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In Reply to: RE: A Mercedes and a bus... posted by tube wrangler on March 13, 2015 at 20:33:06
"Unfortunately, a lot of music is...exactly like the CM items that...balanced....circuits are cancelling out."
You don't have an understanding of what "common" means.BTW Balanced and push pull are not the same thing.
Dennis, you are clearly very confused. I don't think I can help you.
I don't think anyone can say that I haven't tried over the years.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Edits: 03/14/15Follow Ups:
YMMV.
---Dennis---
No. There's no varying when it comes to the principals involved here.This is not subjective or just someones opinion.
You clearly don't understand Common Mode Rejection.
You also don't understand audio signal.
When a microphone is in the presents of sound there will be an output signal.
If the microphone is single ended the signal will be represented by a voltage potential difference between ground and the one signal wire. If the microphone is balanced the signal will be represented by a voltage potential difference between the two signal wires.
Common Mode Rejection is the rejection of any signal that is common to the two balanced wires.
That can never be the audio single generated by the microphone because, as I have already explained, a balanced microphone's audio output is represented by a voltage potential DIFFERENCE between the two signal wires.
CMR does not reject voltage potential differences between the two signal wires, CMR rejects only voltages that are the same in each of the two audio signal wires.
I don't think I could have made this more clear.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Edits: 03/14/15
Correct, and I have never disagreed with this-- as far as it goes.....That same microphone's two balanced output leads would normally be encased in a cable that has a shield around it.
That shield would be grounded-- in some cases, one of the two "hot" leads would also be grounded.
True balanced would have the two leads "hot". S.E. would ground one of the two "hot" leads to the shield in most cases..
In the case of balanced, with a shield NOT connected to either lead, Common-Mode rejection takes place between the two "hot" wires, and to the grounded shield.
In the S.E. example, where one lead is also grounded to the shield, there is no longer a Common-Mode. There's only one "hot" wire. The loss here is "hot lead" energy being absorbed into the shield, and into the grounded lead.
The energy loss in the two balanced "hot" leads with a shield example-- is to load the Common-Mode that is between the two "hot" wires-- into the shield. The Common-Mode is thus attenuated into the shield.
You have losses either way-- into the shield, etc. In the S.E. example, loss is broadband, but not totally-- that depends on what the wire's characteristics are.. In the balanced example, loss is both into broadband and into attenuating the Common-Mode between the two "hot" wires in the shielded cable/interconnect.
Some users prefer to wire the mike as S.E.-- that is, with one "hot" pin grounded with the shield also as ground. For short runs, this sounds more real.
Others don't have time for that and wire everything balanced. For longer runs, and for frequently moved/disturbed studios, this is the way to go.
You are, however, attenuating the Common-Mode mostly because your wire has a shield around BOTH "hot" leads..
What if it didn't have a shield? Just two wires-- balanced. If they're near each other, or twisted together, then the internally radiated Common Mode is also being attenuated by the cable-- even without a shield because the two "hot" leads are close enough to radiate into each other.
There. You have it.
---Dennis---
Edits: 03/14/15
"In the case of balanced, with a shield NOT connected to either lead, Common-Mode rejection takes place between the two "hot" wires, and to the grounded shield."No.
Ground is only connected to the shield to deflect the hum fields, that the cable might encounter, away from the two conductors.
CMR is between the two "Hot" wires only.
Any hum that gets through the shield is impressed on each of the two "Hot" wires equally and is therefore reject by the CMR of the input. (circuit or transformer)
"You have losses either way-- into the shield, etc. In the S.E. example, loss is broadband, but not totally-- that depends on what the wire's characteristics are..
In the balanced example, loss is both into broadband and into attenuating the Common-Mode between the two "hot" wires in the shielded cable/interconnect.You are, however, attenuating the Common-Mode mostly because your wire has a shield around BOTH "hot" leads..
What if it didn't have a shield? Just two wires-- balanced. If they're near each other, or twisted together, then the internally radiated Common Mode is also being attenuated by the cable-- even without a shield because the two "hot" leads are close enough to radiate into each other."
Dennis, you are mixing two different properties/functions.
The shield will interact with the two "Hot" wires to whatever degree (subject to discussion) but that's has nothing to do with the function of CMR.
