Solving n_{n0} Using Charge Neutrality & Mass Action Law

[mzh]

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

I understand I should assume that $N_D > N_A$, but I can't work it out.

Any comments are very welcomed.

 

[mzh]

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

I think I found the solution to this. The important point to note is that we assume relatively high temperatures. Given the relationship for $N_D^+ = \frac{N_D}{1+2\exp\left[\frac{E_F - E_D}{kT}\right]}$, we can assume that $E_F - E_D$ is much lower than zero. Then, when dividing by $kT = 0.025 \mbox{eV}$ at room temperature, the exponential term becomes approximately zero and so $N_D^+ = N_D$. Then, $N_D$ can be inserted into the charge neutrality condition and, after expressing $p=\frac{n_i^2}{n}$, the resulting quadratic equation can be solved for $n$. Great.