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For p-n junction I know that electrons and holes will diffuse to opposite directions forming negatively and positively charged regions. But do these diffusion take place extremely fast that they cannot diffuse further a certain distance from the junction as shown in the above figure? if diffusion occurs slowly then aren't these carriers supposed to diffuse more further on? I mean shouldn't excess negative charges remain diffused all over the p-type semiconductor instead of being constrained within a certain distance of the junction? or is it that the above image is just a analogy to show that such charges form in those region but actually these charges diffuse all over the semiconductor and all I know about this is just wrong? Could it be that at the beginning some electrons do diffuse through out the p-type semiconductor and start to oppose other electrons entering into the p-type to not to diffuse further on and thus create a short ranged concentrated negative charge density near the junction?

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2 Answers 2

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The reason for this behavior is that the electrons are attracted by atomic nuclei. When an electron goes from the n-doped in the p-doped material, it leaves behind it an unshielded positive charge. Therefore, the migration of electrons across the junction gives the n-doped material a total positive space charge. This charge pulls the electrons as close to the n-donated region as they can be. Hence they spread over a thin sheet on the p-side of the junction.

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  • $\begingroup$ Didn't these electrons leave their atoms to fillup the holes of the p-doped material? i mean isn't the tendency of pulling electrons by holes apparently more than the nucleus do? Otherwise why would electrons leave a low energized condition to fill other holes which would raise a bit unstability (because of anion fotmation)? $\endgroup$
    – MSKB
    Commented Sep 13, 2021 at 11:41
  • $\begingroup$ Well, the electrons can actually do both: They can fill holes in the p-doped material and at the same time stay close to the positive charges left behind in the n-doped region. They just have to stay near the junction on the p-side - instead of diffusing all over the material. $\endgroup$
    – Larss 96
    Commented Sep 13, 2021 at 11:45
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The question is a little bit confusing because I would think you’d expect the contrary: the faster the diffusion, the larger the depletion region (DR for short). So I am not sure what your line of thinking is, therefore, I will take a very general approach, and remind some stuff that I’m can’t be certain you’ve heard about.

  Mostly, I want to start by making sure that, if you admit the situation depicted in your drawing as a starting point, you understand why it forms a barrier to further diffusion. As drawn, you have negative net charges on the p-side of the DR and positive charges on the n-side. I assume you know that such a charge distribution will create an electric field pointing form + to -. Now let’s say that an electron from the n-side wants to take a hike on the p-side. The electric field as oriented will push it away (electrons are negatively charged). Same goes for holes from the p-side who would dare try to diffuse to the n-side. So the point to remember: having a depletion region blocks further diffusion.  

Other reminder before answering: how the DR forms. When a free electron arrives in the p side from the n side, it encounters sooo many holes that it immediately wanna get intimate and recombine. This additional electron from the n-side is now stuck in the p-side and therefore starts to build up the negative charge that you see on your drawing.  

Those being established, you should start to see that the size of the DR is a trade-off between the rate of two processes: rate of recombination and rate of diffusion. Imagine if you have a super fast diffusion and really slow recombination: electrons would have time to sprint wherever they want before they recombine, and therefore would recombine anywhere in the p-zone, and your idea would be true (charges all over, in case of super fast diffusion). On the contrary, if recombination happens instantly, then the electrons/holes recombine right at the gate of their new region. This recombination process creates charges and stops further diffusion of other charges as established above. Therefore, if diffusion is very slow (or recombination very fast), the depletion region is smaller. In real life, recombination happens on much smaller timescales than diffusion, so the DR is always pretty thin. But the exact size varies on the materials used and does depend in good part on the tradeoff discussed: slow diffusion, thin DR, fast diffusion, larger DR.

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  • $\begingroup$ "This additional electron from the n-side is now stuck in the p-side and therefore starts to build up the negative charge that you see on your drawing." Does those electron remain stucked at wherever they hole was which they filledm? Don' they move again to another hole? $\endgroup$
    – MSKB
    Commented Sep 13, 2021 at 11:43
  • $\begingroup$ They can (although we find it usually more convenient to think of it as if the holes are moving when they are the majority carriers, it is indeed always a movement of electrons). However, the electrons will not wander further from the junction, because of the field that we mentionned, which tends to pull them back $\endgroup$ Commented Sep 13, 2021 at 11:49
  • $\begingroup$ Okay now I see...... I thought that electrons get greedy due to the presence of loads of holes in the p-type so they almost ignores the attraction of nucleus and diffuses to all over the p-type $\endgroup$
    – MSKB
    Commented Sep 13, 2021 at 11:51
  • $\begingroup$ But it seems like even thougj electrons wants to fill those holes but due to a bunch of electron moving from n-type to p-type at the same time, the net attraction force of positive charges increases on single electron causing it to not diffuse further. I guess I misunderstood because of thinking transition of a electron at a time which does not happen actually $\endgroup$
    – MSKB
    Commented Sep 13, 2021 at 11:54
  • $\begingroup$ Electrons diffusing one by one would not change anything to the process described above (in fact, it is just one way to make the rate of diffusion slower, i.e. having a thinner DR) $\endgroup$ Commented Sep 13, 2021 at 11:58

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