| JasonL |
Well i was wondering if there is a circuit out there that i can build or use for my problem here..
i have a aleph p1.7 preamp that has rca jacks or balanced i also have a aleph 5 balanced amp that im building and want to take advantage of the balanced fetures. now my cd player is rca is there a circuit that will convert rca to balanced ( xlr ).. i have a schematic from my dad that uses a cap to make balanced to rca but i dont know if that is backwards compatable. is it .. ? |
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| Bob2 |
Jason
look at ESP project 51
Bob2 |
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| JasonL |
| id love to but i have not installed all my user files onto my new apple i book what is teh esp webpage address.. ? |
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| skaara |
| Doesnt aleph p accept rca inputs too and provide balanced output?! |
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| Steve Eddy |
| quote: | Originally posted by JasonL
Well i was wondering if there is a circuit out there that i can build or use for my problem here..
i have a aleph p1.7 preamp that has rca jacks or balanced i also have a aleph 5 balanced amp that im building and want to take advantage of the balanced fetures. now my cd player is rca is there a circuit that will convert rca to balanced ( xlr ).. i have a schematic from my dad that uses a cap to make balanced to rca but i dont know if that is backwards compatable. is it .. ? |
Something you might want to consider is the <a href="http://www.jensentransformers.com">Jensen</a> DM2-2RX IsoMax unit. At $200 it's not exactly cheap, but one of the best ways I've found to convert from balanced to unbalanced and vice versa is with a transformer.
You can also buy the raw transformers (either the JT-11-DMCF or the JT-11-DMPC depending whether you want flying leads or PC mount) and do it yourself. The transformers are about $80 each so you don't save a whole lot but it gives you more freedom with the design.
se |
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| Nelson Pass |
The Aleph P is perfectly happy to do this conversion
for you without loss of performance. |
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| Steve Eddy |
| quote: | Originally posted by Nelson Pass
The Aleph P is perfectly happy to do this conversion
for you without loss of performance. |
How will his Aleph P convert the unbalanced output of his CD player into a balanced output?
The way I read his post, he has a CD player with unbalanced outputs and he wants to convert those to balanced so he can feed the Aleph P's balanced inputs.
se |
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| Peter Daniel |
If your CD player doesn't have a true balance output, I think that additional circuit can only degrade the sound and benefit of going from CDP to preamp balanced would be diminished.
So you can either use quality transformer for conversion (if you really want to go full balanced from CDP to Pre) or use Aleph P to do it for you;). I would say this is the main advantage of using preamp in that place, because in addition controlling the volume it does the conversion. Unless you want to build Aleph P inside your CD player and somehow place volume knob on a front panel.;) |
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| GRollins |
The Aleph P is essentially a differential circuit, albeit with two curent sources and a bit of resistance between the sources. Differentials make fine phase splitters as long as there is enough resistance under the cathode/source/emitter.
Grey |
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| eLarson |
The Aleph P ought to be fantastic at converting unbalanced to balanced. If you are interested in building something new, the Bride of Son of Zen with active current sources at the Source pins of the gain MOSFETs also does the job nicely.
I'd just stick with feeding the input with the unbalanced output of the compact disc player.
Erik |
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| Bakmeel |
Jason,
The Aleph Ono uses a cirquit to invert the output signal, so it can be used for Balanced purposes. Couldn't that work for your CD as well?
Bouke |
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| Steve Eddy |
| quote: | Originally posted by eLarson
I'd just stick with feeding the input with the unbalanced output of the compact disc player. |
But he doesn't get the advantage of a balanced interface between his CD player and preamp, which apparently is his goal here and where arguably, a balanced interface would be the most beneficial.
se |
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| Steve Eddy |
| quote: | Originally posted by Peter Daniel
If your CD player doesn't have a true balance output, I think that additional circuit can only degrade the sound and benefit of going from CDP to preamp balanced would be diminished. |
You can achieve a true balanced output from the CD player without additional active circuitry or even a transformer. Just need to mirror the output network of the existing circuit, creating in effect a balanced voltage divider.
