The waveform of the dc-side voltage Va (=Vpn-Vnn)consists of 6o°segments of the line-line voltagesHence, this rectifier is often termed a six-pulse rectifier(六脉波整流器)/2Vucosotd (ot) =2Vu.=1.35VLL(5-68)元人N2VAreaAVao(b)0ot#0元66:Conduction sequence: ab→ac-→bc-→ba→ca→cb→ab
7 The waveform of the dc-side voltage vd (=vPn-vNn) consists of 60°segments of the line-line voltages. Hence, this rectifier is often termed a six-pulse rectifier (六脉波整流器). 1 3 6 2VLL = 1.35VLL 2V (5− 68) LL costd (t) = Vd 0 = 3 − 6 • Conduction sequence: ab→ac→bc→ba→ca→cb→ab
InputLine-Current1.0aLahOt2111131115713(a)(b)Figure 5-33 Line current in a three-phase rectifier in the idealized case with L, = 0 anda constantdc current.sin l1ot+.sin5@1ot-sin1115元
8 5 7 11 s d i = 2 3 I (sin t − 1 sin 5t − 1 sin 7t + 1 sin 11t +.) Input Line-Current
2=0.8161(5-69)3:0.78(5 -70)元1s(5-71)shhDPF=1.0(5 - 72)3DPFPF=0.955(5 -73)I.元. The input power factor is higher than that of single-phaserectifier.9
(5 − 71) (5 −72) = 0.78 (5 −70) DPF = 1.0 h I 6 I = I Is PF = Is1 DPF = 3 = 0.955 (5 −73) = Is1 sh s1 d (5− 69) phase rectifier. 9 s d = 0.816 d I 2 I 3 I = • The input power factor is higher than that of single-
5-6-2EffectofL.oncurrentcommutationFD1太Da太D5UdFigure5-34Three-phaserectifier太 D4 D6 D2with afinite L,and a constant dccurrent.Previously,we assume that the ac-side inductance Ls=0,so the current commutation is finished instantaneouslyNow we consider the existence of Ls: In such case, theinput current needs a finite amount of time to reverse its10E-mail:dhirerCationedm
5-6-2 Effect of Ls on current commutation Previously, we assume that the ac-side inductance Ls=0, so the current commutation is finished instantaneously. Now we consider the existence of Ls . In such case, the input current needs a finite amount of time to reverse its10 E-mail: xjdhuiraengc@tbiojtun.e.du.cn
1CurrentCommutationConsider the commutation ofcurrent from D5/Dto D/D1:During commutation interval:comyid-ld18ia=iuic=I6NLiddidi(a)11VldtdtLIddidi004u0(b)dtdtdiUecnVanUbn2Vu sinotV2uCHAreaA2dt2+下0udi+VcnV1VpnVNan2dt(c)Figure 5-35Current commutation process
11 1 Current Commutation 2 2 Pn an s s s c s a dt 2 di dt di L dt dt dt di di dt di di = van + vcn v = v − L = = van − vcn vLc = L s = − L vLa = L s = L Consider the commutation of current from D 5 /D6 to D 6 /D 1. During commutation interval: ia = i ic = Id − i 2VLL sint