# In a boost converter the duty cycle is 0.5. The inductor current is assumed to be continuous. Capacitor C is assumed to be very large. If the switching frequency is 20 kHz, the peak to peak inductor current ripple is

### Right Answer is:

0.25 A

#### SOLUTION

Input Voltage (V_{in}) = 20 V

Frequency (f) = 20 kHz = 20 × 10^{3} Hz

Duty cycle (α) = 0.5

Inductor (L) = 2 mH = 2 × 10^{−}^{3}

The peak to peak ripple inductor current is given as

ΔI = (α × V_{in})/(f × L)

ΔI = (0.5 × 20)/(20 × 10^{3 }× 2 × 10^{−}^{3})

**ΔI =0.25 A**