se |
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| Peter Daniel |
Steve,
Nothing comes free.;) Do you have an example of such circuit? I've seen the Ono balanced converter, but it looks rather complicated. |
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| Steve Eddy |
| quote: | Originally posted by Peter Daniel
Nothing comes free.;) Do you have an example of such circuit? I've seen the Ono balanced converter, but it looks rather complicated. |
I didn't mean to suggest anything came free. Only that it needn't be terribly complicated. My personal preference is to use transformers (though I just use them at the inputs as they give better common-mode rejection from single-ended sources than active balanced inputs give from balanced sources).
As for a circuit example:
<center>
<img src="http://www.q-audio.com/images/drivera.jpg">
</center>
Let's say that describes the existing output circuit.
All you have to do is mirror it, thus:
<center>
<img src="http://www.q-audio.com/images/driverb.jpg">
</center>
If the original circuit has no shut resistor, just use a pair of 100k resistors. Also, if it has an output coupling capacitor, mirror that as well.
Voila. A true balanced output.
se |
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| Peter Daniel |
| But there must be a price here to pay, otherwise everybody would be using that circuit, and to tell the truth I never seen it in use. But maybe I was looking at the wrong schematics?;) |
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| Nelson Pass |
Actually it's a fine circuit. The - line may not be driven,
but it does experience the same noise pickup as the
+ line, and will give good rejection going into a balanced
input.
I think you don't see it often because using it requires
understanding of the nature of balanced lines, and also
because the public has been educated to expect both lines
to be actively driven. |
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| Steve Eddy |
| quote: | Originally posted by Peter Daniel
But there must be a price here to pay, otherwise everybody would be using that circuit, and to tell the truth I never seen it in use. But maybe I was looking at the wrong schematics?;) |
I think the reason you don't see such a circuit very often is because it seems most people don't quite understand just what constitutes a balanced interface.
Most are of the belief that it means transmitting two signals of opposite polarity. So you typically see designs using some sort of active inverting circuit in conjunction with a non-inverting circuit, such as:
<center>
<img src="http://www.q-audio.com/images/driverc.jpg">
</center>
However all balanced means is that the IMPEDANCE of each line with respect to ground is the same. That's it. And when you take an unbalanced output and create the voltage divider as noted in the previous post, you've got a true balanced output. If there's any price to pay, it's that you don't get the additional 6dB of voltage gain that you get if you use a second active inverting driver outputting a signal of the same magnitude as the non-inverting side.
I've seen designs that maintain separate non-inverted and inverted signal paths all the way from the input to the output. And while such a design is "balanced" you don't get any benefit from it.
The benefit comes from feeding the balanced line into a balanced DIFFERENTIAL input so that any common-mode noise on the line can be rejected. If you maintain separate signal paths from input to output, any common-mode noise on the line will simply be amplified and passed on down the line.
se |
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| Peter Daniel |
| Another audio mystery revealed.;) I will try this circuit in my next project. Thanks Steve for sharing it with us. |
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| Steve Eddy |
| quote: | Originally posted by Nelson Pass
Actually it's a fine circuit. The - line may not be driven,
but it does experience the same noise pickup as the
+ line, and will give good rejection going into a balanced
input. |
Well, the - line isn't SEPARATELY driven as it would be with an active solution. But the - line has the same signal on it as it would if it were, just that the signal level would be half that of the active solution (providing the active solution is separately driving both lines at the same level).
So for example if you had an unbalanced output driving the line with say, 1 volt, and you converted to balanced by driving the second line with the same signal only inverted, you'd measure +1 volt on the + line and -1 volt on the - line.
If you converted to a balanced output using the passive solution, you'd measure +0.5 volts on the + line and -0.5 volts on the - line.
And not to be too much of a stickler here, it's not the line that rejects the noise. That's done by the differential input. What balancing does is help keep common-mode noise common.
