Virtual Earth Summing - Noise Calcs

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ej_whyte

Well-known member
Joined
Nov 12, 2010
Messages
263
Location
Cambridge, UK
I'm trying to do some noise calculations at the minute but just need to clarify a couple of things up and need a bit of help please. So it is for an 8 channel virtual earth mixer, this is what I was working with at the moment:

Calculate noise for individual channel input circuitry (including summing resistor)
RMS addition of 8x that figure for 8 channels (Vn-in)
RMS addition of summing amp voltage noise x gain+1, summing amp current noise and feedback resistor noise (Vn-sum)
RMS addition of (Vn-in x gain) + Vn-sum)

There are three things I am not sure on

1. I assume that there is no attenuation of the channel input's noise as the virtual earth means that there is no real potential divider formed with other channel's summing resistors
2. Does the channel input noise need to be multiplied by the summing amp's gain, or is it treated as unity?
3.Not quite sure on how to treat the summing resistors A) calculate johnson noise for 1 resistor and then do RMS addition for 8 of these, or B) Treat them all as parallel and calculate johnson noise from the parallel value. I would guess A as it is actually the physically correct amount of resistors.

Thanks
 
ej_whyte said:
1. I assume that there is no attenuation of the channel input's noise as the virtual earth means that there is no real potential divider formed with other channel's summing resistors
That's correct. Input-to-output transfer is 1:1.
2. Does the channel input noise need to be multiplied by the summing amp's gain, or is it treated as unity?
Unity, for the same reason.
3.Not quite sure on how to treat the summing resistors A) calculate johnson noise for 1 resistor and then do RMS addition for 8 of these, or B) Treat them all as parallel and calculate johnson noise from the parallel value. I would guess A as it is actually the physically correct amount of resistors.
I bet you'll find that the result is the same  :)
In the end, I believe you'll conclude that channel noise and summing amp noise are dominant. In a proper design, resistor noise is negligible.
 
ej_whyte said:
I'm trying to do some noise calculations at the minute but just need to clarify a couple of things up and need a bit of help please. So it is for an 8 channel virtual earth mixer, this is what I was working with at the moment:
Generally you calculate just the noise contribution of the summing amp alone, then separately ASSume ideal combination with the noise from the sundry input channels, combined as incoherent noise sources. (square root of sum of the sources squared ). 

An 8 input VE sum amp with equal input and feedback Rs, will have a noise gain of 9x.
Calculate noise for individual channel input circuitry (including summing resistor)
Summing resistors are generally considered in the sum amp noise calculation, and considered all in parallel with other summing and feedback resistors for noise contribution (thermal noise and when this combined impedance is multiplied times the summing amps input noise current. 
RMS addition of 8x that figure for 8 channels (Vn-in)
noise is combined as incoherent sources for total noise calculation. so more like 2.8x for 8 stems
RMS addition of summing amp voltage noise x gain+1,
yup, 9x sum amp ein for unity bus gain. Technically noise is considered in series with + input and all input resistors grounded (i.e. driven from low impedance). So noise gain is like a non-inverting gain stage or 1+RF/(Rin's in parallel).

Note: if summing resistors are driven from a non-zero impedance like a pot wiper, this actual source impedance is in series with input resistors for noise gain.  The effect of this can actually be heard in something like an aux send bus, where noise gain actually increases when all sends are turned full down. While this phenomenon usually is more apparent as hum from poor ground quality than hiss. 
summing amp current noise and feedback resistor noise (Vn-sum)
yup, but all R's in parallel. Note, any R in series with + input added in series with other parallel R before thermal and multiply times noise current.
RMS addition of (Vn-in x gain) + Vn-sum)
Not clear what this is referring to? Noise coming into console input?
There are three things I am not sure on

1. I assume that there is no attenuation of the channel input's noise as the virtual earth means that there is no real potential divider formed with other channel's summing resistors
No insertion loss in VE summing, gain is simple -Rfeedback/Rinput
2. Does the channel input noise need to be multiplied by the summing amp's gain, or is it treated as unity?
Channel noise passes through at sum amp gain, but combines incoherently so 8 identical channels combine to 2.8x  more noise.
3.Not quite sure on how to treat the summing resistors A) calculate johnson noise for 1 resistor and then do RMS addition for 8 of these, or B) Treat them all as parallel and calculate johnson noise from the parallel value. I would guess A as it is actually the physically correct amount of resistors.

