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Consider a silicon p-n junction at room temperature having the following parameters: Doping on the n-side = $1\times10^{17}\ \text{cm}^{-3}$ Depletion width on the n-side = $0.1\ \mu\text{m}$ Depletion width on the p-side = $1.0\ \mu\text{m}$ Intrinsic carrier concentration = $1.4\times10^{10}\ \text{cm}^{-3}$ Thermal voltage = $26\ \text{mV}$ Permittivity of free space = $8.85\times10^{-14}\ \text{F cm}^{-1}$ Dielectric constant of silicon = 12 The built-in potential of the junction is

1. 0.70 V
2. 0.76 V
3. 0.82 V
4. cannot be estimated from the data given

Correct answer: 0.76 V

Solution

For an abrupt p-n junction, charge neutrality gives $N_A x_p = N_D x_n$. With $N_D=10^{17}\ \text{cm}^{-3}$, $x_n=0.1\ \mu\text{m}$ and $x_p=1.0\ \mu\text{m}$, we get $N_A=10^{16}\ \text{cm}^{-3}$. Then $V_{bi}=V_T\ln\left(\frac{N_A N_D}{n_i^2}\right)$, which evaluates to about $0.76\ \text{V}$.

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