I thought I wanted a circuit that was a basic pan control, but after mapping it out it isn't so standard after all. The final design will have this exposed: two mono inputs (left and right), one stereo output (left/right), one pan/balance control. When the pan/balance control is set to 50%, one mono is 100% on one side, 0% on the other side. The other mono does the same but on the other side. As the knob is turned to one side, that side gets quieter and the other side louder until only the other side can be heard. Here is a table of what I am looking for. The question is how do I make a circuit to do this?

enter image description here

So some more info. This circuit is going to be used with my high-end Amateur radio that has two bands (think radios) built into it. These are the two inputs. I will be using headphones and monitoring two different channels at one time. When I need to focus on one band I want to be able to hear it on both ears.

I posted the same question in an Amateur Radio group and someone there said he will work on a circuit. I shared the design below with this person. This person liked it and added a wise requirement:

I would prefer full isolation between the two inputs. A traditional balance control provides some amount of output from one radio TO the output of the other radio. I would be happier if a signal wasn’t going into the output of my expensive radio, just to be safe. It’s not hard to do.

So the gentleman on Facebook has come back and added a bit, first canceling what he said above and then bring up an interesting point about how it would sound:

Alright I was thinking about this more. The responder to your question on StackExchange is correct - you don't need to do any further isolation because of the "virtual ground" at the "-" input to the op-amp. I think the one remaining question is how to get a response that's more pleasing to the ear (logarithmic). A pair of linear pots are going to hard to use, especially if they're ganged together. If you keep them separate then you could use "audio" pots. But I think there might be a tricky way to get a near-logarithmic response from linear pots.

Is there such a thing as a dual logarithmic center-tap POT?

  • \$\begingroup\$ If you are sure that is what you want, I think it might be somewhat difficult in a linear, analog circuit. You might have better luck mixing digitally and converting the digital to analog. Do your signals start out analog or digital? Can you relax the requirements somewhat? In particular, can you soften the requirement of a flat 100% response for each channel when the pot is adjusting the other channel? \$\endgroup\$ – Math Keeps Me Busy Dec 27 '20 at 1:19
  • \$\begingroup\$ The input and output are analog. Can the 100% be relaxed, well I see in the comments below that it is down by like 5%, yea that is fine. The main issue is to isolate, I can always turn the volume up a bit if need be :) \$\endgroup\$ – Sam Dec 27 '20 at 12:25

Use center tapped potentiometers into virtual grounds. something like this:


simulate this circuit – Schematic created using CircuitLab

dual center tapped potentiometer:

enter image description here

  • \$\begingroup\$ While this is a good "solution", it doesn't meet the OP's spec. Check what happens when the pot is set to 25% or 75%. That is why I asked the OP if it was OK to relax the flat response requirement. \$\endgroup\$ – Math Keeps Me Busy Dec 27 '20 at 2:32
  • \$\begingroup\$ it's within 5% of what was asked for no one is bothered by a 5% difference (that's 0.4 decibels) , use a lower resistance pot (or a higher input impedance) for better compliance. \$\endgroup\$ – Jasen Dec 27 '20 at 2:57
  • \$\begingroup\$ there may be some trick where you wrap the potentiometer around the op-amp and get better compliance, but I didn't spot it. \$\endgroup\$ – Jasen Dec 27 '20 at 3:00
  • \$\begingroup\$ Yes, I'm sure there is a way to do it with linear circuits, but getting to work exactly as the OP asked I don't think will be trivial. I agree that the OP would probably be happy with your circuit. \$\endgroup\$ – Math Keeps Me Busy Dec 27 '20 at 3:03
  • \$\begingroup\$ So var this is GREAT, but allow me to add a bit more to the story, so please see my additions to the original post. \$\endgroup\$ – Sam Dec 28 '20 at 2:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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