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As we learn that diodes and transistors are made by some materials, and they are called P or N according to having extra electrons or holes.

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If there are extra electrons on n-type silicon (doped), why don't they just flow through ground maybe and N part becomes neutral. And same thing is P side as well. Is there anything special about fabrication of them which keeps those holes and electrons in the package and not let them to be neutral?

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    \$\begingroup\$ In fact a neutral region does form in the junction betwen the P and N material. The diode finds a lower energy state if some of the electrons from the N side jump over and fill holes in the P side. When a diode is reverse-based, this neutral region grows wider, but when it is forward-biased, it disappears. There is a limit to the spontaneous formation of the neutral region because it requires a build up of charge. Every electron that jumps across to fill a hole leaves behind a net positive charge which acts on it to pull it back. \$\endgroup\$
    – Kaz
    Apr 16, 2013 at 23:05

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If there are extra electrons on n-type silicon (doped), why don't they just flow through ground maybe and N part becomes neutral.

The key thing that's glossed over in your description of the pn junction, is that when we say there are "extra electrons" in the n-region, we're not mentioning where they come from. These electrons come from "donor" impurities in the silicon crystal. When these impurity atoms give up their electrons, they become positively charged. However these positive charge sites aren't mobile so they don't contribute to conduction through the diode.

The balance between the mobile free electrons and the immobile positively-charged donor sites actually gives the n-region a neutral charge overall.

Similarly, in the p-region, we have mobile positively charged holes in balance with immobile negatively charged acceptor impurity sites.

If all the "extra electrons" were to drain out of the n-side of the pn junction, the n-region wouldn't be left neutral, it would be left with a large postive charge from the remaining donor sites.

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  • \$\begingroup\$ What I was thinking was if I connect positive side of battery to N-type side, and negative side to P-type, all electrons flow into battery from N-type, and electrons from negative side of battery would fill all holes in P-type, so the diode would act like normal wire. But as far as I understand there is some special things about silicon atom and doping process which does not allow this. \$\endgroup\$
    – tcak
    Apr 17, 2013 at 21:30
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    \$\begingroup\$ This is sort of what happens. But after you take (some of) the free electrons from the n-side and fill (some of) the holes on the p-side, there's still no way for current to flow from the p-side to the n-side. So there's no complete circuit and current is blocked. Which is exactly what you expect for a revesrse-biased diode. But what you're imagining does show why a reverse biased pn diode acts like a capacitor. \$\endgroup\$
    – The Photon
    Apr 17, 2013 at 21:49
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A doped semiconductor doesn't have an excess or a deficit of electrons such that the semiconductor has net electric charge.

Rather, for example, an n-type semiconductor has an excess of electrons in the conduction band, i.e., mobile electrons, compared to a pure semiconductor at the same temperature. Essentially, the impurity makes the semiconductor a good conductor.

Just as you don't expect the conduction band electrons in copper to flow to ground, you shouldn't expect that of a doped semiconductor.

Having said that, there are other factors to consider such as diffusion current but that is, ahem, beyond the scope of this answer.

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There are a couple of extra things going on. The net charge in P and N regions is still zero (at least when they are not part of a PN junction) so electrons and "holes" in P and N type silicone are not extra, they're just not held onto very tightly and are easily displaced.

A funny thing happens when you make a PN junction though. Some of the electrons near the junction move to fill in the holes. This is referred to as the depletion region. Wikipedia has a better explanation than I could give so here's the link: https://en.wikipedia.org/wiki/Depletion_region

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They dont go to ground because ground is negative and the N type is also negative.

When electrons move from the N-type to the P-type, it creates holes in the N-type (the spots where electrons used to be). Since the n-type is usually connected to a negative side, that means there are more electrons willing to fill up those holes quickly. Then those electrons see the spaces in the p-type ( holes), and they rush to them, creating holes in the n-type...and this repeats.

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  • \$\begingroup\$ As far as I know ground is neutral, not negative. Therefore, yes, extra electrons should flow into ground, so it can be neutral. But it is not happening. \$\endgroup\$
    – tcak
    Apr 16, 2013 at 17:25
  • \$\begingroup\$ When I say negative, I was basing it off your description where electrons go to ground. The potential of ground is considered 0V. If your circuit contains all positive voltages, then ground is considered negative. I should have been clear on that. \$\endgroup\$
    – efox29
    Apr 16, 2013 at 19:04

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