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Electric field inside semiconductor diode

  1. Oct 2, 2011 #1
    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: [itex]Q_{tot} = -Ad_pn_A+Ad_nn_D=d_nn_D-d_pn_A[/itex]

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

    Is there some misunderstanding? And please forgive my bad language.
     
    Last edited: Oct 2, 2011
  2. jcsd
  3. Oct 2, 2011 #2

    phyzguy

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    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.
     
  4. Oct 3, 2011 #3
    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).
     
  5. Oct 3, 2011 #4

    phyzguy

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    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).
     
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