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I wanted a circuit I thought was a basic pan control, but after mapping it out it wasn't so standard after all.

The final design will have this exposed: two mono inputs (left and right), one stereo output (left/right), and 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 the same 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?

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  • \$\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
    Commented Dec 27, 2020 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
    Commented Dec 27, 2020 at 12:25
  • \$\begingroup\$ I don't think it's possible in analogue domain with a pot up to two gangs. You are asking to mute/full volume one or other signal depending which side of centre the pot is, but also get some volume functionality. There might be a way with 3 gangs if you (for instance) put Dc on one gang and use a comparator to sense left/right and use the logic signal for switching. Even then I am not sure, and a 3 gang pot is going to be a pain. I would say either do this with a combination of switches and pots, or digitise a pot and use a microcontroller to mute/route/volume control signals. \$\endgroup\$
    – danmcb
    Commented Sep 14, 2023 at 9:03
  • \$\begingroup\$ you can definitely make a single linear pot "quasi log" with a resistor from wiper to ground. It's not perfect but good enough and done a lot in pro gear to avoid log pots, which have a poor life expectancy (because of the thin tracks). \$\endgroup\$
    – danmcb
    Commented Sep 14, 2023 at 9:06

1 Answer 1

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

schematic

simulate this circuit – Schematic created using CircuitLab

dual center tapped potentiometer:

enter image description here

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  • \$\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
    Commented Dec 27, 2020 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 Слава Україні
    Commented Dec 27, 2020 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 Слава Україні
    Commented Dec 27, 2020 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
    Commented Dec 27, 2020 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
    Commented Dec 28, 2020 at 2:40

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