Calculate the width of the space charge region in a pn junction when a reversebiased voltage is applied.

Again consider a silicon pn junction at $T=300 \mathrm{~K}$ with doping concentrations of $N_{a}=$ $10^{16} \mathrm{~cm}^{-3}$ and $N_{d}=10^{15} \mathrm{~cm}^{-3}$ . Assume that $n_{i}=1.5 \times 10^{10} \mathrm{~cm}^{-3}$ and $V_{R}=5 \mathrm{~V}$ .

Question

Calculate the width of the space charge region in a pn junction when a reversebiased voltage is applied.

Again consider a silicon pn junction at [latex]T=300 \mathrm{~K}[/latex] with doping concentrations of [latex]N_{a}=[/latex] [latex]10^{16} \mathrm{~cm}^{-3}[/latex] and [latex]N_{d}=10^{15} \mathrm{~cm}^{-3}[/latex]. Assume that [latex]n_{i}=1.5 imes 10^{10} \mathrm{~cm}^{-3}[/latex] and [latex]V_{R}=5 \mathrm{~V}[/latex].

Accepted Answer

The built-in potential barrier was calculated in Example 7.1 for this case and is [latex]V_{b i}=0.635 \mathrm{~V}[/latex]. The space charge width is determined from Equation (7.34). We have

[latex]\begin{array}{c}W=\left\{\frac{2 \boldsymbol{\epsilon}_{s}\left(V_{b i}+V_{R} ight)}{e}\left[\frac{N_{a}+N_{d}}{N_{a} N_{d}} ight] ight\}^{1 / 2} \ \end{array}[/latex] (7.34)

[latex]\begin{array}{c}W=\left\{\frac{2(11.7)\left(8.85 imes 10^{-14} ight)(0.635+5)}{1.6 imes 10^{-19}}\left[\frac{10^{16}+10^{15}}{\left(10^{16} ight)\left(10^{15} ight)} ight] ight\}^{1 / 2}\end{array}[/latex]

so that

[latex]W=2.83 imes 10^{-4} \mathrm{~cm}=2.83 \mu \mathrm{m}[/latex]

Comment

The space charge width has increased from [latex]0.951 \mu \mathrm{m}[/latex] at zero bias to [latex]2.83 \mu \mathrm{m}[/latex] at a reverse bias of [latex]5 \mathrm{~V}[/latex].

Question 7.3: Calculate the width of the space charge region in a pn junct...

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