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I needed an active low pass filter for my project that can produce an output voltage within the required settling time and ripple voltage ranges. Although I managed to get a good enough output voltage with the shown resistance/capacitance values, if you calculate the cut-off frequency of this Sallen-Key low pass filter it's around ~4250 Hz. The input signal however, is 20 kHz.

The Sallen-Key low pass filter:

enter image description here

The output waveform:

enter image description here

My questions are:

How does a low pass filter with a cut-off frequency of 4250 Hz let a 20 kHz signal through?

Are cut-off frequencies, and therefore, stopbands of filters ignored in LTspice for some reason?

Additional info:

  • Op amp used in the circuit is the ideal single pole amplifier model of LTspice named "opamp" in op amps directive.

  • Input signal is 20 kHz 0-5 V pulse waveform with 50% duty cycle.

  • If the op amp is switched for different models such as LT1001, the output voltage level is around a few pV. This still happens when the resistance and capacitance values are changed and the cut-off frequency is more reasonable. Example: 5k resistors, 1n capacitors, ~32 kHz cut-off.

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    \$\begingroup\$ What is the peak-to-peak voltage of your 20kHz input? Perhaps the filter is performing correctly. \$\endgroup\$
    – Math Keeps Me Busy
    Apr 12, 2021 at 19:59
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    \$\begingroup\$ LTSpice has absolutely no notion of your "cut-off frequency"; that's an emergent property. I think your circuit might just be incorrectly designed (or as Math.. mentions, your expectations are unreasonable). \$\endgroup\$
    – Marcus Müller
    Apr 12, 2021 at 19:59
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    \$\begingroup\$ If your cutoff was 5 kHz you'd expect 12 dB/octave attenuation (2 poles), so 24 dB total at 20 kHz. For a 5V sine you'd expect your signal to be attenuated to around 300 mV at the output. Since you're using a pulse the higher order harmonics will be attenuated more, but you shouldn't expect to see no 20kHz component on the output. \$\endgroup\$
    – John D
    Apr 12, 2021 at 20:08
  • \$\begingroup\$ The problem was that I incorrectly assumed the filter would pass no 20 KHz component since the signal has a frequency of almost 5 times the cut-off frequency. I have been using high order (20-25) filters for digital filtering a lot lately which left the picture of a very fine transition band in my mind and that led to the confusion in the first place. However I really do appreciate the calculations and the comments, thanks everyone. \$\endgroup\$
    – Curious Student
    Apr 12, 2021 at 21:31
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    \$\begingroup\$ @CuriousStudent If those filters were FIRs, then they were not as high as you think they were. Because IIRs of that order have a ton of numerical issues. \$\endgroup\$
    – a concerned citizen
    Apr 12, 2021 at 22:04

1 Answer 1

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Back of the envelope calculation...

20k/4250 = about 4

It's a second order filter, so at a frequency 4x above its -3dB point it should attenuate the signal by 4 squared, or about 16.

Output signal is about 0.3Vpp.

Input signal is 5Vpp.

5/0.3 = about 16.

So it delivers the correct attenuation.

If you want more attenuation, use a lower -3dB point, or a higher order filter, or a much higher PWM frequency so it gets attenuated a lot more.

If you used a high -3dB point of 4250 Hz because you wanted fast settling time, then don't use a PWM for analog output when you want fast settling time, unless your micro supports high PWM frequencies. Use a DAC instead.

if the op amp is switched for different models such as LT1001, the output voltage level is around a few pV.

Did you forget the opamp power supplies?

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  • \$\begingroup\$ I did not forget the supply voltages and I find it strange that the output is still only a few pV. However, the ideal model is the main op amp model I will be using for the design so that doesn't cause me any trouble. I really appreciate the detailed response, thanks. \$\endgroup\$
    – Curious Student
    Apr 12, 2021 at 20:25
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    \$\begingroup\$ Nice answer (+1), just a minor correction: 20k/4.25k is closer to 5 than to 4, and if the number 16 is closely satisfied it's because 5/4~4/pi (for the square wave). \$\endgroup\$
    – a concerned citizen
    Apr 12, 2021 at 22:08
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    \$\begingroup\$ It's engineering math ;) \$\endgroup\$
    – bobflux
    Apr 12, 2021 at 23:30

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