Source follower (common-drain) small-signal gain: What is the voltage gain A_v of a source follower stage (assume linear operation and typical device parameters)?

Introduction / Context:
The source follower (common-drain for FETs, analogous to the BJT emitter follower) is a go-to stage when you need high input impedance, low output impedance, and near-unity voltage gain. It is widely used as a buffer between a high-impedance source and a heavier load, preserving signal amplitude while improving drive capability.

Given Data / Assumptions:

• Small-signal linear operation around a proper DC bias point.
• Transconductance g_m is finite and positive.
• Load connected at the source node together with any source resistor.

Concept / Approach:
For a small-signal model, the source follower gain is A_v ≈ g_m * (R_S || R_L) / [1 + g_m * (R_S || R_L)]. Since the numerator and denominator share the same term, A_v approaches +1 as g_m * (R_S || R_L) becomes large; practical circuits yield a value slightly less than unity (for example 0.9–0.99). Importantly, the stage is non-inverting: the output at the source follows the input at the gate in phase.

Step-by-Step Solution:

Verification / Alternative check:
SPICE simulations of a JFET/MOSFET source follower with g_m around a few mS and R_eq of a few kΩ yield 0.95–0.99 gain, consistent with the formula and bench measurements.

Why Other Options Are Wrong:

• A_V = g_m * R_D: This pertains to common-source stages; source followers do not use drain load for gain setting in this manner.
• A_V = g_m * R_S: Not a correct expression for source follower voltage gain.
• Approximately −1: That is the behavior of a common-source (or common-emitter) inverter, not a follower.

Common Pitfalls:
Expecting voltage gain > 1 from a follower; its purpose is buffering, not voltage amplification. To obtain gain, use a common-source stage or add a preceding gain block.

Final Answer:
approximately +1 (slightly less than unity)