Calrec '8000 sums in groups-of-eight.. and sums these again in groups-of-eight..
Jakob E.
Jakob E.
How's the summing done in very large consoles?
- Thread starter gyraf
- Start date
Help Support GroupDIY Audio Forum:
Samuel Groner
Well-known member
Well, there is no theoretical advantage of sub-mixing. If we have 10k summing resistors, the mixing amp sees 100 ohm with 100 channels. It is not impossible to build an amp that has close-to-zero NF for that impedance. Gain is 40 dB, not impossible either.
The advantages for sub-mixing are probably things like higher RF-immunity, lower bus capacity due to shorter runs etc.
Samuel
NewYorkDave
Well-known member
But for performance approaching "ideal", the open-loop gain of the summing amp needs to be much higher than the closed-loop gain. So we're talking at least 80dB OLG for this hypothetical 100-channel console.
Also, don't forget that the noise gain of the summing amp will be 101x. With such physically long paths from the channels to the summing amp, there's bound to be a fair amount of garbage riding on "ground." And you've already mentioned the issue of high buss capacitance.
Further reading:
http://www.elecdesign.com/Articles/Index.cfm?AD=1&ArticleID=7164
Also, don't forget that the noise gain of the summing amp will be 101x. With such physically long paths from the channels to the summing amp, there's bound to be a fair amount of garbage riding on "ground." And you've already mentioned the issue of high buss capacitance.
Further reading:
http://www.elecdesign.com/Articles/Index.cfm?AD=1&ArticleID=7164
NewYorkDave
Well-known member
100k summing resistors! That's very unusual for a solid-state console. Do you have any schematics of this beast?
FredForssell
Well-known member
Well... there is something else to think about IF you are using an inverting summing amplifier, especially if you want it to run Class A. The summing amp has to source and sink the combined peak signal current of all channels (each driving a 10k ohm resistor). That's about 50 ma for your 100 channel mixer (using 5 volts peak or +16 dBu peak as a signal, and about 6 ma for 12 channels.
Depending on what you do for a summing amp, it may sound better to use small blocks summed again at the output. Something to give some thought to anyway.
Cheers,
pstamler
Well-known member
[quote author="FredForssell"]Well... there is something else to think about IF you are using an inverting summing amplifier, especially if you want it to run Class A. The summing amp has to source and sink the combined peak signal current of all channels (each driving a 10k ohm resistor). That's about 50 ma for your 100 channel mixer (using 5 volts peak or +16 dBu peak as a signal, and about 6 ma for 12 channels.[/quote]
Well, if you were really putting 50mA into the summing junction, then (assuming the feedback resistor is 10k) you'd get 500V pk out if the circuit could do that (+53dBu), or massive clipping in real life.
Things won't really be this bad, since the signals aren't all identical; each input is about 0.5mA pk (5V into 10k), so the total will be sqrt(100) * 0.5 = 5mA. Now, that's still 50V pk out (+33dBu), so you'll have to pull the faders down some.
Peace,
Paul
Well, if you were really putting 50mA into the summing junction, then (assuming the feedback resistor is 10k) you'd get 500V pk out if the circuit could do that (+53dBu), or massive clipping in real life.
Things won't really be this bad, since the signals aren't all identical; each input is about 0.5mA pk (5V into 10k), so the total will be sqrt(100) * 0.5 = 5mA. Now, that's still 50V pk out (+33dBu), so you'll have to pull the faders down some.
Peace,
Paul
Samuel Groner
Well-known member
Which leads to a Q I had in mind for a long time: How do you implement a fader for a balanced summing bus? It would be nice to have a stereo fader before the mixing amp to avoid clipping, but this would degrade CMRR. Any hints?
Samuel
Samuel Groner
Well-known member
That was my first thought as well, but I think it doesn't work with virtual earth summing, as closing the fader reduces the source impedance that the mix amp sees which results in higher make-up gain. This again results in a change in noise figure, but not in gain.
It does work for passive summing, though.
Samuel
FredForssell
Well-known member
Thanks for setting me straight Paul! Your figures make sense to me. Sorry for not thinking this through before posting.
SSLtech
Well-known member
A number of top-flight manufacturers use "bucket-summing". One of John Musgrave's main modifications to the Neve V-series was to bucket-sum in buckets of 12. It also keeps the physical length of the buses shorter.
