (rejects

dc)

0 Hz(rejects acabove 45

Hz)

45 Hz - 20kHz (rejects

dc)

---

Error ±0.1% ofreading, ± 1

digit

±0.5% ofreading ± 2

digits

±0.5% ofreading, ±2

digits

±1% ofreading, ±2

digits

±0.5% ofreading,± 1 digit

Figure l9.10 -- Front panel view of digital power meter, with function block.

to flow through the sensing resistor. An input signal at 0 Hz or 40 Hz-10 kHz will givea true RMS display.

Low-impedance sense resistors can be used for complete current waveforms, butthey can introduce grounding problems in a circuit. Industrial laboratories specializingin power electronics make extensive use of Hall-effect current probes that sense themagnetic fields around conductors. One major drawback is that the probes have aninternal dc offset which might drift as the temperature changes.

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Timing is critical in the operation of most power converters. Since switches arethe only means of control, the exact moment when a switch operates is a key piece ofinformation. The SCR provides a typical example. Consider the waveforms in Figure19.11. A sinusoidal waveform (perhaps the input voltage to a rectifier) serves as a timingreference. It is straightforward to measure the time shift between this wave’s zerocrossing and the turn-on rise of the switched signal below it. The example in the Figureshows two 60 Hz waveforms. The time delayτd can be measured directly from the graphas about 1.25 ms. Since a full 360° degree period lasts 16.667 ms, an angleφd can becalculated fromτd as

One important detail in making this measurement is to make sure both oscilloscope traces

(l9.1)

are triggered simultaneously. The “chop” mode of the scope is useful for this purpose.

19.3.4 Procedure

Figure l9.11 -- Phase delay measurement example.

Part One -- The DiodeIn this first part, a waveform generator is used as the power source so that higherfrequencies and small components can be used for testing. Please use care in connectingand operating this instrument. It is not designed to supply high current.1. Obtain a toroidal or audio transformer. One alternative is a 60 turn primary and

a 60 turn secondary on a Ferroxcube 500XT400-3C8 ferrite toroid core.2. Construct the circuit shown in Figure 19.12. Diodes should be standard 1N4148

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rectifiers, or similar.3. Be careful with ground connections. Set the waveform generator for

approximately 1 kHz output at about 10 V peak. Observe Vout with the scope.

4. Sketch the output waveform under “no load” conditions (the scope probe gives

Figure l9.12 -- Rectifier test circuit.

a slight load).5. Load the bridge with a 500Ω resistor (compute the necessary power rating first).

Observe and sketch the output voltage waveform. Use a current probe or placea 10Ω resistor in series with the transformer output to observe the current outof the transformer. Sketch the current waveform.

6. Connect an inductor (a reasonable value is about 25-50 mH) in series with theresistor. Observe and sketch the resistor voltage waveform and the current intothe bridge. Measure the average resistor voltage with a multimeter.

7. Remove the R-L load. Instead, attach a 47Ω resistor in series with an 0.1 µFcapacitor. Place a 10 kΩ resistor across the capacitor terminals. Observe andsketch the capacitor voltage waveform and the bridge input current waveform.Also measure the average capacitor voltage.

Figure l9.13 -- Rectifier test circuit with RC load.

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Part 2 -- The SCRIn Section 19.2, we saw a simple controlled rectifier example, and the waveforms

which result. The gate control was described as a pulse. In this Section, an R-C delaycircuit for operating an SCR will be examined. It is similar to the circuit used incommercial lamp dimmers.1. Construct the circuit shown in Figure 19.14. The input source is taken from the

25 V transformer set.

2. Use a resistor substitution box or other variable resistor for Rc. Observe the Vg

Figure l9.14 -- SCR rectifier test circuit.

and Vload waveforms for Rc values of 0, 200, 400, 600, 800, and 1000Ω.Sketch the two waveforms for two or three values of Rc. Measure the turn-ondelay by comparing Vg or the gate current with Vload with the oscilloscope setfor line triggering. Also, use a multimeter to measure the average value of Vloadin all cases.


1. For part 1, what waveforms would you expect for R, R-L, and R-C loads? Hint:think in terms of simple filtering.

2. Compare the actual waveforms with those expected. Compute the actual circuittime constants, and discuss how they might affect the waveforms.

3. The R-L and R-C cases of part 1 represent different output filter arrangementsfor a rectifier circuit. Discuss the circumstances under which each of these willbe effective and appropriate.

4. Comment on how the diode forward voltage drop (about 1V) affects thewaveforms.

5. From part 2 data, tabulate and plot the average load voltage vs. the value of Rc.6. Compute the SCR gate circuit R-C time constants. Is the turn-on delay governed

by R-C?

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19.4 Experiment -- Single-phase Rectifiers

19.4.1 IntroductionThis experiment examines the properties of single-phase controlled rectifiers.

Converter concepts such as source conversion and switch types will be illustrated.Popular applications such as battery chargers and dc motor drives will be tested. InSection 19.3, you had a chance to become familiar with the basic action of the SCR, andalso studied a number of non-resistive rectifier load circuits. Now, the SCR boxesdescribed in Chapter 18 will be used to study controlled-rectifier action in more depth.

19.4.2 Basic theory

In a switched converter network, KVL will not allow us to connect the utility acvoltage source directly to a dc voltage source, so a dc current source is needed. Considersome of the loads studied in Section 19.3. A resistive load is a crude example of currentconversion: the incoming voltage produces a proportional resistor current. This is a farcry from an ideal current source, but it does result in a conversion function. An inductiveload is a better conversion example: the inductor voltage VL = L(di/dt) resists any changein current. If L is very large, any reasonable voltage will not alter the inductor current,and a current source is realized. A capacitive load has the opposite behavior. Thecapacitor current iC = C(dv/dt) responds whenever an attempt is made to change thecapacitor voltage. If C is very large, no amount of current will change the voltage, anda voltage source is realized. Many electrical loads, especially motors, are inductive. Asa result, most circuits behave in such a way that current does not change much over veryshort periods of time. In power electronics practice, this behavior, along with the desiredsource conversion function, means that most loads are treated as “short-term” currentsources.

The basic single-phase controlled rectifier is shown in Figure 19.15. If the loadis inductive, there is a KCL problem: the switch cannot be turned off. To see that thisis so, observe what happens to L(di/dt) when an attempt is made to turn the switch off.The current must approach zero almost instantaneously. The value of L(di/dt) is a hugenegative voltage. In practice, this voltage will likely exceed the blocking capabilities ofthe switch, which will be damaged. A simple solution is to provide a second switchacross the load, as shown in Figure 19.16. In simple ac-dc converters, this switch can bea diode, or it can be a second bidirectional-blocking forward-conducting device.

Two typical loads for ac-dc converters are the dc motor and the rechargeablebattery. The dc motor has an inductive model, as shown in Figure 19.17. The motorlooks rather like a voltage source, but its windings provide a significant series inductance.A battery is intended as a good dc voltage source. To provide the desired energyconversion function, something must be placed in series with the battery for currentconversion. A resistor can be used with a loss in efficiency. Alternatively, an inductorcan be used, as in the circuit of Figure 19.18. Analysis of this circuit is not trivial. Thecontrolled switch is on whenever Vin is positive. When it is on, the output current isgiven by

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When the controlled switch is off, inductor current is

Figure l9.15 -- Basic single-phase controlled rectifier.

Figure l9.16 -- Single-phase converter with second switch for KCL.

Figure l9.17 -- Basic dc motor circuit equivalent.

(l9.2)

If L is large, the current is approximately constant. These differential equations can be

(l9.3)

solved to give actual values of current. Charging current is controlled by the resistorvalue and also by the phase delay used to operate the SCR.

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The form of circuit expected for single-phase conversion appears in Figure 19.19.

Figure l9.18 -- Single-phase battery charger with inductive filter.

The switches are implemented with an SCR and a diode. In this case, the output voltageis Vin whenever the SCR is on, and zero when it is off. The SCR turns off automaticallywhen Vin ≤ 0. The voltage waveform is a familiar one: a half-wave rectified waveform,possibly with a delayed turn-on. The voltage has an average value given by

whereα is the angle of delay before the SCR is turned on. This integral can be computed

(l9.4)

Figure l9.19 -- A single-phase ac voltage to dc current converter.

to give

Since the output is a dc current source, average power exists only at dc, and is simply

(l9.5)

Iout·Vout(ave).

19.4.3 Procedure

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Part 1: Ideal R-L load

Figure l9.20 -- Typical output voltage waveforms.

1. Connect phase “A” of the nominal 25 V supply, the SCR labelled “A” in theSCR box, and a resistive load, as shown in Figure 19.21. Estimate the requiredresistor power rating. Record the RMS value of the source voltage.

2. Set the SCR turn-on for zero delay (i.e. set it to act as a diode). Observe the

Figure l9.21 -- SCR test circuit, resistive load.

output and load voltage waveforms (see the figure). Measure the followingoutput parameters and sketch the voltage waveform: Vout(ave), Vout(rms).

3. Repeat #2 for delays of approximately 30˚, 60˚, 90˚, 120˚, and 150˚. Sketch justone or two typical waveforms, rather than the whole series.

4. Place an inductor, for example 25 mH, in series with the resistor, as shown inFigure 19.22. Observe the resistor voltage waveform. Repeat #2 and #3 abovefor this new circuit, and also record the average and rms values of Vload.

Part 2 -- A battery load1. Connect a rechargeable battery in the circuit of Figure 19.23.2. Measure battery currents Iout(ave)and Iout(rms). Do this for, and record the phase

delay angle value, at least four different SCR delay angles. Observe and sketchthe resistor current waveform at one intermediate SCR delay angle.

3. Measure the input power from the ac source.

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Figure l9.22 -- SCR test circuit, series R-L load.

Figure l9.23 -- SCR test circuit with battery charger load.

1. For the simple loads of Part 1, tabulate and plot Vout(ave)and Vout(rms) vs. theSCR phase delay angles. Compute a theoretical result, and compare it to thedata. Do these agree?

2. For the battery load in Part 2, tabulate and plot Iout(ave) and Iout(RMS) vs. theSCR phase delay angle. Again consider whether your results are consistent withtheoretical expectations.

3. Why is the “flyback” diode included in these circuits?4. Compute the efficiency of the battery charger studied here.5. Comment on how the diode and SCR forward voltage drops affect your results.

19.5 Experiment -- Polyphase rectifiers

19.5.1 IntroductionThis experiment will examine the properties of ac-dc converters with polyphase

input voltage sources. The midpoint converter will be the focus of this experiment, andwill be tested with inductive loads including a dc motor. In the preceding experiment, thecircuits had a single SCR in a half-wave rectifier configuration. A flyback diode wasneeded to provide a current path when the SCR turned off. The simple half-wave circuittransferred energy to the load no more than half of the time. The half-wave circuit seems

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ineffective. Polyphase sources offer a much better alternative.A three-phase source, the most common polyphase type, has three voltages equal

in amplitude and 120° apart. The definition of polyphase sources also applies to the “two-phase” case -- two voltages, each 180° apart. While this is consistent, a confusinghistorical nomenclature has arisen. In communication systems, a “quadrature” voltagesource is a set of two signals, spaced by 90°-- a sine and a cosine. Quadrature sourceswere used in many early power systems, and gave rise to an unfortunate use of the term“two-phase” for such a source. We will use the term two-phase to refer to a set of twovoltages 180° apart rather than to a quadrature set. In this experiment, the SCR boxeswill be used to study controlled-rectifier action with two and three-phase sources.

19.5.2 Basic theory

When multiple input sources are available for ac-dc converters, it is natural touse all of them. The circuit of Figure 19.24 shows the most general such converter -- anm-phase to dc converter. The experiment here considers a simplified version of thisconverter -- the one in which there is a common neutral connection between input andoutput. This is called amidpointconverter, and appears in Figure 19.25.

Figure l9.24 -- Polyphase ac to dc converter.

