Fermi level with 10¹⁷ Sb then further doping

Antimony (group V) is a donor, so the wafer is n-type and the Fermi level $E_{Fn}$ lies above the intrinsic level $E_{Fi}$. For a non-degenerate semiconductor $n = n_i\,e^{(E_{Fn}-E_{Fi})/kT}$, so

$$E_{Fn}-E_{Fi} = kT\ln\frac{n}{n_i}$$

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

Case 1: $N_d = 10^{17}\ \text{cm}^{-3}$ antimony

At room temperature all donors are ionised and $N_d \gg n_i$, so $n \approx N_d = 10^{17}\ \text{cm}^{-3}$.

$$\begin{aligned} E_{Fn}-E_{Fi} &= 0.02585\ln\frac{10^{17}}{1.45\times10^{10}} \\ &= 0.02585\times15.75 = 0.407\ \text{eV} \end{aligned}$$

Answer: the Fermi level is 0.407 eV above $E_{Fi}$.

Case 2: further doped with $2\times10^{16}\ \text{cm}^{-3}$ atoms

The type of the second dopant is not printed; the usual reading is that it is an acceptor (e.g. boron), $N_a = 2\times10^{16}\ \text{cm}^{-3}$. The acceptors capture an equal number of the donated electrons (compensation):

$$n = N_d - N_a = 10^{17} - 2\times10^{16} = 8\times10^{16}\ \text{cm}^{-3}$$ $$\begin{aligned} E_{Fn}-E_{Fi} &= 0.02585\ln\frac{8\times10^{16}}{1.45\times10^{10}} \\ &= 0.401\ \text{eV} \end{aligned}$$

Answer: the Fermi level is still above $E_{Fi}$, now 0.401 eV above it, i.e. it moves down by about 0.006 eV towards $E_{Fi}$. The sample stays n-type because $N_d > N_a$.

(If the extra $2\times10^{16}$ atoms were more antimony, $n = 1.2\times10^{17}$ and $E_F - E_{Fi} = 0.412$ eV, a shift of 0.005 eV upward.)

 Ec ---------------------------
 EFn ......... (0.407 eV) 1st
 EFn ......... (0.401 eV) 2nd
 EFi - - - - - - - - - - - - -
 Ev ---------------------------