In a p-type semiconductor there are always free electrons due to thermal generation allowing for conduction. But If such an electron falls to a hole, it can no longer conduct.

So shouldn't more holes slow down conduction? If intrinsic semiconductors have many more electrons in the conductive band at any one point in time, I don't understand why a p-type semiconductor should be more conductive.

(My humble guess is that, when there are less holes, an electron jumping to the conductive band due to thermal generation is far more likely to jump back to the same place, because the other holes are too far away)

  • $\begingroup$ The intrinsic carrier concentrations in Si is order of $10^{10}$ per cc. Doping allows attaining $10^{16}$ or more of the majority carrier, with the minority carrier then being only $10^{4}$ or so. So by doping you have a million times more majority carriers than for intrinsic Si. $\endgroup$
    – Jon Custer
    Commented May 5, 2023 at 16:46
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    $\begingroup$ Valence band electrons (or lack thereof, i.e. holes) also contribute to current. I would suggest taking a look at this question and the answers to it. $\endgroup$
    – Puk
    Commented May 5, 2023 at 18:36

1 Answer 1


You seem to be assuming that the holes are immobile. But the holes can move: they behave like positively charged particles. They can carry a current.

  • $\begingroup$ Well they are in the valence band. So I'm assuming they can move only within the atom. Is that enough to heavily impact the conductivity? As pointed out by Jon Custer, a p-type semiconductor has about $10^{16}$ holes whereas an intrinsic semiconductor has about $10^{10}$. The difference is huge but the hole can only move about one space in the lattice and immediately gets stuck again. $\endgroup$ Commented May 6, 2023 at 18:30
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    $\begingroup$ @LukasLejdar Holes usually don't get "stuck". They behave as mobile charges, similar to electrons. Both holes and electrons can get trapped at lattice defects, but that has little effect on the bulk conductivity of undepleted, doped material. $\endgroup$
    – John Doty
    Commented May 6, 2023 at 20:09

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