If an interference field impinges upon each line identically (and techniques such as twisted pairs helps assure that they do impinge identically), if the impedance of each line with respect to ground is the same, then the voltage induced across each line will be the same in terms of magnitude and polarity and subtracted out at the differential input.
If the impedances are not the same, then the voltage induced across each line will be different, and the differential input will amplify the difference between the two.
| quote: | I think you don't see it often because using it requires
understanding of the nature of balanced lines, and also
because the public has been educated to expect both lines
to be actively driven. |
Yup. Though I wouldn't call that "educated." Mislead would be more like it. And unfortunately in this industry of ours, it happens all too often and is looked upon (sadly quite rightly in many respects) as a technological backwater.
But hey, rightly or wrongly, it's still a lot of fun. :)
se |
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| Steve Eddy |
| quote: | Originally posted by Peter Daniel
Another audio mystery revealed.;) I will try this circuit in my next project. Thanks Steve for sharing it with us. |
My pleasure. Hope it works out for you (and any others)!
se |
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| dice45 |
Steve,
thanks for sharing! :) |
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| Steve Eddy |
| quote: | Originally posted by dice45
Steve,
thanks for sharing! :) |
Oh, I didn't mean to share it with you. MEANIE! :grumpy:
Awwwww, I didn't really mean it. You can use it too. I love ya, ya big lug. :hug:
se |
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| Bob2 |
Steve
Being of limited knowledge when it comes to this stuff, what would the circuit look like to convert an amp to a balanced input using your balanced output.
Bob2 |
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| Nelson Pass |
| quote: | Originally posted by Steve Eddy
If you converted to a balanced output using the passive solution, you'd measure +0.5 volts on the + line and -0.5 volts on the - line.
|
Sorry, I don't see any mechanism on your passive schematic
that would create this effect. The - line does not carry signal,
only noise. |
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| Peter Daniel |
| quote: | Originally posted by Nelson Pass
The - line does not carry signal,
only noise. |
So I new there was a price to pay after all.;) Of course it can't carry a signal since the signal is going through the ground.
So how does it really compare in reality, passive versus active retriving of a balance line? |
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| eLarson |
Why add more circuitry to convert unbalanced to balanced when the preamp in question already does this? If you feed the balanced output of the preamp to the balanced input of a power amp, you are good to go. :)
You could go to http://www.jensen-transformers.com/ln_in.html and get one of those if you absolutely must have the unbalanced to balanced conversion prior to your preamp.
But for the record, I wouldn't bother. ;)
Erik |
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| Steve Eddy |
| quote: | Originally posted by Bob2
Being of limited knowledge when it comes to this stuff, what would the circuit look like to convert an amp to a balanced input using your balanced output. |
Unfortunately, at an input, it's not such a simple matter as it is with an output.
At the input, it's not just a matter of balancing the impedances of the two lines, but also of providing a differential input so that any common-mode noise on the lines can be rejected (which is the raison d'etre of balanced interfaces in the first place) and depending on the existing input circuit could involve much more significant changes in the circuit itself rather than just tacking on a few parts.
So without a specific circuit to work from, there really is no generic answer as with outputs.
se |
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| JasonL |
im going to try the esp projects. and see how they work.. is there a sound difrence between
balanced line in xlr compaired to rca in .. will i hear the difrence ..? |
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| Steve Eddy |
| quote: | Originally posted by Nelson Pass
Sorry, I don't see any mechanism on your passive schematic
that would create this effect. The - line does not carry signal,
only noise. |
Well no, you wouldn't see anything on the - line if you analyze it simply as drawn. That's because as drawn, it's not terminated. There's no load so the - leg is an open circuit with respect to ground.
Tie a load resistor across the two output terminals and then let me know what you see.
se |
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| Steve Eddy |
| quote: | Originally posted by Peter Daniel
So I new there was a price to pay after all.;) Of course it can't carry a signal since the signal is going through the ground. |
Until you get around to plugging it into the component that it's going to be feeding. Until then, it's not going to be of much use anyway. :)
| quote: | | So how does it really compare in reality, passive versus active retriving of a balance line? |
Basically, on the plus side, you're not contributing additional non-linearities due to active circuitry.
