Linearly graded pn junctions

[kn336a]

Hi, my question is:

Given a linearly graded on-junction with the following doping conditions a=10^21

I have to calculate Emax, Vbi, W and the junction capacitance with Va=0.2V and Vr=-5V

This is for Silicon @ T=300k

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From what I understand, I have to use an iterative method but I don't think I'm getting the correct answer.

I'm using this equation Vbi=Vt*ln((a*Xo)/(ni))^2


Vt=0.0259
ni=1.5x10^10
a=10^21

where Xo= depletion region width

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Since I'm not given Vbi or Xo, I tried using an iterative method and Vbi and Xo converge at 20.202

I'm not quite sure what to do with this number or which value it pertains to. I would make the assumption that the potential barrier would be around 0.65-0.7V

Can anyone lead me to the right direction? Thanks.

 

[BigJDubs]

To answer your question, you need to use the Shockley equation. This equation relates the depletion region width (Xo) and the built-in potential barrier (Vbi) of a pn junction at equilibrium. The equation is Vbi = Vt * ln[(a*Xo)/ni], where Vt is the thermal voltage (kT/q), a is the doping concentration of the p-type material, Xo is the depletion region width, and ni is the intrinsic carrier concentration. Once you solve for Xo, you can calculate the maximum electric field (Emax) using Emax = Vbi/(2*Xo), W is the depletion region width (Xo), and the junction capacitance using Cj = epsilon0 * A/Xo. You can also use the following equations to calculate Vbi, W, and the junction capacitance: Vbi = Vr + Va W = 2 * Xo Cj = epsilon0 * A/Xo where Vr and Va are the reverse and forward bias voltages respectively, A is the area of the junction, and epsilon0 is the permittivity of free space.