A 3.3\,V Zener is fed from a 9.0\,V supply at 10\,mA. Calculate the series resistor. [2 marks]

WJEC Eduqas style-style practice: A 5.1\,V Zener diode is supplied through a series resistor from a 9.0\,V supply. The Zener current should be 10\,mA with no load. Calculate the value of the series resistor.

A Component 1 Calculate question. The series resistor takes the **supply voltage minus the Zener voltage**: V_R = 9.0 - 5.1 = 3.9\,V (1 mark for subtracting the Zener voltage). The current through the resistor is the Zener current, I = 10\,mA = 0.010\,A (1 mark). Then R = V_RI = 3.90.010 = 390\, (2 marks for the calculation and the value, a preferred value). Markers reward subtracting the Zener voltage, using the given current and the resistor value. A common error is to use the full 9.0\,V across the resistor.

WalesElectronicsSyllabus dot point

How does a Zener diode hold a steady voltage, and how is it used to regulate a supply?

The Zener diode and its reverse breakdown behaviour, and its use with a series resistor to provide a stable regulated output voltage.

A focused answer to WJEC Eduqas GCSE Electronics on the Zener diode, covering its reverse breakdown behaviour and its use with a series resistor to provide a stable regulated output voltage.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-15

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
The Zener diode
Zener voltage regulation
Sizing the series resistor
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What this topic is asking

WJEC Eduqas wants you to know the Zener diode and its special reverse breakdown behaviour, and how a Zener diode together with a series resistor provides a stable, regulated output voltage from a supply that varies. You should explain the regulating action and calculate the series resistor.

The Zener diode

An ordinary diode would be damaged by reverse breakdown, but a Zener diode is made to work there safely. Its defining feature is that, in breakdown, the voltage across it hardly changes while the current varies over a wide range. This near-constant voltage is exactly what is needed for a reference or a regulated supply. Zener diodes are made for standard voltages such as $3.3\,\text{V}$ , $5.1\,\text{V}$ and $9.1\,\text{V}$ .

Zener voltage regulation

The series resistor is essential: it limits the Zener current and absorbs the changes in supply voltage. If the supply rises, more voltage appears across the resistor and the Zener simply passes more current, keeping the output at the Zener voltage. If the supply falls, less voltage is dropped across the resistor, but the Zener still holds the output. The result is a stable output despite a varying input, useful as a reference voltage or to power a small low-current load.

Sizing the series resistor

Because the resistor and Zener are in series, the same current flows through both (in the no-load case), and the voltage across the resistor is the supply voltage minus the Zener voltage. Dividing gives the resistance. You also check the power the Zener and the resistor must dissipate ( $P = VI$ ) so that neither overheats. The resistor must be small enough to supply enough current to keep the Zener in breakdown at the lowest supply voltage and largest load.

Designing a simple regulator

A $5.1\,\text{V}$ Zener diode is to provide a regulated voltage from a $12\,\text{V}$ supply, with a Zener current of $15\,\text{mA}$ (no load). Find the series resistor and the power it dissipates.

step 1: Find the voltage across the resistor

The resistor takes the supply minus the Zener voltage: $V_R = 12 - 5.1 = 6.9\,\text{V}$ .

step 2: Find the resistance

The same current flows through the resistor and the Zener: $I = 15\,\text{mA} = 0.015\,\text{A}$ . So $R = \dfrac{V_R}{I} = \dfrac{6.9}{0.015} = 460\,\Omega$ . The nearest preferred value is about $470\,\Omega$ .

step 3: Find the resistor power

Use $P = V_R I = 6.9 \times 0.015 = 0.10\,\text{W}$ , so a $0.25\,\text{W}$ resistor is fine.

step 4: State the answers

Use a $470\,\Omega$ resistor, dissipating about $0.10\,\text{W}$ . The output stays at $5.1\,\text{V}$ ; if the supply varies, the extra or missing voltage is absorbed by the series resistor while the Zener holds the output.

Try this

Q1. State what happens to the voltage across a Zener diode once it is in reverse breakdown. [1 mark]

Cue. It stays almost constant at the Zener voltage, even as the current changes.

Q2. A $3.3\,\text{V}$ Zener is fed from a $9.0\,\text{V}$ supply at $10\,\text{mA}$ . Calculate the series resistor. [2 marks]

Cue. $V_R = 9.0 - 3.3 = 5.7\,\text{V}$ ; $R = \dfrac{5.7}{0.010} = 570\,\Omega$ .

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas style4 marksExplain how a Zener diode and a series resistor produce a stable output voltage from a varying supply.

Show worked answer →

A Component 1 Explain question on Zener regulation. The Zener diode is connected in reverse across the output, with a series resistor between it and the supply (1 mark for the arrangement). In reverse, once the voltage reaches the Zener (breakdown) voltage, the Zener conducts and holds the voltage across it almost constant at that value (1 mark). If the supply voltage rises, the extra voltage is dropped across the series resistor, and the Zener takes more current, so the output stays at the Zener voltage; if the supply falls, the opposite happens (2 marks for the resistor absorbing the change while the Zener holds the output). Markers reward the reverse connection, the constant Zener voltage, and the series resistor absorbing supply changes.

Eduqas style4 marksA

5.1\,\text{V}

Zener diode is supplied through a series resistor from a

9.0\,\text{V}

supply. The Zener current should be

10\,\text{mA}

with no load. Calculate the value of the series resistor.

Show worked answer →

A Component 1 Calculate question. The series resistor takes the supply voltage minus the Zener voltage: $V_R = 9.0 - 5.1 = 3.9\,\text{V}$ (1 mark for subtracting the Zener voltage). The current through the resistor is the Zener current, $I = 10\,\text{mA} = 0.010\,\text{A}$ (1 mark). Then $R = \dfrac{V_R}{I} = \dfrac{3.9}{0.010} = 390\,\Omega$ (2 marks for the calculation and the value, a preferred value). Markers reward subtracting the Zener voltage, using the given current and the resistor value. A common error is to use the full $9.0\,\text{V}$ across the resistor.

Related dot points

Sources & how we know this

WJEC Eduqas GCSE Electronics specification (from 2017) — WJEC Eduqas (2017)