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I'm designing a low-pass filter with a passband up to 1 MHz and a stopband starting at 1.8 MHz. I've designed a 4-stage filter using Sallen-Key topology and built it using NE5532P op-amps, supplied with 5V and GND.

enter image description here Here are the first two stages of my filter:

enter image description here And here are the last two stages:

Note: The schematics show OPA350s because I plan to use them in the PCB, but I'm currently testing with NE5532P on breadboards.

The problem arises in the last stage: the output is distorted. After eliminating other possible issues, I believe the problem is related to the 2.79nF capacitor. If I replace this capacitor with a 10pF one, the last stage starts working, but the filter no longer meets my requirements. I've read that op-amps have limited capacitive load capabilities. Is the 2.79nF capacitor acting as a problematic load for the op-amp? I've found suggestions about adding a resistor in series or using Multiple Feedback topology instead of Sallen-Key for the last stage. My questions are:

What's causing this problem, and how can I solve it while maintaining the required filter characteristics? Will I face the same issue with the OPA350 op-amp?

Any insights or suggestions would be greatly appreciated. Thank you!

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  • \$\begingroup\$ Are you sure that 2.75nF is the desired value for the filter? It seems much too large relative to the 10pF at the input. Usually the values are similar compare with the first stage. Also 1MHz is probably too high for NE5532 and you may be slew rate limiting or overloading the first stage to get the output you are expecting. \$\endgroup\$
    – Kevin White
    Commented Jul 31 at 22:16
  • \$\begingroup\$ @KevinWhite I did calculations for frequency response in Matlab and it seemed alright. As you and others have pointed out the problem is with the bandwidth of an op amp. \$\endgroup\$
    – Nicolas Tsagareli
    Commented Aug 4 at 20:01

2 Answers 2

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Is the 2.79nF capacitor acting as a problematic load for the op-amp?

No that is not a problem, the op amp is not loaded with 2.79nF to ground. For high order Chebyshev filter types the last stage will have a relative high Q which results in a high capacitor ratio as seen in your example. Maybe consider switching to a Butterworth filter that exhibits filter stages with lower Q?

enter image description here

The problem arises in the last stage: the output is distorted. After eliminating other possible issues, I believe the problem is related to the 2.79nF capacitor.

The main problem is the op amp used for breadboarding. The NE5532, especially in the last high Q stage, has not the necessary GBW, that is ~60MHz for a min. loop gain factor of 10 (20dB). The NE5532 also needs at least 10V single or +/-5V dual supply and does not work with just 5V. Furthermore, because of the single supply method, I don't see the op amps being biased with Vcc/2 for linear operation.

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  • \$\begingroup\$ Hello, thank you for your quick answer and apologize for my late one. Your answer about GBW has been very helpful. Turns out I had wrong and little understanding of GBW of op amps. I have another question. In one article (ww1.microchip.com/downloads/en/DeviceDoc/adn003.pdf) I found that when picking op amp I should use factor of 100 to calculate correct GBW that I need. For example if I need 2.5 Gain at 1.8MHz GBW that I need is GBW = 100*2.5*1.8 = 450MHz. Is not this too high? \$\endgroup\$
    – Nicolas Tsagareli
    Commented Aug 4 at 19:57
  • \$\begingroup\$ The recommended open loop gain margin factor is 100 (40dB). I used only 10 (gain error < 10%) as an absolute minimum. In your GBW calculation example you didn't take into account the Q of the filter stages, which raises the necessary op amp GBW even more. The (simplified) formula to estimate GBW product is GBW = 100 * Gcl * fc * Q. The filter stage with the highest Q (~5 for a low ripple Chebyshev flter) is the last stage. You end up with a GBW of 2.25GHz! Op amps with such a GBW are typically decompensated and unstable at unity/low gain. \$\endgroup\$
    – Raonoke
    Commented Aug 5 at 1:42
  • \$\begingroup\$ Thank you for clarification. \$\endgroup\$
    – Nicolas Tsagareli
    Commented Aug 5 at 8:07
  • \$\begingroup\$ @Nicolas Thanks for getting going on your question with others. I especially thought that Raonoke wrote well for you. In any case, they suggest a Butterworth because of the lower max-Q for the stages. But (-20*log10(sqrt(Butterworth(8)*conjugate(Butterworth(8))))).subs(omega,1.8).n() gives me -40.8439584238072 at 1.8 MHz with the 8th order Butterworth that they suggested. (Down the obvious -3.0103 dB at omega=1.) What's your stop-band expectations at 1.8 MHz? \$\endgroup\$
    – periblepsis
    Commented Aug 6 at 3:27
  • \$\begingroup\$ Stop-band expectation is d=0.1. I have decided to use Butterworth as Raonoke advised. \$\endgroup\$
    – Nicolas Tsagareli
    Commented Aug 6 at 19:19
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The OPA350 isn't very descriptive on what capacitance will do to the signal other than the amount of overshoot that will happen when loaded: enter image description here Source: https://www.ti.com/lit/ds/symlink/opa350.pdf

And most of their testing is done with 100pFs or less of loading. You already do have some resistance in series with the capacitor with the R7 and R8 resistors

I would consider these things:

  1. Get a different opamp (for at least the last stage) that can handle capacitave loads
  2. Increase R7 and R8 (and adjust C7 and C8 accordingly to maintain the same bandpass
  3. Use the OPA350 and a series resistor with C7. This will reduce the effectiveness and soften the pole and the rolloff (AFAIK), but is best simulated, put the circuit into spice and see what works.
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    \$\begingroup\$ Most opamps are ok driving capacitors with only few 10's of ohms series resistance - this one has 500! That isn't the problem. \$\endgroup\$
    – Kevin White
    Commented Jul 31 at 22:37

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