Question

Consider the circuit shown in the figure below:


If the scaling current then the value of voltage $V_R$ is equal to ______V. $(V_T=25mV)$
$I_{s1}=2I_{s2}=5\times {10}^{-16}A,$

• A. 0.702

Answer 1

The correct option is A 0.702
$\begin{array}{ccc}{I}_C& =I_{C1}+I_{C2}& \\ now,I_C& =I_{s1}exp\left[\frac{V_{BE}}{V_T}\right]+I_{s2}exp\left[\frac{V_{BE}}{V_T}\right]& [\because V_R=V_{BE1}=V_{BE2}]\\ I_C& =[I_{s1}+I_{s2}]exp\left[\frac{V_R}{V_T}\right]& \\ \frac{I_C}{\frac{3}{2}{I}_{s1}}& =exp\left[\frac{V_R}{V_T}\right]& \\ V_R& =V_{T}ln\left[\frac{1.2\times {10}^{-3}}{1.5\times 5\times {10}^{-16}}\right]& \\ V_R& =0.702V& \end{array}$