It is said that P-Type semiconductors' majority carriers are holes. But based on my current understanding, there should be no extra holes except for those generated by heat (i.e. electron-hole pairs like in intrinsic semiconductors).

Lets assume we have a P-Type semiconductor – Silicon doped with Boron (with an Boron atom in the middle in the image below).

enter image description here

Please correct my wrong assumptions:

  • For one chemical bond to take place between 2 atoms we need a pair of electrons (each atom "offers" 1 electron).
  • A Boron atom pairs with exactly 3 Silicon atoms. There is one surrounding Silicon atom (the one under the Boron atom in the image above) whose electron doesn't participate in any chemical bonds.
  • Both the Boron atom and the Silicon atom under it are electrically negative to the outside and won't accept any other valence electrons.

Where does the majority of holes come from then? Based on my understanding, the P-Type semiconductor should have electrons as the majority carriers (because one electron of the Silicon atom (in the image above) does not bond with the Boron atom) instead of majority of holes and thus should behave like an N-Type semiconductor.


1 Answer 1


No, Boron still bonds to 4 Si atoms. Since Boron has only 3 valence electrons, it can only form three covalent bonds with Si. This leaves one of the four silicon atoms with an unsatisfied bond. To compensate, a nearby electron (from an Si-Si bond) will "jump" to complete the B-Si bond leaving the Si-Si bond with a net positive charge (the hole). To fill that broken bond, another nearby electron will jump to fill in the vacancy etc... Thus the "hole" is highly mobile and can conduct when a voltage is applied. The Boron atom does become negatively charged, but that additional negative charge cannot conduct because it takes place in the bonding.

  • $\begingroup$ Thank you for answer. You wrote that Boron can form 3 complete bonds – do you know how is the 4th one – the "incomplete" – bond called (I guess it isn't covalent)? $\endgroup$
    – sarasvati
    Commented Jan 11, 2016 at 0:12
  • $\begingroup$ Yes, see my edited post $\endgroup$ Commented Jan 11, 2016 at 1:32

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