Why does depletion region have a charge? Why isn’t a hole neutral? What balances the barrier potential? What makes the barrier potential?
In: Physics
When you put together the two halves of the junction together you allow electrons and holes from either side to flow to the other by diffusion, which causes a current. The steady state of the system is one in which there is no external source of energy so we can be sure that in the steady state there can be no current which means something has to balance out the diffusion.
The something that balances it out is the buildup of an electric field which drives the charge carriers in the opposite direction of their diffusion, this electric field is the result of the nuclei of the dopants being left electrically imbalanced because of the diffusion, and thus the dopants are ions, at the N side you are left with positively charged dopants and on the P side you have negatively charged ones.
When we talk about the depletion region having charge, it is usually in reference to the charge of the ionised dopants. Although there are still charge carriers in this region it is common to disregard them in most cases since their concentration is so low compared to other regions.
As for why holes are positive, take a look at this lattice comprised of Silicon atoms doped with Boron: https://userscontent2.emaze.com/images/e0864643-cfba-4772-8c39-a9e71fad1094/897109c8-8679-4c91-a99b-fa4dc68c5d9dimage6.png
If the Bororn was instead Silicon, it would be able to share it’s 4 electrons with nearby atoms, but because it has one less electron in it’s final shell, it can’t form this bond and nearby electrons from the silicon can rush in and fill up the “hole” in the lattice, but because that electron left a silicon atom, that atom is now also left with a hole which is then filled and so on and so on.
If you think about it you can see that the electrons are moving towards the boron atom which initially had the missing electron and this is in essence the same as saying a positive charged hole propogated away from the boron atom.
I feel like this is super long so I’ll try to summarize.
1. It has a charge due to ionised dopants left without charge carriers to balance them
2. Holes are absences of electrons moving opposite in direction from them
3. The barrier potential is balanced by the diffusion of charge carriers (the opposite actually)
4. The need to limit diffusion is what creates the potential barrier
A pn junction is a diode. The p part is a semiconductor (Si/Ge) doped with a metal like aluminium, which is electron deficient in its last sub shell, thus a p semiconductor has ‘holes'(empty spaces made by electrons moving away temporarily)
The n part is a semiconductor doped with a metal like arsenic(As) that has a lone electron in its valence shell. So a n semiconductor has ‘electrons’.
Before going to junctions, let’s see how these p and n semiconductors work. P semiconductors, as mentioned above, has more holes compared to electrons, so the major carrier of p semiconductors are ‘holes’. In n semiconductors, there are more electrons than holes, so the major carriers are electrons. Since electrons are more mobile than holes, n semiconductors are more sensitive.
When a p and a n is connected together, and a voltage is supplied to their ends, holes and electrons move. They form a small voltage barrier between them and for Si this value is about 0.7V and for Ge this value is about 0.3V. Generally there are two ways of supplying voltage. If you connect the + end to the p and – to n, then the pn junction is forward biased. If the + is connected to the n and the – to p, then the pn junction is reverse biased.
In forward bias, if the voltage provided is high (just more than 0.3V for Ge and 0.7V for Si), it will break the voltage barrier inside the junction, and the diode will let current flow through it.
In reverse bias a higher voltage supplement (exceeding 0.3V or 0.7V) will broaden the voltage barrier inside the junction, until a certain point, but this barrier eventually breaks (at a very large voltage), and a large amount of current will flow through the diode (for a very small period of time), and it will burn out in an instant.
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