Question

Consider the transistor circuitt shown in the figure below:
The value of $V_{\text{in}}$ is chosen such that the output voltage $V_0$ is equal to 6 V. Then the value of $V_{\text{in}}$ is equal to
$\left(\text{Assume,}{V}_{BE\left(ON\right)}\right)=0.7V$

• A. 3.97 V
• B. 4.21 V
• C. 5.21 V
• D. 3.11 V

Answer 1

The correct option is A 3.97 V


$\text{Now,}{I}_C=\frac{12-6}{2.2}{\times }^{-3}=2.727mA$
$\therefore I_B=\frac{I_C}{\beta }=\frac{2.727}{30}=\frac{1}{11}mA$
$\text{Now,}{I}_2=\frac{0.7-(-12)}{100k\Omega }=0.127mA$
$\therefore I_1=I_2+I_B=0.218mA$
$\text{thus,}{V}_{\text{in}}=I_1\times 15\times {10}^3\Omega +0.7$
$V_{\text{in}}=3.968V=3.97V$