jh6you said:
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
mrfeedback said:
You don't need any particular circuit per se.
Just take an AC-coupled signal source and use it to drive the transformer with a full-swing, low frequency sinusoid, and then slowly decrease the level of the sinusoid until you get down to zero.
Voila.
se
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.
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.
No. 🙂
Your balanced interface explains 50-50 common noise signal
through the hot and cold line. But, signal is only through the hot line.
Your balanced interface explains 50-50 common noise signal
through the hot and cold line. But, signal is only through the hot line.
Please read this.
can have the signal on either line or symmetrically on both lines."
I would like to highlight "High-CMRR balanced interfaces
can have the signal on either line or symmetrically on both lines."
jh6you said:
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
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.
Eric.
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.
Eric.
jh6you said:
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:
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
Nelson Pass said:
You call <b>this</b> all hell breaking loose? You must live a very sheltered life. This is a geriatric ward on Thorazine compared to what goes on elsewhere. 🙂
se
Thanks Steve, that is a fair enough suggestion, however is not greater initial amplitude than line level 0Vu required ?.
Eric.
Eric.
mrfeedback said:
Why? It wasn't magnetized by the signal, but rather the DC offset voltage. Which typically is no more than a few tens of millivolts. 0dBFS for a Red Book CD player is 2 volts RMS which is well more than enough.
se
balanced
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.
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.
Re: balanced
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
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.
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.
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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