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).

 

[paranormal]

Hi, thank you for providing the context and equations for your homework problem. I would approach this problem by first defining and understanding the terms and variables involved. Based on the given information, it seems that the semiconductor diode has a volume charge region at x = 0, and no charge in regions x < -dp and x > dn. The surface area of the diode is A, and the material has a permitivity of ε.

To answer the first question, we need to calculate the total charge in the region -dp < x < dn and find the relation between dp and dn. From the given information, we know that the volume charge distribution is n_A for p-type and n_D for n-type. Therefore, the total charge in the region can be calculated as Q_{tot} = -Ad_pn_A + Ad_nn_D = d_nn_D - d_pn_A. This means that the total charge is dependent on both dp and dn, and we cannot determine the relation between them without more information.

For the second question, we are asked to express the electric field E(x) in terms of dn and the other parameters in the region -dp < x < dn. To do this, we can use Gauss' Law for Dielectrics, which states that D = εE for a linear homogeneous material. In this case, we have a semiconductor material with a permitivity of ε, so we can write the equation as D = εE. Since there is no free charge in the region -dp < x < dn, we can use Gauss' Law to find the electric field in that region. By integrating over the surface area, we can write the equation as: \oint\limits_S\vec{d}\cdot d \vec{a} = 0 = \oint\limits_S\epsilon \vec{E} d\vec{a} = \epsilon E \oint\limits_S d\vec{a}. This means that E = 0 in the region -dp < x < dn.

In conclusion, based on the given information, we can calculate the total charge in the region -dp < x < dn, but we cannot determine the relation between dp and dn without more information. Additionally, we can express the electric field in terms of dn and the other parameters in the region -dp < x < dn, and we find that the electric field is 0 in that region. I hope this