SLIDEREAD

Solutions to Supplemental Problemsto accompany the 3rd Edition ofPower Electronics: Converters, Applications and DesignbyNed Mohan, Tore Undeland, and William Robbins by John Wiley & Sons1

Chapter 1 - Power Electronic SystemsS1.1.In linear electronics, semiconductor devices are used in the middle of their linear amplificationregions where both the voltage across the component and the current thru it are relatively large.This results in high power dissipation.In power electronics, the semiconductor devices are used as switches. When the device is on(approximating a closed switch) the voltage across the device is very low (usually 1-3 voltsmaximum) and the current through it is large. The power dissipation, while substantial, is muchless than operating in the linear amplification region at the same current level. When the deviceis off (approximating an open switch) the voltage across the component is large but the current isvery small and the power dissipation in the off state can usually be considered as zero.S1.2.1. Advances in microelectronics enabling the fabrication of high performance controllers inboth digital and analog forms.2. Revolutionary improvements in the capabilities (voltage, current, power dissipation,switching speeds) of semiconductor devices which operate as the switches in powerelectronic converters.3. Large expansion of the market for power electronic converters.S1.3.The table shown below characterizes the application areas in terms of the relative importance orpriority the power electronics designer must place on each of the listed specifications. Theassesments in the table are highly qualitative.ApplicationPwr RatingDynamicsEfficiencyCostResidential 10kWslowlow priorityhigh priorityCommercial 100 kWfasthigh priorityIndustrialall ecommunications 1MWmoderatemoderateprioritymoderatepriorityhigh priorityall rangesmoderatehigh priority 100kW 1kWmoderatefasthigh prioritymodratepriorityCopyright by John Wiley & tylow prioritymoderateprioritySize andWeightmoderateprioritylow prioritylow prioritymoderateprioritylow priorityhigh prioritymoderatepriority2

S1.4.a) Figure 1-3a is already diagrammed showing that the converter has two rectifiers, oneinverter, a transformer and two energy storage capacitors.b) Block diagram shown below.Safety IsolationRectifier &low; passfilterInverterRectifier &low; passfilterVoUtilityController by John Wiley & SonsLoadVref3

Chapter 2 - Overview of Semiconductor Power SwitchesS2.1.a) Ideal i-v curves for a diode and thyristor are shown below. A more complete figure for thediode is shown in Fig. 2-1 of the text and for the thyristor, Fig. 2-3 of the text.iivIdeal diode i-v curvevIdeal thyristor i-v curveb) Ideal characteristics are used when the basic operation of a converter circuit is beinganalyzed or designed from a top-down (system) view point. In this situation, idealizedcharacteristics greatly simplify the effort with minimal loss of accuracy. The nonidealcharacteristics which are eliminated in the idealizations are second order effects thatonly have minor effects on the overall converter characteristics.Real characteristics are used when the effects of the actual characteristics are to estimated.These effects are usually most important on the device itself and may cause the devicemaximum capabilities to be exceeded. For example the on-state resistance of a diode, anonideal characteristic, causes power dissipation in the diode when it is in the on-state. Anaccurate knowledge of the power losses is necessary to correctly dimension the heat sink. Ifthe power dissipation exceeds the maximum power rating of the diode, the internaltemperature of the diode may exceed the maximum allowable value and at the very least, thereliability of the diode will be significantly reduced.S2.2.a) Diode voltage and load resistor voltage shown below.20 millseconds322 V0 VtInputVoltage-322 V322 V0 V0 VLoadResistorVoltagettDiodeVoltage-322 V

1 0.01322Vd 0.02 322sin(2πx50t)dt π 102 volts0b) Devices in parallel have the same voltage impressed across them. The device with the lowerforward voltage across it for a given current, will conduct a larger current than the otherparalleled devices. This means that the device carrying the most current will dissipate morepower than the other paralled devices. Consequently the temperature will be larger in thisdevice than in the other devices. This higher temperature will lead to an even larger share ofthe current going through this device. The end result may be that the current in this devicemay exceed the rating of the device.

