Doping to raise Fermi level by 0.1 eV

To raise the Fermi level above the intrinsic level, the number of electrons must exceed the number of holes, so a donor (pentavalent, group V) impurity such as phosphorus, arsenic or antimony must be added, making the material n-type. (The semiconductor is taken as silicon.)

Relation used

For a non-degenerate n-type semiconductor with all donors ionised:

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

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

Calculation

$$\begin{aligned} \frac{E_{Fn}-E_{Fi}}{kT} &= \frac{0.1}{0.02585} = 3.868\\ e^{3.868} &= 47.86\\ N_d &= 1.45\times10^{10}\times47.86\\ &= 6.94\times10^{11}\ \text{cm}^{-3} \end{aligned}$$

Answer: a donor (n-type) dopant is needed, with $N_d \approx 6.9\times10^{11}\ \text{cm}^{-3}$ (about 1 donor per $7\times10^{10}$ Si atoms).

Check

$N_d$ is about 48 times $n_i$, so $n \approx N_d$ is a fair approximation. (Solving exactly with $n = N_d/2 + \sqrt{N_d^2/4 + n_i^2}$ changes the result by well under 1%.)

 Ec ---------------------------
 EFn .......................... 0.1 eV above EFi
 EFi - - - - - - - - - - - - - -
 Ev ---------------------------

The Fermi level rises logarithmically with doping: each tenfold increase in $N_d$ raises $E_F$ by $kT\ln10 \approx 0.06$ eV.