In the midpoint converter, the KVL and KCL restrictions require that no morethan one switch may be on at any time, and one switch must be on if the load current isnot zero. This implies that qi ≤ 1. If the load is a current source, qi = 1. We want

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to operate the switches so that the dc value is maximized and the unwanted ac

Figure l9.25 -- Polyphase midpoint ac-dc converter.

components are minimized. It can be shown that if the switching frequency is chosen toequalωin, the best choice of switching functions is to follow the polyphase input: eachswitch is on 1/m of the time, and switching functions are spaced 360°/m apart. The dcoutput component can be controlled by adjusting the phase of the switching functions.The output voltage is given by

This gives a complicated series, in the form

(l9.6)

The dc component appears in the n=1 term, and is given by

(l9.7)

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This, in turn, can be simplified to give

(l9.8)

Notice that Vout(ave)depends onφ, whereφ is defined as the delay angle between any

(l9.9)

voltage and the switching function associated with it. The result is a controlled rectifier.

19.5.3 Procedure

Part 1: Two-phase converter1. Set up the 25 V 60 Hz supply for two-phase output. Remember to connect the

output neutral.2. Connect the SCRs labelled “A” and “B” in the SCR box with a resistive load,

as shown in Figure 19.26. Estimate the required resistor power rating, and abideby it. Set the SCR box for two-phase control.

3. Set the SCR turn-on for zero delay (i.e. set it to act as a diode). Observe the

Figure l9.26 -- Two-phase SCR test circuit.

output voltage waveform. Record Vout(rms)and Vout(ave)and sketch the outputvoltage waveform.

4. Repeat #3 for delays of approximately 30˚, 60˚, 90˚, 120˚, and a delay close to180˚. Sketch just one typical waveform, rather than the whole series.

Part 2 -- Three-phase converter1. Connect the 25 V 60 Hz supply and SCR box for three-phase operation.2. Apply 120/208 V 3φ power to the three-phase setup.3. Set the SCR turn-on for zero delay (i.e. set it to act as a diode). Observe the

output voltage waveform. Measure Vout(ave) and Vout(rms), and sketch thewaveform.

4. Repeat #3 for delays of approximately 30˚, 60˚, 90˚, and 120˚. Sketch just onetypical waveform, rather than the whole series.

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5. Repeat #3 with a dc motor as the converter load, except set the delay at about

Figure l9.27 -- Three-phase SCR converter test circuit.

90˚ initially. Observe and sketch the motor current and voltage waveforms withno motor shaft load.

6. Add a moderate shaft load, and observe how the current and voltage waveformschange. Record your observations.


1. For each of the load and source combinations, tabulate and plot Vout(ave)andVout(rms) vs. the SCR phase delay angle. What would you expect in theory?How well do your results agree with the theory?

2. For one phase delay angle (pick 60°, for instance), plot Vout(ave)and Vout(rms)vs.the number of input phases (you have data for one, two, and three). What doyou expect to happen when more phases are used?

3. Why is the flyback diodenot included in these circuits?4. A bridge converter implements the full switch matrix of Fig. 19. . How would

Vout(ave)be affected by the use of a bridge rather than a midpoint converter?

19.6 Experiment -- One-Quadrant Dc-Dc Conversion

19.6.1 Introduction

In this experiment, some common one-quadrant dc-dc converters will beimplemented and tested. The main tool of the experiment is the FET switch control unit.The circuitry provides direct adjustment of the switching function frequency and dutyratio. The FET box can be used to generate any of the four common dc-dc one- quadrantconverters (buck, boost, buck-boost, or boost-buck). In this experiment, we will set uptwo of these circuits and characterize their operation. The converters we will build are

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similar to commercial applications, with fast switching functions and switchingfrequencies as high as 200 kHz.

19.6.2 Basic theory

Dc-dc converters can be depicted as providing energy transfer between two idealsources. In practice, one of the sources is almost always implemented as an electricalenergy storage element. For example, the buck converter stores energy in an inductorwhen the controlling switch is on, and discharges that energy through the load when thecontrolling switch is off. An objective is to keep energy flow into the load nearlyconstant as the switches operate. Since most converters apply nearly constant voltagesor currents to the storage elements, much of the analysis is straightforward.

The general dc-dc converter is shown in Figure 19.28. The four major one-

Figure l9.28 -- General dc-dc converter matrix.

quadrant converters are shown in Figure 19.29. Notice that the indirect converters (buck-boost and boost-buck) are shown as two matrices connected together. In each of thesecircuits, practical current sources are formed with inductors. Output voltage sources areformed with capacitors. The average values, governed by the switching function dutyratios, are of interest. For example, the buck converter below has Vout = q1Vin, whichis just

The average value of this series is simply D1Vin. Unwanted Fourier components appear

(l9.10)

at the switching frequency fswitch and its multiples. The largest unwanted component is

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In general, as switching frequencies increase, the values of dv/dt and di/dt values also

Figure l9.29 -- Major one-quadrant dc-dc converters.

(l9.11)

increase, and smaller inductors and capacitors can be used without sacrifice inperformance.

19.6.3 Procedure

Figure l9.30 -- The buck converter.

Part 1: Buck converter1. Set up the FET control box as a buck dc-dc converter. Be sure to provide the

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capacitor shown across the input power supply. The capacitor should be placedas close to the FET box as possible so that the inductance of the wires to theFET will be very low.

2. Set the power supply current limit for about 3 A, and the voltage to 12 V. Set

Figure l9.31 -- Buck converter test circuit.

the duty ratio to about 50%. Observe the load voltage, Vload, and the outputvoltage Vout.

3. Turn on the FET box and the input power supply.4. Set fswitch to about 100 kHz. Notice that Vout allows easy measurement offswitch.5. Confirm that the duty ratio is close to 50%. Sketch the Vout and Vload

waveforms.6. Use the oscilloscope to measure the peak-to-peak ripple on Vload at duty ratios

of 10%, 50%, and 90%. The ripple measurement can be performed by setting theoscilloscope input coupling to “ac,” and expanding the voltage scale.

7. Measure average values of Vload and Iin at duty ratios of 10%, 30%, 50%, 70%,and 90%. Use your multimeter for Vload(ave). Record the input and output RMSvoltage and current as well as the power.

8. Change fswitch to 5 kHz. Try to measure the waveform time constant during thevoltage fall. Recall that this will be the time for Vload to fall from its peak valueto 36.8% of that value.

Part 2 -- Buck-boost converter1. Set up the FET control box as a buck-boost dc-dc converter, as shown in Figure

19.32.2. Set the power supply current limit for about 3 A, and the voltage to 12 V. Set

the duty ratio to zero. Connect oscilloscope leads across the load resistor andacross the inductor.

3. Turn on the FET box and the input power supply. Set fswitch to about 100 kHz.4. Confirm that the duty ratio is set to about 50%. Sketch the inductor voltage

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waveform. Use the current probe to sketch the inductor current waveform.

Figure l9.32 -- Buck-boost test circuit.

Observe and sketch the load voltage waveform.5. Measure average and RMS values of Vload and Iin at duty ratios of 20%, 40%,

50%, 60%, and 70%. Record power readings as well.The output voltagebecomes high quickly with D>60%, so be careful.


1. Tabulate your data in an organized fashion. Compare the RMS readings fromthe wattmeters with the various average readings. Do they agree?

2. Compute and tabulate ratios of Vload(ave)/Vin for these converters for your data.Are the results consistent with duty ratio settings?

3. Estimate theaverageinput and output power from theaveragereadings of Vloadand Iin for each operating condition. Compare these results to the wattmeterreadings. Calculate efficiency, Pout/Pin, from the wattmeter readings.

4. Estimate the inductor value from the measurements of fall time in the buckconverter. Use this value to compute an expected load voltage ripple at 100kHz. How do your results compare with the data? If the inductor value wereto double, how would this affect the behavior of these circuits?

19.7 Dc-Dc Converters for Motor Drives

19.7.1 Introduction

This experiment will examine the operation of dc-dc converters in the context ofmotor drives. Motor drives are the most common application of multi-quadrant dc-dcconverters. Dc power supplies are almost always based on one-quadrant circuits: theoutput power is intended to be positive at all times, and there is rarely a need to handlenegative output current. Some special-purpose laboratory supplies are designed to handlemulti-quadrant output, but often do not have efficiency as a high priority.

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Large dc motor drives have energy considerations beyond those we have seen sofar: the rotating kinetic energy of a large motor is quite considerable, and in a brakingsituation, this energy must be removed. The energy can be converted to heat and lost, asit is in gasoline engine vehicles, or it can be directed back to the energy source forrecovery. Thisregenerationenergy can be controlled only with multi-quadrant circuits.

Figure l9.33 -- The dc motor equivalent circuit with mechanical model.

19.7.2 Basic Theory

The equivalent circuit of a dc motor is repeated in Figure 19.33. Rotation of themotor produces aback EMF, equal to kωif, where k is some constant,ω is the motorshaft speed in rad/s, and if is the “field exciting current” or some other variablerepresenting the magnetic field strength inside the motor. If a voltage Vt is applied to themotor terminals, a current (Vt − Vg)/Ra will flow in the steady state. This current sendspower into the motor, which is converted to mechanical energy. The input poweraccelerates the motor until the electrical input power exactly balances any mechanicalinput into a shaft load.

The power input, and hence the operating speed, of a dc motor can be controlledeasily by altering the value of Vt. Since the actual output power is into some mechanicalload, with its associated inertia,ω will not change much over short time periods, and Vgwill be nearly constant. The average electrical power into the motor will be determinedby the average value of Vt, even if the inductance is low. For example, a simple buckconverter could be used to control Vt, as in Figure 19.34. The output can vary from 0to 100% of the input dc source voltage, and thus speed can be altered from near 0 to nearrated speed.

If i f is held approximately constant, as it is inseparately excitedmotors ormotors with permanent magnets, the input power is a very direct function of converterduty ratio, and speed control is made easy. In fact, many converters have feedbacksystems built in for this purpose. Consider the unit of Figure 19.35 in which the transistorduty ratio depends on a low-power input voltage. This voltage could be generated asshown in the figure, with some reference signal being compared to information from atachometer. In this unit, the duty ratio automatically increases if the motor speed drops.This results in additional input power and is intended to keep running speed nearly

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constant. The operating speed can be changed easily by adjusting the reference voltage,

Figure l9.34 -- Buck converter in a dc motor drive application.\

Figure l9.35 -- Dc-dc converter set up for speed control.

and will be held constant at the desired value.In the case of a very high power motor drive, a considerable amount of energy

is represented even during a speeddecrease. An excellent example is the braking of anelectric vehicle -- an operation which must occur often in any transportation application.In mechanical systems, the braking energy is converted to heat, dissipated, and lost. Thisis not an acceptable situation in battery-powered electrical units, and causes substantialwear and tear even when batteries are not an issue. Multi-quadrant dc converters can beused to convert the braking energy back into electrical form. The energy can bedissipated just as it would be by mechanical brakes, or it can be returned to the energysource. The dissipation version is calleddynamic brakingand is often used for very rapiddeceleration of electric motors. The recovery version is calledregenerative braking, orjust regeneration,since it uses the motor as a generator during deceleration.

The buck converter is sometimes called a class-A chopper when used as a motordrive. As we have seen, this converter makes a useful, simple motor control. Whenbraking, the controlling switch turns off, and only the diode conducts. If inductance islow, the output current quickly goes to zero, and the motor coasts to a stop. If inductanceis high, the braking energy is dissipated in Ra. Figure 19.36 shows some typical

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waveforms.

A boost converter can be used to return generated energy from a motor to a

Figure l9.36 -- Buck converter current waveforms.

source, provided that the source voltage Vt is less than Vg. Such a converter, shown inFigure 19.37, can provide regenerative or dissipative braking. It is sometimes referred toas a class-B chopper.

A motor could be attached to both a buck and a boost converter, with the buck

Figure l9.37 -- Boost converter for braking energy recovery.

converter operating (with Ia > 0) during “motoring,” and the boost converter operating(with Ia < 0) during regeneration. Such a unit is called a class-C chopper, and is widelyused for dc motor drives.

To operate a class-C chopper, the boost converter active switch is shut off, andthe buck converter is used to accelerate and provide running power to the motor. Whenregeneration is desired, the buck converter is shut off until Ia becomes negative. At thatpoint, the boost converter can begin operating, and will transfer energy from the motorback to the source. In effect, the buck and boost converters operate independently.

