Open-loop polarity test — In an op-amp with open loop (no negative feedback), if the inverting (−) input is at a lower potential than the noninverting (+) input (i.e., V− < V+), what direction will the output swing?

Difficulty: Easy

Correct Answer: swing positive

Explanation:


Introduction:
Understanding open-loop polarity is foundational for troubleshooting op-amp circuits. It also explains why negative feedback stabilizes the output while positive feedback can lead to latching or oscillation. The sign of the differential input dictates the extreme to which the output saturates in open loop.


Given Data / Assumptions:

  • No feedback (open-loop operation)
  • V+ and V− compared directly by very large gain Aol
  • Supply rails available to allow saturation


Concept / Approach:

The op-amp output is approximately Vout ≈ Aol * (V+ − V−). If V+ > V−, the quantity (V+ − V−) is positive, so the output rushes toward the positive rail. Conversely, if V− > V+, the output rushes negative.


Step-by-Step Reasoning:

Given V− < V+ → (V+ − V−) > 0Large Aol multiplies this small differenceTherefore Vout drives toward + saturationSign determines direction; magnitude set by Aol and rails


Verification / Alternative check:

Introduce a tiny positive offset at V+ and observe simulation or bench result: the output saturates positive in the absence of feedback, matching the sign rule.


Why Other Options Are Wrong:

  • swing negative: Would require V− > V+.
  • close the loop / be balanced: These are not behaviors of an open-loop configuration.
  • oscillate about zero: Not inherently; open-loop DC conditions saturate to a rail.


Common Pitfalls:

  • Assuming the output stays near zero without feedback—it does not.
  • Confusing the labels of + and − inputs, which flips the outcome.


Final Answer:

swing positive

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion