SSP-EXAM Discussion

"The highest the dispersion in a band the highest is the conductivity"

"The highest the dispersion in a band the highest is the conductivity"

Timothee Jamin -
Atsakymų skaičius: 1

Hello,

After doing the exam and looking at the correction, I was uncertain about one of my answer because I don't know if we can qualify it as a "physical interpretation".


Because we know that the conductivity is proportional to the inverse of the mass of a conductor and the inverse of the effective mass for a semiconductor.

Then we can say that the conductivity is proportional to:

\( \sigma_{semiconductor} \propto \frac{1}{\hbar^{2}} \frac{d^{2}E_{n}(\vec{k})}{d\vec{k}^{2}} \)

And,

\( \sigma_{conductor} \propto E_{\vec{k}} \)

So, for the semiconductor, higher is the curvature, higher is the dispersion, higher is the conductivity. And, for a conductor, higher is the dispersion relation, higher is the conductivity.


Thank you in advance for your reply.
Atsakymas į Timothee Jamin

Re: "The highest the dispersion in a band the highest is the conductivity"

Cedric Crespos -
Your reasoning is correct for the case of semiconductor since effective mass of charge carriers m* is indeed inversely proportional to the curvature of the band. Thus, the higher is the dispersion, the higher is the curvature and the lowest is the mass so the highest is the mobility of charge carriers and the higher is the conductivity.

For metals the charge carriers are electrons characterized by a real mass of electron and does not depend on the dispersion. What is depending on the dispersion if the current density calculated as the integral or the sum of the first derivatives of Ek with respect to k. So the higher is the dispersion the highest are the values of the dEk/dk in the Brillouin zone and the highest is the current density.

CC