Question

A high inductive load is fed by a dc-dc converter as shown below. If the duty ratio of converter is 4/5, then value of peak forward blocking voltage $\left(V_{peak}\right)$ across the switch will be

Answer 1

For given converter:

When switch 'S' is off and diode 'D' is in on state, voltage across switch is $V_{peak}$
Applying KVL in input loop,

$-200-\frac{V_0}{5}+V_{peak}=0$

$V_{peak}=\frac{V_0}{5}+200$

For flyback dc-dc converter

$V_0=\frac{N_2}{N_1}\left(\frac{\alpha }{1-\alpha }\right)V_s$

$=5\left(\frac{4/5}{1-\frac{4}{5}}\right)\times 200=5\times \frac{4}{5}\times \frac{5}{1}\times 200$

$=20\times 200=4000V$

$\therefore V_{peak}=\frac{4000}{5}+200=800+200=1000V$