How do I use one pot to mix two signals?

Started by xorophone, November 30, 2016, 04:37:24 PM

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xorophone

Hello!

I'm doing a small side project where I want two line level stereo audio signals from 2 different computers to be mixed together. I was planning on using ONE potentiometer to change the volume of both audio channels (so that if the pot is all the way to the left you'll only be able to hear one channel, if it was all the way to the right you'd only be able to hear the other channel and if it was in the middle both channels would be mixed equally. Basically like a PAN-knob.)

I tried feeding a signal into pin 1 on the pot and another signal into pin 3. Then I took the output from pin 2 on the pot and connected all of the ground pins together. This kind of worked. It seems to mix the signals together like I wanted it to and I can "sweep" between them, but when I get to a certain value on the potentiometer, the output signal becomes VERY loud and when I recorded it using my audio interface, it got really distorted. When I turned down the gain on the audio interface the distortion actually seemed to go away and it sounded normal again, but it's of course not usable if the signal becomes insanely loud when you turn the pot to a certain degree. I tried using a lower value potentiometer, but it was even worse. The highest I tried was a 1M pot.

I was thinking that you might be able to use a dual gang pot and basically connect them like a normal volume pot with one of the signals going in to pin 1 (on "gang1") and the other going in to pin 3 (on "gang2") and then ground the opposite sides and connect both middle pins together to mix the signal. I'm guessing that would work, but I need to be able to mix stereo signals, which I'm guessing would mean I need a quadruple gang pot. That's not possible because I'll never have space for it and I'm guessing they're really hard to find.

What's the best way to do this? I want to make it passive if possible too. I'm not making any fancy mixer or anything. Just want something very simple to play around with. If it's too hard I guess I'll just use two pots instead, but if possible, I want to do it this way.

Thanks for the help!


Edit: The really loud signal is just a constant distorted sound. You can't hear any of the original two signals in it and it's significantly louder than just the two signals together.

anotherjim

The basic idea of a balance control is to feed A to pot1 and B to pot3 and ground pot 2. It works by shorting to ground A or B depending which end of the pot wiper (2) is turned to.
Because it shorts, you need resistors in the feeds to prevent damage to whatever is feeding. 10k resistors might be ok with a 100k balance pot. For balance between 2 stereo feeds, a single dual gang pot should do it. One gang balances the left signals and another the right. There will be a little volume loss, but if you want passive it can't be helped.

You might get ground hum between 2 computer audio outs, especially if one or both are laptop.

Agung Kurniawan

Yep, you may get ground hum. Im prefer to use kind of a triton delay mixing section. Just use setreo pot with pot A be your right chanel and your Pot B your Left. When Pot A3 will be your right input signal, your Pot B1 should be your left input signal.
Both Lug 2 should be you out to opamp. And wire up the other lug to ground.
CMIIW ;)
Multiple gain stage followed by some active EQ is delicious.

xorophone

Quote from: Agung Kurniawan on November 30, 2016, 06:41:47 PM
Yep, you may get ground hum. Im prefer to use kind of a triton delay mixing section. Just use setreo pot with pot A be your right chanel and your Pot B your Left. When Pot A3 will be your right input signal, your Pot B1 should be your left input signal.
Both Lug 2 should be you out to opamp. And wire up the other lug to ground.
CMIIW ;)
Quote from: anotherjim on November 30, 2016, 05:34:43 PM
The basic idea of a balance control is to feed A to pot1 and B to pot3 and ground pot 2. It works by shorting to ground A or B depending which end of the pot wiper (2) is turned to.
Because it shorts, you need resistors in the feeds to prevent damage to whatever is feeding. 10k resistors might be ok with a 100k balance pot. For balance between 2 stereo feeds, a single dual gang pot should do it. One gang balances the left signals and another the right. There will be a little volume loss, but if you want passive it can't be helped.

You might get ground hum between 2 computer audio outs, especially if one or both are laptop.

Great explanations. Thanks!

Jim, where should I put the 10k resistor?

Oh, I didn't think about the ground hum part. That's a shame. Is there a way to eliminate it? I'm guessing it comes from the fact that I'm using multiple ground sources (from both computers).

anotherjim

Resistors go between each input signal feed wire and the pot lugs 1 or 3.
Normally (in the usual left/right balance control), the outputs are direct from the pot lugs. I think you also need to mix these to one output since one gang has to provide the left and the other the right. That means 2 more resistors.
This is the arrangement...