Manufacturers that did it:
Focusrite (the REAL focusrites, the ones designed by Rupert)
Calrec
Manufacturers who didn't do it:
SSL
Neve
Keith
Manufacturers that did it:
Focusrite (the REAL focusrites, the ones designed by Rupert)
Calrec
Manufacturers who didn't do it:
SSL
Neve
Keith
Samuel Groner
Well-known member
Keith, I would be very interested in studying the mix amps of the 90*98 and the 9k; do you mind scanning for me? :roll: Thanks!
Samuel
Samuel
SSLtech
Well-known member
From memory, I think they might have been hybrid devices... Shown as a white block on the schematic, I'd have to check.
Keith
Keith
> How do you implement a fader for a balanced summing bus?
Use a balanced fader. You could use a 2-gang stereo pot.
But what you really do is: the channel is almost invariably unbalanced, going push-pull only at the end. Put your fader in the unbalanced portion. This also means you can have post-fader gain, which you usually want (but avoid to save money).
> The summing amp has to source and sink the combined peak signal current of all channels
Yes, .OR. the sum-amp's peak output voltage divided by the feedback resistor.
And for typical small-board gain structure, if ALL inputs are delivering maximum level, the sum-amp will voltage-clip long before it runs out of current to sink the sum of the input currents.
If a mixer is unity-gain for one input (which seems reasonable), and you drive all inputs hard, the output is much higher than any input. Sum is greater than the parts. Worst-case, all channels same 10V signal, a 100-in "unity gain" summer will output 10+10+10+10... or 10V*100 or 1,000V (or try to). For random signals (and music may be similar), dynamics and phasing will probably give an output more like the square-root of the number of inputs, or 100V, 99.99% of the time.
For many musical mixes, at any one time, there is one Star and the rest have to fit behind. If you pot-up the Star to -6dB, and four to sixteen back-up tracks at -12dB, the sum will be around 0dB. So it often makes sense to run a unity-gain summer, but mark the faders for nominal use around -10dB, with the understanding that the Star track is set a little higher (but not 0dB) and the others somewhat lower. That assumes the track levels are uniform to begin with... since they never are, you have to use some judgement too.
> Well, there is no theoretical advantage of sub-mixing.
No, but there are real practical advantages that may be compelling. A 100-in bus might end up 16 feet long, a very effective taxi-cab antenna and heavy parasitic reactance to the audio. A 100-in mix network in a box has 101 input cables, a real snarl.
And using made-up numbers we can show a significant noise advantage.
Assume the only mix-amp we have has one microVolt of input noise at any source impedance. (I said this was a made-up example, but a TL072 gives roughly this performance.) And assume we want unity-gain from any (single) input. And that all our sources are noise-free.
For the trivial case of a 1-input mixer, the output noise is 2uV. For a 2-input, it is 3uV. For 10 inputs, we get 11uV.
We see that for 100 inputs, we get 101uV.
Now try again but with ten 10-in mixers feeding an eleventh 10-in mixer (still 100 total inputs). Now the noise is only 37uV.
In today's world, we can expect much better than 1uV input noise per mix-amp. As Samuel says, we may be able to noise-match any source impedance at low Noise Figure.
With a noise-less amp, only resistor noises matter.
10-in, 909 ohms noise source, 0.3uV noise, noise gain 11, 3uV output noise
100-in, 99 ohms noise source, 0.1uV noise, noise gain 101, 10uV output noise
10*10-in to 1*10-in, 10uV output noise.
As Samuel says, with perfect amplifiers, sub-mixing has no noise advantage.
Use a balanced fader. You could use a 2-gang stereo pot.
But what you really do is: the channel is almost invariably unbalanced, going push-pull only at the end. Put your fader in the unbalanced portion. This also means you can have post-fader gain, which you usually want (but avoid to save money).
> The summing amp has to source and sink the combined peak signal current of all channels
Yes, .OR. the sum-amp's peak output voltage divided by the feedback resistor.
And for typical small-board gain structure, if ALL inputs are delivering maximum level, the sum-amp will voltage-clip long before it runs out of current to sink the sum of the input currents.
If a mixer is unity-gain for one input (which seems reasonable), and you drive all inputs hard, the output is much higher than any input. Sum is greater than the parts. Worst-case, all channels same 10V signal, a 100-in "unity gain" summer will output 10+10+10+10... or 10V*100 or 1,000V (or try to). For random signals (and music may be similar), dynamics and phasing will probably give an output more like the square-root of the number of inputs, or 100V, 99.99% of the time.
