Difficulty: Easy
Correct Answer: A_V = V_out / V_in
Explanation:
Introduction / Context:
Amplifier “gain” must be specified by type: voltage, current, transconductance, transresistance, or power. Confusion arises when expressions appropriate to one form (e.g., current gain β = ΔI_C/ΔI_B) are mislabeled as voltage gain. For linear small-signal analysis, the signal voltage gain relates output voltage variation to input voltage variation and is dimensionless (ratio of volts to volts).
Given Data / Assumptions:
Concept / Approach:
Voltage gain is defined as A_V = V_out / V_in, often written as ΔV_out / ΔV_in for small-signal derivatives. In a common-emitter BJT stage, A_V can be approximated by −g_m * R_out, but this is a model-based computation, not the definition. Other expressions such as I_C * R_C (voltage across a resistor given a current) calculate a specific voltage, not a ratio of two voltages; ΔI_C/ΔI_B is current gain β; P_out/P_in is power gain; g_m/r_o is not a general definition of voltage gain (though it appears in specific configurations).
Step-by-Step Solution:
Verification / Alternative check:
Data sheets and textbooks consistently specify small-signal voltage gain as a ratio of voltages, often in dB using 20 * log10(|A_V|).
Why Other Options Are Wrong:
I_C * R_C yields a voltage, not a ratio. ΔI_C/ΔI_B is β (current gain). P_out/P_in is power gain, not voltage gain. g_m/r_o is a parameter ratio, not a universal definition of A_V.
Common Pitfalls:
Mixing definitions and computations; using a computed expression as if it were the general definition; ignoring whether gains are specified as magnitude, signed, or in dB.
Final Answer:
A_V = V_out / V_in
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