Assume that the entire AC component of the inductor ripple current (0.6 A peak to peak) flows into the output capacitor of a buck converter. With a switching frequency of 300 kHz and an output capacitance of 25 µF (ignore the ESR), which of the following is the peak-to-peak output voltage ripple?

Answer: A. 10 mV

The current into the capacitor is a triangular wave with an amplitude of ΔI/2. The charge that enters the capacitor during the half period T/2 when this current is positive is the area of a triangle, (1/2) × (T/2) × (ΔI/2) = ΔI × T/8. The voltage ripple is therefore ΔV = ΔI/(8fC) = 0.6/(8 × 300 kHz × 25 µF) = 10 mV. In a real capacitor the voltage ΔI × ESR from the ESR is added, which is why ceramic capacitors with low ESR are often used.

• A. Correct. 0.6 ÷ (8 × 300 × 10³ × 25 × 10⁻⁶) = 10 mV.
• B. This is ΔI/(2fC), which greatly overestimates the charge carried by the triangular current.
• C. This is ΔI/(fC), which assumes the ripple current flows in the same direction for a whole period.