Second degree DE for pn-junction carrier concentration

[rire1979]

Upon some calculation I arrive to the expression:

d²DPn(x)/dx² = DPn(x)/Lh²

Where:

DPn(x) = Pn(x) + Pno - excess minority carriers (holes) concentration in the n-type part of the pn junction.

Now the roots to the characteristic equation are +/- 1/Lh where Lh is the length of the diffusion.

Therefore the solution looks like:

DPn(x) = Ae^-1/Lh + Be^1/Lh

I know for a fact the solution is DPn(x) = DPn(0)e^-x/Lh

The initial conditions would be that:
@ x = 0 we have hole concentration DPn(0)
@ x = Lh we have DPn(Lh) = 0

But I have no idea how to arrive at the solution in bold. I'm missing something and I was thinking you could help.

Thank you.

 

[Lord Crc]

Try writing down the characteristic equation again and verifying the roots.

 

[rire1979]

Try writing down the characteristic equation again and verifying the roots.

r² - 1/Lh² = 0

r = +/- 1/Lh as I've mentioned.

 

[Lord Crc]

I'm sorry, bit late here... misread that and thought the char. equation was r^2 - 1/Lh^2 r = 0, which, from my quick glance, could get you where you wanted. My bad.