A transistor switch carries 200\ mA with h_FE = 100. Find the minimum base current to saturate it. [2 marks]

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How is a transistor used as a switch to turn a load on and off from a small control signal?

Transistor switching: saturation and cut-off, choosing the base resistor, the Darlington pair, and driving output transducers such as lamps, LEDs, buzzers and motors.

An Eduqas A-Level Electronics answer on using a transistor as a switch: the saturation and cut-off states, choosing the base resistor to saturate the transistor, the Darlington pair for high gain, and driving output transducers such as lamps, LEDs, buzzers and motors from a logic signal.

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

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Quick answer

Used as a switch, a transistor is driven between two states: cut-off (no base current, so no collector current, the load is off) and saturation (enough base current that the transistor is fully on, with only a small collector-emitter voltage, the load is on). To saturate, the base current must exceed $\frac{I_C}{h_{FE}}$ , usually with an overdrive factor; the base resistor is then $R_B = \frac{V_\text{drive} - 0.7}{I_B}$ . A Darlington pair cascades two transistors so their gains multiply, letting a tiny control current switch a large load (at the cost of a higher on-voltage). The transistor switch drives output transducers such as lamps, LEDs (with a series resistor), buzzers and motors (with a flyback diode for inductive loads).

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What this dot point is asking

Eduqas wants you to use a transistor as a switch: the saturation and cut-off states, choosing the base resistor to saturate it, the Darlington pair for high gain, and driving output transducers (lamps, LEDs, buzzers, motors). This is the output subsystem that lets a logic or sensor signal control a real load.

The answer

Saturation and cut-off

Choosing the base resistor

The Darlington pair

Driving output transducers

Designing a transistor switch for a buzzer

A buzzer drawing $60\ \text{mA}$ at $6.0\ \text{V}$ is switched by a transistor ( $h_{FE} = 120$ ) from a $5.0\ \text{V}$ logic output. Find a suitable base resistor.

step 1: Find the minimum base current

$I_B(\text{min}) = \frac{I_C}{h_{FE}} = \frac{60 \times 10^{-3}}{120} = 0.5\ \text{mA}$ .

step 2: Apply an overdrive factor

Use about 3 times for hard saturation: $I_B \approx 1.5\ \text{mA}$ .

step 3: Find the base resistor

$R_B = \frac{V_\text{drive} - 0.7}{I_B} = \frac{5.0 - 0.7}{1.5 \times 10^{-3}} = \frac{4.3}{1.5 \times 10^{-3}} = 2.9\ \text{k}\Omega$ . A $2.7\ \text{k}\Omega$ preferred value gives a slightly larger, safe base current.

Examples in context

The transistor switch is the most common output stage in the course: it turns on the lamp, buzzer, LED or motor that a sensing or logic circuit has decided to activate. A Darlington pair lets a feeble light-sensor current switch a powerful load, and the base-resistor calculation is a staple of both the written papers and the non-exam assessment, where a microcontroller or comparator output must reliably drive a real transducer.

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 $200\ \text{mA}$ with $h_{FE} = 100$ . Find the minimum base current to saturate it. [2 marks]

Cue. $I_B = \frac{200 \times 10^{-3}}{100} = 2.0\ \text{mA}$ .

Q3. State why a Darlington pair has a high current gain. [1 mark]

Cue. The two transistors' gains multiply ( $h_{FE} \approx h_{FE1} \times h_{FE2}$ ).

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 20196 marksA transistor switch must turn on a lamp that draws

120\ \text{mA}

from a

6.0\ \text{V}

supply. The transistor has a current gain

h_{FE} = 100

and is driven from a

5.0\ \text{V}

logic output. Calculate the minimum base current to saturate the transistor and a suitable base resistor.

Show worked answer →

Minimum base current (up to 3 marks): to saturate, the base current must supply more than $\frac{I_C}{h_{FE}}$ . With $I_C = 120\ \text{mA}$ , $I_B(\text{min}) = \frac{120 \times 10^{-3}}{100} = 1.2\ \text{mA}$ . Designers use about 2 to 5 times this to ensure hard saturation, so choose roughly $3\ \text{mA}$ .

Base resistor (up to 3 marks): the base resistor drops the logic voltage minus the $0.7\ \text{V}$ base-emitter voltage: $V_R = 5.0 - 0.7 = 4.3\ \text{V}$ . For $I_B = 3\ \text{mA}$ , $R_B = \frac{4.3}{3 \times 10^{-3}} = 1.4\ \text{k}\Omega$ (a $1.2\ \text{k}\Omega$ preferred value gives a slightly larger, safe base current).

Markers reward $I_B(\text{min}) = 1.2\ \text{mA}$ , the use of an overdrive factor for hard saturation, and a base resistor of order $1\ \text{k}\Omega$ found from $\frac{5.0 - 0.7}{I_B}$ .

Eduqas 20224 marksExplain what a Darlington pair is and why it is used to drive a high-current load from a small control current.

Show worked answer →

What it is (up to 2 marks): a Darlington pair is two transistors connected so that the emitter of the first feeds the base of the second; they act as a single transistor with a very high combined current gain, approximately the product of the two individual gains ( $h_{FE} \approx h_{FE1} \times h_{FE2}$ ).

Why it is used (up to 2 marks): the very high overall gain means a tiny base current (such as a weak sensor or logic output can supply) controls a large load current, so it can drive motors, relays or solenoids directly. The trade-off is a higher on-state voltage (about $1.4\ \text{V}$ , two base-emitter drops).

Markers reward the cascade connection, the multiplied current gain, and the ability to control a large load from a small drive current.

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Sources & how we know this

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