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Second degree DE for pn-junction carrier concentration

  1. Dec 6, 2009 #1
    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 + Be1/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.
     
  2. jcsd
  3. Dec 6, 2009 #2
    Try writing down the characteristic equation again and verifying the roots.
     
  4. Dec 6, 2009 #3
    r2 - 1/Lh2 = 0

    r = +/- 1/Lh as I've mentioned.
     
  5. Dec 6, 2009 #4
    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.
     
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