I've been thinking 10k, but soundcard outputs can usually handle lower resistance, since you can also plug 30ohm headphones into them. Try and find the capabilities of yours. So, the values in the diagram might be scaled down by a tenth, giving 1k feed resistors and 4k7 pot and output resistors. The pot type is linear. Lower resistances means it is less loaded by the input impedance of whatever its all feeding.

Ground hum, if a problem... portables are often ok if you can run them only from battery. Otherwise, a low value resistor can be fitted in each ground/screen connection from the inputs. 47ohm might do it.



ElectricDruid

#5
If you wanted an active solution, you can do this with one 8-pin dual op-amp:



Pots 10K, say, Rs all 100K.

The pot controls a mix of the blue signal and an inverted copy of the red signal. At the top of the pot, the red signal is cancelled by the inverted copy, so you get only blue signal. At the bottom, there's no cancellation, and no blue signal, so you get only red signal.

It's a clever trick I saw somewhere once, and I keep finding uses for it.

HTH,
Tom

xorophone

Quote from: anotherjim on December 01, 2016, 05:40:00 AM
Resistors go between each input signal feed wire and the pot lugs 1 or 3.
Normally (in the usual left/right balance control), the outputs are direct from the pot lugs. I think you also need to mix these to one output since one gang has to provide the left and the other the right. That means 2 more resistors.
This is the arrangement...


I've been thinking 10k, but soundcard outputs can usually handle lower resistance, since you can also plug 30ohm headphones into them. Try and find the capabilities of yours. So, the values in the diagram might be scaled down by a tenth, giving 1k feed resistors and 4k7 pot and output resistors. The pot type is linear. Lower resistances means it is less loaded by the input impedance of whatever its all feeding.

Ground hum, if a problem... portables are often ok if you can run them only from battery. Otherwise, a low value resistor can be fitted in each ground/screen connection from the inputs. 47ohm might do it.

Ah thanks! Where do you mean I should place the 47ohm resistor though? Between ground and what? And screen is basically the same as ground, right?

Quote from: ElectricDruid on December 01, 2016, 03:49:32 PM
If you wanted an active solution, you can do this with one 8-pin dual op-amp:



Pots 10K, say, Rs all 100K.

The pot controls a mix of the blue signal and an inverted copy of the red signal. At the top of the pot, the red signal is cancelled by the inverted copy, so you get only blue signal. At the bottom, there's no cancellation, and no blue signal, so you get only red signal.

It's a clever trick I saw somewhere once, and I keep finding uses for it.

HTH,
Tom

Thank you! I'll try making it passive first, but otherwise I'll probably use the schematic you shared. :)

anotherjim

First, try to treat all the ground connections as points of a star, with the output screens at its centre. The 2 pot wipers (lug2) go to the star, the 4 input screens either go to the star or thru low value resistors. The important thing is you get a little resistance between all 3 grounded items - the 2 sources and the destination, but there is no need to fit resistance going to the destination. It is best that the grounds to the pots are well connected to the destination ground.

xorophone

Quote from: anotherjim on December 02, 2016, 12:28:23 PM
First, try to treat all the ground connections as points of a star, with the output screens at its centre. The 2 pot wipers (lug2) go to the star, the 4 input screens either go to the star or thru low value resistors. The important thing is you get a little resistance between all 3 grounded items - the 2 sources and the destination, but there is no need to fit resistance going to the destination. It is best that the grounds to the pots are well connected to the destination ground.

Sorry for the late reply!

Thanks! I'll start trying this on a breadboard! :)

noah.vanhaute

So it looks like it has been a year or two since the last post on this thread. But i'd like to know if the passive blend cuircuit actually worked. Thanks!

merlinb

Quote from: ElectricDruid on December 01, 2016, 03:49:32 PM
If you wanted an active solution, you can do this with one 8-pin dual op-amp:
The pot controls a mix of the blue signal and an inverted copy of the red signal. At the top of the pot, the red signal is cancelled by the inverted copy, so you get only blue signal. At the bottom, there's no cancellation, and no blue signal, so you get only red signal.

