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:
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\$\begingroup\$ PN junction: What does a battery have to do with that?! \$\endgroup\$– Marcus MüllerCommented Dec 10, 2021 at 15:17
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\$\begingroup\$ Did Do you mean why can't the battery swap its polarity to be forward-biased in respect to the PN junction? \$\endgroup\$– WintermuteCommented Dec 10, 2021 at 15:24
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\$\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\$– HearthCommented Dec 10, 2021 at 15:54
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\$\begingroup\$ The electric field of depeletion layer (its direction of action should promote e- movemnt to the right) \$\endgroup\$– DamstridiumCommented Dec 10, 2021 at 15:57
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\$\begingroup\$ gfycat.com/menacingfixedamericancrow \$\endgroup\$– SyedCommented Dec 10, 2021 at 17:52
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2 Answers
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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, 2021 at 15:49
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\$\begingroup\$ So Do all the charges in the N material fade away after biasing \$\endgroup\$ Commented Dec 10, 2021 at 15:52
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\$\begingroup\$ As limited charges get used up right \$\endgroup\$ Commented Dec 10, 2021 at 15:52
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2\$\begingroup\$ They redistribute towards battery positive and stop moving. That means they stop flowing and current rapidly becomes zero amps. \$\endgroup\$– Andy akaCommented Dec 10, 2021 at 15:53
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\$\begingroup\$ Ow so thats why the cause of current through the semi conductor diode is drifting of charges mainly right(after breakdown)? \$\endgroup\$ Commented Dec 10, 2021 at 15:55
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\$\begingroup\$ Drift current i mean to say \$\endgroup\$ Commented Dec 10, 2021 at 15:56
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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, 2021 at 15:18
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\$\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\$ Commented Dec 10, 2021 at 15:31
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\$\begingroup\$ Go through the battery? In which direction? Add a diagram \$\endgroup\$– pjc50Commented Dec 10, 2021 at 15:40
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\$\begingroup\$ I know this can look stupid but i am new to semi conductors so sorry if you feel the frustration \$\endgroup\$ Commented Dec 10, 2021 at 15:43
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\$\begingroup\$ Break the silence bro i am sorry i know but i had to just ask this \$\endgroup\$ Commented Dec 10, 2021 at 15:46