The general dc voltage to dc current converter shown in Figure 19.39 is notespecially useful for dc motor control, since the inductance value is usually too low for

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useful regeneration. The converter cannot provide negative steady-state voltage to the

Figure l9.38 -- Class-C chopper circuit.

Figure l9.39 -- Two-quadrant dc-dc converter.

motor, and so is no more useful than the class-C chopper above. It is called a class-Dchopper. In many applications, it is necessary to reverse the motor direction, whilemaintaining both motoring and regeneration. Two class-C choppers, one for each motorVt polarity, can be assembled for this purpose. The resulting class-E chopper is shownin Figure 19.40.

19.7.3 Procedure

Part 1: Class-A motor drive1. Obtain a small dc motor. Use your multimeter to measure Ra for it.2. Set up the FET control box as a buck dc-dc converter, with a dc motor as the

output load. Be sure to provide the capacitor and resistor shown across the input

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power supply. The resistor is intended take up any regenerated energy.

Figure l9.40 -- Class-E chopper circuit for bidirectional dc motor drive.

3. Set the power supply current limit for about 5.0 A, and the voltage to 24 V

Figure l9.41 -- Class-A chopper test circuit.

(depending on the motor ratings). Set the duty ratio dial to zero. Connect anoscilloscope lead across the motor. Set up the current probe to measure themotor armature current.

4. Turn on the FET box and the input power supply. Set fswitch to about 20 kHz(a period of 50 µs). Monitor and record the power supply current and voltagelevels and shut down if anything unexpected appears.

5. Sketch the motor terminal voltage waveform and current waveform with a dutyratio setting of about 50%. If there are intervals during which Ia = 0, takespecial care to record the value of voltage during such intervals. Measure thepeak-to-peak ripple on Ia.

6. Observe and sketch the motor voltage and current waveforms, and measure themotor average voltage and current, at: (a) the lowest duty ratio for which the

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motor runs steadily, and (b) a duty ratio of about 90%. In each case, observequalitatively the effects of shaft load on the current waveform.

7. Change fswitch to 2 kHz. Measure the time constant of Vt during its fall. Thiswill allow you to estimate La later.

Part 2 -- Class-C chopper circuit1. The two active switches of the class-C chopper operate separately, but they must

act in complement. In this part of the experiment, it will be necessary to usespecial care to insure proper switch operation.

2. Use two FET control boxes. One, designated themasterunit, will determine theswitching functions of both portions of the class-C chopper. The second,designated theslaveunit, will function only as requested by the master unit. TheFET box design in Chapter 18 has a switch to select this function: up for masterand down for slave. Connect the control terminal on the master unit to thecontrol terminal on the slave unit. Adjust the slave duty ratio to the 100 %position. The frequency controls should be in approximately the same position.

3. Wire the circuit shown in Figure 19.42. The 0.2Ω resistor will help preventdamage if the switching functions are not operating exactly in complement.

4. Operate the master unit as a buck converter. Observe operation at approximately

Figure l9.42 -- Class-C chopper test circuit.

50% duty ratio, as in part 1. Sketch the motor voltage and current waveforms,and record the average motor voltage and current. Measure the peak-to-peakripple on Ia.

5. Increase the duty ratio to near the maximum amount, then drop the duty ratioabruptly in an attempt to observe converter action during regeneration. You mayneed to repeat this step, or spin the shaft externally by hand or with anothermotor to observe regeneration behavior (we’ve used an electric drill in the past).Make note of your observations. Try to obtain a situation in which Ia reverses.


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1. During intervals when Ia = 0, the motor voltage is not necessarily zero. Interpretits value during such intervals. (Hint: Since Ia = 0, this is an open-circuitvoltage.)

2. Compute average power into the motor for each operating condition.3. Estimate the motor series inductance value from the buck converter data. Use

this to estimate the voltage ripple at 20 kHz. Does this match yourmeasurements?

4. Is the average motor voltage determined by the duty ratio?5. Why is the class-A chopper incapable of regeneration? What will happen in

such a converter if Vt is suddenly lowered?

19.8 Dc-Ac Conversion -- Voltage-Sourced Inverters

19.8.1 Introduction

This experiment examines inverter circuits. Inverters, or dc-ac converters, areused for generation of backup ac power, for ac motor drives, and for switching amplifiers.Commercial inverters are also used in alternate energy applications such as thedemonstration photovoltaic and battery power plants. Various simple types of inverterswill be studied in this experiment. Emphasis will be placed on phase control techniquesand on simple switching functions. Inverters can be classified according to the propertiesof the dc source. Those with a dc current source are often used for sending energy intoan ac power system. Dc regeneration and HVDC transmission are two examples. Suchinverters are really the same as ac-dc converters with ac voltage inputs. In effect, theonly difference between rectifiers and inverters when dc current sources are involved isthe direction of any unidirectional switches used in the circuits. The ac-dc/dc-acconverters with dc current sources almost always involve a utility ac voltage supply, andhave well-defined values of phase. Any switching function can be altered in phaserelative to this fixed reference frame, resulting in variation of input and output powerlevels.

Ac-dc/dc-ac converters which involve dc voltage sources are, in practice, quitedifferent from the current-sourced versions. Such converters imply an ac current source -- rare in practice. Normally, the “ac current source” simply reflects some electrical load,such as a circuit or a motor, which requires ac power. These loads rarely define a phasereference -- the phase is determined by impedance characteristics, and will always leador lag the input voltage by a definite amount.

The voltage sourced inverters, then, provide a window into the complexities oftrue, independent inversion -- conversion of a dc source into some ac source. Inversionis important in three basic applications:1. Transfer of energy from some dc source into an ac power system. Such uses

include alternate energy source conversion and dc motor regeneration.2. Backup ac power. Since most electrical equipment is intended for ac power

from a utility, switching power converters which produce ac from storage

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batteries have long been important.3. Ac motor drives. Induction motors, synchronous motors, and so-called brushless

dc motors require ac power at variable frequency. Today, this is often achievedby converting power from some dc source.

19.8.2 Basic Theory

Figure l9.43 -- Basic dc-ac converter.

The most general dc-ac converter (with single phase ac) is shown in Figure19.43. It has at most four switches. The switching functions are constrained by KVL andKCL, as usual. They also must have a Fourier component at the ac source frequency, toensure nonzero power flow. An important practical consideration is that most real acsources cannot tolerate a dc component. This further constrains the functions. Areasonable choice is to operate the switches with a duty ratio of 50 % to bring about thehighest symmetry and hence avoid any dc component on the ac source. A generalconverter of the dc voltage source type is shown in Figure 19.44. As in the case of thetwo-quadrant buck converter, Vout can be either Vin, -Vin, or 0, depending on the switchcombination. Notice the switch types to be used in this converter. In general, current canflow in either direction, although there is only one voltage polarity.

Figure l9.45 -- Induction motor equivalent circuit for each phase.

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Consider the induction motor equivalent circuit, shown in Figure 19.45. This

Figure l9.44 -- General voltage-sourced dc-ac converter.

circuit requires ac power, and is an inductive load. For the purposes of KCL a currentsource is an appropriate model over short times. Nonetheless, the phase of this currentsource is linked to the phase of the input voltage. Changing the phase of the inputvoltage will change the current phase so as not to alter power.

Because of the ac current source phase behavior, dc-ac converters are harder to

Figure l9.46 -- Half-bridge inverter with R-L load.

study than dc-dc converters. Properties of the load are important to converter operation.A half-bridge converter with R-L load is shown in Figure 19.46. KVL and KCL requirethe two switching functions to act in complement, so that q1 + q2 = 1. The practicalrequirement of no dc component in Vout requires D1 = D2 = ½. Vout is a simple squarewave at the desired frequency, and has a value of Vout = (2D1 - 1)Vin. This generatesthe Fourier series

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Output current and power depend on the load, in a manner which is actually easy to

(l9.12)

calculate. Consider that Vout = RIout +L(dIout/dt). In steady state, Iout is periodic at thesame frequency as Vout, and has a Fourier series

This series can be differentiated easily. Let Xn = nωL. Then we can solve for Iout term

Figure l9.47 -- Typical output waveforms of voltage-sourced inverter.

(l9.13)

by term, obtaining

This is equivalent to finding each component of Iout separately in terms of the

(l9.14)

corresponding component of Vout. It is easy to automate such a procedure on anycomputer. Sample waveforms are given in Figure 19.47. Some loads of this type willbe tested in this experiment.

19.8.3 Procedure

Part 1: Voltage-sourced inverter, R-L and R-L-C loads

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1. Set up two FET control boxes in a master-slave configuration, to form a half-bridge inverter. Be sure to include the capacitors and resistors shown with thepower supply input. Refer to Figure 19.48.

2. Set the power supply current limit, if there is one, for about 2.0 A, and the

Figure l9.48 -- Voltage-sourced inverter test circuit.

voltage to zero. Set the master and slave duty ratio dials for 50 %. Connectoscilloscope leads and wattmeters across the load resistor and across theconverter output, as shown in Figure 19.48.

3. Turn on the FET boxes and the input power supply. Set fswitch to 10 kHz. Setthe supply to 24 V.

4. Measure the output average voltage. Adjust the duty ratio and fine-tune the inputsupplies to get 50% duty and zero average output.Small average offsets canswing the current to the power supply current limits.

5. Sketch the load resistor voltage and the output voltage waveforms. Record theRMS voltage, current, and power from the wattmeter. Measure the average inputcurrent from the supply.

6. Compute the series capacitance needed for resonance with the inductor at theswitching frequency. Add this capacitor in series with the inductor. Adjust thefrequency as needed to obtain near-resonant operation.

7. Sketch the resistor voltage waveform with the R-L-C load. Record the RMSvalue of the load resistor voltage, the output power, and the average input currentfrom the supplies.

Part 2 -- Isolated converter with ac link (half-bridge forward converter)1. Set fswitch to 50 kHz. Wire a toroidal transformer into the circuit, as shown in

Figure 19.49.

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2. Measure the average output voltage of this circuit. Sketch the waveform.

Figure l9.49 -- Isolated ac link converter.

3. Place a 1 µFcapacitor across the output terminals. Again measure the averageoutput voltage. Also measure the average input current from each source bytaking advantage of the series resistors.


1. How do you expect waveforms for an R-L load to change with frequency?2. An alternate method for control of Vout(wanted)is to add a third switch across the

output combination. This allows zero as a possible output value. The duty ratioscan then be adjusted so that Vout changes. Assuming that the dc componentremains at zero, find Vout(wanted)as a function of D for this control scheme.

3. What is the effect of a resonant load, such as that in Part I?4. What is the efficiency of the converter circuits tested above?

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19.9 Experiment -- PWM Inverters

19.9.1 IntroductionThis experiment introduces pulse-width modulation circuits. The intent is to gain

familiarity with the waveforms and concepts of pulse-width modulation as it applies topower electronics. The PWM inverter will be examined in the context of an ac motordrive. Basic ac drive concepts will be one feature of this experiment. The typical PWMwaveform shown in Figure 19.50 has a switching frequency of 960 Hz.

Control of output voltage in dc-ac converters is critically important in a wide

Figure l9.50 -- Typical PWM waveform (withfswitch = 16fout).

range of applications. Backup ac power supplies require control of voltage so that Voutcan be varied independent of the input source. Ac motor drives require control of bothVout and fout to achieve good performance. There are two methods in common use forcontrolling the output voltage (and power) of an inverter. The first is to adjust the relativetiming of switching functions in a bridge. The load voltage depends on the relative phases.This “phase displacement control” is often used in very high power situations, such aslocomotive controls, where the switches are slow SCRs.

The second method, pulse-width modulation (PWM) divorces the switchingfunction frequency from the intended properties of Vout. This method can provide thedesired results if the switches can be operated much faster than fout. To understandPWM, consider a very fast square wave. The duty ratio can be varied slowly between 0and 100%, in a manner similar to pulse width control in dc-dc converters. In a buckconverter, for example, this would have the effect of slowly changing Vout(ave). This slowchange can be sinusoidal perhaps, which gives a slow sinusoidal change in the averagevalue of Vout. Here “slow” simply means much more slowly than the switching function.Vout can easily change at the rate of 60 Hz if, for example, a switching frequency of10000 Hz is used.