All three wires are in close proximity so there will be capacitance and inductance between each of them but again Dennis, that's not CMR.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Edits: 03/14/15
Any signal that is common to both "hot" leads will be attenuated by any grounded shield that is around them.
This is so basic that no further explanation is needed, and none will be given.
---Dennis---
I hope you were trying to say "Any external hum signal that is trying to get to the two "Hot" wires will be attenuated by the grounded shield"Yes. That's why the shield is there.
But this is not CMR and has nothing to do with CMR.
Again, CMR is what a input signal transformer (that does not have either side of it's primary winding grounded) does.
Any external hum signal that gets through the grounded shield is impressed upon the two "Hot" wires.
That hum signal will be impressed upon the two "Hot" wires equally.
And because the transformer can only respond to the voltage potential differences between each end of the primary winding, the hum signal that came from an external source (a signal that has nothing to do with the output of the microphone and is therefore not music signal) is rejected.
That is CMR and a quick explanation of how and why it works.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Edits: 03/14/15
Yours is an excellent explanation of how major studios normally apply CMR.
One can also apply CMR to an operating pair of triodes in push-pull. Simply tie the two cathodes together into ONE common cathode resistor that goes to ground. Common-Mode Reduction will be applied directly, or indirectly to all of the signals present in the two tubes.
Another application is with two tetrodes (or pentodes) operating in push-pull.
To apply VERY effective CMR to the pair, connect a resistor of equal value in series with each G2. Then, tie the two resistors together at their outboard ends. Take this COMMON POINT, and apply G2 bias into it. You have just installed CMR into the push-pull pair of tetrodes, or pentodes.
CMR can be applied in numerous ways in many, many circuits. Another CMR application is to use a center-tapped choke on push-pull grids on a triode pair (or otherwise also).. The choke isn't necessary, you can also do it with resistors, or with a push-pull transformer.
I've been applying all of these and more CMR methods for many years now.
The more places you can find to apply CMR in a push-pull circuit, the better it will perform. It's only open to your imagination.
It only stands to reason that I would find ways to do it with wiring schemes also.
---Dennis---
"Any signal that is common to both "hot" leads will be attenuated by any grounded shield that is around them.
This is so basic that no further explanation is needed, and none will be given."
But the audio signal is the *difference* between the two "hot" leads; it is not "common" to them. By contrast, a stray signal like hum pickup will be common to the two hot leads, and will be balanced out.
The strange thing that you appeared to be saying was that an actual audio signal could be common to both "hot" leads. This, of course, is not true. Anything that is common to the two hot leads is, by definition, not part of the audio signal, and therefore it should, quite properly, be rejected.
Chris
I understand the theory and agree with it as far as it goes.
Here's the problem: ALL musical signals contain some artifacts that are identical to common (to both leads) signals in some way-- at some time in the musical cascade (for want of a better term admittedly). Difference signals are not the ONLY components in music.
Push-pull operation applies CMR at certain points in the circuitry (it should-- the more the better-- for push-pull). So does balanced signal systems such as studio wiring, etc.
I'm not arguing that this doesn't occur. What I am saying is that S.E. amplifier operation and also S.E. signal wiring cannot reduce the Common Mode.
In MUSIC, that is an ADVANTAGE. In cleaning-up signal, it is a DISADVANTAGE.
No one is arguing how this works. We all agree here. What I am saying is that artifacts of real music have components that are identical with Common-Mode distortion.
The question before us is-- do we wish to reproduce that common-mode signal or do we want it to be attenuated?
Your answer will put you into one of two camps-- S.E. amps or push-pull. S.E. wiring or balanced?
I think that, in practical terms, all of us are using combinations of both.
The item here is that I understand what each is doing musically.
---Dennis---
"Here's the problem: ALL musical signals contain some artifacts that are identical to common (to both leads) signals in some way"
No, music signal does not contain any signal that is common to both "hot" leads.
By definition, music signal is the difference between the two wires (or the one wire and ground in a SE circuit).
There is no signal common to both wires created by the source (microphone/phono cartridge).
Signal common to both wires is, and can only be, noise generated from an outside source that has nothing to do with the music signal generated at the signal source.
Every time you assert the idea that there is signal that is common to both "Hot" wires, in a music signal, you show your ignorance.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Look-- the difference signal is most of what's left after common-mode attenuation has been applied.
It IS NOT----- ALL of the generated signal.
