Topic: Electronic Devices
Electric Field of 1 V/m is applied to a Boron doped Silicon semiconductor slab having doping density of $10^{16}atoms/cm^3$ at 300K temperature. Determine the approximate resistivity of the slab. (Consider intrinsic carrier concentration of Silicon at 300 $K = 1.5\times10^{10}/cm^3$; Hole Mobility $= 500~cm^2/Vs$ at 300 K; Electron Mobility $= 1300~cm^2/Vs$ at 300 K).
- A. 0.48 $\Omega$-cm
- B. 0.35 $\Omega$-cm
- C. 0.16 $\Omega$-cm
- D. 1.25 $\Omega$-cm
Answer: D. 1.25 $\Omega$-cm
Explanation: Boron is a p-type dopant (acceptor), so the hole concentration $p \approx N_A = 10^{16} \text{ cm}^{-3}$. The electron concentration is negligible ($n = n_i^2/p$). The conductivity $\sigma \approx q \times p \times \mu_p = (1.6 \times 10^{-19} \text{ C}) \times (10^{16} \text{ cm}^{-3}) \times (500 \text{ cm}^2/\text{Vs}) = 0.8 \text{ S/cm}$. The resistivity $\rho$ is the inverse of conductivity: $\rho = 1 / \sigma = 1 / 0.8 = 1.25 \text{ } \Omega\text{-cm}$.