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How does a transistor act as an electronic switch driven by a small signal?

The bipolar transistor and MOSFET as switches, base/gate biasing, the Darlington pair and switching inductive loads.

A CCEA A-Level Technology and Design answer on using a bipolar transistor or MOSFET as an electronic switch, base and gate biasing, current gain and the Darlington pair, and protecting against the back-e.m.f. of an inductive load with a flyback diode.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

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

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

What this dot point is asking

CCEA expects you to use a bipolar transistor and a MOSFET as electronic switches in the process/output stage, to bias the base or gate correctly, to use current gain ( $h_{FE}$ ) and the Darlington pair, and to protect against the back-e.m.f. of inductive loads with a flyback diode. Calculations of base current are common.

The answer

The transistor as a current-operated switch

Current gain and the Darlington pair

MOSFETs and inductive loads

Worked example: sizing the base resistor and protecting the load

Driving a 12 V relay from a 5 V logic output

A relay coil needs $I_C = 60\ \text{mA}$ ; the transistor has $h_{FE} = 100$ ; the controlling logic output is 5 V.

step Find the base current

I_B = \frac{I_C}{h_{FE}} = \frac{60\ \text{mA}}{100} = 0.6\ \text{mA}.

Drive a little harder (say 1 mA) to guarantee saturation.

step Choose the base resistor

The base resistor drops the logic voltage minus the 0.7 V base-emitter drop across itself: with about 4.3 V across it and roughly 1 mA wanted, $R_B \approx 4.3\ \text{V} / 1\ \text{mA} \approx 4.3\ \text{k}\Omega$ (a 3.9 kilohm or 4.7 kilohm resistor is fine).

step Protect against back-e.m.f.

Fit a flyback diode in reverse across the relay coil, so when the transistor switches off the coil's induced voltage is safely clamped and the transistor is not destroyed. This sequence, gain to base current to base resistor plus protection, is the standard CCEA driver design.

Examples in context

Example 1. Microcontroller driving a motor. A microcontroller pin cannot supply motor current, so it drives a MOSFET (or transistor) that switches the motor, with a flyback diode across the motor, the standard interface between logic and power.

Example 2. Touch-sensitive lamp. A Darlington pair amplifies the tiny current that flows through skin when a contact is touched, enough to switch a lamp, a use that depends on the pair's very high combined gain.

Try this

Q1. State the approximate base-emitter voltage at which an NPN transistor turns on. [1 mark]

Cue. About 0.7 V.

Q2. A transistor must pass a collector current of 50 mA and has hFE = 200. Find the minimum base current. [2 marks]

Cue. $I_B = 50\ \text{mA} / 200 = 0.25\ \text{mA}$ .

Q3. Why is a diode fitted across a relay coil, and which way round? [2 marks]

Cue. It is a flyback diode, fitted in reverse across the coil, to safely conduct the back-e.m.f. produced when the coil switches off and protect the transistor.

Exam-style practice questions

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

CCEA 20196 marksDescribe how an NPN transistor is used to switch a relay from the output of a sensing circuit, and explain the purpose of the diode connected across the relay coil.

Show worked answer →

The NPN transistor acts as a current-operated switch. The output of the sensing circuit (for example a potential divider) is connected through a base resistor to the base. When the base voltage rises above about 0.7 V and supplies enough base current, the transistor turns on (saturates) and a much larger collector current flows, energising the relay coil between the supply and the collector; the relay contacts then switch the load. When the base voltage is low the transistor is off and the relay is de-energised. A small base current thus controls a large load current (current gain $h_{FE}$ ).

The diode across the relay coil is a flyback (freewheeling) diode. The coil is inductive: when the transistor switches off, the collapsing magnetic field induces a large back-e.m.f. that would otherwise destroy the transistor. The diode is connected in reverse across the coil so that this induced voltage forward-biases it and is safely discharged as a current loop, protecting the transistor.

Markers reward the base resistor and 0.7 V turn-on, the small-current-controls-large-current idea, and a correct explanation of the flyback diode protecting against the coil's back-e.m.f.

CCEA 20214 marksA transistor has a current gain (hFE) of 100 and needs a collector current of 80 mA to operate a load. Calculate the minimum base current required, and state one advantage of a MOSFET over a bipolar transistor for switching.

Show worked answer →

Current gain relates collector and base current:

h_{FE} = \frac{I_C}{I_B} \Rightarrow I_B = \frac{I_C}{h_{FE}} = \frac{80\ \text{mA}}{100} = 0.8\ \text{mA}.

So a base current of at least 0.8 mA is needed (often a little more to ensure full saturation).

One advantage of a MOSFET: it is voltage-controlled with a very high input (gate) resistance, so it draws almost no steady gate current, making it easy to drive directly from logic or a microcontroller output, and it can have a very low on-resistance for efficient high-current switching.

Markers want the correct $I_B = I_C/h_{FE}$ calculation and a valid MOSFET advantage (voltage-controlled / negligible input current / low on-resistance).

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Sources & how we know this

CCEA GCE Technology and Design specification — CCEA (2016)