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In reverse bias , when we go up the breakdown voltage why is there a voltage drop across the diode?.I mean the diode does resist to the flow of charges but when the electrons flow inside the diode the electric field of the diode tries to accelerate them.

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  • $\begingroup$ Why is there a voltage drop in forward bias? Are you expecting no voltage drop at all, and if so, how does material resistivity factor into that? $\endgroup$ – Jon Custer Apr 16 at 23:04
  • $\begingroup$ Can you please clarify your question? As written, it seems that you are asking why does not the reverse breakdown make the junction become superconductive, but there is not a word as to why do you think it should. $\endgroup$ – kkm Apr 17 at 0:59
  • $\begingroup$ Hmm the electric field of the depletion region in forward is opposite to the flow of electrons.But the electric field of the depletion region in reverse bias has the same direction with the flow of charges when we go up to breakdown voltages. $\endgroup$ – Altair Apr 17 at 6:16

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