Thanks
Calculate as if all Rs connected to - input are parallel. Any R between + input and ground in series with composite - feedback R.

Hope that helps... I wasted too many hours on this over past decades.

In the '80s I came up with concept of replacing summing amp input Resistors with current sources. In theory sum amp noise gain becomes unity. In practice I was able to execute large number of stems with only modest real noise gain. But as I am quick to advise, this is all moot when a few open mics at moderate gain are cranking. That said noise gain in bus still matters because high noise gain also causes phase shift and THD to rise, and less of that is always good even if noise floor is dominated by input signal.

FWIW using a modern uber-opamp in sum amp socket can go a long way to making that stage mostly transparent for modest number of stems (or use a digital console).

JR
 
Ok thanks for the help, that has cleared some things up. Im still slightly confused as both of your answers to my 2nd point seem to contradict each other, or that may just be the way I am reading it?

JohnRoberts said:
ej_whyte said:
2. Does the channel input noise need to be multiplied by the summing amp's gain, or is it treated as unity?
Channel noise passes through at sum amp gain, but combines incoherently so 8 identical channels combine to 2.8x  more noise.

abbey road d enfer said:
ej_whyte said:
2. Does the channel input noise need to be multiplied by the summing amp's gain, or is it treated as unity?
Unity, for the same reason.

The other thing I am still unclear on is how to treat the summing resistors. Using 8 channels with 10k resistors for example:

If treated individually each resistor contributes 1.89uV of johnson noise, and 8 of these would sum via RMS to produce 5.34uV

If treated as parallel from the start, the effective resistance would be 1250Ω, resulting in 0.67uV

Which of these figures is correct? The second example seems totally counter-intuitive to me


Thanks again
 
ej_whyte said:
Ok thanks for the help, that has cleared some things up. Im still slightly confused as both of your answers to my 2nd point seem to contradict each other, or that may just be the way I am reading it?

JohnRoberts said:
ej_whyte said:
2. Does the channel input noise need to be multiplied by the summing amp's gain, or is it treated as unity?
Channel noise passes through at sum amp gain, but combines incoherently so 8 identical channels combine to 2.8x  more noise.

abbey road d enfer said:
ej_whyte said:
2. Does the channel input noise need to be multiplied by the summing amp's gain, or is it treated as unity?
Unity, for the same reason.
Both are correct,because in the second case, you have to take into account the x8 gain, so 0.67uV x 8 = Ta Da! 5.34uV
Each channel noise is passed at unity; 8 channels produce 2.8x.
JR and I are saying the same thing, errr...differently.
The other thing I am still unclear on is how to treat the summing resistors. Using 8 channels with 10k resistors for example:

If treated individually each resistor contributes 1.89uV of johnson noise, and 8 of these would sum via RMS to produce 5.34uV

If treated as parallel from the start, the effective resistance would be 1250Ω, resulting in 0.67uV
Which of these figures is correct? The second example seems totally counter-intuitive to me
Both are correct, because in the second case, you have to take into account the x8 gain, so 0.67uV x 8 = Ta Da! 5.34uV
 
Right so here goes. For now I have treated the summing resistors as part if input circuitry's noise but in future I will move it to the summing amp.

