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When P and N type semiconductors are connected together, some of the electrons drift from the N type into the P type to recombine with holes. This leaves positive ions on the N type and negative ions on the P type. Therefore an electric field exists across the depletion zone that stops further drift from electrons. This is also the potential that an external voltage source must exceed in order for current begin flowing.

But once current starts flowing, why does the built in voltage still remain at the depletion zone? Once current begins flowing, don't the electrons that drifted over the PN junction at the beginning now flow forward? Why is there still an electric field across the junction?

  • $\begingroup$ Well, as you noted, the built-in field does not remain. Also note that a diode works by holes (from P) and electrons (from N) flowing into the depletion region and annihilating each other there. $\endgroup$ – Jon Custer Jan 13 '19 at 3:30
  • $\begingroup$ @Jon Custer But the voltage does remain. For silicon diodes, there is approximately 0.7 volts across them when current is flowing. $\endgroup$ – S. Rotos Jan 13 '19 at 10:08
  • $\begingroup$ By that same argument you would have a built-in voltage for resistors... $\endgroup$ – Jon Custer Jan 13 '19 at 17:43
  • $\begingroup$ @Jon Custer I really don't see how. Resistor's principle of operation is completely different, and their current is much more dependent on voltage. $\endgroup$ – S. Rotos Jan 14 '19 at 7:01
  • $\begingroup$ The answers in physics.stackexchange.com/questions/177910/… may help somewhat. You need to apply a voltage to shift the Imrefs of the holes and electrons enough to start getting significant recombination in the depletion region. Just because that process starts does not mean you can take the applied bias off. Just like a resistor (sure, different processes at heart), if you don't apply a bias you don't get any current. $\endgroup$ – Jon Custer Jan 14 '19 at 13:51

The built-in field comes from diffusion of charge carriers, from a region of high concentration into a region of lower concentration. While an applied forward current DOES change those concentrations, the numbers (for silicon diodes at least) do not indicate a zero-thickness depletion/space-charge region at any feasible current density. Calculating the current density that does cause the depletion region to vanish is a standard sort of textbook problem; that much current would melt any normal diode if it didn't vaporize the attachment wires.

The more useful effect is to reverse bias a PN junction to widen the depletion region, which modulates the stray capacitance (making a voltage-variable capacitance, a varicap diode).

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