1
\$\begingroup\$

enter image description hereI'm designing a resonant filter based on a state variable filter topology with sweepable frequency and variable resonance. The filter is designed to sweep from roughly 20 Hz up to 20 kHz (50 kohm linear dual gang potentiometer with 50 ohm limiting resistors and 180 nF capacitors). The prototype works fine and sounds good however, both in the simulation and on the prototype, when turning the potentiometer to sweep the frequency band, 75% of the pot's turn works from 20 Hz to roughly 80 Hz or 100 Hz and the remaining 25% of the pot's turn sweeps up to 20 kHz. Is there a way to sweep the entire audible frequency band using one potentiometer while resolving the pot's sensitivity at high frequency? (cutoff frequency vs pot rotation).

Any help would be greatly appreciated!

\$\endgroup\$
5
  • 3
    \$\begingroup\$ Show us the schematics of your filter - simply paste it into your question. \$\endgroup\$ Jan 9, 2020 at 20:15
  • 1
    \$\begingroup\$ And check whether you've got a linear pot or a log (audio taper). \$\endgroup\$
    – Transistor
    Jan 9, 2020 at 20:32
  • \$\begingroup\$ Sorry about that, I've added the schematic to the initial post. And yes, I'm using a linear pot. I've tried different pots both lin and log and still same outcome. \$\endgroup\$
    – Ryan
    Jan 9, 2020 at 20:45
  • 2
    \$\begingroup\$ Ground symbols facing up triggers my PTSD. Seriously, though, work out the math on those RCs and maybe you will see why this is happening. \$\endgroup\$ Jan 9, 2020 at 20:47
  • \$\begingroup\$ is this why a lot of sweepable audio filters use the rare and mystical anti-log pot? \$\endgroup\$
    – user156429
    Jan 9, 2020 at 21:43

4 Answers 4

2
\$\begingroup\$

Frequency is proportional to 1/(R*C), so it won't be linear with R.

\$\endgroup\$
7
  • \$\begingroup\$ That's correct, it's inversely proportional. But the math of fc = 1/(2piRC) where R is varied from 50 ohm to 50 kohm plots differently from what the simulation and prototype are showing. \$\endgroup\$
    – Ryan
    Jan 13, 2020 at 0:29
  • \$\begingroup\$ Hi, Ryan: The usual schematic for this kind of filter does not have grounded pots connected as you have drawn. Usually, the pots are just variable resistors, acting as the input resistors of the integrators. Did you make a mistake here? Try fixing that. Also, depending on your op amp, At high frequencies loop gain is low and the simple equations won’t be quite right. Is it at high frequencies that you see the problem? \$\endgroup\$
    – user69795
    Jan 13, 2020 at 0:42
  • \$\begingroup\$ Thanks for your reply. I've removed the grounding off the pots in the simulation software and the prototype that I've built has the third terminal of the pots left unconnected. Yes, the issue is at high frequencies. The pot sweeps from 1 kHz up to 20 kHz on the last 25% of the pots turn whereas the first 75% of the pot's turn sweeps 20 Hz up to 100 Hz (roughly). \$\endgroup\$
    – Ryan
    Jan 13, 2020 at 0:47
  • \$\begingroup\$ Hi, Ryan. Doesn’t that make sense if f=1/(2*piRC)? Seems to be working as it should..... \$\endgroup\$
    – user69795
    Jan 14, 2020 at 1:38
  • \$\begingroup\$ I'm with you 100%. Totally makes sense. Do you think it's possible to sweep the entire audible frequency band with one potentiometer while evening out the cutoff frequency vs pot rotation so it won't sweep so fast at high frequencies? \$\endgroup\$
    – Ryan
    Jan 15, 2020 at 2:34
2
\$\begingroup\$

Assume the output impedance of U2 is 0 ohms across the frequency band of interest. If so, then the impedance seen by the U3 circuit is 50 ohms at one end of the FREQ1 pot rotation, 0 ohms at the other end, and approx. 12.5K ohms in the center. Same for the U3 circuit. This introduces a non-linearity into the circuit's response to pot rotation.

Both pots should be connected as variable series resistances rather than voltage dividers. This eliminates the other problem, where 100% of the signal goes away at one end of the pot rotation.

\$\endgroup\$
2
  • \$\begingroup\$ Thank you! I believe this is the issue. Do you have any advice on how to fix this? \$\endgroup\$
    – Ryan
    Jan 13, 2020 at 0:50
  • \$\begingroup\$ Wouldn't leaving the third terminal of the pots unconnected make it a variable resistor rather than a voltage divider? disregard the fact that the pots are grounded in this schematic. In my prototype, the unused terminals are left unconnected. I'm sorry if what i'm saying doesn't make sense haha. \$\endgroup\$
    – Ryan
    Jan 13, 2020 at 0:57
0
\$\begingroup\$

Probably you made mistake in your potentiometers characteristics. Take in mind that in USA key A means log potentiometer, and key B means linear. In Europe this is in opposition: key A means linear potentiometer, and key B means log.

\$\endgroup\$
1
  • 2
    \$\begingroup\$ I've taken that into consideration and double-checked the datasheets and everything tracks. I think the circuit is affecting the pots since the simulation also shows the same discrepancy. \$\endgroup\$
    – Ryan
    Jan 9, 2020 at 20:56
0
\$\begingroup\$

You designed a 1/R controlled value to alter hi/lo freq. BW in a k1/R= f1 for centre frequency in an RC active BPF.

Instead you should have inverted R value in software then sent to Pot and calibrate for those values.

division= use 2's compliment and add with carry removed.

Also you have the Pot's shown as attenuators so you lose the signal at the same time.

Do not ground signal on Pots!

\$\endgroup\$
6
  • \$\begingroup\$ I've added a schematic to my original post to clear things up. I believe you're referring to digital audio? Whereas my design is purely analogue. \$\endgroup\$
    – Ryan
    Jan 13, 2020 at 0:25
  • \$\begingroup\$ correct answer above now \$\endgroup\$ Jan 13, 2020 at 0:28
  • \$\begingroup\$ Thank you! I think I understand what you're saying but could you elaborate more with reference to the schematic to better understand? That would be extremely helpful. \$\endgroup\$
    – Ryan
    Jan 13, 2020 at 0:34
  • \$\begingroup\$ Use only 2 pins on the Pots \$\endgroup\$ Jan 13, 2020 at 4:08
  • \$\begingroup\$ Unlike the schematic above, I am only using 2 pins on the potentiometers. The unused terminals are left unconnected and still, there are non-linearities. \$\endgroup\$
    – Ryan
    Jan 13, 2020 at 8:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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