Since $E_G \gg k_B T$, we can only use the law of mass action.
Since $E_G \gg k_B T$, we can only use the law of mass action.
But the question offers us another piece of information - we are around $|N_D-N_A| \approx n_i$.
But the question offers us another piece of information - we are around $|N_D-N_A| \approx n_i$.
That means that we are near the transition between extrinsic and intrinsic regimes.
That means that we are near the transition between extrinsic and intrinsic regimes.
However, we also know that dopants energies are quite small such that $E_C-E_D \ll E_G$ and $E_A-E_V \ll E_G$.
Because the dopants stop being ionized at very low temperatures (see next exercise), we can neglect $n_D$ and $n_A$ in this exercise, just like in the lecture.
That means that we expect $n_i \ll n_D$ and $n_i \ll n_A$ (since both $n_i$ and dopant ionazition depends exponentially of the corresponding energy differences).
Writing $n_e n_h = n_i^2$ and $n_e - n_h = N_D - N_A$ and solving these together, we obtain
Therefore, we can confidently say that $N_D \gg n_D$ and $N_A \gg n_A$ so we esentially recover the dopant ionization condition.
Thus, neglecting $n_D$ and $n_A$ such that they are both 0, the solutions to the charge balance are:
$$ n_{e} = \frac{1}{2}(\sqrt{D^2+4n_i^2}+D)$$
$$ n_{e} = \frac{1}{2}(\sqrt{D^2+4n_i^2}+D)$$
$$ n_{h} = \frac{1}{2}(\sqrt{D^2+4n_i^2}-D)$$
$$ n_{h} = \frac{1}{2}(\sqrt{D^2+4n_i^2}-D)$$
where $D = N_D - N_A$ and $n_i=n_{e,intrinsic}=n_{h,intrinsic}$.
where $D = N_D - N_A$.
Note that for N_D = N_A results for intrinsic semiconductors are recovered.
For $n_i \gg |N_D - N_A|$ we recover the intrinsic regime, while the opposite limit gives the extrinsic expressions.
