Question

A dc-dc converter as shown below with switch 'S' is switched at a frequency of 20 kHz with duty ratio of 0.6. All components of the circuit are ideal and the initial current in the inductor is zero. The energy stored in inductor at the end of 10 complete switching cycles will be

Answer 1

For buck-boost converter,

During $t_{on}$ i.e, 0.6T inductor stores energy,

$D=\frac{T_{on}}{T}=0.6$

During $t_{off}$ i.e, 0.4T, inductor releases energy.

For one cycle: Rise in current in inductor = 0.2T for time.

For 10 cycles : Rise in current (0.2T) $\times$ 10 = 2T

Using above graph, $i=\frac{100}{L}t$

$i=\frac{100}{L}\left(2T\right)=\frac{100\times 2}{L\times f}$

$=\frac{100\times 2}{10\times {10}^{-3}\times 20\times {10}^{-3}}=1A$

Energy stored $=\frac{1}{2}L{i}^2=\frac{1}{2}\times (10\times {10}^{-3})(1{)}^2=5mJ$