Pure silicon at room temperature has an intrinsic carrier concentration of \(1.5\times10^{16}\ \text{m}^{-3}\). One out of every \(10^7\) silicon atoms is replaced by a phosphorus atom. Predict the effect on the electrical properties: find the conduc…

Answer: \(\sigma\) rises from \(4.39\times10^{-4}\) S/m to 108 S/m (about \(2.5\times10^5\) times); the doped Si is n-type with \(p=4.5\times10^{10}\ \text{m}^{-3}\).

Before doping (intrinsic):

\[\sigma_i=n_ie(\mu_e+\mu_h)=1.5\times10^{16}\times1.6\times10^{-19}\times(0.135+0.048)=4.39\times10^{-4}\ \text{S/m}\]

After doping. Phosphorus is pentavalent, so each P atom donates one free electron and the crystal becomes n-type.

Donor concentration \(N_D=\dfrac{5\times10^{28}}{10^7}=5\times10^{21}\ \text{m}^{-3}\).

At room temperature all donors are ionised, and \(N_D\gg n_i\), so \(n\approx N_D=5\times10^{21}\ \text{m}^{-3}\). Holes are negligible, hence

\[\sigma_n\approx N_De\mu_e=5\times10^{21}\times1.6\times10^{-19}\times0.135=108\ \text{S/m}\]

Hole concentration (mass action law \(np=n_i^2\)):

\[p=\frac{n_i^2}{n}=\frac{(1.5\times10^{16})^2}{5\times10^{21}}=4.5\times10^{10}\ \text{m}^{-3}\]

Effect: \(\dfrac{\sigma_n}{\sigma_i}=\dfrac{108}{4.39\times10^{-4}}\approx2.5\times10^5\). Replacing just one atom in ten million raises the conductivity about 2.5 lakh times (resistivity falls from about 2280 Ω m to about 0.0093 Ω m). Electrons become the majority carriers and holes the minority carriers.