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Why n*p always equal to ni square? (semiconductor)

  1. Jan 28, 2013 #1
    Why n*p always equal to ni square?? (semiconductor)


    For you guys who studied semiconductor physics must be familiar with the equation:


    I can understand why this is true for the intrinsic case (the broken bonds would always provide electron and hole in pairs )

    But why is this still true for dopped semiconductors? Take Si for example, n=p=ni=1010 in intrinsic case (that's we all know). However, if dope 1015 Nd into the material, then n~1015 and p~105. The highlighted part is my confusion! Why p become smaller? Where do the holes go?

    Thanks for helping!
  2. jcsd
  3. Jan 28, 2013 #2


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    Re: Why n*p always equal to ni square?? (semiconductor)

    Dopant brings in extra electrons that quenches the holes?
  4. Jan 28, 2013 #3
    Re: Why n*p always equal to ni square?? (semiconductor)

    Don't think so. I have also thought of that but if this is true there must be some relevant theories to explain the "dopping and quenching relationship", or, do the scientists consider this is a neglectable question?
  5. Jan 28, 2013 #4
    Re: Why n*p always equal to ni square?? (semiconductor)


    You are correct in being suspicious of that explanation. See the attachment for a mathematical explanation.


    Attached Files:

  6. Jan 28, 2013 #5
    Re: Why n*p always equal to ni square?? (semiconductor)

    Thanks a lot Ratch. Actually I saw a similar maths explanation before, you both use the fact of neutrality to form the equation.
    But could you tell me where do the holes go? Are they being eliminate by extra electrons? This is also the only physical explanation i can imagine. But since Nd>>p, the explanation of eliminating seems ... just not right, cause the eliminating efficiency will be too small to be (felt) right.
    Please enlighten me
  7. Jan 28, 2013 #6
    Re: Why n*p always equal to ni square?? (semiconductor)


    In intrinsic semiconductor material, the electron and hole concentration is caused by thermal recombination and generation (R-G). At room temperature, the balance for silicon is Nd = Na =10^10. If more mobile electrons are added to the semiconductor by doping, then the R-G equilibrium changes, and does not need to generate as many electrons thermally to support the equation np=ni^2. Since thermally generated charge carriers come in pairs, then the hole generation is also reduced. So the holes don't go anywhere, it is just that fewer of them are generated by the R-G thermal process, which has slowed down due to the excess of electrons. That is what those equations mean, and it the basis for their derivation. Ratch
  8. Jan 29, 2013 #7
    Re: Why n*p always equal to ni square?? (semiconductor)

    I think I got what you mean, thanks!
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