GATE 2015 EE – SET 2 – Question 32

Solution :

Continuous condcution , the average inductor voltage ,$$V_{L}$$ = 0

$$V_{L} $$= $$V_{P}\frac{T}{T_{s}}$$

$$V_{L} $$= 15 *$$ \frac{T_{ON}}{T_{s}}+(-45)*\frac{T_{OFF}}{T_{s}}$$= 0

$$V_{L} $$ = 15 *$$ \frac{T_{ON}}{T_{s}}-45*\frac{T_{OFF}}{T_{s}}$$= 0

15 $$T_{ON} – 45 T_{OFF}$$ = 0

15 $$T_{ON}  =  45 T_{OFF}$$

$$\frac{T_{OFF}}{T_{ON}}$$ =$$ \frac{1}{3}$$
$$\frac{T_{ON}}{T_{OFF}}$$ = $$ \frac{3}{1}$$

$$\delta$$ =$$ \frac{T_{ON}}{T_{ON}+T_{OFF}}$$ = $$\frac{3}{1+3}$$ = $$\frac{3}{4}$$ = 0.75