Question

A single-phase H-bridge topology below is operated with the input dc voltage Vdc=200 V, modulation index, ma=0.6 and switching frequency fs=1950 Hz. The fundamental frequency of the generated AC voltage is 50 Hz. a) Calculate the rms value of fundamental component b) Calculate the rms values of most significant harmonics (with at least %10 of amplitude of the biggest harmonics) for the pole voltage v10 indicated on the inverter c) Calculate the rms voltage value for the fundamental and for the harmonic number 39 of the output voltage Vsint 91 Vac 2 92 R driver 91 driver 1 > . Vo Voc 2 N 91 92 92 91 Vsina To 92 *91 K 72 driver driver 92 V sint Sin2 AWAMATAN -100 UUTTTTTIILIKO he amplitude of the fundamental and harmonics can be obtained from the gure below that shows the relationship between Vion/(Vdc/2) and the har- ionics. V20 1.00 m,-0.6 -100 100 -100 200 -200 50 -50 mp21 0.60 io 0.37 0.37 2017.10.07 1 1.025 0.13 0.13 07 10 my m2 'mpt2 1.05 0 2m-1 2+1 2m 3 2m5 25 0.20 0.20 0 0.040.08 0.040 my 3m2 3+2 Harmonic 3m 3m 3m 6 3m6 23

Answer 1

Solution a) from the amplitude of harmonie diagram Koch) = 0.6 for fondamental component of Voch pole voltage Vio fundamental -0.6 x Vdcl2 =0.6x2 20.6* = 60 V Rms value of feendamental = 60.5 = 60424V amplitude Vio (21)=1.00 xVdc vode = 1.00x24 RM s Vio (21) 100 =70.7V b) harmonic with mf= = 21 = loov 2 v2 = 2x21 Il - MF + 2 =21+2 VO (23) = Viol19 )= 0.13 X 200 = 13 V Rms V10 (19) = RMS VO(23) = 13/12 = 9-2 Qm6+ V10 (41) = Vo(43) = 0.3 7X200 RMS Vio (41) = V10(43) - 37 ป 3 mf = 3x21563 Rm 3 Vio (63) = 0.2 x 200 2 all other harmonic are less than 10% of biggest harmonic ie, 21 harmonic = 26 V =14 V CSScanned with CamScanner

c output voltage Vo=40-120=2010 so fundamental &ms voltage Rm s Vo (1) = 2x Rms Vo= 2x42:4 = 84.8V for 39 numbles harmonic Rms Vo (39) = 2x Rms Vio (39) 2 x 0.07* 200 2 - 14V CS$canned with CamScanner

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