There's a better way. It still uses two opamps, but you can use them more independently for whatever you need them to do:



xorophone

Quote from: noah.vanhaute on June 28, 2018, 03:29:41 PM
So it looks like it has been a year or two since the last post on this thread. But i'd like to know if the passive blend cuircuit actually worked. Thanks!

Sorry, but I actually never got around to trying it. I might have a bit of time left over to give it a try next week, but I'll see.

xorophone

Quote from: noah.vanhaute on June 28, 2018, 03:29:41 PM
So it looks like it has been a year or two since the last post on this thread. But i'd like to know if the passive blend cuircuit actually worked. Thanks!

I can confirm that the circuit does indeed work, but expect a big drop in volume. I tried changing the 10k resistors to 1k, the 50k resistors to 4.7k and the 50k potentiometer to 5k. That greatly increased the output volume, so I suggest making those changes. However, if it's possible for you to use an active circuit I highly recommend that instead.

noah.vanhaute

Quote from: xorophone on July 03, 2018, 01:59:13 PM
Quote from: noah.vanhaute on June 28, 2018, 03:29:41 PM
So it looks like it has been a year or two since the last post on this thread. But i'd like to know if the passive blend cuircuit actually worked. Thanks!

I can confirm that the circuit does indeed work, but expect a big drop in volume. I tried changing the 10k resistors to 1k, the 50k resistors to 4.7k and the 50k potentiometer to 5k. That greatly increased the output volume, so I suggest making those changes. However, if it's possible for you to use an active circuit I highly recommend that instead.
I'll go for the active solution then, thanks.
Quote from: merlinb on June 29, 2018, 03:01:33 AM
Quote from: ElectricDruid on December 01, 2016, 03:49:32 PM
If you wanted an active solution, you can do this with one 8-pin dual op-amp:
The pot controls a mix of the blue signal and an inverted copy of the red signal. At the top of the pot, the red signal is cancelled by the inverted copy, so you get only blue signal. At the bottom, there's no cancellation, and no blue signal, so you get only red signal.

There's a better way. It still uses two opamps, but you can use them more independently for whatever you need them to do:



I assume that I have to use the non-inverting inputs here?

PRR

> I assume that I have to use the non-inverting inputs here?

Why??
  • SUPPORTER

noah.vanhaute

Quote from: PRR on July 15, 2018, 07:37:31 PM
> I assume that I have to use the non-inverting inputs here?

Why??
Because I'm new to op-amps and don't know which I should use. That's why I asked.

PRR

> I'm new to op-amps

As are we all.

Explain why you think you "have to use the non-inverting inputs". Others with some more experience may be able to confirm your assumption; or point-out why it is incorrect.
  • SUPPORTER

ElectricDruid

Here's a nice example using that pan/blend circuit with an op-amp to make up the gain:

http://www.geofex.com/Article_Folders/panner.pdf

You want the circuit in the box on the left.

Kipper4

#18
Quote from: PRR on July 16, 2018, 01:47:44 PM
> I'm new to op-amps

As are we all.

Explain why you think you "have to use the non-inverting inputs". Others with some more experience may be able to confirm your assumption; or point-out why it is incorrect.


I on the other hand would assume they are non inverting op amps.
There are many examples on the web.
I assume this because it would be called -A and -B if it was inverting.


Edit.
For what it's worth, my opinion is that the set up advocated by Electric Druid in reply 5 is the way to go.
I wired it up wrong a couple of times before I got it right. It works a treat.
I've tried several times with a passive Pot mixer and was dissapointed with most of them.
Couldnt quite get the mix I wanted.

Ma throats as dry as an overcooked kipper.


Smoke me a Kipper. I'll be back for breakfast.

Grey Paper.
http://www.aronnelson.com/DIYFiles/up/

ElectricDruid

Quote from: Kipper4 on July 17, 2018, 09:56:48 AM
Edit.
For what it's worth, my opinion is that the set up advocated by Electric Druid in reply 5 is the way to go.
I wired it up wrong a couple of times before I got it right. It works a treat.
I've tried several times with a passive Pot mixer and was dissapointed with most of them.
Couldnt quite get the mix I wanted.

AnotherJim's pan/blend is good too, but it is very lossy in the passive form. It's much better if you use it as the input to an inverting op-amp mixer, so that you get out the same level you put in. That's why I posted the Geofex article - it includes that same circuit, but in an active form without the problems.

T.