Why can't electrons leave the N side and enter the P side in reverse bias?

Question

Why can't electrons (excess electrons) of the N side of a semiconductor diode enter the P side via the battery in reverse bias mode of PN junction? like this:

Did Do you mean why can't the battery swap its polarity to be forward-biased in respect to the PN junction? — Wintermute, Dec 10 '21 at 15:24
They do, to a small extent. Then the growing depletion region produces enough of an electric field to balance it out, and the conduction stops. (except for a small leakage current) — Hearth, Dec 10 '21 at 15:54
The electric field of depeletion layer (its direction of action should promote e- movemnt to the right) — Damstridium, Dec 10 '21 at 15:57

score 3 · Answer 1 · answered Dec 10 '21 at 15:18

3

They can, if you apply a high enough voltage. We call that "breakdown". Normally the electrons do not have enough energy to cross the depletion region.

answered Dec 10 '21 at 15:18

Elliot Alderson

31,192
5
29
67

No I mean There are exess electrons on the n side right? So if we apply reverse biasing ie connect the positive terminal of the battery with diode then why cant electrons just go through the battery and reach the holes on other side? why go through the pain of going in depletion region.I mean the electric field will of the battery will make them move towards positive terminal – Damstridium Dec 10 '21 at 15:31
Go through the battery? In which direction? _Add a diagram_ – pjc50 Dec 10 '21 at 15:40
I have added it – Damstridium Dec 10 '21 at 15:42
I know this can look stupid but i am new to semi conductors so sorry if you feel the frustration – Damstridium Dec 10 '21 at 15:43
Break the silence bro i am sorry i know but i had to just ask this – Damstridium Dec 10 '21 at 15:46
Why don't the electrons just flow backward through the battery? Because -- battery? – TimWescott Dec 10 '21 at 16:04

score 3 · Accepted Answer · answered Dec 10 '21 at 15:49

3

With battery positive on N and battery negative on P you have a reverse biased PN junction. That forms a wide depletion layer that is basically non conductive unless the breakdown voltage is exceeded.

The majority carrier electrons in the N material are attracted to the battery positive terminal but, there is only a limited supply of them and this produces a really tiny glitch of charge movement into the battery that is unsustainable.

It's a similar story for the majority carriers in the P material: -

Image from here.

answered Dec 10 '21 at 15:49

Andy aka

434,556
28
351
777

So Do all the charges in the N material fade away after biasing – Damstridium Dec 10 '21 at 15:52
As limited charges get used up right – Damstridium Dec 10 '21 at 15:52
2

They redistribute towards battery positive and stop moving. That means they stop flowing and current rapidly becomes zero amps. – Andy aka Dec 10 '21 at 15:53
Ow so thats why the cause of current through the semi conductor diode is drifting of charges mainly right(after breakdown)? – Damstridium Dec 10 '21 at 15:55
Drift current i mean to say – Damstridium Dec 10 '21 at 15:56
clarify your point please. – Andy aka Dec 10 '21 at 15:56
After breakdown voltage is achieved ,in the depletion region Electrons get stripped off of their bonds right? and create electron hole pair generation which is the cause of current after breakdown voltage right – Damstridium Dec 10 '21 at 15:59
Is this the valid reason? because i read from a source that this is why conduction increases after breakdown is that correct? – Damstridium Dec 10 '21 at 16:01
Yup its on the internet thankyou so much – Damstridium Dec 10 '21 at 16:04
It's correct @Damstridium – Andy aka Dec 10 '21 at 16:22
Thankyou so much – Damstridium Dec 11 '21 at 00:16

Why can't electrons leave the N side and enter the P side in reverse bias?

2 Answers2