Question

A single phase full converter is made to deliver a constant load current. For zero degree firing angle, the overlap angle is ${15}^{\circ }.$
The overlap angle for firing angle of ${30}^{\circ }$ is

Answer 1

The dc load current is given by

$I_0=\frac{V_m}{\omega L_s}[\cos \alpha -\cos (\alpha +\mu \left)\right]$

Let ${\mu }_1$ be the overlap angle for firing angle ${\alpha }_1$, then

$I_0=\frac{V_m}{\omega L_s}[\cos {\alpha }_1-\cos ({\alpha }_1+{\mu }_1\left)\right]$

or, $\cos {\alpha }_1-\cos ({\alpha }_1+{\mu }_1)=\cos {\alpha }_2-\cos ({\alpha }_2+{\mu }_2)$

or,$\cos 0^{\circ }-\cos {15}^{\circ }=\cos {30}^{\circ }-\cos ({30}^{\circ }+{\mu }_2)$

or, $0.0341=0.866-\cos ({30}^{\circ }+{\mu }_2)$

or, $\cos ({30}^{\circ }+{\mu }_2)=0.8319$

${30}^{\circ }+{\mu }_2={33.7}^{\circ }$

or, ${\mu }_2={3.7}^{\circ }$