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Boost Converter Calculator

Design a boost (step-up) converter: duty cycle, inductor value and ripple, peak switch current, input current and minimum output capacitor from input/output voltage, load current and frequency.

Input voltage (minimum)

Output voltage

Output current

Switching frequency

Inductor ripple current

Output ripple voltage target

Estimated efficiency

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How it works

D = 1 − V_in × η ÷ V_out · ΔI_L = ripple % × I_out × V_out ÷ V_in · L = V_in × (V_out − V_in) ÷ (ΔI_L × f_s × V_out) · I_L(avg) = I_out ÷ (1 − D) · I_peak = ΔI_L ÷ 2 + I_out ÷ (1 − D) · C_out(min) = I_out × D ÷ (f_s × ΔV_out)

A boost converter stores energy in an inductor while the switch is on and releases it, stacked on top of the input voltage, through a diode or synchronous switch while it is off. The duty cycle sets the voltage step-up; including the efficiency makes it slightly larger than the ideal 1 − V_in/V_out. The equations follow Texas Instruments application report SLVA372C. Design at the minimum input voltage, where the duty cycle and the switch current are highest, and pick an inductor whose saturation current exceeds the peak current with margin.

Worked example

Boosting 5 V to 12 V at 1 A with 500 kHz switching, 30% ripple, a 50 mV ripple target and 90% efficiency: D = 1 − 5 × 0.9 ÷ 12 = 62.50%. ΔI_L = 0.3 × 1 × 12 ÷ 5 = 720 mA, L = 5 × 7 ÷ (0.72 × 500,000 × 12) = 8.102 µH, the average input current is 2.667 A, the peak switch current is 3.027 A and the output capacitor needs at least 1 × 0.625 ÷ (500,000 × 0.05) = 25 µF.

Sources: Texas Instruments SLVA372C — Basic Calculation of a Boost Converter's Power Stage · Wikipedia — Boost converter

Frequently asked questions

How do I calculate the duty cycle of a boost converter?

Ideally D = 1 − V_in ÷ V_out. Including efficiency, D = 1 − V_in × η ÷ V_out: 5 V to 12 V at 90% efficiency needs 62.5% instead of 58.3%.

How do I choose the inductor?

Pick a standard value at or above the minimum inductance and check that its saturation current is above the peak switch current with 20–30% margin. A larger inductor lowers ripple but is bigger and slower to respond.

Why is the input current higher than the output current?

Power is conserved: P_in = P_out ÷ η. Raising 5 V to 12 V at 1 A needs about 12 W ÷ 0.9 = 13.3 W from the source, so 2.67 A at 5 V.

What limits how far a boost converter can step up?

Losses grow quickly as the duty cycle approaches 100%. Above roughly 5–6× step-up (D above 80–85%) consider a flyback, SEPIC or a coupled-inductor design.

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This calculator is for educational purposes. Double-check important results before acting on them.