Gain calculation on R.G. Panning for Fun

[melangston]

Greetings,

I am having difficulty determining how to calculate the output level of the 2-input:1-output panning circuit described in Panning for Fun (page 1, top left boxed circuit).

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

It states for unity gain 3.41R is the appropriate feedback resistor value is appropriate, but when using resistances of 1, 0.707, and 3.41, and voltage of 1V for both sources I do not get Vout of -1 (or 1). I am using the Scaling summing amplifier equation [-Vout=Rfeedback*((V1/R1)+(V2/R2)+...(Vn/Rn))].

As referred to in another post, I've reviewed the 1976 National Semiconductors Audio Handbook where the "panning for fun" circuit is described and it references Wye-Delta transformations to determine Rin for gain calculation. I am not sure of which R to use (and which V for that matter).

http://bitsavers.trailing-edge.com/components/national/_dataBooks/1976_National_Audio_Handbook.pdf  - Schematic Page 2-60. Wye-Delta on Page 6-10

Can anyone offer guidance on which values are used in these calculations?

Thank you.

[Edit 1] Added links to references per request

[antonis]

Hi & Welcome..

Unfortunately, we too are having calculation difficulty w/o a schematic.. 
(or, an attachement, at least..)

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

P.S.
Just a welcome tease..

[melangston]

I can see how that would help. I've included to the original post. Let me know if anything else would help. 

[ElectricDruid]

Welcome!

I can *mostly* explain it, although I can't get the numbers to come out right on the nose. Anyway, bear with me.

First, imagine one side only, and then look at it as two parts: an input divider, followed by the inverting op-amp. For the input divider we have 15K, with a 10K variable resistor to ground. When we're panned fully over to that side, we have the full 10K. So it's a basic voltage divider of 15K at the top and 10K at the bottom.

However, that 10K to ground appears in parallel with the 15K input resistor of the mixer (since it terminates at a virtual ground). By my calculations, 10K and 15K in parallel is 6K.

So our final input divider is 15K at the top, and 6K (10K || 15K) at the bottom. That gives a loss at fully-panned-one-way of 6 / (15+6) = x0.2857. To get that back up to unity needs a gain of 1/0.2857 = x3.5.

That's what I mean about the numbers - I get x 3.5, not x3.41. 3.41 gives 0.293 for the loss. There's probably something less significant that I'm forgetting about, but I think the general principle is right.

Still, 0.007 out is close enough for engineering, right?!?

[melangston]

Thanks so much for your time!

Your explanation makes sense and doing the math I get to the same result. (BTW your 0.2857 vs original 2.93 is due to your using approximate resistances ie 10k and 15k vs the theoretical 1 and 0.707). And seeing your explanation of deriving 3.5 is very helpful.

However, I'm still struggling with making sense of the ouput voltage using these.
For example, given:
   -Vout = Rf((V1/R1)+(V2/R2))
and assuming the blend potentiometer is panned to the top as in your example.

51k would be Rf.
The quantity V2/R2 would be 0.
Would R1 be the top 2 resistors in Panning for fun (15k + 15k), The top right resistor plus the variable resistor (15k + 10k))? I'm getting Vout of -0.499 and -0.706 for those values respectively.
Would V1 be the voltage of the original source (eg 1V) or the one we calculated (0.293)?

Thanks again. I'm not too concerned about the precision but I like to know I'm at least understanding the math correctly.

[PRR]

> For the input divider we have 15K

Pot is 0.707 times R. But pots only come in round numbers. So take pot=10k, R is 1/0.707 of that or 10k/0.707 or 14,142 Ohms. The "15k" is significantly rounded so it can be found in a 2008 part-bin. (Even today not everybody has all 2% resistors in stock.)

The loss is given as 3 digits but above the math is taken to 5 digits.

Here is more-exact math limited by both digits and amplifier gain. (And apparently pot virtual end resistance.)

[melangston]

Very cool. I appreciate your taking the time to put that together.

I got the same Vout as your simulation by using the voltage we calculated (0.293) along with the only resistor between that potential and the op-amp input (15k). I thought the other resistors would play a role in the resistance used in the formula, which I think is where my confusion came from. I appreciate your helping me understand it.

By these calculations, as we sweep the pot from 0%, 50%, 100%. The output voltage goes from 1V to 1.4V to 1V, respectively. With audio AC signals, this represents a gain of 3dB at the middle compared to ends I believe. My understanding was that this circuit eliminated the change in overall volume as you change the ratio of the source signals present. Am I missing something here as well?

This has been very enlightening.

[PRR]

First: a terminology clarification. Despite the name of that PDF, the "boxed" circuit on page 1 is not "Pan", it is cross-fade. It blends TWO signals to one output. Two *different* signals. (What would be the point of splitting one signal and blending it together?)

Yes, it was convenient to simulate with one source.

> The output voltage goes from 1V to 1.4V to 1V

Simulate/math for one source and one silence. At mid-point the signal is 0.707.

If you have two *identical* signals then yes this adds-up to 1.414. But two *different* non-coherent signals of nominally identical level will generally sum as RMS: 0.707+0.707= 1.0

This 0.707/each at center is not universally agreed "optimum". D. Self and I think Dennis Bohm have written on this.

https://en.wikipedia.org/wiki/Fade_(audio_engineering)#Crossfading
https://math.stackexchange.com/questions/4621/simple-formula-for-curve-of-dj-crossfader-volume-dipped
https://dsp.stackexchange.com/questions/14754/equal-power-crossfade

[melangston]

That also makes more sense. There's still a change in overall signal level with pot rotation, but with your help in the vocabulary, using the search phrase "constant power crossfade" I was able to find an answer (coincidentally by you on the diyaudio.com) that answered some follow up questions I had.

I hope posting a link to diyaudio.com is okay, I didn't see anything about it in the rules for posting.

https://www.diyaudio.com/forums/analog-line-level/316362-constant-power-crossfader.html

Based on this and the information in the three links you provided in the previous post (thank you for those), A) the 3dB change in signal level I get is correct, B) my signals (the 2 pickups of a guitar) are likely partially coherent so its going to be complicated, and C) from a practicality standpoint the circuit as is will do fine (which was never really in question).

Thank you everyone for your help. I am glad that I feel I have a better understanding of how all this works thanks to you.

-Michael

[ElectricDruid]

BTW your 0.2857 vs original 2.93 is due to your using approximate resistances ie 10k and 15k vs the theoretical 1 and 0.707

Pot is 0.707 times R. But pots only come in round numbers. So take pot=10k, R is 1/0.707 of that or 10k/0.707 or 14,142 Ohms. The "15k" is significantly rounded so it can be found in a 2008 part-bin.

Ah, yes! Thanks both! I *knew* there was something simple I'd overlooked.