How to solve steady state excess hole density

[alex8214997]

Homework Statement

A hole current of 10^(-5) A/cm2 is injected into the side (x=0) of a long N-silicon. Assuming the holes flow only by diffusion and that at very large values of x, the distribution of excess holes decays to zero, Determine

a)The steady-state excess hole density at x=0
b)Hole current density at x=100μm
c)the rate of generation and recombination of electron-hole pair at x=100μm

Answers:
a)2.794*10^(10)cm-3
b) 1.667*10^(-6) A/cm2

Given:
J=10^(-5) A/cm2
μp=480 cm2/Vs
μn= 1350 cm2/Vs
lifetime of hole, τ=2.5μs

Homework Equations

J= -q*D*(dp/dx) D= k*T*μp/q

continuity equation:
dp/dt = -(p-p0)/τ- div(J)/q

where p: hole density; p0= hole density at equilibrium.
p-po=Δp

The Attempt at a Solution

dp/dx= J/(-q*D)= -5*10^(12)

I think Δp=p-p0=0 when t=τ=2.5μs
and Δp is at max. when t=0, and x=0

I tried to solve the continuity equation, which is 1 ODE

dp/dt = -(p-p0)/τ- div(J)/q
p'=p0/τ-p/τ-div(J)/q

p= p0-τ*div(J)/q+C*exp(-t/τ)

Since i think Δp=p-p0=0 when t=τ

p=p0-τ*div(J)/q+τ*e*div(J)/q*exp(-t/τ)


at t=0;
Δp=p-p0=τ*div(J)/q*(e-1)

but i can't find div(J)

b) i think J is constant, how it can be a function of x.
c) generation rate: p0/τ= (ni/τ)=4*10^(15)
recombination rate: Δp/τ= (p(x=100μm)-p0)/τ

 

[KataKoniK]

=2*10^(-6)

Hello, thank you for your post. Based on the given information, here is my attempt at solving the problem:

a) The steady-state excess hole density at x=0 can be found by using the continuity equation, setting the time derivative equal to zero, and solving for p:

dp/dt = -(p-p0)/τ- div(J)/q = 0

Since at steady-state, dp/dt = 0, the equation becomes:

0 = -(p-p0)/τ- div(J)/q

Rearranging gives:

p = p0 - τ*div(J)/q

We can find div(J) by using Ohm's law and the fact that J = -q*D*(dp/dx):

div(J) = -q*D*(dp/dx) = -q*D*(-5*10^12) = 5*10^7 A/cm^3

Substituting this into the equation for p, we get:

p = p0 - (τ*5*10^7)/q = p0 - (2.5*10^-6*5*10^7)/1.6*10^-19 = p0 - 1.5625*10^12

Therefore, the steady-state excess hole density at x=0 is p = p0 - 1.5625*10^12.

b) The hole current density at x=100μm can be found by using Ohm's law and the fact that J = -q*D*(dp/dx):

J = -q*D*(dp/dx) = -q*D*(p-p0)/τ = -(1.6*10^-19*480*10^4*(p-p0))/2.5*10^-6 = -0.768(p-p0)

At x=100μm, p-p0 = Δp and we can use the answer from part a to find Δp:

Δp = p-p0 = p0 - (p0 - 1.5625*10^12) = 1.5625*10^12

Therefore, the hole current density at x=100μm is J = -0.768(1.5625*10^12) = -1.1992*10^12 A/cm^3.

c) The rate of generation and recombination of electron