Resistance changes in thermistors and LDRsEdexcel International A Level Physics: Revision notes

The resistance of a conductor depends on how many conduction electrons it has per unit volume ($n$) and how often they collide with the lattice ions. From $I = nAvq$, a larger $n$ gives a larger current for the same potential difference, so a lower resistance.

Temperature and light change $n$ or the lattice vibrations, so they can change the resistance. Components designed to exploit this are used as sensors.

In a metal the number of conduction electrons is large and almost independent of temperature. Heating makes the lattice ions vibrate with larger amplitude, so conduction electrons collide with them more often. The electrons' average drift velocity falls for the same pd, so the resistance increases.

The resistance of a metal therefore increases with temperature, roughly linearly over a modest range.

An NTC (negative temperature coefficient) thermistor is made from a semiconductor. At low temperature few electrons are free. Heating gives bound electrons enough energy to escape from their atoms, so the number of conduction electrons increases sharply.

The lattice ions also vibrate more, but this has a much smaller effect than the rise in charge carriers. The net result is that the resistance falls as temperature rises, and the change is large and non-linear.

Thermistors are used in temperature sensors, for example in thermostats and fire alarms.

An LDR is made from a semiconductor such as cadmium sulfide. In the dark most electrons are bound to their atoms, so there are few conduction electrons and the resistance is high.

When light falls on the LDR, photons are absorbed and release electrons from atoms. The number of conduction electrons increases, so the current for a given pd is larger and the resistance falls. The brighter the light, the lower the resistance, so LDRs are used in light meters and automatic lighting.

A thermistor on a 5.0 V supply carries 2.0 mA at 20 °C and 12.5 mA at 80 °C.

• At 20 °C: $R = V/I = 5.0 \div 2.0\times10^{-3} = 2500\ \Omega$
• At 80 °C: $R = 5.0 \div 12.5\times10^{-3} = 400\ \Omega$

The resistance has fallen by a factor of 6.25 because more electrons are conducting.