Chapter 3 - Review of Basic ConceptsS3.1.a) Input voltage shown below.Vi30 V04 µs2 µst6 µsIn the steady state, there is no average voltage across the inductor so the average of Vi equalsthe average of Vo. Vi b) 250 W (30)(4) 20 V Vo.6(20)2400, R 250 1.6 Ω ; (Io)2(1.6) 250 ; Io R2501.6 12.5Ac) Inductor voltage vi(t) - Vo where it is assumed that the time-varying portion of the outputvoltage is very small (C very large). Hence inductor voltage and current waveforms are asdiL(t)shown below. For the inductor vL(t) L dt . During the time intervals when the inductorvoltage is constant, the inductor will change linearly with time.10 vDuring the 4 µs interval the inductor current I1 5 µH t I1 2x106 tAt the end of the 4 µsec interval I1 8 I2 or I2 - I1 8 AI1(6µs) (8A)(4µs)(0.5) (8A)(2µs)(0.5)Average current Io 12.5 A I1 4A6 µsHence I1 8.5A and I2 16.5 A

Vi10 Vt-20 V2 µs4 µsi (t)L6 µsI2Io 12.5 AI1t [iL(t)]2 [Io iripple(t)]2 [Io]2 2 Io iripple(t) [iripple(t)]2iripple(t) is shown below. As is clear from the diagram it has no average value.Hence iripple(t) 0. Thus [iL(t)]2 [Io]2 [iripple(t)]2 ;Mean square value of a triangular wave {base-to-peak}2/3 ; See solutions to problem 3-3ein existing solutions manual.[I2 - I1]2/464/4 [iripple(t)]2 : 3 5.33 A23To find IL,rms [Io]2 [iripple(t)]2 (12.5)2 5.33 12.71 A(20)240080, R 80 5Ω ; (Io)2(5) 80W ; Io R5 4AThe inductor voltage waveform is unchanged from part b). The current waveform is asshown below.d) 80 W

4 µsi (t)L2 µs6 µs8 A I24A I o0 I1Q1Q1 Q2Q2tThe ripple current iripple(t)]is unchanged from part b) since it is governed by the inductorvoltage. Thus [iripple(t)]2 5.33 A2.IL,rms (4)2 5.33 4.62 AIL,rmsIL,rms12.71 12.5 1.017 : For part c)I)L,avg4.62e) For part b) 4 1.16IL,avgThe ripple current in the inductor is independent of the load. Thus as the load resistanceincreases, the average current is reduced and the ratio of the rms current to the averagecurrent gets larger.f) The current in the capacitor is equal to the inductor current minus the current in the load.The inductor current IL,avg iripple(t) . The capacitor cannot conduct any dc current suchas IL,avg. Hence the capacitor current equals the ripple current through the inductor. Thecapacitor currents for parts b) and c) are the same and are shown below.The rms ripple current in the inductor was found to be 5.33 2.31 A in parts b) and c).Thus the rms capacitor current in both situations is equal to 2.31 A.

4 µsi (t)C2 µs6 µs4At-4 AQ1Q2Q1 Q2S3.2.a) The waveform in Fig. 3-3a is a square wave. The rms value of the fundamental is given by(see solutions to prob. 3-3 in the solutions manual of the second or third edition, both are the4Asame) F1 (1.414)(π) 100 amps where A is the base-to-peak amplitude of the squarewave.Solving for A 110.06 amps.The rms value of a square wave is equal to its ampliude. Thus Irms 110.06 amps.4A cos(π/6)b) For the waveform of Fig. 3-3b, F1 (1.414)(π) (see solutions to prob. 3-3. in the solutions4A cos(π/6)of the second or third edition). (1.414)(π) 100 amps. Solving for the amplitudeA 126.9 amps.The waveform in Fig. 3-3b is rectangular waveform which the signal is equal to zero for 1/3(126.9)2 (2)of the time. Thus the rms current Irms 103.9 amps.3c) The larger the fraction of a period that a periodic waveform has a value of zero, the smallerthe rms value of the waveform will be compared to its base-to-peak amplitude.S3.3.N2a) Inductance L R where R is the reluctance of the magnetic path thru the core and gapcombined. See Eq. 3-69 in text.

lgapIf the reluctance of the core can be neglected, then R µ A ; lgap 1 mm is the airgapo clength, µo 4πx10-7 is the magnetic permeability of free space, and Ac 1 cm2 is the areaof the core and gap (ignoring flux fringing in the gap).µo Ac N2 (4πx10-7)(10-4)(100)2Thus L 1.26 millihenrieslgap10-3b) LI Nφ : Εq. 3-67 of text. Imax N Ac Bmax (100)(10-4)(0.3) 2.38 AL1.26x10-3L (Imax)2(1.26mH)(2.4)2c) Maximum energy stored in inductor 0.0036 joules22(200)( 10-4)(0.3)d) L (0.00126)(4) 5 millihenries ; Imax 1.2 amps5x10-3(5x10-3)(1.2)2Maximum energy stored in inductor 0.0036 joules2(0.3)2(10-4)(10-3) 0.0036 joules. This method of estimating the energy stored in the(2)(4πx10-7)airgap agrees with the value obtained using the inductance and maximum currents.e) W S3.4.The figure reference in the problem statement should be Fig. P3-3b.a) Transformer equivalent circuit (no leakage inductance) shown below. When there is no load,isec 0, and the primary current is composed of only magnetizing current im.dimLm dt vpri ; Ignoring reluctance of core, magnetizing inductance given byµo Ac N2 (4πx10-7)(3x10-4)(12)2Lm 54.4 µhenrieslgap 10-3

