Passive Balanced Attenuator (between Pre & Power)

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Hi,

I need to attenuate the overall gain of my pre + power amp.
The simplest solution is of course to purchase a in-line attenuator like the Rothwell Audio "balanced in-line attenuators".
balanced in-line attenuators
smallbalatten2.jpg


But am think might as well DIY-ing it & combine them into the XLR interconnect itself.

Just to check from the experts here which circuits is correct (using the U-pad design).

Many thanks in advance.

An externally hosted image should be here but it was not working when we last tested it.
 
Thanks.

My power amp is DIY NC400. Pin 1 is connected to the Chassis which is also connected to the XLR socket casing (I think).

Audio ground is connected to XLR socket casing.

I think that will mean that Pin 1 is also connected indirectly to audio ground of the NC400 board, right ??
 
From all the posts here, seem like both can work (ie will not blow up my pre, power or speakers).

But circuit B is idea provided ........... audio ground no good, but only chassis ...................... this I'm confuse.
Also, if I can get this correct, do I need to half my R2 value (like in the case of R1) for circuit B ??

The resistors are cheap, if needed to, can always do 2 to compare any different. Most important is not to blow up anything.


My pre-amp is ARC DS-pre. (which I think their output impedance is 500 Ohms)
My power amp is DIY mono Hypex Ncore NC400 (which I think their input impedance is 104 K-Ohms).

An externally hosted image should be here but it was not working when we last tested it.
 
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It is not that circuit A is incorrect, it does its job but does not provide common mode rejection.

It is very easy to understand the difference between the two circuits.
Imagine there is an identical voltage (referenced to ground) induced by some parasitic sources on hot and cold lines of both circuits.
In B, this common mode voltage is equally attenuated in each line, the resistors networks being identical, and as much as the useful signal which is a differential mode voltage.
In A, the common mode voltage is not attenuated at all.
 
It is not that circuit A is incorrect, it does its job but does not provide common mode rejection.

It is very easy to understand the difference between the two circuits.
Imagine there is an identical voltage (referenced to ground) induced by some parasitic sources on hot and cold lines of both circuits.
In B, this common mode voltage is equally attenuated in each line, the resistors networks being identical, and as much as the useful signal which is a differential mode voltage.
In A, the common mode voltage is not attenuated at all.

So circuit B basically gives current created as a result of common mode voltage a path to ground?

Trying to wrap my head around this - thanks for the explanation.
 
It is not that circuit A is incorrect, it does its job but does not provide common mode rejection.

It is very easy to understand the difference between the two circuits.
Imagine there is an identical voltage (referenced to ground) induced by some parasitic sources on hot and cold lines of both circuits.
In B, this common mode voltage is equally attenuated in each line, the resistors networks being identical, and as much as the useful signal which is a differential mode voltage.
In A, the common mode voltage is not attenuated at all.

Hmm, I'm not sure that is correct. Circuit B would inject common mode noise in to the circuit via the R2 resistors connected to ground unless they are very precisely matched in value. Even if they were precisely matched, they would reduce the common mode impedance of the input stage because they would be in parallel with it. That would make the CMRR of the interface more sensitive to mismatch in the value of R1. In circuit A, the precision of R2 has no effect on interface balance, and so, cannot degrade the CMRR. The objective of a balanced interface is that common mode noise be made equal in both signal paths. That objective is different than for common mode rejection over an unbalanced circuit interface, which doesn't depend on the precise matching of R2, and which is where circuit B would be superior. However, perhaps, I'm missing something here.
 
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