Balanced line

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jh6you said:
I have thought about the half signal more.
But, I am afraid that I can not understand it yet.
Even if the ground are not tied as you say, how the impedance-matching
cold line could carry the half signal...? I am really confused.

If the impedance is matched, the cold line would work for common mode rejection only.
This is all I could understand.

It's simply a matter of perspective and where you take your reference point from.

The signal exists across the two "hot" and "cold" lines. So let's say that you put your reference probe on the "cold" line and your measurement probe on the "hot" line. And let's say that the voltage you measure across those two lines is +1 volt.

Now, instead of placing your reference probe on the "cold" line, place it at the node between the two balancing resistors at the recevier end. Then if you place your measurement probe on the "hot" line, you'll measure +0.5 volts. And if you place your measurement probe on the "cold" line and you'll measure -0.5 volts.

It's simply a consequence of the voltage divider formed by the two balancing resistors is all.

Does this help?

se
 
Disabled Account
Joined 2002
Steve

I know how to measure the voltage.
That was not my question.

You are saying,

The signal exists across the two "hot" and "cold" lines.

Is it true with the simple impedance matching arrangement
you are referring to? This is my question.
 
Disabled Account
Joined 2002
Please read this.
http://www.jeffrowland.com/tectalk6.htm

“Rejecting common-mode noise does not require symmetrical
signal swings on the balanced lines. High-CMRR balanced interfaces
can have the signal on either line or symmetrically on both lines.
The presence or absence of a normal-mode signal has nothing to
do with rejection of common-mode noise. You need to consider
signal symmetry only in the context of cable shields or crosstalk.”
I would like to highlight "High-CMRR balanced interfaces
can have the signal on either line or symmetrically on both lines.
"
 
jh6you said:
Please read this.
I would like to highlight "High-CMRR balanced interfaces
can have the signal on either line or symmetrically on both lines.
"

If you tie together the reference grounds of the sending and receiving circuits of the passive balancing circuit I presented, then yes, from the perspective of the ground reference at the receiving circuit (which again is tied to the ground reference of the sending circuit), the full signal will be seen on the "hot" lead and nothing on the "cold" lead.

But if the grounds are not connected, then the signal exists between the "hot" and "cold" leads and from the perspective of the ground reference of the receiving circuit, you'll see + half the signal on the "hot" line and - half the signal on the "cold" lead.

And it was in the context of the reference grounds not being connected that I have been speaking and which you said:

<i>Even if the ground are not tied as you say, how the impedance-matching
cold line could carry the half signal...?</i>

se
 
jh6you said:
Steve

I have never heard about or seen any balanced interface
with two lines.

Well, any piece of pro gear with the ground lift switch set to "lift" or any balanced interface using a cable with the shield tied only to one end would be a balanced interface with only two lines. Or cables with no shielding at all.

The fact remains that there is no inherent requirement of tying together the reference grounds of the two circuits.

se
 
mrfeedback said:
Steve, thankyou and yes I know the principle of demagnetising.
More so I meant a suitable oscillator circuit with decaying output,
along the lines of for example the once made TDK tape head demagnetiser cassette.
A box with cable connectors so cable/transformer assemblies can be demagnetised is what I have in mind.

Ah. No, I don't recall having seen any purpose designed circuits out there like that. Wouldn't it be easier to just use something like CoolEdit to create the signal and just burn it onto a CD and use a CD player as the signal source?

se
 
Re: balanced

AMPMAN said:
Jason,go to www.dself.demon.co.uk/balanced.htm go to fig.15 there is a picture showing clearly how to solve your problem. why nobody else could draw a diagram of an rca plug to xlr beats me.:xeye:

How does that help? From what I read, he wants to convert the unbalanced output of his CD player to a balanced output so he can drive the balanced input on his Aleph P, not just make up an RCA to XLR adaptor cable. Such an adaptor cable doesn't make the unbalanced RCA output a balanced output. And his Aleph P already has RCA inputs, so what's the point?

se
 
steve

The point is while your drawings are fine for experienced people ,interpreting them can be perplexing to some. what is needed is a diagram showing how we can use our existing cd players with their rca plugs which are grounded. lets imagine a coil in a phono player, we take the leads straight to the inputs of a balanced preamp, its obvious that the current oscillating backwards and forwards will turn on the pos and then the neg inputs.The ground wire is then the shield,what is harder to grasp is when you ground one wire how can it turn on a transistor. the diagram I referred to shows how, I for one would like to see how its done in the p1.7 to recap most everybody has unbalanced equipment we want to be shown how to connect it to balanced in the simplest way possible, surely this not to much to ask.
 
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