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PN junction shorted

  1. Sep 12, 2011 #1
    Hi friends, here's my question.
    The two terminals of a PN junction are at different potentials so, if I connect a them with a wire i.e. short them, What would happen? Would the potential difference between the two ends becomes zero or what? As far I'm understanding, there would be a flow of electrons from the P region to the N region because of potential difference. But as soon as the PD becomes less, the barrier potential becomes low and therefore the diffusion process will start, in which electrons would move from N to P and so on....
    But, this would constitute a current without any battery, etc.
    And why can't we measure inbuilt potential difference of the PN junction in Lab using voltmeters.
    I'd greatly appreciate any help.
    Thank you :)
     
  2. jcsd
  3. Sep 12, 2011 #2
    From my view, if you short it out, a small current will flow until that potential difference is gone. After that, diode will behave just like a wire.
     
  4. Sep 12, 2011 #3
    Thank you Bassalisk.
    But, after the potential difference is gone i.e. the barrier potential becomes zero then the diffusion of electrons would start again since there is no electric field now that could oppose there flow.
    Again, a pd would be created and it would lead to an another current of electrons moving from low potential to higher potential and so on...
    So, we have a continuous current!!!!!
    I know it won't work like that but, I'm not able to reach to any other conclusion.
     
  5. Sep 12, 2011 #4
    It does make sense what you are talking. Interesting. I will see this through and come back to you. If cabraham or sophiecentaur come by, they will answer it for sure.
     
  6. Sep 12, 2011 #5
    Thanks Bassalisk, I'm eagerly waiting for the solution. And cabraham or sophiecentaurc, sounds like they are the experts here.
     
  7. Sep 12, 2011 #6

    jim hardy

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    there's something called a "Depletion Region" where holes in P region get filled by electrons from N region.. and vice versa

    http://en.wikipedia.org/wiki/Depletion_region

    When enough charge (a minute amount) has moved to complete that process, current ceases.

    The potentials are internal and they cancel so external voltage is zero.

    You MIGHT be able to measure the charge with an electrometer - i dont know - would make an interesting lab experiment.

    Something related happens in a vacuum tube diode. Shorting its terminals will result in a small current - electrons are shaken off the hot cathode and will migrate to the plate. It's called (if i remember right from 1961) contact potential and can produce a couple tenths of a volt.
    But in a vacuum tube diode there is an external energy source, the cathode heater.

    hope i'm right - keep me honest, centaur?

    old jim
     
  8. Sep 12, 2011 #7
    Could you explain it in details. I didn't get how the potentials can cancel each other.
     
  9. Sep 12, 2011 #8

    jim hardy

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    the Wilipedia link has a drawing showing the gradient inside the silicon - a picture is worth a thousand words

    think of an analogy: two capacitors in series, both charged to maybe 0.1 volt, connected + to +

    between their negative leads there is zero volts
     
  10. Sep 13, 2011 #9
    I think you are talking about this picture 576px-Pn-junction-equilibrium-graphs.png

    And I got the concept of the capacitors but, I don't see any analogy. Here, it is clearly shown that the two regions p and n are not at the same potential, the n is at a higher potential and the p at a lower one. So, there is a potential difference between the two terminals.
    Please point out if I'm making any mistake in the analysis.
    Thank you.
     
  11. Sep 13, 2011 #10
    Current will not flow!

    Its called the junction potential, and NO, you cannot measure it. First of all it is established across the depletion width at equilibrium. We are not talking of conduction here but rather equilibrium attained during the formation of the depletion width itself. The depletion width is formed (by diffusion across the heterogenenous metallurgical junction) in consistency with Poisson's equation. (Another viewpoint is that it is formed so that Maxwell's equations remain satisfied.)

    Remember that the assumption made here is that there is no electric field outside the depletion width! This is an idealization, but it is a reasonably good model at low voltages and for reasonably long devices (with lengths greater than the diffusion length -- the so called long base assumption, a term which comes from BJT terminology).

    So, please understand that any act of measuring the junction potential even under the assumption that the contacts are Ohmic and ideal will disturb equilibrium. The diode in this condition is not a voltage source. The potential difference across the depletion width cannot be measured without also accounting for the contact potential at the contacts. A careful calculation will tell you that the contact potential difference exactly equals the junction potential in magnitude with the polarity reversed. So since the 'electrode measurement' will measure the algebraic sum of the contact potential difference and the junction potential, the result will be 0.

