Question

Consider a silicon p-n junction at $T = 300 \, \text{K}$ with doping concentrations of acceptor $N_A = 10^{16} \, \text{cm}^{-3}$ and donor $N_D = 10^{15} \, \text{cm}^{-3}$. Assume that intrinsic concentration $n_i = 1.5 \times 10^{10} \, \text{cm}^{-3}$, relative permittivity $= 11.7$, and $V_{bi} = 0.635 \, \text{V}$. The width of the space charge region in the p-n junction is:

• A. 9.51 m
• B. 0.951 m
• C. 95.1 m
• D. 5.91 m

Hint:

The width of the space charge region depends on doping concentrations and the builtin potential of the pn junction.

Answer 1

The Correct Option is A

The width of the space charge region $W$ in a pn junction is given by the formula:
$W = \sqrt{\frac{2 \epsilon V_{bi}}{q} \left( \frac{1}{N_A} + \frac{1}{N_D} \right) }$
where $\epsilon$ is the permittivity, $V_{bi}$ is the built-in voltage, and $N_A$ and $N_D$ are the doping concentrations. Substituting the values:
$W \approx 9.51 \ \mu \text{m}$