Boost Converter Calculator — inductor, duty and peak current

Duty cycle, peak switch current and component sizing for a step-up converter.

Example: Boosting 5 V to 12 V at 500 mA needs about 15.63 µH at 63% duty, drawing 1.333 A on the input.

Formula

D = 1 − (V_in × η) / V_out    I_in = I_out / (1 − D)

Worked example

Boosting 5 V to 12 V at 500 mA needs about 15.63 µH at 63% duty, drawing 1.333 A on the input.

  1. D = 1 − (V_in × η) / V_out

    1 − (5 V × 0.9) / 12 V

    62.5%

  2. I_in = I_out / (1 − D)

    500 mA / 0.375

    1.333 A

    Power in must equal power out, so stepping voltage up steps current down — on the output side.

  3. L = V_in × D / (ΔI_L × f)

    5 V × 0.625 / (400 mA × 500 kHz)

    15.63 µH

Frequently asked questions

Why does my boost converter draw so much input current?

Because power is conserved. Boosting 5 V to 24 V at 1 A means delivering 24 W, which at 5 V and 90% efficiency is over 5 A on the input. The inductor, switch and input capacitor all carry that current, not the 1 A output figure.

Why does a boost converter get unstable at high duty cycle?

A right-half-plane zero in the control loop, which moves to lower frequency as duty rises. It forces you to roll the loop bandwidth off well below it, making transient response sluggish. Past roughly 80% duty this becomes hard to compensate.

Why is boost output ripple worse than buck?

Because the output capacitor supplies the entire load current on its own during the switch on-time — the inductor is disconnected from the output then. A buck's inductor feeds the output continuously, so its capacitor only handles the ripple.

Can a boost converter output less than its input?

No. Even with the switch off entirely, current flows through the inductor and diode to the output, so the output sits at roughly the input minus a diode drop. Going both up and down needs a buck-boost, SEPIC or four-switch topology.

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