Diode and doping on N and P regions

  • Thread starter Thread starter fisico30
  • Start date Start date
  • Tags Tags
    Diode Doping
AI Thread Summary
In a diode semiconductor, the number of extra electrons in the N region is typically larger than the number of holes in the P region due to doping levels, which can be controlled by the device designer. Majority carriers in the N region are electrons, while minority carriers are holes, which can exist in small quantities despite being less prevalent. Even at non-zero temperatures, both electrons and holes are present in both regions due to equilibrium conditions. The relationship between electron and hole densities is governed by an equilibrium constant equation, indicating that holes can be injected into the N region temporarily before recombining with electrons. Understanding these dynamics is crucial for the functionality of semiconductor devices.
fisico30
Messages
362
Reaction score
0
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
 
Physics news on Phys.org
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.
 
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.
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top