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Solid state depletion region

  1. Jul 3, 2013 #1
    how does letting the electrons diffuse to the P material/region lead to a "depletion" region in the center where the N and P material meet? please go into detail
     
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  3. Jul 4, 2013 #2

    Simon Bridge

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    http://hyperphysics.phy-astr.gsu.edu/hbase/solids/pnjun.html
    Electrons fall into holes leading to a region of net-neutrality.

    It would be better if you explained your understanding of the process so we could figure out where the trouble lies... at least some idea of the level you are doing this at? What we don't do is actual detailed physics courses ... you have school for that. What we can do is help you find the answers to your questions :)
     
    Last edited: Jul 4, 2013
  4. Jul 4, 2013 #3
    simon bridge: okay then, whenever i say "current" i will be referring to electron current, NOT the hole current.

    here's my understanding; the PN junction comes in contact, the electrons diffuse across the junction into the P material (diffusion current), then there is an E-field set up across the junction since the materials are no longer neutral. the electric field forces electrons back to the N material and now we have our drift current. My professor said that the depletion region was the region where the electrons are mostly gone but not completely. he stated that when the diffusion current = drift current, then the system is in equilibrium and this is when the depletion region is set up. but whether or not there is a net current of zero, does not mean there is NO current at all. if there are currents flowing from the P to N, and N to P materials, then that still implies that there exists current; ie.. charge still has to flow from one side to the other. so how then, is the depletion region depleted of electrons? i am not convinced because i don't have any reason to believe that there should be any less charge carriers in the depletion region than there are anywhere else in the entire semiconductor.
     
  5. Jul 4, 2013 #4

    Simon Bridge

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    Why would they do that?

    How indeed - in a completely depleted depletion region, all the acceptor and donor sites will be ionized, so there will be no surplus charge carriers there. (You know how materials get to be P-type of N-type?)
    Are there any other sources of charge in the material?
    If the current dropped to zero, not just the "net current but all movement of charge, would you feel better about this?

    You mean apart from empirical observations supporting this claim?
    You appear to be happy that electrons, for instance, can move from their donor site into the P region? Then what does it leave behind?

    Have a look at:
    https://en.wikipedia.org/wiki/Deple...of_depletion_region_in_a_p.E2.80.93n_junction
    It may help you to distinguish between (mobile) charge carriers and the space charge associated with the material lattice.
     
  6. Jul 16, 2013 #5
    well.. as i understand, the P-materials do not have fully filled valence shells, and the N- have relatively have one additional electron to give away if you will, ie the donor ions.
    it is my understanding that alone/separately, these materials have their holes and "excess" electrons
    constantly moving around within the material. and when the two come in contact, it would make sense that the electrons diffuse across the junction into the P-material, and same for the holes moving across into the N-material.

    yes, i know how materials get to be N & P type
    any other sources of charges? no, just the donor and acceptor ions.
    and not really, even if the charges were still (ie no current), i still don't see how any of this leads to the formation of a depletion zone.
    let me then ask.. if i were to place a charge in the depletion region, what would happen to the charge? would it initially start heading to the N region due to the electric field that is set up?

    i don't care about empirical evidence; and when i say that i mean, that doesn't help me understand the theory any more. the empirical evidence is what im already working with. I have a P-type material and an N-type material, and you stick the two together, charge diffuses, charge drifts due to set up E-field, then a depletion region somehow sets up. and my goal is to understand this not just regurgitate the process. i want to know the physics behind it.
    electron moves from N→P, then it leaves behind a hole. i'm not sure where to go with this though.

    i am aware of the difference.. i'm just wanting to know how the space charge region comes to into formation. the wiki page you provided says that that after the transfer, the holes and electrons recombine and this somehow leaves behind a depletion region. how?... lol this is what i am confused with

    thanks
     
  7. Jul 17, 2013 #6

    Simon Bridge

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    ... so you don't think that the atoms that make up the main body of the crystal have electrons and protons in them?
     
  8. Jul 17, 2013 #7
    i fail to see why this is relevant, please make the connection for me?

    also, i never said that they didn't have charged particles (how in the world are protons a charge carrier for semiconductors?...), but they are still no sources of charged particles, Si is not a source of an excess electron nor is it the source of an unfilled valence shell, ie a hole; ie, it is not the source of any charge carrier. the Si only acts as a medium for which the charge carriers donated by the donators and acceptors can travel through, but this does not make it a source of electrons, or a source of holes..... the Si is just a medium. but again, i don't see the connection from this to the formation of the depletion region. would you be so kind as to make it for me please
     
    Last edited: Jul 17, 2013
  9. Jul 17, 2013 #8
    This may be part of the confusion.
    It does not leave behind a hole but a fixed positive ion.
    So the concentration of charge carriers decreases by the diffusion of the electrons.
    It does not mean that the concentration of electrons goes to zero in this "depletion region.
    The fixed positive ion will contribute to the positive space charge of the region.

