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Power & Energy

How does a linear voltage regulator work?

A linear voltage regulator holds a steady output voltage by controlling a pass transistor between the input and the output. A reference, like a zener diode, sets the voltage; the transistor carries the load's current and drops whatever voltage is spare as heat. It's the circuit inside the classic 7805 and every LDO.

Why a zener alone isn't enough

A zener with a series resistor holds its voltage only while the resistor can feed both the zener and the load. From 12 V, 470 Ω feeds a 5.6 V zener about 13 mA. A 50 Ω load wanting 100 mA takes all of that and more: the zener turns off, and the output collapses to about 1.15 V, where the 23 mA the resistor can pass is all the load gets.

Let a transistor carry it

Put an NPN transistor between the supply and the load. Its collector goes to the supply, its base sits on the zener, and its emitter feeds the load. The emitter follows the base, one diode drop lower:

Vout ≈ VZ − 0.7 V = 5.6 − 0.7 ≈ 4.9 V

Now the load's current comes from the supply through the collector. The zener only has to supply the base, about a hundredth of the load current, so its own current hardly changes and its voltage stays put. This is an emitter follower, and it works like power steering: the zener points the way with almost no effort, and the transistor does the pushing.

Live experiment

Let the transistor carry the load

A 12 V battery, 470 Ω and a 5.6 V zener feed loads from 1 kΩ down to 50 Ω. On the zener alone, only the lightest load holds and 50 Ω sags to about 1.15 V. Add Q1, a pass transistor, and every load sits near 4.9 V.

Try it yourself: free lesson

“The pass transistor” is lesson 1 of 6 in Buck Bay, Power & Energy World 2. Free with an account. Lessons unlock in order, so you’ll do the 7 before it first.

Real regulator chips

A chip like the 7805 adds an error amplifier: it compares a fraction of the output with a precise reference and drives the pass transistor to correct any difference. That holds 5 V far more tightly than a bare follower, as the load and the input change. It also adds current limiting and thermal shutdown, so a short circuit doesn't destroy it.

Every regulator needs headroom, its dropout voltage: room across the pass transistor for it to control anything. A 7805 needs about 2 V, so at least 7 V in. An LDO (low-dropout regulator) uses a different pass transistor and can work with a few hundred millivolts or less.

The price: heat

A linear regulator turns every spare volt into heat:

Pheat = (Vin − Vout) × I

From 12 V to 4.8 V at 100 mA, that's 0.72 W, more than the load's own 0.48 W. Its efficiency can never beat Vout ÷ Vin. That's fine for small currents or a small gap, and linear regulators stay popular where low noise matters, like audio and sensors. For big drops at big currents, a switching regulator wastes far less.

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

“The pass transistor” is lesson 1 of 6 in Buck Bay, Power & Energy World 2. Free with an account. Lessons unlock in order, so you’ll do the 7 before it first.

Or start from the very first lesson, no account needed

Work it out

Practice questions

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

  1. Question 1 of 3In a pass regulator, the load current comes through…
    Show the answer

    Answer: The transistor. The zener only sets the base; the transistor carries the load.

    • Why not “The series resistor”? The series resistor only feeds the zener and the transistor's base, a few milliamps. It couldn't carry a big load.
    • Why not “The base”? The base takes a tiny current, about a hundredth of the load's. The rest flows from collector to emitter.
    • Why not “The zener”? The zener only sets the base voltage, carrying a milliamp or so. The load's current flows through the transistor.
  2. Question 2 of 3A linear regulator from 12 V to 4.8 V at 100 mA wastes about…
    Show the answer

    Answer: 0.7 W. (12 − 4.8) × 0.1 = 0.72 W, as heat.

    • Why not “Nothing”? A linear regulator always turns the spare volts times the current into heat. It can never waste nothing.
    • Why not “7 W”? 7 W would need 1 A. At 100 mA, 7.2 V across the regulator is about 0.7 W.
    • Why not “0.07 W”? That's a tenth of the real figure. Heat is (12 − 4.8) V × 0.1 A = 0.72 W.
  3. Question 3 of 3A transistor's base-emitter junction behaves like a silicon diode. About how much does it drop?
    Show the answer

    Answer: 0.7 V. It's a silicon PN junction, so it drops about 0.7 V once it conducts, like any silicon diode.

    • Why not “0.3 V”? About 0.3 V is a Schottky or germanium junction. A silicon base-emitter junction drops about 0.7 V.
    • Why not “5 V”? 5 V is far more than any forward-biased silicon junction drops. It conducts well by about 0.7 V.
Keep practising: free lesson

“The pass transistor” is lesson 1 of 6 in Buck Bay, Power & Energy World 2. Free with an account. Lessons unlock in order, so you’ll do the 7 before it first.

Questions people ask

How does a linear regulator work?

It compares its output with a fixed reference and controls a pass transistor between the input and the output, so the transistor drops whatever voltage is spare. The simplest version is a zener on a transistor's base: the emitter follows one diode drop below the zener.

What is dropout voltage?

The smallest gap between input and output at which the regulator still holds its voltage. A 7805 needs about 2 V of headroom, so at least 7 V in for 5 V out. A low-dropout regulator, an LDO, can work with a few hundred millivolts or less.

How much heat does a linear regulator make?

(V_in − V_out) × I. From 12 V to 5 V at 0.5 A, that's 7 V × 0.5 A = 3.5 W, enough to need a heatsink. Its efficiency can never beat V_out ÷ V_in, here about 42%.

Why do linear regulators need capacitors?

To stay stable and to supply fast bursts of current. Datasheets give the values: a 7805 asks for about 0.33 µF on its input and 0.1 µF on its output, placed close to its pins. Many LDOs need a specific output capacitor to avoid oscillating.

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