Fermi level of Si with 10¹⁶ Sb

Antimony is a group V donor, so the wafer is n-type and the Fermi level moves above the intrinsic Fermi level $E_{Fi}$.

Relation used

For a non-degenerate semiconductor

$$n = n_i\exp\left(\frac{E_{Fn}-E_{Fi}}{kT}\right) \;\Rightarrow\; E_{Fn}-E_{Fi} = kT\ln\frac{n}{n_i}$$

Data used (Si, 300 K): $n_i = 1.45\times10^{10}\ \text{cm}^{-3}$, $kT = 8.617\times10^{-5}\times300 = 0.02585\ \text{eV}$, $\mu_e = 1350$, $\mu_h = 450\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$, $e = 1.6\times10^{-19}\ \text{C}$.

At room temperature all donors are ionised and $N_d = 10^{16} \gg n_i$, so

$$n \approx N_d = 10^{16}\ \text{cm}^{-3},\qquad p = \frac{n_i^2}{N_d} = 2.1\times10^{4}\ \text{cm}^{-3}$$

Calculation

$$\begin{aligned} E_{Fn}-E_{Fi} &= 0.02585\ln\frac{10^{16}}{1.45\times10^{10}}\\ &= 0.02585\ln(6.90\times10^{5})\\ &= 0.02585\times13.44\\ &= 0.348\ \text{eV} \end{aligned}$$

Answer: the Fermi level lies about 0.348 eV above the intrinsic Fermi level (about 0.2 eV below $E_c$, since $E_c - E_{Fi} \approx 0.55$ eV).

 Ec -----------------------------
         ~0.20 eV
 EFn ............................
         0.348 eV
 EFi - - - - - - - - - - - - - - -
         ~0.55 eV
 Ev -----------------------------

The level is still well below $E_c$ (more than $3kT$), so the non-degenerate formula used is valid.