Question

The value of emitter current in the given circuit is ............ μA. 

Hint:

For transistors, the emitter current is the sum of the base and collector currents. Use Kirchhoff’s laws and current gain to find the desired current.

Answer 1

Correct Answer: 444.8

Step 1: Apply Kirchhoff’s Voltage Law (KVL) to the input loop.
In the given circuit, the emitter current \( I_E \) can be found by first analyzing the base current \( I_B \). According to the problem, the voltage drop across the emitter resistor \( R_E \) is given as 0.3 V. \[ V_E = I_E \times R_E \] From this, the emitter current is: \[ I_E = \frac{V_E}{R_E} = \frac{0.3 \, \text{V}}{2 \, \text{k}\Omega} = 0.15 \, \text{mA} = 150 \, \mu\text{A} \]
Step 2: Use the current gain to find the collector current.
The current gain \( \beta \) of the transistor is given as 100, and the base current \( I_B \) is related to the collector current \( I_C \) by the relation: \[ I_C = \beta \times I_B \] To calculate \( I_B \), we can use the voltage drop across the base resistor \( R_B \), which is \( 0.3 \, \text{V} \). The voltage across \( R_B \) is: \[ V_B = I_B \times R_B \] where \( R_B = 2 \, \text{M}\Omega \). Solving for \( I_B \), we find: \[ I_B = \frac{V_B}{R_B} = \frac{0.3 \, \text{V}}{2 \times 10^6 \, \Omega} = 0.15 \, \mu\text{A} \]
Step 3: Conclusion.
The emitter current \( I_E \) is the sum of the base current and the collector current, and using the relation for \( I_C \), the total emitter current is approximately 2.5 μA.