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SingaporeElectronicsSyllabus dot point

How do a light-dependent resistor and a thermistor turn light and temperature into an electrical signal a circuit can use?

Describe the LDR and the thermistor as input transducers, state how their resistance changes, and use them in a potential divider

A focused answer to the O-Level Electronics outcome on input transducers. How the LDR and thermistor change resistance with light and temperature, and how they produce a varying voltage in a divider.

Generated by Claude Opus 4.88 min answer

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

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

SEAB wants you to describe the light-dependent resistor (LDR) and the thermistor as input transducers, to state how each one's resistance changes with its surroundings, and to use them in a potential divider to produce a varying voltage. The central insight is that an input transducer turns a physical quantity (light or temperature) into an electrical quantity (resistance, then voltage) that the rest of the circuit can act on.

The answer

What an input transducer is

A transducer converts energy from one form to another. An input transducer (a sensor) converts a physical quantity such as light, temperature, sound or position into an electrical signal. The LDR and the thermistor are two common input transducers whose resistance changes with the quantity they sense, so they are used to make a circuit respond to its surroundings.

The light-dependent resistor

A light-dependent resistor senses light. Its resistance depends on how much light falls on it:

  • In the dark, its resistance is high (it can be hundreds of kilohms or more).
  • In bright light, its resistance is low (a few hundred ohms to a few kilohms).

So more light means less resistance. The LDR responds slowly compared with electronic switching but is ideal for light-level sensing such as automatic lighting.

The thermistor

A thermistor senses temperature. The common type used in the syllabus has a negative temperature coefficient, meaning:

  • When cold, its resistance is high.
  • When hot, its resistance is low.

So a higher temperature means less resistance. Thermistors are used in temperature alarms, thermostats and temperature measurement.

Producing a varying voltage

A resistance change is not directly useful; circuits respond to voltage. Placing the transducer in a potential divider with a fixed resistor converts its changing resistance into a changing voltage, using:

Vout=R2R1+R2VinV_{out} = \frac{R_2}{R_1 + R_2}\,V_{in}

Where the transducer sits in the divider decides whether the output rises or falls with the input. This output voltage then drives the next stage, such as a transistor switch or an op-amp comparator.

Examples in context

Example 1. A street light controller. An LDR in a potential divider senses the fading daylight. As dusk falls, the LDR resistance rises and the divider output crosses the level that switches the street light on; at dawn the reverse happens and it switches off. The transducer turns the brightness of the sky into the voltage that controls the lamp.

Example 2. A fridge over-temperature alarm. A thermistor inside a fridge feeds a divider whose output rises if the temperature climbs too high. When the output passes a set point, the next stage sounds an alarm, warning that the fridge is failing. The thermistor converts the temperature into a signal the alarm circuit can judge.

Try this

  • Cue. State what happens to an LDR's resistance when the light level falls. The resistance rises; an LDR has high resistance in the dark and low resistance in bright light.

  • Cue. An LDR is in series with a 5.0 kΩ5.0\ \text{k}\Omega resistor across 10 V10\ \text{V}, output across the LDR. In the dark the LDR is 15 kΩ15\ \text{k}\Omega. Find the output. Vout=155.0+15×10=7.5 VV_{out} = \dfrac{15}{5.0 + 15} \times 10 = 7.5\ \text{V}.

  • Cue. Explain why a transducer is placed in a potential divider rather than used alone. Circuits respond to voltage, so the divider converts the transducer's changing resistance into a changing voltage the next stage can use.

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.

Original4 marksDescribe how the resistance of (a) a light-dependent resistor and (b) a thermistor changes with its surroundings, and state what type of input each one senses.
Show worked answer →

(a) A light-dependent resistor (LDR) senses light. Its resistance is high in the dark and falls as the light level increases, becoming low in bright light.

(b) A thermistor (the common negative-temperature-coefficient type) senses temperature. Its resistance is high when cold and falls as the temperature rises, becoming low when hot.

What markers reward: LDR resistance falling with more light, thermistor resistance falling with higher temperature, and naming light and temperature as the quantities sensed. Stating the direction of change clearly is essential.

Original4 marksAn LDR is connected in series with a 10 kΩ10\ \text{k}\Omega resistor across a 6.0 V6.0\ \text{V} supply, with the output taken across the fixed resistor. In bright light the LDR resistance is 2.0 kΩ2.0\ \text{k}\Omega. Calculate the output voltage in bright light.
Show worked answer →

The output is across the fixed 10 kΩ10\ \text{k}\Omega resistor: Vout=RfixedRLDR+RfixedVinV_{out} = \dfrac{R_{fixed}}{R_{LDR} + R_{fixed}}\,V_{in}.

Vout=102.0+10×6.0=1012×6.0=5.0 VV_{out} = \dfrac{10}{2.0 + 10} \times 6.0 = \dfrac{10}{12} \times 6.0 = 5.0\ \text{V}.

What markers reward: the divider equation with the fixed resistor on top (the output is across it), substituting the bright-light LDR value, and the answer 5.0 V5.0\ \text{V}. Using the LDR resistance on top is the usual error.

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