What is sizing the base resistor (BJT)?

To saturate the transistor you must supply at least the base current I_B = I_Ch_FE. The base resistor drops the difference between the driving voltage and the 0.7\,V base-emitter voltage. In practice a smaller resistor is chosen so that more base current flows than the minimum, driving the transistor hard into saturation for reliable switching.

A transistor switch has h_FE = 200 and must pass a collector current of 40\,mA. Find the minimum base current to saturate it. [2 marks]

WalesElectronicsSyllabus dot point

How do bipolar and MOSFET transistors act as electronic switches, and how do you design a switching stage?

Transistors as switches: the bipolar junction transistor and MOSFET, cut-off and saturation, the base (or gate) resistor, switching a load, and the Darlington pair.

A focused answer to WJEC A-Level Electronics transistor switching, covering the bipolar junction transistor and MOSFET as switches, cut-off and saturation, sizing the base or gate resistor, switching output loads, and the Darlington pair for high current gain.

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 transistor switch is the standard interface between a low-power processing block and a high-power output. WJEC expects you to explain how a bipolar junction transistor (BJT) and a MOSFET switch between cut-off and saturation, size the base or gate resistor, drive a real load such as a relay or motor, and recognise the Darlington pair for high gain. Base-resistor calculations and the comparison of BJT with MOSFET are dependable exam questions.

The answer

The two switching states

Keeping the transistor in one of these two states (rather than the linear region used for amplifiers) means it dissipates very little power, because either the current or the voltage across it is nearly zero.

Sizing the base resistor (BJT)

To saturate the transistor you must supply at least the base current $I_B = \frac{I_C}{h_{FE}}$ . The base resistor drops the difference between the driving voltage and the $0.7\,\text{V}$ base-emitter voltage. In practice a smaller resistor is chosen so that more base current flows than the minimum, driving the transistor hard into saturation for reliable switching.

The MOSFET as a switch

Switching real loads

Output loads (relays, motors, lamps) are placed in the collector (or drain) circuit. Inductive loads must have a flyback diode across them to absorb the back-EMF when the transistor switches off, or the voltage spike will destroy the transistor.

Switching an LED array with a bipolar transistor

A transistor must switch an LED array drawing $150\,\text{mA}$ . It has $h_{FE} = 120$ and $V_{BE} = 0.7\,\text{V}$ , driven from a $5.0\,\text{V}$ logic output. Find the maximum base resistor.

step Find the base current to saturate

$I_B = \dfrac{I_C}{h_{FE}} = \dfrac{150 \times 10^{-3}}{120} = 1.25\,\text{mA}$ .

step Find the voltage across the base resistor

$V = V_{in} - V_{BE} = 5.0 - 0.7 = 4.3\,\text{V}$ .

step Apply Ohm's law

$R_B = \dfrac{4.3}{1.25 \times 10^{-3}} = 3.4\,\text{k}\Omega$ . A standard value somewhat below this, such as $2.7\,\text{k}\Omega$ , is used to overdrive the base and guarantee saturation.

Examples in context

Example 1. A microcontroller driving a relay: A PIC output pin supplies only a few milliamps, far too little for a relay coil. A MOSFET switch (or a BJT with a base resistor) lets the pin control the coil, with a flyback diode across the coil. This is the standard output interface in almost every microcontroller project.
Example 2. A Darlington pair for a high-current solenoid: When the load current is very large and the gain of one transistor is not enough, two transistors are connected as a Darlington pair so their gains multiply, giving an effective gain of several thousand. A tiny base current then switches a heavy solenoid.
Example 3. PWM motor speed control: A MOSFET switched rapidly on and off (pulse-width modulation) controls a motor's average power without dissipating much heat, because the MOSFET is always either fully on (low voltage across it) or fully off (no current). This switching efficiency is exactly why MOSFETs dominate power control.

Try this

Q1. A transistor switch has $h_{FE} = 200$ and must pass a collector current of $40\,\text{mA}$ . Find the minimum base current to saturate it. [2 marks]

Cue. $I_B = \frac{I_C}{h_{FE}} = \frac{40 \times 10^{-3}}{200} = 0.20\,\text{mA}$ .

Q2. Explain why a transistor used as a switch dissipates little power in both of its states. [2 marks]

Cue. In cut-off the current is nearly zero, so $P = VI$ is small; in saturation the collector-emitter voltage is nearly zero, so $P = VI$ is again small. Power is only large in the linear region between.

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.

WJEC Eduqas 20205 marksAn NPN transistor switches a relay coil that draws

80\,\text{mA}

. The transistor has a current gain

h_{FE} = 100

, a base-emitter voltage of

0.7\,\text{V}

, and is driven from a

5.0\,\text{V}

logic output. Calculate a suitable base resistor and state why a smaller value than the calculated maximum is normally chosen.

Show worked answer →

Find the base current needed to saturate the transistor, then size the base resistor with Ohm's law.

Base current to just saturate: $I_B = \dfrac{I_C}{h_{FE}} = \dfrac{80 \times 10^{-3}}{100} = 0.80\,\text{mA}$ .

Voltage across the base resistor: $5.0 - 0.7 = 4.3\,\text{V}$ .

Maximum base resistor: $R_B = \dfrac{V}{I_B} = \dfrac{4.3}{0.80 \times 10^{-3}} = 5.4\,\text{k}\Omega$ .

A smaller resistor is chosen so that more base current flows than the bare minimum, driving the transistor hard into saturation. This keeps the collector-emitter voltage low and the switching reliable even if the gain is lower than quoted or the load current rises.

Markers reward the base current, the resistor voltage, $R_B \approx 5.4\,\text{k}\Omega$ , and the overdrive justification.

WJEC Eduqas 20183 marksState two advantages of using a MOSFET rather than a bipolar transistor to switch a load, and one advantage the bipolar transistor retains.

Show worked answer →

A MOSFET is voltage-controlled at its gate and draws virtually no steady gate current, so it places almost no load on the driving circuit. It also typically has a very low on-resistance, so it dissipates little power when fully on.

The bipolar transistor's advantage is that it can be switched with a simple base resistor and is cheaper and adequate for many low-power switching jobs; it is also less easily damaged by static than a MOSFET gate.

Markers reward two valid MOSFET advantages (negligible gate current, low on-resistance, fast switching) and one valid bipolar advantage (simple drive, low cost, static robustness).

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Sources & how we know this

WJEC Eduqas GCE A-level Electronics specification — WJEC Eduqas (2017)