Using input & interstage transformers in a balanced passive mix bus design

[quantyk]

> So did a 50-60

That's not something to copy.

The 68-2700-68 networks are just 1db pads. Someone thought this was a Good Idea. A 6db pad will control in/out impedance variations. A 1db pad limits impedance to between 136 and 2768 ohms.... not much of a "limit".

Are you mixing for mono or stereo? Are your sources true 600 ohms or something more modern? (If they are 600, how will you put two 600 loads on one 600 output?)

If stereo is involved, you must consider CrossTalk. A matched network will back-leak all inputs to all other sources, which will reduce L-R separation. The simplest way to kill crosstalk is to have a "zero" impedance somewhere. Could be the source. Could be the mix-bus. Does not have to be dead-zero. 100 ohm sources feeding 10K mix resistors to a 100 ohm bus will have -80db L-R crosstalk which is musically acceptable. Much higher crosstalk is often very fine.

I think you need Voltage Divider theory, and draw-out your full plan including source and load impedances. You may omit transformers for a first approximation.

Where are your mix-level controls? They usually introduce impedance issues.

Still absorbing some of the info shared here, but wanted to answer these valid questions;

This will be for stereo, and the source will be an 8-channel D/A, probably nominal 100 ohms.

For this iteration, I'm going to skip mix controls and deal with levels and panning in the DAW, as I want to focus on the transformer utilization. Once I get a sense of how best to coordinate the bussing, then I would look at some simple switching options. The modus operandi is to create a hybrid working environment, where an analog soundstage can be pushed and orchestrated from a digital interface.

The math of parallel vs series 2ndary stacking is escaping me, so I'm going to do some breadboard experiments & compare the sounds.  Leaning towards the series approach, as it's been well recommended, and could be fewer components, but will try an iteration with the UTC parallel approach.

gee PRR, i would prefer to be a voltage uniter, not a divider... or was that a voltage decider...razz.. but I have been catching up on attenuators and recognize the pads in the UTC gizmo.  I'm obviously not an EE, but enjoy the DIY process, even with the steep learning curves.

Seems not much is simple regarding the application of transformers, but I know I like their impact on the sound so far.

thanks again for all the insights & information.

cheers & happy TG

 

[abbey road d enfer]

Please note that the series-mixing app picture is dated 1931. The audio quality expectations in those times would not have seriously challenged the transformer design.
Today, if you expect 20kHz BW and good x-talk in stereo, you would not want to use just any old transformer in there.
IMO, series-mixing in 2011 is just a recreation of an obsolete technique, like building a new oil lamp, something that works, but has no practical justification.

 

[Marik]

Here is a VERY rough sketch of the concept, for further discussion. I've sketched a couple of versions, with an additional set of mix resistors, and another with a ZFT arrangement for the mixbuss TFs

This one can be done with two 5 chambered transformers, instead of 10 single ones.

Best, M

 

[abbey road d enfer]

Here is a VERY rough sketch of the concept, for further discussion.

Unfortunately, TOO rough to make the discussion possible. As it is, all these Zobels are in parallels, x-talk is -6dB and input impedance is impossibly low.

 

[JohnRoberts]

Spoil sport..... Just think of all the transformer sound it would make?    8)

JR

 

[Marik]

As it is, all these Zobels are in parallels, x-talk is -6dB and input impedance is impossibly low.

Of course, none of that applies to a single/multi-chambered transformer per channel. 4 summing channels can easily be made on one EI transformer, or some 8-10 channels on UI. 150 Ohm, or even 600 Ohm input with pretty low winding DCR should not be a problem.

Best, M 

 

[abbey road d enfer]

Of course, none of that applies to a single/multi-chambered transformer per channel. 4 summing channels can easily be made on one EI transformer, or some 8-10 channels on UI. 150 Ohm, or even 600 Ohm input with pretty low winding DCR should not be a problem.

No. When all windings share a single core, they are in parallels AC-wise, so, on a 4 input thingy, the impedance of an input is equal to the three others in parallels. Assuming all sources are 100 ohms (typical of many converters), the input impedance will be 33 ohms + 4/3 the CDR of one winding, for a total of about 90 ohms. And x-talk is about -8dB.

