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About fermi levels in extrensic semiconductors.

  1. Oct 12, 2012 #1
    Why does the Fermi level level drop with increase in temperature for a n type semiconductor.? What's the basic idea behind Fermi level?
    How does it get affected by temprature?
  2. jcsd
  3. Oct 12, 2012 #2


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    In an n-type semiconductor at absolute zero, all levels up to the Fermi level are filled, above all are empty. Hence it must be just above the upper band gap. At finite temperature, the Fermi energy is a parameter of the Fermi distribution. It has to be chosen so that the total density of electrons is equal to the nuclear charge density. When you rise the temperature starting from zero, it is very easy to occupy states above the band gap, while only few holes will be generated as they are separated by the gap from the T=0K Fermi energy. Hence the occupation of the conduction band would increase and with it the charge density. To keep charge density constant, the Fermi energy has to get smaller so that more holes are created.
  4. Oct 12, 2012 #3
    Thank you!
  5. Oct 13, 2012 #4

    Do you mean "How is the Fermi level defined"? It is the energy level at which all occupancy states at that energy level have an equal probability of being either empty or occupied. In other words, the energy level where the occupancy or vacancy probability is 0.5.

    An increased temperature causes thermally generated electron-hole pairs to increase until they swamp out the doner or acceptor doping concentrations. The semiconductor becomes more intrinsic as the temperature rises until it becomes completely intrinsic.

    Because the semiconductor is turning intrinsic. The Fermi level is just about the same as the intrinsic energy level value for intrinsic semiconductors. For a temperature increase, the Fermi level will rise toward the middle of the forbidden band for p-type semiconductors, and fall toward the middle for n-type semiconductors.

    For T-->0°K or freezeout, an n-type semiconductor will have all its donor levels occupied, and the conduction band empty. The doner level is 0.045eV lower than the conduction band if the n-type is doped with phosphorus. The Fermi level will have to be above the doner band level to show that the band is completely occupied, and below the conduction band to to show it is empty. By the same reasoning, the Fermi level for a p-type semiconductor will be just slightly above the valance band for that freezeout temperature.

  6. Oct 18, 2012 #5
    Far before the semiconductor gets intrinsic, the Fermi level goes away from the conduction band as the number of electrons due to the dopant is nearly constant, as is the equivalent density of states in the conduction band, hence exp[(Ec-Ef)/kT] is constant, which implies that Ec-Ef widens proportionally to T.
  7. Oct 18, 2012 #6

    I assume you are responding to my post?

    There are two regions before the intrinsic region, the freezout region and the extrinsic region. You have not specified to which region you are referring, but as the temperature rises from the freezout to the extrinsic region, the Fermi level moves toward the midpoint of the forbidden gap, and settles at room temperature Fermi level. As the temperature level rises into the intrinsic region, the Fermi level moves closer to the midpoint level and eventually settles at that point. When the temperature rises from the freezeout region to the extrinsic level, the electrons jump from the donor sites to the conduction band, thereby increasing the n concentration and lowering the Fermi level. When the temperature increases into the intrinsic region, the thermally generated holes and electrons become so much that they swamp out the previous n concentration and the semiconductor turns intrinsic.

    I never heard of "the equivalent density of states" phrase.

  8. Oct 21, 2012 #7
    As the original post gives no exotic temperature and tells "N-type" I consider it's at normal temperature range, that is, when the dopant concentration determines primarily the carrier concentration. Extrinsic, yes.

    And at this temperature range, there is no single Fermi level, precisely because the carrier concentration is in a first approximation constant. The direct consequence of a constant electron concentration is that the Fermi level drops away from the conduction band when temperature increases. No intrinsic carrier density is needed for it.
  9. Oct 21, 2012 #8

    The OP did not ask about freezeout or high temperatures directly, but he does ask about why the Fermi level drops. Also a subsequent post referred to freezeout temperatures. But we can discuss the Fermi level in only the extrinsic region if you wish.

    Throughout this extrinsic temperature range, the electrons are provided mostly by the ionized donor level dopants picking up thermal energy and making the small jump to the conduction band. There are only so many of these donor dopants, so after they are all ionized, the conduction band contains a constant amount of electrons for a wide temperature range called the extrinsic region. During this time the Fermi level is fairly constant because the conduction level electrons are constant. When the temperature increases high enough, electrons from the silicon atoms themselves are able to jump the gap from the valance level to the conduction band in quantity, and finally swamp out the number of electrons from the donor sites. This lowers the Fermi level toward its intrinsic level, thereby making the semiconductor more intrinsic, and the temperature enters into the intrinsic region.

    Well, that is where we disagree. The conduction electrons remain constant during the extrinsic range, until enough thermal energy is available to ionize the silicon atoms, flood the conduction band with electrons that jumped the forbidden gap, and turn the semiconductor intrinsic.

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