A red LED (V_F = 2.0\ V, I = 10\ mA) runs from a 6.0\ V supply. Find the series resistor. [2 marks]

WJEC Eduqas Eduqas 2020-style practice: A red light-emitting diode with a forward voltage of 2.0\ V and a recommended forward current of 15\ mA is to be driven from a 5.0\ V supply. Calculate the value of the series resistor required, and the power it dissipates.

Series resistor (up to 3 marks): the resistor drops the supply voltage minus the LED forward voltage: V_R = 5.0 - 2.0 = 3.0\ V. By Ohm's law R = V_RI = 3.015 × 10^-3 = 200\. Power in the resistor (up to 1 mark): P = V_R I = 3.0 × 15 × 10^-3 = 4.5 × 10^-2\ W = 45\ mW. Markers reward the voltage across the resistor 3.0\ V, the resistance 200\, and the power 45\ mW (a quarter-watt resistor is fine).

EnglandElectronicsSyllabus dot point

How does a diode conduct in only one direction, and how is that used to rectify an alternating supply?

Diodes and rectification: the diode characteristic and forward voltage, light-emitting and Zener diodes, half-wave and full-wave (bridge) rectification, and reservoir smoothing.

An Eduqas A-Level Electronics answer on diodes and rectification: the diode current-voltage characteristic and forward voltage, light-emitting diodes and the series resistor calculation, the Zener diode as a voltage reference, half-wave and full-wave bridge rectification, and reservoir-capacitor smoothing with ripple.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-02

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

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Quick answer

A diode conducts in one direction only: it passes current in forward bias once the voltage exceeds its forward voltage (about $0.7\ \text{V}$ for silicon, $1.8$ to $3.3\ \text{V}$ for an LED), and blocks current in reverse bias. A light-emitting diode (LED) emits light when forward biased and needs a series resistor to set its current. A Zener diode is run in reverse breakdown at a fixed voltage and is used as a voltage reference or regulator. Rectification turns AC into DC: a single diode gives half-wave rectification (one half-cycle), while a bridge of four diodes gives full-wave rectification (both half-cycles). A reservoir capacitor across the output charges on each peak and discharges between peaks, smoothing the pulsed output to a near-steady DC with a small ripple.

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What this dot point is asking
The answer
Examples in context
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What this dot point is asking

Eduqas wants you to describe the diode current-voltage characteristic and forward voltage, use light-emitting and Zener diodes, distinguish half-wave from full-wave (bridge) rectification, and explain reservoir-capacitor smoothing. Rectification is how an AC mains supply becomes the DC rail that powers electronics.

The answer

The diode characteristic and forward voltage

Light-emitting and Zener diodes

Half-wave and full-wave rectification

Reservoir smoothing and ripple

Sizing an LED series resistor for a different supply

A blue LED ( $V_F = 3.2\ \text{V}$ , $I = 20\ \text{mA}$ ) is driven from a $9.0\ \text{V}$ supply. Find the series resistor and check its power rating.

step 1: Find the voltage across the resistor

The resistor takes the difference: $V_R = V_S - V_F = 9.0 - 3.2 = 5.8\ \text{V}$ .

step 2: Apply Ohm's law for the current

$R = \frac{V_R}{I} = \frac{5.8}{20 \times 10^{-3}} = 290\ \Omega$ . The nearest preferred value is $300\ \Omega$ .

step 3: Check the power rating

$P = V_R I = 5.8 \times 20 \times 10^{-3} = 0.116\ \text{W}$ , so a quarter-watt ( $0.25\ \text{W}$ ) resistor is adequate.

Examples in context

Rectification and smoothing are the heart of every mains power supply, turning the $230\ \text{V}$ RMS AC into the smooth low-voltage DC that electronics needs. Diodes also protect circuits against reverse-polarity connection, steer current in logic, and free-wheel the current of an inductive load such as a relay or motor. LEDs are the standard indicator and now the standard light source, and Zener diodes set reference voltages throughout analogue circuits.

Try this

Q1. State the approximate forward voltage of a silicon diode. [1 mark]

Cue. About $0.7\ \text{V}$ .

Q2. A red LED ( $V_F = 2.0\ \text{V}$ , $I = 10\ \text{mA}$ ) runs from a $6.0\ \text{V}$ supply. Find the series resistor. [2 marks]

Cue. $R = \frac{6.0 - 2.0}{0.010} = 400\ \Omega$ .

Q3. State how many diodes a full-wave bridge rectifier uses. [1 mark]

Cue. Four.

Exam-style practice questions

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

Eduqas 20204 marksA red light-emitting diode with a forward voltage of

2.0\ \text{V}

and a recommended forward current of

15\ \text{mA}

is to be driven from a

5.0\ \text{V}

supply. Calculate the value of the series resistor required, and the power it dissipates.

Show worked answer →

Series resistor (up to 3 marks): the resistor drops the supply voltage minus the LED forward voltage: $V_R = 5.0 - 2.0 = 3.0\ \text{V}$ . By Ohm's law $R = \dfrac{V_R}{I} = \dfrac{3.0}{15 \times 10^{-3}} = 200\ \Omega$ .

Power in the resistor (up to 1 mark): $P = V_R I = 3.0 \times 15 \times 10^{-3} = 4.5 \times 10^{-2}\ \text{W} = 45\ \text{mW}$ .

Markers reward the voltage across the resistor $3.0\ \text{V}$ , the resistance $200\ \Omega$ , and the power $45\ \text{mW}$ (a quarter-watt resistor is fine).

Eduqas 20225 marksExplain the difference between half-wave and full-wave rectification, and state two advantages of full-wave rectification with a reservoir capacitor.

Show worked answer →

Difference (up to 3 marks): half-wave rectification uses a single diode that conducts on only one half of each AC cycle, so the output is a series of one-sided pulses with gaps. Full-wave rectification (a bridge of four diodes) conducts on both halves of each cycle, inverting the negative half, so the output has twice as many pulses and no gaps.

Advantages of full-wave with a reservoir capacitor (up to 2 marks): the ripple is smaller because the capacitor is topped up twice as often, and the average (DC) output voltage is higher and the supply is used more efficiently because energy is delivered on both half-cycles.

Markers reward the one-diode versus bridge distinction, conduction on one versus both half-cycles, and two valid advantages (lower ripple, higher average output / better efficiency).

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Sources & how we know this

Eduqas GCE AS/A Level Electronics specification (A410QS) — WJEC Eduqas (2017)