How does a buck converter work?

A buck converter steps a DC voltage down by switching the input on and off very fast and smoothing the result with a coil and a capacitor. The output is about the input times the duty cycle, the fraction of time the switch is on: V_out ≈ D × V_in. Because the switch is always fully on or fully off, almost nothing is burnt as heat.

Chop, then smooth

Switch 12 V on and off, on for half of every cycle, and you get a square wave from 0 to 12 V whose average is 6 V. A coil between the switch and the load smooths it towards that average, and a capacitor across the load holds it steady.

On a quarter of the time, about 3 V; three quarters, about 9 V. Real converters lose a little in the diode and the coil's wire, so 12 V at half duty gives about 5.8 V.

Live experiment

Chop and smooth

12 V is switched on and off at 1 kHz by M1, then smoothed by a 0.1 H coil and 100 µF into a 60 Ω load. Keep M1 on a quarter, half or three quarters of each cycle, and the load settles near 2.8, 5.8 or 8.8 V, with 92 to 97% of the power arriving.

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.

Two halves of each cycle

1. Switch on: the input pushes current through the coil to the load, and the coil's current climbs.
2. Switch off: the coil's current can't stop at once, so it keeps flowing round through the freewheel diode, from ground, into the load. The current falls.

The coil's current rises and falls a little each cycle around the load's current, and the capacitor absorbs that wobble.

Gears for electricity

A buck converter trades volts for amps, like a low gear on a bicycle. Power in is nearly power out, so half the voltage out means nearly twice the current out, and the input supplies only about half the load's current. That's why phone chargers, laptop bricks and the supplies beside every processor are switchers, typically 85–95% efficient, where a linear regulator would burn the difference.

Ripple

The output still wobbles a little at the switching frequency:

ΔI is how far the coil's current swings each cycle. Example: 12 V at half duty, 1 kHz, a 0.1 H coil and 100 µF. The coil's current swings by about 30 mA, so the ripple is about 0.03 ÷ (8 × 1000 × 0.0001) ≈ 0.04 V.

Pick the coil first, big enough that its current never runs dry; then a capacitor for the ripple you can accept. Make the coil too small and its current falls to zero each cycle, and the output drifts above D × V_in. Real converters switch at hundreds of kilohertz or more, so their coils are microhenries and their capacitors tens of microfarads.