Question

A three-phase full converter is supplied from a three phase 230 V, 60 Hz supply. The load current is continuous and negligible ripple. If the average load current is $I_{dc}=150A$ and the commutating inductance $L_C=0.1$ mH and the firing angle is $\alpha =20°$, then the overlap angle is

Answer 1

$V_{mph}=\frac{230\sqrt{2}}{\sqrt{3}}=187.79V$
$V_{d0}=\frac{3\sqrt{3}{V}_{mph}}{\pi }=310.61V$
$\Delta V_{d0}=\frac{V_{d0}}{2}[cos\alpha -cos(\alpha +\mu \left)\right]=6f_s{L}_s{I}_0$
$Where,V_{d0}=\frac{3V_{mL}}{\pi }=\frac{3\sqrt{3}{V}_{mph}}{\pi }$
$\frac{310.61}{2}[cos\alpha -cos(\alpha +\mu \left)\right]=6\times 60\times 150\times 0.1\times {10}^{-3}$
$or\frac{5.4\times 2}{310.61}=cos\alpha -cos(\alpha +\mu )$

$Given,\alpha =20°$
$cos\alpha -cos(\alpha +\mu )=0.03477$
$cos20°-0.03477=cos(20°+\mu )$
$cos(20°+\mu )=0.939669-0.03477$
$cos(20°+\mu )=0.9042$
$20°+\mu =25.187°$
$\mu =5.187=5.19°$