Why Does a Zener Diode Hold Its Voltage, and When Does It Stop?

Reviewed by Loaf on

October 29, 2026•Last verified October 29, 2026•7 min read

Because in reverse breakdown its current can rise steeply for almost no rise in voltage, so the series resistor absorbs changes in the input and the Zener absorbs changes in the load. It stops regulating when its own current falls below the minimum IZmin: with the lowest input and the heaviest load, too little current is left over for the Zener, it leaves breakdown, and the output falls to whatever the resistors divide it to.

What you'll learn

  • What happens inside a Zener diode at breakdown, and why below about 5 V and above about 7 V it is two different mechanisms
  • How a shunt regulator shares the input current between the Zener and the load
  • A worked design: choosing RS, checking the worst cases and finding where regulation is lost
  • How much the output still moves, from the Zener's dynamic resistance
  • The mistakes that break a regulator: checking the wrong corner and forgetting the Zener's power

Conventions and assumptions

The Zener is modelled in breakdown as a fixed voltage VZ0 in series with a small dynamic resistance rz, valid while its reverse current is at least IZmin, taken here as 5 mA. The worked example uses an input from 10 V to 14 V, a 5.1 V Zener (5.1 V at 20 mA, rz = 10 Ω) and a load drawing 0 to 40 mA. Currents in the Zener are reverse currents, flowing from cathode to anode.

What happens at breakdown?

Reverse-biasing a p-n junction normally passes only a tiny leakage current. At a high enough reverse voltage the junction breaks down, and the current rises steeply. In heavily doped junctions with breakdown below about 5 V, the depletion region is so thin that the field lets electrons tunnel straight from the valence band on the p side to the conduction band on the n side: this is Zener breakdown. In more lightly doped junctions, breaking down above about 7 V, carriers gain enough energy crossing the wider depletion region to knock out more carriers, which knock out more: avalanche breakdown. Both are reversible, and both give the steep curve a regulator needs. They differ in temperature coefficient: tunnelling breakdown voltage falls with temperature and avalanche voltage rises, which is why diodes near 5 to 6 V drift least.

Reverse characteristic of the 5.1 volt ZenerZener current against voltage in reverse. Almost no current flows until about minus 4.3 volts, then the curve bends at the knee and becomes a steep straight line with a dynamic resistance of 10 ohms: 5.1 volts at 20 milliamps and 5.79 volts at 89 milliamps. The operating band runs from the 5 milliamp minimum to the 89 milliamp worst case of the worked example.-6-5-4-3-2-10-80-60-40-200voltage across the Zener (V)current (mA)operating band5 to 89 mA5.1 V at 20 mAknee
Reverse characteristic of the example's 5.1 V Zener: almost no current until the knee, then a steep line with rz = 10 Ω. The shaded band, 5 to 89 mA, is where the worked design keeps it.

How does the regulator share the current?

In a shunt regulator, the input drives a series resistor RS into a node that holds the Zener and the load in parallel. While the Zener is in breakdown, the node sits near VZ, so the series current is IS = (Vin − VZ)/RS, set by the input alone. That current splits: the load takes IL = VZ/RL, and the Zener takes the rest, IZ = IS − IL. When the input rises, IS rises and the extra goes into the Zener. When the load draws more, the Zener gives up the same amount. The voltage barely moves as long as IZ stays in the band between IZmin and the current where the Zener's power rating is reached.

Worked example: a Zener shunt regulatorAn input between 10 and 14 volts feeds a 100 ohm series resistor. At the output node a 5.1 volt Zener diode, cathode up, and a load drawing 0 to 40 milliamps both go to ground. The series current IS splits into the Zener current IZ and the load current IL.+−10–14 VRS = 100 Ω5.1 VRLISIZIL: 0–40 mAregulates while IZ ≥ 5 mA
The worked example: 10 to 14 V through RS = 100 Ω into a 5.1 V Zener and a load drawing 0 to 40 mA. IS splits into IZ and IL.

Worked example: designing the regulator

  • The tightest case for regulation is the lowest input with the heaviest load, 10 V and 40 mA. RS must still pass IZmin + IL = 45 mA there: RS ≤ (10 − 5.1)/0.045 = 108.9 Ω. Choose RS = 100 Ω.
  • Check that corner: IS = (10 − 5.1)/100 = 49 mA, so IZ = 49 − 40 = 9 mA, above the 5 mA minimum.
  • The hardest case for the Zener is the highest input with no load: IZ = (14 − 5.1)/100 = 89 mA, so it dissipates 5.1 × 0.089 = 454 mW. A 500 mW part is right at its limit; use a 1 W Zener.
  • The resistor in the same case dissipates (14 − 5.1)²/100 = 792 mW, so it needs at least a 1 W rating as well.
  • Where regulation is lost: at 10 V, the Zener keeps 5 mA only while IL ≤ 49 − 5 = 44 mA, that is RL ≥ 5.1/0.044 = 115.9 Ω.

What happens when it stops?

Load the regulator with 80 Ω at the 10 V input and the numbers no longer work: 80 Ω would take 64 mA at 5.1 V, more than the 49 mA available. The Zener's current falls to zero and it leaves breakdown, so it might as well not be there. The output is then a plain divider: 10 × 80/(100 + 80) = 4.44 V. Regulation fails gradually as IZ drops through the knee and then completely, and the output follows the input and the load instead of holding near 5.1 V.

How much does the output still move?

The dynamic resistance means the output is never perfectly fixed. Changing the input by 4 V moves the output by 4 × rz/(RS + rz) = 4 × 10/110 = 0.364 V, and changing the load by 40 mA moves it by 0.04 × (RS ∥ rz) = 0.364 V. A diode with smaller rz, or an IC regulator, does better; this is why a bare Zener suits references and light loads.

Where this sits in the GATE syllabus

The GATE 2027 EC syllabus lists the Zener diode alongside the P-N junction in Section 3, Electronic Devices, and diode circuits in Section 4, Analog Circuits. Questions test the breakdown mechanism, the current split in a regulator, and the energy or power a Zener absorbs.

Common mistakes

  • Checking the design at the nominal input only. Regulation fails at the lowest input with the heaviest load; the Zener's power peaks at the highest input with no load.
  • Assuming the Zener's current is fixed. It is whatever is left after the load takes its share.
  • Forgetting the minimum current. Below IZmin the Zener is near the knee and no longer holds its voltage.
  • Rating the parts at nominal values. Both the Zener and RS see their largest power with no load.
  • Calling every breakdown 'Zener breakdown'. Above about 7 V it is avalanche breakdown, with the opposite temperature coefficient.

Check your understanding

  • Q1. In the worked example, what is the Zener current at 12 V with a 20 mA load? Answer: IS = (12 − 5.1)/100 = 69 mA, so IZ = 69 − 20 = 49 mA.
  • Q2. If RS were 150 Ω, what is the largest load current that still regulates at the 10 V input? Answer: IS = 4.9/150 = 32.7 mA, so IL ≤ 32.7 − 5 = 27.7 mA.
  • Q3. Which breakdown mechanism dominates in a 3.3 V Zener, and which way does its voltage drift as it warms? Answer: tunnelling (Zener breakdown); its breakdown voltage falls as temperature rises.

Related

Official source

Last verified

The GATE 2027 EC syllabus lists the Zener diode in Section 3, Electronic Devices, and diode circuits in Section 4, Analog Circuits (checked against the official PDF from IIT Madras on 29 October 2026). Every current, power and voltage in the worked example was computed in docs/verification/ece/zener_regulator.py.

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