Your microphone-- let's assume it's balanced, is transmitting accurate total information from it's two equal but opposite outputs.. These two outputs truly are equal but opposite. BOTH contain both Common_Mode and difference signal information.
Some would argue that the mike element itself cancels any Common_mode across it. Theoretically, yes, actually it's only partial cancellation because of the mike's internal generator resistance.
However, when you apply a twisted wire pair (your mike's output cable) to the balanced signal, you have now installed an attenuator-- a Common_Mode attenuator.
Anything that the mike delivered that was common to both output leads is now attenuated-- so that mostly, only the difference signal remains..
We can argue all day about who prefers to listen to this sanitized mostly difference-only signal, or who prefers to have his mike cable not attenuate what Common-Mode existed inside the mike-- before the output wiring was applied.., but we cannot deny that a real difference exists.
There's more than just a difference signal that the mike is generating.
In practice, push-pull amps and balanced circuitry is clean because it's been cleaned-up-- mostly all that's left is the difference signal.
S.E. stuff sounds FAR more "real".
It does this because nothing that occurred naturally in the Common_mode is filtered out, sadly leaving only a difference signal to listen to.
A truly balanced difference signal is the Gold Standard for both balanced and push-pull circuitry. The less of anything else it contains, the cleaner and "better" it is.
The problem is that it WILL NOT convey a sense of reality to the listener. It is fully incapable of it.
That leaves S.E.-- distortions, hum and all the rest-- if you want music to sound real.
---Dennis---
" That leaves S.E.-- distortions, hum and all the rest-- if you want music to sound real."
Total, unadulterated, SE-amp-marketer, bull cookies.
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
Dennis, I didn't even read your whole post.The microphone is not capable of generating common mode signal.
If the two "hot" wires are at the same potential, one vs. the other, you have silence. No signal.
This would happen when there is no signal and at zero crossing. In other words, when there is no voltage potential between the two "Hot" wires it's because, in that moment, there is no signal, nada, nothing.
I think it's OK to cancel nothing.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Edits: 03/15/15 03/15/15
> Every time you assert the idea that there is signal that is common
> to both "Hot" wires, in a music signal, you show your ignorance.
Incompetence, to be precise.
Like other insanely dumb "idea" to bias cable & interconnects with hum to make them respond quickly to musical transients.
I am not sure the exact source of these claims, but consider the marketing required for the amps he sells at the price he commands. There has got to be some stated reason to go and buy his v. something else. Examine the target market. Is it comprised of engineers capable of design and building their own vision? I don't think so.
Now the entertaining question is( IMO), does dennis actually believe the stuff he is saying?
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
> Now the entertaining question is( IMO), does Dennis actually
> believe the stuff he is saying?
1000% he does.
He is not looking at it as a system, rather he is focusing on the cable. Signal we want is 180 degrees out of phase and one leg is inverted by circuitry and becomes additive. Anything we dont want will be common mode , in phase, and when the one leg is inverted by said circuitry , it will cancel. This inherent operation will take care of everything we can possibly hear and the shield/foil thickness will take care of stuff cats couldnt even think about hearing. Much more explanation will be necessary. None will be heeded.
"Signal we want is 180 degrees out of phase and one leg is inverted by circuitry..."
It can be done with active circuitry but many sources are inherently balanced.
A dynamic microphone is like a speaker used in reverse.
A coil of wire sitting in a magnetic field being moved by a diaphragm by sound pressure.
When the wire is moved, signal voltage is generated because magnetic flux lines are being crossed.
That coil wire has two end, neither end has to be grounded.
A moving coil phono cartridge (or a moving magnet phono cartridge for that matter) is an inherently balanced signal source.
When used with a step up transformer there's no need to ground either end of the coil or either end of the primary winding of the step up transformer.
When we hook it up this way using one "2 conductor shielded cable" for each channel, (looking a one channel)
we get CMR for any noise picked up in the tone arm cable.
Many studio units (micpres, EQs, compressors, mixing consoles, etc) are balanced in and balanced out using input and output transformers.
The circuitry in between is single ended but the primary winding of the input transformer and the secondary of the output transformer are inherently balanced as long as you don't ground either end of the windings.
The secondary of the output transformer on your tubed power amp is inherently balanced and so are your speakers. The only reason to ground them (if your not using global negative feedback) is safety. BTW, I don't ground mine.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
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