Operational Transconductance Amplifier (OTA) voltage gain For an OTA driving a resistive load RL (single-ended), the small-signal voltage gain can be expressed in terms of transconductance gm as:

Introduction / Context:
An Operational Transconductance Amplifier converts an input voltage to an output current with proportionality gm (amps per volt). When that current is sent into a resistive load, it develops a voltage. Recognizing this relationship helps when designing voltage-controlled filters, oscillators, and automatic gain control circuits that use OTAs.

Given Data / Assumptions:

• OTA modeled as io = gm * vi.
• Load is a linear resistor RL to ground (or to a reference node), producing vo = io * RL.
• Small-signal, linear operation around a bias point (gm may be set by a control current).

Concept / Approach:
By definition, the OTA emits an output current proportional to input voltage. That current flowing through RL generates an output voltage. Hence, voltage gain is the product of gm and RL. If the load is reactive or frequency dependent, the gain becomes frequency dependent, but the basic proportionality remains.

Step-by-Step Solution:
Start with io = gm * vi.Voltage across load: vo = io * RL = (gm * vi) * RL.Therefore, Av = vo / vi = gm * RL.Sign conventions depend on the particular OTA pinout and load connection; magnitude is gm * RL.

Verification / Alternative check:
If RL doubles, Av doubles, which matches intuition: more resistance creates more voltage from the same current. If gm increases via bias current, gain increases proportionally.

Why Other Options Are Wrong:
1/gm or RL/gm: imply inverse dependence, incorrect for a simple resistive load.gm/CL: mixes dimensions (per farad), relating to a pole frequency rather than voltage gain.

Common Pitfalls:
Forgetting that real OTAs have limited output compliance and may need a buffer; high RL or large signals can lead to distortion or clipping.

Final Answer:
Av = gm * RL