DiodewithanRLCLoad-UniversityofMassachusetts…：与RLC负载二极管-麻州大学…

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1、DiodewithanRLCLoadvL(t)vC(t)VCoClosetheswitchatt=0VCoKVLaroundtheloopCharacteristicEquation3CasesCase1=ω0“criticallydamped”s1=s2=-rootsareequali(t)=(A1+A2t)es1t3Cases(continued)Case2>ω0“overdamped”rootsarerealanddistincti(t)=A1es2t+A2es2t3Cases(continued)Case3

2、<ω0“underdamped”s1,2=-+/-jωrωr=the“ringing”frequency,orthedampedresonantfrequencyωr=√ωo2–α2i(t)=e-t(A1cosωrt+A2sinωrt)exponentiallydampedsinusoidExample2.6DetermineanexpressionforthecurrentDetermineanexpressionforthecurrentDeterminetheconductiontimeofthediodeT

3、heconductiontimewilloccurwhenthecurrentgoesthroughzero.ConductionTimeFreewheelingDiodesFreewheelingDiodeFreewheelingDiodesD2isreversebiasedwhentheswitchisclosedWhentheswitchopens,currentintheinductorcontinues.D2becomesforwardbiased,“discharging”theinductor.Analyz

4、ingthecircuitConsider2“Modes”ofoperation.Mode1iswhentheswitchisclosed.Mode2iswhentheswitchisopened.CircuitinMode1i1(t)Mode1(continued)CircuitinMode2I1i2Mode2(continued)Example2.7InductorCurrentRecoveryofTrappedEnergyReturnStoredEnergytotheSourceAddasecondwinding

5、andadiode“Feedback”windingTheinductorandfeedbackwindinglooklikeatransformerEquivalentCircuitLm=MagnetizingInductancev2/v1=N2/N1=i1/i2ReferSecondarytoPrimarySideOperationalMode1Switchclosed@t=0DiodeD1isreversebiased,ai2=0Vs=vD/a–Vs/avD=Vs(1+a)=reversediodevoltage

6、primarycurrenti1=isVs=Lm(di1/dt)i1(t)=(Vs/Lm)tfor0<=t<=t1OperationalMode2Begins@t=t1whenswitchisopenedi1(t=t1)=(Vs/Lm)t1=initialcurrentI0Lm(di1/dt)+Vs/a=0i1(t)=-(Vs/aLm)t+I0for0<=t<=t2Findtheconductiontimet2Solve-(Vs/aLm)t2+I0=0yieldst2=(aLmI0)/VsI0=(Vst1)/Lmt1=

7、(LmI0)Vst2=at1WaveformSummaryExample2.8Lm=250μHN1=10N2=100VS=220VThereisnoinitialcurrent.Switchisclosedforatimet1=50μs,thenopened.Leakageinductancesandresistancesofthetransformer=0.DeterminethereversevoltageofD1Theturnsratioisa=N2/N1=100/10=10vD=VS(1+a)=(220V)(1+

8、10)=2420VoltsCalculatethepeakvalueoftheprimaryandsecondarycurrentsFromabove,I0=(Vs/Lm)t1I0=(220V/250μH)(50μs)=44AmperesI’0=I0/a=44A/10=4.4AmperesDeterminethec