Inverted output from panning control to mix wet/dry signals

[ewd]

I'm looking at making a panning control for mixing the wet & dry signals outputted from a chorus pedal, for a harmonica-playing friend. Plan on basing it on RG Keen's "Panning for Fun" article on geofex.com (the two-to-one example). RG's example uses an inverting OpAmp circuit to mix the signals, resulted in an inverted output.

My questions are:
1) Does the inverted audio output have any consequence regarding the receiving power amplifier, or any other possible subsequent effects pedal in a pedal chain?

2) Should I pass the signal through another inverting OpAmp to restore the initial phase?

3) Would "correcting" the signal phase have any effect on the audio dignal, other than the actual re-inverting?

I'm intending on using an RC4558 dual OpAmp (with a +9v/-9V powering the OpAmp), so I have another half of the IC to use, if necessary.

[Enzo]

Absolute phase really doesn;t matter in your signal chain. About the same as whether you pick a string going up or down.

[ewd]

Thank you Enzo!

[R.G.]

True/inverted phase only matters when the signal will somehow get mixed with the other phase. As long as that does not happen, you're fine.

Interesting cases of mixing happen in things like parallel amplifiers where one inverts; paralleled speakers with wires mixed up, or speakers with internally different polarities; splitter/mixer effects chains; speaker and PA with different inversions; speaker cabs with open backs where the front and back wave mix and "cancel" lower bass, things like that. For single source/single output, it does not matter at all.

That being said, inverted-phase issues can't happen when all the things in the chain are non-inverting. So if you have a spare opamp, it only costs you a couple of resistors, maybe a cap or two, and some copper trace to make the whole thing non-inverting and side step any possible issues.

[ewd]

Thank you for the insight R.G. Think I will go the extra step for good measure.

[ewd]

As a follow-on, I now have another question(s): What do I use as the input resistance for calculating my negative feedback resistor for the 2nd inverting Op Amp stage? I have been researching all day and what I've been able to come up with is that the output impedance of the 1st inverting stage is extremely low as is inherent in all inverting op amps (PLEASE correct me if this is wrong!). If I am correct, almost any negative feedback resistor in the 2nd inverting stage will result in a large output gain (A=Rf/Rin), but I have to have a feedback resistor, even in an inverting voltage follower application(??).
-or alternatively, should I just substitute a small pot for the feedback resistor on the 2nd inverting stage and use it as a gain control?

[R.G.]

As a follow-on, I now have another question(s): What do I use as the input resistance for calculating my negative feedback resistor for the 2nd inverting Op Amp stage? I have been researching all day and what I've been able to come up with is that the output impedance of the 1st inverting stage is extremely low as is inherent in all inverting op amps (PLEASE correct me if this is wrong!).

Hmmm. I get a strong smell of trouble here, maybe more than I can root out without a schematic.

Inverting stages always have a series input resistor and a feedback resistor, you're right there. But using the output impedance of the preceding stage for that resistance is nearly always a mistake. With a low output impedance from the first stage driving a second stage, you'd assume the first stage output impedance was zero (not too far wrong) and put in a resistor of your choosing.

So I might make a unity gain second inverting stage be a 10K series input resistor and a 10K feedback resistor for a nominal unity gain, but inverting. The stage driving this would only negligibly change things with an output impedance of zero up to about 1K ohms.

You can still dink with the feedback resistor for an overall gain control, but stick in a resistor of your choosing to force the second inverting stage to do what you want, not worry about the imponderables from the first stage.

This is kind of the first rule of engineering: what cannot be controlled must be made irrelevant some way.

A proposed, even crude schematic would let me make this a lot clearer.

[ewd]

Hope this works, I've attempted to upload my schematic as a jpg.

I've used the values from your article as nominal resistor values to start with, and tied in the other half of the dual Op Amp as the 2nd inverting stage.

[ewd]

So add another resistor in series between R8 and pin 1, making R8 the input resistor and the added resistor the feedback resistor (call it R9, taking the final output from between R8 and R9)?

[ewd]

Or after a closer look, would the R8 resistor be the feedback resistor and an added R9 be the input resistor?

[Dave H]

[ATTACH=CONFIG]25014[/ATTACH] Hope this works, I've attempted to upload my schematic as a jpg.

I've used the values from your article as nominal resistor values to start with, and tied in the other half of the dual Op Amp as the 2nd inverting stage.

That won't work. You need a resistor from pin 1 to pin 6, then a resistor from pin 6 to pin 7 and take the output directly from pin 7. I'll redraw it so it's easier to follow.

[Dave H]

Here's a crude schematic.

[R.G.]

Good one, Dave. That's the way to do it.

@ewd: you need to drive inputs A and B from sources with impedances much lower than the resistors in the network. An opamp buffer before each one or an emitter or source follower would be fine.

[ewd]

Dave H/RG~ Thanks so much for the advice/info/education. Will be working on adding in the buffers to each input now.

[ewd]

Had to take a break due to other priorities, but finally got back to this project. Here is my (hopefully) final schematic incorporating buffered inputs and a final non-inverted output:
Would appreciate any glaring errors being pointed out.