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Silicon junction diode

  1. Mar 10, 2014 #1
    1. The problem statement, all variables and given/known data

    In a silicon junction diode, the region of the planar junction between n-type and p-type
    semiconductors can be approximately represented as two adjoining slabs of charge, one
    negative and one positive. Away from the junction, outside these charge layers, the
    potential is constant, with a value of Vn in the n-type material and Vp in the p-type
    material. Given that the difference between Vp and Vn is 0.3 V, and that the thickness
    of each of the two slabs of charge is 10^-4 m, find the charge density in each of the
    two slabs, and make a graph of the potential V as a function of position through the
    junction. What is the strength of the electric field at the midplane?


    2. Relevant equations

    I using the relationship that the second derivative of the potential is equal to σ/ε.
    I have certain conditions. Putting the middle of the bar at x=0, the potential must be continuous through the middle and I have conditions set at the endpoints

    phi_1(-10^-4)=0
    phi_2(10^-4)=.3
    phi_1(0)=phi_2(0)
    σ1+σ2 = 0

    3. The attempt at a solution

    I get two expressions

    phi_1(x)= σ1/(2ε)x^2+Ax + B
    phi_2(x) = σ2/(2ε)x^2+Cx+D

    I found that A=C and B=D using the conditions, but now I have too many unknowns and not enough equations.
     
  2. jcsd
  3. Mar 10, 2014 #2
    Update 1. Found that B = .15
    by substituting σ1 for -σ2 and plugging in the values of 10^-4 and -10^-4 into the respective equations and added them.
     
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