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During forward bias : An electron from the negative terminal of battery enters the conduction band of $n$ side, runs toward the junction, waits for a hole for $\tau$ seconds, when it finds one, it falls into the hole in the valence band of $p$ side, then the electron hops along the $p$ side holes and finally reaches the positive terminal of the battery.

Question1 : Why does the electron in the conduction band fall into the hole in the valence band ? Is it because the valence band has lower energy ? If so, kindly keep this in mind when looking into my second question.

During reverse bias, during recovery phase : An electron from the negative terminal of battery enters the $p$ side valence band, runs toward the junction, then what ? There are no empty energy states available in the valence band of $n$ side, so it seems the electron must climb up to the conduction band to complete the loop. ( Note that if we abruptly change from forward bias to reverse, initially, during recovery phase, there exist many diffused electrons on the $p$ side near the junction. )

Question2 : Does the electron in the valence band really climb up to the conduction band at the junction ?

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    \$\begingroup\$ Perhaps a better question is : when you switch instantly from forward to reverse bias, there are a lot of electrons hopping along the $p$ side holes. What happens to them? \$\endgroup\$ – Brian Drummond May 5 '18 at 10:57
  • \$\begingroup\$ @BrianDrummond I see.. for simplicity lets say the diode is forward biased to start with and now we disconnect the battery. In this situation with zero bias, I think those excess electrons on the \$p\$ side fall into valence band holes (recombine) in \$\tau\$ seconds. But these recombinations still leave excess electrons on the \$p\$ side ? Somehow I feel that those excess electrons should go back to the \$n\$ side, but I don't see how. Recombinations essentially lock those excess electrons in the valence band instead of the conduction band. Hmm.. Any thoughts ? \$\endgroup\$ – AgentS May 5 '18 at 11:06
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    \$\begingroup\$ No, consider not simply disconnecting the battery but reversing it. \$\endgroup\$ – Brian Drummond May 5 '18 at 11:07
  • \$\begingroup\$ Okay, reversing the battery pushes the electrons back to the \$n\$ side ? \$\endgroup\$ – AgentS May 5 '18 at 11:09
  • \$\begingroup\$ But that doesn't happen instantly - I think some of those electrons still fall into the holes in the valence band on the \$p\$ side, leaving excess electrons in the valence band on the \$p\$ side ? \$\endgroup\$ – AgentS May 5 '18 at 11:10

When a PN junction is forward biased there will be excess minority electrons in the P Region due to diffusion of majority electrons in the N region into the P Region.Conversely, there well be minority holes in the N region.

When you reverse bias the PN junction the force(forward voltage) that allows for Electrons and holes to move to the P and N regions,respectively,is gone.

The question is where do these excess electrons and holes go?

Because of the electric field in the depletion region(from the N side to the P Side),excess electrons in the P side will be swept back to their original position in the N side(Electrons move in the opposite direction of the E-field).In the same Manner,holes(moving with the direction of the E-field)in the N side will be swept back To the P region.

This process causes the large reverse recovery current the moment a PN junction is Reverse biased.

As holes and electrons move back to the P and N regions,respectively,the depletion Region starts to respond the reverse voltage and widens to prevent further migration of holes and electrons and only small reverse current(leakage current) will pass Through the diode.

  • \$\begingroup\$ Ty :) That means the barrier potential itself drives the excess minority electrons back to the \$n\$ side. Why must this draw current from the battery ? The excess minority carriers on the \$p\$ side simply cross the junction and stay in \$n\$ side, right ? I feel this process happens with out a battery also. What is the use of battery here ? \$\endgroup\$ – AgentS May 5 '18 at 12:17
  • \$\begingroup\$ The battery allows for faster neutralization of electrons (migrated back to the N region) and holes (migrated back to the P region) to re-establish the depletion region again. \$\endgroup\$ – Mohammed Hisham May 5 '18 at 13:23
  • \$\begingroup\$ Of course,this process has to happen without the need of a battery to provide the reverse bias... If you simply remove the diode from the circuit after being forward biased then thermal equilibrium has to be established if this process depends on the existence of a battery then how thermal equilibrium could be established in this case? \$\endgroup\$ – Mohammed Hisham May 5 '18 at 13:27
  • \$\begingroup\$ so when the diode is physically removed from a forward biased circuit, the barrier voltage makes the diode reach equilibrium. I think I see now. Can I ask one more question.. (next comment) \$\endgroup\$ – AgentS May 5 '18 at 13:33
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    \$\begingroup\$ Np :) You've already explained in detail about the reverse recovery current. Thank you for that. Hope somebody explains how depletion region shrinks when reverse bias is removed..I'll wait some more time and maybe open a new question. \$\endgroup\$ – AgentS May 5 '18 at 14:25

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