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 − 2Ls Id The base impedance is 230 = =10.58() rated base I 21.74 = Vrated Z Therefore base L s = 0.05Z = 0.0510.58 = 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