Im working on a Dual LFO with "Depth" and Rate controls. I already got the "root" LFO signal working and if I do this, I can generate the LFO swinging from 0v to 9v... and with Less Depth from, say, 8v to 9v: enter image description here enter image description here

Now I need the same signal but with the direction inverted (when the first one goes up, the second one goes down) BUT having the same Maxs and Mins.

If I do this (img below), with Depth at minimum I get, say, 8v to 9v in the first one, but 0v to 1v in the second one... enter image description here enter image description here

and I want the same 8v to 9v, but with the orientation being the inverse, like this: enter image description here

Things I already know how to do:

  • Invert a signal
  • Add DC Offset to signal (but the offset always changes depending on the Depth)
  • Remove DC Offset to signal
  • Substract two signals
  • Sum two signals

I need to figure this out!! There has to be a simple way!


I came up with this math: 9v - 2 * Signal(without offset, centered on 0)

Based on that I made this simplification for testing the concept where I generate the two signals: LFO1 goes from 7 to 9 AUX_SIM goes from -1 to 1 (just LFO1 without the offset)

And it works: enter image description here

SO, then I generate the AUX_SIM from LFO1 removing the offset and is ok: enter image description here

BUT when I put it all together, The LFO1 without the offset is not centered on Cero and everything shifts, because life is HARD :D

enter image description here

I thought I was close this time... Any ideas??

  • \$\begingroup\$ This would be easier if you have a dual-gang potentiometer. Do you? \$\endgroup\$
    – Hearth
    Jul 20 at 15:41
  • \$\begingroup\$ Yes, I though about that. But I want to make it with simpler components and now I want to solve the challenge! :D \$\endgroup\$ Jul 20 at 15:45
  • \$\begingroup\$ I'm not actually sure it's possible without significant complexity, but I don't have time right now to give it much thought. \$\endgroup\$
    – Hearth
    Jul 20 at 15:49
  • \$\begingroup\$ @Hearth I edited the original question with new insight! \$\endgroup\$ Jul 20 at 17:51

1 Answer 1


This may do what you want but, I would run it through a simulator first if I were you: -

enter image description here

You should also make sure that you have power rails on your op-amp that are in excess of 9 volts and preferably 12 volts (for most bog-standard op-amps). I mention this because you didn't say what power rails the op-amp receives.

  • \$\begingroup\$ Hi Andy. I tried that but didn't work. It doesn't even invert the signal. Im connecting the opamp to 9v and ground. That's why I put VREF on the Positive pin of it, with 4.5v \$\endgroup\$ Jul 20 at 17:01
  • \$\begingroup\$ What op-amp did you use? Wrong op-amp and it won't work. Even with the best op-amp you might need a positive supply slightly higher than 9 volts such as 9.1 volts. \$\endgroup\$
    – Andy aka
    Jul 20 at 17:19
  • \$\begingroup\$ I'm using UniversalOpamp2 in Eagle (the one in the image was the generic one just for simplicity). I edited the original post with new insights and the exact schematic and plot Im using. I always use that opamp without problems. \$\endgroup\$ Jul 20 at 17:46
  • \$\begingroup\$ Those additions don't help and contradict your original schematic. You should repeat the circuit using the potentiometer (as originally disclosed) but with a higher positive rail for the op-amp like 12 volts @ElectronOne \$\endgroup\$
    – Andy aka
    Jul 20 at 17:52
  • \$\begingroup\$ Im going to power the device with 9v battery, and the LFO that I already designed (it also has a Shape control) swings from 0 to 9v :/ \$\endgroup\$ Jul 20 at 17:55

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