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Solid State: Fermi surface

  1. Oct 28, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi all

    I can't seem to figure out, why semicodunctors do not have Fermi surfaces. At T=0K, there are no electrons in the conduction band, and thus there is no Fermi surface - all OK here.

    But at T > 0K, there are electrons in the conduction band. Why is it then that it is said that "semiconductors do not have Fermi surfaces"?
     
  2. jcsd
  3. Oct 28, 2009 #2
    If the Fermi energy level falls in the bandgap (where no electron states can exist) of a semiconductor, there is no Fermi surface.
     
  4. Oct 28, 2009 #3
    But the Fermi energy is the energy of the highest occupied state - so if we have T>0 K, there will be thermal excitation up to the conduction band. This implies that the Fermi energy lies in the conduction band?
     
  5. Oct 28, 2009 #4
    I'm sorry, I wrote Fermi energy level but it should be just Fermi level which is not the same thing.

    Also,the Fermi energy is the energy of the highest occupied state at T=0K.
     
  6. Oct 28, 2009 #5
    Oh, you mean the chemical potential. But does it make a difference that it lies in the energy gap? I cannot see the connection between the chemical potential and the Fermi surface. And thank you for helping me; I really appreciate it.
     
  7. Oct 28, 2009 #6
    Not quite, the chemical potential is just [tex]\mu[/tex]. The Fermi level comes from the Fermi-Dirac distribution:

    [tex]
    f(E)=\frac{1}{1+\exp\left[\frac{E-\mu}{k_BT}\right]}
    [/tex]

    In relation to the conduction band, this is expressed as

    [tex]
    f(E)=\frac{1}{1+\exp\left[\frac{K-\zeta}{k_BT}\right]}
    [/tex]

    where [tex]K[/tex] is the difference between the Fermi and conduction energies and [tex]\zeta[/tex] is the Fermi level:

    [tex]
    \zeta=\zeta_0\left[1-\frac{\pi^2}{12}\left(\frac{k_BT}{\zeta_0}\right)^2-\frac{\pi^4}{80}\left(\frac{k_BT}{\zeta_0}\right)^4+\cdots\right]
    [/tex]

    with [tex]\zeta_0[/tex] being the Fermi energy at 0K. So, depending on the temperature, the Fermi level can/will be in the bandgap, hence saying that semiconductors do not have Fermi surfaces.
     
  8. Oct 28, 2009 #7
    According to http://en.wikipedia.org/wiki/Fermi_level, then [itex]\zeta[/itex] is the difference between the Fermi energy and the conduction band energy. Is this wrong?

    I haven't heard of the "Fermi level" before, and it is not intuitive for me, unfortunately. Is there a physical explanation of why semiconductors do not have Fermi surfaces (apart from the fact that their Fermi level is in the band gap), just like there are explanations of why metals have Fermi surfaces (i.e. they have free electrons)?
     
  9. Oct 28, 2009 #8
     
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