Question

In the circuit shown below, if the switching frequency is 250 Hz, then the ripple current in the inductor $\left(\Delta I\right)$ is

Answer 1

The given converter is a step up chopper
$\therefore V_0=\frac{1}{1-\alpha }{V}_s$
$\frac{12}{5}=\frac{1}{1-\alpha }$
$\alpha =0.5834$
Ripple in the inductor current is,
$\Delta I=\frac{V_s\alpha }{fL}$
$\Delta I=\frac{5\times 0.5834}{250\times 2.5\times {10}^{-3}}=4.6A$