Power Electronics, 1st Edition Solution Manual

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CHAPTER 1 SOLUTIONS(1-1)

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(1-2)

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T i m e0 s2 u s4 u s6 u s8 u s1 0 u s1 2 u s1 4 u s1 6 u sV ( D 1 : 2 )- 5 V0 V5 V1 0 V1 5 V2 0 V2 5 VT i m e0 s2 u s4 u s6 u s8 u s1 0 u s1 2 u s1 4 u s1 6 u sV ( S 1 : 4 )- 5 V0 V5 V1 0 V1 5 V2 0 V2 5 V( 4 . 0 8 3 3 u , - 8 5 1 . 6 9 0 m )( 1 . 4 3 3 3 u , 2 3 . 8 0 0 )In part (b), the voltage across the current source is reduced from 24 V by the switch resistance and diodevoltage drop.

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(1-3)Time0s5us10us15usV(V2:-)-20V0V20V40V(3.150u,-1.052)(3.150u,-1.052)96.46n,23.94)

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(1-4)T i m e0 s2 u s4 u s6 u s8 u s1 0 u s1 2 u s1 4 u s1 6 u sV ( V 2 : - )- 5 V0 V5 V1 0 V1 5 V2 0 V2 5 V( 3 . 8 3 3 3 u , - 1 . 0 5 1 7 )( 8 0 0 . 0 0 0 n , 2 3 . 9 2 4 )

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CHAPTER 2 SOLUTIONS2/21/102-1) Square waves and triangular waves for voltage and current are two examples._____________________________________________________________________________________2-2)a)( )( ) ( )( )()222[170sin 377]2890sin 377.10vttp tv t i tt WR====b)peak power = 2890 W.c) P = 2890/2 = 1445 W._____________________________________________________________________________________2-3)v(t) = 5sin2πt V.a) 4sin2πt A.; p(t) = v(t)i(t) = 20 sin22πt W.; P = 10 W.b) 3sin4πt A.; p(t) = 15sin(2πt)sin(4πt) W.; P = 0_____________________________________________________________________________________2-4)a)( )( ) ( )0050405070070100tmsp tv t i tmstmsmstms==b)( ) ( )7005011408.0.100msTmsPv t i t dtdtWTms===c)( )()()7005040800.;8100800.msTmsWp t dtdtmJorWPTWmsmJ======_____________________________________________________________________________________2-5)a)( )( ) ( )70.0650.61040.101401420WtmsWmstmsp tv t i tWmstmsmstms−== b)( )()61014006101170504019.20msmsmsTmsmsPp t dtdtdtdtWTms==+−+=c)

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( )()()()61014006107050400.38.;1920380.msmsmsTmsmsWp t dtdtdtdtJorWPTmsmJ==+−+====_____________________________________________________________________________________2-6)()()()())2.,12224.)3.1.,123.137.2.dcavgavgavgPV IaIAPWbIAPW=======_____________________________________________________________________________________2-7)a)( )( )( )( ) ( )()()()( )2025sin 377.25sin 3771.0sin 37725sin 37712.5 1cos 754.112.5.RTRvti t Rt Vp tv t i tttttWPp t dtWT======−==b)( )( )()()()( )( ) ( )()()()()( )3010 103771.0 cos 3773.77 cos 377.3.771.03.77 cos 3771.0sin 377sin 7541.89sin 754.210LLTLdi tvtLtt Vdtptv t i ttttt WPp t dtT−=========c)( )( ) ( )()()( )0121.0sin 37712sin 377.10Tdcp tv t i ttt WPp t dtT=====_____________________________________________________________________________________

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2-8)Resistor:( )( )( )( ) ( )()()( )21/601/601/6020000824sin 260.824sin 26026sin 2601696sin 260144sin 260.111696sin 260144sin 2601/ 60167288.Tv ti t Rt Vp tv t i ttttt WPp t dtdtt dttTW==+==++=++==++=+=Inductor:0.LP=dc source:( )( )2612.dcavgdcPIVW===_____________________________________________________________________________________2-9)a) With the heater on,()()( )()()( )()222150021500.12.5221202sin120212.52 sin3000sinmax3000.mmmmmV IPWIp tV Itttp tW==→======b) P = 1500(5/12) = 625 W.c) W = PT = (625 W)(12 s) = 7500 J.(or 1500(5) = 7500 W.)_____________________________________________________________________________________2-10)( )( )()()()()03119090004.0.149004103.6.tLLLitvt dtdttmsLimsA−=====a)()()2211 0.13.60.648.22WLiJ===b) All stored energy is absorbed by R: WR= 0.648 J.c)0.64816.2.4016.2.RRSRWPWTmsPPW=====d) No change in power supplied by the source: 16.2 W._____________________________________________________________________________________

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2-11)a)()( )()2002 1.212,15.49.20.01011141400.0.01015.49140011.1ttononWWLior iALi tvddt ALttms=========b) Energy stored in L must be transferred to the resistor in (20-11.1) = 8.9 ms.Allowing fivetime constants,8.9101.7.;5.6251.71.7LmsLmHmsRRmsms====_____________________________________________________________________________________2-12)a) i(t) = 1800t for 0 < t < 4 msi(4 ms) = 7.2A.; WLpeak= 1.296J.b)Time0s20ms40ms60ms80ms100ms-W(Vcc)-1.0KW0W1.0KWSourceinst.power(supplied)W(L1)-1.0KW0W1.0KWInd.inst.power-I(Vcc)-10A0A10ASourcecurrentI(L1)0A5A10ASEL>>Inductorcurrent_____________________________________________________________________________________

