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 (Vin) = 20 V
Frequency (f) = 20 kHz = 20 × 103 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 × Vin)/(f × L)
ΔI = (0.5 × 20)/(20 × 103 × 2 × 10−3)
ΔI =0.25 A