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Donor and recipient energy bands in doped semiconductor

  1. Dec 21, 2011 #1
    Are the electrons in the donor and recipient energy bands in doped material involved in conduction of electricity, and why?
     
  2. jcsd
  3. Dec 21, 2011 #2

    berkeman

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    Sounds like a schoolwork question. Please tell us what you know so far about electrons in the conduction band...
     
  4. Dec 23, 2011 #3
    No it's not a class question. I just found my self asking this question and not really getting a compromise. Look here it's clear that not only the conduction band but also the valence band is involved in electric conductivity as it the usual current in conduction band and the hole current in valence band. Now what about the electrons residing in the donor and the recipient bands residing in the doped semiconductors? got it?
     
  5. Dec 26, 2011 #4

    es1

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    I am assuming you mean the donor and acceptor levels.

    You mention it's clear that not only the conduction but also valence band is involved in electric conductivity. So why is it not also clear that the donor and acceptor levels are involved as they reside between the conduction and valence band? It takes even less energy to ionize from (or to) them, which is their point.

    Unless a donor and acceptor band is something else entirely... In which case I'll need you to define it for me.
     
  6. Dec 28, 2011 #5
    @Esi... It's not clear as we all see that they are not included in the calculations. They are actually assumed not to be involved... For the reason am looking for.. Got it?
     
  7. Dec 28, 2011 #6

    dlgoff

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  8. Dec 28, 2011 #7

    es1

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    Which calculations? The free carrier count?
    Some models do include them. See this link for example:
    http://ecee.colorado.edu/~bart/book/book/chapter2/ch2_6.htm#2_6_4

    If your book does not include them then they likely made some simplifying assumptions. Things like the level is small, fully ionized, donor and acceptors are balanced. Or more simply, the levels all contain no carriers and therefore do not contribute to the count.
     
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