When a material is doped with a donor density of $N_d$, I can easily calculate the concentration of electrons and holes in the following way:

$$n\approx N_d\\p=\frac{n_i^2}{n}=\frac{n_i^2}{N_d}$$

Similarly, when a material is doped with an acceptor density of $N_a$: $$p\approx N_a\\n=\frac{n_i^2}{p}=\frac{n_i^2}{N_a}$$

But what happens when I dope a material with both donors and acceptors? Do the donor electrons recombine with the acceptor holes? This would result in:$$n\approx N_d-N_a\space \space , N_d>N_a$$ $$p\approx N_a-N_d\space \space , N_a>N_d$$

Is this the case?

  • 1
    $\begingroup$ You're correct, there will be recombination of electrons from donors and holes of the acceptors. Your formulae also seem correct. The most abundant type of dopants will determine whether it is p- or n-type $\endgroup$
    – Simon
    Jan 18, 2020 at 13:02

1 Answer 1


You are right. Let me write down its derivation for the sake of completion.

Denote the carrier concentration in the intrinsic case as $n_i \equiv n(T) = p(T)$ at a given temperature $T$. The law of mass action holds regardless of presence of impurities prompting us \begin{equation} n \cdot p = n^2_i \end{equation}

What we also know is the relative difference of carriers. $$n - p = N_d - N_a$$ Substitute for $n \text{ or } p$ to arrive at $$n = \frac{(N_d - N_a) + [(N_d - N_a)^2 + 4n_i^2]^{1/2}}{2} \\ p = \frac{(N_d - N_a) - [(N_d - N_a)^2 + 4n_i^2]^{1/2}}{2} $$

In predominantly extrinsic regime with $n_i \ll |N_d -N_a|$ we have,

Case 1: $N_d > N_a$ $$n = N_d - N_a \ \ \quad p = \frac{n_i^2}{N_d-N_a}$$

Case 2: $N_a > N_d$ $$ n = \frac{n_i^2}{N_a-N_d} \ \ \quad p = N_a - N_d $$


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