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In practice, implementing this slow variation of pulse width is exactly the sameas designing a dc-dc converter. The only issue is to place Vout in two quadrants. And,in fact, the full-bridge inverter operating under PWM is identical to a two-quadrant buckconverter. The pulse width is intentionally varied, rather than held nearly constant as inthe dc-dc case.

Convenient speed control for ac motors has been a desired goal almost since suchmachines were invented. The ac induction motor, for example, is mechanically simpleand rugged, requires no electrical connections between stationary and rotating parts, andis inexpensive. Unfortunately, the induction motor lacks the easy control of a dc motor.Ac motor speed control in general requires adjustment of the motor input frequency. Thespeed depends on frequency in a direct manner, and the ability to vary f solves much ofthe problem. Unfortunately, it is not entirely trivial to vary the frequency applied to amotor. Ac motors are basically inductive, and have a frequency-dependent impedance.As the frequency is lowered, input current and internal magnetic flux rise. To counteractthese effects, the voltage must change along with frequency. If the voltage is altered inthe correct manner as a function of frequency, an ac motor can be made to look almostlike a dc motor in terms of speed and torque control.

Until PWM converters were feasible, the process of altering fout and Vouttogether was exceedingly difficult. Ac motor controllers were almost unknown twentyyears ago, and those which did exist were expensive, complicated, and used additionalmotors for energy conversion. Today, ac motor drives are beginning to displace dcmotors in some applications. These drive units are nearly always based on PWM.

19.9.2 Basic TheoryThe half-bridge inverter in Figure 19.51 has output Vout = (2q1 - 1)Vin. If the

duty ratio is varied with time as somemodulating functionM(t), Vout becomes

Remember that D must be between 0 and 1. With ½[k·cos(ωoutt)+1] for M(t), Vout is

(l9.15)

Notice that this isnot in the form of a Fourier series -- time appears in several places, as

(l9.16)

does sin[nπcos(ωoutt)]. This can be decomposed into a Fourier series by using propertiesof Bessel functions. The necessary relations appear among the trigonometric identitiesin the Appendix. The Fourier components appear at frequencies of nωswitch ± mωout.The components drop in amplitude quickly for increasing m, but slowly for increasing n.It is easy to use a series R-L load as a low-pass filter (just as in the dc-dc buck converter)in order to separate fout from the unwanted components. This function can be performedby a deliberate R-L load, or by a motor winding or transformer. The experimentprocedure below illustrates both PWM and the filtering process.

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19.9.3 Procedure

Figure l9.51 -- Half-bridge inverter for PWM control.

Part 1: Voltage-sourced PWM inverter, R-L load1. Set up two FET control boxes in a master-slave configuration, to form a half-

bridge inverter. Be sure to include the capacitors and resistors shown with thepower supply input.

2. Set the power supply current limit, if there is one, for about 1.5 A, and the

Figure l9.52 -- Voltage-sourced inverter test circuit.

voltage to zero. Set the master duty ratio to about 50 % and the slave to 100 %.

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3. Set up the waveform generator at your bench to produce a sinusoidal voltage,with frequency of about 50 Hz, such that the lowest voltage is 1 V and thehighest is 3 V (sine wave with 2 Vp-p and 2 V dcoffset). Connect this voltageto the duty ratio input on the master converter unit.

4. Connect oscilloscope leads across the load resistor and across the converteroutput (refer Figure 19.52). Turn on the FET boxes and the input power supply.Set fswitch as low as possible.

5. Monitor the average value of Vout, and observe the power supply currents.Adjust the function generator to get near zero average output.

6. Sketch the Vout waveform. Can you see the modulating process? Record theRMS voltage current, and power into the load resistor. Measure average dcvoltage across the 1Ω series input resistors.

7. Increase fswitch to about 50 kHz.8. Decrease the amplitude of the function generator modulating waveform by about

50%. Sketch the Vload waveform. Record meter data as in step 6.

Part 2 -- Induction motor drive1. Turn off the power supply. Replace the R-L load with a series combination of

5 Ω and a transformer winding of 12.6 V rating.2. Place a 1000Ω load across the transformer secondary terminals. Turn on the

power supply, and measure the RMS and average voltage across this resistor, andobserve and sketch the waveform.Be aware of the relatively high voltagelevel.

3. With the supply off, wire a single-phase motor into the circuit, as shown inFigure 19.53. Apply power. Sketch the converter output voltage and resistorvoltage waveforms (refer to the Figure).

4. Change the sine wave frequency and amplitude to observe an ac motor controlfunction.


1. For this type of converter, how do you expect waveforms for an R-L load tochange with switching frequency? With modulating frequency?

2. Why is PWM advantageous in ac motor control?3. Draw the full-bridge inverter for PWM. What relationships would you expect

among the various switching functions?4. Compute and tabulate the converter’s efficiency from your part 1 data.5. Compare PWM with the simpler inverter scheme of the previous experiment.

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19.10 Ac-Ac Conversion

Figure l9.53 -- Ac motor drive test circuit.

19.10.1 Introduction

Direct frequency conversion, as introduced in Chapter 7, is one framework inwhich to study the ac-ac converter. In this experiment, we will examine converters of thistype. The bilateral switch is beginning to gain in importance, and one device set whichperforms this function will be examined. Many applications which require ac output,especially motor drives and related transportation areas, would appear to benefit if powercould be converted directly from the ac mains. Today, a few workers in powerelectronics would dispute this, and claim that conversion via a dc link is superior to directac-ac conversion. In principle, however, the reduction in switching network complexity,along with the possibility of extending ac-ac conversion into such areas as poweramplifiers, makes a direct ac-ac converter a kind of “ultimate dream.” A generic ac-acunit would serve for any possible conversion function, including rectification andinversion. In the long run, a power bilateral switch will open up a wide range of newprocesses and applications for power conversion, just as the SCR did during the 1960s.

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19.10.2 Realizing the bilateral switch

Any converter can be built from a bilateral switch. While this makes it apowerful tool, its properties are not always advantageous. For example, it is easier to builda reliable dc-dc buck converter with a transistor and a diode than with bilateral switches.There is no alternative for ac-ac conversion, and in fact a bilateral switch bringsadvantages to certain other converter types.

The TRIAC approximates full bilateral capabilities. This device, in effect, is aset of inverse-parallel SCRs, built together, and sharing a single gate. The TRIAC iscontrolled just like an SCR -- once on, it remains on until the current becomes zero. Thisprovides several characteristics:• Simple ac regulators can be realized with TRIACs, since they will perform phase

control or integral cycle control on a resistive load without trouble.• The TRIAC is difficult to use when dc currents or voltages are involved, since

conditions needed for turn-off may not appear.• A TRIAC of given size does not use semiconductor material very effectively,

and has much lower ratings than two SCRs which are half as big.TRIACs are common in consumer products which benefit from simple ac regulators.Some such devices are used in certain ac motor control applications, particularly thosewhich resemble phase controlled ac regulators. TRIACs have limitations on levels ofexternally applied dv/dt and di/dt. The devices are also relatively slow to turn off. Theselimitations confine them mainly to power mains frequencies.

The TRIAC is a very limited bilateral switch. An alternative is to combinemultiple switches of other types to form a true bilateral device. For example, twoswitches of type can be placed in inverse series to form an equivalent . Thegates of the two sub-units are operated from the same switching function. Possibleimplementations appear in Figure 19.54. It is actually relatively convenient to implementthe FET version shown in the Figure, since the gate drive can be used without change inmost cases. The BJT version is more problematic, since applied base current mustsupport both devices without any balance troubles.

Figure l9.54 -- Implementations of the bilateral switch.

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19.10.3 Basic Theory

Once a bilateral switch is available, it is relatively simple to operate an ac-acconverter. The requirement is that switching functions contain a Fourier component ata frequency fswitch = fin ± fout. The most straightforward way to accomplish this is to setfswitch to either the sum or difference of input and intended output frequencies. Thusconversion of 60 Hz to 400 Hz would result if a switching frequency of either 340 Hz or460 Hz were applied, for example.

Very often, the input frequency is known and fixed, and variation of the outputfrequency is important to an application. This is true of ac motor drives, for example.It is sometimes useful to consider the switching frequency as fin plus some modulatingfunction. For example, if the switching function phase is made to be a function of time,M(t), a switching function will take the form

This is “phase modulation,” since phase is a function of time. Phase modulation reduces

(l9.17)

trivially to the form fswitch = fin ± fout if M(t) is set to ±ωoutt. This form of switchingfunction is easy to implement, and is referred to as “linear phase modulation” since M(t)is a linear function of time. The choice M(t) = +ωoutt is given the special name“Universal Frequency Converter” (UFC) since it works for any choice of fin or fout. Thechoice M(t) = -ωoutt gives rise to a “Slow-Switching Frequency Converter” (SSFC), andhas the constraint that fin ≠ fout. Both of these types are statements of the simplest choicesof switching frequencies. Other choices of M(t) can be made. These are nonlinear phasemodulation methods, with almost unlimited possibilities. Many nonlinear phasemodulation techniques allow control of power flow even with passive loads.

It should be pointed out that phase modulation is not the only way to performac-ac conversion. Pulse-width modulation can also be used, since PWM allows thecreation of nearly any Fourier component, given a fast switching frequency. PWMmethods are being discussed for use in ac output converters which allow ac or dc input.

19.10.4 Procedure

Set up a two-phase to one-phase ac-ac converter based on dual FET bilateral switches, asshown in Figure 19.55. Apply a switching function at some frequency, and observeresults at the output terminals. Record your observations.

Pay special attention to the waveforms which appear when the switchingfrequency is exactly equal to the input frequency. This case is particularly easy tosimulate and study.

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19.10.5 Study Questions

Figure l9.55 -- Ac-ac converter test circuit.

1. Draw some expected waveforms for two-phase to one-phase ac-ac conversion.Do the experimental results go along with these?

2. What would have been the effect of a resonant load?3. How would PWM provide advantages in this ac-ac converter?

19.11 Frequency Characteristics of Passive Components


The next two experiments will study the behavior of real passive components.In the first experiment, basic operation of realistic capacitors, inductors, and resistors willbe examined. The capacitors, inductors, and resistors used in circuit analysis have idealproperties: ic = C dvc/dt, vL = L diL/dt, and vR = R iR. However, real devices are notideal. In the context of power electronics, these effects are often important. Wires haveresistance and inductance, coils of wire have capacitance between the turns, and so on.In order to use passive components in power electronic circuits, it is important tounderstand what the effects are, how they are characterized and measured, and theimplications for design.

19.11.2 Basic theory, capacitorsWhen two conductors of any shape are placed in an electric field, a charge

develops on each one. In a system which consists only of perfect conductors and perfectinsulators, the charge Q depends on voltage in a linear fashion, so that Q = CV. Ifcapacitance is constant (e.g. the conductors are stationary), this leads to the simplerelationship i = C dv/dt. A real capacitor must provide conductors to hold the charge,

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wires to allow application of voltage, insulation which physically supports and separatesthe conductors, and a protective package to prevent damage. This naturally complicatesthe real behavior.

There are two common classes of commercial capacitors. The first consists oftwo flat metal conductors, separated by a dielectric layer. The second type, known as the“electrolytic capacitor,” consists of an oxidized metal conductor and a nonmetallicconductor. Both types can be modelled with a parallel-plate geometry, shown in Figure19.56. The capacitance value depends on the plate spacing d, the plate area A, and the

Figure l9.56 -- Parallel-plate capacitor.

dielectric constant of the insulator,ε,according to the relation C =εA/d.A large value of capacitance requireslarge plates, small spacings, and highdielectric constants.

The nonideal effects arerelatively clear: the wires and platesintroduce series inductance andresistance, and an imperfect dielectric could allow some current flow between the plates.A candidate circuit model emerges, given in Figure 19.57.

This circuit model can be simplified if the capacitor is operated at some specific

Figure l9.57 -- Proposed circuit model for simple dielectric capacitor.

