# Silicon junction diode

1. Mar 10, 2014

### Slightly

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. Mar 10, 2014

### Slightly

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.