Op-amp virtual ground concept — If ground is applied to the noninverting (+) terminal of an inverting amplifier, what is the voltage condition at the inverting (−) terminal under linear operation?

Difficulty: Easy

Correct Answer: be virtual ground

Explanation:


Introduction:
The idea of “virtual ground” is central to analyzing inverting op-amp circuits. It enables simple current and gain calculations and explains why the inverting topology presents a predictable input impedance and precise closed-loop gain.


Given Data / Assumptions:

  • Ideal or high-gain op-amp
  • Negative feedback from output to inverting input
  • Noninverting input tied to ground (0 V)
  • Linear (unsaturated) operation with sufficient supply rails


Concept / Approach:

With negative feedback, the op-amp drives its output so that the differential input voltage (V+ − V−) is approximately zero. If V+ is fixed at 0 V (ground), then V− is held very near 0 V. The point is called a “virtual ground” because it is at (approximately) ground potential without being physically connected to ground.


Step-by-Step Solution:

Set V+ = 0 V (physical ground)With Aol ≫ 1 and feedback present, enforce V+ − V− ≈ 0 → V− ≈ 0 VNo significant current flows into op-amp input; currents go through external resistorsThis allows simple current summation at the inverting node for gain derivations


Verification / Alternative check:

Write node equation at V− with ideal op-amp input current ≈ 0 and solve; solution converges to V− ≈ V+ = 0 V under closed-loop operation.


Why Other Options Are Wrong:

  • not need an input resistor: The input resistor sets gain and input impedance; it is required.
  • have high reverse current: Ideal input current is ~0; reverse conduction is not the concept.
  • not invert the signal: Inverting stage by definition inverts.
  • float at an undefined potential: Feedback fixes the node near 0 V.


Common Pitfalls:

  • Assuming virtual ground equals a physical short to ground—there is no direct connection.
  • Applying virtual ground logic when the op-amp is saturated or feedback is broken.


Final Answer:

be virtual ground

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