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A hole is the lack of an electron.

Thus, if we consider for example a N doped semiconductor. The so-called "majority charge carriers" are the electrons. But whenever there is a mobile electron that moves from a site to another site, the former site will have a hole, thus I don't see why there would not be exactly the same number of holes as the number of electrons.

Thus, I don't catch why the electrons would be really "majoritary", for example in a N semiconductor.

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    $\begingroup$ I'm not sure that this is the explanation. Let's see other comments. $\endgroup$
    – Mathieu Krisztian
    Nov 3, 2019 at 9:42
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    $\begingroup$ I am with you Mathieu. I won't buy that cheap argument, because electrons aren't particles but quasiparticles in solids, too. $\endgroup$
    – untreated_paramediensis_karnik
    Nov 3, 2019 at 12:21
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    $\begingroup$ If a semiconductor is doped n-type (for example), then there are (mobile) electrons in the conduction band even if the valence band is full (no holes) so I don't understand your argument. $\endgroup$
    – Alfred Centauri
    Nov 3, 2019 at 14:00
  • $\begingroup$ Thank you Alfred. I believe that your explanation is the best one. Thank you. $\endgroup$
    – Mathieu Krisztian
    Nov 3, 2019 at 15:51
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    $\begingroup$ @Alfred thank you. I was not aware of that. $\endgroup$
    – Mathieu Krisztian
    Nov 6, 2019 at 17:55

1 Answer 1

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Your argument only holds for undoped (intrinsic) semiconductors. Doping introduces one type of carriers and creates a strong asymmetry in the concentration of both carrier types.

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  • $\begingroup$ All right thank you. Actually I feel that the best explanation was the one from Alfred Centauri. $\endgroup$
    – Mathieu Krisztian
    Nov 3, 2019 at 15:51
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    $\begingroup$ Really? They are they same except that I pointed out your error in more detail. $\endgroup$
    – my2cts
    Nov 3, 2019 at 15:56
  • $\begingroup$ ok, actually, Alfred gives an extra information, which is useful. Maybe you may merge his information in your answer, for people that may ask themselves the same question. Thank you. $\endgroup$
    – Mathieu Krisztian
    Nov 3, 2019 at 15:57
  • $\begingroup$ Which extra information in which answer ? I only see a comment. $\endgroup$
    – my2cts
    Nov 3, 2019 at 16:00
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    $\begingroup$ I cannot follow your reasoning. You should read up on semiconductors. $\endgroup$
    – my2cts
    Nov 3, 2019 at 16:51

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