Discussion
Consider an n-type silicon in which the fully ionized dopant concentration is \(10^{17}\ cm^{-3}\). The intrinsic electron density is \(1.5 \times 10^{10}\ cm^{-3}\). Which of the following options correctly states the equilibrium hole concentration in \(cm^{-3}\)?
Answer: A
The equilibrium hole concentration is \(2.25 \times 10^{3}\ cm^{-3}\).
With all donors ionised, the electron concentration in the n-type silicon is the donor concentration, \(n = 10^{17}\ cm^{-3}\). At equilibrium the mass-action law fixes the product of electron and hole concentrations:
\[ \begin{aligned} np &= n_i^{2} \\ p &= \frac{(1.5 \times 10^{10})^{2}}{10^{17}} \\ &= \frac{2.25 \times 10^{20}}{10^{17}} = 2.25 \times 10^{3} \end{aligned} \]
Holes are the minority carriers, so their number is tiny compared with the \(10^{10}\) of intrinsic silicon: doping raises one carrier type and suppresses the other. The remaining options do not follow from \(n_i^2/N_D\) with these numbers.
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