A Zener as a regulator
One resistor and one Zener make a fixed rail out of a supply that wanders. Whether it works comes down to two sums — the lowest input with the biggest load, and the highest input with no load at all — and the resistor has to satisfy both at once.
The circuit is two parts. A resistor from the unregulated supply to a node, a Zener from that node to ground, and the load in parallel with the Zener.
The resistor has two jobs at once, and every difficulty on this page comes from that. It drops whatever is left over between the supply and VZ, and in doing so it sets the total current the node has to share out. What the load does not take, the Zener takes:
IR = IZ + ILOAD, always. The resistor decides the total, the load helps itself first, and the Zener passes the remainder to hold the node still. A shunt regulator is subtraction, which is why it behaves so badly when the load is turned off — that is the case where the Zener takes everything.
Move both sliders. The supply one is what a datasheet calls line regulation; the load one is load regulation; they are the same circuit failing in two directions.
Sizing the resistor
There is no formula, because there are two constraints pulling opposite ways.
The resistor must be small enough that at the lowest input voltage, with the biggest load, there is still enough current left to keep the Zener above its knee — R no larger than (VIN,min − VZ) divided by (ILOAD,max + IZ,min).
And large enough that at the highest input, with the load disconnected, the Zener is not asked to burn more than it can — R no smaller than (VIN,max − VZ) divided by IZ,max.
IZ,min is a few milliamps rather than the datasheet's 1 mA knee — right at the knee the voltage has already sagged. IZ,max is the power rating divided by VZ, halved for a part in free air.
A worked one. An 8–12 V supply, a load that draws 0 to 20 mA, a 1N4733A:
| Sum | Answer | |
|---|---|---|
| Small enough | (8 − 5.1) ÷ (20 mA + 5 mA) | R ≤ 116 Ω |
| Large enough | (12 − 5.1) ÷ 100 mA | R ≥ 69 Ω |
100 Ω sits in the middle of that window, so the design works. Check both ends before believing it: at 8 V with the full load, the resistor passes 29 mA and the Zener gets 9 mA — above the knee. At 12 V with the load unplugged, the Zener takes all 69 mA and burns 350 mW, inside a derated 1 W part.
The useful part of the exercise is what happens when the window closes. Widen the supply to 8–20 V, or the load to 100 mA, and the two inequalities cross: there is no resistor that satisfies both, and no amount of care will find one. That is the circuit telling you to use a different one.
What it costs
It draws the same current whatever the load does. The 100 Ω resistor above pulls 69 mA from a 12 V supply — 830 mW — to deliver 20 mA at 5 V, which is 100 mW. The other 730 mW is heat in a resistor and a diode, and it is spent identically when the load is asleep.
The dangerous state is no load, not full load. Every other regulator on the site is stressed by its load. This one is stressed by the absence of one, which is exactly backwards from the intuition people arrive with, and is how a Zener that worked on the bench dies the first time a connector falls out.
The output is not flat. ZZT is 7 Ω, so the rail moves about 7 mV for every extra milliamp through the Zener — and the load changing by 20 mA changes the Zener current by 20 mA. Add ±5 % part tolerance on top and the "5 V rail" is somewhere between about 4.8 and 5.4 V, moving as the load breathes.
Nothing is protected. Short the output and the only thing limiting the current is the resistor. A Zener regulator has no current limit, no thermal shutdown, and no idea anything has gone wrong.
Where it is still the right answer
| Job | Why the Zener wins |
|---|---|
| A bias or reference of a few mA | Two parts, nothing to start up, nothing to oscillate |
| Clamping a MOSFET gate | It does nothing at all until the voltage is too high |
| Protecting an input pin | Same, and it fails safe: a dead Zener usually fails short |
| A fixed offset in a signal path | Nothing else subtracts a constant that cheaply |
| Setting a comparator threshold | One part, and the threshold is a part number |
The offset one is worth drawing out, because it is the same two parts in the other order. Put the Zener in series between the input and the output, stripe towards the input, and a resistor from the output down to ground. The resistor pulls just enough current through the Zener to keep it in breakdown, and the output sits 5.1 V below the input at any input voltage — a level shifter made of one diode. It holds only while that current keeps flowing, which is the same knee as everywhere else on this page: draw much from the output and the shift collapses.
For anything that has to hold a rail up under a load that changes, the answer is a regulator with feedback in it. That is what an LDO does, for about the same money and a tenth of the wasted current. The Zener's place is beside it — holding a gate down, clamping a pin, or being the one part that has a voltage written on it.
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