A transistor switch carries 50\ mA with h_FE = 100. Find the minimum base current. [2 marks]

State why the base resistor drops only V_drive - 0.7\ V, not the whole drive voltage. [1 mark]

WJEC Eduqas Eduqas 2022-style practice: The transistor above is driven from a 5.0\ V logic output and needs a base current of about 1.0\ mA. Calculate a suitable base resistor, taking the base-emitter voltage as 0.7\ V.

The base resistor drops the drive voltage minus the base-emitter voltage: V_R = 5.0 - 0.7 = 4.3\ V. Apply Ohm's law: R_B = V_RI_B = 4.31.0 × 10^-3 = 4.3 × 10^3\ = 4.3\ k. Markers reward subtracting the 0.7\ V base-emitter drop, the use of Ohm's law, and a base resistor of about 4.3\ k (a 3.9\ k preferred value gives a slightly larger, safe base current). The common error is dropping the whole 5.0\ V across the resistor.

EnglandElectronicsSyllabus dot point

How is a bipolar transistor used as a switch to turn a load on from a small input current?

Transistor switching: the bipolar transistor as a switch, cut-off and saturation, current gain, and choosing the base resistor to drive a load.

An Eduqas GCSE Electronics answer on using a bipolar transistor as a switch: the cut-off and saturation states, current gain relating collector and base current, and choosing the base resistor so a small input current turns on a larger load current.

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 bipolar transistor used as a switch is driven between cut-off (no base current, so no collector current, load off) and saturation (enough base current that it is fully on, with only a small collector-emitter voltage, load on). The current gain $h_{FE}$ relates the currents:

I_C = h_{FE}\,I_B, \qquad I_B = \frac{I_C}{h_{FE}}.

A small base current controls a much larger collector (load) current. The base resistor sets the base current; it drops the drive voltage minus the

0.7\ \text{V}

base-emitter voltage:

R_B = \frac{V_\text{drive} - 0.7}{I_B}.

Choose

I_B

a little above the minimum

\frac{I_C}{h_{FE}}

so the transistor saturates fully.

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

What this dot point is asking

Eduqas wants you to use a bipolar transistor as a switch: the two states (cut-off and saturation), the current gain that relates the small base current to the larger collector current, and how to choose the base resistor so a small input from a sensor or logic gate reliably turns on a load. The transistor switch is the bridge from a low-current control signal to a real output device.

The answer

The transistor as a switch

Current gain

Choosing the base resistor

Driving a load

Designing a transistor switch for a lamp

A lamp drawing $120\ \text{mA}$ is to be switched by a transistor ( $h_{FE} = 150$ ) from a $5.0\ \text{V}$ logic output. Find a suitable base resistor.

step 1: Find the minimum base current

$I_B(\text{min}) = \dfrac{I_C}{h_{FE}} = \dfrac{120 \times 10^{-3}}{150} = 0.8\ \text{mA}$ .

step 2: Choose a base current for hard saturation

Use about twice the minimum so the transistor is firmly on: $I_B \approx 1.6\ \text{mA}$ .

step 3: Find the base resistor

$R_B = \dfrac{V_\text{drive} - 0.7}{I_B} = \dfrac{5.0 - 0.7}{1.6 \times 10^{-3}} = \dfrac{4.3}{1.6 \times 10^{-3}} = 2.7\ \text{k}\Omega$ . A $2.7\ \text{k}\Omega$ preferred value is ideal.

Examples in context

The transistor switch is the most common output stage in the course. A light sensor in a potential divider feeds a comparator, whose output drives a transistor switch that turns on a lamp at dusk; a temperature sensor switches a fan; a microcontroller pin (which can supply only a few milliamps) drives a transistor to switch a buzzer or motor. The base-resistor and current-gain calculations recur throughout both written papers and the non-exam assessment, where a control decision must reliably drive a real load.

Try this

Q1. State the two states a transistor is switched between when used as a switch. [2 marks]

Cue. Cut-off (off) and saturation (fully on).

Q2. A transistor switch carries $50\ \text{mA}$ with $h_{FE} = 100$ . Find the minimum base current. [2 marks]

Cue. $I_B = \frac{50 \times 10^{-3}}{100} = 0.5\ \text{mA}$ .

Q3. State why the base resistor drops only $V_\text{drive} - 0.7\ \text{V}$ , not the whole drive voltage. [1 mark]

Cue. The base-emitter junction is a forward-biased diode dropping about $0.7\ \text{V}$ .

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 transistor with a current gain of

100

must switch on a lamp drawing

80\ \text{mA}

. Calculate the minimum base current needed to saturate the transistor.

Show worked answer →

Use the current-gain relationship $I_C = h_{FE}\,I_B$ , rearranged for the base current: $I_B = \dfrac{I_C}{h_{FE}}$ .

Substitute: $I_B = \dfrac{80 \times 10^{-3}}{100} = 0.8 \times 10^{-3}\ \text{A} = 0.8\ \text{mA}$ .

Markers reward the rearrangement, the conversion of the collector current to amps, and the answer $0.8\ \text{mA}$ . In practice a designer would use a little more than this (an overdrive factor) to ensure the transistor is hard on.

Eduqas 20224 marksThe transistor above is driven from a

5.0\ \text{V}

logic output and needs a base current of about

1.0\ \text{mA}

. Calculate a suitable base resistor, taking the base-emitter voltage as

0.7\ \text{V}

Show worked answer →

The base resistor drops the drive voltage minus the base-emitter voltage: $V_R = 5.0 - 0.7 = 4.3\ \text{V}$ .

Apply Ohm's law: $R_B = \dfrac{V_R}{I_B} = \dfrac{4.3}{1.0 \times 10^{-3}} = 4.3 \times 10^{3}\ \Omega = 4.3\ \text{k}\Omega$ .

Markers reward subtracting the $0.7\ \text{V}$ base-emitter drop, the use of Ohm's law, and a base resistor of about $4.3\ \text{k}\Omega$ (a $3.9\ \text{k}\Omega$ preferred value gives a slightly larger, safe base current). The common error is dropping the whole $5.0\ \text{V}$ across the resistor.

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Sources & how we know this

WJEC Eduqas GCSE (9-1) Electronics specification (C490) — WJEC Eduqas (2017)