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B Existence of depletion layer in contact surface

  1. Mar 4, 2017 #1
    in a P-N junction diode, in an equilibrium state, a depletion layer forms in the surface of contact of N type and P type. But the problem is, this layer is formed by diffusion of electron to a lower electron density zone. Why does it always have to form right in the contactt surface? when electron comes from N side to P side, why does it havve to stay in the contact surface area? It should diffuse further away from the N type.
     
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  3. Mar 5, 2017 #2

    ZapperZ

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    Charges will move when there's an electric field, and an electric field will form when there's a potential gradient.

    A potential gradient forms due to the contact potential difference between n-type and p-type semiconductor. Since the chemical potential/Fermi levels for both sides will reach equilibrium, this will set up this contact potential difference. See below:

    diode4.gif

    How wide this depletion zone is depends the magnitude of this potential difference. This means that this doesn't extend all the way through each side. That is why the charges that move are only around the junction.

    Zz.
     
  4. Mar 5, 2017 #3
    That's not what I meant. when electron moves from n type to p type for diffusion, why doesn't the electron go further inside the p type to fill holes further away than contact surface?
     
  5. Mar 5, 2017 #4

    ZapperZ

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    But that is not what is happening even in the depletion zone, i.e. the electrons are not filling up holes in the p-type semiconductor!

    The electrons are in the conduction band, while the holes are in the valence band. There is no recombination. The electrons and holes that migrated in the depletion zone is due to the existence of the initial electric field due to the contact potential difference. It is not due to wanting to electrons from the n-type semiconductor filling up holes in the p-type semiconductor.

    Zz.
     
  6. Mar 5, 2017 #5
    ok, why doesnt the holes and electrons simply diffuse further away from the contact surface?
     
  7. Mar 5, 2017 #6

    ZapperZ

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    Why should they? There's no field to cause such a drift.

    Zz.
     
  8. Mar 5, 2017 #7
    There's ann electrron density gradient which may cause the drift?
     
  9. Mar 5, 2017 #8

    ZapperZ

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    Note that there would have been no diffusion from n to p had there been no potential gradient across the junction. Why? Because electrons leaving n-type would cause the material to have a net-charge, and conversely for the p-type. These two semiconductors started out being neutral. They don't spontaneously become charged. That's why you don't get such diffusion of one type of charge into the other material. There has to be a field gradient for the majority charge carriers to migrate.

    Zz.
     
  10. Mar 6, 2017 #9

    DrDu

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    I don't agree here. The force driving the electrons is not the electric field but the electrochemical potential gradient. When you bring 2 semiconductors into contact, at the beginning, they are charge neutral, but, as electron densities in the conduction and valence bands are vastly different in n and p type, the chemical potential is different whence the current is diffusion driven. Due to this current, the semiconductors charge up and the electrochemical potential levels out, i.e. the diffusion current is reduced due to the repulsion of the electrons/holes by the negative/positive charge which builds up.
     
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