On the minus side, you don't get the 6dB of voltage gain you can with an active circuit so your signal to noise ratio will be a wee bit higher with the passive solution.
In other words, the common-mode voltages due to induced noise on the line will be of the same magnitude regardless of whether you go the active or passive route, so the higher the absolute signal level, the better the signal to noise ratio.
se |
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| eLarson |
| quote: | Originally posted by JasonL
im going to try the esp projects. and see how they work.. is there a sound difrence between
balanced line in xlr compaired to rca in .. will i hear the difrence ..? |
You may well hear a difference. The question is, will it be an improvement? (Then again, you may not. Or you may think you hear a vast improvement simply because you built the thing. I can't be the only one who's fallen for that.) ;)
I can't see how adding more circuitry between the CD player and the Aleph P 1.7 would do anything beneficial. It would add a bit more distortion to the CD player's output than you had before.
I'd rest comfortably knowing that the Aleph P can do a wonderful job converting a single-ended input to a balanced output without adding anything else to what is a neatly simple circuit.
Erik |
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| JasonL |
| that is a verry true fact.. i think ill stick to a rca input and have all my amps to balanced.. and when i can afford a nice marantz balanced cd player then ill buy one.. : O ) |
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| Steve Eddy |
| quote: | Originally posted by eLarson
I'd rest comfortably knowing that the Aleph P can do a wonderful job converting a single-ended input to a balanced output without adding anything else to what is a neatly simple circuit. |
But that arrangement has no benefit whatsoever with regard to the line between the CD player and the Aleph P in terms of rejecting any noise that may be induced into that line.
Running unbalanced from the CD player to the Aleph P, the induced noise will be differential, rather than common-mode which means the Aleph P will simply amplify it and pass it on down the line.
And since the noise is differential, the fact that the Aleph P's output is balanced is irrelevant. It will feed the noise to the amplifier as a differential signal as well which means it will not be rejected by the amplifier's differential input.
se |
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| JasonL |
| so i guess your saying that i will hear alot of noise..? even if i have good hi quality shelded rca cables. |
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| Peter Daniel |
| quote: | Originally posted by Steve Eddy
But that arrangement has no benefit whatsoever with regard to the line between the CD player and the Aleph P in terms of rejecting any noise that may be induced into that line.
Running unbalanced from the CD player to the Aleph P, the induced noise will be differential, rather than common-mode which means the Aleph P will simply amplify it and pass it on down the line.
And since the noise is differential, the fact that the Aleph P's output is balanced is irrelevant. It will feed the noise to the amplifier as a differential signal as well which means it will not be rejected by the amplifier's differential input.
se |
Why don't we consider CDP and Aleph P as one unit. A player with Aleph as a buffer and the line between them very short (in my case 10"). So maybe the chances of noise pickup are very slim. I've seen this length of cables inside the players and preamps alone.;)
And one could assume that the line from preamp to the amps is more noise sensitive (in my case is more like 10 ft long). So there is still advantage in having balanced outputs on the preamp in terms of rejecting any noise that may be induced into that line.;) |
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| JasonL |
| speaking of mail packages i still havent recieved mine : O ( PETERRRRRRR... |
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| Steve Eddy |
| quote: | Originally posted by JasonL
so i guess your saying that i will hear alot of noise..? even if i have good hi quality shelded rca cables. |
No, that's not quite what I was saying. What I was addressing was the notion that simply because the Aleph P has a balanced output that this will amoeliorate the fact that the interface between CD player and the Aleph P is unbalanced as seemed to be implied by the post I was responding to.
How much if any noise you may ultimately hear depends on any number of factors. I'm simply saying that all else being equal, balanced interfaces will result in less noise than unbalanced interfaces, irrespective of whether or not there's any actual audible noise to begin with.
