Topic: Electronic Devices and Circuits
A small amount of phosphorus is added to silicon so that there are $2.5\times10^{17}$ conduction electrons per cubic meter and $9.0\times10^{14}$ holes per cubic meter. Let $\mu_{e}=0.13~\text{m}^{2}/\text{V.s}$ and $\mu_{h}=0.05~\text{m}^{2}/\text{V.s}$. The resistivity in $\Omega\text{-m}$ will be
- A. 5.2072
- B. 192.042
- C. 129.02
- D. 0
Answer: B. 192.042
Explanation: Phosphorus makes the material n-type.
$n = 2.5 \times 10^{17}~\text{m}^{-3}$ and $p = 9.0 \times 10^{14}~\text{m}^{-3}$.
Conductivity $\sigma = nq\mu_e + pq\mu_h \approx nq\mu_e$ (since $n \gg p$).
Resistivity $\rho = \frac{1}{\sigma} = \frac{1}{2.5 \times 10^{17} \times 1.602 \times 10^{-19} \times 0.13}$.
$\rho \approx 192.067~\Omega\text{-m}$. Option (b) is the closest.