# For a silicon diode with long P and N regions, the acceptor and donor impurity concentrations are $1 \times {10^{17}}{\rm{c}}{{\rm{m}}^3}{\rm{\;and\; 0 votes 58 views in General closed For a silicon diode with long P and N regions, the acceptor and donor impurity concentrations are \(1 \times {10^{17}}{\rm{c}}{{\rm{m}}^3}{\rm{\;and\;}}1 \times {10^{15}}{\rm{c}}{{\rm{m}}^3}$ respectively. The lifetimes of electrons in the P-region and holes in the N-region are both 100 μs. The electron and hole diffusion coefficients are $49{\rm{c}}{{\rm{m}}^3}$ and $36{\rm{\;c}}{{\rm{m}}^2}/{\rm{s}}$ respectively. Assume $\frac{{{\rm{kT}}}}{{\rm{q}}} = 26{\rm{\;mV}}$, the intrinsic carrier concentration is $1 \times {10^{10}}{\rm{\;c}}{{\rm{m}}^{ - 3}}$ , and ${\rm{q}} = 1.6 \times {10^{ - 19}}{\rm{C}}.$ When a forward voltage of 208 mV is applied across the diode, the hole current density (in nA / cm2)  injected from P region to N region is

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Concept:

The reverse saturation hole current is given by,

${{\rm{J}}_{{\rm{po}}}} = {\rm{q\eta }}_{\rm{i}}^2 \cdot \left( {\frac{{{{\rm{D}}_{\rm{p}}}}}{{{{\rm{L}}_{\rm{p}}}{{\rm{N}}_{\rm{D}}}}}} \right)$

Application:

We have ${{\rm{L}}_{\rm{p}}} = \sqrt {{{\rm{D}}_{\rm{p}}}{{\rm{\tau }}_{\rm{p}}} = \sqrt {36} \times 100 \times {{10}^{ - 6}}}$

$\begin{array}{l} \Rightarrow {{\rm{L}}_{\rm{p}}} = 60 \times {10^{ - 3}}{\rm{cm}}\\ {{\rm{J}}_{{\rm{po}}}} = 1.6 \times {10^{ - 19}} \times {\left( {{{10}^{10}}} \right)^2} \times \left[ {\frac{{36}}{{60 \times {{10}^{ - 3}} \times {{10}^{15}}}}} \right]\\ \Rightarrow {{\rm{J}}_{{\rm{po}}}} = 9.6\frac{{{\rm{pA}}}}{{{\rm{c}}{{\rm{m}}^2}}} \end{array}$

Now, current density at V = 208 mV

$\begin{array}{l} {{\rm{J}}_{\rm{p}}} = {{\rm{J}}_{{\rm{po}}}}\left( {{{\rm{e}}^{\frac{{\rm{V}}}{{{{\rm{V}}_{\rm{T}}}}}}} - 1} \right)\\ = 9.6 \times {10^{ - 12}}\left( {{{\rm{e}}^{\left( {\frac{{208}}{{26}}} \right)}} - 1} \right)\\ \Rightarrow {{\rm{J}}_{\rm{p}}} = 28.61\frac{{{\rm{nA}}}}{{{\rm{c}}{{\rm{m}}^2}}} \end{array}$