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Degeneracy factor for holes
You calculate the occupation of donor states in your book by requiring that only (00, 01, 10) states are possible. Therefore, the probability that donor states are unoccupied is:
ND+ = ND*f_00 = ND/ [1 + 2 exp( EF-ED)/kT] so degeneracy factor is 2 -- this works fine ...
However, if I apply the same formulation to holes and require that only (0000, 0001, 0010, 0100, 1000) states are possible, then I find ...
NA- = NA[1-f_0000] = NA* 4exp(-(EA-EF)/kT]/ [ 1 + 4 exp(-(EA-EF)/kT)]
= NA/[ 1+ (1/4)exp(EA-EF)/kT].
When I compare with textbooks, the sign and everything else is fine, but people write gA=4, but I find gA=1/4.
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Asked by Anonymous - 1 year ago - 1 response
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Answers (1)
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Posted on 11 June, 2007 by Supriyo Datta
+2 0 Login to vote I would write NA- = NA * f_0000, rather than NA * [1 - f_0000]
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since 0000 is the state with one electron more than 0001, 0010, 0100 or 1000.
Also, E_0000 = 0, N_0000 = 0
E_0001 = E_0010 = E_0100 = E_1000 = - EA
N_0001 = N_0010 = N_0100 = N_1000 = - 1
so that I get f_0000 = 1 / 1 + 4 exp(EA - Ef)/kT
.. this seems to be in agreement with what you said the textbooks say ..
Makes sense?