Is it true that for an even number of donors and acceptors every extra electron given by the donors would fill a hole provided by the acceptors and the net extra charge carrier density by impurities would vanish?

If so, then there should be no more holes at all in a $n$-doped semiconductor, no? But I saw in derivations that there still considered $n+p$ to be the total carrier density.

  • $\begingroup$ I don't know what you saw, and your question is a bit vague, but note that impurity atoms of either kind will be ionized so that there will always be both positive and negative carriers present. $\endgroup$
    – garyp
    Feb 11, 2017 at 22:29

1 Answer 1


Yes it is possible, and in fact common, to dope a semiconductor with both acceptors and donors. When this is done it's called (partially) compensated doping.

The effect is that the carriers produced by the more concentrated dopant are reduced by the number of available impurities of the other type.

For example, it's common to form a p-n junction by applying a moderate dopant concentration (say 1015 cm-3) of one type, say donors, in such a way that they diffuse deeply into the bulk material. And then to apply a much stronger concentration (say 1017 cm-3) of the other type (acceptors), but only allowing them to diffuse shallowly.

Then the surface region has both acceptors and donors, but because the acceptors are at higher concentration, the net effect is to produce p-type material, with effective concentration $N_A-N_D$. Since $N_A \gg N_D$, this is effectively just $N_A$.

You can also see compensation being used in this diagram of a typical npn BJT structure, where the emitter is formed by high-concentration donor doping in a region previously (or subsequently) doped with acceptors to form the base region:

enter image description here

If so, then there should be no more holes at all in a n-doped semiconductor, no?

If the concentration of the two dopants is equal, the number of carriers will be reduced, but they won't be reduced below the intrinsic carrier concentration $n_i$, because the Law of Mass Action tells us

$$np = n_i^2$$

  • $\begingroup$ Could you elaborate on the last sentence a little more? I have read about the Law of Mass Action but I still don't understand it from an intuitive point of view. Why should the electrons not eat up ALL the holes? Does is have to do with statistics and rates? $\endgroup$
    – Marsl
    Feb 12, 2017 at 11:26
  • $\begingroup$ Yes, basically statistics. The hole concentration is governed by the location of the Fermi level. The Fermi level will never be higher than just slightly above the conduction band edge. Even at this level there will still be a small concentration of holes (assuming temperature above absolute zero). $\endgroup$
    – The Photon
    Feb 12, 2017 at 18:18

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