Physics - Magnetism Question with Solution | TestHub

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 $\ln \left(1+\frac{{\mathrm{I}}_{\mathrm{D}}}{{\mathrm{I}}_{\mathrm{S}}}\right)=\frac{{\mathrm{V}}_{\mathrm{D}}}{{\mathrm{nV}}_{\mathrm{T}}}$ $\therefore {\mathrm{V}}_{\mathrm{D}}={\mathrm{nV}}_{\mathrm{T}}\ln \left(1+\frac{{\mathrm{I}}_{\mathrm{D}}}{{\mathrm{I}}_{\mathrm{S}}}\right)={\mathrm{V}}_{\mathrm{T}}\ln \left(1+\frac{{\mathrm{I}}_{\mathrm{D}}}{{\mathrm{I}}_{\mathrm{S}}}\right)$ $={\mathrm{V}}_{\mathrm{T}}\ln \left(\frac{{\mathrm{I}}_{\text{majority }}}{{\mathrm{I}}_{\mathrm{S}}}\right)$ $\begin{array}{l}\text{ Here, }{\mathrm{n}}_{\mathrm{e}}=\left(10^{17}-10^{16}\right){\mathrm{cm}}^{-3}\\ =10^{16}(10-1){\mathrm{cm}}^{-3}=9\times 10^{16}{\mathrm{cm}}^{-3}\end{array}$ We know that, $n_e\times n_h=n_i^2$ $\therefore {\mathrm{n}}_{\mathrm{h}}=\frac{{\mathrm{n}}_{\mathrm{i}}^2}{{\mathrm{n}}_{\mathrm{e}}}=\frac{{\left(1.4\times 10^{10}\right)}^2}{9\times 10^{16}}{\mathrm{cm}}^3$ Also, $\frac{{\mathrm{I}}_{\text{majority }}}{{\mathrm{I}}_{\mathrm{S}}}\propto \frac{{\mathrm{n}}_{\mathrm{e}}}{{\mathrm{n}}_{\mathrm{h}}}$ $\begin{array}{rl}\therefore {\mathrm{V}}_{\mathrm{D}}=& \frac{\mathrm{kT}}{\mathrm{e}}\ln \left[\frac{9\times 10^{16}}{{\left(1.4\times 10^{10}\right)}^2}\right].\\ & =\frac{\mathrm{kT}}{\mathrm{e}}\ln \left(4\times 10^{16}\right)\end{array}$