Attenuating output TX with a strapped pot- balanced?

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boji

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Jan 6, 2010
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Location
Maryland, USA
Would anyone mind confirming that the below remains balanced?

HWWcsEJ.jpg


The reason I am thrown into doubt is from an old post from Ian:
https://groupdiy.com/index.php?topic=70339.msg899621#msg899621
"With the 1K pot method the load impedance is fairly constant but the source impedance varies with the pot position with a maximum when the pot is close to half resistance. It is also unbalanced. With a 1K pot, the source impedance will be a lot lower than a true 600 ohm source so there is no need to worry about HF losses unless you  have very long cables..."

However I moved on his latest advice when implementing:
https://groupdiy.com/index.php?topic=31472.msg783462#msg783462
"I think that is where you are misunderstanding. You can do a pot on the secondary of the transformer without unbalancing it. It is what in the past I have called the 'Neve trick'."

Edit: Apologies Ian, if it sounds like I'm putting you on the spot! Just want to be extra sure before I let go of doing a bridged-t pad.
 
I believe it is actually because it's an isolated network with only two terminals so whatever current is sourced from one leg must be supplied by the other and visa versa. But as explained in your citation, the problem is that the impedance will vary. And 1K might load the output stage a little more than desired. Especially if the downstream load is 600 ohms like an older 1176 or an EQP1.
 
Balanced basically means the total impedance of the the hot leg is the same as the total impedance of the cold leg so that an interference current flowing in hot and cold induces equal an opposite voltages which get cancelled by the receiving input stage..  This usually means both the source and load impedances in each leg should be the same. If they are not, the effect is to worsen the common mode rejection ratio (CMRR). However, the common mode input impedance of a well designed balanced input will be very high so that small changes in the driving source impedance n each leg have only a small effect on CMRR

Clearly in the 'Neve trick' circuit does not have identical hot and cold source impedances but the errors are small so for many circumstances there will still be a reasonable CMRR. If I rememeber correctly it was only used for monitor outputs at Neve.

If you want a truly balanced attenuator there are a couple of ways you can do it, one using a dual pot an another using a single pot (see attached schematic).

Balanced-Attenuator.png


Cheers

Ian

 
ruffrecords said:
Clearly in the 'Neve trick' circuit does not have identical hot and cold source impedances
To clarify, it is only the source impedances due to transformer stay capactitance which become unbalanced in this situation, but they are unlikely to matter. The resistances do remain balanced, or rather, there is only one source resistance since it's a floating transformer winding.

It's a different story if your signal comes from a couple of opamps ('electrically balanced') rather than a transformer. With opamps you really do need to use one of the circuits in Ian's post.
 
> remains balanced?

Balanced to what??

Ground? There is no ground shown on that drawing. (But what is the stray arrow top-right?)

That's a drawing. In a Real World there are capacitances everywhere. In a simple winding they are already unbalanced to core (and chassis common) so you are never "Balanced" at the top of the audio band. A good winding will have balanced capacitances. Put an off-center load on it, now you are poorly balanced.

Does it matter? For 600 Ohms, probably not. 200pFd is small re: 600r up to a MHz. a 1K pot probably does not hurt unless you have to pass a Government Test. At 10K it may matter.

So what if it is "unbalanced"? You are surely going to a differential input? If it is 1/3rd on one side and 2/3 on the other, the diff-input will extract the signal.

Anyway: TRY IT! Run the line from that pot around your arc-welder, SCR dimmers, etc, to a proper diff-input. Diddle the send pot and make-up the gain at the destination. Does frequency response change? Does noise/buzz/hash change? (You can always pad-down and gain-up and find crap; I mean at reasonable levels.)
 
You could also make two levels of "proper" balanced attenuation with a center off DPDT (in addition to no attenuation). And the switch positions would be in natrual order (like out, -10 dB, -20 dB). Just use one pole to bypass the series resistors and the other pole to add a parallel cross-shunt. Center pos is only one cross-shunt. The load impedance would vary however.
 
Wow, thank you everyone.  Very helpful!

If you want a truly balanced attenuator there are a couple of ways you can do it
Thanks for the suggestion! I may very well go with the "low cost" version or stick with the original.

Balanced to what??
In my case, usually 500 slot devices off the group acn very close by, or occasionally feeding an insert to gear like square mentioned.  If you think 600ohm devices may be as picky as Square warned, I can cave and buy the expensive 600 ohm t-pad pots.

TRY IT! Run the line from that pot around your arc-welder.
Yessr.  I have a pultec clone I should test it on. I just want to keep drafting out the group card.  ::)
 

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