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Power & Energy

Build a clean 5 V rail: a buck, then a linear regulator

To get a clean, cool 5 V from a much higher voltage, use two stages. A buck converter drops most of the voltage efficiently, to a couple of volts above the target. A linear regulator after it drops the rest and holds the output steady against the buck's ripple. Each covers the other's weakness: the buck wastes little power, and the regulator leaves almost no ripple.

The spec

  • A 15 V battery
  • A USB load wanting about 4.8 V at 100 mA, and staying within 4.6 to 5.0 V
  • No more than a quarter of a watt burnt in the regulator, and no ripple at the port

Plan 1: the regulator alone

A pass regulator, a transistor whose base is held by a 5.6 V zener, gives a steady 4.88 V straight from 15 V. But the transistor drops the difference at the full load current:

P = (15 − 4.8) V × 0.1 A ≈ 1 W

Four times the budget, and only about a third of the battery's power reaches the load.

Plan 2: the buck alone

A switch, diode, 0.1 H coil and 100 µF capacitor bring 15 V down to about 7.2 V with little waste. But it isn't regulated: the output sits wherever the duty cycle and the load put it, here 7.1 to 7.3 V, far too high for USB, and it still ripples.

Plan 3: buck, then regulator

Feed the buck's 7.2 V into the pass regulator. The output is 4.79 to 4.80 V across the whole switching cycle, and the transistor now drops only the last 2.4 V:

P = (7.2 − 4.8) V × 0.1 A ≈ 0.24 W

Live experiment

Three plans for one rail

A 15 V battery must feed a 100 mA USB load at 4.6 to 5.0 V with nothing overheating. Try the pass regulator alone (4.88 V, but 0.98 W in its transistor), the buck alone (7.1 to 7.3 V, far too high), or the buck and then the regulator: 4.79 to 4.80 V, with 0.23 W.

Try it yourself: free lesson

“Project: A clean USB rail” is the project that ends Buck Bay, Power & Energy World 2. Free with an account. Lessons unlock in order, so you’ll do the 13 before it first.

Parts list

StageParts
BuckSwitch (MOSFET), diode, 0.1 H coil, 100 µF capacitor
Regulator470 Ω resistor, 5.6 V zener, NPN pass transistor
LoadUSB port on the transistor's emitter

Why the regulator needs headroom

A pass regulator's output sits about 0.7 V below its transistor's base, and the zener's resistor needs volts across it to feed the base. Give it too little and it drops out: from 5 V in, the output sags to about 3.9 V. That's why the buck aims for about 7 V, not 5. A modern low-dropout regulator needs only a few hundred millivolts, so the buck can sit closer and waste even less.

The same two-stage pattern powers precision converters, radio receivers and audio boards everywhere: switch for efficiency, then regulate linearly for quiet.

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

“Project: A clean USB rail” is the project that ends Buck Bay, Power & Energy World 2. Free with an account. Lessons unlock in order, so you’ll do the 13 before it first.

Or start from the very first lesson, no account needed

Practice questions

Three quick questions from the lessons. Pick an answer to check it.

  1. Question 1 of 3Why did the zener alone sag to about 1.2 V with a 50 Ω load?
    Show the answer

    Answer: 470 Ω from 12 V can pass only about 23 mA, and the load wanted 100. Everything the load takes has to come through the series resistor first. With too little current to share, the zener turns off and the voltage collapses.

    • Why not “The zener broke”? Nothing broke. Below 5.6 V the zener simply stops conducting, and the resistor and load form a voltage divider.
    • Why not “The battery was flat”? The battery still gave 12 V. 470 Ω and 50 Ω divide it down to about 1.2 V.
    • Why not “Zeners only work backwards”? It was fitted reverse-biased, the right way. The problem was too little current through 470 Ω.
  2. Question 2 of 3At 5 V in, the load…
    Show the answer

    Answer: Drops out to about 3.9 V: no headroom. The zener needs current through R1 and Q1 needs some volts across it. At 5 V there's room for neither.

    • Why not “Goes to 5 V”? The transistor always drops something, at least its base-emitter voltage, so the output can't reach the input.
    • Why not “Stays at 4.8 V”? The output sits about 0.7 V below the transistor's base, and the zener feeding that base needs volts to spare. From 5 V there isn't enough.
    • Why not “Turns off”? It doesn't switch off; it just can't regulate, and the output follows the input down to about 3.9 V.
  3. Question 3 of 3Without the coil, wiring M1's source straight to the load gives…
    Show the answer

    Answer: Nearly the full 12 V, chopped. Without a coil the capacitor charges to the switch's peak on every on-pulse and has no reason to stop at the average.

    • Why not “A short”? Nothing connects the battery straight to ground. The load just sees the switch's pulses.
    • Why not “0 V”? The switch still connects the battery to the load half the time, so the load gets 12 V pulses.
    • Why not “Half: 6 V”? The average of the chopped wave is about 6 V, but with nothing to smooth it the load sees the full 12 V pulses.
Keep practising: free lesson

“Project: A clean USB rail” is the project that ends Buck Bay, Power & Energy World 2. Free with an account. Lessons unlock in order, so you’ll do the 13 before it first.

Questions people ask

Why put a linear regulator after a switching regulator?

The switcher drops most of the voltage efficiently but leaves ripple and switching noise. The linear regulator after it holds its output steady against that ripple, and because it only has to drop a volt or two, it barely warms up.

How much headroom does the linear regulator need?

Enough to cover its own dropout plus the ripple's lowest dip. A zener-and-transistor pass regulator like this one needs a couple of volts; a modern low-dropout regulator needs only a few hundred millivolts, so the buck can be set closer to the output.

How much power does the pass transistor waste?

The voltage across it times the current through it. From 15 V to 4.8 V at 100 mA that's about 1 W. From a 7.2 V buck output it's 2.4 V × 0.1 A, about 0.24 W.

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