Input channel noise: 3.46uV, -107dBu
Summing resistor noise (10k): 1.89uV, -114.4dBu
RMS summation of the above: 3.94uV, -108dBu      (shouldn't this be just under -107dBu? i've been over the numbers and it seems ok :S)

RMS summation of the 3.94uV for 8 channels: 11.14uV, -99.1dBu

Opamp voltage noise (rough ballpark between 2520 and 990c): 245nV, -130dBu
Opamp voltage noise multiplied by noise gain of 9: 2.21uV, -113.1dBu
Feedback resistor noise (10k): 1.89uV, -114.4dBu
Opamp current noise through 10k feedback resistor: 1.48uV, -116.6dBu      (1pA/√Hz, 22K BW)
RMS summation of (opamp voltage noise x noise gain), feedback resistor noise and opamp current noise: 3.26uV, -109.7dBu

RMS summation of 11.14uV for input channels and 3.26uV for summing amp: 11.61uV, -98.7dBu

I think (hope) that is correct :) The final thing that I am slightly confused about is the use of the nV/√Hz unit. The / indicates a 'per' and indeed that is how it is said, however I am finding it strange that to get a voltage number you have to multiply the nV/√Hz by the √BW rather than divide it as the 'per' would indicate? :S

Thanks
 
After re-reading your earlier comments I am now not quite sure if i have treated the noise current correctly. At the moment I have just used the current to find the voltage through the feedback resistor, should the current actually be applied to the parallel of all the summing and feedback resistors, and then multiplied by the summing amp gain?

Cheers
 
ej_whyte said:
Right so here goes. For now I have treated the summing resistors as part if input circuitry's noise but in future I will move it to the summing amp.

Input channel noise: 3.46uV, -107dBu
Summing resistor noise (10k): 1.89uV, -114.4dBu
RMS summation of the above: 3.94uV, -108dBu      (shouldn't this be just under -107dBu? i've been over the numbers and it seems ok :S)
More like -105.5 dBu.  = SQRT (3.46uV^2+1.89uV ^2)
RMS summation of the 3.94uV for 8 channels: 11.14uV, -99.1dBu
I get same 11uV but that looks to me like -96.8 dBu (below >0.775V) 
Opamp voltage noise (rough ballpark between 2520 and 990c): 245nV, -130dBu
Opamp voltage noise multiplied by noise gain of 9: 2.21uV, -113.1dBu
Feedback resistor noise (10k): 1.89uV, -114.4dBu
Opamp current noise through 10k feedback resistor: 1.48uV, -116.6dBu      (1pA/√Hz, 22K BW)
RMS summation of (opamp voltage noise x noise gain), feedback resistor noise and opamp current noise: 3.26uV, -109.7dBu

RMS summation of 11.14uV for input channels and 3.26uV for summing amp: 11.61uV, -98.7dBu
I didn't check this but opamp noise current times total feedback R impedance gets added to opamp noise voltage before it gets multiplied by noise gain 9x.

Still going to be modest compared to one mic preamp at decent gain in a real room with people breathing.  Multiply the noise of a perfect 200 ohm microphone times 60 dB of gain, just for a floor below the actual room noise.
I think (hope) that is correct :) The final thing that I am slightly confused about is the use of the nV/√Hz unit. The / indicates a 'per' and indeed that is how it is said, however I am finding it strange that to get a voltage number you have to multiply the nV/√Hz by the √BW rather than divide it as the 'per' would indicate? :S

Thanks
The noise per root Hz is the noise for 1 Hz of bandwidth. This typically gets multiplied by 19,980 (20k-20) for full audio bandwidth calculation.

JR
 
Aah I see the problem, I was converting the voltages to dBv instead of dBu. That is as I though with the current, I will re-adjust the calcs. Thanks for all the help it has really cleared things up :)

Cheers
 
OK I have redone the numbers with the current noise being multiplied by the gain. The current noise now creates 13.32uV, -95.3dBu and is the dominant source. Final spec for the whole sum amp + input channel combination is 17.6uV, -92.9dBu, does that seem reasonable for a 8ch, 990C summing amp with 10k resistors all round?

