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Consider a silicon p-n junction with a uniform acceptor doping concentration of 1017 cm3 on the p side and a uniform donor doping concentration of 1016 cm−3 on the n-side. No external voltage is applied to the diode. Given: kT/q = 26 mV, ni =1.5 ×1010 cm−3, εSi = 12ε0, ε0 = 8.85 × 10−14 F/m, and q = 1.6 ×10−19 C. The charge per unit junction area (nC cm−2) in the depletion region on the p-side is ___________.

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We have

\({{\rm{x}}_{\rm{p}}} = {\left\{ {\frac{{2ϵ{_{{\rm{si}}}}.{{\rm{V}}_{{\rm{bi}}}}}}{9}{\rm{\;}}\left[ {\frac{{{{\rm{N}}_{\rm{d}}}}}{{{{\rm{N}}_{\rm{a}}}}}} \right]\left[ {\frac{1}{{{{\rm{N}}_{\rm{a}}} + {{\rm{N}}_{\rm{d}}}}}} \right]} \right\}^{\frac{1}{2}}}\)

Now \({{\rm{V}}_{{\rm{bi}}}} = \frac{{{\rm{KT}}}}{4}\ln \left( {\frac{{{{\rm{N}}_{\rm{a}}}{{\rm{N}}_{\rm{d}}}}}{{{\rm{n}}_{\rm{i}}^2}}} \right)\)

\(\Rightarrow {{\rm{V}}_{{\rm{bi}}}} = 26 \times {10^{ - 3}}\ln \left( {\frac{{{{10}^{16}} \times {{10}^{17}}}}{{11.5 \times {{10}^{10}}}}} \right) = 0.757{\rm{\;V\;}}\)

Now,

\({{\rm{x}}_{\rm{p}}} = {\left\{ {\frac{{2 \times 12 \times 8.85 \times {{10}^{ - 16}} \times {{10}^{16}}}}{{\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {{{10}^{17}}} \right)\left( {{{10}^{16}} + {{10}^{17}}} \right)}}} \right\}^{\frac{1}{2}}}\)

\(= {\left\{ {\frac{{2 \times 12 \times 8.85 \times {{10}^{ - 16}} \times {{10}^{16}}}}{{\left( {1.6 \times {{10}^{ - 19}}} \right)\left( {{{10}^{17}}} \right) \times 11 \times {{10}^{16}}}}} \right\}^{\frac{1}{2}}}\)

\(\Rightarrow {{\rm{x}}_{\rm{p}}}{\rm{\;}} = 3 \times {10^{ - 7}}{\rm{cm}}\)

Now charge per unit volume in the depletion region is

\(\begin{array}{l} -{\rm{e}}{{\rm{N}}_{\rm{a}}} = - \left( {1.6 \times {{10}^{ - 19}}} \right) \times {10^{17}}{\rm{C}}/{\rm{c}}{{\rm{m}}^3}\\ \Rightarrow - {\rm{e}}{{\rm{N}}_{\rm{a}}} = - 1.6{\rm{\;}} \times {\rm{\;}}{10^{ - 2}}{\rm{\;C}}/{\rm{c}}{{\rm{m}}^3} \end{array}\)

Charge per unit area \({\rm{Q}}\) in depletion region on p-side is:

\(Q=-{\rm{e}}{{\rm{N}}_{\rm{a}}}.{{\rm{x}}_{\rm{p}}}\)

\({\rm{Q}} = - 1.6 \times {10^{ - 2}} \times 3 \times {10^{ - 7}}C/cm^2\)

\( {\rm{Q}} = - 4.8{\rm{\;nC}}/{\rm{c}}{{\rm{m}}^2} \)

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