Cross fading two audio signals.. what taper dual gang pot?

[eyAyXGhF]

Hi all,
I have two equal amplitude audio signals and I want to use a single dual-gang potentiometer to cross fade between the two so that the amplitude remains constant.

I'm looking at this series of potentiometers:
https://www.alphapotentiometers.net/html/16mm_pot_22.html

That gives me a choice of tapers A B C D. And on their "tapers" page ( https://www.alphapotentiometers.net/html/taper_curves.html ) it doesn't list a taper D, but does show a taper "AM - CM" that looks to be what I'd want. Taper A on a dual-gang pot I assume would be for a stereo volume control, not a cross-fade.

 

[crutschow]

Perhaps you could just use one pot with each end of the pot connected to its respective signal, and the wiper being the output. With that, with the wiper at the pot ends, you would have full volume of either signal and a 50/50 mix with the wiper in the middle.

 

[eyAyXGhF]

Perhaps you could just use one pot with each end of the pot connected to its respective signal, and the wiper being the output. With that, with the wiper at the pot ends, you would have full volume of either signal and a 50/50 mix with the wiper in the middle.

That's a lot simpler for sure. Maybe I'm completely being confused by the whole amplitude vs. perceived amplitude of audio stuff, thinking that the 50/50 mix should actually be 20/20 like it would be using an audio and reverse audio taper. But maybe that's just for panning a single source to two separate sources, not mixing two sources into one output like I'm doing here. Definitely too confused

 

[crutschow]

That's a lot simpler for sure. Maybe I'm completely being confused by the whole amplitude vs. perceived amplitude of audio stuff, thinking that the 50/50 mix should actually be 20/20 like it would be using an audio and reverse audio taper. But maybe that's just for panning a single source to two separate sources, not mixing two sources into one output like I'm doing here. Definitely too confused

Perhaps the simplest is just to try it and see how it works.

 

[Mr RB]

Perhaps you could just use one pot with each end of the pot connected to its respective signal, and the wiper being the output. With that, with the wiper at the pot ends, you would have full volume of either signal and a 50/50 mix with the wiper in the middle.

Because of the very high input impedance of the audio amp, even with a 1M pot there will still be a significant signal strength coming across the 1M from the audio source that is supposed to be off. So it will mix to an extent, but it won't properly kill the sound at the ends of travel.

You could use a dual gang linear pot, as two pots but wired in reverse. That should work pretty well. On mixer desks they use linear slide pots, and you can cross fade two channels by manually cross fading the two linear slide pots.

It will still need some kind of mixer after the two pots, either connect the two pot outputs (wipers) to two separate amp channels, or mix them through a couple of high value resistors (like use 10k pots, and mix via two 100k resistors).

 

[crutschow]

Because of the very high input impedance of the audio amp, even with a 1M pot there will still be a significant signal strength coming across the 1M from the audio source that is supposed to be off. So it will mix to an extent, but it won't properly kill the sound at the ends of travel.

................

My assumption (which I should have stated) was that the two sources driving the pot were low impedance (compared to the pot resistance) such as an op amp output. In that case the high impedance load at the pot wiper would see only one of the sources at either end of the pot rotation, and would see both signals equally at the middle of the rotation.

 

[Nigel Goodwin]

Because of the very high input impedance of the audio amp, even with a 1M pot there will still be a significant signal strength coming across the 1M from the audio source that is supposed to be off.

Why are you assuming a high impedance input, the chances of that are extremely low - not impossible, but probably well under 1%.

Sensible way though, in either case, would be a simple opamp mixer (one opamp, five resistors and a pan pot).

 

[eyAyXGhF]

Thanks for all the suggestions guys,

Just to make things a bit clearer. Both of the outputs that need to be blended are already coming straight out of op-amps (Although one has a 10k resistor going to ground on the output for another reason - this signal also goes somewhere else). After this "blend" pot, the signal will then be going to a separate volume control and then to another op-amp buffer before final output. I also have a pair of free op-amps that I can use if necessary...

 

[Nigel Goodwin]

Have a look at the balance circuit in figure 7 below

**broken link removed**

Replace the volume control pots with two 10K resistors joined together, and take the output from there - directly in to the inverting input pin of an opamp (as a mixer) would be good.

 

[crutschow]

Why are you assuming a high impedance input, the chances of that are extremely low - not impossible, but probably well under 1%.

OK, a further definition. By high impedance I mean larger than the pot resistance. The circuit you proposed in post #9 is basically the same circuit with two added resistors (equal to the pot resistance).

 

[Mr RB]

My assumption (which I should have stated) was that the two sources driving the pot were low impedance (compared to the pot resistance) such as an op amp output. In that case the high impedance load at the pot wiper would see only one of the sources at either end of the pot rotation, and would see both signals equally at the middle of the rotation.

Thanks Crutschow I understand what you meant now! That does rely on the pot being a high value and the two audio sources both being an extremely low source impedance. Normally you can't rely on that in audio as even things like audio line outputs can have 1k+ output impedance.

If the OP is using custom circuitry to drive the fader pot it might be ok but I still think a dual gang pot will give the best performance.

