《电力电子技术》课程教学课件（PPT讲稿）Topic 17 Three-phase Thyristors Rectifier

Topic 17Main contentsThree-phase Thyristors Rectifier

Main contents • Three-phase Thyristors Rectifier Topic 17

6-4 Three-Phase Full-BridgeConvertersNa大。大(b)(a)·Common-cathode group and common-anode group ofthyristors·Numbering of the 6 thyristors indicates the trigger12345612sequence:

6-4 Three-Phase Full- Bridge Converters •Common-cathode group and common-anode group of thyristors •Numbering of the 6 thyristors indicates the trigger sequence: 1→2→3→4→5→6→1→2→

6-4-1 ldealized Circuit with Ls=0 and ia(t)=ld3TT达T5DaT天T6ATNN22(a)(b)Figure 6-19 Three-phase thyristor converter with L, = O and a constant dc current

6-4-1 Idealized Circuit with Ls=0 and id (t)=Id

1 anlysisthe instant of natural conduction.the crossover pointof phase (or line) voltage (wty,wt2,wt3,wt4,wts,wt)UdUPnVeaUbnTVanUen3iaVdoot(a)0Vdwt=0ia66Thewaveformsofα=O°ia = iT1-iT43/2V.sinotdot=1.35VLL(6 -36)doTE-mail:xihuang@bitu.edBu.cn

1 anlysis • the instant of natural conduction.——the crossover point of phase (or line) voltage (ωt1 ,ωt2 ,ωt3 ,ωt4 ,ωt5 ,ωt6 ) π 1 3 2π E-mail: xjhuang@bjtu.ed3u.c3n d0 ia = iT 1 −iT 4 V = (6 −36) π 2VLL LL sintdt =1.35V The waveforms of  = 0º vd

UenUbnVanUpn-t(b)0wt=0αVenUNnia+ot0(c)iaia-.-UcbUcaUbaUbeUacVabUebUdOcnUbnDanVenuda@(d)0ot=oaWaveforms in the converter of Fig. 6-19.Figure6-202瓜+α13/2Vusin otdot= 1.35Vucosα = Vao cosα(6- 40)1元da元+α373

31π 3 3π V = (6 − 40)  2π +  +  d  2VLLsin t d t = 1.35 VLL cos  = Vd0 cos 

·The conducting intervals of the thyristors are120°Therefore, the rms current through thyristor:13.The conducting intervals of each phase are 240°Therefore, the rms AC input current2I. = I, = I.V3ia= iT1- iT4it = iT3 - iT6i= it5- iT2

•The conducting intervals of the thyristors are 120º. Therefore, the rms current through thyristor: •The conducting intervals of each phase are 240º. Therefore, the rms AC input current 3 T I = I d c d 2 I 3 I a = Ib = I = ia = iT 1 − iT4 ib= iT 3 − iT 6 ic = iT 5 − iT2

.Thefrequencyα =30 °Vdof the outputVabcbripple is six timesthe supplyfrequency.0al27yL sinotdot=1.35l cosα = o cosα2F(6-40)23

π 1 LL 2π+ π + d LL d0 α α=30° 0 ωt vd V = 2V sintdt =1.35V cos = V cos (6 − 40) 3 3 3 vab vcb vab vac vbc vba vca vcb •The frequency of the output ripple is six times the supply frequency

α =60 °VacVbeVbaVeaVebVabVab0wt12VL sinotdot=1.35l cosα = xo cosα(6-40)2

π 1 LL 2π+ π + d LL d0 0 ωt vd V = 2V sintdt =1.35V cos = V cos (6 − 40) 3 3 3 vab vac vbc vba vca α α=60° vcb vab

α=90 ·When α=90°Va=0, therefore, inVcaVbaVcbVacVbcVabchrectifier mode,thetheoretical rangeXNNNNXof α is 0 0 ~ 90°.0t(In practical, whenVa-→0, la-→0 andia discontinuous)VabTheoreticalwavePracticalwave2元3/2V. sin otdot =1.35Vu, cosα = Vao cosα(6-40)a元+373

π 1 LL 2π+ π + d LL d0 •When α=90º, Vd=0, therefore, in rectifier mode, the theoretical range of α is 0 º ～ 90º. (In practical, when Vd→0, Id→0 and id discontinuous). 0 ωt vd V = 2V sintdt =1.35V cos = V cos (6 − 40) 3 3 3 vab α α=90 ° vcb vab vac vbc vba vca vcb Theoretical wave Practical wave

fa)a=0t(e)a=120g(6)Q=30020(D)a=150°(c)a=60°D01(g)a=180%(d)a=900Figure 6-21 The dc-side voltage waveforms as a-ofunction of a where Vaa= A/(/3). (From ref. 2withpermission.)·lnrectifiermode,thetheoretical·lninvertermode,thetheoreticalrangeofαis0~90°rangeof α is 90°~180°

•In rectifier mode, the theoretical range of α is 0 º ～ 90º •In inverter mode, the theoretical range of α is 90 º ～ 180º