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 '19 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 '19 at 18:17
  • \$\begingroup\$ Just tried it with an ideal opamp. It does not make any difference. \$\endgroup\$ – jusaca May 20 '19 at 18:27

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 '19 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 '19 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 '19 at 20:02

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 this is a voltage amplifier (or attenuator) and the cut-off frequency is defined as the frequency where the voltage is half that of the passband. In decibels that means -6 dB.

-3 dB is the frequency where output power is half of the passband.

  • \$\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 '19 at 8:08
  • \$\begingroup\$ No, it has nothing to do with the filter order. See this question for example: electronics.stackexchange.com/questions/71358/… \$\endgroup\$ – TemeV May 21 '19 at 12:49
  • \$\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 '19 at 15:50
  • \$\begingroup\$ Quite commonly yes, but cut-off frequency is not that unambiguous term. That calculator of yours seems to define it as -6 dB. \$\endgroup\$ – TemeV May 21 '19 at 16:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.