Audio Blending Between Opamps?

[Brandon]

Hi everyone. Let me start off by saying that I am a hobbyist when it
comes to electronics so hopefully this question won't sound too
stupid. In a little audio project I am working on, I am using two
opamps (one dual opamp actually) as buffers for incoming audio
signals. I want to blend the output of the two opamps using a 100K
pot and then "feed" that blended signal into a third opamp. Will that
work like I am expecting or are there any concerns that I don't know
about?

I've seen circuits where the output of an opamp goes directly into the
input of another omamp (some bandpass filters for example) but is this
any different because the signals come together in a Y and there are
two paths for a signal to take? Because opamps have relatively high
input resistance (about 2M ohms typical in this case) and a low output
resistance (about 75 ohms), is it possible that the signal from one of
the "input opamps" will affect the other? Or is that never much of a
concern?

-Brandon

 

[Phil Allison]

"Brandon"

In a little audio project I am working on, I am using two
opamps (one dual opamp actually) as buffers for incoming audio
signals. I want to blend the output of the two opamps using a 100K
pot and then "feed" that blended signal into a third opamp.

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

What is the function of the pot ????




....... Phil

 

[Brandon]

In a little audio project I am working on, I am using two

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

What is the function of the pot ????

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

 

[Phil Allison]

"Brandon"

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."

** Blending is not the same as crossfading.

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?

** Blending is not the same as crossfading.

You used the wrong term.

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.

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

http://www.rane.com/note146.html



....... Phil

 

[JosephKK]

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

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.

 

[Brandon]

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

So you are recommending something like this?
http://spaaarky21.home.comcast.net/~spaaarky21/Blend.png

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
http://www.muzique.com/lab/imp.htm 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

 

[Brandon]

I think that is basically what the crossfader is a DJ mixer is - a

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.

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?

 

[[email protected]]

On May 16, 5:41 pm, "Tam" <[email protected]> wrote:

So you are recommending something like this?
http://spaaarky21.home.comcast.net/~spaaarky21/Blend.png

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.

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.

See above.

 

[Brandon]

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?

 

[Martin Griffith]

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

This may be of interest to you

http://es.geocities.com/mart_in_medina/bendypot.PDF


martin

 

[JosephKK]

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?

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.

 

[YD]

Hi everyone. Let me start off by saying that I am a hobbyist when it
comes to electronics so hopefully this question won't sound too
stupid. In a little audio project I am working on, I am using two
opamps (one dual opamp actually) as buffers for incoming audio
signals. I want to blend the output of the two opamps using a 100K
pot and then "feed" that blended signal into a third opamp. Will that
work like I am expecting or are there any concerns that I don't know
about?

I've seen circuits where the output of an opamp goes directly into the
input of another omamp (some bandpass filters for example) but is this
any different because the signals come together in a Y and there are
two paths for a signal to take? Because opamps have relatively high
input resistance (about 2M ohms typical in this case) and a low output
resistance (about 75 ohms), is it possible that the signal from one of
the "input opamps" will affect the other? Or is that never much of a
concern?

-Brandon

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.

 

[YD]

So you are recommending something like this?
http://spaaarky21.home.comcast.net/~spaaarky21/Blend.png

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
http://www.muzique.com/lab/imp.htm 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

He means the input impedance affecting passive pickups. Impedance wise
they're a mess. Load it too heavily and you lose the high end. Load it
too lightly and you may get a peak at the high end making it sound too
bright and kind of scratchy. Most guitar amps have about one megohm
input impedance.

- YD.

 

[Brandon]

So you are recommending something like this?

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

It might have been a mistake mentioning a DJ crossfader because I
really do want a linear fade. That was just an example. It seems to
be the consensus that running a resistor to the ground will give me a
"dip" (which I don't actually want.)

The gain on the second stage is something I kind of skipped over I
guess. I have it setup as a voltage follower in a non-inverting
configuration so the gain will be 1+Rinput/Rfeedback and since the
feedback loop has no resistor, I didn't think about the input
resistance. I agree that I definitely need a fixed resistor but will
adding one prevent the mix from reaching 100% A and 0% B? Won't the
sources' "presence" at the second stage's input be determined by the
ratio of the resistance along their path to the 2nd stage? Suppose
that both of the opamps in the 1st stage had a 100K resistor between
them and the 100K pot and then the wiper of the pot goes to the input
of the 2nd stage's opamp. If the pot were in the middle position,
both of the input signals would follow a path with a resistance of 50K
+100K so the two would mix 1:1. But if the pot was positioned toward
one end, one of the signals's paths would have a resistance of 100K
+100K=200K and the other would have a resistance of 0K+100K=100K so
they would mix 2:1, not 1:0. And I think the same would be true if
the two fixed resistors were replaces with one resistor between the
pot and the 2nd stage input.

Maybe I would be better off just making the two opamps in the 1st
stage inverting and using the pot to control their gain from 0 to 1.
Then the two outputs could easily be mixed 1:1 with fixed resistors.

