Doping to raise Fermi level by 0.7 eV

To raise the Fermi level above $E_{Fi}$, electrons must be made majority carriers, so a donor (n-type, group V) dopant such as phosphorus, arsenic or antimony must be used.

Doping level

For a non-degenerate n-type semiconductor

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

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

$$\begin{aligned} \frac{E_{Fn}-E_{Fi}}{kT} &= \frac{0.7}{0.02585} = 27.08\\ N_d &= 1.45\times10^{10}\times e^{27.08}\\ &= 1.45\times10^{10}\times5.75\times10^{11}\\ &= 8.34\times10^{21}\ \text{cm}^{-3} \end{aligned}$$

Answer: donor doping, $N_d \approx 8.3\times10^{21}\ \text{cm}^{-3}$ (by the formula).

Important remark

For silicon $E_g \approx 1.1$ eV, so $E_c - E_{Fi} \approx 0.55$ eV. Raising $E_F$ by 0.7 eV puts it about 0.15 eV inside the conduction band. The semiconductor is then degenerate:

• $N_d$ would exceed $N_c \approx 2.8\times10^{19}\ \text{cm}^{-3}$, and would be about 17% of the Si atomic density ($5\times10^{22}$), far beyond normal doping (and beyond the solubility of most donors).
• The Boltzmann formula is no longer exact, so the number above is only an estimate; the true requirement is "very heavy doping, $N_d > N_c$".

In practice this degenerate n⁺ doping behaves like a metal and is used for ohmic contacts and tunnel diodes.