Question

On increasing temperature of a semiconductor, what will be effect on its conductivity?

Hint:

Use \( \sigma = n e \mu \): heating breaks covalent bonds and creates many more electron-hole pairs, so the carrier number wins over the drop in mobility.

Answer 1

Step 1 (Concept): The conductivity of a semiconductor is given by \( \sigma = n e \mu \), where \( n \) is the number of free charge carriers (electrons and holes) per unit volume, \( e \) is the electronic charge and \( \mu \) is the mobility. In a semiconductor, \( n \) is very sensitive to temperature.

Step 2 (Effect of heating): At low temperature almost all the valence electrons are locked in covalent bonds, so very few carriers are free. On raising the temperature, thermal energy breaks a large number of covalent bonds and creates many extra electron-hole pairs. Hence \( n \) increases rapidly (roughly exponentially) with temperature.

Step 3 (Net result): The sharp rise in the carrier number \( n \) dominates over the small fall in mobility \( \mu \). Therefore \( \sigma = n e \mu \) increases, i.e. the resistance falls. A semiconductor thus has a negative temperature coefficient of resistance.

\[\boxed{\text{Conductivity increases (resistance decreases) as temperature rises.}}\]