For many musical mixes, at any one time, there is one Star and the rest have to fit behind. If you pot-up the Star to -6dB, and four to sixteen back-up tracks at -12dB, the sum will be around 0dB. So it often makes sense to run a unity-gain summer, but mark the faders for nominal use around -10dB, with the understanding that the Star track is set a little higher (but not 0dB) and the others somewhat lower. That assumes the track levels are uniform to begin with... since they never are, you have to use some judgement too.
> Well, there is no theoretical advantage of sub-mixing.
No, but there are real practical advantages that may be compelling. A 100-in bus might end up 16 feet long, a very effective taxi-cab antenna and heavy parasitic reactance to the audio. A 100-in mix network in a box has 101 input cables, a real snarl.
And using made-up numbers we can show a significant noise advantage.
Assume the only mix-amp we have has one microVolt of input noise at any source impedance. (I said this was a made-up example, but a TL072 gives roughly this performance.) And assume we want unity-gain from any (single) input. And that all our sources are noise-free.
For the trivial case of a 1-input mixer, the output noise is 2uV. For a 2-input, it is 3uV. For 10 inputs, we get 11uV.
We see that for 100 inputs, we get 101uV.
Now try again but with ten 10-in mixers feeding an eleventh 10-in mixer (still 100 total inputs). Now the noise is only 37uV.
In today's world, we can expect much better than 1uV input noise per mix-amp. As Samuel says, we may be able to noise-match any source impedance at low Noise Figure.
With a noise-less amp, only resistor noises matter.
10-in, 909 ohms noise source, 0.3uV noise, noise gain 11, 3uV output noise
100-in, 99 ohms noise source, 0.1uV noise, noise gain 101, 10uV output noise
10*10-in to 1*10-in, 10uV output noise.
As Samuel says, with perfect amplifiers, sub-mixing has no noise advantage.
dale116dot7
Well-known member
Noise might be less of an issue but the amount of drive required by the mix bus amp gets to be substantial as the channel count goes up. Voltage summing does not have this problem but virtual ground summing does. Basically it takes a bunch of current from the summing amp to maintain the bus at ground.
-Dale
-Dale
> the amount of drive required by the mix bus amp gets to be substantial as the channel count goes up.
In practice it never happens. The feedback current is never more than the peak output divided by the feedback resistor. For typical 10K resistors, and 10V-20V peaks, 1mA-2mA max.
No matter how many input channels.
If the feedback resistor is no lower than the opamp's minimum load (and the next-stage loading is small), then it can't run out of current. For 5532 amp we could use 600Ω feedback, no sweat.
In fact we rarely see virtual-earth feedback resistors smaller than 10K or 5K. Why don't we use smaller resistors for lower thermal noise? Of course at some point the noise impedance becomes awkward, but we could always use a step-up transformer. No, the real problem with low-low-impedance mixing is that, one way or another, huge currents get dumped into ground, and thus into the input of any ground-referenced amplifier. Say we picked 1K mix resistors. We would probably drive them from a 500Ω fader or pan-pot. Even if the mix-resistor is disconnected from the bus, there are 10mA peaks through the 500Ω fader/panpot to ground. If ground resistance is 1Ω, that's 10mV of crap in our ground. Even if we got the ground system down to 0.001Ω (which may be awful hard), the ground-crap of "live but muted" channels is not hidden in noise.
In practice it never happens. The feedback current is never more than the peak output divided by the feedback resistor. For typical 10K resistors, and 10V-20V peaks, 1mA-2mA max.
No matter how many input channels.
If the feedback resistor is no lower than the opamp's minimum load (and the next-stage loading is small), then it can't run out of current. For 5532 amp we could use 600Ω feedback, no sweat.
In fact we rarely see virtual-earth feedback resistors smaller than 10K or 5K. Why don't we use smaller resistors for lower thermal noise? Of course at some point the noise impedance becomes awkward, but we could always use a step-up transformer. No, the real problem with low-low-impedance mixing is that, one way or another, huge currents get dumped into ground, and thus into the input of any ground-referenced amplifier. Say we picked 1K mix resistors. We would probably drive them from a 500Ω fader or pan-pot. Even if the mix-resistor is disconnected from the bus, there are 10mA peaks through the 500Ω fader/panpot to ground. If ground resistance is 1Ω, that's 10mV of crap in our ground. Even if we got the ground system down to 0.001Ω (which may be awful hard), the ground-crap of "live but muted" channels is not hidden in noise.
