Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How a fermi level can lies between valence and conduvtion band?

  1. Jul 11, 2012 #1
    As we know, in intrinsic semiconductors the fermi level is halfway between the valence and conduction band.
    and
    The fermi level is the level in which the most energetic electron settles at T=0.
    but my question:
    How can the fermi level be in a forbidden region in intrinsic semiconductors? or How can most energetic electron be in a forbidden region?
     
  2. jcsd
  3. Jul 11, 2012 #2

    DrDu

    User Avatar
    Science Advisor

    What is somewhere in the band gap (usually not in the middle) is the Fermi energy (and not level) which is the limit of the chemical potential as T->0. So there is no paradox.
     
  4. Jul 11, 2012 #3
    The Fermi level is not defined as the level in which the most energetic electron settles at any temperature. The most energetic carrier in an electronic level at T=0 settles to the Fermi level when that band is occupied by at least one electron. However, that is a special case. In the limit of an intrinsic semiconductor at T=0, the conduction band is empty.
    When there is a high density of carriers in a free carrier band, then your statement is valid. When the density of carriers is low or the temperature very high, then your statement is not true. If the density of free carriers is low, the Fermi level for those carriers is going to be in the forbidden region of the energy spectrum.
    The distribution of electrons in a semiconductor at thermal equilibrium is determined by fermion statistics. The density of electrons is the product of the density of states and the Fermi-Dirac function.
    f((E,T,E_F)=[1+exp({E-E_F}/kT)]
    where f is the Fermi-Dirac function, E is the energy of the electron, k is Boltzmann's constant, T is the temperature and E_F is the Fermi energy. This is given in:
    Jacques I. Pankove, "Optical Processes in Semiconductors" (Dover, 1971) page 7.
    Although not stated, the density of states also determines the Fermi energy. Your definition is only valid for a single band with a parabolic density of states.
    Your statement about the Fermi level is true for free carriers in a metal at T=0. For free carriers in a metal, the free carriers are governed by a single band with a parabolic dispersion. For instance, copper and silver have a large density of conduction electrons and no holes. Only the conduction band is important in copper and silver. Aluminum and beryllium have a large density of holes and no conduction electrons. Only the valence band is important in aluminum or beryllium.
    However, semiconductors contain two free carrier bands: conduction and valence. Therefore, the density of states function is a bit complex compared to the density of states function in a semiconductor.
    A graphical way to calculate the Fermi level given a carrier concentration is shown in:
    Jacques I. Pankove, "Optical Processes in Semiconductors" (Dover, 1971) pages 414-415. You will notice that when T>0, the Fermi level can be negative. In fact, Pankove says on the bottom of page 414:
    "If η is negative, ε lies outside the parabolic band (i.e., inside the energy gap."
    If you are interested in either intrinsic or compensated semiconductors, then I recommend "Optical Processes in Semiconductors".
    Remember that the Fermi level is just a statistical parameter. There doesn't have to be any electron at the Fermi level. Also, illumination complicates this picture.
     
  5. Sep 23, 2012 #4
    HI
    Thanks for your replies, However I saw some other interesting justification and prefer to put them here. They are taken from
    http://askville.amazon.com/Fermi-En...py-Absolute/AnswerViewer.do?requestId=4906977

    1) The source of this confusion is probably because there are a lot of notational inconsistencies in the literature. It's also because some discussions are talking about metals and some are talking about semiconductors. In the case of metals, the Fermi energy is in the valence band, so it's simple to say "the Fermi energy level is defined by the highest energy level that a fermion can occupy at absolute zero." The Fermi energy level is actually just defined as the value of the chemical potential at zero temperature in the context of Fermi-Dirac Statistics.

    f(E) = 1 / (exp((E - u)/t) + 1)

    f(E) is the thermal average number of particles in an orbital of energy E.
    u is the chemical potential, which is temperature dependent in general
    t is temperature

    The Fermi energy level is u(t = 0) = E_f

    In the case of semiconductors, there's a bandgap between the conduction and valence bands, and this Fermi energy level happens to lie in the bandgap. At absolute zero for the semiconductor, none of the orbitals in the valence band have energy equal to the Fermi energy level, so they're all filled up just with lower energy.


    2) If you solve the fermi-dirac equation (you have above) for E=Ef, you get one half; this means there is some probability that there exists a particle at the fermi energy (which is obviously impossible in a semiconductor).
    But when you solve to get the total concentration of particles at E=Ef you have to integrate
    [f(Ef)D(Ef) dE] with D(Ef)-density of states at the fermi level, which is obviously zero, since no particle can exist in the band gap. So really the Fermi energy level is just a reference level.
     
    Last edited by a moderator: May 6, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How a fermi level can lies between valence and conduvtion band?
Loading...