Question

In the following common emitter circuit, $\beta = 100$ and $V_{CE}=7\text{ V}$. If $V_{BE}$ is negligible, then the base current is

• A. $0.015\ \text{mA}$
• B. $0.045\ \text{mA}$
• C. $0.025\ \text{mA}$
• D. $0.035\ \text{mA}$

Hint:

In a CE transistor circuit, first find $I_C$ from the collector resistor using $V_{CC}-V_C$, then use $I_B=I_C/\beta$.

Answer 1

The Correct Option is B

Step 1: From the circuit, the supply voltage is: \[ V_{CC} = 15\ \text{V} \] Collector resistance: \[ R_C = 2\,\text{k}\Omega \]
Step 2: Given: \[ V_{CE} = 7\ \text{V} \] Since the emitter is grounded, the collector voltage is: \[ V_C = 7\ \text{V} \]
Step 3: Voltage drop across the collector resistance: \[ V_{RC} = V_{CC} - V_C = 15 - 7 = 8\ \text{V} \]
Step 4: Collector current: \[ I_C = \frac{V_{RC}}{R_C} = \frac{8}{2000} = 4\times10^{-3}\ \text{A} = 4\ \text{mA} \]
Step 5: Using the relation $\beta = \dfrac{I_C}{I_B}$: \[ I_B = \frac{I_C}{\beta} = \frac{4\ \text{mA}}{100} = 0.04\ \text{mA} \]
Step 6: The closest value among the given options is: \[ \boxed{0.045\ \text{mA}} \]