Differential amplifier behavior — common-mode versus differential: If the two inputs to a differential amplifier are exactly the same signal (pure common-mode), is the output equal to the input signal multiplied by 2, or does it ideally cancel to zero?

Introduction / Context:
Differential amplifiers respond to the difference between their two inputs while rejecting signals that are common to both. This behavior underpins instrumentation amplifiers and noise rejection strategies. The statement under test claims that identical inputs produce an output equal to “signal × 2,” which would imply summing rather than differencing. We examine whether that is correct under ideal assumptions.

Given Data / Assumptions:

• Two amplifier inputs receive exactly the same waveform (common-mode input).
• Ideal differential amplifier characteristics: infinite common-mode rejection ratio (CMRR).
• Linear operation and no saturation.

Concept / Approach:
An ideal differential amplifier multiplies the difference (v+ − v−) by the differential gain Ad. When the inputs are identical, v+ = v−, so v+ − v− = 0. Therefore the ideal output is 0 V, not twice the input. Real amplifiers have finite CMRR, so small residual output may appear, but it should be minimal compared with the differential response.

Step-by-Step Solution:

Verification / Alternative check:
Instrumentation amplifier datasheets specify CMRR (often 80–120 dB). High CMRR means large rejection of common-mode signals; measured outputs under identical inputs are near zero, not doubled.

Why Other Options Are Wrong:

• Correct: Incorrect because identical inputs should cancel in a differential stage.
• Only true when CMRR is infinite: Even with infinite CMRR the output is zero, not doubled.
• Only true for square waves: Waveform shape is irrelevant; difference is zero for any waveform.

Common Pitfalls:
Confusing summing amplifiers with differential amplifiers; misinterpreting common-mode behavior as additive gain rather than rejection.

Final Answer:
Incorrect