02 Negative Temperature Coefficient

Aim¶

To show how the resistance of a semiconductor (P-Ge) depends on temperature.

Subjects¶

• 5D20 (Resistivity and Temperature)

Diagram¶

Equipment¶

• Bar of P-Ge

• Power supply

• Current meter with large display

• Gas flame

Presentation¶

Set the Ammeter at a $1\mathrm{~A}$-scale. The voltage of the power supply is raised until a current of about $0.05\mathrm{~A}$ flows in the circuit. The bar of $\mathrm{P}-\mathrm{Ge}$ is heated by the gas flame and soon the current rises to a much higher value. After a short time of heating the gas flame can be removed and the current continues to rise, faster and faster, only limited by the power supply.

Explanation¶

The resistance of a semiconductor drops with temperature because at a higher temperature there are more free charge-carriers in it.

The current flowing in the material heats it up: $P_{e l}=\frac{V^{2}}{R}$ The heat leaving the piece of material is proportional to $\Delta T: P_{\text {out }} \propto \Delta T$ (Newton cooling). When $P_{\text {out }}=P_{e l}$ there will be thermal equilibrium and the temperature is constant. Reaching such an equilibrium takes some time.

In this demonstration $R$ lowers due to a rise in temperature and so $P_{e l}$, rises due to a rise in temperature. When this rise is faster than the rise of $P_{\text {out }}$ an ever faster rising of $\Delta T$ (like an avalanche) will result.