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How do we combine a sensor, a potential divider and a transistor to make a circuit that switches a load automatically?

Explain a sensor-driven transistor switching circuit and design one to turn a load on in the dark or when it is hot

A focused answer to the O-Level Electronics outcome on transistor switching. Combining a sensor potential divider with a transistor and base resistor to switch a load on automatically.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

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

SEAB wants you to explain a sensor-driven transistor switching circuit and to design one that turns a load on automatically when it gets dark or hot. The central insight is that three building blocks combine: a potential divider with a sensor produces a voltage that changes with the surroundings, a base resistor feeds that voltage to a transistor, and the transistor switches the load. Each part has a clear job.

The answer

The three building blocks

A sensor switching circuit has three stages:

A sensing potential divider: a sensor (LDR or thermistor) and a fixed or variable resistor in series across the supply. Its output voltage changes as the light or temperature changes.
A base resistor: connects the divider output to the base of the transistor, limiting the base current to a safe value.
A transistor and load: the transistor switches the load (a lamp, buzzer or relay) placed in its collector.

How the switch is triggered

The transistor turns on only when the voltage at its base is about $0.7\ \text{V}$ above its emitter. The divider is arranged so that, at the wanted light or temperature, its output crosses $0.7\ \text{V}$ :

For a dark sensor, the LDR is placed so the junction voltage rises as it gets darker. When darkness pushes the junction past $0.7\ \text{V}$ , the transistor switches the load on.
For a heat sensor, the thermistor is placed so the junction voltage rises as it gets hotter, switching the load on when it is hot enough.

Setting the trigger point

A variable resistor in the divider lets you choose the exact light or temperature at which switching occurs. Adjusting it changes the divider ratio, so the junction reaches $0.7\ \text{V}$ at a different point. This is how a dusk-sensing lamp is set to come on at just the right level of darkness.

Why each part matters

The divider converts the physical quantity into a voltage. The transistor converts that small base signal into a large switched current. The base resistor protects the transistor. Remove any part and the circuit fails: with no base resistor the transistor can be destroyed; with no divider there is nothing to sense; with no transistor the weak sensor signal cannot drive a real load.

Worked example

A dark-sensing circuit uses an LDR and a fixed resistor in a divider on a $9.0\ \text{V}$ supply, feeding an npn transistor through a base resistor. The transistor switches a lamp needing $90\ \text{mA}$ . The transistor gain is $200$ . Find the base current needed and a suitable base resistor, taking the divider output at switch-on as $0.7\ \text{V}$ above the emitter and the base-emitter drop as $0.7\ \text{V}$ .

Step 1: Find the base current to switch the lamp

The collector current is the lamp current, $90\ \text{mA}$ . Using $I_B = \dfrac{I_C}{\beta} = \dfrac{90}{200} = 0.45\ \text{mA}$ .

Step 2: Identify the voltage across the base resistor

At switch-on the divider drives the base. The base-emitter junction takes $0.7\ \text{V}$ , and to drive the wanted base current the divider must supply a little more. Taking the available driving voltage above the $0.7\ \text{V}$ junction as about $4.3\ \text{V}$ for this design, that is the voltage across the base resistor.

Step 3: Find the base resistor with Ohm's law

$R_B = \dfrac{V}{I_B} = \dfrac{4.3}{0.45 \times 10^{-3}} \approx 9.6\ \text{k}\Omega$ , so a $10\ \text{k}\Omega$ resistor is a sensible choice.

Step 4: Interpret the design

A base current of $0.45\ \text{mA}$ , set by a $10\ \text{k}\Omega$ resistor, lets the transistor switch the $90\ \text{mA}$ lamp on when darkness raises the divider voltage past the turn-on point. The small sensing signal commands the large lamp current.

Common traps

Putting the load in the base, not the collector: The load goes in the collector circuit, where the large switched current flows; the base carries only the small control current.
Omitting the base resistor: Without it the base current is uncontrolled and can destroy the transistor.
Getting the sensor orientation wrong: Whether the circuit triggers in the dark or the light depends on which arm holds the LDR. Trace the junction voltage as the light changes.
Forgetting the $0.7\ \text{V}$ turn-on threshold: The transistor stays off until the base is about $0.7\ \text{V}$ above the emitter; the divider must reach this at the wanted condition.
Ignoring the variable resistor's role: It is there to set the exact trigger level, not just as an extra component; adjusting it tunes the switching point.

Examples in context

Example 1. An automatic night light. An LDR divider feeding a transistor switches on a lamp when the room gets dark, then switches it off again at dawn. The variable resistor lets the user choose how dark it must be before the light comes on. This is the classic first project that ties together sensing, switching and tuning.

Example 2. A fire or overheat alarm. A thermistor divider feeding a transistor switches on a buzzer when the temperature rises past a set point. The same three-stage circuit, with the sensor changed and the trigger adjusted, becomes a temperature warning instead of a light switch, showing how general the design is.

Try this

Cue. State where the load is connected in a transistor switching circuit and why. In the collector circuit, because that is where the large switched current flows; the base carries only the small control current.
Cue. Explain how to change a dark-switching LDR circuit into a light-switching one. Swap the positions of the LDR and the fixed resistor in the divider, so the junction voltage now rises in bright light instead of in the dark.
Cue. A lamp needs $50\ \text{mA}$ and the transistor gain is $100$ . Find the minimum base current to switch it on. $I_B = I_C / \beta = 50 / 100 = 0.50\ \text{mA}$ .

Exam-style practice questions

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

Original5 marksA circuit has an LDR and a variable resistor forming a potential divider, with the junction connected through a base resistor to an npn transistor that switches a lamp in its collector. Explain how the circuit turns the lamp on when it gets dark, and state the purpose of the variable resistor.

Show worked answer →

As it gets dark, the LDR resistance rises. The divider is arranged so the voltage at the junction rises in the dark. When this voltage reaches about $0.7\ \text{V}$ above the emitter, a base current flows.

The base current turns the transistor on, allowing a large collector current to flow through the lamp, so the lamp lights. The variable resistor sets the light level at which switching happens, by adjusting the divider so the junction reaches $0.7\ \text{V}$ at the chosen darkness.

What markers reward: dark raising the LDR resistance and the junction voltage to $0.7\ \text{V}$ , the base current switching the transistor and lamp on, and the variable resistor setting the trigger point.

Original3 marksState the purpose of the base resistor in a transistor switching circuit and explain what could happen if it were omitted.