Figure l9.58 -- Standard series equivalent circuit of a capacitor

frequency. Then the parallel portion can be transformed into a series equivalent. Thisgives thestandard modelshown in Figure 19.58. Many manufacturers use this standardmodel as a basis for describing their capacitors. Theequivalent series resistance(ESR,another name for the Rs of the circuit) is often given at some frequency, such as 120 Hz.The equivalent series inductance(ESL) is often given in the form of a “self-resonantfrequency,” fr. The properties of the standard circuit include:

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• Voltage drop across the ESR, which reduces the stored charge relative toexpected values.

• The resonant effect of the series R-L-C circuit means that a plot of impedancevs. frequency will fall at first (capacitive impedance), reach a minimum, and thenrise (inductive impedance).

• Above fr, the device is an inductor!

• Any current flow produces loss in the ESR. The higher the frequency of theapplied voltage, the higher the current and losses.

• The circuit does not model real behavior well at very low frequencies. Forinstance, the original circuit of Figure 19.57 shows that some leakage current willflow when a dc voltage is applied. Real capacitors show this effect, but thestandard model does not include it.

The ESR mainly represents the dielectric properties. Dielectrics do not usuallyact in accordance with Ohm’s Law. Because of this, the ESR is generally a kind ofnonlinear resistance; for example, it varies with frequency. One traditional way tocharacterize such materials is with the “loss tangent.” The loss tangent, also called tanδ or “dissipation factor” (df), is defined as the ratio of resistance to reactance for a seriesR-X circuit. For the standard model of the capacitor, tanδ = ωRC. The loss tangent hasa characteristic value for a given capacitor dielectric material, regardless of the plategeometry. Many dielectric materials are characterized by a loss tangent which is roughlyconstant over a wide frequency range. For these reasons, loss tangent or df is often givenin capacitor specification sheets. The ESL is related to packaging and lead structure,since wire inductance is the most important factor.

In electrolytic capacitors, the insulating layer is formed through anelectrochemical reaction between the two conductors. If voltage of the wrong polarity isapplied, the reaction reverses, and the insulating layer is destroyed. The device becomesa resistor, and usually overheats and fails quickly. When the correct polarity is applied,the electrolytic capacitor shows the same general properties as the simple dielectricversion, with two changes: the leakage current levels are much higher, which means ESRis smaller and df is higher; and the effective plate surface area is very high, which allowshigh values of capacitance per unit volume.

19.11.3 Basic theory, inductorsInductors are formed simply by wrapping a coil around a magnetic material.

Current in the coil creates magnetic flux in the material. If the material is linear, the fluxis proportional to current, so thatλ = Li. The constant of proportionality in this casedefines inductance. By Faraday’s Law, if this flux varies with time, it gives rise to avoltage vL = L di/dt. Magnetic materials in general arenot linear; these effects will bestudied in more detail in the next later experiment.

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Even without considering the effects of nonlinearities, some aspects of realinductors should be clear. The wire coil has resistance. Incoming and outgoing wires acta bit like parallel plates. A simple circuit model might be that in Figure 19.59.

The symbol ESR is often used for the effective resistance in inductors. As in the

Figure l9.59 -- Proposed circuit model for basic inductor

capacitive case, the loss tangent is defined as the ratio of resistance to reactance. It ishard to build inductors which handle high energy levels and still have very low losstangents.

Inductor types are defined mainly by the magnetic material. This can be a linearmaterial, such as air, or any magnetic material. Good linearity is a desirable feature, somost practical inductors have an air gap. In this experiment, two inductors will be testedfor their ESR values and resonant frequencies.

19.11.4 Basic theory, resistorsResistors are generally taken for granted in electrical networks. However, they

are subject to nonlinear effects like other components. Consider that Ohm’s Law is reallyan approximation, based on a highly simplified model of electron behavior in metals.Nonmetals do not usually follow this behavior. The general tendency is that the higherthe resistivity of a material, the less linear its V vs. I behavior is likely to be. The partsmust have connections, which implies a series inductance. A basic model is shown Figure19.60.

Even in materials which follow Ohm’s Law, other effects are important. All

Figure l9.60 -- Proposed circuit model for basic resistor

materials, including metals, vary in resistivity as a function of temperature. For most

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metals, the variation is approximately linear. In power electronics, heat loss andtemperature effects are almost always significant. Since resistance values change withheating, it can be hard to predict the actual value of a given resistor.

There are several major types of resistors:

• Compositionand metal oxideresistors are formed as a block (or cylinder) ofnonmetal. The material resistivity and stability are important considerations.Carbon composition resistors were once the most common. Most materials of thistype are relatively sensitive to temperature shifts.

• Film resistors are usually formed by depositing a metal or carbon film on aceramic substrate. In industry, they are often preferred over composition types.Precision resistors are usually formed with metal films.

• Wirewoundresistors are common at high power levels. A coil of wire, usuallynichrome, is wrapped around a ceramic rod or tube in the simplest arrangement.While this is a convenient way to spread out heat, it adds a significantinductance. Special versions which are double wound to avoid excess inductancecan be obtained. These “non-inductive” versions are substantially more costlythan the simple types.

In this experiment, we will briefly examine the characteristics of standard wirewoundresistors. This will serve as an introduction to the overall problems with real resistors.

19.11.5 ProcedurePart 1: Reference measurements1. Your instructor will assign a set of capacitors, inductors, and resistors to each

team. If a bridge or similar instrument is available, gather data at a specificfrequency for Cs, Cp, Rs, Rp, Ls, Lp, and D = df for each of your assigned parts.Special emphasis should be placed on the values relevant to the various models.

Part 2: Frequency sweeps1. Insert one capacitor in the circuit in Figure 19.61. If the capacitor is electrolytic,

be sure to observe the polarity marks, and add a dc offset to Vin so that Vc > 0always. If it is not electrolytic, use an offset of zero. If the capacitor is lessthan 0.5 µF, substitute a 1 kΩ resistor for Rs.

2. Set Vin at 1000 Hz, and observe Vin and Vc on your oscilloscope. Measure thepeak-to peak amplitudes and the phase shift between them in degrees. (Hint:Phase measurements can be made more easily by adjusting the oscilloscope tomeasure time differences.) Use the highest available Vin value.

3. Look for the resonant frequency of the part by adjusting fs until the phase shiftis close to zero. The amplitude will be close to minimum for a capacitor.Record amplitudes of Vin and Vc at this frequency.

4. Choose at least three frequencies below resonance and three above it; also make

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a measurement at 1 kHz. The frequencies you choose should cover roughly a

Figure l9.61 -- Capacitive impedance test circuit.

factor of 100 in frequency. Record amplitudes of Vin and Vc, and the phase shiftbetween them, at each of these frequencies.

5. Repeat parts 1-4 for your other assigned capacitors and inductors. In somecases, you may want to change the resistor value to allow better measurementsof Vc. If you do this, be sure to use R≥ 50 Ω, and be sure to record the Rvalue actually used for each part measured. A value R = 1000 is suggested forinductors.

Part 3: Power resistor tests (these can be performed during frequency sweeps)1. Connect a wirewound resistor to a variable dc supply.2. Adjust the voltage and current to draw about 10% of the rated power, or 1 W,

whichever is less. Record the voltage and current, then let the part warm up forat least 5 minutes. Record the voltage and current again.

3. Adjust the voltage and current to draw about 50% of rated power, or 10 W,whichever is less. Wait at least five minutes, then record the voltage andcurrent.

4. Repeat step 3 for 100% of rated power.

19.11.6 Study questions1. Use data for Vin, Vc, and phase to compute the impedance for each tested

capacitor and inductor.2. Plot impedance magnitude and phase vs. frequency for each capacitor and

inductor.3. Plot resistance vs. power for the wirewound resistor.4. Calculate ESL for each capacitor based on the resonant frequency you measured.5. Calculate ESR at 1000 Hz and at a frequency near resonance for each capacitor

and inductor from your frequency sweep data. Compare the 1000 Hz results tothe RLC meter data.

6. Discuss how your data conform to the proposed simple models.19.12 Magnetics

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Figure l9.62 -- Possible test circuit for magnetic core.

This experiment provides an overview of many concepts of magnetic componentdesign. It concentrates on inductor and transformer design for high-speed converters, andon magnetic properties of materials. In power electronics, both the energy storagecapabilities of inductors and the potential-shifting properties of transformers are importantin converter designs. But the requirements are much different from those in otherelectrical engineering areas. High power levels are often an issue at relatively highfrequencies. Losses must be as low as possible.

Modern switching converters and their applications take full advantage of thelatest magnetic materials. High-performance motors use advanced permanent magnetmaterials, as well as PWM inverters. The magnetic amplifier and saturable reactor --power amplifier technologies of the 1940s -- are making a strong comeback in powerelectronics because of new materials. Material properties are inextricably linked tomagnetic components, even more than for capacitors. Two important reasons for thisinvolve the behavior of “magnetic conductors.” Highly permeable materials with theability to direct magnetic flux are only a few orders of magnitude better than air orvacuum. They have limited capacity for carrying magnetic flux. A second difficulty isthat the most permeable materials are metals -- they carry magnetic flux, but also carryelectric flux, so that a time-changing magnetic flux will create current flow and losses inthe materials.

19.12.2 Basic theoryThe important concepts of magnetic circuits, magnetic domains, the hysteresis

loop, and saturation lead to design considerations for devices. In the laboratory, we will

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examine candidate designs for inductors and transformers. Any real magnetic materialexhibits a maximum value of B, called the saturation flux density Bsat, beyond which itdisplays permeability close to that of vacuum. Since the fluxφ in a material is found asNi/ℜ, whereℜ is reluctance, the current ultimately determines whether the saturation fluxdensity has been reached. Alternatively, the fluxφ is given byλ/N, or ∫v dt/N. Theintegral of voltage (“volt seconds”) determines the flux, within some integration constant.

In a transformer, saturation must be avoided so that leakage flux stays low. Forac applications, this means that only the volt second integral is relevant. If a dc currentis imposed on a transformer, an additional flux contribution will appear. In most power-frequency transformers, the number of turns is high, and dc current must be avoided. Inan inductor, dc current is almost always needed, so that a net energy can be stored.Inductors must be designed to tolerate high levels of dc current, while providingconsistent, linear, inductance. These two requirements are in conflict with the need forinductors of substantial value.

Another important issue in modern magnetics design is the current capacity ofthe wire used to wind the magnetic device. For instance, a design might call for a greatmany turns and a small core. While this is possible when small wire is used, the wiremight melt if the intended currents are applied. It is not uncommon to require hundredsor even thousands of windings in power frequency devices, and the temptation to useever-smaller wire is great. To realize a design, it is helpful to realize that only a fractionof any core opening can actually be filled with wire (some must be allowed for insulationand for air spaces around loose windings). It is also helpful to have some guidance aboutcurrent capabilities of copper. While no absolute rules can exist, it is often safe to imposecurrents of up to 500 A/cm² on a copper wire. This number gives some rough assistancein picking the wire sizes needed.

A transformer design procedure might be as follows:• Identify the frequency, voltage, current, and power requirements.• Choose a wire size adequate for the current.• Find the volt second level needed to meet ratings.• Identify a magnetic core with low losses over the intended frequency range, and

proper , A, and winding opening size to support the necessary flux.• Compute winding and core losses to see if they are at acceptable levels.• Build and test the device.

An inductor design would seek to:• Identify the frequency, voltage, and dc current requirements.• Identify the energy storage or L requirement.• Choose a wire size as needed.• Find the volt second level needed.• Compute a gap length, if necessary to meet energy and idc requirements.• Build and test the device.

19.12.3 ProcedurePart 1: Core hysteresis loops

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1. You will be assigned several test cores of various types. Use a ruler to obtaingeometric data for each one, or use manufacturer’s data.

2. Wrap approximately 25 turns of wire around each toroid test core, and 40 turnsaround each pot core.

3. Measure the inductance of each device with the bridge or equivalent instrument,if available.

4. Place each core, in turn, in the circuit shown in the figure 19.64, with properchoices of R and C. Be sure to include the 50Ω resistor Rs.

5. Use the oscilloscope in the X-Y mode to observe the hysteresis loop at somefrequency. Sketch the loop in your notebook. Measure the slope of either sideof the loop near the origin (B=0 and H=0). Don’t forget to make note of theslope units.