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2-13)a) The zener diode breaks down when the transistor turns off to maintain inductor current.b) Switch closed: 0 < t < 20 ms.( )()()12.12160/0.07520,1600.023.2.LLLLLditvVLdtdivA sdtLat tmsiA========Switch open, zener on:12208.8106.7/0.075:3.230106.7106.7LLLvVdivA sdtLt to return to zeroitms=−= −−=== −− ===−−Therefore, inductor current returns to zero at 20 + 30 = 50 ms.iL= 0 for 50 ms < t < 70 ms.c)Time0s10ms20ms30ms40ms50ms60ms70msW(D1)0W40mW80mWSEL>>Zenerinst.powerW(L1)-40mW0W40mWInductorinst.power

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d)( )()()00.111 0.036413.73.0.072LTZZPPpt dtWT====_____________________________________________________________________________________2-14)a) The zener diode breaks down when the transistor turns off to maintain inductor current.b) Switch closed: 0 < t < 15 ms.( )()()20.20400/0.05015,4000.0156.0.LLLLLditvVLdtdivA sdtLat tmsiA========Switch open, zener on:203010.10200/0.050:6.030200200LLLvVdivA sdtLt to return to zeroitms=−= −−=== −− ===−−Therefore, inductorcurrent returns to zero at 15 + 30 = 45 ms.iL= 0 for 45 ms < t < 75 ms.c)Time0s20ms40ms60ms80msW(D1)0W100W200WSEL>>Zenerinst.powerW(L1)-200W0W200WInductorinst.power

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d)( )()()00.111 0.0318036.0.0752LTZZPPpt dtWT====_____________________________________________________________________________________2-15)Examples are square wave (Vrms = Vm) and a triangular wave (Vrms = Vm/√3)._____________________________________________________________________________________2-16)Phase conductors:()22120.572.PI RW===Neutral conductor:()()221230.5216.NPI RW===()()223 72216432.720.167123totalNNNPWPRI=+====_____________________________________________________________________________________2-17)Re: Prob. 2-4100.78.37.40.52.83.rmsmrmsmVVDVIIDA======_____________________________________________________________________________________2-18) Re: Prob. 2-5()0.0060.010.0222200.0060.0114108.36.20175427.75.26.0.02rmsmrmsVVDVIdtdtdtA====+−+==_____________________________________________________________________________________

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2-19)()()()()()()2222220015324.58.2221.11.52.2.22cos25231.12.01.5cos20cos1157.0.2222cos(46045 )cos 460135rmsrmsmmnnnVVIAV IPV IWNote thattist==++==++==+−=+− +− =−+−_____________________________________________________________________________________2-20)()()()()( )()()()()()01111122222001:3 100300.260 :1/0.010.01894018762.10.010.0189460 :1/0.010.03776015375.10.010.0377cos218741536300 5cos 62.1cos 722mmnnndcVVYRjCjIVYjYRjCjIVYjV IPV I=====+=+=== −+==+=+=== −+=+−=+ +()5.115001751181793.W=++=_____________________________________________________________________________________2-21)dc Source:501212114.4dcdcavgPVIW−===Resistor:

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()()()()222201,2,01222229.5.303.51.44600.01100.641.48600.013.510.6419.59.83.22386.rmsrmsrmsrmsrmsRrmsPIRIIIIIAIAjIAjIAPIRW==++===+==+=++===_____________________________________________________________________________________2-22)()()()()() ()200122222260.375.1650.269.162600.02530.0923.166600.0250.2690.09230.3750.426.220.623.;0.426162.9.rmsrmsrmsrmsPIRVIARIAjIAjIAIAPIRW======+==+=++=====_____________________________________________________________________________________2-23)()001cos2mmnnnV IPV I==+−nVnInPn∑Pn02051001001205501502101.256.25156.2536.670.5561.85158.1450.31250.781158.9Power including terms through n = 4 is 158.9 watts._____________________________________________________________________________________

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2-24)()001cos2mmnnnV IPV I==+−nVnInθn-ϕn°Pn050.000010.00500.0150.000010.026.6223.6225.00002.545.022.1316.66671.1156.35.1412.50000.62563.41.7Through n = 4, ∑Pn= 753 W._____________________________________________________________________________________2-25)()001cos2mmnnnV IPV I==+−()()()00220,0050360.7200.7209.8()0.73625.20dcRVdcdcLVVIARPI RWdc component onlyPI VWP−−==========Resistor Average PowernVnZnInanglePn050.0020.000.70.009.81127.3225.435.010.67250.66263.6637.241.711.0029.22342.4451.160.831.176.87431.8365.940.481.262.33525.4681.050.311.320.99PR= ∑ Pn≈ 300 W._____________________________________________________________________________________2-26)a) THD = 5% → I9= (0.05)(10) = 0.5 A.b) THD = 10% → I9= (0.10)(10) = 1 A.c) THD = 20% → I9= (0.20)(10) = 2 A.d) THD = 40% → I9=(0.40)(10) = 4 A._____________________________________________________________________________________2-27)a)()17010cos 3000736.22nPPW== ++=

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