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Why are electron and hole currents zero at thermal equilibrium?

  1. Feb 17, 2008 #1
    Hi, I'm currently studying in an introductory semiconductor course where we use the following equations (numbers 1-5 on the first page):

    http://web.mit.edu/kimt/www/6.012/TheFiveEquations.pdf

    as a model of the underlying physics.

    Now, it is claimed that at thermal equilibrium, we can take J_e and J_h to be identically zero. As I understand it, the requirement of T.E. allows us to state that the derivatives w.r.t. time are zero. However, I'm not certain as to why we need to mandate that the currents are zero.

    Of course there's the intuitive notion that if there are currents, then things are moving and thus are not "at equilibrium"; but what I'm looking for is an algebraic derivation from the physical equations that restrict the current to be zero. However, I have not had too much success so far, and most resources just seem to assume that J_i = 0 at thermal equilibrium as a trivial result.

    Can anyone give me some suggestions on how to begin?
     
  2. jcsd
  3. Feb 17, 2008 #2

    Mapes

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    Here's a justification approach: Any flux [itex]J[/itex] (energy, matter, momentum, etc.) can be related to the gradient of a generalized potential:

    [tex]J\propto-\nabla\Phi[/tex]

    For charge carriers, we want to use the electrochemical potential:

    [tex]\Phi=\mu+q\phi[/tex]

    where [itex]\mu[/itex] is the chemical potential (related to concentration), [itex]q[/itex] is the carrier charge, and [itex]\phi[/itex] is the electric field. Now, if we assume that the electric field is zero and the carrier concentration is uniform, then the flux must be zero. Does this sound reasonable?
     
  4. Feb 17, 2008 #3
    Actually, I'm not sure I'm convinced with that argument (if I understand you correctly).

    We use the claim that the flux is zero in order to obtain a relation between carrier concentration and the electric field. (Equations 1 and 2 in the pdf.) In particular, we use this to analyze the case of an abrupt p-n junction where at equilibrium neither the concentration nor the electric potential are uniform, but the current densities are zero.

    This is identical (I think) to your formalism, where the conclusion is that [tex]\Phi[/tex] may be uniform while each of its components are not. Given this example I'm not so sure about the generality of your argument.
     
  5. Feb 17, 2008 #4

    Mapes

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    Well, another way of looking at it is, if the electron flux isn't zero at thermal equilibrium, why aren't all the electrons piled up on one side of the sample? That just isn't experimentally observed.

    One could argue that electrons and holes are forming on the left side of a sample, diffusing, and recombining at the right side. But you could just as well argue that they're diffusing similarly towards the left side. Thermal generation and recombination do occur, but the diffusion process is undirected, and thus the electron and hole fluxes are zero.

    I realize these are "softer" arguments than you're looking for. But the fact is that all diffusion relations are phenomenological. Fick's Law, whose general form I wrote above, is phenomenological.

    The only other approach I can think is thermodynamical: the coordinated motion of carriers (a non-zero flux) in the absence of a driving force would decrease the entropy of the system, which is prohibited by the Second Law.
     
  6. Feb 24, 2008 #5
    Look at the pn-junction : at thermal equilibrium, both of the Fermi levels are aligned. So the Fermi level of the p type semiconductor is aligned with that one of the n type semiconductor. This implies that it's going to cost energy to :

    1)get electrons from the n side to the p side
    2)get holes from the p side to the n side.

    So, without providing that energy, ie by a forward bias, there will be no current because of the potential difference over the junction.

    More here :
    http://hyperphysics.phy-astr.gsu.edu/hbase/solids/pnjun2.html#c1

    marlon
     
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