LCR mix bus assignment questions

[juniorhifikit]

Hi,

I found NYDave's balanced LCR switching circuit drawing, and had a question.  I'd like to do the same thing, but unbalanced, and am worried about loading the mix bus in weird ways.  If I had 15K mix bus resistors for each bus (L & R), but then wanted to assign a channel to both busses, wouldn't I then need 30K resistors in parallel to each bus to keep the load the same?  Like this?

http://www.scottgreiner.com/LCR_assgn.jpg

Most drawings I've seen use, say, 10K summing resistors, and then 10K resistors switched in parallel to each buss (to assign to the center)  Doesn't this change the load on the mix bus?

Also, should I keep the number of 15K resistors constant on the bus, and thus switch the unassigned points to ground through 15K resistors?

Thanks!

 

[PRR]

> Doesn't this change the load on the mix bus?

If you have more than "a few" sources, then a change in one source makes only very-small difference.

If your assign is switches, you probably won't be switching "live"; you will assign, then set/reset levels.

If your mix-stage is either a high-gain (mike amp), or a zero-impedance stage, the number of source resistors is largely swamped by either your ~~300 ohm shunt or the zero-impedance.

> switch the unassigned points to ground through 15K resistors?

Your diagram shows this.

 

[juniorhifikit]

> Doesn't this change the load on the mix bus?

If you have more than "a few" sources, then a change in one source makes only very-small difference.

There will be 32 channels

If your assign is switches, you probably won't be switching "live"; you will assign, then set/reset levels.

True.  Maybe.  Sort of...

If your mix-stage is either a high-gain (mike amp), or a zero-impedance stage, the number of source resistors is largely swamped by either your ~~300 ohm shunt or the zero-impedance.

I'm basing this on an early SSL summing amp fed by 5534 voltage follower stages through 15K resistors.  No pan circuit, just switches, and I'm unclear how to best treat the "center" situation.  If someone put all 32 channels in "center", 15K parallel resistors (centered to L & R busses) would change the source impedance compared to using 30K paralleled for "center", no?.  Similarly, all channels assigned "left" or "right" through series 15K resistance would also be different.  I'm confused and have just enough knowledge to get in over my head! ;D

> switch the unassigned points to ground through 15K resistors?

Your diagram shows this.

Yes, I drew it that way, but is it kosher oh guru?  

 

[abbey road d enfer]

I'm basing this on an early SSL summing amp fed by 5534 voltage follower stages through 15K resistors.  No pan circuit, just switches, and I'm unclear how to best treat the "center" situation.  If someone put all 32 channels in "center", 15K parallel resistors (centered to L & R busses) would change the source impedance compared to using 30K paralleled for "center", no?.

You are mixing two different issues. With equal bus injection resistors (15k), the load seen by the buffer stage may change from 15k to 7.5k, which has no consequence at all.
The source impedance seen by the summing amp with your arrangement may change from 1k to 500ohms, which is a problem if you don't use a virtual-ground summing amp.  That would be a minor problem anyway, since it only affects gain (depends how much importance you attach to the accuracy of the level diagram).

 

[juniorhifikit]

You are mixing two different issues. With equal bus injection resistors (15k), the load seen by the buffer stage may change from 15k to 7.5k, which has no consequence at all.
The source impedance seen by the summing amp with your arrangement may change from 1k to 500ohms, which is a problem if you don't use a virtual-ground summing amp.  That would be a minor problem anyway, since it only affects gain (depends how much importance you attach to the accuracy of the level diagram).

It is a virtual ground summing amp, so that takes care of that issue.  As far as using the 15K resistors in parallel for "center", would this give a level jump as compared to 15K left-only or right-only?

 

[abbey road d enfer]

As far as using the 15K resistors in parallel for "center", would this give a level jump as compared to 15K left-only or right-only?

You would have the same level on both L and R bus, so, yes, the level would somehow jump acoustically. If you want to avoid that, you have to tune the resistors that define the center position. Unfortunately, there is not a single answer to that question. Constant power summing (giving the same acoustic level requires 1.414 times value for the center position. That's 3dB attenuation on both L & R, so each side produces 1/2 the power. But for mono compatibility, you want half the voltage, -6dB, so you're back to 2x the value, as per your original schemo. That has been a subject of discussion on pan-pots for years. Most mixer manufacturers ended up opting for -4.5 dB as a compromise. Now in many DAW's you have the choice for -3, -6 or 0dB (DJ style).
The issue there is that when switching, whatever your choice, level will jump anyway.

 

[juniorhifikit]

As far as using the 15K resistors in parallel for "center", would this give a level jump as compared to 15K left-only or right-only?

You would have the same level on both L and R bus, so, yes, the level would somehow jump acoustically. If you want to avoid that, you have to tune the resistors that define the center position. Unfortunately, there is not a single answer to that question. Constant power summing (giving the same acoustic level requires 1.414 times value for the center position. That's 3dB attenuation on both L & R, so each side produces 1/2 the power. But for mono compatibility, you want half the voltage, -6dB, so you're back to 2x the value, as per your original schemo. That has been a subject of discussion on pan-pots for years. Most mixer manufacturers ended up opting for -4.5 dB as a compromise. Now in many DAW's you have the choice for -3, -6 or 0dB (DJ style).
The issue there is that when switching, whatever your choice, level will jump anyway.

Thanks.

If your assign is switches, you probably won't be switching "live"; you will assign, then set/reset levels.

This is most likely the case, so it would be easiest to just join the two 15K resistors with a switch and have -6dB.