P-N junction voltage under polarization

  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. Greg Bernhardt

    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. 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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