Understand it in 60 seconds · Power electronics
The buck converter
How a switch, a diode, an inductor and a capacitor turn 12 V into 5 V with almost no loss — and why the output is the duty cycle times the input.
Intermediate · published 27 September 2026
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What you will understand
- Q1 on: the source pushes current into the inductor, and iL rises. Q1 off: the inductor keeps the current going through the diode, and iL falls.
- The duty cycle D sets the output: ideally Vout = D Vin; with the diode drop, D ≈ 0.44 for 12 V → 5 V.
- At 200 kHz, the inductor current ripples around Iout and the capacitor smooths Vout.
- Nothing is burnt on purpose: typically 85 to 95 % efficient. At low Vout, the diode drop is no detail.
- At light load (ΔIL/2 > Iout) it goes into discontinuous mode; Q1 must hold Vin + Vf plus the switching spikes.
Inside the lesson
The cheat sheet
Every equation on one page, with the circuit drawn. Export it to PDF to keep it.
The animated presentation
The idea step by step, with the currents flowing in the circuit. Press Present (F5) to play it.
Your own copy
It opens as a Circuit Notebook project: change the values, reuse the circuit in your report.
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