Fermi level with 10¹⁶ B then 10¹⁷ Sb

Boron is an acceptor and antimony a donor. Use

$$E_{Fi}-E_{Fp} = kT\ln\frac{p}{n_i},\qquad E_{Fn}-E_{Fi} = kT\ln\frac{n}{n_i}$$

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

Step 1: Only boron, $N_a = 10^{16}\ \text{cm}^{-3}$

The wafer is p-type, $p \approx N_a = 10^{16}\ \text{cm}^{-3}$, so $E_F$ lies below $E_{Fi}$:

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

Step 2: Add antimony, $N_d = 10^{17}\ \text{cm}^{-3}$

Now $N_d > N_a$, so the wafer is converted to n-type (compensated):

$$n = N_d - N_a = 10^{17} - 10^{16} = 9\times10^{16}\ \text{cm}^{-3}$$ $$\begin{aligned} E_{Fn}-E_{Fi} &= 0.02585\ln\frac{9\times10^{16}}{1.45\times10^{10}}\\ &= 0.02585\times15.64 = 0.404\ \text{eV} \end{aligned}$$

Shift

$$\Delta E_F = 0.348 + 0.404 = 0.752\ \text{eV (upwards)}$$

 Ec -------------------------------
 EFn ........ 0.404 eV above EFi (after Sb)
 EFi - - - - - - - - - - - - - - - -
 EFp ........ 0.348 eV below EFi (B only)
 Ev -------------------------------

Answer: with boron alone the Fermi level is 0.348 eV below $E_{Fi}$; after adding antimony it moves to 0.404 eV above $E_{Fi}$, a total upward shift of about 0.75 eV.