 

[quantyk]

Quick query on crosstalk: if the left and right busses are not electrically connected and galvanically isolated, why would there be any crosstalk between L & R? I take it you mean 'vertical' crosstalk between channels on the same bus, yes? and that resistors 'between' each 2ndary would affect/improve this?

 

[abbey road d enfer]

Quick query on crosstalk: if the left and right busses are not electrically connected and galvanically isolated, why would there be any crosstalk between L & R? I take it you mean 'vertical' crosstalk between channels on the same bus, yes?

Correct.

and that resistors 'between' each 2ndary would affect/improve this?

Yes, definitely. In passive summing, x-talk is directly related to the transfer efficiency, i.e. the higher the mixing loss the higher the separation between sources.

 

[PRR]

> if the left and right busses are not electrically connected and galvanically isolated, why would there be any crosstalk between L & R?

Transformers work both directions.

Signal comes out of source through left transformer to ch1 mix and appears on the bus. It sneaks back out ch2 mix through transformer to ch2 source. From here it gets to ch2 Right trans and mix.

Draw out all your impedances. Assume transformers are perfect both-ways. Apply 1V DC at an input and used voltage-divider concepts to find the stray DC at the "other" side.

The fix is a "ZERO" impedance at source or bus. If signal passes through 10K mix resistance to 1K bus to 10K mix and 1K source, crosstalk is -40db for 2-input mix, -22 for 8 ins. If source or bus is 10 ohms -80db.

Alternatively the mix impedances may be infinite, current sources. I think JR had some thoughts once. OTAs, some VCAs, and pentodes can be worked this way.)

Alternatively L and R sources may be separate amplifiers. There still can be some crosstalk, amplifiers always leak from output to input, but this is typically very small.)

 

[quantyk]

Hmmm. late night, so maybe (definitely) I'm missing something...

I understand they work both ways, but it seems that ch1 signal would never reach ch2 transformer, as ch1 is dedicated L and ch2 is dedicated R, and L & R are separate busses.

This is part of why I'm skipping panpots and controls which could provide an opportunity for L & R to meet.  Separate +/- busses for L & R, with separate transformers on each bus, and a separate (and discrete) amp for each channel.

I can see how ch1 and ch3 and other such channels on the same bus could interact ("vertical crosstalk"), and have been drawing and redrawing this circuit while absorbing the research, attached is the latest draft.

Preliminary breadboard tests seem to confirm the series/stacked secondary approach (va-va-voom), so will take that path a bit further.

More questions!

Could one consider the stacked xfmr secondaries as AC voltage sources? How would voltage division apply here?

If I calculate the equivalent Z value for the parallel combination of the secondary and the Zobel, I get about 1846 ohms, and 4 of these in series adds to 7385.

Trying to work out how to connect the Zobels in series, and trying to understand approaches to controlling crosstalk between channels on the same bus; this goes back to my original query of how the xfrmr termination and the mix bus

Since this design uses series xfrms rather than parallel resistors, would the same techniques apply (using resistors to control crosstalk).

Or, would I use attenuators , like L pads, to isolate the 2ndaries from each other? Seems that I want each xfrmr to contribute signal only to the bus and not to each other.

Thanks again for the help, and appreciate the patience and continued engagement.

 

[abbey road d enfer]

Could one consider the stacked xfmr secondaries as AC voltage sources?

Yes. With a source impedance.

How would voltage division apply here?

Since they are in series, voltages add, impedances too. If one of the sources is V and Z, the combined 4 sources equivalent source will be 4V and 4Z.

If I calculate the equivalent Z value for the parallel combination of the secondary and the Zobel, I get about 1846 ohms, and 4 of these in series adds to 7385.

Wrong. The Zobel is there to compensate for the increase of impedance due to the leakage inductance (which appears in series with the reflected impedance. The impedance reflected at the secondary depends on the impedance of the source that is connected to it. Are you sure that the DCR is 1000r? Seems rather high to me... Remember that the primary's DCR appears also in the reflected output, so basically for a 1:1 xfmr, the reflected impedance is equal to Zsource + primary DCR + secondary DCR.

Trying to work out how to connect the Zobels in series,

Forget about them for now.

and trying to understand approaches to controlling crosstalk between channels on the same bus; this goes back to my original query of how the xfrmr termination and the mix bus. Since this design uses series xfrms rather than parallel resistors, would the same techniques apply (using resistors to control crosstalk).