Sorry for all the questions!

Cheers
 
ej_whyte said:
Aah I see the problem, I was converting the voltages to dBv instead of dBu. That is as I though with the current, I will re-adjust the calcs. Thanks for all the help it has really cleared things up :)

Cheers
Not to be overly pedantic but the sundry dB(variants) are popular usage deviations for convenience away from what is strictly a power reference.

0dBm = 1 mW, which coincidentally translates to 0.7746 V into 600 ohms.  Decibels are so convenient for describing ratios that we routinely insist on using it for voltage ratios, even heaven forbid transformer ratios, who's power ratio is always 0dB (actually slight power loss). 

dBu generally means Voltage wrt 0.775V
dBv AFAIk does not have a precise definition and is sometimes used interchangeably with dBu
dBV as popularly been considered voltage wrt 1V, so roughly 2.2 dB different from dBu.


[/pedantry]

JR
 
JohnRoberts said:
ej_whyte said:
Aah I see the problem, I was converting the voltages to dBv instead of dBu. That is as I though with the current, I will re-adjust the calcs. Thanks for all the help it has really cleared things up :)

Cheers

dBu generally means Voltage wrt 0.775V
dBv AFAIk does not have a precise definition and is sometimes used interchangeably with dBu
dBV as popularly been considered voltage wrt 1V, so roughly 2.2 dB different from dBu.

Yeah that what I was doing, referencing it to 1v instead of 0.7746v, explains why all my levels were all about 2.2db out from yours :)

Should the opamp current noise be calculated across the parallel of all the summing and feedback Rs or just the feedback resistor alone?

Thanks
 
ej_whyte said:
JohnRoberts said:
ej_whyte said:
Aah I see the problem, I was converting the voltages to dBv instead of dBu. That is as I though with the current, I will re-adjust the calcs. Thanks for all the help it has really cleared things up :)

Cheers

dBu generally means Voltage wrt 0.775V
dBv AFAIk does not have a precise definition and is sometimes used interchangeably with dBu
dBV as popularly been considered voltage wrt 1V, so roughly 2.2 dB different from dBu.

Yeah that what I was doing, referencing it to 1v instead of 0.7746v, explains why all my levels were all about 2.2db out from yours :)

Should the opamp current noise be calculated across the parallel of all the summing and feedback Rs or just the feedback resistor alone?

Thanks
Composite of all Rs in parallel (except for any R in + input that is series).

JR

PS: the dBu vs dBV was a common source of mistakes when interfacing between Pro +4 dBm (u) gear and -10dBV semi-pro gear. 
 
OK re-worked the numbers with the current drawn across the parallel of all Rs, brings it back down to 164.8nV (before being multiplied by gain) and the overall noise level is now at -96.5dBu.

Thanks for everyone's help, this place is great :)
 
ej_whyte said:
The / indicates a 'per' and indeed that is how it is said, however I am finding it strange that to get a voltage number you have to multiply the nV/√Hz by the √BW rather than divide it as the 'per' would indicate?
How many miles do you do in 8 hours at 10mph? 10 mph x 8h = 80 miles
How much noise do you get over a 20kHz BW for 10nV/sqrtHz? 10 nV/sqrtHz x 141 sqrtHz = 1410 nV or 1.41uV
 
Just one small point. Several times you refer to the RMS sum of several noise voltages. What you really mean to say (and do) is RSS (root of the sum of the squares). That might be why your addition of 3.46uV and 1.89uV came out wrong - it should be ~3.94uV or -105.8dBU.

I would be very surprised if your channel amp noise was as low as 107dBu except under the most favourable conditions. Under most normal operating conditions it is likely to be no better than -90dBu. In other words it is likely to be in excess of 20uV which just demonstrates how relatively unimportant is bus resistor noise.

Cheers

Ian
 

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