 

[Brandon]

Maybe I would be better off just making the two opamps in the 1st

stage inverting and using the pot to control their gain from 0 to 1.
Then the two outputs could easily be mixed 1:1 with fixed resistors.

Oh, wait. What was I thinking? That would be really nice but I can't
do it because the the whole point of those opamps in my circuit is to
act as a buffer (with a high input impedance) for signals coming from
external sources with unknown resistances. Otherwise, that would be
the perfect solution. Maybe I'll have to consider adding another set
of opamps and do my blending that way if there isn't a satisfactory
way of blending all the way to 0% : 100% using just a pot and
resistors between two opamps.

 

[whit3rd]

...using two
opamps (one dual opamp actually) as buffers for incoming audio
signals.  I want to blend the output of the two opamps using a 100K
pot and then "feed" that blended signal into a third opamp.

There are three concerns here. First, the 100K value is a little
high, 10K is a good value for circa 1V signals. High impedances cause
pickup and noise, while low impedance can contribute to bass
distortion (your op amps actually heat and cool and the thermal
drift causes a bass audio distortion source).

Second, the 'third opamp' should be connected as an inverting
amplifier, feedback resistor from output to the summing junction.
That makes the pot wiper a 'pseudo-ground' and prevents
the buffer amplifiers from seeing variances in load due to the
other buffer's signal. It also allows you to largely ignore CMRR
for that third amplifier. Note this means a minimum resistance
should be maintained, you want to connect the buffers to the
potentiometer through a 1k fixed resistor.

Third, there are situations where (for instance) right/left channel
signals have a significant phase relationship, and the
wrong blending would cancel the signal instead of reinforcing it.
It's useful to have a relative-polarity-reversal switch in this kind
of signal blender.

Often for audio, an input can go open circuit (wire pulled loose); if
your
buffers are inverting op amps, that's OK. If they're followers, that
makes the input likely to hum loudly at 60 Hz... we've all heard that
happen.

 

[Brandon]

Maybe I would be better off just making the two opamps in the 1st

Pot yes.

Resistors no.

0 to 100% both ways.

Simple enough?

John

I hate to say it but not entirely. When you say "pot yes, resistors
no" do you mean it will works as expected with the pot ("pot yes") but
not when you add the fixed resistors ("resistors no")? If that IS
what you meant, then it looks like I a might need to redesign things a
bit since 0-100% is what I am after but a fixed resistor is needed to
keep the second stage's gain in check when the pot is all the way to
one end.

If that is not what you meant (I could get 0-100% using both a pot and
fixed resistors), I'll believe it (you guys are the experts) but I am
still curious how the numbers work out. In one of my last posts, I
was working under the assumption that the ratio of signal reaching the
2nd stage's input was the the same as the ratio of the resistance
along each signal's path from the 1st stage's outputs to the 2nd
stage's input. Is that right?

 

[Brandon]

I hate to say it but not entirely. When you say "pot yes, resistors

OK, you win, I am defeated. I give up.

John

Oh no! Don't give up! I was just wondering. I'm sort of new at
this.

I'm really not trying to be difficult. I'm just trying to figure this
stuff out.

 

[JosephKK]

<snip>

Now you get to teach Brandon how to run SwCAD III.

 

[[email protected]]

There are three concerns here. First, the 100K value is a little
high, 10K is a good value for circa 1V signals. High impedances cause
pickup and noise, while low impedance can contribute to bass
distortion (your op amps actually heat and cool and the thermal
drift causes a bass audio distortion source).

You really have to drive an op amp hard to see thermal distortion.
This is more of an issue with circuits that have low amounts of
feedback.

Second, the 'third opamp' should be connected as an inverting
amplifier, feedback resistor from output to the summing junction.
That makes the pot wiper a 'pseudo-ground' and prevents
the buffer amplifiers from seeing variances in load due to the
other buffer's signal. It also allows you to largely ignore CMRR
for that third amplifier. Note this means a minimum resistance
should be maintained, you want to connect the buffers to the
potentiometer through a 1k fixed resistor.

I'm not sure I approve of this scheme. FIrst of all, yes, the
inverting amplifier configuration has less issues with the CMRR. But
putting the wiper direction to the negative input means you have the
standard inverting configuration, Say op amp 1 has resistor R1 going
to the center tap Op amp 2 has resistor R2. [The potentiometer is of
value R1 + R2. For the third op amp, you have feedback resistor RF.
Using your scheme, we can consider R1 and R2 to have minimum values of
Rmin, which you set to 1k. Now at the extreme position, isn't the gain
of the 3rd op amp RF/Rmin? But as you add resistance (move the pot
positon) the gain drops to RF/R1 from the op amp 1 path. The other
resistor should fall out of the picture in the gain equation since it
has no potential across it in a superposition sense.

Maybe I interpretted your circuit description incorrectly.