An op-amp is connected with its non-inverting input receiving a 1 V signal and no feedback network connecting the output back to any input. What will the output be?
AApproximately 1 V, since the op-amp amplifies the input with unity gain by default
BApproximately 100,000 V, limited only by the op-amp's current capacity
CSaturated at one of the supply rail voltages, because without feedback any differential input drives the output to its limit
DZero, because the absence of a feedback path prevents amplification
Without negative feedback, the op-amp operates open-loop with its enormous open-loop gain (typically 10⁵ to 10⁶). Any nonzero differential input — even from offset voltages — is multiplied by this gain, driving the output to the positive or negative supply rail. This is the comparator configuration: the output indicates which input is larger, not a proportional amplification. The virtual short principle only applies when negative feedback is present to constrain the differential input near zero.
Question 2 Multiple Choice
An engineer builds an ideal inverting op-amp integrator and applies a sinusoidal AC signal. After extended operation, the output slowly drifts toward the supply rail. The most likely explanation is:
AThe signal frequency is above the op-amp's gain-bandwidth product
BA small DC offset at the input is being integrated continuously, producing a growing ramp that eventually saturates the output
CThe feedback capacitor is discharging through the input resistor
DThe virtual short approximation breaks down at high output voltages
An ideal integrator integrates everything at its input — including any small DC offset from input bias currents, offset voltage, or the signal source. Even a 1 mV DC component produces an output ramp of (1 mV)/(RC) per second, which accumulates without bound until the output hits a supply rail. Real integrator designs include a large feedback resistor in parallel with the capacitor to limit DC gain and prevent this runaway.
Question 3 True / False
The 'virtual short' at an op-amp's inputs means the two input terminals are physically connected, allowing current to flow between them.
TTrue
FFalse
Answer: False
The virtual short is a consequence of negative feedback, not a physical connection. With very high open-loop gain and negative feedback, the output adjusts until the differential input voltage is driven essentially to zero. The inputs are NOT connected — no current flows between them (that is the 'virtual open' rule). Virtual short means the voltages are equal; virtual open means no current enters the input terminals. Conflating these two rules leads to incorrect KCL analysis.
Question 4 True / False
An op-amp differentiator and an op-amp integrator are both analyzed using the same virtual short and virtual open rules, despite performing opposite mathematical operations.
TTrue
FFalse
Answer: True
Both circuits use negative feedback and are therefore analyzed identically: apply virtual short (V⁻ = V⁺), virtual open (no current into inputs), then KCL at the inverting node. The difference is only the capacitor placement: in the feedback path for the integrator (Z_f = 1/sC), or in the input path for the differentiator (Z_in = 1/sC). The gain formula G = −Z_f/Z_in gives −1/(sRC) for the integrator and −sRC for the differentiator. Same analysis, opposite placement.
Question 5 Short Answer
Why does a comparator (op-amp without feedback) saturate at a supply rail voltage for almost any input, while the same op-amp in an inverting amplifier configuration gives a controlled, proportional output?
Think about your answer, then reveal below.
Model answer: Without feedback, the op-amp multiplies the differential input by its open-loop gain (~10⁵ to 10⁶). Even a 10 μV difference drives the output to 1–10 V — exceeding the supply rails — so the output saturates. With negative feedback, part of the output is fed back to the inverting input, continuously adjusting until the differential input is driven near zero. This self-correcting loop constrains the gain to the feedback network ratio (−Z_f/Z_in) and keeps the output in the linear region.
Negative feedback trades raw gain for stability and precision. The op-amp's enormous gain is 'spent' enforcing the virtual short condition — the feedback network does most of the work, and the op-amp simply ensures zero differential input. Remove the feedback, and there is nothing to constrain the gain, so the output saturates.