Buck Converter Calculator

Duty cycle, inductor, output capacitor and peak currents for a step-down converter.

What it does

A buck converter steps a DC voltage down efficiently by switching. The calculator gives the duty cycle, inductor value, output capacitor and peak currents for continuous conduction.

Where it's used

12 V → 5 V and 5 V → 3.3 V supplies, battery-powered devices and point-of-load regulators.

How to use it

Enter input and output voltage and the output current.
Enter the switching frequency from your controller’s datasheet.
Choose the ripple current (20–40 % is typical) and output ripple voltage.
Pick an inductor and capacitor at or above the results.

The formula explained

D: duty cycle ≈ Vout / (Vin·η)
L: inductance for the chosen ripple current ΔI
Cout: capacitance for the chosen ripple voltage ΔV
Ipk: peak inductor current = Iout + ΔI/2
fsw: switching frequency

Worked example

24 V down to 3.3 V at 3 A.

Inputs:

Input voltage: 24 V
Output voltage: 3.3 V
Output current: 3 A
Switching frequency: 400k Hz
Inductor ripple (ΔI/Iout): 30 %
Output voltage ripple: 30m V
Efficiency estimate: 88 %

The calculator gives:

Duty cycle D: 15.62 %
Minimum inductance: 8.984 µH
Minimum output capacitance: 9.375 µF

Tips and common mistakes

Choose an inductor whose saturation current is comfortably above Ipk.
The capacitor formula ignores ESR; ceramic capacitors keep ripple close to the result.
Higher frequency allows a smaller inductor but raises switching losses.

Questions people ask

How is the duty cycle worked out?

Ideally D = Vout / Vin. Allowing for losses it is about Vout / (Vin × efficiency): 3.3 V from 24 V at 88 % gives 3.3 / (24 × 0.88) ≈ 15.6 %. Very small duty cycles mean very short on-times, which limits how high the switching frequency can go.

How is the inductor value calculated?

L = (Vin − Vout) × D / (f × ΔI), where ΔI is the ripple current you allow. In the example, 20.7 V × 0.156 / (400 kHz × 0.9 A) ≈ 9 µH, so a standard 10 µH inductor rated above the peak current fits.

Why use a buck converter instead of a linear regulator?

A linear regulator burns the difference as heat: from 24 V to 3.3 V at 3 A that is 62 W. A buck converter at 88 % efficiency loses only about 1.35 W for the same 9.9 W output.

Reference: CCM

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