Why does the specific resistance of a semiconductor decrease with an increase in temperature, as an increase in temperature should cause the specific resistance to increase?

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    $\begingroup$ It is better if you do some research. These can constitute a whole chapter or even a book! Semiconductors are poorer conductors than metals but better than insulators. Impurities are mixed with them (called doping) so that the conductivity increases. It depends on the orbital structure and many more...... buy a book on this topic and read, it's very interesting. $\endgroup$ – Wrichik Basu Apr 4 '17 at 16:28
  • $\begingroup$ Why should an increase in temperature cause the resistivity to increase? In metals the simplest explanation is that increasing temperature causes increasing carrier-phonon scattering. That happens in semiconductors as well, but the increase in carrier concentration with temperature successfully competes to result in increasing conductivity. $\endgroup$ – Jon Custer Apr 4 '17 at 17:02
  • $\begingroup$ It's all about the carriers. Furthermore, if you decrease the temperature of a semiconductor too much, it's resistance will increase exponentially. Freezout. Read up. $\endgroup$ – daFireman Jan 25 '18 at 1:55

The electrical resistance in a material is a property determined by electrical charges in motion. So, there are two sources of temperture dependence: the charges can change (basically, become more or less numerous), or the motion can change.

Electrical resistance in metals is almost entirely due to scattering (thermal interaction with the motion of the charge carriers), because the number of charge carriers is just the population of electrons in the conduction band.

So, for conduction in a metal, where charge carriers are constant in number,

increase in temperature should cause the specific resistance to increase

is correct. In the Drude approximation, a metal's resistance rises proportionally to absolute (Kelvin) temperature.

Semiconductors, on the other hand, have no charge carriers except a few that are CREATED by thermal excitation. So, under any conditions where the temperature has a larger effect on charge-carrier-density than 3000 parts per million per degree K, we expect the temperature dependence of resistance to change sign. In semiconductors, it does. In some materials (semimetals, like graphite at room temperature) the two effects just balance; we don't call those materials semiconductors or metals.


The relaxation time t , as well as to some extent the charged carrier density, n, changes with temperature in a semiconductor. The density for a semiconductor at any arbitrary temperature,T can be given as

$n(T) = n'e^{-\frac{E}{k_{B}T}}$

In the above equation E is the energy gap between the upper portion of the valence band and lower portion of conduction band. $k_{B}$ is Boltzmann's constant

Equation suggests that the density of the charged carrier increases with temperature hence conductivity increases so finally resistivity decreases

Why the last part?

Here is the equation I hope you know this

Conductivity = $\dfrac{(n q^{2}t)}{m}$


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