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So I have this simple OP-AMP circuit that is supposed to be a low-pass filter for which I made the calculations so it passes all frequencies below 1591 Hz (approx). The values of the components are in the image.

However, the frequency response in my simulation program (LTSpice) is way higher (two orders of magnitude!) above my theoretical ones.

To make it even funnier, empirically I got the correct frequency response (in a bread board with an LT1013 Amp) so what's wrong in my simulation?

[![LTSpice Simulation.

LTSpice Simulation

For Circuitlab I got the expected output as well. Same circuit. Is LTSpice that bad?

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

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You specified a linear frequency sweep, with 50 points from 1 Hz to 120 MHz. That means the simulation is run at 1 Hz, 2.400001 MHz, 4.80001 MHz, etc., up to 120 MHz.

Most importantly, it means that the part of the output curve between 1 Hz and 2.4 MHz is generated by interpolation rather than by doing any simulations at the intermediate points. Apparently it did a linear interpolation on a linear scale, which translated to a curve when you changed your plotting axes to be logarithmic.

I'd recommend to switch to a log sweep, which is called either "decade" or "octave" sweep in LTSpice, depending whether you want to specify the number of points per decade or per octave.

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  • \$\begingroup\$ Thank you so much! \$\endgroup\$ – Javier González Apr 18 at 0:40
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    \$\begingroup\$ Agree that a decade sweep is the way to go. When you choose a decade sweep, you get the specified number of points per decade instead of just total. As a result a linear sweep is easily severely undersampled. In this case you would have gotten 300 points (over 6 decades) just by changing to decade mode. Alternatively you could use 300 points in linear sweep mode and obtain a similar result. On a log-log plot like this (dBs are logs) the decade sweep should produce a superior curve. \$\endgroup\$ – Jim Apr 18 at 0:44
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Javier....the simulation shows the correct response. Use an IDEAL opamp model - and the result will be as expected by you.

So - what is the problem with the REAL opamp model? Answer: It is real - with a finite output impedance. What happens?

(1) For rising frequencies, the gain drops (can be seen for the ideal model)

(2) At the same time, the capacitive impedance of the feedback path becomes smaller and smaller....and a rising part of the input signal arrives DIRECTLY (that means: NOT trough the opamp) at the output - across the FINITE output impedance!.

(3) Hence, the output voltage cannot drop as expected ....due to the parasitic path through the feedback circuits.

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