Question

If $V_s$ is the input dc voltage in single PWM inverter, then the pulse width required for eliminating third harmonic and corresponding rms value of fundamental component of output voltage are

• A. $60{}^{\circ },\frac{\sqrt{6}}{\pi }{V}_s$
• B. $120{}^{\circ },\frac{2\sqrt{2}}{\pi }{V}_s$
• C. $60{}^{\circ },\frac{2\sqrt{2}}{\pi }{V}_s$
• D. $120{}^{\circ },\frac{\sqrt{6}}{\pi }{V}_s$

Answer 1

The correct option is D $120{}^{\circ },\frac{\sqrt{6}}{\pi }{V}_s$

$V_{0n,rms}=\frac{4V_s}{n\pi \sqrt{2}}\sin nd\cdot \sin \frac{n\pi }{2}$

$nd=\pi$
For eliminating third harmonic,

$3d=\pi$

$d=\frac{\pi }{3}=60{}^{\circ }$

$\therefore \text{pulse width,}2d=120{}^{\circ }$

$V_{01,rms}=\frac{2\sqrt{2}{V}_s}{\pi }\times \sin 60{}^{\circ }=\frac{\sqrt{6}{V}_s}{\pi }V$