In other words, I'm not insisting that any interface must or even should necessarily be balanced in order to provide a satisfying result. Only the end user can decide that for themself.
se |
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| Steve Eddy |
| quote: | Originally posted by Peter Daniel
Why don't we consider CDP and Aleph P as one unit. A player with Aleph as a buffer and the line between them very short (in my case 10"). So maybe the chances of noise pickup are very slim. I've seen this length of cables inside the players and preamps alone.;)
And one could assume that the line from preamp to the amps is more noise sensitive (in my case is more like 10 ft long). So there is still advantage in having balanced outputs on the preamp in terms of rejecting any noise that may be induced into that line.;) |
That's fine.
As I explained to Jason, I'm not insisting that a balanced interface is the only way to go. Ultimately I'm just trying to explain what it is that they are, and what it isn't that they aren't. :)
se |
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| Nelson Pass |
| quote: | Originally posted by Steve Eddy
Well no, you wouldn't see anything on the - line if you analyze it simply as drawn. That's because as drawn, it's not terminated. There's no load so the - leg is an open circuit with respect to ground.
Tie a load resistor across the two output terminals and then let me know what you see.
se |
Only if you terminate it in a transformer. |
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| Steve Eddy |
| quote: | Originally posted by Nelson Pass
Only if you terminate it in a transformer. |
How would terminating it in a transformer be any different than terminating it with a resistor?
se |
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| Val Lewton |
"How would terminating it in a transformer be any different than terminating it with a resistor?"
You are just kidding right? A transformer breaks the galvanic current (noise and signal return) conducted through the ground lead. I have reread this thread several times and I fear that it has created more confusion than edification (well maybe lots of eddyfication...) on the subject of balanced interfaces. There are factors that have not even been touched upon concerning signal currents on the ground lead and what actually causes common mode noise to begin with. Am afraid that the interface under discussion is not the only thing unbalanced around here. Do a web search or peruse a good engineering text book on your next trip to the bookstore for the real lowdown on balanced interfaces.
Val |
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| Peter Daniel |
| Maybe a trip to the bookstore is not such a bad idea. I might get the "Steppenwolf" book as well.;) |
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| AMPMAN |
| so I take a wire from the centre pin of the rca jack to the + in onmy p 1,7 I then disconnect the outer case from ground and go from the case to the - in on the preamp is this correct |
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| Steve Eddy |
| quote: | Originally posted by Val Lewton
"How would terminating it in a transformer be any different than terminating it with a resistor?"
You are just kidding right? A transformer breaks the galvanic current (noise and signal return) conducted through the ground lead. |
Great. Now answer the question in the context it was asked.
| quote: | | I have reread this thread several times and I fear that it has created more confusion than edification (well maybe lots of eddyfication...) on the subject of balanced interfaces. |
If you read it several times and still could not catch the context of my question above, then there's something more than any "eddyfication" going on here.
| quote: | | There are factors that have not even been touched upon concerning signal currents on the ground lead and what actually causes common mode noise to begin with. |
Signal currents on the ground lead? What are you talking about? There is no ground lead. This is a balanced interface we're talking about here. Not an unbalanced interface.
<center>
<img src="http://www.q-audio.com/images/balanced2.jpg">
</center>
| quote: | | Am afraid that the interface under discussion is not the only thing unbalanced around here. Do a web search or peruse a good engineering text book on your next trip to the bookstore for the real lowdown on balanced interfaces. |
I'm at a loss as to why those who opposed your being banned did so claiming that you brought knowledge to these forums. Is this an example of the knowledge you bring here? "Go read a book"?
se |
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| eLarson |
| quote: | Originally posted by Steve Eddy
Running unbalanced from the CD player to the Aleph P, the induced noise will be differential, rather than common-mode which means the Aleph P will simply amplify it and pass it on down the line.
se |
Why wouldn't the ground conductor carry the same interference/noise as the signal conductor? If the ground conductor of the unbalanced interconnect is attached to the negative input of the amp, I think you are still in business when the signals are summed.