 

[eyAyXGhF]

Okay, here's the basic solution I'm trying. I haven't had a chance to breadboard it yet, but it simulates fine. The source signals and the op-amp stuff right after them are just part of the existing circuit. The blend pot is a 100k linear and at either end of the range it seems to be 100% of the appropriate signal at 100% amplitude. The half way point is a mix of two, each at about 45% amplitude. So the only issue I'm seeing is that the 50/50 mix is actually 45/45 so a slight decrease in overall amplitude.. but I don't think that will be an issue.

 

[crutschow]

Okay, here's the basic solution I'm trying. I haven't had a chance to breadboard it yet, but it simulates fine. The source signals and the op-amp stuff right after them are just part of the existing circuit. The blend pot is a 100k linear and at either end of the range it seems to be 100% of the appropriate signal at 100% amplitude. The half way point is a mix of two, each at about 45% amplitude. So the only issue I'm seeing is that the 50/50 mix is actually 45/45 so a slight decrease in overall amplitude.. but I don't think that will be an issue.

I don't understand why the 50% point mix is 45/45 when it should be 50/50. Are you sure you measured it correctly?

 

[Nigel Goodwin]

Okay, here's the basic solution I'm trying. I haven't had a chance to breadboard it yet, but it simulates fine. The source signals and the op-amp stuff right after them are just part of the existing circuit. The blend pot is a 100k linear and at either end of the range it seems to be 100% of the appropriate signal at 100% amplitude. The half way point is a mix of two, each at about 45% amplitude. So the only issue I'm seeing is that the 50/50 mix is actually 45/45 so a slight decrease in overall amplitude.. but I don't think that will be an issue.
View attachment 70757View attachment 70757

The middle opamp is completely wrong, it needs to be an inverting opamp to act as a mixer, and needs a feedback resistor.

Check the link I posted earlier for wiring the pan pot correctly.

 

[Nigel Goodwin]

Thanks Crutschow I understand what you meant now! That does rely on the pot being a high value and the two audio sources both being an extremely low source impedance. Normally you can't rely on that in audio as even things like audio line outputs can have 1k+ output impedance.

There seems a lack of understanding on audio here

1K is a low impedance, and greater than 10K isn't a high impedance, a high impedance would usually be 1M or more, or at least in the 100K's.

If the OP is using custom circuitry to drive the fader pot it might be ok but I still think a dual gang pot will give the best performance.

Not really an option, log/anti log pots are extremely difficult to find, they are expensive, and the tracking between the two pots is awful. This is why the pan pot type systems are almost universally used.

 

[crutschow]

The middle opamp is completely wrong, it needs to be an inverting opamp to act as a mixer, and needs a feedback resistor.

................

It's not completely (or even partially) wrong at all. At the midpoint of the pot the signal at the op amp non-inverting input will be a 50/50 (voltage) mix of the two input signals. You don't need to use an inverting amp.

 

[Mr RB]

There seems a lack of understanding on audio here

1K is a low impedance, and greater than 10K isn't a high impedance, a high impedance would usually be 1M or more, or at least in the 100K's.
...

Yes I know that, I own a lot of studio equipment and spent many years servicing pro audio equipment.

Even a 1k source impedance to 100k pot is only 100:1 attenuation, once that is amplified it WILL NOT give a complete cut to the audio on that side. Some will still bleed through.

(re using a dual gang pot)... Not really an option, log/anti log pots are extremely difficult to find, they are expensive, and the tracking between the two pots is awful. This is why the pan pot type systems are almost universally used.

I already stated you can use a dual gang linear pot, in studio equipment (mixer desks etc) they use linear slide fader pots, and cross fading works just fine if you even turn one slider up while the other slider is turned down.

I thought you played guitar Nigel? I'm surprised you didn't know this stuff.

 

[Nigel Goodwin]

Yes I know that, I own a lot of studio equipment and spent many years servicing pro audio equipment.

So you shouldn't be misusing 'high impedance' then

Even a 1k source impedance to 100k pot is only 100:1 attenuation, once that is amplified it WILL NOT give a complete cut to the audio on that side. Some will still bleed through.



I already stated you can use a dual gang linear pot, in studio equipment (mixer desks etc) they use linear slide fader pots, and cross fading works just fine if you even turn one slider up while the other slider is turned down.

I thought you played guitar Nigel? I'm surprised you didn't know this stuff.

No I don't play, I did PA - and built the mixer for the band back in the day

And while commercial mixers use linear pan pots (as mine did) they don't usually use dual ones, for the stated reasons.

I've repaired various professional mixers over the years, never seen one with dual pan pots?.

 

[alec_t]

If you want virtual silence from one channel while the other channel is active then feeding both channels through a single pot won't cut the mustard. Instead, have you considered using two voltage-controlled amplifiers? The SSM2018T for example gives 140dB gain range with ~5V control voltage change. A simple inverting opamp could be used to provide the inverse control voltage for the second channel, so a single pot would control both channel amps for the fade.

 

[Nigel Goodwin]

If you want virtual silence from one channel while the other channel is active then feeding both channels through a single pot won't cut the mustard.

Again, refer to the link I posted earlier.