Duty cycle and inductor ripple in a buck converter

In CCM, V_out = D × V_in. During the ON time, the voltage across L is V_in − V_out.

Worked answer.

Step 1: Duty cycle. For an ideal buck converter in continuous conduction mode (CCM), V_out = D V_in, so
D = 12 / 24 = 0.5.

Step 2: ON time. The switching period is T = 1/f_s = 1/50,000 = 20 µs. The ON time is t_on = D T = 0.5 × 20 = 10 µs.

Step 3: Ripple current. While the switch is ON, the inductor sees V_L = V_in − V_out = 24 − 12 = 12 V. From V = L di/dt:
ΔI = V_L × t_on / L = (12 × 10 × 10⁻⁶) / (100 × 10⁻⁶) = 1.2 A peak-to-peak.

Check using the OFF interval. While the switch is OFF, the inductor sees −V_out = −12 V for t_off = 10 µs. The current falls by 12 × 10 µs / 100 µH = 1.2 A. The rise and fall are equal, which is the volt-second balance that must hold in steady state.

The general formula is ΔI = V_out(1 − D) / (L f_s) = 12 × 0.5 / (100 × 10⁻⁶ × 50,000) = 1.2 A, which gives the same result.

Interpretation. If the load current is 5 A, the inductor current swings between 4.4 A and 5.6 A. The converter stays in CCM as long as the average load current is greater than ΔI/2 = 0.6 A. Below that, the current touches zero and the converter enters discontinuous mode, where V_out = D V_in no longer holds.

Common mistake. Using V_in instead of (V_in − V_out) across the inductor during the ON time.

Likely follow-up. "How would you halve the ripple?" Double either the inductance or the switching frequency. Higher frequency shrinks the inductor but raises switching losses, so this is a design trade-off.