A depletion region is formed by electron hole combination at the junction and it creates positive ions on the $N$ side and negative ions on the $P$ side.
Can someone let me know why wouldn't the electron just beside the $(+)$ move towards it and neutralize everything. Aren't electrons mobile? Electrons are mobile so it sounds odd that such a positive ion region can be created? Is it because of the effect of $(-)$ which is farther to the right? The immediate $(+)$ should take precedence right? And that also leads me to this question, if there is a $+-$ region like this, would there be any electric field felt outside that region?!
Strangely, on the other side it all makes, holes besides $(-)$ ions which lack electrons, have no electrons to pull and there is a negative charged region that won't allow any electron migration.
I would have expected a symmetrical behavior but that doesn't seem to be so.
N side P Side
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─●──●──●──●──+──+──-──-──○──○──○──○─
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─●──●──●──●──+──+──-──-──○──○──○──○─
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─●──●──●──●──+──+──-──-──○──○──○──○─
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─●──●──●──●──+──+──-──-──○──○──○──○─
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[● is a valence electron and ○ is a hole]
I am just trying to get a better feel of depletion region because it gives me a way to visualize or model things in that region. It all sounds very easy to understand but when you go deeper it doesn't seem so obvious. Probably I have to go into band theory and such to really see what is happening?
Best Answer
An $n$-type semiconductor is made up of impurity atoms that could donate electrons to the lattice, which causes conduction. Hence they are called donor atoms. Since they donate electrons, these donor atoms become positively ionized. On the other hand, in a $p$- type semiconductor, the impurity atoms are called acceptor atoms, as they create a deficiency in the sea of electrons and hence appear as vacant sites unoccupied by electrons. These entities are called holes and are assumed to be positive. Hence the acceptor atoms are negatively ionized as these vacant sites have a high probability of capturing electrons nearby.
First of all, don't think that a $pn$ junction is not formed by taking individually a $p$-type and an $n$ type and combining them. They are developed on a single semiconductor crystal. Now, the answer to your question.
The concentration of electrons is higher than that of holes in the $n$-type semiconductor and reverse for the holes. In other words, electrons are majority carriers in an $n$-type semiconductor and holes are majority carriers in a $p$-type semiconductor. So, at the junction (or interface) between the two types, there is a concentration gradient of charge carriers such that they would diffuse across the junction and recombine, thereby depleting the electrons in the $n$- region and holes in the $p$-region. Hence the name depletion region.
Then why don't all charges just recombine?
Because the flow of carriers across the junction is stopped at some point. The positive ions in the $n$-side and negative ions in the $p$-side, as discussed above, are immobile (unlike the majority carriers).
Suppose our crystal is free from all defects. The electrons from the $n$-side will periodically encounter the negative potential of the ions in the $p$-side. It's like a potential well problem: only those electrons having sufficient energy to overcome the potential barrier that the negative ions create for them will diffuse across the junction and recombine. So not all charge carriers will diffuse across the junction and neutralize the polarity. To overcome this potential barrier, we'd need to supply energy to the electrons, which is what we do by forward biasing the $pn$ junction. You need to overcome the electric field in the junction, by an external field.
So, the next question would be: then do the electrons in the $n$-side neutralize the positive ions there?
Of course; the electrons are mobile.
Consider an intrinsic semiconductor. At a given temperature, electron-hole pairs are formed via thermal excitation. When you dope it and construct an $n$-type material (an extrinsic semiconductor), then the so called "$5^{th}$" (extra) valence electron of each atom is loosely bound to their parent atom and can be excited to higher energy states at much lower temperatures than that required for the intrinsic carriers. Hence at ordinary room temperatures, the "$5^{th}$" electrons of each atom will detach from their atoms (becoming unbound) and the atoms will become ions. These detached (unbound) electrons will then contribute to conduction.
So, in other words, at ordinary temperatures it is not possible for the positive ions in the $n$-side to "re-capture" these electrons and neutralize back to atoms. At ordinary temperatures, the electron energy level is much higher than the bound energy level of the valence electron. So such a possibility of "re-capturing" is not there. But that doesn't mean the electrons will not interact with the ions; the electrons have a probability of scattering by the positive potentials, moving from one positive ion to the next. A similar argument can be made at the $p$-side also, which is left to the reader as an exercise.