The P1.7 schematic from the service manual only shows XLR connectors, so I can't really say what the connection would be.
JasonL: Why not just buy a balanced output CD player and make the whole discussion moot? :)
Erik |
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| Steve Eddy |
| quote: | Originally posted by JoeKnect
Sorry I guess I am a little confused about the context too. Anyone else? An active balanced circuit (as opposed to transformer) requires a ground reference. |
The output circuit and the input circuit each reference their own local grounds. There is no requirement that those two reference grounds be tied together.
The only reason you'd tie them together is as a band-aid to help amoeliorate problems caused by AC powered power supplies. Not for any requirement of the balanced interface. Personally I prefer to just get rid of the problem in the first place and leave the band-aids to Johnson & Johnson. But that's another story.
| quote: | | Most of the audio balanced interfaces have used have a plus, minus and ground conection using XLR connections. The ground connection is often used as a sheild also. |
It's intended for a shield and is tied to the equipment chassis, not to fulfill any grounding requirement.
Have you had any experience with balanced professional audio gear? Much of it includes what's called a ground lift switch which allows the signal reference ground of the piece of gear to be disconnected from its chassis which is connected to the shield on the XLR.
| quote: | | Chapter 6 shows a balanced transformerless audio circuit simular to your last diagram WITH A GROUND REFERENCE CONNECTION for the input. Check it out. |
Fine. However that doesn't change the fact that such a connection isn't any inherent requirement of a balanced interface.
se |
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| Steve Eddy |
| quote: | Originally posted by eLarson
Why wouldn't the ground conductor carry the same interference/noise as the signal conductor? If the ground conductor of the unbalanced interconnect is attached to the negative input of the amp, I think you are still in business when the signals are summed. |
They're not summed, they're differentiated.
And unless each line has the same impedance with respect to ground, then the voltage across it will not be the same and when the two lines are differentiated, the difference of the voltages across each line will not be rejected but amplified.
se |
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| pmkap |
I just don't get it. I'll accept your statement that your passive implementation of your 'balanced' output circut will present a balaced impedence to a balanaced input circut, and externally imposed noise will indeed be common mode.... but your contention that one will have a balaced signal eludes me.
As I see it from my naive perspective, we have the positive signal, the output from the opamp passing through a voltage divider going to ground. We also have a mirror image of that voltage divider, originating at the ground leg of that op amp circut, going trough a series resistor (same value as the series resitor from the + opamp output), and shunted ( with the same value as its mirror shunt) to that 'signal' ground, i.e. we've got a voltage divider starting at ground and ending at ground.... so how the heck does a voltage divider starting at ground and ending at ground, have anything but ground at the midpoint of a 2 resistor voltage divider. I will admit its been 35+ years since I leaned Kirshoff's Law, but has it changed in the meantime? I'm not disputing that this termination does provide benefits in interfacing to a true balanced termination with regards to noise rejection (common mode rejection requiring equal impedences), but you previous contention that this will provide a balanced signal still eludes me. |
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| Steve Eddy |
| quote: | Originally posted by pmkap
I just don't get it. I'll accept your statement that your passive implementation of your 'balanced' output circut will present a balaced impedence to a balanaced input circut, and externally imposed noise will indeed be common mode.... but your contention that one will have a balaced signal eludes me. |
It eludes you because I didn't make such a contention.
As I said previously, "balanced" only refers to the impedances of the two lines with respect to ground. Whether the lines are balanced or unbalanced, the <b>signal</b> exits differentially between the two lines.
Now if you're referring to my having mentioned bipolar signals, i.e. +v and -v, that's easy enough to explain. You just seem to be looking at it from a different perspective than I am.
As far as I'm concerned, "signal" only matters at the receiving end, and that's the perspective I'm looking at it from. In other words, I'm looking at what the balanced, differential input at the receiving end "sees."
