Zero doping calculation in amset

Dear @alex

I want to calculate the transport properties of the pristine system. In this regard, I would sincerely request you to let me know whether zero doping calculation can be performed using amset.
Please let me know.

Thanks and regards
Anustup Mukherjee

Dear Anustup,

If you have no doping then there will be no free carriers to contribute to transport so, so there will be no transport (unless your system is a metal or very small band gap semiconductor where thermal excitations can result in free carriers).

Using a doping of ~1e13 should give you the intrinsic transport properties. You can also turn off IMP scattering to remove the effects of impurity scattering.

Best,
Alex

Dear @alex

Thank you for the information.

If I correctly understood, the doping tag is for the concentration of the charge carriers which are already present in the pristine system, right ?

Now if that’s the case, how should once proceed to perform actual doping calculations, where the system is extrinsically doped?

Please let me know.

Thanks and regards
Anustup Mukherjee

Dear @alex

Please clear the above doubt for me. Thank you.

Thanks and regards
Anustup Mukherjee

There will never be a completely pristine system. Thermodynamics dictates that there will always be some defects present.

The doping tag sets the number of free carriers. It doesn’t matter whether these come from defects or thermally excited carriers. You set a carrier concentration and the Fermi level position that produces this carrier concentration (e.g., doping and temperature dependent) is solved.

The doping tag also indirectly controls the number of ionised impurities for IMP scattering. See this equation here: Scattering Rates - AMSET Documentation

This is only a rough model for impurity scattering assuming some defect compensation but it is hard to do much better without actually calculating the defect concentrations from first-principles defects calculations.

Dear @alex

Thank you for the information. It was really helpful.