《电力电子技术》课程教学课件（PPT讲稿）Topic 10 Half-Bridge and Full-Bridge Converter、Push-Pull Converter、Comparison of isolated converter

Topic 10Main contentsLast lecture reviewHalf-Bridge and Full-Bridge ConverterPush-Pull Converter Comparison of isolated converterSummary

Main contents • Last lecture review • Half-Bridge and Full-Bridge Converter • Push-Pull Converter • Comparison of isolated converter • Summary Topic 10

1.1Half-BridgeConverter1convertercircuitODi(t)i0D30888200Vs(t)R880本CbD4DOHalf-Bridge converter requires two split capacitors Ca and Cp toform a dc input mid-point.The switches Q, and Qz are turned on alternatively, each for aninterval ton

1.1 Half-Bridge Converter Half-Bridge converter requires two split capacitors Ca and Cb to form a dc input mid-point. The switches Q1 and Q2 are turned on alternatively, each for an interval ton. 1 converter circuit

2.Principlesof Half bridge converter(1)WhenT,isONandT,isOFF十1本TC12+NuoiNiVV1YY1N,Va/2NVoiTCN,Va1CLou21ot2N.2T/2N, VaVAVN.2I。ip2=0ipi=it

2. Principles of Half bridge converter 1 21 1 2 oi L o N 2 v N1 2 v = Vd ,u = v = N2Vd = N2Vd −V iD1 = iL iD 2 =0 (1) When T1 is ON and T2 is OFF

(2) T, is OFF and T, is ON公+DI.+CL本DVCNuoi03N,3LViT2本C2T2N,V./2NVoiN,VaVaWoi=-V22V2N, 2tonN2VaT/2VAV2N,ipi= 0ip2=it

T2 1 22 1 oi L o N 2 v N1 2 v = − Vd u = −v = N2Vd 2 = N2Vd −V iD1 = 0 iD 2 =iL (2) T1 is OFF and T2 is ON

(3) Both T, and T2 are OFFD本Ti+CV+Voi = 0Vi=-V.UoiNiVND2ipr=iTD2本2C(a)TD =-Ian<0.5S+△：onN2Va/2NVoi2T,?In the condition of steady-state:oN.0+(-V)(s-tar0=0AtonTA-+OrN, 2T/2VN.NL2DVdNiNi T,I。T.is definedgenerallywhereD0Keep in mind!Ts(b)

(a) voi 00 tt N 2 V d / 2 N 1 ton  Ts /2 iL Io Vo (b) D1 D2 C1 L i L Io +v1 - + R uoi - +Vd Vd /2C2 - + Vo - N1 N2 T 1 T 2 N2 D 1 D 2 i voi = 0 vL = − Vo i = i = L2 (3) Both T1 and T2 are OFF 2 on s t +  = Ts , D = ton  0.5 T In the condition of steady -state: 1 2 o on o on o 2 on N V T N 2 2 D Vd N1 Ts N1 + ( −V )( s − t ) = 0  V = N t = N （ 2 d − V ） t D = Ton TS is defined generally. Keep in mind! where

tDiT,peak =Ti本+CiL+UoiVadiodessec ondaryNVVd22C2V D1,2,peakN1(a)N2Va/2N1diodesprimaryVD,peak = VdWhat is the role of anti-parallel diode?T,/2To provide a path for the currentrequired due to leakaaeflux oflthe transformer.1(b)

pri mary diod e s N diod e s = Vd V = V VD , peak N1 D 1 , 2 , p eak d T , p eak d V = ( 2 ) V sec ondary What is the role of anti - parallel diode? To provide a path for the current required due to leakage flux of the transformer. (a) v o i 00 tt N 2 V d / 2 N 1 ton  Ts/2 i L Io Vo (b) R D 1 D 2 C 1 L i L Io Vd/2 C2 +1 v- +uoi - +Vd - + Vo - N1 N2 T 1 T 2 N2

2.1 An practical Half-Bridge dc-dc supplyaIC1本D3本D1220VAC320VDCAS1X120-220VACV2120VACB12本02本04C202DWhat is the role of capacitor Cp?

2.1 An practical Half-Bridge dc-dc supply What is the role of capacitor Cb?

1.2Full-Bridge Converter1convertercircuitOODLi(t)iDs(t)i07080Ryt)0E不D,FDD6Full-bridgeisolatedbuckconverter

9 1.2 Full-Bridge Converter 1 converter circuit Full-bridge isolated buck converter

2OperationprincipleD本T大.T.(Similarto that of push-N1pull converter)V(1) When T3, T4 are on and T1, T2DKare off (O<t<ton): D, conducts(a)and D, gets reverse biased. iN2Va/N1through D, increase linearly1N2V2V211otN,T/2NV-V(10-31)N,Loip2=0ipi=il(b)

2 Operation principle (Similar to that of push - pull converter) (1) When T 3 , T4 are on and T 1 , T2 are off (0<t<ton ): D 1 conducts and D2 gets reverse biased. iL through D1 increase linearly. voi 00 tt N 2 V d/N 1 ton  Ts/2 iL Io Vo (b) R D 1 N1 iL Io (a) T 1 + N2 u 2 - N2 D2 L +vo i - +Vd - +Vo - +v1 - T 2 T 3 T 4 1 (10 −31) 1 d oi 21 d − Vo iD 2 = 0 N1 iD1 = iLN vL =  2 Vd N v = V v = v = N2 V

(2) When T1, T2 are on and T3WDZ.本T本.T4 are off (0<t<ton): D1 getsNuV.reverse biased and D,NDconducts. i, through D2T4本Tincreaselinearly(a)N2Va/NN.10VVi = :VotNIV.hV7.-V(10 -31)VN,T/2=0ip2 =iD110t(b)

(2) When T 1 , T2 are on and T 3 , T 4 a o n 1 re o ff ( 0 < t<t ): D gets reverse biased and D 2 conducts. iL through D2 increase linearly. L L d o N v N − V D 1 D 2 1 =  2 V 1 v = − V v = − v = N2 1 d oi 22 Vd N i = 0 i = i (10 −31) voi 0 t N 2 V d/N 1 ton  Ts /2 iL Vo R Io (b) t 0 D 1 D2 N1 2 N2 iL Io (a) T 1 +- N u 2 L +vo i - +Vd - +Vo - +v1 - T 2 T 3 T 4