Yet another Passive Summing Mixer project question

[ytsestef]

Every once in a while another passive summing box is built, and there are a lot of people who don't want to make things complex by placing switches or panpots for positioning the signal [L/R/Center] as this requires a lot of wiring and is way more expensive than a control-less box.
This is the case with me as well, and I don't like the idea of wasting a channel to place something in the center (by feeding it to two inputs).

So I came up with the idea of a summing box that has 6 Stereo Pairs and 4 Mono (total 16 channels). I concluded to this ratio between mono and stereo because I don't want to limit drastically the number of stereo signals and I hardly use more or less than four channels (Vocal track, Bass, Kick, Snare) panned dead centre.

I guess It would be like bypassing the switches altogether and wiring the thing as if they were set permanently in preset positions. I prepared a schematic but my knowledge in electronics is poor (though I am skilled at soldering), so I need you to check it, because I have a serious suspicion it is all wrong. To be exact I don't even understand in the first schematic (the one upon which mine is based) how do the mono channels (which are sent to both the L and R bus) manage NOT to function as conductors for all the channels to cross both busses. I guess it has something to do with the current flow or they dont!  :-\ Anyway, here is the original schematic:

http://www.twin-x.com/groupdiy/albums/userpics/balancedmixnetwork.pdf

And here is mine. Please if someone who knows about that stuff reads it, check the resistor values as well as the topology and comment on it.
I couldn't find a project like mine and that is why I started a new thread. MODS, If it is in the wrong place, feel free to move it.

Thanks a lot in advance!

 

[PRR]

> how do the mono channels (which are sent to both the L and R bus) manage NOT to function as conductors for all the channels to cross both busses.

They do.

Compute "how much".

I'm lazy. I re-drew as un-balanced. Taking half the plan, divided down the line of symmetry, will give the same answer, with less head-scratch.

A 1V signal comes in on a left channel. Right away it gets whacked by 10K against 118.5 ohms (lazily computed as 120). 0.012V on bus.

That sneaks "out" through a 14K resistor to a both-sides source. If the source is zero impedance, the sneakage is killed. If the source is typical/practical, it has some impedance. 15r to 470r resistors are often found, 100 is easy to compute. Our 0.012V on L bus is whacked by 14K against 100, 0.000,084V at source pin.

This gets over to right bus, through 14K against 120, and is whacked yet again, to 0.000,000,99V.

(We could round our ***** off and just compute 10K:100, 10K:100, 10K:100, or 100:1 three times, and get the rough answer on one hand.)

Rounding, a left-only input sneaks to the right output, but a million times weaker, -120dB cross-talk. You could, as radio stations used to, feed football in one bus and flower-hour in the other bus, with emergency source to both, and the sneakage would not be audible. When the two channels are a single stereo program, L and R played in the same room, -120dB crosstalk is utterly insignificant.

If the both-sides source is far from "low" impedance, crosstalk is higher. Say it were infinite impedance. We'd still have 0.012V on L bus, then 14K+14K= 28K against 120r, 0.000,05V sneakage to R bus, -85dB. Not so impressive, but still insignificant.

You may see a pattern. Things get cut by ~~100:1. That is because the plan expects a mike-amp at the output, with gain near 100, and sum output similar to each input. So the 10K+10K:237 ratio was selected for 84:1 loss. Also the 10K-14K mix resistors are selected to be a couple orders of magnitude (100:1) higher than typical source impedances, for low-low loading. If the sneakage has to cross three whack-down dividers, 120dB loss; if the both-side source is omitted (infinite Z) then sneakage only passes two whacks and only 80dB.

A minimum-loss mixer might have all 1K-1.4K resistors without the 237 ohm whack. Recovery amp gain would be more like 5 or 10, each sneak-whack-path has loss of only about 10, worst-case crosstalk might be 100:1 or -40dB. This is fine for Stereo, could be just-audible if the footballers are cursing loudly in a quiet part of flower-hour.

 

[ytsestef]

Thanks for your answer!

So if I got it right, the design is correct, but the resistor values are a matter of taste.
It is a compromise of losing level vs. getting rid of crosstalk.
more loss = less crosstalk
less loss = more crosstalk

Am I right?
You mentioned that with the plan I posted, I would need a mike-amp with gain near 100. Are we talking db here??? 100 db gain is impossible! I'd have to push my millenia's BOTH input and output to the maximum!! 

 

[JohnRoberts]

100:1 =40dB

JR

 

[ytsestef]

Makes much more sense now  ;D ;D

All good then, I'm off to order the parts!!
Will post progress here, hopefully

Thanks guys!!

 

[ytsestef]

I thought I might as well do it on a veroboard instead of fat copper wires. I guess that would result in a cleaner job because I won't have to straighten the wires and figure out a way to mount them. Good ol' veroboard is mounted like any pcb and if it is a little messy on the solder side (only visually, of course) it won't be visible. Is there any downside, sound-wise?

 

[qweetz]

Hi Stefanos.

How did you get on with your summing unit?  :  )