For a given semiconductor at a given temperature $T$ with doping levels of acceptors $N_A$ and donors $N_D$ and intrinsic carrier concentration $n_i$ without any external biases, how can I calculate the densities of holes and electrons.

I know that, at equillibrium, $np=n_i^2$, which is one equation. I also know that $n$ and $p$ can be calculated directly with the difference between the fermi level and the intrinsic energy level $E_f - E_i$ with the formulas $n=n_i \exp{(\frac{E_f-E_i}{kT})}$ and $p=n_i \exp{(\frac{E_i-E_f}{kT})}$.

For heavily doped scenarios in which $N_D \gg N_A$ or vice versa, I could calculate $E_f-E_i$ by assuming $n \approx N_D$. However, for situations in which the semiconductor isn't heavily doped or for situations where $N_A$ and $N_D$ are close, how could I calculate the the difference between the fermi energy after doping and the intrinsic fermi energy?

  • $\begingroup$ I'm not sure but I think for heavy doping there's no single formula. The formula you mentioned assumes a Boltzmann distribution which is a good approximation at low doping but for heavy doping effects of Pauli exclusions should be considered $\endgroup$
    – Azad
    May 17, 2015 at 18:18
  • $\begingroup$ The equations are a little complicated, but Sze's Physics of Semiconductor Devices covers this in section 1.4.3 Calculation of the Fermi Level. He shows it explicitly for n or p doping, you would have to generalize for mixed (which looks straightforward but lots of terms). $\endgroup$
    – Jon Custer
    May 18, 2015 at 16:18

1 Answer 1


I can't comment yet so I'll put this in as an answer.

When acceptor and donor dopant concentrations are similar (less than ten times) we simply take the difference as the 'net doping'. This type of semiconductor is called a compensated semiconductor.

We can then approximate thermal equilibrium majority carrier concentration at room temperature to the net doping. You can then use that to calculate the Fermi level with respect to the intrinsic Fermi level.


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