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Due to the applied voltage, electrons from n-side cross the depletion region and reach p-side (where they are minority carries). Similarly, holes from p-side cross the junction and reach the n-side (where they are minority carries). This process under forward bias is known as minority carrier injection. At the junction boundary, on each side, the minority carrier concentration increases significantly compared to the locations far from the junction.

Due to this concentration gradient, the injected electrons on p-side diffuse from the junction edge of p-side to the other end of p-side. Likewise, the injected holes on n-side diffuse from the junction edge of n-side to the other end of n-side (Fig. 14.14). This motion of charged carriers on either side gives rise to current. The total diode forward current is sum of hole diffusion current and conventional current due to electron diffusion. The magnitude of this current is usually in mA. From

In the big picture, I understand that forward biasing a p-n junction makes it more conducting and allows current through but the mechanism makes no sense to me. In the quoted passage above, it is said that due to applied voltage, the negative charge carriers from n side move to the p side which makes the middle depletion region larger. If this happens, wouldn't the resistance keep increasing and net current become zero?


Related questions:

Working of PN junction

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Wait, first one comment: when you forward bias a junction the depletion region gets smaller, not larger. I am not really sure if you really meant to write that.

About the mechanism of injection... without going into the math. I will talk about electrons here but a complementary and symmetric discussion can be done with holes.

  • First, let me talk about the equilibrium condition at $V=0$. You have many (a lot many) more free electrons in the n-type material rather than in the p-type. The obvious expectation is that random motion (diffusion) will spread the electrons, and thus transfers a lot of them from N to P... and indeed that happens until there is such a negative charge build-up on the P side edge (and symmetrically a positive one on the N side edge) that you have the formation of an electric field that balances this trend. In the end, the electric field is caused by the ionized donors/acceptors in the depletion region, you can imagine them as the leftover of this equilibration process. So the equilibrium is reached: you have many more electrons on the N side but they cannot diffuse to the P side because there is a barrier (the integral effect of the electric field) preventing them to do so.

  • When you directly bias your diode you are making it "more convenient" for electrons to stay on the P side with respect to the N side. This is in practice simply reducing the electric field in the depletion region and the height of the barrier cited above. So the barrier is not able anymore to keep electrons in N from flooding the p-type region.


[Edit] Addition after comment, on why the barrier gets thinner when we forward bias the junction.

Ok, I see the tricky argument: we transfer even more electrons from N to P, we should thus expect even more charge build-up and a wider (and higher) barrier. So let's see what is wrong in this line of thought.

  1. First note that there is some sort of contradiction here: (1) the bias reduces the barrier, (2) electrons now overcome the barrier and move from N to P, (3) this increases the barrier... looks like a sort of self-terminating or self-contradicting process. You probably already noticed.

  2. The argument for the increased barrier has some merit, but it fails because - differently from the equilibration process imagined above - now we have a battery and a circuit with some current that can flow: electrons are transferred from N to P, the diffuse there, eventually recombine and current will be carried by holes... and in the end some electrons will exit into the wire on the left side of the device. The details of all these processes are not so obvious but the key observation is that charge is flowing. Sure, some electrons are going from N to P but also some electrons are exiting P to go into the outer wiring. Who will win? Will the P region get more or less negatively charged? Not obvious to tell from this argument.

  3. A useful evocative picture could be to look at the junction region as a capacitor (there are some similarities indeed, and variable capacitors can be built using diodes: see varicaps), when you apply the direct bias you are charging that capacitor and increasing the charge on the P side and decreasing it on the N side. This simply means that electrons on the N side and holes on the P side are now able to reach closer to the junction and leave less space to ionized donors/acceptors.

  4. Injected electrons could still be a source of confusion. I see here a possible source of doubts: in the end we are injecting electrons from N to P, so it looks completely natural to expect that the P edge gets MORE negatively charged. Well thought but... no, that is not what happens. I am not sure I can help much intuition here, and I do understand it is confusing. I can just tell you that this is not the case. Electrons that are injected and diffuse deep into in the P type region before recombination (quite deep, typically for a length which is orders of magnitude larger than the depletion region), are not associated with any significant net charge build up. What happens in practice is that the negative charge of the injected electrons is neutralized by the (massive amount of) holes that populate the P type region, which will react to any minuscule amount electric field and re-establish local charge neutrality on a very short time scale. Admittedly, I think this is one of the most subtle aspects of PN junctions.

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  • $\begingroup$ If the n side electrons move to p side, then wouldn't the barrier become wider? Or am I missing something $\endgroup$ Commented Jun 5, 2021 at 14:23
  • $\begingroup$ Mh, ok, I think I see your point. The short answer is that, no, it is not getting wider but thinner. But I see: we just come from saying that electrons go from N to P until a barrier builds up and stops this flow... now we transfer even more electrons => the barrier gets even larger. However, now we have a battery and a circuit, electrons go from N to P but also proceed and exit the P region into the outer wire. It is not obvious if/where charge will build up. Rather, you apply a bias to the PN and it will fall on the most resistive region, which is the junction => smaller barrier. $\endgroup$
    – Ste
    Commented Jun 5, 2021 at 14:45
  • $\begingroup$ Could you explain what you mean by the most resistive region? @Ste I think your answer is the one I was looking for but I think some more detail on why what I said is wrong would help. $\endgroup$ Commented Jun 5, 2021 at 15:15
  • $\begingroup$ I added more details above in the answer! By "most resistive region" I just mean that the depleted region has a very little amount of free carriers so its conductivity $\sigma = ne\mu$ is very small, because $n$ is so minuscule. $\endgroup$
    – Ste
    Commented Jun 5, 2021 at 15:57
  • $\begingroup$ Thank you @Ste I think I got the basic idea: Theer is mroe chance to cross between the gap than to build up the gap. I also like to say that I appreciate you taking out the time to actually understand what I was trying to ask ^^ $\endgroup$ Commented Jun 5, 2021 at 17:15

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