Timeline for In pn diode, how does an electron flow from p to n during reverse recovery phase
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 5, 2018 at 14:25 | comment | added | AgentS | 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. | |
| May 5, 2018 at 14:20 | comment | added | Mohammed Hisham | I'm afraid,I couldn't answer this question.May be someone will see the comments and answer it. | |
| May 5, 2018 at 13:38 | comment | added | AgentS | If the diode were in reverse bias, it will have a long depletion region. Also there will exist extra electrons in the valence band of the \$p\$ side due to capacitor action. Now if we physically remove this diode, how do these extra electrons on the \$p\$ side go back to the \$n\$ side ? (Please note that the electrons are in valence band, so they are not free. Also there is no electric field outside the junction. Also the electrons don't feel electric field even in the depletion region because they are in the valence band which is full) | |
| May 5, 2018 at 13:33 | comment | added | AgentS | 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) | |
| May 5, 2018 at 13:27 | comment | added | Mohammed Hisham | 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? | |
| May 5, 2018 at 13:23 | comment | added | Mohammed Hisham | 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. | |
| May 5, 2018 at 12:17 | comment | added | AgentS | 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 ? | |
| May 5, 2018 at 11:51 | history | answered | Mohammed Hisham | CC BY-SA 4.0 |