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A p-n junction has acceptor impurity concentration of $10^{17} \mathrm{~cm}^{-3}$ in the $\mathrm{p}$-side and donor impurity concentration of $10^{16} \mathrm{~cm}^{-3}$ in the $\mathrm{n}$-side. What is the contact potential at the junction (kT = thermal energy, intrinsic carrier concentration $\mathrm{n}_{\mathrm{i}}=1.4 \times 10^{10} \mathrm{~cm}^{-3}$ ) ?
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Verified Answer
The correct answer is:
$(\mathrm{kT} / \mathrm{e}) \ln \left(4 \times 10^{12}\right)$
We have, $\mathrm{I}_{\mathrm{D}}=\mathrm{I}_{\mathrm{S}}\left(\mathrm{e}^{\mathrm{V}_{\mathrm{D}} / \mathrm{V}_{\mathrm{T}}-1}\right)$
where, $V_{\mathrm{T}}=\frac{\mathrm{kT}}{\mathrm{e}}$
$\therefore\left(\frac{\mathrm{I}_{\mathrm{D}}}{\mathrm{I}_{\mathrm{S}}}+1\right)=\mathrm{e}^{\left(\frac{\mathrm{V}_{0}}{\mathrm{~V}_{\mathrm{T}}}\right)}$ or $\ln \left(1+\frac{\mathrm{I}_{\mathrm{D}}}{\mathrm{I}_{\mathrm{S}}}\right)=\frac{\mathrm{V}_{\mathrm{D}}}{\mathrm{nV}_{\mathrm{T}}}$
$\therefore \mathrm{V}_{\mathrm{D}}=\mathrm{nV}_{\mathrm{T}} \ln \left(1+\frac{\mathrm{I}_{\mathrm{D}}}{\mathrm{I}_{\mathrm{S}}}\right)=\mathrm{V}_{\mathrm{T}} \ln \left(1+\frac{\mathrm{I}_{\mathrm{D}}}{\mathrm{I}_{\mathrm{S}}}\right)$
$\quad=\mathrm{V}_{\mathrm{T}} \ln \left(\frac{\mathrm{I}_{\text {majority }}}{\mathrm{I}_{\mathrm{S}}}\right)$
$\begin{array}{l}
\text { Here, } \mathrm{n}_{\mathrm{e}}=\left(10^{17}-10^{16}\right) \mathrm{cm}^{-3} \\
\quad=10^{16}(10-1) \mathrm{cm}^{-3}=9 \times 10^{16} \mathrm{~cm}^{-3}
\end{array}$
We know that, $\quad n_{e} \times n_{h}=n_{i}^{2}$
$\therefore \mathrm{n}_{\mathrm{h}}=\frac{\mathrm{n}_{\mathrm{i}}^{2}}{\mathrm{n}_{\mathrm{e}}}=\frac{\left(1.4 \times 10^{10}\right)^{2}}{9 \times 10^{16}} \mathrm{~cm}^{3}$
Also, $\frac{\mathrm{I}_{\text {majority }}}{\mathrm{I}_{\mathrm{S}}} \propto \frac{\mathrm{n}_{\mathrm{e}}}{\mathrm{n}_{\mathrm{h}}}$
$\begin{aligned} \therefore \mathrm{V}_{\mathrm{D}}=& \frac{\mathrm{kT}}{\mathrm{e}} \ln \left[\frac{9 \times 10^{16}}{\left(1.4 \times 10^{10}\right)^{2}}\right].\\ &=\frac{\mathrm{kT}}{\mathrm{e}} \ln \left(4 \times 10^{16}\right) \end{aligned}$
where, $V_{\mathrm{T}}=\frac{\mathrm{kT}}{\mathrm{e}}$
$\therefore\left(\frac{\mathrm{I}_{\mathrm{D}}}{\mathrm{I}_{\mathrm{S}}}+1\right)=\mathrm{e}^{\left(\frac{\mathrm{V}_{0}}{\mathrm{~V}_{\mathrm{T}}}\right)}$ or $\ln \left(1+\frac{\mathrm{I}_{\mathrm{D}}}{\mathrm{I}_{\mathrm{S}}}\right)=\frac{\mathrm{V}_{\mathrm{D}}}{\mathrm{nV}_{\mathrm{T}}}$
$\therefore \mathrm{V}_{\mathrm{D}}=\mathrm{nV}_{\mathrm{T}} \ln \left(1+\frac{\mathrm{I}_{\mathrm{D}}}{\mathrm{I}_{\mathrm{S}}}\right)=\mathrm{V}_{\mathrm{T}} \ln \left(1+\frac{\mathrm{I}_{\mathrm{D}}}{\mathrm{I}_{\mathrm{S}}}\right)$
$\quad=\mathrm{V}_{\mathrm{T}} \ln \left(\frac{\mathrm{I}_{\text {majority }}}{\mathrm{I}_{\mathrm{S}}}\right)$
$\begin{array}{l}
\text { Here, } \mathrm{n}_{\mathrm{e}}=\left(10^{17}-10^{16}\right) \mathrm{cm}^{-3} \\
\quad=10^{16}(10-1) \mathrm{cm}^{-3}=9 \times 10^{16} \mathrm{~cm}^{-3}
\end{array}$
We know that, $\quad n_{e} \times n_{h}=n_{i}^{2}$
$\therefore \mathrm{n}_{\mathrm{h}}=\frac{\mathrm{n}_{\mathrm{i}}^{2}}{\mathrm{n}_{\mathrm{e}}}=\frac{\left(1.4 \times 10^{10}\right)^{2}}{9 \times 10^{16}} \mathrm{~cm}^{3}$
Also, $\frac{\mathrm{I}_{\text {majority }}}{\mathrm{I}_{\mathrm{S}}} \propto \frac{\mathrm{n}_{\mathrm{e}}}{\mathrm{n}_{\mathrm{h}}}$
$\begin{aligned} \therefore \mathrm{V}_{\mathrm{D}}=& \frac{\mathrm{kT}}{\mathrm{e}} \ln \left[\frac{9 \times 10^{16}}{\left(1.4 \times 10^{10}\right)^{2}}\right].\\ &=\frac{\mathrm{kT}}{\mathrm{e}} \ln \left(4 \times 10^{16}\right) \end{aligned}$
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