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Buck converter calculator

A buck converter steps a voltage down by switching it on and off very fast, then smoothing it with a coil and a capacitor. The fraction of each cycle the switch is on, the duty cycle, sets the output. Type your voltages, frequency and parts to get the duty cycle, the ripple and the coil you need.

The battery or adapter.

Less than Vin: a buck only steps down.

1k = 1 kHz. Real chips switch at 100k–2M.

100m = 100 mA.

100m = 0.1 H, 22µ = 22 µH. Leave empty for the duty cycle only.

µ or u = micro.

Duty cycle, D = Vout ÷ Vin

50%
Switch on each cycle
500 µs
Input current (no losses)
50 mA
Ripple current, ΔI
30 mA
Ripple ÷ load current
30%
Coil current swings
85 mA to 115 mA
Output ripple, ΔV
37.5 mV
Coil for 30% ripple
100 mH
Smallest coil that never runs dry
15 mH

This is the ideal converter. A real one loses a little in the diode and the coil's wire, so it runs a touch longer on each cycle than D says. Its feedback loop sets the duty for you.

The formulas

The output is the input times the duty cycle, so:

D = Vout over Vin

While the switch is on, the coil has Vin − Vout across it and its current climbs. How far it swings each cycle is the ripple current:

ΔI = (Vin − Vout) × D over f × L

The output capacitor soaks up that swing, leaving a small ripple voltage:

ΔV ≈ ΔI over 8 × f × C

A worked example

World 26's buck: 12 V in, 6 V out, switching at 1 kHz into a 100 mA load, with a 0.1 H coil and 100 µF.

D = 6 over 12 = 50%
ΔI = (12 − 6) × 0.5 over 1000 × 0.1 = 0.03 A = 30 mA

That's 30% of the load, so the coil's current swings between 85 and 115 mA and never runs dry. The output ripple is 0.03 ÷ (8 × 1000 × 0.0001) ≈ 0.04 V. With 10 µF it would be ten times that, about 0.4 V.

Real converters

World 26's buck switches at 1 kHz, so it needs a big coil. Real converter chips switch at hundreds of kilohertz to a few megahertz, which is why their coils are microhenries: 12 V to 5 V at 500 kHz and 1 A, with 30% ripple, needs only about 19 µH. Losses in the diode and the coil's wire mean the real duty cycle runs a little above Vout ÷ Vin, but even so a buck is typically 85–95% efficient, where a linear regulator would burn the difference as heat.

Free lesson

Now build it yourself

Each lesson comes in three short parts: see the idea, try it on a live board, then build it yourself, with quick checks as you go.

Try it yourself: free lesson

“Chop it down” is lesson 3 of 6 in Buck Bay, Power & Energy World 2. Free with an account. Lessons unlock in order, so you’ll do the 9 before it first.

Or start from the very first lesson, no account needed

The ideas behind it

Questions people ask

How do I calculate a buck converter's duty cycle?

Divide the output voltage by the input: D = Vout ÷ Vin. 6 V from 12 V is 50%, and 3.3 V from 12 V is about 28%. A real converter runs a little longer each cycle to make up for its losses, and its feedback loop sets that for you.

How do I choose a buck converter's inductor?

Pick a ripple current of about 20–40% of the load current, then L = (Vin − Vout) × D ÷ (f × ΔI). Smaller coils swing more. Below (Vin − Vout) × D ÷ (2 × f × I_load) the current falls to zero each cycle and the output drifts up.

How big should the output capacitor be?

Big enough for the ripple you can accept: ΔV ≈ ΔI ÷ (8 × f × C). Switching faster shrinks both the coil and the capacitor you need. In practice the capacitor's ESR often adds more ripple than its capacitance, so use low-ESR ceramic or polymer parts.

Can a buck converter step voltage up?

No. The output is the input times a duty cycle between 0 and 1, so it's always less. To go up, use a boost converter. To go either way, use a buck-boost.

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