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Calculation of Conduction Electron Population for Semi Conductor

  1. Feb 8, 2008 #1
    Hi everyone,

    From this website (link: http://hyperphysics.phy-astr.gsu.edu/HBASE/solids/fermi3.html#c1 ), we get the following expression:

    [tex]Population Density of Conduction Electron = \int ^{\infty}_{Egap}N(E)d(E) = \frac{2^{5/2}(m \pi k T)^{3/2} exp (-E gap/2kT)}{h^3}[/tex]

    My questions are based on the derivation given by the website as follow:-

    1. EF = Egap / 2
    Is this a defined property of semiconductors?

    2. Is there an error in the following expression?
    [tex]N(E)d(E) = \frac{8\sqrt{2} \pi m^{3/2}}{h^3} \sqrt{E-Egap}\ e^{-(E-Egap/2)}[/tex]
    I mean originally it has kT in the power of exponential term. Is it mistakenly dropped?

    3. Is "integration by parts" applied on the following expression?
    [tex]N(E)d(E) = \frac{8\sqrt{2} \pi m^{3/2}}{h^3} \sqrt{E-Egap}\ e^{-(E-Egap/2)}[/tex]
    Somehow, I can't solve the integration.

  2. jcsd
  3. Feb 8, 2008 #2
    Formula for density of electrons in conduction band is a good approximation for a classical gas of carriers in intrinsic semiconductors. Only if electron gas is classical you can approximate Fermi - Dirac statistic with Boltzmann one and only if the semiconductor is intrinsic Fermi's level is in the middle of energy gap. In a more general
    way Fermi's level depends on doping level, temperature and effective density of states.
    Regarding point 2 the exponential factor must be dimensionless.

    For more details it's better if you'll see on solid state physics's books. You' ll find clearer derivations than the one you found on the link you posted.

  4. Feb 8, 2008 #3


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    Just to address your first question: in intrinsic semiconductors (no impurities present), the Fermi level is often approximated as lying at the center of the band gap. It doesn't lie exactly at the center at non-zero temperatures, however, because the effective masses of holes and electrons are different. A more precise equation is

    [tex]E_F=E_V+\frac{E_g}{2}+\frac{3}{4}k_B T \ln\frac{m^\star_h}{m^\star_e}[/tex]
  5. Feb 8, 2008 #4
    Thanks Mapes for the explanation, I finally understood the r/s.

    Thanks matteo. I'm gng to hit those books in my lib. :)
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