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Fermi Level Statistics

  1. Oct 14, 2005 #1
    This is sort of a simple question. In a steady-state semiconductor doped n-type, is it possible for the quasi-Fermi level for the holes (F sub p) to be above the intrinsic level for the conductor (E sub i)? That is, E - F = (negative value)?
     
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  3. Oct 15, 2005 #2

    ZapperZ

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    Maybe there's some special terminology being used in EE, but this question is rather odd.

    1. n-type semiconductor has electrons as the majority charge carrier. So why is there a Fermi level for holes?

    2. Usually, the Fermi level is the outcome of the WHOLE population of charge carriers, and not a combination of Fermi level for electrons plus Fermi level for holes, i.e. they can't be separated.

    3. What is "the intrinsic level for the conductor"? What conductor? Where did this conductor come from since you are talking about a semiconductor. There isn't an intrinsic level for a conductor.

    Zz.
     
  4. Oct 15, 2005 #3

    Gokul43201

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    Zz, I'm not terribly familiar with this, but the EE folks do have such a thing as a Quasi-Fermi Level for electrons and holes separately. I believe it is some non-equilibrium parameter when a current is flowing through a SC.
     
  5. Oct 15, 2005 #4

    ZapperZ

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    Well, this person did say that it is in a steady state. So that can't be it.

    Zz.
     
  6. Oct 15, 2005 #5
    Yes there is a special terminology used in EE

    1.See in EE we use the quasi Fermi levels (both 'F sub n' for electrons and 'F sub p' for holes) as a replacement for the equilibrium Fermi level (E sub F) in the steady state analysis
    since the equilibrium Fermi level can only be used to described a doped material in an equilibrium state the quasi Fermi levels are used to describe the electron and hole concentration in the steady state with respect to the equilibrium Fermi level
    an example of this is when the semiconductor is optically exited
    when the the generation rate reach the steady state,
    both the generation of electron and holes can be studied by computing the derivation of the quasi Fermi levels from the original Fermi level.

    2.I think 1 answers that.

    3.Again in EE we seem to refer to semiconductors as conductors, I know it's wrong but that's just the way it goes.




    I think you have mistaken that with thermal equilibrium a semiconductor can be under current and in a steady state if the current in it, is in a steady state i.e there is no change in the generation or recombination rate of carriers


    I am not sure of this answer so don't take it as your final, but i think that if the material is heavily doped and that the excitation source applied to it does not give that much energy there is no reason why this could not happen but as i said don't take my answer for granted try to ask someone with a PhD and come back to say if i was right or wrong.
     
  7. Oct 15, 2005 #6

    ZapperZ

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    Eeeewww... I should have just stayed away and not looked in here!

    :)

    Zz.
     
  8. Oct 15, 2005 #7
    Thanks for the replies guys. I think it is possible since I can't find any outstanding errors in my math or my assumptions.
     
  9. Oct 15, 2005 #8
    Why?


    Like i said don't take that for granted sometimes there is more things to take into account than you've learned so you'd better ask someone who has in depth knowledge of the subject more than you.....
     
  10. Oct 15, 2005 #9
    Well, my solid state textbook (Streetman's) defines the quasi-Fermi level for holes with the following statement:
    [tex]E_i - F_p = k T \ln{\frac{p}{n_i}}[/tex]
    So, for the difference to be negative, the number of holes would have to be less than the intrinsic number of holes. In an n-type semiconductor, this is definitely possible, so I would say that yes, the quasi-Fermi level for holes can be above the intrinsic level.

    For example, consider steady-state optical generation. As the optical generation rate approaches zero, the semiconductor behaves more and more like it is at equilibrium. In that case, [itex]p[/itex] approaches [itex]p_0[/itex], the equilibrium value, and the above equation becomes equivalent to the relation for semiconductors at equilibrium. Hence, in this case, both the quasi-Fermi level for holes and electrons approach the equilibrium Fermi level, which for n-type material, is clearly above the intrinsic value.
     
  11. Oct 15, 2005 #10
    Nice, thanks.
     
  12. Oct 16, 2005 #11
    I don't know why you care about separation of Fp and Ein. In reverse bias, gap between Fp and Fn increases by q*Vapplied (Fp is above Fn). In forward bias, it depends on how much voltage is applied. Usually you only look at the quasi-fermi levels (in steady state), Fp and Fn. Just draw an energy band diagram for forward and reverse bias and it should be clear.

    edit: i guess if you want to calculate carrier density on p or n side you need to know Fn and Ein or Fp and Eip. But usually you should just use n*p = (ni)^2*exp((Fn-Fp)/kT). i should type this in latex... sorry about that. ni is intrinsic density level (ie. not doped)
     
    Last edited: Oct 16, 2005
  13. Oct 16, 2005 #12
    Why are you talking about a P-N Junction when he's asking about a doped n-type semiconductor ?
     
  14. Oct 16, 2005 #13
    whoops, didn't see it's just n-type. nevermind about what i said then, no wonder i thought it was an odd question. sorry about the confusion. in my electronics class, we only looked at quasi fermi levels for junctions. didn't realize this applies for single type semiconductor.
     
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