Question

A single-phase bridge converter is used to charge a battery of 200 V having an internal resistance of 2 $\Omega$ as shown in figure. The SCRs are triggered by a constant dc signal. If SCR 2 gets open circuited, what will be the average charging current?

Answer 1

$T_{1}and{T}_4$ gets forward biased, when

$V_m\sin {\theta }_1\le E$




$I_{avg}$ = (average current)

$=\frac{1}{2\pi R}{\int }_{{\theta }_1}^{\pi -{\theta }_1}(V_m\sin \omega t-E)d\theta$

$\therefore I_0\left(avg\right)=\frac{1}{2\pi R}(2V_m\cos \theta -E(\pi -2{\theta }_1\left)\right]$

$=\frac{1}{2\pi \times 2}[2\times (230\times \sqrt{2})\cos {\theta }_1-200(\pi -2{\theta }_1\left)\right]$

where, ${\theta }_1={\sin }^{-1}\left(\frac{E}{V_m}\right)$

$={\sin }^{-1}\left(\frac{200}{230\times \sqrt{2}}\right)$

= 38° = 0.66 rad

$\therefore I_0\left(avg\right)=\frac{1}{2\pi \times 2}[2\sqrt{2}\times 230\cos 38{}^{\circ }$ $-200(\pi -2\times 0.66)]=11.9A$