6. Decrease the applied frequency until a saturation effect is obvious. Again, sketchthe waveforms. Measure the slope of the loop as far into saturation as possible.Also record the vc value at saturation.

7. Apply a small dc offset from the input supply, and observe the change in theloop. Record your observations. Keep in mind that vc measures the integral ofvoltage, and notλ directly.

Part II -- Transformer design and testing1. Create a transformer by adding a second 25 turn winding on one of your test

cores specified by your instructor.2. Observe transformer performance by using the circuit in Figure 19.63, with Rload

= 1 kΩ. Vary the source voltage frequency and amplitude so that saturationeffects can be noted. You should sketch waveforms which demonstrate bothsaturated action and correct transformer action.

4. Use a 1N4004 diode in series with 50Ω as a load, and repeat your observations.

Figure l9.63 -- Core test circuit -- transformer.

19.12.4 Study questions1. Calculate the permeability of each core at 1000 Hz. Do your numbers agree

with bridge or equivalent measurements?2. Calculate Bsat from your hysteresis curves. What volt second rating do you

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expect for your 25 turn inductors? How do the converter and transformer resultscompare to your expectations?

3. How would a dc current load affect the output of a transformer?4. How many turns would be required to create a 10 mH inductor with each of your

cores? What would the dc current rating be, based on saturation limits?

19.13 Converter Design Project

19.13.1 IntroductionThe intent of the last experiment (normally conducted over a period of about a

month) is to provide overall experience with a converter design problem. Each group willdesign, build, and test a different dc-dc converter. The initial effort tests a basic designwith lab boxes, and considers effects of capacitor ESR and switch forward voltage drop.

The first consideration is to create a paper design as the basis for starting theproject:

• Begin with one of the specification sets listed at the end of this Section.For a dc-dc converter, start with a switching frequency of 100 kHz.

• Draw a converter circuit capable of performing the intended function.

• Given your specifications, find the switching frequency for which L >Lcrit.

• Compute the range of load resistances which correspond to the statedoutput power levels.

• Find the value of capacitor necessary to meet the intended ripplerequirements.

In the final experiment, our previous work will culminate in the design of anactual converter. The first activity will test the basic design values, with the objective ofdeveloping a tentative circuit. Effects of capacitor ESR and switch voltage drops will beexamined. The second activity will concern implementation of the actual switches, alongwith their switching functions. The last effort will examine power loss and heat transferconsiderations, and will include tests of the finished converter.

19.13.2 Refining the models

Voltage drops can be integrated into converter design in a straightforwardmanner. Consider the circuit shown in Figure 19.64, in which an ideal diode and avoltage source have been used to model a real diode. The output voltage Vout can bewritten in terms of switching functions as:

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Vout = q1Vin - q2Vforward(diode), Vout(ave)= D1Vin - D2Vfor

If V for is assumed to be roughly 1 V, the average value of Vout can be found as

Figure l9.64 -- Buck converter with simple model for real diode.

Vout(ave)= D1Vin + D1 - 1

The input current is still Iin = q1Iout, so Pin(ave) = D1VinIout. But now,

Pout(ave)= D1VinIout - D2VforIout,

which gives an efficiency of

η = 1 - (D2Vfor)/(D1Vin).

This is certainly less than 100 %, and represents the loss in the real diode while it is on.A series resistor could also have been included as part of the diode or transistor models.

Capacitor ESR means that capacitors do not really show any ideal voltage

Figure l9.65 -- Boost converter with capacitor ESR included.

characteristics. For example, an electrolytic capacitor with tanδ = 0.10 and C = 200 µFwould have ESR of 4 mΩ at 20 kHz. Consider the effect in the boost converter shown

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in Figure 19.65. When switch #1 is on, the output current charges the capacitor, but thereis a voltage drop across the ESR which appears at the output. As switch #1 turns off, theinternal capacitor voltage Vc stays quite constant, but the capacitor current changesabruptly, which in turn reverses the voltage drop across the ESR, and causes an abruptshift in Vout. The output voltage for this converter can be calculated without too muchtrouble, if the capacitor voltage is assumed to change in a linear manner. The Voutwaveform is shown in Figure 19.66.

As indicated in Figure 19.66, the ESR voltage drop can be a significant fraction

Figure l9.66 -- Output voltage waveform for boost converter of Figure 19.66.

of the output voltage ripple. If a maximum output ripple is specified, the value of ESRat the switching frequency will have to be taken into account when designing theconverter. For example, the above converter displays output ripple of about 240 mV,although a ripple of less than 160 mV would have been expected if the capacitor were aperfect 200 µF device.

19.13.3 Procedure, part I1. Obtain an inductor, or design a pot-core inductor to meet your requirements. Use

this inductor, along with an FET control box, to build a converter as designedin your preparation. Be sure to place a capacitor across Vin, so that it willappear as a voltage source.

2. Observe Vout under a range of conditions. Be sure to test for a variety of valueswithin your required converter operating range. Sketch Vout under what you findto be roughly worst-case Vout ripple. Be sure to record duty ratio data, alongwith sufficient data to calculate efficiency.

3. Observe voltage drops across the FET and the diode under your range ofoperating conditions. The intent is to gather data to support circuit models ofyour actual diode and FET.

4. If the converter you have built does not meet the specifications for output ripple,try other capacitors in an effort to solve the problem. Experience with yourconverter circuit is probably more important than meeting all specifications in

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detail.5. In the next part of the design project, you will select actual FET and diode parts,

in place of the FET box. Be sure you have collected all data needed for modelsof your devices and for the capacitors. Also, observe inductor current to confirmthe critical inductance value.

19.13.4 Study questions -- part I1. Model theon condition of the FET as a 0.3Ω resistor, and theon condition of

the diode as a 1 Vdrop. What value of Vout vs. duty ratio would be expectedfor your converter? Does this agree with measured data?

2. Will you need an FET with lower resistance to meet the requirements?3. Compute the ESR of your capacitors from the waveform data.4. Find the peak currents and voltages in your inductor and capacitor.5. Two common problems in dc-dc converters are: (a) the user accidently connects

the input voltage in the reverse direction, and (b) a short circuit is accidentlyconnected at the output. Would the converter you are designing be able tohandle these faults? If not, how might it be altered?

19.13.5 Gate drive implementationAt this point, it is logical to fully implement the converter design with discrete

components in place of the laboratory boxes. An ideal switch requires a third gateterminal which controls its state. In the ideal case, a switching function is applied directlyto this gate (without regard for electrical ground or isolation). The switch isinstantaneously on whenever the gate is high, and instantaneously off whenever the gateis low. While the concept of a gate terminal carries over to real switches, there is a hostof new considerations introduced by a real gate.

The potential at the gate normally has restrictions relative to the potentials on theswitch terminals. Operation of a real gate requires energy (and hence power and time);these must be supplied from some drive circuit. As you might guess, there is a timedifference between the application of a gate signal and the actual switching action. Inaddition to gate characteristics, semiconductor switches have internal properties, such asresistance and capacitance, which limit time rates of change of voltage or current evenwith a perfect delay-free gate signal.

A gate drive typically tries to minimize commutation time while not addingsignificantly to power requirements. Gate drives are low-power electronic circuits, andare designed much differently from the converters in which they operate. They are stillswitching circuits, however, so many concepts of switch networks still apply.

19.13.6 Basic theory -- gate and base drivesThe major classes of gated power semiconductors -- thyristors, BJTs, and FETs -

- have distinct gate drive requirements. For example, a BJT can be modelled as a current-controlled current source, and requires current into the base when it is to be on. Thiscurrent is supplied at low voltage (so that power is low), but can be as high as a tenth ormore of the switch on-state current. In the case of the FET, the steady-state gate current

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is essentially zero, and the gate voltage determines switch operation. The SCR, as atypical thyristor requires only a pulse at the gate in order to turn on. Gate turn-offthyristors (GTOs) also require only pulses, although a substantial negative gate currentmust be applied to turn the devices off.

The basic requirements are simple enough, but details add great complexity. Letus first examine the FET, which has relatively simple gate characteristics. To turn theFET on, the gate region must be charged. The region represents a capacitance, and thegate drive must be designed to charge this capacitance sufficiently. The channel willinvert when the gate voltage exceeds a threshold level, Vth. In fact, the gate voltage mustbe maintained at a level considerably greater than Vth, so that enough charge will bepresent in the channel for low effective channel resistance. The gate drive must thereforeapply an “overdrive” gate voltage, for switching.

To turn the FET off, the gate region must be discharged, and the gate voltagemust be maintained below Vth. Real devices have wide error tolerances on Vth. Forexample, the MTM15N35 FET can have any Vth between 1.5 and 4.5, depending on thedevice and the operating temperature.

Some gate drive design considerations are apparent:• For turn-on, the drive must rapidly charge the gate to a voltage much

higher than Vgs, while not exceeding the dielectric breakdown limit ofthe gate insulator. Typical gate-source capacitance values range fromseveral hundred to several thousand picofarads.

• For turn-off, the drive must rapidly discharge the gate to a voltagelower than Vth, again without exceeding dielectric limits.

Several device specifications are important in the FET gate drive design. Referring to theIRF521, a typical power FET, notice the following relevant information:• Vth between 2.0 and 4.0 V.• Limits on gate-source voltage (gate dielectric breakdown limit):

-20 V < Vgs < +20 V.• Maximum gate current level: 1.5 A (absolute value).• Gate-source input capacitance: 600 pF.With these data, it is possible to design many different gate drive circuits. Consider theperformance of a simple gate drive, consisting of a nearly ideal switch operated from avoltage source of 50Ω output impedance. At turn-on, the gate capacitance is uncharged,and the maximum gate current is simply Vin/Rs, or 0.24 A in this example. The RC timeconstant is 30 ns so this gate drive will charge the gate to more than 10 V in about 60 ns.The actual device turn-on time of 110 ns is reasonably consistent with this number.Slightly faster turn-on could be achieved by using a smaller Rs. A value as low as 8Ωcan be used without excessive gate current.

At turn-off, the gate must discharge through the second resistor. This resistor hasthe undesirable effect of drawing power whenever the FET is on, so is kept relativelyhigh. It also acts as a voltage divider for Vgs. For the circuit of Figure 19.67, turn-offis quite slow.

More specific requirements on the gate drive are becoming apparent:• The gate drive must provide a voltage between the gate and source of about 10

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to 18 V, with low source impedance, for fast turn-on.

Figure l9.67 -- Sample FET gate drive circuit.

• The gate drive should provide a low impedance discharge path during turn-off.This path can apply zero volts across the gate and source, or could use Vgs < 0if a negative voltage source is available.

• The gate drive can be designed so that current is drawn only in a brief pulse.In principle, most of the ½CV² energy in the gate can be recovered, so thatpower requirements are very low.

A typical high-performance gate drive circuit appears in Figure 19.68.

Figure l9.68 -- High-performance gate drive circuit for FET.

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The main drawback of FET gate drives is the need to apply a substantial voltage

(a) (b)Figure l9.69 -- The buck converter with FET switch.

between the gate and source as long as the device is to be on. In many converters, thisis not a trivial manner. The familiar buck dc-dc converter is shown in Figure 19.69. Forthe unit of Figure 19.69 to work, a voltage equal to Vin + 10 V must be applied to theFET gate whenever switch #1 is to be on. This second voltage source, higher than theavailable source, can be a major stumbling block in building a converter. If the switchis relocated as in Figure 19.69b, the problem of higher voltage level is alleviated, but theoutput is no longer referred to ground potential. The voltage requirements of FETssometimes make them difficult to use in many situations.

Bipolar transistors need only a fraction of a volt on the gate (base) terminal. Thebase behavior is far more complicated than the FET’s gate behavior. A BJT base drivemust supply enough current to guarantee that the transistor will be fully on. This meansthat Ib must be chosen to be larger than the expected collector current Ic/β. In powertransistors,β can be quite low -- values of 5 or less are not unusual for devices rated overten amps. Furthermore, in order to operate the device at the highest possible speed, anoverdrive current pulse should be applied at turn-on. This pulse is usually chosen to benearly equal to Ic. Such high currents are difficult to achieve, since they call forimpedance levels well under 1Ω.

