Power Electronics GATE 2002

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