# Charge density in an abrupt p-n junction

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1. Mar 19, 2017

### lampCable

1. The problem statement, all variables and given/known data
In an abrupt p-n junction we consider the junction between one side p-doped with $N_A$ acceptor atoms and another side n-doped with $N_D$ donor atoms. Initially the chemical potential is different in the two sides, but thermal equilibrium requires that the chemical potential be uniform. This causes the annihilation of electrons and holes, from the n side to the p side, giving rise to a region called the depletion zone where there are very few charge carriers. In the depletion zone, we consequently get a nonzero charge density as shown in the figure below. What I do not understand is why the magnitude of the charge density becomes $qN_A$ and $qN_D$ for the p side and n side, respectively? Why do we not get for example $qN_A/2$ and then just twice the distance $-x_p*2$?

2. Relevant equations

3. The attempt at a solution
I understand that my reasoning is incorrect. If I would continue to divide $N_A$ I would eventually end up with the initial situation, and the chemical potential would not be uniform. So I need to annihilate enough electrons and holes in order to create a uniform chemical potential, but how do I know that this happens when $\rho = -qN_A$ and $\rho = qN_D$?

2. Mar 19, 2017

### lampCable

Turns out that the answer to this question was simple. The charge density at any point is $\rho=e(p+N_D-n-N_A )$. But as I mentioned in the question, in the depletion layer $n=p=0$. And on the p side we have $N_D=0$ since there are no donor atoms there, which gives $\rho=-eN_A$. Similarly on the n side we have $N_A=0$ since there are no acceptor atoms there, which gives $\rho=eN_D$. (Note that $N_D$ and $N_A$ refers to the ionized number of donor and acceptor atoms, respectively.)