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Electric field inside PN junction in equilibrium

  1. Jun 7, 2013 #1
    I need help to understand this solution:
    attachment.php?attachmentid=59361&stc=1&d=1370665607.jpg
    attachment.php?attachmentid=59362&stc=1&d=1370665607.jpg
    Why electric field is zero at x < - b?
    I think it should be non-zero because the electric field from positive and negative ions don't completely cancel each other.
    How can you know E rises as x approaches zero?
     

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  3. Jun 8, 2013 #2

    ehild

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    It is only a schematic plot. The electric field is not zero just at the boundary of the depletion region. (And the boundary of the depletion region is not sharp, either). But it tends to zero quite fast, as the forces from the positive and negative regions tend to cancel.

    Imagine you are a positively charged particle. You walk into the depletion region from left, from the n part. When you are inside the depletion region, still in the n part, there are some positive ions both behind you (pushing you forward) and the other positive ions in front of you, pushing you backwards. But all the negative ions on the other side pull you forward. When you are exactly at the interface between the p and n region, (x=0) all the positive ions are behind you, pushing forward and all the negative ions are in front of you, pulling you forward.

    ehild
     
  4. Jun 8, 2013 #3
    Thanks, the example is great!
    I am not quite understand it. Can you explain it more detail?
     
  5. Jun 8, 2013 #4

    ehild

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    The space charge region around the p-n junction is quite thin, the positive and negative "poles" are close, the electric field is similar to that of a dipole, q and -q charges d distance apart. At far away from the junction, in the line of the dipole at distance x , the electric field is E=kq/(x-d/2)-q(x+d/2)≈2kq/x3.

    ehild
     
  6. Jun 8, 2013 #5
    Thanks, got it now!
     
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