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
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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