Are ideal opamps being sold? By ideal opamps, I mean those whose output can be calculated from the appropriate equations (negative and positive feedback.)

I am asking this because I am running simulations on Falstad and the real opamps (those whose supply pins are not initially connected) give very different results from the theory I know.

For example I put some negative feedback on a real opamp and I put the feedback resistor and the input resistor to be the same connected VEE to GND and VCC to 10V and input value of 3V. I expected an output value of 2V however it showed basicaly a voltage divider. Isn't the simulator good or what?

I have designed many circuits using ideal opamps and now I need to translate those circuits to circuits with a real opamp.

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    \$\begingroup\$ Can you show us the circuit you modeled? \$\endgroup\$ – Math Keeps Me Busy Dec 2 '20 at 19:06
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    \$\begingroup\$ No theoretical mathematical calculation matches up with actual values. It just gets you in the ballpark. \$\endgroup\$ – StainlessSteelRat Dec 2 '20 at 20:32
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    \$\begingroup\$ Related (new question) from same OP, not duplicate: electronics.stackexchange.com/questions/535215/… \$\endgroup\$ – MarkU Dec 2 '20 at 21:08
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    \$\begingroup\$ Actually I am going to differ with everyone else. I mean, of course there are no ideal op-amps or resistors or capacitors or anything. But for simple textbook circuits, and if you only use a 3 digit multi-meter, most op-amps are ideal enough over some portion of their range. To figure out what is wrong with your circuit, please post your WHOLE CIRCUIT instead of describing it in words. \$\endgroup\$ – mkeith Dec 3 '20 at 6:27
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    \$\begingroup\$ If you don't already have a schematic of your circuit to include, there's a schematic editor built in to this site. Edit your question, and find the schematic button in the toolbar or press ctrl-m. \$\endgroup\$ – Phil Frost Dec 3 '20 at 18:17

While there are no ideal op amps, real-world parts approximate well within limits, including but not limited to: input voltage limits, output voltage and current, gain, bandwidth and noise. The practical limits for real-world parts vary from one device to another, of course.


There are no ideal opamps in the real world (they would have to be superconducting, draw zero current on the input terminals and source infinite amounts of current on the output), use LT spice (or other spice package) with real opamps to simulate real world opamps.

All real opamps have offsets and common mode ranges and limits on the current that can be sourced. They also have limits on voltage rails. Lt spice will simulate almost all of those.

  • \$\begingroup\$ Voltage Spike the results of the simulation on falstad are wrong correct and 2V is the correct answer huh? \$\endgroup\$ – The Force Awakens Dec 2 '20 at 18:55
  • \$\begingroup\$ @TheForceAwakens I have run into particular circuits that do not simulate correctly in Falstead but do simulate correctly in LT Spice. I never figured out why. This circuit was a circuit that oscillated based on hysteresis and the oscillation frequency in real life did not match what Falstead was simulating. \$\endgroup\$ – DKNguyen Dec 2 '20 at 19:09
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    \$\begingroup\$ Falstads solver probably isn't that great, LT spice has some of the best (if not the best) solvers out of any software package. \$\endgroup\$ – Voltage Spike Dec 2 '20 at 19:12
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    \$\begingroup\$ Is that a problem? \$\endgroup\$ – Voltage Spike Dec 2 '20 at 19:33
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    \$\begingroup\$ Open loop gain still matters in closed loop, it can never be exceeded and openloop gain affects the bandwidth when combined with GPBW \$\endgroup\$ – Voltage Spike Dec 2 '20 at 20:55

For any engineering problem, if you hear someone talk about an "ideal <something>", that something they're talking about does not exist in its ideal form. This kind of comes from Plato's theory of forms, where there's this ideal world out there, and everything in this world is just an imperfect shadow of it.

This counts in spades for op-amps, which is probably why today's DigiKey website lists 38,955 different part numbers under "Linear - Amplifiers - Instrumentation, OP Amps, Buffer Amps".

The answer to any question of the form "can I buy an ideal <something>" is no.


Ideal op amp model is a heuristic, which exists only to simplify the math for designing a useful feedback system.

In a real op amp system with negative feedback, there is a small but important voltage difference between the input terminals. Given the output equation for voltage gain with an offset voltage:

$$ V_{out} = A_{v} (V_{+} - V_{-} + V_{os}) $$

Solving for \$V_{+}\$ yields

$$ V_{+} = (V_{out} / A_{v}) + V_{-} - V_{os} $$

This equation is useful for estimating the range of \$V_{+}\$ given a worst-case range of offset voltage \$V_{os}\$ and worst-case voltage gain \$A_{v}\$. This is a secondary effect that should be considered when open-loop gain is limited (such as near the op amp's frequency limits) or when considering temperature variations that can affect offset voltage. But for initial design of the feedback network, these effects are negligible -- if the system is properly converging, then the errors get corrected by the feedback. During initial design it's hard to estimate what that voltage difference might be, but we know if the system is working then that voltage difference is small and negligible. So we design with the ideal op amp model assuming \$V_{+} = V_{-}\$ (with no current between the two nodes) and only evaluate \$V_{+}\$ after the initial design is complete.

