P-N Junction Diode and its applications: Practice Problem & Solution
Pure Si at 500 K has equal number of electron ($n_e$) and hole ($n_h$) concentrations of $1.5 \times 10^{16} m^{-3}$. Doping by indium increases $n_h$ to $4.5 \times 10^{22} m^{-3}$. The doped semiconductor is of: (2011 Mains)
Solution Explained:
To solve this problem, we apply the core principles of P-N Junction Diode and its applications. Understanding the underlying formula is key to arriving at the correct answer below:
Using the mass action law $n_e n_h = n_i^2$, we have $n_e = \frac{n_i^2}{n_h} = \frac{(1.5 \times 10^{16})^2}{4.5 \times 10^{22}} = \frac{2.25 \times 10^{32}}{4.5 \times 10^{22}} = 0.5 \times 10^{10} = 5 \times 10^9 m^{-3}$. Indium is trivalent, so it forms a p-type semiconductor.[cite: 1]
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