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PN junction diode

  1. Apr 9, 2014 #1
    If p-type semiconductor and n-type semiconductor of a diode are equally doped, and if the diode is forward biased, then holes will move toward the n-type semiconductor and electrons will move toward the p-type semiconductor and they will diffuse with each other and the ions become neutral atoms since the hole and electron have disappeared. Then will there be any electron that will go to the positive terminal of the battery if all of them have diffused with each other? I can't understand, please help me!
  2. jcsd
  3. May 9, 2014 #2
    The amount of doping in the P and N junctions determines whether ALL or SOME of them recombine and become neutral. Also, the width of the depletion region gets thinner as forward bias voltage is applied across the diode, and this process too determines whether all or some of the dominant charge carriers in the semiconductor junction actually recombine to neutrality.
  4. May 11, 2014 #3


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    Doping concentrations are tiny compared to the total number of atoms - there are always electrons left.

    It is the other way round, the atoms become ions. Holes are not ions - they are spots where an additional electron can be bound. And free electrons come from atoms where one electron is not bound, so it will easily form an ion.
    Those ions are then the source for the electric fields and therefore the voltage difference, by the way.
  5. Jun 11, 2014 #4
    then it becomes neutral
  6. Jun 12, 2014 #5
    Last edited by a moderator: Sep 25, 2014
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