In short: The ideal op-amp model assumes infinite open-loop gain, infinite input impedance and zero output impedance, which lets you collapse almost any linear feedback circuit to two rules — no current enters the inputs, and the inputs sit at the same voltage (the "virtual short").
The operational amplifier is the workhorse of analog design, yet nobody analyses one by solving its full transistor-level netlist by hand. Instead, engineers lean on a deliberately simplified model — the ideal op-amp — that captures the behaviour that matters in a feedback loop while discarding second-order detail. Understanding exactly which assumptions you are making, and what they buy you, is the difference between confidently sketching a transfer function on a whiteboard and getting stuck. This article lays out the three core assumptions, derives the virtual-short rule from them, and works a concrete example.
The three ideal assumptions
An operational amplifier is a differential-input, single-ended-output voltage amplifier. Its open-loop output is Vout = A(V+ − V−), where A is the open-loop gain. The ideal model pins down three properties:
- Infinite open-loop gain (A → ∞). A real part such as a Texas Instruments OPA189 has a DC gain near 140 dB — that is roughly 107 V/V — large enough that treating it as infinite introduces negligible error at DC.
- Infinite input impedance (Zin → ∞). No current flows into either input terminal. A CMOS-input part draws input bias currents in the femtoamp-to-picoamp range, so for most circuits "zero input current" is an excellent approximation.
- Zero output impedance (Zout → 0). The output can source or sink any current demanded without its voltage sagging, so the load does not perturb the gain equation.
A 140 dB open-loop gain means that to drive a 10 V output, the differential input voltage need only be about 10 V ÷ 107 = 1 µV. That vanishingly small differential is the physical seed of the virtual-short idea.
From infinite gain to the virtual short
The virtual short is not a separate assumption — it is a consequence of infinite gain operating inside negative feedback. Rearrange the open-loop relation: V+ − V− = Vout / A. If the output is finite (say a few volts) and A is infinite, then the differential input voltage must be zero. In other words, negative feedback drives the amplifier until its two inputs are at the same potential:
V+ = V−
This is the virtual short: the inputs behave as though shorted together, but no current flows between them (because of the infinite input impedance). When the non-inverting input is tied to ground, the inverting node sits at 0 V even though it is not physically connected to ground — a "virtual ground". These two facts — equal input voltages and zero input current — are all you need to analyse the vast majority of linear op-amp circuits by inspection.
Worked example: the inverting amplifier
Consider the classic inverting configuration: input Vin through resistor R1 to the inverting node, feedback resistor Rf from output back to that node, non-inverting input grounded. Apply the two rules:
- Non-inverting input is at 0 V, so by the virtual short the inverting node is also at 0 V (virtual ground).
- No current enters the inverting input, so all current through R1 continues through Rf: Vin/R1 = −Vout/Rf.
Solving gives the familiar closed-loop gain Vout/Vin = −Rf/R1. With Rf = 100 kΩ and R1 = 10 kΩ the gain is exactly −10, set purely by a resistor ratio and completely independent of the op-amp's own gain. That independence is the whole point of feedback: it trades away raw, unpredictable open-loop gain for a precise, component-defined closed-loop gain. The same two rules yield the non-inverting gain (1 + Rf/R1), the integrator, the summing amplifier and the difference amplifier in a line or two of algebra each.
Where the assumptions break — and why it still matters
Each ideal property has a real-world limit, and the virtual short only holds while feedback is active. The table below contrasts the idealisation with a representative real part.
| Property | Ideal | Real (typical precision op-amp) |
|---|---|---|
| Open-loop gain | ∞ | 120–140 dB at DC, rolling off above ~10 Hz |
| Input impedance | ∞ | 1012–1013 Ω (CMOS/JFET) |
| Output impedance | 0 Ω | tens of ohms open-loop, reduced by loop gain |
| Bandwidth | ∞ | Finite GBW (e.g. 1–40 MHz) |
The virtual short fails the moment the output saturates against a rail or the loop runs out of gain at high frequency. A comparator (open loop, no feedback) never establishes a virtual short at all — which is exactly why you must not analyse it with these rules.
The single most important caveat is that infinite gain is a DC fiction. Real gain falls at 20 dB/decade, so the virtual-short approximation degrades as frequency rises — the topic we explore in depth in our companion piece on the gain–bandwidth product. When you move from hand analysis to verification, SPICE-class simulators (Cadence Spectre, Synopsys PrimeSim, Siemens Eldo) use full device models that expose exactly these departures, and understanding the ideal baseline is what lets you read a slew-rate and settling plot and know whether it looks right.
Why interviewers love this topic
"State the ideal op-amp assumptions and derive the inverting gain" is a near-universal analog warm-up question. Interviewers are checking that you can name the three assumptions crisply, explain that the virtual short is a result of infinite gain plus feedback (not an axiom), and apply it without reaching for nodal matrices. If you want structured, interview-focused practice with these derivations and their silicon-level reality, our analog design courses walk through dozens of worked topologies, and our partner Avecas analog design services builds production amplifiers on exactly these principles.
FAQ
Is the virtual short a short circuit?
No. The two inputs are forced to the same voltage by feedback, but because input impedance is infinite, no current flows between them. It is "virtual" precisely because it mimics a short in voltage without being a conductive path.
Does a virtual ground exist without feedback?
No. The virtual ground only appears while negative feedback is active and the output is unsaturated. Remove the feedback resistor, or saturate the output, and the inverting node is no longer held at 0 V.
Why assume infinite gain when real gain is finite?
Because closed-loop behaviour becomes independent of the exact gain once it is large. With 140 dB of open-loop gain, the error from assuming infinity is parts-per-million at DC — far smaller than resistor tolerances — so the simplification costs almost nothing.
Do these rules apply to comparators?
No. A comparator runs open-loop (or with positive feedback), so no virtual short is established. Analyse it as a switching device comparing two inputs, not with the ideal-op-amp feedback rules.
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