Fermi level of n-type semiconductor Nd>Na>0

  • #1
girlinphysics
25
0

Homework Statement


This isn't actually a homework question, but in my semiconductors textbook, the following equation has been given:

[tex]E_f = E_g - k_BTln(\frac{n_0}{N_d - N_a})[/tex]

This is for the limiting case Nd>Na>0. I got a little confused as to where that equation has come from.

Homework Equations


[tex]N_d - N_a[/tex] (Where Nd = donor level and Na = acceptor level)
Also: [tex]E_F = E_g - E_d[/tex]

The Attempt at a Solution



I think this means the fermi level is at donor level so I can write:

[tex]n = n_0e^{\beta(E_f - E_g)} = n_0e^{-\beta E_d}[/tex]

where beta = 1/kT

Is this how they got to that equation? Just by rearranging for Ef?
 

Answers and Replies

  • #2
18,935
9,219
Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 
  • #3
girlinphysics
25
0
Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
Unfortunately I haven't solved this problem, but I will try re-working it in case I have made it confusing. Thanks for the advice!
 

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