Fermi level with 10¹⁶ Sb then 2×10¹⁷ B

For a non-degenerate semiconductor, the Fermi level shift from the intrinsic level is

$$E_F - E_{Fi} = kT\ln\frac{n}{n_i} \ (\text{n-type}), \qquad E_{Fi} - E_F = kT\ln\frac{p}{n_i} \ (\text{p-type})$$

Assume: $T = 300$ K, $kT = 0.02586$ eV, $n_i = 1.45\times10^{10}$ cm$^{-3}$, full ionisation.

(a) Doped with $10^{16}$ Sb cm$^{-3}$

Sb is a donor, so $n = N_d = 10^{16}$ cm$^{-3}$.

$$\begin{aligned} E_F - E_{Fi} &= 0.02586\ln\frac{10^{16}}{1.45\times10^{10}} \\ &= 0.02586\times13.44 = 0.348\ \text{eV} \end{aligned}$$

$E_F$ is 0.348 eV above $E_{Fi}$.

(b) Further doped with $2\times10^{17}$ B cm$^{-3}$

B is an acceptor. Since $N_a > N_d$, the sample becomes p-type:

$$\begin{aligned} p &= N_a - N_d = 2\times10^{17} - 10^{16} = 1.9\times10^{17}\ \text{cm}^{-3} \\ E_{Fi} - E_F &= 0.02586\ln\frac{1.9\times10^{17}}{1.45\times10^{10}} \\ &= 0.02586\times16.39 = 0.424\ \text{eV} \end{aligned}$$

$E_F$ is now 0.424 eV below $E_{Fi}$; it has moved down by $0.348 + 0.424 = 0.772$ eV.

 Ec ---------------------------
     EF (a) ....  0.348 eV above
 Ei - - - - - - - - - - - - - -
     EF (b) ....  0.424 eV below
 Ev ---------------------------

Answer: (a) $E_F - E_{Fi} = +0.348$ eV; (b) $E_F - E_{Fi} = -0.424$ eV (p-type).