**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

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