then there's no reason to add any capacitors
Oh Contrair, as I stated earlier:
If you are going to use a transformer in the interface: The capacitors are important to prevent DC currents in transformer windings which will affect the linearity and sonics of many small transformers. I have had personal experience in this matter and was quite suprised how little DC current it took to effect the sonics. Why do I have the feeling another Eddytorial is looming in my near future?
Joe
Oh Contrair, as I stated earlier:
If you are going to use a transformer in the interface: The capacitors are important to prevent DC currents in transformer windings which will affect the linearity and sonics of many small transformers. I have had personal experience in this matter and was quite suprised how little DC current it took to effect the sonics. Why do I have the feeling another Eddytorial is looming in my near future?
Joe
Re: Do you think the capacitors are not important?
Obviously you didn't get around to AN-002. 🙂
"Active circuits having high <i>differential</i> input impedance (sometimes called "bridging" since several can be bridged across a line with minimal level loss) are widely used, but the importance of the <i><b>common-mode</i></b> input impedances of these circuits have long been ignored by most designers. Almost all such "electronically balanced" inputs have common-mode input impedances in the 5 kohm to 20 kohm range, which seriously degrades their performance as a balanced line receiver. The common-mode input impedances of an input transformer are <u>inherently</u> about 1000 times that of most "actively balanced" inputs, <i><b>giving the transformer about 60 dB better ground noise rejection</i></b> in the real world, where significant source impedance unbalances almost always exist."
All the italics, bold and underlining are from the original article, not any additions on my part.
But the only connection made between components by the interconnect is a chassis connection. That's because the whole purpose of pin 1 on an XLR connector is one of SHIELDING. Again, there is no inherent requirement of the signal grounds being tied together.
As I said previously, it can be advantageous when you have AC voltage differences between the two signal reference grounds such as those caused by AC-based power supplies. But there is no inherent requirement.
se
Joseph Knecht said:
Obviously you didn't get around to AN-002. 🙂
"Active circuits having high <i>differential</i> input impedance (sometimes called "bridging" since several can be bridged across a line with minimal level loss) are widely used, but the importance of the <i><b>common-mode</i></b> input impedances of these circuits have long been ignored by most designers. Almost all such "electronically balanced" inputs have common-mode input impedances in the 5 kohm to 20 kohm range, which seriously degrades their performance as a balanced line receiver. The common-mode input impedances of an input transformer are <u>inherently</u> about 1000 times that of most "actively balanced" inputs, <i><b>giving the transformer about 60 dB better ground noise rejection</i></b> in the real world, where significant source impedance unbalances almost always exist."
All the italics, bold and underlining are from the original article, not any additions on my part.
But the only connection made between components by the interconnect is a chassis connection. That's because the whole purpose of pin 1 on an XLR connector is one of SHIELDING. Again, there is no inherent requirement of the signal grounds being tied together.
As I said previously, it can be advantageous when you have AC voltage differences between the two signal reference grounds such as those caused by AC-based power supplies. But there is no inherent requirement.
se
Re: Nope....
Obligatory grain of salt and backpedaling?
The context of this thread is about someone who wants to convert the unbalanced output of his CD player to balanced so he can feed the <b>truly balanced input</b> of his preamp.
No one made any mention of its use as a phase inverter with a DPDT switch until you brought it up just now, which would have taken the discussion into a completely different context. And since no one made any such mention, I had no reason to believe that their comments were related to any other context than the context of this thread. Which again is the issue of converting an unbalanced output to a balanced output in order to feed a <b>truly balanced input</b>.
se
pmkap said:
Obligatory grain of salt and backpedaling?
The context of this thread is about someone who wants to convert the unbalanced output of his CD player to balanced so he can feed the <b>truly balanced input</b> of his preamp.
No one made any mention of its use as a phase inverter with a DPDT switch until you brought it up just now, which would have taken the discussion into a completely different context. And since no one made any such mention, I had no reason to believe that their comments were related to any other context than the context of this thread. Which again is the issue of converting an unbalanced output to a balanced output in order to feed a <b>truly balanced input</b>.
se
another Eddytorial
Yes I was right..... And it was a quite a show too. Such a strong proponent of transformer coupling should introduce a product to exploit the virtures of such an interface.
Whoops, it appears I am a little late with that suggestion....
http://www.q-audio.com/home/index.html
Yes I was right..... And it was a quite a show too. Such a strong proponent of transformer coupling should introduce a product to exploit the virtures of such an interface.
Whoops, it appears I am a little late with that suggestion....
http://www.q-audio.com/home/index.html
Attachments
Peter Daniel said:
Yes, same thing. You're just mirroring the output impedance of the driven output to give each line the same impedance to ground.
se
Re: then there's no reason to add any capacitors
IF you are going to use a transformer in the interface AND your output hasn't significant DC offset AND the input the trasnformer's feeding hasn't significant input bias current. Then you may want to consider capacitive coupling.
However DC output offset can be dealt with by means other than coupling capacitors. From a DC servo in the feedback loop to a simple resistor if you're using say a discrete, single-ended output stage. Also, if the input the transformer's driving has a significant amount of input bias current, such as with bipolar inputs, a bit of DC output offset on the primary can cancel its effects.
