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Semiconductor physics

  1. Apr 10, 2010 #1
    1. The problem statement, all variables and given/known data

    I am asked to show that the conduction electron density in an n type semiconductor is given by:

    [tex]n=\frac{1}{2} \left( -n_0 e^{-(E_c-E_d)/kT} + \sqrt{n_0^2 e^{-2(E_c-E_d)/kT}+4 n_0 N_d e^{-(E_c-E_d)/kT} } \right)[/tex]

    where
    [tex]n_0=2(m^* kT/2 \pi \hbar^2 )^{3/2}[/tex]


    For low temp [itex]p=0[/itex] and [itex] E_c-\mu >> kT[/itex]


    2. Relevant equations



    3. The attempt at a solution

    Start with electrical neutrality:
    [tex]\Delta n=n_e-n_h=n_d^+ - n_a^-[/tex]
    Using the mass action law:
    [tex]n_h=n_i^2/n_e[/tex]
    Thus we get:
    [tex]n_e^2-n_e\Delta n=n_i^2[/tex]
    solving the quadratic:
    [tex]n_e=\frac{1}{2}\left( \Delta n+ \sqrt{(\Delta n)^2 +4n_i^2} \right) [/tex]

    From this, I gather that:

    [tex] \Delta n=n_e-n_h=n_d^+ - n_a^- =n_0 e^{-(E_c-E_d)/kT}[/tex]
    [tex]n_i =\sqrt{n_0 N_d e^{-(E_c-E_d)/kT}}[/tex]

    But I've been thinking about this for a while now and I dont see how to derive this. Any help is appreciated.

    I'm also concerned about my understanding for what exactly is the chemical potential in the Fermi distribution and why does it physically mean to say [itex]e^{-(E_c-E_d)/kT}[/itex] or [itex]e^{-(E_c-\mu)/kT}[/itex]

    My understanding is that the chemical potential is the fermi energy, which is basically the highest energy level at T=0. However, how could the fermi energy/chemical potential change w.r.t. T if it is defined w.r.t. T=0?

    Thank you for the help!
     
  2. jcsd
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