ivpri -priLmimiisecnsecload vsec-n:1ideal tranformerL m magnitizing inductanceVoltage, flux density, and magnetizing current waveforms shown below. Amplitudes ofcurrent and flux density calculated in part b).20 µsecvpri6.67µsec3.33µsec6.67µsec3.33µsec300 V0-300 Vtim18.5 A0t-18.5 ABcore0.28 T0t-0.28 Tb) During the 6.67 µsec time interval when the primary voltage is 300 V, the magnetizingdimcurrent is found from the equation Lm dt vpri Integrating this equation , assuming t 0is at the start of the interval when the primary voltage steps up to 300 v, yields by John Wiley & Sons11

vpri tim(t) im(0) L; Over the 6.67 µsec time interval, the total change in currentm(300v)(6.67µsec)Δ im 54µhenries 37 A. Hence the amplitude of the magnetizing current is 18.5 A.Estimate the flux density using LmIm Npri Acore Bcore where Im the base-to-peakvalue of the magnetizing current. Using this yields for the base-to-peak flux density(54µH)(18.5A)Bcore 0.28 T(12)(3x10-4 m2 )Note: This flux density exceeds the 0.2 T specified in the problem statement. The valueof saturation flux density should be increased to 0.3 T.c) The waveforms for the magnetizing current and the square of the magnetizing current versustime are shown below. The square of the current versus time are easily derived from thecurrent versus time waveform. Also shown on the waveform are the time functions that applyto the square of the current during the time intervals that the current is changing linearly intime.im18.5 A0t-18.5 A3.33 3.33 3.33 3.33 3.33 3.33µsec µsec µsec µsec µsec µsec2m2342 Ai0t20 µsec2342 At3.33µs22342 A1-t23.33µsThe area under one of the flat segments is (342 A2 )(3,33µs) 0.00114 A2-sec.The area under one of the parabolic segments is (342 A2 )(3,33µs)/3 0.00038 A2-sec.

(2)(0.00114) (4)(0.00038) 190 A2 ;20 µsecRMS value of the magnetizing current 190 13.8 A (im)2 d) New component of primary current shown below.20 µsecim6.67µsec6.67µsec3.33µsec20 A0t-20 ARMS value of the new primary current component is given by(20)2(13.33 µsec) 16.33 A20 µsece) The square of the total primary current can be written as(im1(t) im2(t))2 (im1(t))2 2im1(t) im2(t) (im2(t))2 ;im1(t) is the magnetizing current shown in part a). im2(t) is the new primary currentcomponent shown in part d). When the time average (im1(t) im2(t))2 is computed, thecross-product term 2im1(t) im2(t) averages to zero. Hence the rms value of the total primarycurrent is; (im1(t) im2(t))2 190 267 21.4 Aµo Ac N2(4πx10-7)(3x10-4)(12)2f) Lm 544 µhenrieslgap10-4The magnetizing current waveform remains the same shape, but the amplitude is now 1.85 A.The flux density waveform remains unchanged in both shape and amplitude.The rms value of the magnetizing current is now 1.38 A.The 20 A component of the primary current is unaffected by the change in gap length.Thus the rms value of the total primary current is now (im1(t) im2(t))2 1.9 267 16.4 A

g) For the voltage waveform shown in part a) vpri 300 V Npri Acore 2Bsat /(ΔT) where1ΔT 6.67 µsec 3f where f is the frequency of the voltage waveform (50 kHz in part a).Letting f vary, but keeping the interval that the voltage is positive and equal to 300 V at 1/3300of the period yields at the edge of core saturation Npri Acore 2Bsat 3fSolving for f 300 46.3 kHz.(12)(3x10-4)(2)(0.3)(3)h) This lowest frequency does not depend on the length of the airgap. The derivation in part g)does not depend on the value of inductance and hence the length of the airgap. by John Wiley & Sons14

Chapter 5 - Line-Frequency Diode RectifiersS5.1.a) Voltage and current plots shown below.v (t)dvs (t)tis (t)I 10Ad0T1Ts2T2Tstb) For vs(t) 0, D1 is off and D2 conducts. Inductor current equals zero. As vs(t) crosses zerogoing positive, it takes a finite time for the current is(t) to build up to 10 A. During thisbuildup