The 2N3055 can serve as a guide for base drive design. This device allows abase current as high as 7 A. The voltage at the base terminal will not exceed roughly 1.5V. The gain valueβ can be as low as 5, so that base current of about 3 A might beneeded if Ic approaches the maximum value of 15 A. A candidate base drive circuit isshown in Figure 19.70. The major difference between this and the FET drive is that itinvolves much lower impedances. Many base drive circuits include transformers to obtainhigh currents and low impedances.

Thyristors, as latching devices, are much different from transistors. A signalmust be applied to the gate only until sufficient anode current flows. Then the gate signalcan be removed. For example, the 2N6508 requires a gate signal of 75 mA or more ata voltage of up to 1.5 V. The gate signal will not be needed once the anode currentreaches a "holding current" of 40 mA. This can occur as little as 2 µs after the start ofthe gate pulse. A pulse transformer can serve as a suitable driver for such requirements.

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This has the important advantage of isolation of the gate.

Figure l9.70 -- Candidate BJT base drive circuit.

Most SCR applications are tied to an ac power line frequency, so that turn-onspeed is often not a major consideration. With this in mind, it is relatively easy to buildan SCR gate drive, provided certain details are considered. A controlled rectifier circuitis shown in Figure 19.71. The load inductance normally makes the operation consistentwith periodic steady state. However, take the case of circuit start-up. Inductor currentis initially zero, and three-phase voltages are applied to the SCRs. While a gate signalis applied to SCR A, the inductor current changes as vL/L. A large inductor will meana slow change, and a long time may pass before the SCR anode current exceeds the“holding” value. To alleviate this, a small resistor can be added as a ballast load, toensure SCR turn-on latching. Otherwise, the circuit might not operate.

Figure l9.71 -- SCR controlled rectifier for three-phase input.

19.13.7 Basic theory -- snubbersFor some converter designs, very high momentary voltages appear across the

FET during the Part I procedures, especially if the layout is loose. Power FETs can easilybe destroyed when a Vds value beyond their limit appears. High momentary voltages canoccur even in relatively low voltage converters. Each circuit connection, wire, orcomponent, has inductance, and thus behaves as a current source over short time intervals.When one attempts to reduce any current by switching action, a negative voltage vL =L(di/dt) will be produced. The faster the switch, or the larger the inductor, the more

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extreme this negative voltage. In an FET switching 10 A in 100 ns, even with a circuitinductance as low as 1.0 µH, a voltage of 100 V will be generated, independent of theconverter voltage levels. This voltage will appear across the switch as it turns off,causing commutation loss and possibly approaching the switch voltage limits.

This effect is detrimental, since it leads to extra losses and even switch failure.It is necessary to minimize the circuit inductance, and to keep di/dt sufficiently small.Leads should be short and of proper size. Loops or other features which enhancemagnetic field coupling must be avoided. It is desirable to act upon the switchingtrajectory itself. After all, the problem here is a switching trajectory which reachesvoltages much higher than any circuit voltage when current is still high.

Circuits which act directly on the switching trajectory are calledsnubbers. Thefunction of a snubber is to shape the switching trajectory and hence protect the switchingdevice. For this reason, most snubbers are wired directly to a switch, and would beconsidered an actual part of the switch itself for purposes of converter design. Forexample, a capacitor can be placed across the semiconductor, so that some of the energyneeded by the load during commutation is provided by the capacitor. A resistor might beused to discharge the capacitor one the switch has reached the new operating state.

A simple RC snubber causes the switch voltage to change relatively slowly(beginning at about zero) during turn-off. Unfortunately, the opposite occurs during turn-on. In this case, the transistor voltage changes slowly, beginning at Vin, during switching.The switching power loss will actually increase. It is apparent that a turn-off snubber anda turn-on snubber have different requirements. To avoid this difficulty, it is necessary toadd circuitry so that the simple RC snubber discussed above will operate only duringtransistor turn-off. A switch is needed! This switch needs to carry current only duringcommutation, so that its ratings can be based on momentary performance, rather than oncontinuous capabilities. A possible snubber circuit is shown in Figure 19.72. Duringturn-off, it causes voltage to change relatively slowly. While the transistor is off, thesnubber capacitor charges up to Voff. But at transistor turn-on, the snubber circuit diodeprevents current flow from the snubber capacitor. The second resistor, R2, discharges Cduring the transistor’s on-time so that the snubber will again be ready for turn-off.

This simple snubber is very popular, and has been used successfully in a wide

Figure l9.72 -- A simple snubber for the transistor in a dc-dc converter.

range of circuits. This snubber is known as alossy snubber, since energy is dissipatedin its resistance components. The intent is to decrease the total switching losses by

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adding a lossy snubber. Lossless snubbers, which use only switches and energy storageelements, are feasible, and have been discussed in the literature.

19.13.8 Procedure -- part II1. We will study an FET gate drive in the context of your converter. Obtain an

FET and diode with ratings appropriate to your converter. Measure thecharacteristics of these devices on the curve tracer.

2. Assemble the drive circuit shown in Figure 19.73. The SG3526 integrated circuitis a controllable pulse generator for use in dc-dc converters. Notice that thecircuit includes a snubber.

3. Insert the FET and diode into your converter circuit. Pay attention to grounds

Figure l9.73 -- SG3526 based gate drive system.

and to isolation requirements for voltage sources.4. Observe the FET voltage and current, and the gate voltage. Choose a reasonable

value of frequency and duty ratio, based on your converter requirements. Sketchthe waveforms. What are the turn-on and turn-off times of your switches?

5. Vary the duty ratio and frequency. Again observe the FET signals. Take datanecessary to find effects of D and f on rise and fall times. Be sure to observethe effect of very high switching frequency.

6. Use the X-Y mode of your oscilloscope to observe and sketch the switching

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trajectories. Think about how the snubber affects them.7. Check your converter to see if it meets the specifications. Try making

adjustments where necessary. What are the input voltage and power limitationsof your converter?

19.13.9 Study questions -- part II1. Compute (approximately) the power used in your gate drive circuit. How is

converter efficiency affected?2. Include the curves gathered in the lab on the curve tracer for your devices. What

is Vth for your FET? What voltage drops and resistances model yoursemiconductors?

3. Calculate a value of gate capacitance from your measured fall time. Is the valueconsistent with the data sheets?

4. Use the data to predict the performance of your converter. Can the specificationsbe met with these parts?

19.13.10 Thermal performance and feedback control -- introductionNow that the converter has been constructed (and presumably operated with some

success), key operational issues such as feedback control need to be considered. Thedesigner of a converter has relatively little control over how the unit is used. The mostobvious unknowns are the input power source and the output load, but additionalconditions such as ambient temperature, residual voltages, time-rate-of-change of the load,response to a short-circuit or to an open-circuit applied at the load terminals, deviceprotection, reliability, cost, and electromagnetic performance often depend on customerrequirements rather than on the designer’s intent. At least some of these difficulties canbe mitigated through the use of feedback control concepts. The basic idea, as taught ina first Control Systems course, is to subtract some fraction of the circuit output from anintended “reference” value, and use the result as a new input signal. When this is done,the output can be made almost arbitrarily close to the desired value even in the presenceof load changes, supply changes, residual drops across switches, and other disturbances.

19.13.11 Basic theory -- converter control

Figure l9.74 -- Simple feedback control block diagram.

Of the many ways to implement feedback control in a dc-dc converter, perhaps

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the simplest is calledoutput voltage control. Here, the output voltage is sensed andsubtracted from a given reference value. The result is amplified, and applied to the dutyratio input of a PWM circuit. A block diagram is shown in Figure 19.74. In a buckconverter which uses this circuit, the input-output relationship can be written:

Vout = k(Vref - Vout)Vin.Simplifying this,

Vout(1 + kVin) = kVrefVin.

If k is large so that kVin >> 1, then both sides can be divided by kVin to give

Vout = Vref.

It is easy to show that this result applies even when residual switch voltages are included:a large gain will produce output equal to the reference value, in spite of changes insupply, load, or other uncontrolled parameters, as long as the converter duty ratio is withinthe allowed limits.

While this style ofproportional gainfeedback is simple and appealing, it has twoimportant drawbacks. The first is that the output is not exactly equal to the input, evenin steady state, because k cannot be infinite. This should be clear when you consider thatthe converter duty ratio D is given by D = k(Vref - Vout). Since we expect D to havesome value greater than zero, the value Vref - Vout must also be greater than zero. Asecond drawback is that if k is too large, then any slight disturbance will drive D to thelimits 0 or 1, and the converter output will jump around almost at random. Onealternative is to add an integrating element D = k(Vref - Vout) + ki∫vref - vout to thecontrol. This eliminates steady-state error without requiring extreme gain.

Converter control can also be implemented with series regulators for output

Figure l9.75 -- Buck converter with series regulator at output.

filtering. Consider the regulator circuit below. The zener diode has been used to provide

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a reference signal for the series regulator. The output transistor exhibits a small voltagedrop (typically less than 1 V) between the base and the emitter. If the base is set to 6 V,for example, we might expect about 5 V at theemitter. Consider the effect of an externalload change which attempts to decrease Ve. Then more current will flow into the base,and Ic will increase since it is given byβIb. The emitter current will rise, which wouldbe expected to result in an increase in Ve. Similarly, an attempt to increase Ve willproduce lower emitter current.

This simple circuit will work as long as the outputpass transistoris in its activestate. The transistor must not act as a switch. This, in turn, requires that there besufficient voltage at the collector so that saturation is always avoided, and that there besufficient base current to avoid cutoff. One way to assure both of these is to keep Vc >Vout + 2 V, and to use a high-gain transistor with a high bias current through the zenerdiode. For a pass transistor withβ = 25, and Iout < 2 A, the required bias current is 80mA. The zener diode thus requires power of 0.08 × 6, or about ½ W. A final circuit,designed for Vout + 2 < Vc < Vout + 5, is given in Figure 19.76. With a 10 W load, thisregulator is about 68 % efficient with 7 V input and 43 % efficient with 10 V input.Even though the efficiency is not good, the output quality is very high.

Figure l9.76 -- Series regulator designed for Vc of 7 to 10 V and Vout ∼ 5 V.

19.13.12 Basic theory -- thermal effectsPower lost in a semiconductor (or anything else) is converted to heat. Naturally,

this will cause a temperature rise in the semiconductor. Semiconductor devices have amaximum temperature at which they can function -- typically between 100°C and 200°C.We must ensure that the semiconductor temperature will remain below its limit underoperating conditions. To do this, it is important to understand some basic concepts ofheat flow.

Temperature can be defined as a potential associated with heat flow, just asvoltage can be defined as a potential associated with current flow. In solid substances,the flow of heat follows Fourier’s Law of Heat Conduction, which states that the heat fluxper unit areaq (the units are W/m²) is proportional to the temperature gradient in the

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material. The constant of proportionality is defined as thermal conductivity, and is giventhe symbolk, so that

Fourier’s Law represents the effect of heat conduction (heat flow between solid objects

(l9.18)

in physical contact).When fluids are involved, the heat flow is referred to as “convection,” in which

motion of the fluid carries away much of the heat. Again, the heat flux depends on atemperature difference, and aheat transfer coefficient, h can be defined so that

The heat transfer coefficient is much more complicated than thermal conductivity, since

(l9.19)

it includes flow rates, directions, the surface area of the heat generator, and a variety ofother parameters.

The third type of heat transfer, radiation, is more complicated still. In this case,heat flux is proportional to the fourth power of the temperature difference. The result is

where σ is the Stefan-Boltzman constant,σ = 5.670 10-8 W/(m2K4), and e is the

(l9.20)

“emissivity” of the object,e = qactual/qideal. Typical values ofe are:

Material emissivitypolished metal 0.05

oxidized aluminum 0.10flat black paint 0.90

A flat black heat sink with temperature of 50°C ande = 0.9 will transfer 0.04 W/m² tosurroundings at 20°C. In many situations, radiation heat transfer can be neglected.

