Answer :

Solution :

${\mathrm{I}}_{\mathrm{D}}={\mathrm{I}}_0\left[{\mathrm{e}}^{\frac{{\mathrm{V}}_{\mathrm{D}}}{\eta {\mathrm{V}}_{\mathrm{T}}}}-1\right]$

I_D1 = I₁, I_D2 = 1.5I1

${\mathrm{V}}_{\mathrm{D}}=-0.03,\eta =\frac{15}{13},{\mathrm{V}}_{\mathrm{T}}=26\mathrm{m}\mathrm{V}$

$\eta \times {\mathrm{V}}_{\mathrm{T}}=\frac{15}{13}\times 26\times {10}^{-3}=0.03$

$\Rightarrow \frac{{\mathrm{I}}_{\mathrm{D}2}}{{\mathrm{I}}_{\mathrm{D}1}}=\frac{{\mathrm{e}}^{\frac{{\mathrm{V}}_{\mathrm{D}2}}{\eta {\mathrm{V}}_{\mathrm{T}}}}-1}{{\mathrm{e}}^{\frac{{\mathrm{V}}_{\mathrm{D}1}}{\eta {\mathrm{V}}_{\mathrm{T}}}}-1}$

$\Rightarrow \frac{1.5\times {\mathrm{I}}_1}{{\mathrm{I}}_1}=\frac{{\mathrm{e}}^{\frac{{\mathrm{V}}_{\mathrm{D}2}}{0.03}}-1}{{\mathrm{e}}^{\frac{-0.03}{0.03}}-1}$

$\Rightarrow 1.5\times -0.6321={\mathrm{e}}^{\frac{{\mathrm{V}}_{\mathrm{D}2}}{0.03}}-1$

$\Rightarrow -0.9481+1={\mathrm{e}}^{\frac{{\mathrm{V}}_{\mathrm{D}2}}{0.03}}$

$\Rightarrow 0.0518={\mathrm{e}}^{\frac{{\mathrm{V}}_{\mathrm{D}2}}{0.03}}$

$\Rightarrow \ln 0.0518=\frac{{\mathrm{V}}_{\mathrm{D}2}}{0.03}$

$\Rightarrow {\mathrm{V}}_{\mathrm{D}2}=-2.9599\times 0.03=-0.09$