It seems like all the explanations for diode mechanics simply state that if N-type semiconductor is connected to V+ of the battery, the depletion region grows so current cannot pass.

It seems, however, that the few free electrons (minority carriers) on the P-type semiconductor can move across the layer and vice versa for holes on the N-type semiconductor. This creates the reverse saturation current. This is logical since the electrons move to high potential and holes move to low potential. However, why can't we produce a proper current from this? Electrons emitted from the battery can flow through the P-type semiconductor and across the depletion layer.

Is this main problem that there just aren't enough free electrons on the P-type side to make this happen?

  • $\begingroup$ I don't follow your reasoning here. Are you thinking that electrons from the negative terminal of the battery flow to the P-type region and remain valence band electrons there? $\endgroup$ Commented Nov 20, 2019 at 1:51
  • $\begingroup$ well, electrons in the P-type semiconductor (the few that are free) can move over to the N type semiconductor. The P-type semiconductor is connected to V- of the battery. Electrons should be able to flow from the V- side, through the P-type semiconductor, and across the depletion zone. Why can't this produce current? $\endgroup$ Commented Nov 20, 2019 at 2:04
  • $\begingroup$ Increase the voltage difference...In above some critical value electron getting more kinetic energy to cross depletion layer to create more higher current. $\endgroup$
    – baponkar
    Commented Nov 20, 2019 at 2:27
  • $\begingroup$ that would only be relevant if the battery were reversed $\endgroup$ Commented Nov 20, 2019 at 2:55
  • $\begingroup$ What you need to realise is that the depletion region is a steady-state equilibrium. These processes are continuously happening to maintain the balance. So the minority carriers are indeed pairing up. And that’s what causes the depletion zone. $\endgroup$ Commented Nov 20, 2019 at 3:14

1 Answer 1


Yes, you are right. The origin of the current in the reverse and the forward regime is the same.

In reverse regime (as you said), the minority carriers are pushed through the barrier. In a doped semiconductor, the minority concentration is very small compared to the majority concentration. Hence, a little number of charges can be pushed through the depletion region giving a small reverse current. Furthermore, when you increase the reverse voltage, the reverse current will saturate. This corresponds to the point when the minority carrier concentration is zero at the edges of the depletion region. From this moment, no more minorities can be pushed through the depletion region, thus limiting the current.

In the forward regime, it is the same. However, now, the majority carriers are pushed through the depletion region. The majority concentration is multiple orders of magnitude higher than the minority concentration. Therefore, many more carriers can be pushed through the depletion region. For the forward current to reach saturation, very high voltages are required.


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