# Electric field inside semiconductor diode

1. Oct 2, 2011

### dikmikkel

1. The problem statement, all variables and given/known data
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

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

3. 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

Last edited: Oct 2, 2011
2. Oct 2, 2011

### 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.

3. Oct 3, 2011

### 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).

4. Oct 3, 2011

### 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).