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
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.
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
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.
I used to recommend circuit A
D.Self explains why circuit B is the better.
I now don't recommend circuit A.
BUT !!!!!!!
Pin1 goes to Chassis. Pin1 NEVER goes to Audio Ground.
Change the labels to audio ground.
D.Self explains why circuit B is the better.
I now don't recommend circuit A.
BUT !!!!!!!
Pin1 goes to Chassis. Pin1 NEVER goes to Audio Ground.
Change the labels to audio ground.
Am I correct to say that both circuit will work ?
(ie it will not blow up my pre or power amp ?)
For circuit B, do I need to half the R2 value ?
(ie it will not blow up my pre or power amp ?)
For circuit B, do I need to half the R2 value ?
Use circuit A as it's easier to implement. If you are not totally clued up on balanced lines using circuit B could result in ground loop problems.
circuit B is the correct way to do it. Provided Pin1 is NOT connected to Signal Return/Ground.
Self tells us why Circuit A is not correct.
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 ??
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 ??
Can you elaborate on why Circuit A is worse than Circuit B?
Circuit B requires tight matching of R2 for CMR.
The papers that discuss balanced impedance connections.
It came up again in a Thread very recently. I think the posts were relating to a Linear Audio article and a reprint in one of the electrical webzines.
It came up again in a Thread very recently. I think the posts were relating to a Linear Audio article and a reprint in one of the electrical webzines.
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).
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 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.
Am I correct to say that in circuit B, I'll need to half the R2 value like the case of R1 ?
Seem like I've problem getting this specific question answer :-(
Seem like I've problem getting this specific question answer :-(
Yes.
Let's say
A : R1 = 500 Ohm, R2 = 1000 Ohm,
B : R1 = 500 Ohm, R2 = 500 Ohm
output voltage is half of the input (differential) voltage.
In B, removing the connection to ground of the common node of the R2s makes the circuit equivalent to A.
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.
Another questions (hopefully the last b4 I start purchase these resistor for the circuit A & circuit B build).
Formula used - is this correct ?
Intend to purchase the following resistor from IRC/Welwyn) - http://www.farnell.com/datasheets/12254.pdf - are they good ??
Formula used - is this correct ?
Intend to purchase the following resistor from IRC/Welwyn) - http://www.farnell.com/datasheets/12254.pdf - are they good ??
An externally hosted image should be here but it was not working when we last tested it.
An externally hosted image should be here but it was not working when we last tested it.
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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