Question

A three-phase half wave converter is operated from a three phase star- connected 400 V, 50 Hz supply and load resistance is R = 15 $\Omega$. If it is required to obtain an average output voltage of 60% of the maximum possible output voltage, then the delay angle $\alpha$ in degrees is

• A. 57.76

Answer 1

The correct option is A 57.76
For half wave converter,

Average output voltage = $\frac{3V_{ml}}{2\pi }cos\alpha$

Maximum possible average output voltage,

$V_{0,max}=\frac{3V_{ml}}{2\pi }=\frac{3\times 400\sqrt{2}}{2\pi }=270.09V$

Required voltage = $0.6\times 270.09=162.054V$

For resistive load, the load current is continuous if $\alpha \le \frac{\pi }{6}$ and average output voltage for $\alpha \le \frac{\pi }{6}$ is given by

$V_0=\frac{3V_{mL}}{2\pi }cos\alpha$

To meet the requirement, delay angle should be greater than ${30}^{\circ }$

With resistive load and $\alpha >{30}^{\circ }$, the average output voltage is given by,

$V_0=\frac{1}{\frac{2\pi }{3}}{\int }_{\frac{\pi }{6}+\alpha }^{\pi }{V}_{mph}sin\omega t.d\left(\omega t\right)$

$V_0=\frac{3V_{mph}}{2\pi }[1+cos(\frac{\pi }{6}+\alpha )]$

$162.054=\frac{3\times \frac{400}{\sqrt{3}}\times \sqrt{2}}{2\pi }[1+cos(\frac{\pi }{6}+\alpha )]$

$162.054=155.94[1+cos(\frac{\pi }{6}+\alpha )]$

$1+cos(\frac{\pi }{6}+\alpha )=1.0392$

$cos(\frac{\pi }{6}+\alpha )=0.0392$

$\frac{\pi }{6}+\alpha ={87.76}^{\circ }$

$\alpha ={87.76}^{\circ }-{30}^{\circ }$

$\alpha ={57.76}^{\circ }$