And a wee bit of DC current in the transformer isn't always such a bad thing. It can add a bit of "warmth" to an overly bright circuit by adding a bit of even-order distortion. In fact, you could use the output offset trim pot in a simple opamp output stage as something of a dial-in "warmth" control.
se
Joseph Knecht said:
IF you are going to use a transformer in the interface AND your output hasn't significant DC offset AND the input the trasnformer's feeding hasn't significant input bias current. Then you may want to consider capacitive coupling.
However DC output offset can be dealt with by means other than coupling capacitors. From a DC servo in the feedback loop to a simple resistor if you're using say a discrete, single-ended output stage. Also, if the input the transformer's driving has a significant amount of input bias current, such as with bipolar inputs, a bit of DC output offset on the primary can cancel its effects.
And a wee bit of DC current in the transformer isn't always such a bad thing. It can add a bit of "warmth" to an overly bright circuit by adding a bit of even-order distortion. In fact, you could use the output offset trim pot in a simple opamp output stage as something of a dial-in "warmth" control.
se
Well, I understand that the balanced impedance is the key issue
from you. I however do not understand the +/-0.5v signals yet.
Pls have a look at Peter Daniel's Fig.10 (Equivalent balanced circuit).
I can not figure it out how to get the divided signals.
Would you explain me once more about the signals?
from you. I however do not understand the +/-0.5v signals yet.
Pls have a look at Peter Daniel's Fig.10 (Equivalent balanced circuit).
I can not figure it out how to get the divided signals.
Would you explain me once more about the signals?
Jesen AN-003 explains the impedance balance and coresponding
CMRR only. But, nothing about the divided signals.
CMRR only. But, nothing about the divided signals.
jh6you said:
Sure.
You won't get identical bipolar voltages from the circuit in Fig. 10. That's because that circuit ties together the reference grounds of the source and receiving circuits. In that case, when you measure the voltage on the - line, you'll only be measuring the voltage drop across the RSOURCE resistor on the - line, which will be much smaller than 1/2 the voltage you'd measure directly across the + and - lines.
The perspective I'm coming from, which is that of there being no inherent requirement for the reference grounds of the source and receiving circuits to be tied together, is what I posted previously:
<center>
<img src="http://www.q-audio.com/images/balanced3.jpg">
</center>
Here the reference grounds of the source and receiving circuits are not tied together. And from the perspective of the receiver, referencing its ground reference, which is at the node between the two resistors (which are of identical value by the way), you'll measure the equal, bipolar voltages I was referring to, which will each be half the value of the voltage measured directly across the two lines.
This help?
se
jh6you said:
The signal, in a system where the reference grounds are not tied together, is divided by the voltage divider formed by the balanced pair of resistors at the receiving end.
se
If the reference grounds are not tied together, then we get each half signal.
But, if tied together, then unequal signals.
Steve, after all, we will tie them together between the source and the
receiving circuit in the real world. If I am right, then my understanding
is right--the idea is only for the impedance matching. What would be
your response to this?
But, if tied together, then unequal signals.
Steve, after all, we will tie them together between the source and the
receiving circuit in the real world. If I am right, then my understanding
is right--the idea is only for the impedance matching. What would be
your response to this?
jh6you said:
In the context of that particular passive solution, yes.
Yes, the primary goal is impedance balancing.
As for whether the reference grounds of the source and receive circuit are tied together in the real world, that depends on your particular realworld situation.
Not tying them together is typically only problematic in cases where you have problems with chassis leakage currents caused by the evils of AC power supplies. 🙂
<center>
<img src="http://www.q-audio.com/images/noac3.jpg">
</center>
se
I find all the info on the Jensen transformers here interesting. Many do not know it but all the original 2 drawer Cinema DTS digital playback processors all used Jensen transformers on the output. These particuluar DTS units also have much better sound quality as compared to the later 3 drawer capacitor couple output stage.
Mark Gulbrandsen
Mark Gulbrandsen
"And a wee bit of DC current in the transformer isn't always such a bad thing. It can add....... "
DC current can permanently magnetise the transformer core, causing permanent distortion, bandwidth limiting and maximum level limiting.
It is bad practice to allow DC through audio transformer windings.
Eric.
DC current can permanently magnetise the transformer core, causing permanent distortion, bandwidth limiting and maximum level limiting.
It is bad practice to allow DC through audio transformer windings.
Eric.
mrfeedback said:
Any residual magnetization isn't permanent as it can be trivially demagnetized.
It is if all one is concerned with are objective specs. Which is fine. It just doesn't happen to be my particular concern. My only concern is my personal enjoyment while listening to reproduced music. And if a little DC on my transformers ultimately results in greater enjoyment, so be it.
se
planet10 said:
Absolutely.
Or none at all if they happen to be of the parafeed or OTL variety. The equipment should serve the user. Some prefer their food as prepared. Others may prefer adding a bit of salt and pepper to taste. Vive la difference!
se
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.
🙂
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.
🙂
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