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I am trying to design a butterworth low pass filter with corner frequency at 2khz as seen below.

enter image description here

And my problem is when I find its transfer function It is as follows:

\$H(s)=\dfrac{1}{1+1.414\times (\frac{s}{2\ 000})+(\frac{s}{2\ 000})^2}\$

And bode plot of this function on octave is as folows:

enter image description here

The cutoff frequency occurs at -3 dB on this graph but when I simulate the circuit on LTSpice I found it around -6 dB.

enter image description here

Unfortunately it is around -6 dBs.

I did some research online and I found following graph:

enter image description here

(Source) Following to the graph above it must be around -6 dB because it is second order filter.

I got so much confused. One of them must be wrong. Either transfer function or simulation.

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2 Answers 2

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Whew. Several problems here.

First, you accidentally labeled your C2 reference designator as 16n. I don't think it affects the simulation (since you also set the value to 16n), but it's better to fix it so people looking at your schematic can better understand it and make clear comments on it.

Second, you're doing a linear sweep on the frequencies and trying to compare it to a Bode plot which is a decade sweep. Simply change the lin to dec such that you get: .ac dec 10000 1 5k

Last, your schematic doesn't represent a 2nd order Butterworth, which has complex conjugate poles to acheive the "maximally flat" amplitude characteristic. What you have is two 1st order RC sections cascaded and then buffered with the opamp at the end. Not only do each of these have a single real pole, but they "load" each other down so their transfer functions can't be multiplied together without introducing a buffer in between. If you were trying to match the topology of the circuit shown in the website you referenced (duplicated below), you didn't quite get it 100% correct. Your C2 should be in feedback from the opamp output and not tied to ground.

enter image description here

For more information regarding the last point, see here:
Is this cascaded RC filter transfer function explanation wrong?

and here:
Transfer function of three cascaded RC filters?

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  • \$\begingroup\$ For a Butterworth response, if you include RA & RB then the resistor and capacitor values can be equal value but the gain must be equal to 1.586 or as close to it as you can get. Then the -3dB frequency = 1/(2 * PI * R * C). If you make the op amp into a buffer then R1 = R2 and C1 = 2 * C2 and then the -3dB frequency is equal to 1/(Root2 * 2 * PI * R * C2). \$\endgroup\$
    – user173271
    Commented Dec 6, 2020 at 23:42
  • \$\begingroup\$ @James Dasrite. \$\endgroup\$
    – Ste Kulov
    Commented Dec 7, 2020 at 2:41
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Design of 2 pole (2nd Order) Sallen & Key unity gain and equal value Butterworth Filters.

Butterworth Filters

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