In the Fig. below , the ideal switch S is switched on and off with a switching frequency f = 10 kHz. The switching time period is $T=t_{ON}+t_{OFF}$ μs. The circuit is operated in steady state at the boundary of continuous and discontinuous conduction, so that the inductor current $i$ is as shown in Figure. What are the values of the on-time $t_{ON}$ of the switch and peak current $i_{p}$.
Solution:
When the switch is ON the Diode is reverse biased and the voltage across the inductor is $100$ V. Hence, the current through the inductor is given by.
$\boxed{i(t)=10^6t}$
Therefore peak current as seen from the graph $i_{p}=10^6T_{ON}$
At the instant of switching OFF the inductor back emf polarity is reversed and the diode starts conducting.
The volatge across the inductor now is $500$ V and the initial current being $i_{p}=10^6T_{ON}$ the current equation during OFF time is given by:
$\boxed{i(t)=10^6T_{ON}-5\cdot10^6t}$
From the graph @ $T_{OFF}$ current through inductor is zero
$\boxed{\implies i(T_{OFF})=10^6T_{ON}-5\cdot10^6T_{OFF}=0}$
Therefore $T_{ON}=5\cdot T_{OFF}$
$T=T_{ON}+T_{OFF}=\displaystyle\frac{1}{f}=100$ μs
$T_{ON}=63.33$ μs
$i_{p}=63.33$ Amp
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