This depends very much on the load that is applied to the strings of secondaries. If the load was infinite, there would be no current in the string, so no transfer of energy from one xfmr to the other so no x-talk. If the load was zero, as in a curent-to-voltage converter, current would be maximum and transfer would be 1:4 => -12dB. In your case that would depend on what's connected at the secondary of the "bus" xfmr.

  Or, would I use attenuators , like L pads, to isolate the 2ndaries from each other?

You could but every dB you'll gain in x-talk will be a dB lost in the signal transfer, so you'll have to increase the gain of the make-up amp. Use very high load impedance for zero current.

Seems that I want each xfrmr to contribute signal only to the bus and not to each other.

But you are confronted to the laws of physics. One can only wish for a compromise in this fight. That's the reality of "passive".
You don't need the "bus" output xfmr; it's redundant. You may use an unbalanced high-Z make-up amp there for no-current in the string. Then it's becoming active...Aarrgh.
BTW, what's the problem with vertical x-talk, since each signal coming into the box seems to be used for that and only that?

 

[JohnRoberts]

Yup, stacking the secondaries is like stacking up a bunch of AC batteries in series.  I will second the suggestion for a active buffer between stack and output transformer to minimize the reflected common load.

For hot line level audio signals, the linear multiplication of signal by number of stems suggests you could have voltage headroom issues with any solid state active buffer stage, This is more of an issue on paper than in practice, but I would meter this point and perhaps add protection clamps to any active buffer JIC. 

JR

PS: Yes PRR I did a bunch of work with using current sources to execute lossless summing structures.

 

[quantyk]

Hi, thanks for clarifying some of these points, really appreciate the insights here.

Could one consider the stacked xfmr secondaries as AC voltage sources?

Yes. With a source impedance.

How would voltage division apply here?

Since they are in series, voltages add, impedances too. If one of the sources is V and Z, the combined 4 sources equivalent source will be 4V and 4Z.

If I calculate the equivalent Z value for the parallel combination of the secondary and the Zobel, I get about 1846 ohms, and 4 of these in series adds to 7385.

Wrong. The Zobel is there to compensate for the increase of impedance due to the leakage inductance (which appears in series with the reflected impedance. The impedance reflected at the secondary depends on the impedance of the source that is connected to it. Are you sure that the DCR is 1000r? Seems rather high to me... Remember that the primary's DCR appears also in the reflected output, so basically for a 1:1 xfmr, the reflected impedance is equal to Zsource + primary DCR + secondary DCR.

OK: the LL1531 (http://www.lundahl.se/pdfs/datash/1531_32.pdf) has 2 primaries with "static resistance" of 500 ohms each, which I interpret as DCR.  Connecting the primaries in series, as recommended would give a primary DCR of 1000r.
The secondary DCR is given as 1.3k, so w/o Zobel termination, the reflected impedance, assuming that the DA output/line source is 100r, would be 2.4k (1k+1.3k+100r).

So 4 of these in series would be 9.6k; yes?

For the overall bus Z, I would also have to account for the reflected Z of the bus xfmr, and probably in series with others, right?

holding that thought...

If the input xfrms need to see a 8k load on the secondary, could I connect a 50k resistor in parallel/shunt to bring the bus resistance to 8K? 

Then, would placing a single 1.2nF capacitor in series with this loop meet the capacitance requirements for loading each xfmr? (8k in series with 1.2 nF)

Trying to work out how to connect the Zobels in series,Forget about them for now. 

OK, Just hoping to get a decent frequency response from the input xfmrs...

and trying to understand approaches to controlling crosstalk between channels on the same bus; this goes back to my original query of how the xfrmr termination and the mix bus interact. Since this design uses series xfrms rather than parallel resistors, would the same techniques apply (using resistors to control crosstalk).

This depends very much on the load that is applied to the strings of secondaries. If the load was infinite, there would be no current in the string, so no transfer of energy from one xfmr to the other so no x-talk. If the load was zero, as in a curent-to-voltage converter, current would be maximum and transfer would be 1:4 => -12dB. In your case that would depend on what's connected at the secondary of the "bus" xfmr.

 

hmmm...aren't xfmrs like current-to voltage converters in a way? anyway...

