The two 100k voltage divider resistors are in parallel to the signal then are a 50k resistance to ground.

The simulation does not work since the input is 1V but the output which also should be 1V measures only a few hundredths of a volt.
Your frequency response shows only high frequencies, not the low frequencies you want cut.

My simulation program works perfectly:

So does mine in post #30. It might be helpful to learn more about this particular simulator by starting with one or more simpler circuits.

 

Here's the simulation with a gain of 1.7 (4.66dB) so that the two resistors controlling the response (R2 and R3) are equal:

 

This seems to work.
First plot is 47k pot, 93 Hz @ -3dB
2nd plot is 0k pot, 611 Hz @ -3dB
Now I just need to figure out how to do a parameter list in Eagle like you guys do.

 

but the fc=1/(2*Pi*R*C) doesn't seem to fit the simulation. For those values above, I get 68 Hz and 1129 Hz.

 

but the fc=1/(2*Pi*R*C) doesn't seem to fit the simulation. For those values above, I get 68 Hz and 1129 Hz.

You've got the capacitor (C4) in the wrong place. Although you get a believable simulation output, the circuit won't actually work. Pay careful attention to @crutschow 's circuit.

 

Specifically, C4 should be parallel with R4.

Better yet, buffer the virtual ground with an opamp voltage follower, then it will not interfere with the circuit at all.

 

Ah, got it. Thanks! Using a 100k pot gives me 50-1000 Hz range.

 

You've still got the gain at unity, so that means your Q will be 0.5 - your filter will be critically damped, not Butterworth.

 

You've still got the gain at unity, so that means your Q will be 0.5 - your filter will be critically damped, not Butterworth.

Hmmm, yes. The slope is not 6 db/octave, sometimes it's more, sometimes it's less. But it is smooth up at the transition region. How do I fix that?

 

Here's the voltage follower that gives a gain of 1.45, right?

 

Here's the voltage follower that gives a gain of 1.45, right?

No.
Remove the connection between the output and the (-) input.

 

?

 

With a gain of one, a Butterworth response with its sharp corner uses the resistor to ground having double the resistance of the feedback resistor.

But your two resistors have the same resistance causing a droopy Bessel response.
You might not hear the difference:

 

?

Close, but no cigar.
R2 should connect to the op amp output, not the (-) input.
Otherwise the filter does not see the added gain.

Edit: But you also need to add a capacitor at the bottom of R8 to block the DC bias, as I showed in my simulation, otherwise the gain will cause the op amp output to saturate at its rail.

 

That’s interesting. That Grace Designs Felix preamp I mentioned with the 12dB hpf, says it uses a Thomson-Butterworth. Thomson is synonymous with Bessel I think. I asked them about it and he said it’s a hybrid between Bessel and Thomson. And he suggested that for guitar, a Bessel curve might be better.

 

Close, but no cigar.
R2 should connect to the op amp output, not the (-) input.
Otherwise the filter does not see the added gain.

 

Did you see my edit about also adding a capacitor?

 

Ah. Thanks!

 

Here is the missing capacitor and why it is needed:

 

Here is the result of my simulation.