    The fact that the junction potential cannot be measured directly (and can only be inferred) is very well explained in most standard books on semiconductor physics and devices. Please see for instance the book Operation and Modeling of the MOS Transistor by Yannis Tsividis, or the book Solid State Electronic Devices by Ben Streetman and Sanjoy Banerjee.

    PS (to the Original Poster Bhargava2011): Please do go through the books I've suggested, and do not rely on Wikipedia over a standard textbook on the subject -- there are many mistakes!
     
  12. Sep 13, 2011 #11
    I'd try to refer the books you've mentioned.
    But, could you please explain in detail about the contact potential and the junction potential and their sum. I think its the main point here. If I'll understand it, my problem would be solved.
    Thank you so much.
     
  13. Sep 13, 2011 #12

    es1

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    But if the potential difference is gone what pushes the charge over the barrier. The height of the barrier might be zero but if there is no field to push over it then there will still be no current.
     
  14. Sep 13, 2011 #13
    The full explanation is somewhat tedious to write down and will not be illuminating without a diagram. I know that Tsividis has it, and I shouldn't deprive you of the thrill in figuring it out yourself (since he explains every detail carefully anyway). If you are interested in device physics, you should definitely check out his book...its a classic.
     
  15. Sep 13, 2011 #14

    jim hardy

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    ""A careful calculation will tell you that the contact potential difference exactly equals the junction potential in magnitude with the polarity reversed. ""'

    So Maverick

    do i understand there's an effect from dissimilar affinity for electrons between the silicon and the metal leads? Like in thermocouples?

    It's that simple?

    I'd have thought that right at the outside edges of depletion region there'd be a gradient due to immediate proximity of excess charge, which at a few depletion region widths would fade. Once you're a few microns away from depletion region, distances to its two sides + and - dont look much different so cancel.. That's where i thought the equal and opposite potentials were located, right in edge of D R..

    learn something every day maybe i'll know domething one day.

    old jim
     
  16. Sep 14, 2011 #15
    There is something called diffusion current. The electrons diffuse into the p region which gives the current. But, my concepts are not clear enough so, not absolutely sure!!
     
  17. Sep 14, 2011 #16

    jim hardy

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    ""There is something called diffusion current. The electrons diffuse into the p region which gives the current. But, my concepts are not clear enough so, not absolutely sure!! ""

    Mr Bhargava,
    i too am struggling to make my mental picture a clear one.

    I do not understand what happens right at edge of depletion region.
    Seems to me electrons diffuse into p region, attracted by p donors' affinity for electrons, until the electrostatic pull from the ions they left behind just balances the tug from p donors.
    (of course it's a symmetric push from n and pull from p donor affinities balanced by coulombic attraction but i simplify for sake of brevity)
    Hence the depletion region width is calculable from doping density and a few other numbers.
    Seems there's a wall of charge at each edge of depletion region so why a gradient on only one side???

    i am puzzled.




    This link seems to support what Maverick said about the careful calculation;
    that where the metal leads join the silicon is the location of the equal and opposite voltages required to satisfy Kirchoff and not at edge of the depletion region like i thought.
    http://ecee.colorado.edu/~bart/book/book/chapter3/ch3_3.htm


    So be advised this reply that i prepared last night probably has a glaring mistake.
    Maybe we can tweak it into something useable for a teaching tool.
    i think i should repeat my thought experiment, but start my walking tour inside the wire not in the silicon. I thnk i will find a region with the potential gradient i need right where the wire joins the silicon, at extremities of the silicon not extremities of depletion region..
    it seems logical an undepleted region may not support a gradient for charge would move. That could force the gradient to appear instead in another region where there's dissimilar electron affinities in proximity, like where the electron rich metal leads attach (as Maverick said) .

    Bhargava - i post my thought experiment above uncorrected, with apology, please do not think me discourteous..
    What if i'm wrong on both ?

    maybe some of the more mathematically facile denizens like you and Maverick and Centaur and Yungman will help me out.

    old jim
     
    Last edited: Sep 14, 2011
  18. Sep 15, 2011 #17

    es1

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    This is a little sloppy in terminology but the minority carrier diffusion current gets balanced by the drift current in steady state.

    For a quick proof, assume the drift current did not match the diffusion current in steady state. Then there would be a build up of a charge which would mean that the width was changing. This is a contradiction, so they do match.
     
  19. Sep 16, 2011 #18
    Last edited by a moderator: Apr 26, 2017
  20. Sep 16, 2011 #19
    hmm...I didn't think about that. I'll refer some books, anyways thanks :)
     
  21. Sep 16, 2011 #20

    es1

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