It perhaps wasn't quite so obvious from the last schematic I posted so let me simplify it it a bit on the receiving end:
<center>
<img src="http://www.q-audio.com/images/balanced3.jpg">
</center>
Does that make it a bit more clear?
se |
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| AMPMAN |
| Steve" coould you please draw a diagram showing an rca plug going to a p1.7 input this has to be the clearest way to solve this problem. |
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| Joseph Knecht |
jh6you. thanks for a great link! It is ironic that a company selling transformers has such a good app note on active balanced interfaces. It is also interesting to note the use of a signal ground connection, even for one balanced interface containing a transformer. Note there is a difference in signal ground and chassis ground. The capacitors are important to prevent DC currents in transformer windings which will affect the linearity and sonics of many small transformers.
From
http://www.dself.demon.co.uk/balanced.htm
"Renders innocuous ground-loops, so that people are not tempted to start "lifting grounds" This tactic is only acceptable if the equipment has a dedicated ground-lift switch, that leaves the metalwork firmly connected to mains safety earth. In the absence of this facility, the optimistic will remove the mains earth (not quite so easy now that moulded plugs are standard) and this practice must be roundly condemned as DANGEROUS"
Both these links are required reading and should clear up misunderstandings about in what "context" some of the previous post were meant to be taken .
Joe |
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| pmkap |
I still don't get it.
.It eludes you because I didn't make such a contention.
Stuff and nonsense....
If you converted to a balanced output using the passive solution, you'd measure +0.5 volts on the + line and -0.5 volts on the - line.
You certainly did make such a contention. And if you wish to rationalize that comment by stating that it is only valid on the recieving end for a truly balanced input , I'll grant you that.
But please realize some folks might actually accept your statements without that obligatory grain of salt and backpedaling. Used as a phase inverter, with a dpdt switch, it will simply provide useless common mode noise to a single ended input as others, far more compentent the me, have pointed out.
Whatever..... |
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| Peter Daniel |
| It seems like Mr. Pass suggested similar approach (passive balanced line) in A75 article. |
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| Peter Daniel |
| And here is Fig.10 |
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| pmkap |
SE wrote-
If you converted to a balanced output using the passive solution, you'd measure +0.5 volts on the + line and -0.5 volts on the - line.
You certainly did make such a contention. And that contention is simply not true. if you wish to rationalize that comment by stating that it is provides a pseudo-balanced signal on the recieving end for a truly balanced input , I'll grant you that.[for the purpose of rejecting common mode noise only!] |
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| Steve Eddy |
The capacitors are there simply because that's how Bill chose to represent a "typical unbalanced output." If the unbalanced output you're wanting to convert to balanced is DC coupled (i.e. uses no series output capacitors) then there's no reason to add any capacitors. If it's AC coupled, the example only serves to show that you need to mirror them on the other side of the circuit.
se |
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| Joseph Knecht |
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 |
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| Steve Eddy |
| quote: | Originally posted by Joseph Knecht
jh6you. thanks for a great link! It is ironic that a company selling transformers has such a good app note on active balanced interfaces. |
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.
| quote: | | It is also interesting to note the use of a signal ground connection, even for one balanced interface containing a transformer. Note there is a difference in signal ground and chassis ground. |
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 |
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| Steve Eddy |
| quote: | Originally posted by pmkap
You certainly did make such a contention. And if you wish to rationalize that comment by stating that it is only valid on the recieving end for a truly balanced input , I'll grant you that.
But please realize some folks might actually accept your statements without that obligatory grain of salt and backpedaling. |
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.
| quote: | Used as a phase inverter, with a dpdt switch, it will simply provide useless common mode noise to a single ended input as others, far more compentent the me, have pointed out.
