15.4.2 – Thermistor | Physics with Mr Shone

A thermistor is a thermal resistor whose resistance changes (either increases or decreases) with temperature.

Thermistor Symbol

The I-V Characteristic graph for a thermistor looks like this:

I-V graph for a thermistor

At higher voltages, the resistance of the thermistor decreases.

This can be explained by the larger currents raising the temperature and thus creating more free electrons in the thermistor.

Negative Temperature Coefficient (NTC) thermistors are more commonly used than Positive Temperature Coefficient (PTC) thermistors. NTC thermistors have a resistance that decreases as temperature increases, while PTC thermistors have a resistance that increases as temperature increases.

Note
Some thermistors show an increase in resistance with temperature (ie PTC thermistors). The I-V graph for these will obviously curve in the opposite direction to the one shown above. You will need to read the question carefully to be sure which type of thermistor you are dealing with.

Note

Some thermistors show an increase in resistance with temperature (ie PTC thermistors). The I-V graph for these will obviously curve in the opposite direction to the one shown above.

You will need to read the question carefully to be sure which type of thermistor you are dealing with.

Example 1
The circuit shows a lamp L and a thermistor connected in series. This thermistor has a negative thermal coefficient. Does the lamp L get brighter or dimmer if the thermistor is heated? Explain. (Answering in terms of Current) • As the thermistor is heated, the resistance of thermistor decreases. • This will cause the effective resistance in the circuit to decrease. • The current increases, hence, the lamp gets brighter. OR (Answering in terms of Potential Difference) • As the thermistor is heated, the resistance of thermistor decreases. • Applying the potential divider concept, p.d. across the thermistor will decrease. • Since sum of p.d. of thermistor and p.d. across lamp is the e.m.f. of source, p.d. across lamp will increase, so lamp will become brighter.

Example 1

The circuit shows a lamp L and a thermistor connected in series. This thermistor has a negative thermal coefficient.

Does the lamp L get brighter or dimmer if the thermistor is heated? Explain.

(Answering in terms of Current)

• As the thermistor is heated, the resistance of thermistor decreases.
• This will cause the effective resistance in the circuit to decrease.
• The current increases, hence, the lamp gets brighter.

(Answering in terms of Potential Difference)

• As the thermistor is heated, the resistance of thermistor decreases.
• Applying the potential divider concept, p.d. across the thermistor will decrease.
• Since sum of p.d. of thermistor and p.d. across lamp is the e.m.f. of source, p.d. across lamp will increase, so lamp will become brighter.

Example 2
A thermistor is connected to a d.c. supply. The voltage V across and the current I through the thermistor are measured as the voltage V is increased. At each stage, the temperature of the thermistor is allowed to reach a steady value before the meter readings are recorded. The graph below shows the resulting variation of current I with voltage V. (a) Determine the resistance of the thermistor when the p.d. across it is 1.3 V. R = V/I = 1.3/0.160 = 8.1 Ω The resistance of the thermistor is given by the following chart: (b) State the temperature that corresponds to the value calculated in (a). Reading from the graph, a resistance of 8.1 Ω corresponds to a temperature of 39 ºC.

Example 2

A thermistor is connected to a d.c. supply. The voltage V across and the current I through the thermistor are measured as the voltage V is increased. At each stage, the temperature of the thermistor is allowed to reach a steady value before the meter readings are recorded. The graph below shows the resulting variation of current I with voltage V.

(a) Determine the resistance of the thermistor when the p.d. across it is 1.3 V.

R = V/I = 1.3/0.160 = 8.1 Ω

The resistance of the thermistor is given by the following chart:

(b) State the temperature that corresponds to the value calculated in (a).

Reading from the graph, a resistance of 8.1 Ω corresponds to a temperature of 39 ºC.

Example 1
In the circuit shown, a thermistor is connected in series with a 5.0 kΩ fixed resistor. When the temperature is high, the resistance of the thermistor is 1.0 kΩ and when the temperature is low, the resistance of the thermistor increases to 8.0 kΩ. (a) Determine the potential difference across the thermistor measured by the voltmeter, when the temperature is high. High temperature implies that the resistance of the thermistor will be 1.0 kΩ. $\frac{V_{thermistor}}{e.m.f.} = \frac{R_{thermistor}}{R_{effective}}$ $\frac{V_{thermistor}}{6.0} = \frac{1.0}{6.0}$ $V_0 = 1.0 V$ (b) Determine the potential difference across the thermistor measured by the voltmeter, when the temperature is low. High temperature implies that the resistance of the thermistor will be 8.0 kΩ. $\frac{V_{thermistor}}{e.m.f.} = \frac{R_{thermistor}}{R_{effective}}$ $\frac{V_{thermistor}}{6.0} = \frac{8.0}{13.0}$ $V_0 = 3.7 V$

Example 1

In the circuit shown, a thermistor is connected in series with a 5.0 kΩ fixed resistor.

When the temperature is high, the resistance of the thermistor is 1.0 kΩ and when the temperature is low, the resistance of the thermistor increases to 8.0 kΩ.

(a) Determine the potential difference across the thermistor measured by the voltmeter, when the temperature is high.

High temperature implies that the resistance of the thermistor will be 1.0 kΩ.

$\frac{V_{thermistor}}{e.m.f.} = \frac{R_{thermistor}}{R_{effective}}$

$\frac{V_{thermistor}}{6.0} = \frac{1.0}{6.0}$

$V_0 = 1.0 V$

(b) Determine the potential difference across the thermistor measured by the voltmeter, when the temperature is low.

High temperature implies that the resistance of the thermistor will be 8.0 kΩ.

$\frac{V_{thermistor}}{e.m.f.} = \frac{R_{thermistor}}{R_{effective}}$

$\frac{V_{thermistor}}{6.0} = \frac{8.0}{13.0}$

$V_0 = 3.7 V$

Example2
The circuit below shows a thermistor connected in series with a 100 Ω resistor to a 6.0 V battery with negligible internal resistance. The graph shows the variation of resistance R with temperature T of the thermistor. Determine the difference between the minimum and maximum voltages across the thermistor as its temperature T increases from ice point (0 ºC) to steam point (100 ºC). At the ice point: Resistance of the thermistor, R_ice = 600 Ω Total resistance of the circuit is therefore 100 + 600 = 700 Ω potential difference across the thermistor, V_ice, is given by: $\frac{V_{ice}}{e.m.f.} = \frac{R_{ice}}{R_e}$ $\frac{V_{ice}}{6.0} = \frac{600}{700}$ $V_{ice} = 5.1 V$ At the steam point: Resistance of the thermistor, R_steam = 100 Ω Total resistance of the circuit is therefore 100 + 100 = 200 Ω potential difference across the thermistor, V_steam, is given by: $\frac{V_{steam}}{e.m.f.} = \frac{R_{steam}}{R_e}$ $\frac{V_{steam}}{6.0} = \frac{100}{200}$ $V_{steam} = 3.0 V$ Thus the difference between the maximum and minimum voltages is 5.1-3.0 = 2.1 V

Example2

The circuit below shows a thermistor connected in series with a 100 Ω resistor to a 6.0 V battery with negligible internal resistance.

The graph shows the variation of resistance R with temperature T of the thermistor.

Determine the difference between the minimum and maximum voltages across the thermistor as its temperature T increases from ice point (0 ºC) to steam point (100 ºC).

At the ice point:

Resistance of the thermistor, R_ice = 600 Ω

Total resistance of the circuit is therefore 100 + 600 = 700 Ω

potential difference across the thermistor, V_ice, is given by:

$\frac{V_{ice}}{e.m.f.} = \frac{R_{ice}}{R_e}$