Topic: Electronics Engineering
Assume that $V_{i}=2V;$ $R3=10~\Omega;$ $\beta$ of $Q1=50;$ $V_{CC}=15~V$
Find $I_{out}$
- A. 3.92 mA
- B. 4 mA
- C. 6.84 mA
- D. 2.8 mA
Answer: A. 3.92 mA
Explanation: Here $V_{in}=2V$ and same voltage will appear at the emitter of Q1 due to Op-Amp feedback. So, $I_{E}=\frac{2}{10}A=0.2~A$.
$I_{out}=I_{b}=\frac{I_{E}}{1+\beta}=\frac{0.2}{1+50} = \frac{0.2}{51}=3.92~mA.$