Zener Regulator Playground

Set the input voltage, the series resistor, the load and the Zener voltage, and the schematic shows every current and both power dissipations live — including the moment the load gets heavy enough that regulation collapses into a plain voltage divider.

$V_{out}$: 5.10 V
$I_S$: 10.15 mA
$I_L$: 5.10 mA
$I_Z$: 5.05 mA
$P_{R_S}$: 70.0 mW
$P_Z$: 25.7 mW

V_{in}

12.00 V

R_S

680 Ω

R_L

1000 Ω

V_Z

preset

V_out as V_in sweeps 0–25 V (others held at the slider values):

A diode that is meant to break down

Run an ordinary diode backwards hard enough and it breaks down, usually destructively. A Zener is built so that breakdown happens at a precise, repeatable voltage and does no damage as long as the power is kept in bounds. Past that voltage it clamps: the current through it can change enormously while the voltage across it barely moves. That is the whole regulator.

The series resistor does the real work

R_S absorbs the difference between the input and the clamped output, and it must pass both the load current and enough Zener current to keep the diode in breakdown. With the defaults above — 12 V in, 680 Ω, a 1 kΩ load and a 5.1 V Zener — about 10.1 mA leaves the resistor, 5.1 mA goes to the load and the remaining 5.0 mA passes through the Zener. Every milliamp the load stops taking, the Zener takes instead.

Where regulation stops

If the load draws so much that the divider formed by R_S and the load can no longer reach the Zener voltage, the diode falls out of breakdown and stops regulating — the output simply follows the divider. With those defaults that happens once the load drops below about 503 Ω. Below that you are not regulating at all, and the widget says so.

Why it is only good for small loads

A shunt regulator wastes the difference continuously — worst-case dissipation happens with no load, when the Zener swallows the entire supply current. That makes it cheap and simple for a reference or a few milliamps of bias, and a poor choice for anything substantial, where a series pass regulator or a switching converter is the right answer.

Learn more → Special-Purpose Diodes — Learn

Quick experiments

Watch the Zener absorb what the load does not take. With the defaults, the resistor delivers about 10.1 mA, the load takes 5.1 mA and the Zener takes the remaining 5.0 mA. Now increase the load resistance: load current falls and Zener current rises by the same amount. The total never changes.
Find the dropout point. Lower the load resistance towards 500 Ω. Just below about 503 Ω the widget stops reporting regulation — the divider can no longer reach 5.1 V, so the Zener turns off and the output sags with the load.
Discover the worst case is no load at all. Take the load to its maximum resistance. Load current goes to almost nothing and the Zener absorbs the full supply current, so its dissipation peaks. Always size a Zener for the unloaded condition, not the loaded one.
Vary the input and watch the output refuse to move. Sweep the input from 8 V to 15 V. The output stays pinned at the Zener voltage while the Zener current changes considerably — that indifference to input is exactly what a regulator is for.
Shrink the series resistor and watch dissipation climb. Halve R_S. More current flows, the load still takes what it needs, and the surplus goes through the Zener — so its power dissipation roughly doubles. R_S sets the headroom, and picking it too small cooks the diode.

Formula reference

Series resistor current: $I_S = \frac{V_{in} - V_Z}{R_S}$
Everything the circuit draws passes through R_S.
Zener current: $I_Z = I_S - I_L$
The Zener takes whatever the load leaves.
Zener power dissipation: $P_Z = V_Z \times I_Z$
Worst case at no load, where I_Z equals I_S.
Minimum load for regulation: $R_{L(min)} = \frac{R_S V_Z}{V_{in} - V_Z}$
About 503 Ω for 12 V in, 680 Ω and a 5.1 V Zener.
Choosing the series resistor: $R_S \leq \frac{V_{in(min)} - V_Z}{I_{L(max)} + I_{Z(min)}}$
Worst case: lowest input with the heaviest load.
Load regulation from Zener impedance: $\Delta V_{out} = Z_Z \times \Delta I_Z$
A real Zener is not a perfect clamp; Z_Z is a few ohms.

Symbol	Meaning	Unit
$V_Z$	Zener breakdown voltage	V
$I_Z$	Current through the Zener	A
$I_L$	Current drawn by the load	A
$R_S$	Series dropping resistor	Ω
$Z_Z$	Zener dynamic impedance in breakdown	Ω

Common mistakes

Sizing the Zener's power rating for the loaded condition.
The worst case is no load, when the Zener carries the entire series current instead of sharing it. Calculate P_Z with the load disconnected, then choose a part with margin on top of that.
Fitting a Zener without a series resistor.
Nothing then limits the current and the diode is destroyed almost immediately. R_S is not optional trim — it is what converts a fixed voltage drop into a survivable current.
Expecting regulation to hold at any load.
Below a minimum load resistance the divider cannot reach V_Z, the Zener leaves breakdown, and the output collapses into a plain divider. With 12 V, 680 Ω and 5.1 V that limit is around 503 Ω.
Treating the Zener as a perfect voltage source.
It has a dynamic impedance of a few ohms, so the output moves slightly as Zener current changes. That is the load and line regulation figure, and it is why a Zener makes a rough reference rather than a precise one.
Assuming the marked voltage holds across temperature.
Below about 5 V the tempco is negative and above it positive, crossing near 5.6 V where a Zener is most stable. Parts also carry a ±5 % tolerance as standard, so a nominal 5.1 V device may sit anywhere near 4.8 to 5.4 V.

Frequently asked questions

How does a Zener diode regulate voltage?

It is designed to break down at a precise reverse voltage without damage. Past that point the current through it can change greatly while the voltage across it barely moves, so it clamps the output while the series resistor absorbs the difference.

How do I choose the series resistor for a Zener regulator?

It must pass both the load current and enough Zener current to keep the diode in breakdown, under the worst case of lowest input voltage and heaviest load. Too large and regulation drops out; too small and the Zener overheats at light load.

What power rating does the Zener need?

Size it for the no-load condition, which is the worst case. With no load the Zener carries the entire series current instead of sharing it, so its dissipation is highest exactly when you might expect it to be idle.

When does a Zener regulator stop regulating?

When the load is heavy enough that the divider formed by the series resistor and the load can no longer reach the Zener voltage. The diode leaves breakdown and the output simply follows the divider. For 12 volts in, 680 ohms and a 5.1 volt Zener, that happens below about 503 ohms of load.

Why is a Zener regulator only suitable for small loads?

Because it is a shunt regulator: it wastes the difference between input and output continuously, and dissipation is worst at no load. That is acceptable for a reference or a few milliamps of bias, but a series pass regulator or a switching converter is the right choice for real current.

Related tools

BJT Simulator

Switch (cutoff / active / saturation) + Q-point stability across β and temperature.

Open →

FET Simulator

JFET (Shockley) and MOSFET (square-law) transfer and output curves side by side.

Open →

Op-Amp Calculator

9 configurations — gain formula, circuit schematic, and live waveform in one card.

Open →

Transistor Bias Explorer

DC bias → small-signal equivalent — hybrid-π (BJT) and FET models with Q-point coach.

Open →

CE / CS Amplifier Bench

Design a common-emitter or common-source stage — bias, load line, gain, impedance.

Open →

Differential Pair Playground

Sweep the differential input and watch current redistribute — gains and CMRR live.

Open →

Browse the full circuit toolkit or start a guided lesson in topics.