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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

 |  |  |  |  |  |  |  |  |  |  |  |
─●──●──●──●──+──+──-──-──○──○──○──○─
 |  |  |  |  |  |  |  |  |  |  |  |
─●──●──●──●──+──+──-──-──○──○──○──○─
 |  |  |  |  |  |  |  |  |  |  |  |
─●──●──●──●──+──+──-──-──○──○──○──○─
 |  |  |  |  |  |  |  |  |  |  |  |
─●──●──●──●──+──+──-──-──○──○──○──○─
 |  |  |  |  |  |  |  |  |  |  |  |

[● 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?

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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 appears a s a vacant site left by a electron. 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.

Can someone let me know why wouldn't the electron just beside the $(+)$ move towards it and neutralize everything......

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 the interface between the two types, there is a concentration gradient of charge carriers and hence they would diffuse across the junction and recombine, thereby depleting the electrons in the $n$- region and hols in the $p$-region. Hence the name depletion region.

Then why all charges just don't recombine?

Because the flow of carriers across the junction is stopped sometime. The ions are immobile and you can see positive ions in the $n$-side and negative ions in the $p$-side as discussed above.

Suppose our crystal is free from all defects. Then the electrons from the $n$ side has to encounter with the periodic negative potential of the ions in the $p$-side. It's like a potential well problem. Only those electrons having energy sufficient to overcome that potential barrier will diffuse across the junction and recombine. So all charge carriers will not diffuse across the junction and neutralize the polarity. To overcome this potential barrier we 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 why the electrons could neutralize the ions in the $n$-side. Of course, the electrons are mobile.

Consider an intrinsic semiconductor. At a given temperature, electron-hole pairs are formed in it and it is called thermal excitation. When you dope it and construct an $n$-type material, then the so called $5^{th}$ valence electron is loosely bound to the parent atom and can be activated at very low temperatures than that required for thermal activation of intrinsic carriers. Hence at ordinary room temperatures, obviously, the $5^{th}$ electron will be detached from the atom and the atom becomes ionized and the electron contributes to conduction.

So, for the ion to re-capture these electrons, and neutralize, it is no possible at ordinary temperatures. At ordinary temperatures, the electron energy level is very higher than the bound energy level of the valence electron. So such a possibility is no 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. A similar argument can be made at the $p$-side also, which is left as a homework.

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  • $\begingroup$ Will it make sense to say that there is a force that pulls electrons to holes to recombine which actually is higher than the force of its own ion, and then it continues until the ions exert too much force to stop this recombination? $\endgroup$ – Nishant Sep 18 '16 at 7:47
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    $\begingroup$ No. It is better that you may read about electron-hole recombination in some books on solid state physics like Solid State Physics by Charles Kittel $\endgroup$ – UKH Sep 18 '16 at 13:12
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The depletion zone in a pn-junction, i.e., positive charge density due to ionized donors in the n-region, negative charge density in the p-region due to ionized acceptors, forms to create an electric field to counteract the diffusion of electrons and holes from the respective regions of high concentration (n- and p-region, respectively). This happens so that the total current, which is the sum of drift current in the field and diffusion current due to carrier concentration gradients, becomes zero when no voltage is applied. You can also view it as a result of the difference in work functions for the n- and p- region which has to be balanced by the potential drop created by the charges in the depletion zone in order to prevent currents in equilibrium (zero applied voltage).

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Why am I missing the Coulomb's forces of a charge particle here? As far as I understand, the negative ions collect at the junction until there's enough force to repel the free electrons on the N side by Coulomb's force (both have like charges), this is the equilibrium (balanced) state.

Here's a good link: PN Junction

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