Electric field inside semiconductor diode

[dikmikkel]

Homework Statement

A semiconductor diode is made of acceptor and donor dotted regions. In equilibrilum there is a volume charge region at about x = 0. In that region there is no free charge carriers, only single charged. In x<-dp and x > dn, there is no charge at all. The surface area of the diode is A. And the semi conductor material has the permetivity ε. The volume charge distribution is for p: nA and for n: nD

1) calculate the total charge in the region -dp<x<dn and find the relation betweeen dp and dn.

2) Assume you know dn and express the electric field E(x) in therms of dn and the other parameters in -dp<x<dn

Homework Equations

Gauss Law for Dielectrics.
D = εE (Assuming Linear homogeneous material)

The Attempt at a Solution

1) I've integrated over the entire volume and found: $Q_{tot} = -Ad_pn_A+Ad_nn_D=d_nn_D-d_pn_A$

2) I read from the assingment, that the free charge is 0 in the region -dp<x<dn, and therefore : $\oint\limits_S\vec{d}\cdot d \vec{a} = 0, \oint\limits_S\epsilon \vec{E} d\vec{a}$ = 0

Is there some misunderstanding? And please forgive my bad language.

 

[phyzguy]

Yes, you have a misunderstanding. It says there is no charge in the region x<-dp and x>dn, but you are asked to find the E field in the region -dp<x<dn, where there is charge. The E-field is certainly not zero in this region.

 

[dikmikkel]

ahh, i see. So i should solve possions equation in the charged region the positive and negative charged seperatly? When i do that i get 2 linear functions in cartesian coordinates(symmetry in y,z directions).

 

[phyzguy]

Because of the symmetry you know E points in the x direction, so it just becomes a 1D problem, dE/dx = rho/epsilon0. Rho is a constant in each of the two regions, so you should be able to solve for E(x).