So for n-type Si with donor density $N_d$ and donor energy level $E_d$, $N_d^+ = N_d(1+\frac{1}{1+e^{\beta (E_d - E_f)}/2})$ is the number of ionized donors, so we get an relation between the number of electrons in the conduction band and holes in the valence band, $n = N_d^+ + p$, and you can plug in the terms for $p$ and $n$ from here, which gives you a big equation.

Now, my textbook says that for low to moderate concentration doping with shallow donor impurities, an approximation makes this equation simplify to $E_f = E_c - kTln(N_c/N_d)$, but otherwise you'll have to solve it numerically to get $E_f$.

Taking $E_c = 0$ because it's the bottom of the conduction band and $T=300K$ and $N_c \approx 10^{19}/cm^3$ for Si, a doping concentration of $N_d = 10^{15}$ gives $E_f = -265meV$. However, for $N_d = 10^{20}$, this equation gives me $E_f = +32meV$, which says that the Fermi level is above the conduction band minimum.

This strikes me intuitively as wrong for some reason, but I'm not actually sure why it couldn't be above it.

I checked to see if the approximation you need to use to get that simplified equation for $E_f$ is valid for $N_d = 10^{20}$ and it indeed fails. Solving numerically for it (i.e., not using the approximation) gives $E_f = -17meV$, which is below the conduction band (but above the donor energy level). Solving numerically for $N_d = 10^{15}$ (where the approximation is very accurate) gives the same exact answer as for the approximation, so I know that at least works.

My question is, can the Fermi level ever be above the conduction band minimum (in a semiconductor like Si, barring any really weird things)? And relatedly, if my assumption of what's going on here is right, is there any meaning to the fact that for $N_d = 10^{20}$, $E_f$ is above the impurity energy level?

edit: To expand on this a little, trying $N_d = 10^{21}$ gives $E_f = 13.5meV$, so at least according to the non-approximated equation, it can go above the Fermi level. However, a) I'm still not sure my solver is correct, and b) I'm pretty sure $10^{20}$ is already a really high doping level, and I've never really heard of $10^{21}$.

Alright, I've got my answer, had I read the next paragraph of my book. It says that if $N_d$ exceeds $N_c$, it is "degenerately doped" and $E_f$ can go above $E_c$.


1 Answer 1


Your first equation is using the effective density of states approximation which is valid when the Fermi level is far from the band. As the Fermi level gets closer to the band this is not accurate and the full equation needs to be solved. This is the reason for the difference you observed.

But to answer your specific question, "can the Fermi level be higher than the band", the answer is a big yes. Extreme doping is used in many semiconductor devices, and this can also happen by injecting current into the devices such as laser diode.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.