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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 |
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| Joseph Knecht |
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 |
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| Steve Eddy |
| quote: | Originally posted by Peter Daniel
It seems like Mr. Pass suggested similar approach (passive balanced line) in A75 article. |
Yes, same thing. You're just mirroring the output impedance of the driven output to give each line the same impedance to ground.
se |
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| Steve Eddy |
| quote: | Originally posted by Joseph Knecht
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 |
Thanks for the plug. Beautiful, isn't she?
se |
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| Steve Eddy |
| quote: | Originally posted by Joseph Knecht
Oh Contrair, as I stated earlier:
If you are going to use a transformer in the interface |
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 |
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| jh6you |
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? |
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| jh6you |
Jesen AN-003 explains the impedance balance and coresponding
CMRR only. But, nothing about the divided signals. |
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| Steve Eddy |
| quote: | Originally posted by jh6you
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? |
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 |
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| Steve Eddy |
| quote: | Originally posted by jh6you
Jesen AN-003 explains the impedance balance and coresponding
CMRR only. But, nothing about the divided signals. |
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 |
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| jh6you |
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? |
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| Steve Eddy |
| quote: | Originally posted by jh6you
If the reference grounds are not tied together, then we get each half signal.
But, if tied together, then unequal signals. |
In the context of that particular passive solution, yes.
| quote: | 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? |
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 |
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| Mark A. Gulbrandsen |
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 |
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| mrfeedback |
"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. |
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| Steve Eddy |
| quote: | Originally posted by mrfeedback
"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. |
Any residual magnetization isn't permanent as it can be trivially demagnetized.
| quote: | | It is bad practice to allow DC through audio transformer windings. |
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.
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| planet10 |
| quote: | Originally posted by Steve Eddy
And if a little DC on my transformers ultimately results in greater enjoyment, so be it. |
Or a lot in case of a single eneded tube amp.
dave |
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| Steve Eddy |
| quote: | Originally posted by planet10
Or a lot in case of a single eneded tube amp. |
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 |
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| jh6you |
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.
:) |
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| mrfeedback |
Steve, do you have an appropriate demagnetiser circuit ?.
Eric. |
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| Steve Eddy |
| quote: | Originally posted by jh6you
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 |
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| Steve Eddy |
| quote: | Originally posted by mrfeedback
Steve, do you have an appropriate demagnetiser circuit ?. |
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 |
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| jh6you |
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. |
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| Steve Eddy |
| quote: | Originally posted by jh6you
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. |
Yes. :)
se |
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| jh6you |
No. :)
Your balanced interface explains 50-50 common noise signal
through the hot and cold line. But, signal is only through the hot line. |
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| Steve Eddy |
| quote: | Originally posted by jh6you
No. :)
Your balanced interface explains 50-50 common noise signal
through the hot and cold line. But, signal is only through the hot line. |
Yeah? How do you transmit a signal on only one line?
se |
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| jh6you |
Please read this.| quote: | 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." |
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| Steve Eddy |
| quote: | Originally posted by jh6you
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 |
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| jh6you |
Steve
I have never heard about or seen any balanced interface
with two lines. |
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| mrfeedback |
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. |
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| Steve Eddy |
| quote: | Originally posted by jh6you
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 |
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| Steve Eddy |
| quote: | Originally posted by mrfeedback
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 |
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| Nelson Pass |
Geez, I go away for a couple of days and all
hell breaks loose.
:bigeyes: |
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| Steve Eddy |
| quote: | Originally posted by Nelson Pass
Geez, I go away for a couple of days and all
hell breaks loose. |
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 |
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| mrfeedback |
Thanks Steve, that is a fair enough suggestion, however is not greater initial amplitude than line level 0Vu required ?.
Eric. |
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| Steve Eddy |
| quote: | Originally posted by mrfeedback
Thanks Steve, that is a fair enough suggestion, however is not greater initial amplitude than line level 0Vu required ? |
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 |
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| AMPMAN |
| 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: |
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| Steve Eddy |
| quote: | Originally posted by AMPMAN
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?
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| AMPMAN |
| 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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| jh6you |
| quote: | Geez, I go away for a couple of days and all
hell breaks loose.
:bigeyes:
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Wapooo, wapooo,wapooo... swimming in the pool of hell.
Sometimes, could be better than a big yawn in the boring heaven.
:yummy: |
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