Verified Solution • 2025–26 Session moderate

QPages 419-421: Q32: HC Verma: Concepts of Physics - Volume 2 Semiconductors and Semiconductor Devices Pages 419-421: Q32 Solution: A load resistor of is connected in the collector branch of an…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often multiply the voltage gain (200) by the current gain (50) but forget to keep the correct magnitude from unit conversions (mV to V and µA to A), leading to a result of 10 instead of the true power gain of 10,000. Always convert all quantities to base units before calculating power gain, or explicitly include the factor of $10^4$ that arises from $\frac{\text{V}}{\text{mV}}\times\frac{\text{A}}{\mu\text{A}}$.

A load resistor of 2 kΩ2 \text{ k}\Omega is connected in the collector branch of an amplifier circuit using a transistor in common-emitter mode. The current gain β=50\beta = 50. The input resistance of the transistor is 0.50 kΩ0.50 \text{ k}\Omega. If the input current is changed by 50 �A. a) by what amount does the output voltage change, b) by what amount does the input voltage change and c) what is the power gain?

Step-by-Step Verified Solution

**Answer:** (a) _V_out = 5.0 V (b) _V_in = 25 mV (c) Power gain = 10_. **Step-by-Step Explanation:** Given: R_L = 2 k�, _ = 50, R_in = 0.5 k�, _I_B = 50 _A. (a) _I_C = __I_B = 50 _ 50 _ 10__ = 2.5 mA. _V_out = _I_C _ R_L = 2.5 _ 10__ _ 2000 = 5.0 V. (b) _V_in = _I_B _ R_in = 50 _ 10__ _ 500 = 25 mV. (c) Voltage gain = _V_out/_V_in = 5.0/0.025 = 200. Current gain = _ = 50. Power gain = 200 _ 50 = 10_ .

All Questions in this Chapter

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