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:enter image description here

  • \$\begingroup\$ PN junction: What does a battery have to do with that?! \$\endgroup\$ Dec 10, 2021 at 15:17
  • \$\begingroup\$ Did Do you mean why can't the battery swap its polarity to be forward-biased in respect to the PN junction? \$\endgroup\$
    – Wintermute
    Dec 10, 2021 at 15:24
  • \$\begingroup\$ 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) \$\endgroup\$
    – Hearth
    Dec 10, 2021 at 15:54
  • \$\begingroup\$ The electric field of depeletion layer (its direction of action should promote e- movemnt to the right) \$\endgroup\$ Dec 10, 2021 at 15:57
  • \$\begingroup\$ gfycat.com/menacingfixedamericancrow \$\endgroup\$
    – Syed
    Dec 10, 2021 at 17:52

2 Answers 2


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: -

enter image description here

Image from here.

  • \$\begingroup\$ So Do all the charges in the N material fade away after biasing \$\endgroup\$ Dec 10, 2021 at 15:52
  • \$\begingroup\$ As limited charges get used up right \$\endgroup\$ Dec 10, 2021 at 15:52
  • 2
    \$\begingroup\$ They redistribute towards battery positive and stop moving. That means they stop flowing and current rapidly becomes zero amps. \$\endgroup\$
    – Andy aka
    Dec 10, 2021 at 15:53
  • \$\begingroup\$ Ow so thats why the cause of current through the semi conductor diode is drifting of charges mainly right(after breakdown)? \$\endgroup\$ Dec 10, 2021 at 15:55
  • \$\begingroup\$ Drift current i mean to say \$\endgroup\$ Dec 10, 2021 at 15:56

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.

  • \$\begingroup\$ 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 \$\endgroup\$ Dec 10, 2021 at 15:31
  • \$\begingroup\$ Go through the battery? In which direction? Add a diagram \$\endgroup\$
    – pjc50
    Dec 10, 2021 at 15:40
  • \$\begingroup\$ I have added it \$\endgroup\$ Dec 10, 2021 at 15:42
  • \$\begingroup\$ I know this can look stupid but i am new to semi conductors so sorry if you feel the frustration \$\endgroup\$ Dec 10, 2021 at 15:43
  • \$\begingroup\$ Break the silence bro i am sorry i know but i had to just ask this \$\endgroup\$ Dec 10, 2021 at 15:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.