Sze Eq. 42

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mzh
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Dear PF users
Would be great if somebody could point me out how to arrive at [itex]n_{n0} = \frac{1}{2} \left[ (N_D - N_A) + \sqrt{ (N_D - N_A)^2 + 4n_i^2} \right][/itex] (n-type charge carrier concentration at thermal equilibrium) by using the expression for the charge neutrality [itex]n+N_A = p+N_D[/itex] and the mass action law [itex]np=n_i^2[/itex].

I understand I should assume that [itex]N_D > N_A[/itex], but I cant work it out.

Any comments are very welcomed.
 

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mzh
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Dear PF users
Would be great if somebody could point me out how to arrive at [itex]n_{n0} = \frac{1}{2} \left[ (N_D - N_A) + \sqrt{ (N_D - N_A)^2 + 4n_i^2} \right][/itex] (n-type charge carrier concentration at thermal equilibrium) by using the expression for the charge neutrality [itex]n+N_A = p+N_D[/itex] and the mass action law [itex]np=n_i^2[/itex].
I think I found the solution to this. The important point to note is that we assume relatively high temperatures. Given the relationship for [itex]N_D^+ = \frac{N_D}{1+2\exp\left[\frac{E_F - E_D}{kT}\right]}[/itex], we can assume that [itex]E_F - E_D[/itex] is much lower than zero. Then, when dividing by [itex]kT = 0.025 \mbox{eV}[/itex] at room temperature, the exponential term becomes approximately zero and so [itex]N_D^+ = N_D[/itex]. Then, [itex]N_D[/itex] can be inserted into the charge neutrality condition and, after expressing [itex]p=\frac{n_i^2}{n}[/itex], the resulting quadratic equation can be solved for [itex]n[/itex]. Great.
 

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