Transconductance application: An input change of 2 V produces a transconductance gm = 1.5 mS. What is the resulting change in drain current (ΔID)?

Introduction / Context:Transconductance (gm) links an input voltage change to an output current change in field-effect transistors. It is a core small-signal parameter for amplifiers, directly influencing gain and bandwidth. Here we perform a straightforward calculation using the definition of gm to find the current change.

Given Data / Assumptions:

• Input voltage change ΔV_in = 2 V (e.g., ΔV_GS for a FET).
• Transconductance gm = 1.5 mS = 1.5 × 10^-3 S.
• Linear small-signal approximation applies around the operating point.

Concept / Approach:By definition: gm = ΔI_out / ΔV_in. For a FET, gm = ΔI_D / ΔV_GS. Rearranging gives ΔI_D = gm * ΔV_GS. We simply multiply the transconductance by the input voltage change, keeping units consistent (S * V = A).

Step-by-Step Solution:

Verification / Alternative check:Unit check: siemens * volts = amperes, confirming dimensional correctness. Magnitude (milliamp-level) is reasonable for small-signal changes in many JFET/MOSFET amplifier stages.

Why Other Options Are Wrong:

• 666 mA: Orders of magnitude too large for the given gm and ΔV.
• 0.75 mA: Would correspond to gm = 0.375 mS or ΔV = 0.5 V, not our values.
• 0.5 mA: Would require gm = 0.25 mS at 2 V; not given.

Common Pitfalls:Forgetting to convert mS to S; mixing up ΔI_D = gm / ΔV_GS instead of multiplication; or assuming gm is constant over very large signal swings (it is a small-signal parameter around bias).

Final Answer:3 mA