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Buck Converter Calculator

Size the inductor and output capacitor of a step down (buck) DC to DC converter in continuous conduction mode.

Duty cycleInductor valueOutput capacitorPeak current
V
V
% of Iout
mV
%
Duty cycle46.3 %
Inductor9.722 µH
Ripple current ΔI600 mA
Peak inductor current2.3 A
Minimum output capacitance7.5 µF
Input power11.11 W
How this result was calculated
  1. D = Vout ÷ (Vin × η) = 46.3 %
  2. L = Vout (Vin − Vout) ÷ (ΔI × f × Vin) = 9.722 µH
  3. C = ΔI ÷ (8 × f × ΔV) = 7.5 µF
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Buck Converter Equations

D = Vout ÷ (Vin × η)
L = Vout × (Vin − Vout) ÷ (ΔIL × fsw × Vin)
Cout = ΔIL ÷ (8 × fsw × ΔVout)
Ipeak = Iout + ΔIL ÷ 2

These are the standard continuous conduction equations used in regulator application notes. The capacitor value ignores ESR; with electrolytic capacitors the ESR term (ΔI × ESR) often dominates.

Worked example

12 V to 5 V at 2 A, 500 kHz, 30 % ripple (0.6 A), 20 mV ripple, 90 % efficiency: D = 46.3 %, L = 35 ÷ 3.6 × 10⁶ = 9.7 µH (use 10 µH), Cout ≥ 7.5 µF, peak current 2.3 A.

Key insight: choose an inductor whose saturation current is above the peak current with margin; a saturated inductor makes the current spike and the converter fail.
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Frequently Asked Questions

How do I calculate the inductor for a buck converter?

L = Vout × (Vin − Vout) ÷ (ΔI × f × Vin), with ΔI usually 20 to 40 % of output current.

What is the duty cycle of a buck converter?

Ideally Vout ÷ Vin, slightly higher in practice because of losses.

Why does switching frequency matter?

Higher frequency allows smaller inductors and capacitors but increases switching losses.

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