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Boost Converter Calculator

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

Duty cycleInductor valueOutput capacitorSwitch current
V
V
% of IL
mV
%
Duty cycle62.5 %
Inductor14.58 µH
Average inductor (input) current1.333 A
Peak switch current1.533 A
Minimum output capacitance12.5 µF
How this result was calculated
  1. D = 1 − Vin × η ÷ Vout = 62.5 %
  2. IL = Iout ÷ (1 − D) = 1.333 A, ΔI = 400 mA
  3. L = Vin (Vout − Vin) ÷ (ΔI × f × Vout) = 14.58 µH
  4. C = Iout × D ÷ (f × ΔV) = 12.5 µF
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Boost Converter Equations

D = 1 − (Vin × η) ÷ Vout
IL = Iout ÷ (1 − D)
L = Vin × (Vout − Vin) ÷ (ΔIL × fsw × Vout)
Cout = Iout × D ÷ (fsw × ΔVout)

Worked example

5 V to 12 V at 0.5 A, 500 kHz, 30 % ripple, 50 mV ripple, 90 % efficiency: D = 62.5 %, IL = 1.33 A, L = 14.6 µH (use 15 µH), Cout ≥ 12.5 µF, peak switch current 1.53 A.

Key insight: the input and switch carry far more current than the output: here 1.33 A average to deliver 0.5 A, so size the inductor and switch for the input side.
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Frequently Asked Questions

What is the duty cycle of a boost converter?

Ideally 1 − Vin/Vout; for 5 V to 12 V that is 58 %, slightly more with losses.

Why is the input current higher than the output current?

Power is conserved, so stepping voltage up by a ratio divides current by the same ratio, plus losses.

Can a boost converter be turned off completely?

Not in the basic form; the input passes through the inductor and diode to the output even when the switch is off.

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