Topic: Analog Circuits
For a silicon NPN transistor circuit with its base grounded, emitter connected to $V_{EE} = -8\text{ V}$ through $R_E = 2.4\text{ k}\Omega$, and collector connected to $V_{CC} = 10\text{ V}$ through $R_C = 1.9\text{ k}\Omega$, the value of base current $I_B$ is (Assume $\beta = 100$ and $V_{BE} = 0.7\text{ V}$)
- A. $26.47\text{ }\mu\text{A}$
- B. $52.94\text{ }\mu\text{A}$
- C. $13.235\text{ }\mu\text{A}$
- D. $30.11\text{ }\mu\text{A}$
Answer: D. $30.11\text{ }\mu\text{A}$
Explanation: Applying KVL to the base-emitter loop: $V_B - V_{BE} - I_E R_E - V_{EE} = 0$. Since the base is grounded ($V_B = 0\text{ V}$), we have $0 - 0.7 - I_E(2.4\text{ k}\Omega) - (-8) = 0$. Solving for $I_E$, we get $I_E = \frac{7.3\text{ V}}{2.4\text{ k}\Omega} \approx 3.04\text{ mA}$. The base current is $I_B = \frac{I_E}{1 + \beta} = \frac{3.04\text{ mA}}{101} = 30.11\text{ }\mu\text{A}$.