《电力电子技术》课程教学课件（PPT讲稿）Topic 18 Three-phase Thyristors Rectifier

Topic 18Main contentsThree-phase Thyristors Rectifier

Main contents • Three-phase Thyristors Rectifier Topic 18

Example6-1In the converter circuit of Fig. 6-9, Ls is 5% with the ratedvoltage of 230V at 60 Hz and the rated volt-amperes of5kVA. Calculate the commutation angle μ and V. with therate input voltage, power of 3kW,and α=30°DdFigure 6-9Single-phasethyristor converterwitha finiteL.and a constantdc current

Example 6-1 In the converter circuit of Fig. 6-9, Ls is 5% with the rated voltage of 230V at 60 Hz and the rated volt-amperes of 5kVA. Calculate the commutation angle µ and Vd with the rate input voltage, power of 3kW, and α=30°

Solution2L,Ia(6-24)cos(α+ μ) = cosαJ2VS5000rated21.74(A)1The rated current isratedV230rated230Vrated710.58()Thebase impedance isbaseIrated21.740.05Z0.05x10.58Therefore= 1.4(mH)aS2×元×600The average power through the converterP, =Val -(0.9V, cos α - 2 oL,Ja)la = 3(w)元

Solution The rated current is V 230 I rated rated = 5000 = = 21.74(A) Srated d d d s s d d = 3(kW)  P =V I = (0.9V cos − 2 L I )I The average power through the converter (6 − 24) 2Vs cos( + ) = cos − 2Ls Id The base impedance is 230 = =10.58() rated base I 21.74 = Vrated Z Therefore base L s = 0.05Z = 0.0510.58 = 1.4(mH)  2 60

SolutionUsing the given values in the above equation givesI3 533.531 +8928.6= 0ThereIa= 17.3(A)Using this value of Ig in Eqs. 6-24 and 6-26 results inμ=5.9° and Va=173.5 (V)

Solution Using the given values in the above equation gives I 2 d − d 533.53I +8928.6 = 0 There Id = 17.3( A) Using this value of Id in Eqs. 6-24 and 6-26 results in µ=5.9°and Vd=173.5 (V)

6.InverterModeofOperation(a)α3, 4(iG1,2人o/3,4t-0t(b)Figure 6-15(a) Inverter, assuming a constant dc current. (6) Waveforms

6. Inverter Mode of Operation

idLd(0thy/3+EdUd20L,I dEα =Va = 0.9V. cosα(a)元Vaa3Id=Id002αiEd=Edl(b)Figure 6-16(a) Thyristor inverter with a dc voltage source. (b) Va versus la

s d s 2 L I Ed = Vd = 0.9 V cos  −

=180°-(α+u)(uthy'32wtoα3,4(iq1,2(iq)3, 4otn-wt0Figure 6-17 Voltage across a thyristor in the inverter mode.·Importance of extinction angle in inverter mode

 =180 − ( +u) o • Importance of extinction angle in inverter mode

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