Question

For the full-bridge inverter in Fig. 6, the switching frequency f, is 1,500 Hz. The DC voltage, V, is 600 V. Output voltage is sinusoidal voltage with a frequency equal to 60 Hz. The load is connected between leg A and B 5. a. Prove that the peak voltage of the fundamental frequency component, (Vo)m (V). b. Find the angle between the phasors vo and eafor-0.8, P-1,000 W and Q-500 w peak a Lla A + B+ VO B+ L= 15 mH ea DB- 2 Fig. 6. Full-bridge inverter for problem 5

Answer 1

Refer to the given problem Figure 6," Full bridge inverter" Consider the given data 600 V f - 1500 Hz Frequency of output volt-60 Hz L=15 mH S (a) Consider the Fourier series of the waveform ,()- Σν, sin (no,f) (1) nodd Here, V () is the output voltage and, The expression for the amplitude V, using half-wave symmetry is given as, 2 V D- ()p(Pou) us 2 2V cos (na) n 4V cos (na nt 4V cos (na for in equation (1) Substitute nTt 4V cos (na) |sin (na,t) (2) nT nodd

Thus, if PWM is used Consider the expression for the amplitude modulation m Since m is the amplitude modulation index and it is the ratio of the amplitudes of the reference and the carrier signal. Vd Vearrier If ma 1, the amplitude of the fundamental frequency of the output voltage V is linearly ma C proportional to ma, thus, equation (2) becomes, VmVd This is the peak value That is V . (3) peak V = Thus, the fundamental frequency amplitude of the inverter output is controlled by m Therefore, the required expression is, Vo,peak mad Hence proved.

(b) Consider the given data, ma 0.8, P 1000 W, Q==-500 W Find the rms value of the output voltage V Vorms o,peak m&Vi 0,rms Substitute 600 V for V, and 0.8 for m (0.8) (600) 0,ms VOms 339.411 V Consider the expression, Vrms ms Z V. rms joL rms j(27f)L -339.411 V, 15 mH for L, and 60 Hz for f Substitute V ms 339.411 J(27 (60)(15x103) Ipms 60.0524-90° Ims 60.052 A Here, Angle between voltage and current is 0-90°

Consider the expression for the power P P = V ms!rms COs(0-4) Here, is the angle between the voltage phasors Vo, and ea Substitute P 1000 W, V 339.411 V, and 0= -90°, and Ims 60.052 A 1000 (339.4) (60.052) cos (-90°-) 1000 cos (-90°- 20381.65 (-(90° +) = 0.049 cos cos (90° 0.049 On further simplification 90° cos + cos(-0) = cos 0) (0.049) 90°+87.19° 87.19° -90° -2.81° Therefore, the required angle between the voltage phasors V, and e is -2.81°