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Module 6 : PHYSICS OF SEMICONDUCTOR DEVICES

Lecture 34 : Condition of Charge Neutrality

$\displaystyle q(N_d^++p) = q(N_a^-+n)$

If all the donors and acceptors are ionized, then,

$\displaystyle q(N_d+p) = q(N_a+n)$

Using

, we get

$\displaystyle n$	$\displaystyle =$	$\displaystyle (N_d - N_a) + p$
	$\displaystyle =$	$\displaystyle (N_d-N_a) + \frac{n_i^2}{n}$

Thus we get a quadratic equation for the electron density

$\displaystyle n^2 - (N_d-N_a)n - n_i^2 =0$

with solution

$\displaystyle \boxed{n=\frac{N_d-N_a}{2}+\sqrt{\frac{(N_d-N_a)^2+4n_i^2}{4}}}$