For a silicon PN junction at $T=300\,\text{K}$ with $N_A = 10^{17}\,\text{cm}^{-3}$, $N_D = 10^{15}\,\text{cm}^{-3}$, $n_i = 1.5\times10^{10}\,\text{cm}^{-3}$, the built-in potential $V_{bi} = \frac{kT}{q}\ln\!\left(\frac{N_AN_D}{n_i^2}\right)$ is approximately:

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Electrical Engineering QUESTION #2172

Question 1

$0.694\,\text{V}$✔️
$0.026\,\text{V}$
$1.12\,\text{V}$
$0.3\,\text{V}$

Correct Answer Explanation

$V_{bi} = 0.026\ln\!\left(\frac{10^{32}}{(1.5\times10^{10})^2}\right) = 0.026\ln(4.44\times10^{11}) = 0.026 \times 26.7 \approx 0.694\,\text{V}$.

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