Question

A single phase mid point converter as shown in figure is connected to a highly inductive load. Turns ratio of transformer is 1 : 2. If $\alpha$ is the firing angle, then current though diode is

Answer 1

Average output voltage is;
$E_{dc}=\frac{1}{\pi }{\int }_{\alpha }^{\pi }{E}_{m1}\sin \omega td\left(\omega t\right)$


$\frac{E_2}{E_1}=\frac{2}{1}$
$E_2=2_{E_m}\sin \omega t$
Since transformer is tapped in middle
$\therefore E_{m1}=E_{m2}=\left(\frac{2E_m}{2}\right)=E_m$
$E_{dc}=\frac{1}{\pi }{\int }_{\alpha }^{\pi }{E}_m\sin \omega td\left(\omega t\right)=\frac{E_m}{\pi }(1+\cos \alpha )$
$I_{dc}=\frac{E_{dc}}{R}=\frac{E_m}{\pi R}(1+\cos \alpha )$
Freewheeling diode conducts when voltage goes negative for a period of $\alpha$
$\therefore I_{Df}=\frac{\alpha }{\pi }{I}_{dc}=\frac{E_m(1+\cos \alpha )}{\pi R}\frac{\alpha }{\pi }$
$I_{Df}=\frac{E_m}{{\pi }^2}\frac{(\alpha +\alpha \cos \alpha )}{R}$