Power Electronics Circuits
For an ideal buck converter with an input voltage of 12 V, an output voltage of 3 V, an inductance of 15 µH and a switching frequency of 300 kHz, which of the following is the peak-to-peak inductor ripple current in continuous conduction mode?
- A0.25 A
- B0.5 A
- C2 A
Show answer and explanation
Answer: B. 0.5 A
The duty cycle is D = 3/12 = 0.25, so the on-time is D/f ≈ 0.833 µs. During the on-time 12 − 3 = 9 V appears across the inductor, so the current rises by ΔI = (Vin − Vout) × D/(L × f) = 9 × 0.25/(15 µH × 300 kHz) = 2.25/4.5 = 0.5 A. During the off-time it falls by Vout × (1 − D)/(L × f) = 3 × 0.75/4.5 = 0.5 A, confirming that it returns to its starting value every period. Doubling either the inductance or the frequency halves the ripple.
- A. This is half the ripple (the swing from the average).
- B. Correct. 9 × 0.25 ÷ 4.5 = 0.5 A.
- C. This forgets to multiply by the duty cycle D. The 9 V appears only during the on-time.