Please can you check my concepts??

Under a normal PN junction diode before connecting any batteries we know that there comes a "potential barrier" due to intermixing of holes and electrons because of diffusion effects. But

Btw I also got a doubt on why current was 0 in a reverse bias and Referred to this sites answers so do check my understanding and let me know if you find any faults:
In a reverse bias we know that because of extensive concentration of holes towards the cathode(where the -ve terminal of battery connects the p part of PN junction diode) and electrons towards the anode which induces an opposing electric field which counteracts the external electric field which got induced because of the battery inside the diode resulting in 0 net electric field inside the diode because of which current stops.
Here now as there is no electric field inside the diode at this point and this process is increasing the diffusion forces because the holes are even more concentrated(near the anode)and so are the electrons(towards anode aka where +ve terminal of battery connects the diode) hence because of diffusion effect more extensive the electrons present in holes also move a little bit towards the left end and holes to the right end till their electric field counters the effect of increased diffusion.
Is all of this correct? I have attached pics to illustrate my thoughts:enter image description here

  • \$\begingroup\$ You might try cleaning up your diagram to make it readable and double check your written work. The P part of the PN junction is the anode so "the cathode(where the -ve terminal of battery connects the p part of PN junction diode)" is unclear. \$\endgroup\$ Commented Dec 9, 2021 at 11:55

1 Answer 1


Forgive me if I rehash things you have already learned but I notice you never mentioned drift in your description of the junction. Let's think about the junction in equilibrium: due to the different relative concentrations of holes and electrons in the P and N regions, you will get a diffusion current as holes move across the junction from the P side to the N side and electrons move from the N side to the P side. This movement of charge sets up an electric field "pointing" from the N side to the P side which pulls holes back to the P side and electrons back to the N side. This is your drift current.

With no applied voltage, the drift and diffusion currents will reach equilibrium and there will be no net flow of charge carriers across the junction, so no net current. If you apply a forward bias, the external voltage opposes drift and current flows as charge carriers diffuse across the junction and do not get pulled back. If you apply a reverse bias, the external voltage aids drift, but since there are very few free holes on the N side and very few free electrons on the P side, you do not see a large increase in current.

  • \$\begingroup\$ First of all Sir thankyou so much for the answer, Secondly, cant we say that electrons from n side can travel via the battery(under influence of electric field produced by the battery) and end up at the holes on the p side thereby filling up the holes? What's stopping the exess electrons on n side reaching the p side, because of which we say that conduction should happen only because of drift in a semi conductor nothing else.This is what is bugging me \$\endgroup\$ Commented Dec 10, 2021 at 5:23
  • \$\begingroup\$ Sir I also believe that drift current is caused when minority carrier from both n and p sides enter depletion region ''the pulling back of holes cannot be called drift current" it is the drift of electrons from p side(which are in minority) and holes from n side which are the cause of drift current \$\endgroup\$ Commented Dec 10, 2021 at 8:58
  • \$\begingroup\$ Sir am I able to explain you my point??? \$\endgroup\$ Commented Dec 10, 2021 at 15:03

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