I used this online tool to play around and design a Sallen-Key low pass filter with a cutoff frequency of 0.3Hz. When I now put the resulting values for R1, R2, C1 and C2 in the formula I get exactly my desired 0.3Hz. But when simulating the circuit with LTspice, the cutoff frequency was unexpectedly at 0.46Hz. That is more than 50% off!

enter image description here

Can anybody tell me where the difference is coming from? Am I missing something obvious? The only thing I can think of is that Spice is using numerical solutions, but the error of the calculation should never be that high! And the OpAmp shouldn't make any difference at these low frequencies either... What frequency would I probably get when acutally building the circuit?

  • \$\begingroup\$ I guess the site is assuming ideal opamp (same degree for num. and den.) whereas ltspice is not. \$\endgroup\$
    – edmz
    May 20, 2019 at 18:05
  • \$\begingroup\$ Can you try simulating with an ideal opamp instead? The gain-bandwidth product is not the only non-ideality that might be affecting you. \$\endgroup\$
    – Justin
    May 20, 2019 at 18:17
  • \$\begingroup\$ Just tried it with an ideal opamp. It does not make any difference. \$\endgroup\$
    – jusaca
    May 20, 2019 at 18:27

3 Answers 3


There's no difference in the simulation and calculation, at least none that I can see. Here's what the calculator gives: enter image description here

The "problem" is that in the calculator the cut-off frequency seems to mean the frequency where there is a pole. And with this given topology at that frequency there is 6 dB of attenuation. You need to find the -3 dB cut-off frequency from the graph.

  • \$\begingroup\$ Oh wow, I checked for -3db... This is embarrassing. But it is -6db because the Sallen-Key is second order. It has nothing to do with voltage and power. \$\endgroup\$
    – jusaca
    May 21, 2019 at 8:08
  • \$\begingroup\$ But the cutoff frequency is defined as the point, where power is -3 dB, or voltage is 0.7*Vin... Or am I completly off here? \$\endgroup\$
    – jusaca
    May 21, 2019 at 15:50
  • \$\begingroup\$ No - certainly, the calculator has not assumed -6dB.. The calculator (like any calculator) needs the POLE data which are not identical to the cut-off. The sentence below your drawing is wrong! \$\endgroup\$
    – LvW
    Feb 4, 2020 at 12:13

The online calculator is assuming ideal components. If I use an ideal op amp in LT spice, I get pretty close to the same answer as the tool:

Freq, Mag, Phase 0.30432198871077,-5.8043099990704,89.001051267349

This is the answer that I get from LT spice with an ideal opamp:

enter image description here

They are off only by 0.1dB and that is probably because I can't get the exact same point on the plot (0.304321 vs 0.304735 )

I suspect the input bias current is part of the problem on the OPA2337 , it being 40nA and the current through the resistor being ~2uA (really depends on the waveform coming in, Mine is at 1VAC). Either way with a resistor that high, input bias current can change things albeit in this case maybe 1 or 2%.

Another problem is some manufacturers models don't accurately show real world performance, so double check the model with the data sheet. Especially input bias current.

  • \$\begingroup\$ The OPA2337 datasheet specifies input bias and offset currents in picoamps, not nanoamps. If the simulation model has nA input currents, that could be part of the problem (The fact it's a TI part in an Analog Devices simulator makes me suspect of the available models) \$\endgroup\$
    – The Photon
    May 20, 2019 at 18:48
  • \$\begingroup\$ I think 40000pA is a 40nA, but that is off of their website, so it could mismatch the datasheet \$\endgroup\$
    – Voltage Spike
    May 20, 2019 at 20:00
  • \$\begingroup\$ The datasheet says 10pA, so I'd believe that. Either way, the input bias current on the model is suspect, I've seen many TI models that don't get things right \$\endgroup\$
    – Voltage Spike
    May 20, 2019 at 20:02

Jusaca...the error is on your side. EDIT: See the end of this answer.

You have mixed the pole frequency wp withthe 3dB-cutoff frequency wo.

The parts values give a pole frequency fp=wp/6.28=0.3 Hz and a pole quality factor Qp=0.504.

Please note that the pole frequency is identical to the 3dB cutoff for a BUTTERWORTH response only (Qp=0.7071). Your design with Qp=0.5 gives a filter characteristic worse than Butterworth and even worse than Bessel-Thomson (as far as the transition region between passband and stopband is concerned)

In your case, the 3dB cutoff must be larger (than the pole frequency) by a factor of app. 1.5 .

Fazit: You have designed a filter for a pole frequency of fp=0.3 Hz (and NOT for a cutoff at fo=0.3 Hz). This can be verified very easily: For any second-order filter the phase shift at f=fp must be 90 deg. Look at the phase response - and this will be confirmed at f=0.3 Hz.

  • Here are the corresponding formulas:

wp=1/[R1C1SQRT(kckr)] and Qp=SQRT(kckr)/(1+kc) with kc=C2/C1, kr=R2/R1.

Correction: The error you have made is: For calculation of the resistors you have selected damping factor =1 (which means Qp=0,5).

For a correct calculation select "quality factor" Qp=0.707 and the resistors are 240k and 560k. Now you have a BUTTERWORTH response with fp=fo=0.3 Hz.


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