Samuel Groner
Well-known member
Not sure we're talking about the same thing; I was talking about the master fader, not channel fader. It would be nice to have the master fader before the make-up gain (or somehow controlling make-up gain) for best overload protection - without eating CMRR. I can think of some solution if we allow a transformer, but not yet without one.
Samuel
pstamler
Well-known member
[quote author="PRR"]> With a noise-less amp, only resistor noises matter.
10-in, 909 ohms noise source, 0.3uV noise, noise gain 11, 3uV output noise
100-in, 99 ohms noise source, 0.1uV noise, noise gain 101, 10uV output noise
10*10-in to 1*10-in, 10uV output noise.
As Samuel says, with perfect amplifiers, sub-mixing has no noise advantage.[/quote]
Well, that brings up an issue which has always left me a little puzzled. We're talking about active summing here. into the - input of an active circuit. There's no doubt in my mind that, as far as current noise is concerned, what you say is correct; the input resistors can be treated as a parallel network, so that for 100 10k inputs plus a 10k feedback resistor the resistance is 99 ohms for the purpose of calculating noise generated by input noise current on the active device.
What about Johnson noise, though? The resistors are isolated from one another by the virtual ground of the summing junction. Each is a Johnson noise generator in its own right. Would not the noise voltages generated by Johnson noise in the multiple inputs be summed by the network? So that if a 10k resistor has Johnson noise of 1.8uV, 100 of them in a summing network would have a combined Johnson noise of 1.8uV * sqrt(100), or 18uV?
Or am I totally off-base here?
Peace,
Paul
10-in, 909 ohms noise source, 0.3uV noise, noise gain 11, 3uV output noise
100-in, 99 ohms noise source, 0.1uV noise, noise gain 101, 10uV output noise
10*10-in to 1*10-in, 10uV output noise.
As Samuel says, with perfect amplifiers, sub-mixing has no noise advantage.[/quote]
Well, that brings up an issue which has always left me a little puzzled. We're talking about active summing here. into the - input of an active circuit. There's no doubt in my mind that, as far as current noise is concerned, what you say is correct; the input resistors can be treated as a parallel network, so that for 100 10k inputs plus a 10k feedback resistor the resistance is 99 ohms for the purpose of calculating noise generated by input noise current on the active device.
What about Johnson noise, though? The resistors are isolated from one another by the virtual ground of the summing junction. Each is a Johnson noise generator in its own right. Would not the noise voltages generated by Johnson noise in the multiple inputs be summed by the network? So that if a 10k resistor has Johnson noise of 1.8uV, 100 of them in a summing network would have a combined Johnson noise of 1.8uV * sqrt(100), or 18uV?
Or am I totally off-base here?
Peace,
Paul
NewYorkDave
Well-known member
Paul,
Since the noise is non-correlated, the noise voltages would not simply add but would sum in a power fashion--unless I'm missing something. :roll:
Since the noise is non-correlated, the noise voltages would not simply add but would sum in a power fashion--unless I'm missing something. :roll:
> The resistors are isolated from one another by the virtual ground of the summing junction. Each is a Johnson noise generator in its own right. Would not the noise voltages generated by Johnson noise in the multiple inputs be summed by the network? So that if a 10k resistor has Johnson noise of 1.8uV, 100 of them in a summing network would have a combined Johnson noise of 1.8uV * sqrt(100), or 18uV? Or am I totally off-base here?
Hmmmm... a brain twister.
As my brain is fried tonight, I asked an idiot who does numbers better than me and who never leaps to "obvious" conclusions (can leap off-track, a lot or a little...).
I hope I did this analysis right: SPICE's noise card always baffles me.
