Difficulty: Easy
Correct Answer: Correct
Explanation:
Introduction / Context: The ideal op-amp model is foundational for analyzing analog interface circuits such as active filters, amplifiers, and DAC/ADC signal conditioning. Key idealizations include infinite input impedance (no input current) and infinite open-loop gain (A→∞), which simplify nodal equations and justify feedback assumptions.Given Data / Assumptions:
Concept / Approach: With very high input impedance, the op-amp does not load the preceding stage. With extremely high open-loop gain, even a tiny differential input drives the output until feedback forces near-zero differential input, enabling precise closed-loop behavior defined by passive components. These two assumptions underpin superposition and linear small-signal analyses used throughout mixed-signal interfacing.Step-by-Step Solution:
Adopt ideal model: input currents ≈ 0 A; differential input voltage ≈ 0 V under feedback.Use A→∞ to justify V+ ≈ V− when feedback enforces linear operation.Analyze networks (inverting/non-inverting) using these simplifications for quick, accurate results.Verification / Alternative check:
Compare to datasheets: FET-input and precision op-amps exhibit mega-ohm to giga-ohm input impedance and open-loop gains of 10^5–10^6+, approximating the ideals.Why Other Options Are Wrong:
Incorrect: Conflicts with the standard ideal model taught universally.True only for FET-input op-amps: Bipolar devices can also have very high gain; “ideal” is a model, not a device-type restriction.Depends on output swing: Output swing limits are separate from input Z and open-loop gain ideals.Common Pitfalls:
Applying ideal assumptions outside the linear region (saturation, slew rate limits).Ignoring bias currents and finite A in precision calculations.Final Answer:
Correct
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