In the case of both conduction and convection, heat flux depends linearly on atemperature difference. The thermal conductivity can be used to define a “thermalresistance,” analogous to electrical resistance. The total heat flow in watts can bemultiplied by the thermal resistance to give the temperature difference, so that thetemperature of a semiconductor Tj is equal to Ploss(θja) + Ta, whereθja is the thermalresistance in K/W between the semiconductor and the surrounding ambient, and Ta is theambient temperature. (Please realize that only temperaturedifferenceis involved, so thatunits of K become equivalent to units of °C. Most manufacturers giveθ in °C/W.)

19.13.13 Procedure -- part IIIConverter checkout (open loop):1. Operate your converter over its specification range. Measure its efficiency under

several operating conditions.2. Check the operation of your converter over its full range. Make note of any

combinations of Vin and Pout for which you could not meet the specifications,and determine the issues associated with the limitations.

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3. Measure and record the temperatures of the switching devices under conditionsof maximum load and minimum Vin. Let the circuit stabilize for about tenminutes, but shut it down if your measured values go beyond 120°C. If thetemperature reaches steady state, record it and record electrical information toallow you to calculate power loss in the FET and the diode.

3. Take temperature measurements at two other combinations of Pout and Vin.

Feedback control checkout:4. During the temperature stabilization time, build the circuit of Figure 19.77 for

use in a feedback loop. Notice that the circuit output is 10(Vref - Vout).

5. Insert the feedback control unit into the converter, as follows:

Figure l9.77 -- Op-amp feedback control test circuit.

a) The circuit output attaches to the "sense" input of the SG3526.b) The “voltage” input attaches to the converter output.c) The ground node attaches to the converter output ground.d) For a buck converter, connect the “current” input to ground.e) For a buck-boost converter, place a 0.1Ω resistor in series with the

inductor, on the ground side. The voltage on this resistor goes to the“current” input.

f) For a boost converter, place a 0.1Ω resistor in series with the converterinput source, on the ground side. Connect the voltage on this resistorto the “current” input.

6. Set up the converter with minimum load. Turn it on, and determine whether thecontrol circuit is functioning.

7. Vary Vin, and observe the control action on Vout. What is the circuit lineregulation?

19.13.14 Study questions -- part III1. Estimate the thermal resistancesθjc andθja from your temperature data and from

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measurements of power loss in your switches.2. Discuss your converter design project results in some depth.

CONVERTER SPECIFICATION SETSThe basic specification sets here cover a wide range of dc-dc converter designs andapplications.A.

Input Voltage Range: 5 V to 9 VOutput Voltage: 12 V ± 1 %Output Load Range: 5 W to 20 W

B.Input Voltage Range: 10 V to 25 VOutput Voltage: -12 V ± 2 %Output Load Range: 5 W to 20 W

C.Input Voltage Range: -4.5 V to -16 VOutput Voltage: 5 V ± 1 %Output Load Range: 10 W to 25 W

D.Input Voltage Range: 10 V to 35 VOutput Voltage: +5 to +12 (adjustable) V ± 50 mVOutput Load Range: 0 W to 20 W

E.Input Voltage Range: 12 V to 40 VOutput Voltage: either +80 or -80 V ± 3 %Output Load Range: 10 W to 20 W

F.Input Voltage Range: 10 V to 15 VOutput Voltage: -16 V ± 2 %Output Load Range: 2 W to 20 W

G.Input Voltage Range: 7.0 V to 14 VOutput Voltage: 5 V ± 1 %Output Load Range: 0 W to 50 W

(battery powered)H.

Input Voltage Range: 14 V to 25 VOutput Voltage: 12 V ± 1 %Output Load Range: 0 W to 20 W

I.Input Voltage Range: 9 V to 15 VOutput Voltage: 24 V ± 3 %Output Load Range: 12 W to 35 W

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J.Input Voltage Range: 3 V to 8 VOutput Voltage: 5 V ± 1 %Output Load Range: 0 W to 25 W

(battery powered)K.

Input Voltage Range: 8 V to 15 VOutput Voltage: 6 V ± ½ %Output Load Range: 0 W to 20 W

L.Input Voltage Range: 10 V to 25 VOutput Voltage: -12 V ± 1 %Output Load Range: 0 W to 35 W

M.Input Voltage Range: 25 V to 40 VOutput Voltage: 12 V ± 1 %Output Load Range: 50 W to 100 W

N.Input Voltage Range: 12 V to 18 VOutput Voltage: 24 V ± 1 %Output Load Range: 50 W to 100 W

O.Input Voltage Range: 25 VRMS, 50-60 HzOutput Voltage: 13 V adjustable ± 1 V, with current limited to 5 AOutput Load Range: 0 W to 70 W

P.Input Voltage Range: 25 V to 50 V RMS, 60 HzOutput Voltage: 12 V ± 1 %Output Load Range: 50 W to 100 W

Q.Input Voltage Range: 22 V to 26 VOutput Voltage: 12 2 cos(120πt) V ± 0.2 VOutput Load: 50 W

R.Input Voltage Range: 5 V to 12 VOutput Voltage: 1 to 2 V ± 0.02 V (mean is adjustable)Output Load: 20 to 40 A

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Summary Specification Sheets for Parts

IRF 520 and IRF 521 (MOSFET):Rating Symbol Maximum limitDrain-source voltage VDSS 100 V (520), 60 V (521)Gate-source voltage VGS ±20 VDrain current ID 8 A continuous, 32 A peakJunction temperature TJ 150°C

Characteristic Symbol Typical valueThermal resistance, junction-case ΘJC 3.12 K/W

junction-ambient ΘJA 62.5 K/W (no sink)Residual current, VGS=0, VDS=max IDSS 0.2 mAGate threshold voltage VGS(th) 2 to 4 VOn-state resistance, VGS=10 RDS(on) 0.3 ΩInput capacitance Ciss 600 pFTurn-on delay time td(on) 40 nsRise time tr 70 nsTurn-off delay time td(off) 100 nsFall time tf 70 ns

IRF 351 and MTH15N35 (MOSFET):

Rating Symbol Maximum limitDrain-source voltage VDSS 350 VGate-source voltage VGS ±20 VDrain current ID 15 A continuous, 60 A peakJunction temperature TJ 150°C


junction-ambient ΘJA 30 K/W (no sink)Residual current, VGS=0, VDS=max IDSS 0.25 mAGate threshold voltage VGS(th) 2 to 4.5 VOn-state resistance, VGS=10 RDS(on) 0.3 ΩInput capacitance Ciss 3000 pFTurn-on delay time td(on) 35 ns (351), 60 ns (15N35)Rise time tr 65 ns (351), 180 ns (15N35)Turn-off delay time td(off) 150 ns (351), 450 ns(15N35)Fall time tf 75 ns (351), 180 ns (15N35)

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2N3055A (BJT):

Rating Symbol Maximum limitCollector-emitter voltage VCE0 60 VBase-emitter voltage VBE -7 VCollector current IC 15 A continuousBase current IB 7 AJunction temperature TJ 200°C

Characteristic Symbol Typical valueThermal resistance, junction-case ΘJC 1.52 K/WResidual current, IB=0, VCE=30 ICE0 0.7 mADc current gain hFE 40 (IC=4 A), 5 (IC=10 A)Collector-emitter saturation voltage VCE(sat) 1.1V (IC=4A), 3V (IC=10A)Turn-on delay time td(on) 0.1 µsRise time tr 2 µsStorage time ts 2 µsFall time tf 2 µs

2N6508 (SCR):

Rating Symbol Maximum limitBlocking voltage (forward or backward) VAK(off) 600 VPeak gate current IGM 2 AForward current IT 25 A RMS, 16 A ave,

300 A ½ cycle surgeJunction temperature TJ 125°C

Characteristic Symbol Typical valueThermal resistance, junction-case ΘJC 1.5 K/WResidual current, IG=0, VAK=max IT(off) 2 mAGate trigger current IGT 25 mA (75 mA worst case)Gate trigger voltage VGT 1 VHolding current IH 35 mAOn-state residual voltage VAK(on) 1V (IT=3A), 1.8V (IT=50A)Turn-on time tgt 1.5 µsTurn-off time tq 35 µsCritical rate of rise of off-state voltage dv/dtC 50 V/µs

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MUR3040PT (ultra-fast power rectifier):

Rating Symbol Maximum limitPeak reverse voltage VR 400 VForward current, per leg IF(ave) 15 A, 150 A ½ cycle surgeJunction temperature TJ 175°C


junction-ambient ΘJA 40 K/WReverse current, VR=max IR 500 µAForward voltage, IF = 15 A VF 1.2 VReverse recovery time trr 60 ns (max)

SG3526 (PWM control integrated circuit):

Rating Symbol Maximum limitSupply voltage VS 40 V (max), 7 V (min)Logic input voltage (pins 5, 8, 12) Vin -0.3 to +5.5 VAnalog input voltage (pins 1, 2, 6, 7) Vin -0.3 V to VSJunction temperature TJ 150°C

Characteristic Symbol Typical valueThermal resistance, junction-ambient ΘJA 66 K/WFrequency range f 1 Hz to 500 kHzReverse current, VR=max IR 500 µAForward voltage, IF = 15 A VF 1.2 VReverse recovery time trr 60 ns (max)

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SG3526 Operation

The SG3526 contains several modules:• A +5V series regulator for logic power and to provide a reference signal. the

regulator works over an input range of +8V to +35V.• An undervoltage lockout circuit to shut the IC down and reset it if the +5V level

falls too much.• An oscillator to provide a triangle waveform and a clock pulse train over the

range 1 Hz to 400 kHz. The period is approximately ½(RC), where R and C arethe timing components connected at pins 9 and 10, respectively.

• An amplifier (a bit like an op-amp) with gain-bandwidth product of 1 MHz.This amplifier is actually a transconductance device, and compensatingcapacitance of 100 pF or so must connected from pin 3 to ground to make itwork properly.

• A comparator for pulse-width modulation. The oscillator triangle is comparedto the amplifier output to generate the main PWM square wave.

• Dual totem-pole outputs. These outputs can drive up to 200 mA loads at up to35 V. They are designed primarily for MOSFET gate drives.

• Pulse-steering logic. To allow tightly balanced operation of two converterswitches, this logic directs PWM output pulses to alternate gate drives. Forexample, one pulse goes to the top gate drive, the second to the bottom gatedrive, and so on. While this balances the two outputs, it also limits the outputduty ratio to just under 50%. The outputs can be or-ed to give 0-100% duty.

• Double-pulse lockout logic. If the amplifier output crosses the triangle wavemore than once per clock period, multiple PWM pulses are produced. Thedouble-pulse logic ensures that only the first pulse is accepted. No others willhave any effect until the logic resets with the next clock pulse.

• Current comparator. This separate op-amp allows an external current shunt tobe used for protection. If more than about 100 mV is imposed between pins 6and 7, the circuit will shut down until the current falls again.

Some other features:

• Each pin actually serves as both input and output.• The clock can be synchronized externally by imposing a 150 ns square wave at

pin 12. This simply overrides the internal clock.• A resistor at pin 11 (normally 0Ω) can adjust the output dead time, i.e. the time

between turn-off of the top output and turn-on of the bottom output. RD = 0Ωgives about 1.5 µs of dead time, while 10Ω gives about 5.5 µs when theoscillator is at 40 kHz.

• To drive a single MOSFET over the full 0-100% duty range, it is necessary touse both IC outputs in an OR configuration.

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19.14 References

S. B. Dewan, G. R. Slemon, A. Straughen,Power Semiconductor Drives. New York:John Wiley, 1984.

Donald M. Trotter, Jr., “Capacitors,”Scientific American, vol. 259, no. 1, July, 1988, pp.86-90B.

Data sheets are summarized from various Motorola Semiconductor Corporation DataBooks (IRF 500, 2N3055, 2N6508, MUR3040PT).

SG3256 data sheets are summarized from Signetics, Inc. data books.

IRF 351 data sheets are summarized fromHEXFET Databook, 3rd edition, InternationalRectifier Corp., El Segundo, Calif., 1985, pp. D-67 - D-72.

B. Williams, Power Electronics: Devices, Drivers and Applications. New York: Wiley,1987, pp. 90-108.

J. Kassakian, G. Verghese, M. Schlecht,Principles of Power Electronics. New York:Addison-Wesley, 1991, pp. 693-710.

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A Sequence of Power Electronics Experiments

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