Question

2) For a full bridge inverter square wave voltage with harmonic control, Determine the de source. The fundamental frequency of the output is 50 Hz for a motor. Determine the value of a to produce an output with an amplitude of 600 V at the fundamental frequency, It is desired to eliminate the 5th harmonics for this operation

Answer 1

D1 - T It å based on sageli single pulse harmonic control. from single pulse modulation technique. we have expression ession for output voltage of for n-1,3,5 VE E { u Vacsiny. sin cond) ? sin (nust) LO * and also given that amplitude of fundamental component of output voltage = 6oov ie, Voi = Peak value l amplitude of fundamental - output voltage. for Vois fundanental op voltage pse writing expression liveno sub inco Voy a fuvac. sin(%). Sin (d)} svalut) 9n K; Vos = A sinut? now, Ivoi = 4Vdc sin (1) sind = 600 V I real value]

* now also given that there should not be 5th harmonic component ie vog zo is ordere to → Vos = fuvde, sin (59) .sio (5) } sin (swt) 571 = 0 In order to make this possible sin (5d) =0 . 5d - TJ Sub in ear ② - Voi = Vec vl. Sin (%) Cooruvdex 0.587 vode = 801.83 ut De voltage source. Now finding t., ad WH Vdc -- 20 i da T-20 1. = Tod L = T - TIS G15 ņ . Wt= 4T to your = 8x102 sec