Question

A single phase full bridge inverter has RLC load. The dc input voltage is 230 V and the output frequency is 50 Hz. Find the expression for the load voltage up to the fifth harmonic.

(a) 292 sin 314t + 97.62 sin 314t + 58.57 sin 318t + 28.31 sin 318t + 3.686 sin 318t

(b) 292 sin 314t + 97.62 sin (3 x 314t) + 58.57 sin (5 x 318t)

(c) 292 sin 314t + 97.62 sin (2 x 314t) + 58.57 sin (3 x 318t) + 28.31 sin (4 x 318t) + 3.686 sin (5 x 318t)

(d) 292 sin 512t + 25.62 sin 249t + 6.74 sin 508t

I got this question in a national level competition.

My question comes from Single Phase VSI-4 topic in portion Inverters of Power Electronics

Answer 1

Right option is (b) 292 sin 314t + 97.62 sin (3 x 314t) + 58.57 sin (5 x 318t)

To explain I would say: In a single phase full bridge inverter only odd harmonics are present. i.e. 1,3,5 etc.

Vo = (4Vs/π) sin ωt + (4Vs/3π) sin 3ωt + (4Vs/5π) sin 5ωt

(4Vs/π) = 292 V

ωt = 2 x f x π x t = 2 x 3.14 x 50 x t = 314t.