# Diode and doping on N and P regions

1. Mar 28, 2013

### fisico30

Dear forum,

in a diode semiconductor, the N region is rich in electrons. The P region is rich in holes.

Is the number of extra electrons in the N region equal, larger or smaller than the number of holes in the P region? If so, why?

Also, in the N region, it is said that the majority carriers are the electrons. I get that. The minority carriers are the holes....but there are no holes in the N region so how can there be any minority carriers at all?

Thanks,
fisico30

2. Mar 28, 2013

### marcusl

The number of electrons/holes is a function of doping and can be controlled by the device designer. The doping saturates at around 10^19 cc-1, which is called "extrinsic" doping. It gives the highest conductivity possible which is useful near the electrode connections.

You can inject holes into the n region, where they live for a time (the mean recombination time) before meeting electrons and annihilating. Transistor action is caused by minority carriers injected across the base region into the collector.

3. Mar 28, 2013

### DrDu

No, at non-zero temperature, even in equilibrium there are both electrons and holes present in both the n and p region. They fulfill a equilibrium constant equation $n_en_h=K(T)$. On the n side $n_e=n_{0e}-n_h\approx n_{0e}$ and thus $n_h\approx K/n_{0e}$ where $n_e$ is the number density of electrons $n_h$ the number density of holes and $n_{0e}$ the number density of donor atoms.
A similar equation holds for the p side.