Buck Converter Calculator
Size the inductor and output capacitor of a step down (buck) DC to DC converter in continuous conduction mode.
- D = Vout ÷ (Vin × η) = 46.3 %
- L = Vout (Vin − Vout) ÷ (ΔI × f × Vin) = 9.722 µH
- C = ΔI ÷ (8 × f × ΔV) = 7.5 µF
Buck Converter Equations
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.
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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