Difficulty: Medium
Correct Answer: 4.55 V rms
Explanation:
Introduction:
Differential amplifiers respond to both differential and common-mode components of their inputs. This problem assesses correct use of CMRR to find differential gain and then compute the output for a single-ended stimulus, which contains both a differential part and a common-mode part.
Given Data / Assumptions:
Concept / Approach:
From CMRR, compute the differential gain Ad = CMRR * Acm. Then form the output as v_out = Ad * v_d + Acm * v_cm, noting that signs depend on which input is designated as non-inverting; we report the magnitude here, which is conventional for such problems.
Step-by-Step Solution:
1) Compute Ad: Ad = 3250 * 0.2 = 650.2) Compute differential and common-mode inputs: v_d = 7 mV − 0 = 7 mV; v_cm = (7 mV + 0)/2 = 3.5 mV.3) Differential contribution: Ad * v_d = 650 * 0.007 V = 4.55 V.4) Common-mode contribution: Acm * v_cm = 0.2 * 0.0035 V = 0.0007 V (0.7 mV), which is negligible compared to 4.55 V.5) Total output magnitude ≈ 4.55 V + 0.0007 V ≈ 4.5507 V, which rounds to 4.55 V RMS.
Verification / Alternative check:
Since CMRR is high, the differential term dominates. Ignoring the tiny common-mode term gives 4.55 V, confirming the same answer within 0.02%.
Why Other Options Are Wrong:
1.4 mV rms and 9.1 mV rms: Much too small; they ignore the large differential gain.
650 mV rms and 0.455 V rms: Off by factors of approximately 7 or 10; they misapply CMRR or confuse Ad with Acm.
Common Pitfalls:
Confusing CMRR as a dB quantity when a linear ratio is given, neglecting that single-ended drive creates both v_d and v_cm, or forgetting that Ad = CMRR * Acm.
Final Answer:
4.55 V rms
Discussion & Comments