If \( R_C \) and \( R_B \) are respectively the resistances of in collector and base sides of the circuit, and \( \beta \) is the current amplification factor, then the voltage gain of a transistor amplifier in common emitter configuration is:
Show Hint
- \( \beta R_C R_B \)
- \( \frac{\beta}{R_C R_B} \)
- \( \frac{\beta R_B}{R_C} \)
- \( \frac{\beta R_C}{R_B} \)
The Correct Option is D
Solution and Explanation
Step 1: Understanding the Voltage Gain Formula
For a transistor in a common emitter (CE) configuration, the voltage gain (\( A_V \)) is given by: \[ A_V = \frac{\text{Output voltage}}{\text{Input voltage}} = \frac{V_{\text{out}}}{V_{\text{in}}} \] The voltage gain in CE configuration is expressed as: \[ A_V = \frac{\beta R_C}{R_B} \] where: - \( R_C \) is the resistance in the collector circuit,
- \( R_B \) is the resistance in the base circuit,
- \( \beta \) is the current gain of the transistor.
Step 2: Verifying the Correct Option From the formula: \[ A_V = \frac{\beta R_C}{R_B} \] Comparing with the given options, the correct answer is Option (4): \( \frac{\beta R_C}{R_B} \).
Final Answer: The voltage gain of a transistor amplifier in CE configuration is \( \frac{\beta R_C}{R_B} \), which matches option (4).
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