Difficulty: Easy
Correct Answer: Incorrect
Explanation:
Introduction / Context:An operational transconductance amplifier (OTA) is a specialized amplifier whose output is a current proportional to its input voltage, characterized by a transconductance parameter gm. This behavior makes the OTA a voltage-controlled current source. The statement that an OTA is primarily a current-to-voltage amplifier misidentifies its fundamental role.
Given Data / Assumptions:
Concept / Approach:Unlike a voltage op-amp that produces a low-impedance voltage output, an OTA produces a current. Designers select load resistors, current mirrors, or active networks to translate that current into a voltage. Because gm can be made programmable (e.g., via a bias current), OTAs enable voltage-controlled filters, voltage-controlled amplifiers, and modulators.
Step-by-Step Solution:
State the OTA law: Iout = gm * Vin (or gm * Vdiff).Note device nature: this is voltage-to-current conversion, not current-to-voltage.Recognize Vout creation: Vout = Iout * Rload if a resistor is attached, providing a voltage only through an external element.Conclude: describing the OTA as primarily current-to-voltage is incorrect.Verification / Alternative check:Compare with a transimpedance amplifier (TIA). A TIA converts input current to output voltage (Vout = Iin * Rf). That is the proper term for a current-to-voltage amplifier, highlighting the distinction from an OTA’s voltage-to-current behavior.
Why Other Options Are Wrong:
Only with resistor load: the OTA remains a V-to-I device; the load merely turns current into voltage.True only at unity gain: OTAs are not defined by closed-loop voltage gain but by gm.Acts mainly as current-to-voltage under all conditions: that describes a TIA, not an OTA.Correct: contradicts the device definition.Common Pitfalls:Confusing OTA with op-amp or TIA; forgetting that gm, not open-loop voltage gain, is the key parameter; ignoring the need for an external load to produce a voltage.
Final Answer:Incorrect
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