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In a p-n junction diode, if electrons leave the n type semiconductor, and go to the p type semiconductor and this contributes to the formation of depletion layer. But my doubt is, if we are getting a positive charge on the n type semiconductor's side of depletion layer, and a negative charge on the p type semiconductor's side of depletion layer, wo why don't the electrons from n type semiconductor just neutralise the positive charge? I mean electrons are major charge carriers in n type semiconductor, so why don't they just move to neutralize the positive charge? I tried looking on YouTube, but all the explanations I found on the topic, didn't go into this question. I asked my physics teacher about this, but he simply ignored me. I know that this won't happen, but I am unable to explain why it won't happen.

An image to make my query a bit more clear-https://en.m.wikipedia.org/wiki/File:Pn_Junction_Diffusion_and_Drift.svg

--In this, why don't the electrons from n type semiconductor, neutralize the "uncovered space charges" ?

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The main reason is that an electric field is produced in the depletion region.

Once the ion cores are exposed, the positive ion cores on the n-type side and the negative cores on the p-type side, an electric field is formed. This electric field in the depletion region tends to keep carriers in their respective sides, i.e. it keeps the electrons in the n-type region and the holes in the p-type region, as there is now an energy barrier to crossing the junction. The field created prevents carriers from simply 'neutralising' the charge on the ion cores

Some carriers can have enough energy (without external input) to get into the depletion region, but as a result of the electric field they are rapidly swept away.

I hope this helps a bit.

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  • $\begingroup$ But like in diagrams of my textbook, the positive charges are shown immediately next the n type semiconductor on side of depletion layer. So the negative charges are on the other side of n type semiconductor (as in if we say the n type semiconductor is on the right of positive charges in a diagram, then the depletion layer is on the left), and they cause the development of an electric field from positive to negative charges. But this electric field in within the depletion layer, so how can it prevent the majority charge carriers of n type semiconductor, to neutralize the positive charge? $\endgroup$
    – Krish Vasa
    Nov 26, 2020 at 12:39
  • $\begingroup$ Just for clarification do you mean electrons neutralising a postive charge in the n-type region? Rather than the exposed cores left in the depletion region? $\endgroup$
    – Sam Pering
    Nov 26, 2020 at 12:42
  • $\begingroup$ I think the way I framed my comment is a bit confusing. I edited the question to attach an image. I think that'll make answering my doubt more easier. $\endgroup$
    – Krish Vasa
    Nov 26, 2020 at 13:02
  • $\begingroup$ I would say that diagram is not a particularly helpful one - those positive and negative charges they use in the illustration are the depletion region (in which case my original answer stands), to make it clearer they should probably change the colour of that region. I find this diagram/animation quite good. pveducation.org/pvcdrom/pn-junctions/formation-of-a-pn-junction - that website as a whole does a decent job of explaining some of the basics (I tend to recommend it quite a lot) $\endgroup$
    – Sam Pering
    Nov 26, 2020 at 13:06
  • $\begingroup$ Like in this link, they talk about positive ion cores, so why don't electrons from n type semiconductor neutralize these positive cores? $\endgroup$
    – Krish Vasa
    Nov 26, 2020 at 13:10

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