Audio Blending Between Opamps?

"Brandon"

** It is not at all clear what you mean by blend with a 100K pot.

What is the function of the pot ????

...... Phil

Assume that amp A drives one end of the pot, and amp B drives the other end, and the wiper feeds amp C, probably as a high-impedance follower. That's a blender. If the pot resistance is higher than the closed-loop output impedance of A and B (use a 100K pot maybe) then A and B won't crosstalk into one another much.

John

Well, at a high level, I basically just want to be able to crossfade two audio sources (we'll call them A and B) so that as A becomes "louder," B becomes "softer." I was under the impression that could easily be accomplished with a pot. In this case, input A (coming directly from the output pin of an opamp) would be connected to the

1st lug of the pot, input B (coming from the output of another opamp) would be connected to the 3rd lug, and the shared 2nd lug would be connected directly to the input of a third opamp which would act as a buffer and send the crossfaded signal on its way. Since you questioned it, is there any reason why that sort of setup wouldn't accomplish what I want? Or did I just not explain the scenario well enough up front?

I think that is basically what the crossfader is a DJ mixer is - a dual gang (for stereo) slider-style potentiometer. Assuming that approach would work, what I was really asking about was whether there are any complications I hadn't thought about. I don't think I should need any pull down resistors or anything of the sort.

-Brandon

"Brandon"

** Blending is not the same as crossfading.

** Blending is not the same as crossfading.

You used the wrong term.

** The complication is in how each signal varies at the output as the crossfade pot is moved.

...... Phil

"For performing DJs, the days of the application specific crossfader were over."

Gosh. Such thrilling stuff.

John

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One man\'s trash is another man\'s treasure, troublemaker.

JF

Connect one end of the 100K pot to one output of the dual opamp, the other end to the output of the other output. Take the output from the center connection of the pot. It will work better if you add a ~100K fixed resistor from output to ground. If you can not DC couple the outputs, add a coupling capacitor between the pot output and the following input/bias network.

Tam

I suppose that it is the way it was done in pro equipment for decades might ease your mind. The modern way is digital, but that has a higher startup cost.

So you are recommending something like this?

That was actually sort of what I was expecting but can I ask what purpose the added resistor is serving? I'm here to learn.

Also, since this is an audio application, I was wondering if this will have a negative impact on sound quality. In an article at

Jack Orman talks a lot about the importance of guitar effects pedals having high impedance to preserve high frequencies. Having read that article, adding that 100K resistor raises a red flag but is that not a concern in this case? And if not, why? Is it because of differences in the properties of an opamp and guitar pickup?

-Brandon

Yeah, that would ease my mind. And digital is definitely out of the question. I just do this recreationally and digital might be a little over my head right now. Not to mention that I'm just building a single unit right now. This isn't a prototype for something that will be mass produced or anything.

This is going to be a piece of guitar equipment anyway and the guitar community is a funny group when it comes to tubes vs solid state vs digital. Concerns over digital sampling rate and resolution make sense but still...

How would you compare the difficulty of developing a digital circuit to developing an analog one? Is it any more difficult or is it just more to learn?

One remark, when the pot is on the top,or bottom, then the amplitude is smaller then when the pot is in the middle. This is because the 2 channels sum via 50k + 50 k in parallel to 2 x the amplitude. Perhaps the function of the resistor to ground is to compensate for this, it should then be 25 k. I dunno how important the amplitude change is to you though. Also I did not plot the other slider positions versus output amplitude.

See above.

No. When the pot is at half rotation, the voltage at the wiper is

(0.5 * Va + 0.5 * Vb)

Adding the resistor reduces this mid-point level. That will change the "curve" of the fader, giving a dip in the middle. So...

One end would be 1.0 * Va

The other end would be 1.0 * Vb

Midpoint might be dipped to 0.25 * Va + 0.25 * Vb, something like that, depending on the resistor value.

John

Why would the added resistor reduce the mid-point level without affecting the two "ends" of the pot's rotation?

On a sunny day (Sat, 17 May 2008 11:30:51 -0700) it happened John Larkin wrote in :

OK, you are right, bit rusty on that stuff...

On a sunny day (Sat, 17 May 2008 11:53:16 -0700 (PDT)) it happened Brandon wrote in :

Because the 'ends' drive from a low impedance (opamp output).

This may be of interest to you

martin

At your current level, analog is fine. There was plenty of studio and broadcast equipment built that way. Hunt around you may be able to find some really cheap (near the level of haul it away and it is legally yours). Decent digital is getting much cheaper though. It mostly leverages your home PC for compute power. You might look at RME and similar that you can find on any search engine.

I didn't work out the math, but think the fading will be more linear with respect to rotation of the pot. You might want to work this out for various values of load resistors, or just try it - only costs you 10 cent per resistor value.

Note that you can also do this if the second stage is an inverting feedback amplifier with a fixed value feedback resistor. In fact, the math will be easier (trivial), but you need fixed resistors in series with the two first stage outputs or the second stage gain will try to go to infinity with the pot at either extreme.

Tam

Tam

Late at night, by candle light, Brandon penned this immortal opus:

The output of one amp will merrily absorb whatever little trickles into it "backwards" from the other amp. Make the third amp a unity gain voltage follower and you're all set up.

- YD.

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