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D.C. Circuits

Sensing circuits with LDRs and thermistors

  • 3 things to know
  • Component 2

Overview

Recall questions on how the resistance of each sensor varies, how the output of a divider containing one changes with light or temperature, the effect of swapping the sensor and the fixed resistor, and worked calculations for a light or frost sensing circuit.

What you need to know

Answer each one before you reveal it.

Your recallNot started
Key term

Describe how an LDR in a potential divider can be used to switch a lamp on at dusk.

Key term

How does the resistance of a light dependent resistor change with the light falling on it?

Key term

Why does the resistance of a light dependent resistor fall in bright light?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    How does the resistance of a light dependent resistor change with the light falling on it?

    Model answer

    The resistance of an LDR falls as the illumination increases, and rises again in the dark.

  2. 2

    Why does the resistance of a light dependent resistor fall in bright light?

    Model answer

    The absorbed light frees extra charge carriers in the semiconducting material, so the carrier number density rises and the material conducts more readily.

  3. 3

    How does the resistance of a negative temperature coefficient thermistor change as it warms up?

    Model answer

    Its resistance falls as its temperature rises.

  4. 4

    Why does a negative temperature coefficient thermistor conduct better when hot?

    Model answer

    Thermal energy releases many more charge carriers in the semiconductor, and this increase in carrier number density outweighs the extra scattering caused by lattice vibration.

  5. 5

    A thermistor of 5.0 kΩ is in series with a fixed 10 kΩ resistor across a 9.0 V supply. Calculate the potential difference across the fixed resistor.

    Model answer

    The output is 9.0 × 10 ÷ 15 = 6.0 V. The thermistor then takes 3.0 V, and the two add to the 9.0 V supply.

  6. 6

    A warmed thermistor of resistance 2.0 kΩ is in series with a fixed 10 kΩ resistor across a 9.0 V supply. Calculate the potential difference across the fixed resistor.

    Model answer

    The output is 9.0 × 10 ÷ 12 = 7.5 V. The thermistor takes 1.5 V, and the two add to the 9.0 V supply.

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