《电力电子技术》课程教学课件（PPT讲稿）Topic 15 Line-Frequency Diode Rectifiers

Topic 15Main contentsLine-Frequency Diode Rectifiers

Main contents • Line-Frequency Diode Rectifiers Topic 15

5-6Three-Phase,Full-BridgeRectifiersIn industrial applications, it is preferable to use three.phase rectifier, compared to single-phase rectifier,because of their lower ripple content in the waveformsand a higher power-handing capabilityADeC.太D4±D6±D2·Commonly used3

3 5-6 Three-Phase, Full- Bridge Rectifiers In industrial applications, it is preferable to use three￾phase rectifier, compared to single-phase rectifier, because of their lower ripple content in the waveforms and a higher power-handing capability. • Commonly used

·The circuit consists of atop group (positive) and abottom grope (negative)(5-64)Vd = Vpn-VNn·Four types of load:UPnnUdDCUNnD6ND4

4 •The circuit consists of a top group (positive) and a bottom grope (negative). = vPn −vNn vd (5−64) • Four types of load:

5-6-1ldealizedcircuitwithLs=0Consider rectifier model with large inductor load (R-L load)D太D5UPnUdUdDUNnD6太D4太D6D2ND4NldK(b)(a)Figure 5-31 Three-phase rectifier with a constant dc current

5 5-6-1 Idealized circuit with Ls=0 Consider rectifier model with large inductor load (R-L load)

Voltageand CurrentWaveformsThewaveformsforthevoltages Vpn and VNn,12VLlconsist of 120°segmentsArea(b)of the phase voltages;Va(t) = /2Vu cosot120°-609元元ot<66When DconductingWhen D, conducting0neitherD,orD4conductingFigure 5-32Waveforms in the circuit of Fig.5-31

6 Voltage and Current Waveforms   0 6 a d Id i =  − I (−   t   ) The waveforms for the voltages vPn and vNn, consist of 120°segments of the phase voltages; vd (t) = 2VLL cost 6 When D1 conducting When D conducting 4 neither D1 or D4 conducting

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 costd (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 5t − 1 sin 7t + 1 sin 11t +.)  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 sint