Calculate the built-in potential barrier in a pn junction.

Consider a silicon pn junction at $T=300 \mathrm{~K}$ with doping concentrations of $N_{a}=$ $2 \times 10^{17} \mathrm{~cm}^{-3}$ and $N_{d}=10^{15} \mathrm{~cm}^{-3}$ .

Question

Calculate the built-in potential barrier in a pn junction.

Consider a silicon pn junction at [latex]T=300 \mathrm{~K}[/latex] with doping concentrations of [latex]N_{a}=[/latex] [latex]2 imes 10^{17} \mathrm{~cm}^{-3}[/latex] and [latex]N_{d}=10^{15} \mathrm{~cm}^{-3}[/latex].

Accepted Answer

The built-in potential barrier is determined from Equation (7.10) as

[latex]\begin{array}{c}V_{b i}=\frac{k T}{e} \ln \left(\frac{N_{a} N_{d}}{n_{i}^{2}} ight)=V_{t} \ln \left(\frac{N_{a} N_{d}}{n_{i}^{2}} ight) \ \end{array}[/latex] (7.10)

[latex]\begin{array}{c}V_{b i}=V_{t} \ln \left(\frac{N_{a} N_{d}}{n_{i}^{2}} ight)=(0.0259) \ln \left[\frac{\left(2 imes 10^{17} ight)\left(10^{15} ight)}{\left(1.5 imes 10^{10} ight)^{2}} ight]=0.713 \mathrm{~V}\end{array}[/latex]

If we change the doping concentration in the p region of the pn junction such that the doping concentrations become [latex]N_{a}=10^{16} \mathrm{~cm}^{-3}[/latex] and [latex]N_{d}=10^{15} \mathrm{~cm}^{-3}[/latex], then the built-in potential barrier becomes [latex]V_{b i}=0.635 \mathrm{~V}[/latex].

Comment

The built-in potential barrier changes only slightly as the doping concentrations change by orders of magnitude because of the logarithmic dependence.

Question 7.1: Calculate the built-in potential barrier in a pn junction. C...

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