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Semiconductor Physics : Charge carrier concentration change on doping

  1. Oct 21, 2012 #1
    Hi,

    I have a question regarding the change in charge carrier concentration change.
    For a given semiconductor, say Silicon, when it is not doped,
    it is easy to understand that [tex]{n_0} \times {p_0} = n_i^2[/tex],
    however, on doping with donors to form a n-type semiconductor,
    we have
    [tex]{n_0} \approx {N_D} > > {p_0}[/tex],
    the concentration of free electrons in conduction band increases.

    The question is why does the equation
    [tex]{n_0} \times {p_0} = n_i^2[/tex]
    still hold?

    Afaik, n0's increase is due to the donated electrons from donor, there shouldn't much to do with the holes in valence band, i.e. the free holes concentration (p0) in valence band should remain unchanged on the above doping.

    While if [tex]{n_0} \times {p_0} = n_i^2[/tex] still holds, it actually implies an decrease in p0).

    Probably something I have been missing, would you mind sharing me with your idea?biggrin:
    Many thanks.
     
  2. jcsd
  3. Oct 21, 2012 #2
    The number of holes in the valence band does decrease, yes. You may represent it in a number of ways.

    For instance, the abundance of electrons fills the remaining available states in the valence band.

    Or the Fermi level raises, nicreasing the filling probability in the valence band.

    Or as more electrons fly around, a hole has a shorter life expectancy before it recombines. At the same pair production speed, it means fewer holes.
     
  4. Oct 21, 2012 #3
    Thanks a lot for the kind reply!
    The reasons are very convincing to me.:tongue:
     
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