Ideal op amp model contains a subtle contradiction:

  • No input offset voltage error (\$V_{os} = 0\$)

  • Infinite open-loop voltage gain (\$A_{v} = {infinity}\$)

With negative feedback, an ideal op amp would drive its inverting input to exactly the same voltage as the non-inverting input. Yet, with zero volts between \$V_{+}\$ and \$V_{-}\$, the ideal op amp output would always be zero, making it useless.

This is just an effect of dividing by 'infinite gain'; like dividing by zero, it's just an area where the model breaks down. Sometimes in tutorials you may see references to a "virtual short" between \$V_{+}\$ and \$V_{-}\$, meaning that they are at (nearly) the same voltage -- but if you replaced that with a real, physical connection, then current would be able to flow, and the feedback system would no longer work. The "virtual short" idea is just a simplification that makes it easier to analyze the system and choose component values.

Ideal op amp also has a few other impossible characteristics:

  • no external power supply rails, so output swing is unlimited, and it delivers an unlimited amount of energy to whatever kind of load is attached to the output, from a hidden forever magic power source

  • no output impedance, so it can drive 1000000A without losing signal or overheating

  • no internal PN junctions or any other internal organs, so input common-mode range is unlimited, and it will never be damaged even if there is 1000000V between its inputs. Also, no PN junctions means no shot noise, and no resistive elements means no Johnson noise.

  • costs nothing to manufacture, since it exists only in the minds of circuit designers

Like all models, the ideal op amp model breaks down outside its useful domain.

  • \$\begingroup\$ Re "Schott noise": Do you mean "shot noise"? Or perhaps Schottky noise? \$\endgroup\$ – Peter Mortensen Dec 4 '20 at 12:17
  • \$\begingroup\$ @PeterMortensen Yes, that was a mistake, I think "shot noise" is the correct term (though Schottky did publish work on that effect) \$\endgroup\$ – MarkU Dec 4 '20 at 12:55

Sure, depending on your bandwidth requirements, what you are trying to do, and the precision of your ability to measure.

If frequency is low, meaning you do not have square wave style transitions, and your signals are strong relative to noise or the charge of the electron, and the time scale of your measurements, you can be as close to ideal as you want. Or indistinguishable from ideal.

In a real circuit you also need feedback elements like resistors and capacitors that behave close enough to ideal to be indistinguishable. This is why analog computation and control works.

However, you can not have an ideal op-amp in the theoretical sense. That requires infinite gain, perfect linearity, and other infinities and perfections.


You really don't want an ideal op amp. I'll just quote from one of my favorite books, 'OP AMPS for EVERYONE' (Bruce Carter and Ron Mancini):

The name \${\it ideal\ op\ amp}\$ is applied to this and similar analyses because the salient parameters of the op am are assumed to be perfect. An engineer may wish that an ideal op amp existed at times, but if such a component actually did exist, it would destroy the known universe! See the end of the chapter for an explanation...

An ideal op amp has the following specifications:

\$\bullet\$ It draws no supply current and therefore has no power supplies. Hence, it doesn't even have to be turned on to be dangerous!

\$\bullet\$ It has no \$ V_{OH}\$ and \$V_{OL}\$ limitations because it has no power supplies. Therefore, its output voltage swings from \$\pm \infty V\$.

\$\bullet\$ It has zero output resistance, and therefore it is capable of supplying infinite current at each voltage extreme.

\$\bullet\$ It has infinite gain, and therefore the slightest input signal would allow it to swing to positive and negative infinite voltage (without feedback components, that is).

\$\bullet\$ It has infinite slew rate and therefore would swing to either rail-- both equally destructive-- instantly.

Therefore an ideal op amp, just lying on the table with no power applied, would instantly take a quantum difference between its positive and negative terminals and amplify that difference to an infinite voltage output at infinite current. The resulting surge of power would be a sphere of destruction radiating out from the op amp at the speed of light!

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    \$\begingroup\$ No, it is not that bad. It could only destroy the known universe at the speed of light. As the known universe is expanding (and even accelerating), some parts would never be affected. \$\endgroup\$ – Peter Mortensen Dec 4 '20 at 12:27
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    \$\begingroup\$ @PeterMortensen What a relief! \$\endgroup\$ – mguima Dec 4 '20 at 13:35

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