10-input 10K summers, active and passive:
SPICE's answers:
10-in 10K active summer, gain="1"
**** RESISTOR SQUARED NOISE VOLTAGES (SQ V/HZ)
_____ R_R2_____ R_R3_____ R_R4_____ R_R5_____ R_R6_____ R_R7____
TOTAL 1.658E-16 1.658E-16 1.658E-16 1.658E-16 1.658E-16 1.658E-16
_____ R_R8_____ R_R9_____ R_R10____ R_Rsum___ R_R1____
TOTAL 1.658E-16 1.658E-16 1.658E-16 1.658E-16 1.658E-16
**** TOTAL OUTPUT NOISE VOLTAGE = 1.823E-15 SQ V/HZ = 4.270E-08 V/RT HZ
TRANSFER FUNCTION VALUE: V(Out)/V_V1 = 1.000E+00
EQUIVALENT INPUT NOISE AT V_V1 = 4.270E-08 V/RT HZ
10-in 10K passive summer plus gain=10 amplifier
**** RESISTOR SQUARED NOISE VOLTAGES (SQ V/HZ)
_____ R_R14____ R_R15____ R_R16____ R_R17____ R_R18____ R_R19____
TOTAL 1.658E-16 1.658E-16 1.658E-16 1.658E-16 1.658E-16 1.658E-16
_____ R_R20____ R_R21____ R_R22____ R_R23____ R_R24____ R_Rsum2__
TOTAL 1.658E-16 1.658E-16 1.658E-16 1.658E-16 1.343E-15 1.492E-16
**** TOTAL OUTPUT NOISE VOLTAGE = 3.149E-15 SQ V/HZ = 5.612E-08 V/RT HZ
TRANSFER FUNCTION VALUE: V(Out2)/V_V2 = 1.000E+00
EQUIVALENT INPUT NOISE AT V_V2 = 5.612E-08 V/RT HZ
Hmmmmm.... the passive summer is a couple dB higher noise. And R24 (the gain-set shunt resistor) appears to dominate. We could probably scale-down that impedance: that's "cheating" because it does force more opamp output current, but in a real-world case this cheat is allowed.
I realize that isn't the question you asked, but I have to put-out the dog and go to bed.
Hmmmm... a brain twister.
As my brain is fried tonight, I asked an idiot who does numbers better than me and who never leaps to "obvious" conclusions (can leap off-track, a lot or a little...).
I hope I did this analysis right: SPICE's noise card always baffles me.
10-input 10K summers, active and passive:
SPICE's answers:
10-in 10K active summer, gain="1"
**** RESISTOR SQUARED NOISE VOLTAGES (SQ V/HZ)
_____ R_R2_____ R_R3_____ R_R4_____ R_R5_____ R_R6_____ R_R7____
TOTAL 1.658E-16 1.658E-16 1.658E-16 1.658E-16 1.658E-16 1.658E-16
_____ R_R8_____ R_R9_____ R_R10____ R_Rsum___ R_R1____
TOTAL 1.658E-16 1.658E-16 1.658E-16 1.658E-16 1.658E-16
**** TOTAL OUTPUT NOISE VOLTAGE = 1.823E-15 SQ V/HZ = 4.270E-08 V/RT HZ
TRANSFER FUNCTION VALUE: V(Out)/V_V1 = 1.000E+00
EQUIVALENT INPUT NOISE AT V_V1 = 4.270E-08 V/RT HZ
10-in 10K passive summer plus gain=10 amplifier
**** RESISTOR SQUARED NOISE VOLTAGES (SQ V/HZ)
_____ R_R14____ R_R15____ R_R16____ R_R17____ R_R18____ R_R19____
TOTAL 1.658E-16 1.658E-16 1.658E-16 1.658E-16 1.658E-16 1.658E-16
_____ R_R20____ R_R21____ R_R22____ R_R23____ R_R24____ R_Rsum2__
TOTAL 1.658E-16 1.658E-16 1.658E-16 1.658E-16 1.343E-15 1.492E-16
**** TOTAL OUTPUT NOISE VOLTAGE = 3.149E-15 SQ V/HZ = 5.612E-08 V/RT HZ
TRANSFER FUNCTION VALUE: V(Out2)/V_V2 = 1.000E+00
EQUIVALENT INPUT NOISE AT V_V2 = 5.612E-08 V/RT HZ
Hmmmmm.... the passive summer is a couple dB higher noise. And R24 (the gain-set shunt resistor) appears to dominate. We could probably scale-down that impedance: that's "cheating" because it does force more opamp output current, but in a real-world case this cheat is allowed.
I realize that isn't the question you asked, but I have to put-out the dog and go to bed.
