How a buck converter works
Explain how a buck converter works in synchronous and non-synchronous form. Draw the voltage and current waveforms at the switch node, the inductor and the output capacitor.
| Level | ● screening |
| Track | Power |
| Really testing | The most-asked power question. Draw it without hesitating. |
Free full answer — one of 12 questions answered in full on this site.
Show the answer
Say this first: A switch chops Vin into a square wave at the switch node; an LC filter averages it. In steady state the inductor volt-seconds balance: (Vin − Vout)·D·T = Vout·(1 − D)·T → Vout = D·Vin (ideal, steady-state CCM; with a diode, Vout ≈ D·Vin − (1 − D)·Vf). A non-synchronous buck uses a diode for the off-time freewheel path; a synchronous buck replaces it with a low-side FET.
Waveforms
- Switch node: square wave from Vin to −Vf (diode) or ≈0 V (sync).
- Inductor current: a triangle around Iout; slope +(Vin − Vout)/L when the switch is on, −Vout/L when off; ΔI = peak-to-peak ripple.
- Output cap: carries i_L − Iout (the AC part of the inductor current); ripple estimate ≈ ΔI/(8·fsw·C) + ΔI·ESR (ESL spikes not included).
- Input cap: pulsed current — the worst loop for EMI (see PW-09).
Numbers to quote: 20 V → 5 V gives D = 25%.
Going deeper
Worked example — 12 V → 3.3 V, 2 A, 500 kHz, L = 4.7 µH.
- D = 3.3/12 = 0.275.
- ΔI = (Vin − Vout)·D/(L·fsw) = 8.7 × 0.275 / (4.7 µH × 500 kHz) ≈ 1.0 A — 51% of the load. Usually too much: 10 µH gives 0.48 A.
- Output ripple with 2 × 22 µF (about 30 µF left after DC bias): ΔI/(8·fsw·C) = 1.0/(8 × 500 kHz × 30 µF) ≈ 8.5 mV, plus ESR × ΔI ≈ 2 mV.
Why the switch node goes negative. The inductor current can't stop. When the high-side switch opens, it pulls the switch node down until something conducts — the diode (−0.4 to −0.7 V) or, in a synchronous buck, the low-side FET's body diode during dead time. That short dip is normal; a large ring below it is not (see PW-18).
Where people go wrong
- Drawing the inductor current as a square wave.
- Getting the output-cap ripple from the switch-node waveform instead of from ΔI.
- Not being able to derive Vout = D·Vin from volt-second balance — that derivation is the question.
More like this in the book
This is PW-01 from Hardware Engineer Interview Questions. The book has 20 questions on power conversion and 141 in all, each answered the same way: what to say first, how to explain it, the numbers worth quoting and the follow-ups that come next.
Errata
No corrections known for this question. All errata
Ask about this question
Stuck on a step, or think something is wrong? Send it with PW-01 in the subject. Good questions get answered on this page, and corrections go into the next printing.