Question

A pure Si crystal having \( 5 \times 10^{28} \) atoms m\(^{-3}\) is doped with 1 ppm concentration of antimony. If the concentration of holes in the doped crystal is found to be \( 4.5 \times 10^9 \) m\(^{-3}\), the concentration (in m\(^{-3}\)) of intrinsic charge carriers in the Si crystal is about:

• A. \( 1.2 \times 10^{15} \)
• B. \( 1.5 \times 10^6 \)
• C. \( 3.0 \times 10^{15} \)
• D. \( 2.0 \times 10^6 \)

Hint:

The intrinsic carrier concentration \( n_i \) in semiconductors is important for understanding charge transport. In doped semiconductors, the carrier concentration is influenced by both the intrinsic carriers and the dopants.

Answer 1

The Correct Option is B

Given: - Concentration of silicon atoms in pure Si: \( 5 \times 10^{28} \) m\(^{-3}\). - Doping concentration of antimony: 1 ppm, where \( 1 \, \text{ppm} = \frac{1}{10^6} \). - Concentration of antimony in doped Si: \[ \text{Concentration of Sb} = \frac{1}{10^6} \times 5 \times 10^{28} = 5 \times 10^{22} \, \text{m}^{-3} \] - Concentration of holes in doped crystal: \( 4.5 \times 10^9 \) m\(^{-3}\). The intrinsic carrier concentration \( n_i \) is defined as: \[ n_i = \sqrt{n_e \times n_h} \] where \( n_e \) is electron concentration and \( n_h \) is hole concentration. In the doped Si, electron concentration \( n_e \) is approximately equal to the dopant concentration: \( n_e \approx 5 \times 10^{22} \, \text{m}^{-3} \). Hole concentration \( n_h \) is \( 4.5 \times 10^9 \, \text{m}^{-3} \). Calculate the intrinsic carrier concentration \( n_i \): \[ n_i = \sqrt{5 \times 10^{22} \times 4.5 \times 10^9} \] \[ n_i = \sqrt{2.25 \times 10^{32}} = 1.5 \times 10^{16} \, \text{m}^{-3} \] The concentration of intrinsic charge carriers is approximately \( 1.5 \times 10^{16} \, \text{m}^{-3} \).