Noise testing - mixer

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ej_whyte

Well-known member
Joined
Nov 12, 2010
Messages
263
Location
Cambridge, UK
When testing the noise floor of a virtual earth mixer, how should the inputs be terminated?
Left completely unconnected, terminated with 50 ohm resistor, or muted (disconnected just after summing resistor)?

Thanks
 
Yes....  ;D

It kind of depends how the VE bus will be used. For a sub bus that may only get one or two stems assigned perhaps measure with just one input.

For a master sum bus that you plan to routinely mix using 16 or 24 stems, measure as-used.

The input resistors need to be fed from low impedance, unless there is a passive pan circuit in series with those input resistors.

It will matter where you ground or terminate these input resistors since any voltage difference between that ground and the ground on the VE sum amp + input, will get multiplied by the noise gain (N+1 number of stems).

In my judgment noise is over inspected as a sound quality constraint for sum buses, since the noise floor of typical source material will typically be much higher. Phase shift and distortion, also associated with high noise gain bus amps, does not get masked by program material.

JR
 
 
Ah ok thanks for the info. I am testing the unit I built here http://www.groupdiy.com/index.php?topic=50358.0

So 8ch in to 2 buses then into a master bus, I guess I will route four to each bus for the testing. The summing resistors are fed from THAT 1200s, should I leave the input to the THATs unconnected or terminated with 50Ω?

What would you estimate as a decent noise figure to aim for in such a setup?

Cheers
 
ej_whyte said:
What would you estimate as a decent noise figure to aim for in such a setup?
With 4 injection resistors connected to the bus, the summing amp's noise gain is about 15dB.
With the 990C, which boasts -133dBu EIN, it would compute to -118dBu
With the 2520, which is close to a 5534 in terms of noise, that would be about -110
In these conditions, it seems the dominant source of noise is the THAT 1200's, which are rated at -107; four of them would worsen this by 6dB, resulting in -101.
This is valid only if the bus-injection resistors are of low enough value to make the noise current term negligible.
For the 2520 anything below 10k is suitable.
For the 990c, the optimum source impedance is about 1100 ohms. That would suggest the use of injection resistors of less than 4.7k (unbalanced bus) or 2.2k (balanced bus). Using 10k res in an unbalanced bus configuration would make the contribution of noise current dominant, with noise from the summing amp measuring at about -110, which, again is sufficiently below the 1200's noise to be negligible.
 
abbey road d enfer said:
ej_whyte said:
What would you estimate as a decent noise figure to aim for in such a setup?
With 4 injection resistors connected to the bus, the summing amp's noise gain is about 15dB.
With the 990C, which boasts -133dBu EIN, it would compute to -118dBu
With the 2520, which is close to a 5534 in terms of noise, that would be about -110
In these conditions, it seems the dominant source of noise is the THAT 1200's, which are rated at -107; four of them would worsen this by 6dB, resulting in -101.
This is valid only if the bus-injection resistors are of low enough value to make the noise current term negligible.
For the 2520 anything below 10k is suitable.
For the 990c, the optimum source impedance is about 1100 ohms. That would suggest the use of injection resistors of less than 4.7k (unbalanced bus) or 2.2k (balanced bus). Using 10k res in an unbalanced bus configuration would make the contribution of noise current dominant, with noise from the summing amp measuring at about -110, which, again is sufficiently below the 1200's noise to be negligible.

OK thanks, currently it is using 10k resistors. These were used to prevent the 1200s being loaded to heavily. From my schematic with all 8 inputs connected, the load on each 1200 would be:

10K || (10K + 1/( (1/10K)*7 ) ) = 5.333KΩ

Correct?

The only realisation I have just had is that maybe due to the virtual earth the 1200 would not be loaded by the other channels at all. I feel like I may have just made a big mistake :/

Cheers
 
ej_whyte said:
OK thanks, currently it is using 10k resistors. These were used to prevent the 1200s being loaded to heavily. From my schematic with all 8 inputs connected, the load on each 1200 would be:

10K || (10K + 1/( (1/10K)*7 ) ) = 5.333KΩ

Correct?
I don't think so. The VE bus has zero impedance. the impedance presented to the source (the 1200) is 10k; if you route the output of the 1200 to both sides, you have two 10k res in parallels, resulting in a 5k load, which is perfectly adequate.
The only realisation I have just had is that maybe due to the virtual earth the 1200 would not be loaded by the other channels at all.
That's one of the major points of VE mixing; almost perfect isolation between channels.
 
abbey road d enfer said:
ej_whyte said:
OK thanks, currently it is using 10k resistors. These were used to prevent the 1200s being loaded to heavily. From my schematic with all 8 inputs connected, the load on each 1200 would be:

10K || (10K + 1/( (1/10K)*7 ) ) = 5.333KΩ

Correct?
I don't think so. The VE bus has zero impedance. the impedance presented to the source (the 1200) is 10k; if you route the output of the 1200 to both sides, you have two 10k res in parallels, resulting in a 5k load, which is perfectly adequate.
The only realisation I have just had is that maybe due to the virtual earth the 1200 would not be loaded by the other channels at all.
That's one of the major points of VE mixing; almost perfect isolation between channels.

Aah ok so it is as I thought: I wrongly treated the loading as if it was voltage summing. The switching doesn't give the option to route to both buses, so the 10k can definitely come down in value.

I don't know why i didn't realise this earlier really, slightly embarrasing, I guess i have a better grasp on the actual concept of virtual earth now than when i did this section. So the current set up from the output of each 1200 is a 100uF cap, followed by a 10k resistor to (real) ground to form an RC network, then the 10k resistor to virtual earth. Am I right in saying that the 10k resistor to (real) ground is unnecessary and the resistor to virtual earth completes the RC network? Although with a switch between the summing resistor and the virtual earth point this may not be the case if I want to avoid pops when switching.

Thanks for all your help!
 
ej_whyte said:
Am I right in saying that the 10k resistor to (real) ground is unnecessary and the resistor to virtual earth completes the RC network? Although with a switch between the summing resistor and the virtual earth point this may not be the case if I want to avoid pops when switching.
It is necessary for avoiding switching clicks, but can be of much higher value; 47k is a safe bet.
 

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