Voltage gain dependence: Is a transistor’s small-signal voltage gain inversely proportional to the change in output current, or is that claim incorrect for standard BJT amplifier stages?

Introduction / Context:
Voltage gain in a BJT amplifier arises from transconductance and load impedance. In a common-emitter stage, a small change in input voltage produces a proportional change in collector current, which then develops an output voltage across the load. The relationship between gain and output current change is not “inversely proportional” as the statement claims; rather, more transconductance or more load impedance generally increases gain magnitude.

Given Data / Assumptions:

• Linear small-signal operation around a bias point.
• Common-emitter example with load resistance R_C (or R_L at the collector).
• Hybrid-π model parameters: g_m and r_o.

Concept / Approach:
For a CE stage, the small-signal voltage gain is approximately A_v ≈ −g_m * R_out, where R_out includes the collector resistor in parallel with the transistor’s output resistance r_o and any external load. Since g_m = ΔI_C / ΔV_BE, a larger change in output current per unit input voltage (higher g_m) increases |A_v|, not decreases it. Therefore, voltage gain is directly related to how effectively the device converts input voltage to output current and how that current is turned into voltage by the load.

Step-by-Step Solution:

Verification / Alternative check:
Lab measurements show that increasing bias current (within limits) raises g_m and gain magnitude; SPICE simulations confirm A_v grows with g_m and load resistance.

Why Other Options Are Wrong:
Special cases like saturation or emitter followers have different gain expressions but do not support the claimed inverse relationship.

Common Pitfalls:
Confusing current gain β with transconductance g_m; forgetting that voltage gain depends on both device transconductance and load impedance.

Final Answer:
Incorrect