Question

Consider the battery charging circuit with V_m = 20 V, R = 10 ohms and V_B = 14 V. Find the peak current assuming an ideal diode. Also, find the percentage of each cycle in which the diode is in on state. Sketch V_s(t) and i(t) to scale against time.

Answer 1

The diode is on when

$V_msin(\omega t)>V_B$ or $20sin(\omega t)>14$

This diode goes to on state at

$20sin(\omega t)=14$

$\omega t=arcsin(14/20)=arcsin0.7$

$\omega t=45\degree;135\degree$

The diode is on for $45\degree\le\omega t\le135\degree$ or for $90\degree$ of the phase angle. The whole period is $360\degree$, so the diode is on for

$90\degree/360\degree=0.25=25\%$ of the time.

The peak current is when the ac voltage is at the peak and is

$I_m=\frac{V_m-V_B}{R}=\frac{20-14}{10}=0.6$ A