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