Topic: Analog Circuits
The voltage V in the circuit given below is :
- A. 7.47 V
- B. 8.43 V
- C. 4.47 V
- D. 0.47 V
Answer: C. 4.47 V
Explanation: The base is grounded ($V_B = 0V$). The emitter is forward-biased, so $V_E = -0.7V$.
The emitter current is $I_E = \frac{V_E - (-V_{EE})}{R_E} = \frac{-0.7 - (-9)}{4k\Omega} = \frac{8.3}{4} = 2.075$ mA.
The collector current is $I_C = \alpha I_E = 0.992 \times 2.075 = 2.0584$ mA.
The collector voltage is $V_C = V_{CC} - I_C R_C = 9 - (2.0584 \times 2.2) = 9 - 4.528 = 4.472$ V.
The magnitude of the base-collector voltage is $|V_{BC}| = |V_B - V_C| = |0 - 4.472| = 4.47$ V.