Power8 hours total4 h on the project

Switching Regulators and Their Layout

Specify a buck converter from first principles — inductor, capacitors, feedback, compensation — and lay out the hot loop so it does not radiate or oscillate.

Learning material0/22
Project0/14
Competence checklist0/7

Two ways in. Work the material, or if you already know this, go straight to the project and prove it.

How this skill is structured

  1. 1. Concepts — the ideas stated plainly, with the equations worth memorising.
  2. 2. Worked example — one real problem solved end to end, numbers included.
  3. 3. Tool demonstration — do the thing in a real open-source tool.
  4. 4. Resources — the specific free readings and videos, and what part of each to use.
  5. 5. Project — built alone, producing something a reviewer can check.
  6. 6. Competence checklist — what you must be able to do. This is also the audit rubric.

1. Concepts

Read these first. Tick each one when you could explain it to someone else without notes.

0/8

2. Worked example

Designing a 12 V to 5 V, 2 A buck from first principles

ProblemDesign the power stage for a 12 V (9-15 V range) to 5 V, 2 A synchronous buck switching at 500 kHz. Target 30% ripple current and 50 mV output ripple.

  1. 1
    Duty cycle: D = Vout/Vin. At nominal 12 V, D = 5/12 = 0.417. At 9 V, D = 0.556. At 15 V, D = 0.333.
  2. 2
    Inductor ripple is worst at maximum input voltage (longest off-time). Use Vin = 15 V, D = 0.333.
  3. 3
    Target ripple: dI = 0.30 x 2 A = 0.6 A.
  4. 4
    Inductance: L = Vout x (1 - D) / (dI x f_sw) = 5 x (1 - 0.333) / (0.6 x 500e3) = 5 x 0.667 / 300000 = 11.1 uH.
  5. 5
    Choose the nearest standard value: 10 uH. Recompute actual ripple: dI = 5 x 0.667 / (10e-6 x 500e3) = 0.667 A, which is 33% of 2 A. Acceptable.
  6. 6
    Peak inductor current: I_peak = 2 + 0.667/2 = 2.33 A. The inductor's Isat must exceed this with margin — target Isat >= 3.5 A to cover transients and current limit.
  7. 7
    Irms through the inductor is essentially the DC output current, 2 A, so the inductor's Irms rating must exceed 2 A with margin for its own heating.
  8. 8
    Inductor DCR: choose a part with DCR around 20-30 milliohm. Conduction loss = 2^2 x 0.025 = 100 mW.
  9. 9
    Output capacitor from ripple: the capacitance term gives V = dI/(8 x f x C), so C = 0.667/(8 x 500e3 x 0.050) = 3.3 uF. The ESR term with ceramics (say 3 milliohm) gives 0.667 x 0.003 = 2 mV, negligible.
  10. 10
    But transient response usually dominates. For a 1 A load step with a loop bandwidth of 50 kHz, the dip is roughly dI_load/(2 x pi x f_bw x C). For a 100 mV dip: C = 1/(2 x pi x 50e3 x 0.1) = 32 uF.
  11. 11
    Choose 2 x 22 uF 0805 X5R 16 V. Check DC bias derating at 5 V: an 0805 22 uF 16 V part may fall to about 12 uF each, so 24 uF effective. Close to the 32 uF target — add a third, or accept a slightly larger dip. Add a third: 3 x 22 uF nominal, ~36 uF effective.
  12. 12
    Input capacitor RMS ripple current: I_rms = Iout x sqrt(D x (1 - D)). Worst at D = 0.5: 2 x 0.5 = 1.0 A RMS. Choose ceramics with combined ripple rating well above 1 A — typically 2 x 10 uF 25 V X7R plus a 100 nF, and check the parts' ripple ratings.
  13. 13
    Feedback divider: if the reference is 0.8 V, the ratio is 5/0.8 = 6.25. With a bottom resistor of 10 k, the top is 52.5 k; use 52.3 k (E96) giving 5.03 V. Check the divider current is well above the feedback pin's bias current — 0.8/10k = 80 uA, versus a typical 50 nA bias. Fine.
  14. 14
    Switching losses and thermals: for a synchronous part with 30 milliohm FETs, conduction loss ~ I^2 x Rds(on) x D for the high side plus the same for the low side over (1-D): 4 x 0.03 = 120 mW total. Switching loss depends on the transition time and dominates at high Vin. Estimate total losses at 400-600 mW and check the package's Rth.
  15. 15
    Efficiency estimate: output power 10 W, losses ~0.7 W including the inductor, so about 93%. Compare to an LDO's 42% for the same conversion — 7 W of heat versus 0.7 W. That is why this converter exists.

Answer10 uH inductor with Isat >= 3.5 A and Irms >= 2.5 A, 3 x 22 uF ceramic output (36 uF effective after derating), 2 x 10 uF 25 V input with 1 A RMS ripple capability, 52.3 k / 10 k feedback divider. Peak inductor current 2.33 A, output ripple well under 50 mV, efficiency about 93%.

3. Tool demonstration

Lay out the critical loop correctly and verify your hand calculations against the vendor's own design tool.

0/10

Tool: KiCad layout of a buck converter hot loop, plus a vendor design tool for cross-checking

4. Resources

Free and, wherever possible, openly licensed. The note tells you which part to actually use — do not read them cover to cover.

0/4

5. Project — build this on your own

About 4 hours. This is the artifact that proves the skill. Work it without a walkthrough.

0/14

Design a complete synchronous buck converter — schematic, component selection with full calculations, and a laid-out power stage — for a specification you set (for example 12 V nominal to 3.3 V at 3 A), and defend the layout against a written checklist.

Deliverables

Acceptance criteria — how you know it is good enough

If you want to push further

  • Design the compensation network for an externally compensated controller: compute the crossover frequency and phase margin, and show the Bode plot from a simulation.
  • Add input hot-plug damping (an electrolytic or a series RC across the ceramic input capacitance) and calculate the resonance you are damping.

6. Competence checklist

Tick these honestly. If you are auditing this skill, this is your rubric — you should be able to demonstrate every line from the project you just built.

0/7
Tick every line above first — 7 remaining.