    On the other side, in the p region, the diffused electron recombine with a hole and so contributes to the "depletion" of the p region too.
     
  10. Jul 17, 2013 #9

    Simon Bridge

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    I'm trying to get you to think of the whole material instead of just concentrating on a tiny part of it.
    I'd like you to distinguish between mobile and fixed charge carriers.

    But it may be your understanding is different - are you thinking that n-type material has a net negative charge?

    Use the empty seat model ... n-type material has all the seats full and p-type material has all empty seats. Remove a wall between the two and people from the full-side of the auditorium start to shift over into the empty seats ... when someone moves, they leave an empty seat behind them. But they are unsure about whether this is OK so they don't go very far from their assigned seat.

    You end up with scattered occupied seats close to the boundary on the p-side and scattered unoccupied seats close to the boundary on the n-side.

    That help?
     
    Last edited: Jul 17, 2013
  11. Jul 22, 2013 #10
    no, i'm thinking it has a net positive charge after diffusion, and a net zero charge before diffusion

    not to be too spoiled, but i don't work very well with analogies. can you explain what actually is happening? why are the electrons 'unsure' about going too far from their donor ions? and i thought it was supposed to be a band spread over the crystal, as opposed to electrons being stuck to its donor ion.


    "scattered filled dopant valences close to the boundary on the p-side", meaning that you have unfilled dopants (otherwise why would the filled ones be said to be 'scattered') surrounding the filled ones; ie the majority of the p-dopants near the boundary are unfilled and a few of them are filled.
    and
    "scattered unfilled dopant valences close to the boundary on the n-side", meaning that you have filled dopants surrounding the unfilled ones; ie the majority of the n-dopants near the boundary are filled and a few of them are unfilled.

    did i interpret correctly?

    should i be familiar with solid state physics before i learn about transistors?
     
  12. Jul 23, 2013 #11

    Simon Bridge

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    It's a bit unfortunate that you don't do too well with analogies since pretty much all of solid state is done by some analogy-ish model ... to avoid the analogies, you'll have to do the math - invest in a good solid state college text book. There should be several suggested in the text-books discussion.

    But you seem to be getting it.
    There are mobile and lattice charges - when a mobile charge moves a lot, it leaves the opposite static charge behind. Opposite charges attract - so the mobile charges won't go too far. Mobile charges diffuse into regions of lower concentrations of their own kind (look up "diffusion".)
     
  13. Jul 26, 2013 #12
    when you say lattice charge are you referring to the N and P materials? ie the lattice charge of the N material being positive?

    i still don't understand; because the depletion region is said to be the region that is depleted of all free charge carriers by wikipedia.
    however, i took an electronics course and my prof said that the depletion region is the region that is depleted of the majority charge carriers. if this is the case, then i roughly understand the formation of the depletion region, however if what wikipedia says is true, i still don't understand how all charge carriers are depleted
     
  14. Jul 26, 2013 #13

    Simon Bridge

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    In either material you have mobile and fixed charges. When a mobile charge moves, it leaves the fixed charge behind. The mobile and fixed charges are a model used to approximate the behavior of the bulk material. They are derived from the presence of donors and acceptors - the fixed charges, the lattice charges, come from the fixed-in-place donor and acceptor ions.

    The charge carriers in the depletion region are not free. The depletion region is formed only after recombination of holes and electrons... so in the depletion region there are only and only immobile positive and negative ions... hence,there is no charge carrier...

    http://www.asdn.net/asdn/physics/p-n-junctions.shtml [Broken]
     
    Last edited by a moderator: May 6, 2017
  15. Jul 26, 2013 #14
    This is an exaggeration or simplification, at least.
    There is a gradient of carrier concentration (both types) over the depletion region.
    Neither p or n concentration goes to zero.After all, at equilibrium the product p*n should be the same over the crystal. The concentrations of the minority carriers are actually increased in the depletion region.

    Example of concentration of carriers over the junction can be seen for example in Kittel chapter 19. Or here:
    http://en.wikipedia.org/wiki/P–n_junction
     
    Last edited by a moderator: May 6, 2017
  16. Jul 26, 2013 #15

    Simon Bridge

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    Well sure - perhaps I should have made that clear :)
    It is a common teaching model that illustrates why text books etc say that there are no free charge carriers in the depletion region and motivates the choice of the word "deplete" to describe this region... which was the question I was answering.

    It is probably more helpful to think of the word as jargon: it's a label.
    The originators could have named the region "bob", and talked about the charges in the bob-region... would students then expect the region to jump up and down?

    I can't tell if the problem is understanding the nomenclature (semantics) or the physics.
     
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