Doping ex. 2.3, when not to neglect n_D

Physics Q&A
4

It’s a bit trickier.

We have: E_F = E_C - kT \log[N_C / (N_D - N_A)] and n_D = N_D \exp[-(E_D - E_F)/kT] \ll N_D.

Substituting E_F into the second expression gives N_C / (N_D - N_A) \gg \exp[(E_C - E_D)/kT], or N_D - N_A \ll N_C \exp[-(E_C - E_D)/kT].

Then we can take kT \sim 25 meV, which roughly just gives N_D \ll N_C since E_C - E_D < kT.