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P-N junction voltage under polarization

  1. Jul 22, 2014 #1
    All the books that deal with the p-n junction under applied bias asume that the same equations used for the equilibrium case (no bias) can be used for the biased case provided that the juntcion voltage [itex]V_{ic}[/itex] is substituted by [itex]V_{ic}-V[/itex], where [itex]V[/itex] is the applied bias. Although this can seem plausible I don't see any reason for why the junction voltage drop should be the [itex]V_{ic}-V[/itex], i.e., why the voltajes should obey a sort of superposition principle.

    Is there any reason on the ground of fundamental electrostatic / thermodynamics?

    Thanks.
     
  2. jcsd
  3. Jul 28, 2014 #2

    Greg Bernhardt

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    Staff: Admin

    I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
     
  4. Aug 21, 2014 #3
    First of all, you must be aware of that the built-in voltage is the one that is defined between the two edge of the depletion region, we do not care about other parts of pn-junction in our textbook for simplicity. When you have to consider bias voltage, in practice, two semiconductor-metal junction are formed at the two terminals of pn-junction, that exactly offset the built-in voltage when the bias is zero. When the bias is not zero, from the view of the depletion region, it will see a circuit where the bias voltage and the two semi-metal junctions are connected in series, that is the superposition "Vic-V". Even in an ideal case where ideal conducting wire (regardless of material, metal or not, whatever) is considered, you will find you have to face another pn-junction which is formed in opposite polarization between the two terminals when any contact is made.

    Fernsanz, I wish you not be more confused by my words, :)
     
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