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Origin of the depletion region in a p-n junction

  1. Sep 12, 2012 #1
    Hi the first step to this is usually described as diffusion of electrons from the n type to the p type. However I have encountered sites and books that state that holes diffuse from p type to n type as well. I am of the opinion that only the electrons diffuse, and that there is no diffusion of holes. Can anyone help clarify?
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  3. Sep 12, 2012 #2

    Simon Bridge

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    Physically there are only electrons doing the moving - the "hole" phenomenon is emergent behavior and is mathematically identical.
  4. Sep 12, 2012 #3
    Oh yes i know what hole movement actually is, but the elctrons that diffuse over dont come from the covalent bonds. they are already delocalised and hence when they diffuse over they dont leave behind extra holes. ( i know there are already holes in the ntype side due to the normal intrinsic behavious of the semiconductor).

    thus no extra holes appear on the n side during this diffusion process yes?
  5. Sep 12, 2012 #4
    But there is also a large number of "empty" valence band states on the p-side (these are your holes); this is compared to the large number of filled valence band states on the n-side. Thus valence band electrons will diffuse from the n-side to the p-side in order to fill the empty states. Of course, in this process, empty states are created on the n-side. This entire mechanism is emergant as hole diffusion.

    Note that the conduction band electrons that diffuse to the p-side do not take part in the above because they are at a different energy level.
  6. Sep 12, 2012 #5
    I see what you are saying, I think the problem I am having is that, will valence electrons diffuse over? I was under the impression that valence electrons move to cover holes only via electrostatic attraction. With what you are saying then electrons move from covalent bond to covalent bond via diffusion due to different in concentration as well?

    If this is the case then I wonder why this happens since diffusion would seem to be something that bound electrons would tend not to do.
  7. Sep 12, 2012 #6


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    If the orbitals on e.g. Si would be perfect sp^3 hybrid orbitals, then the bonds would be really very localized and the resulting valence would be nearly flat. The resulting effective mass of the holes would be very high and they would hardly diffuse. However, due to the energy splitting between the s and p orbitals, the bonds are delocalized and the valence band has considerable curvature corresponding to lower effective mass of the holes which allows them to diffuse.
  8. Sep 12, 2012 #7
    That sorta makes sense

    Thanks all for your input :)
  9. Sep 12, 2012 #8
    Picture a single acceptor atom (with it's associated hole) placed in the semiconductor latice. That picture is electrically neutral. Any movement of the holes (i.e. the valence-band electrons) is thus due to diffusion.
  10. Sep 12, 2012 #9
    That's an interesting question. For the sake of completeness let me paint the whole picture. You have (say) phosphorus forming four bonds on the n-side with an extra electron. Say that we are at a temperature where the ionization energy of the extra electron in phosphorus has been overcome such that this extra electron is no longer bound to the donor, and is delocalized. At the same time we have (say) boron forming four bonds on the p-side, with one electron missing from one of the four bonds. I think that you found the hole diffusion a little strange because of the question: why would a silicon, on the n-side, be willing to remove an (valence) electron from one of its four bonds to give it to the incomplete fourth bond of the boron atom on the p-side? Here are my thoughts on this:

    I think such a process occurs because it is energetically favorable. In case you care to know, I arrived at this explanation from the Landauer approach to quantum transport, where you compute conduction based on gradients in the chemical potential. This approach is a powerful technique and accounts for transport of charge carriers both by an electric field and via diffusion. But for this discussion I will stick to physical arguments and not get into the Landauer formalism. I will, however, make a reference to the density of states without getting into too many details.

    You can see a qualitative profile of the density of states ##D(E)## in the conduction and valence bands on the n- and p-sides in the attached figure, where the vertical axis is obviously energy ##E##. The shaded regions represent states filled up with electrons. One thing to note is that this figure is not shown in equilibrium. In equilibrium the chemical potentials ##\mu_1## and ##\mu_2## will be the same. The process of leveling the chemical potentials involves the charge transfer process which gives rise to the depletion layer.

    Now, you can see that on the n-side you can consider the valence band to be almost full (ignoring intrinsic holes). These valence electrons on the n-side see a lot of empty states at a lower energy on the p-side. This is what effectively drives the valence electrons from the n-side to the p-side, which is evidently enough to make silicon, on the n-side, remove an (valence) electron from one of its four bonds to give it to the incomplete fourth bond of the boron atom on the p-side.

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