Question

Consider the circuit shown in the figure below,

Assuming $V_{BE}\left(on\right)=0.7V$,
then the value of output voltage $V_0$ is equal to

• A. -2.12 V
• B. -3.45 V
• C. -1.52 V
• D. -3.15 V

Answer 1

The correct option is B -3.45 V


$I_E=\frac{8-0.7}{3.3K\Omega }$ = 2.212 mA
$I_c=\alpha I_E=\frac{110}{111}\times 2.212mA$
= 2.192 mA
Now
$V_0=-12+I_c\times 3.9\times {10}^3$
= - 3.45 V