The bus xfmr will be connected to the inverting/non-inverting inputs of a discrete differential amp  (neve 10852)  for each bus and I'm wondering how to work out what impedance that amp would prefer to see, and what it would look like as a load.

The amp card was designed as a high level output stage, so perhaps by reviewing the console specs I could find the impedance of the mix bus and infer from there?

Dual Line Amp EV 10852

If it's OK with an source (from the bus) impedance of 8-10k, then should be fine, yes? This would be like a matching impedance linkage.


The bus xfmr is a 600r center-tapped deal with 2 coils on each primary/secondary at 150r, which can be wired in parallel for 150r, or in series for 600r

http://www.tamura-ss.co.jp/en/electronics/trance03/pdf/audio_tp.pdf


If the amp stage is OK with an source (from the bus) impedance of 8-10k, then should be fine, yes? This would be like a 'matching' impedance linkage If it prefers a lower impedance.

However, John's point on the high voltage is well taken, and perhaps I can use the mix bus xfmr to step the voltage down by wiring the primaries in series and the secondaries in parallel, which I think would give a 2:1 ratio, if the impedance ratio (600 pri, 150 sec) is 4:1, then the turns/voltage ratio should be 2:1

Assuming a mix bus/primary load of 8k, this would reflect about 2k to the secondary, right? with the DCRs (150 + 150), that would be 2.3k reflected at the bus xfmr secondary...

And the 4V voltage would be stepped down to 2V, which should be OK for the amplifier, yes? (V as input Value, not actual volts) roughly if each input is at +4 dbu, this would be about 1.22 volts rms, which added in series would be just about 5 volts... i think. so if the step-down is appropriate, then the bus input to the amp would be 2.5v

Or, would I use attenuators , like L pads, to isolate the 2ndaries from each other?

You could but every dB you'll gain in x-talk will be a dB lost in the signal transfer, so you'll have to increase the gain of the make-up amp. Use very high load impedance for zero current.Seems that I want each xfrmr to contribute signal only to the bus and not to each other.  But you are confronted to the laws of physics. One can only wish for a compromise in this fight. That's the reality of "passive".
You don't need the "bus" output xfmr; it's redundant. You may use an unbalanced high-Z make-up amp there for no-current in the string. Then it's becoming active...Aarrgh.[/quote]

This is intended as a hybrid design, there is definitely active amplification of the summed signals, but passive combining...I was hoping to avoid using active elements before the mix bus.

BTW, what's the problem with vertical x-talk, since each signal coming into the box seems to be used for that and only that?

Hmmm...I guess the concern is whether input signal from ch1 would affect the primary loading  & signal transfer from the other channels, as PRR seemed to imply (or as i misunderstood). The point is to mix these signals, after all. It seems unlikely that the signals on the bus would counteract the xfmr action/input on each channel, to the point where it would cause any relevant problems. could the induced voltages from the other secondaries on the bus interfere? It seems that they are additive in this case, yes?

Relevant problems would involve imaging, phasing, frequency response, the little things in life...

And it also seems that a bit of crosstalk is part of the 'analog mojo' factor; I'm reminded of the airwindows plug-in designer who emulates the variable input impedance & crosstalk anomalies of analog consoles...

 

[JohnRoberts]

hmmm...aren't xfmrs like current-to voltage converters in a way? anyway...

Transformers are more like a voltage to magnetic field back to voltage converters. The primary voltage is applied to a winding resistance that draws current and creates a magnetic field. The magnetic field flowing through the common core and through the secondary winding generates current flow and voltage across it. So arguably a voltage to (different) voltage converter, or an impedance convertor, when turns ratios other than one are involved (since input and output power are similar, different voltage means different current and apparent impedance transform).

When the secondaries are in parallel, all the different magnetic cores are shorted together. When the secondaries are in series, only the common load across them all gets coupled back to the primaries. This load and sharing is impacted by differences between the windings and perhaps signals in each.

Making the load across the whole string as light as possible, reduces the significance of matching etc.

JR

, 

 

[abbey road d enfer]

OK: the LL1531 (http://www.lundahl.se/pdfs/datash/1531_32.pdf) has 2 primaries with "static resistance" of 500 ohms each, which I interpret as DCR.  Connecting the primaries in series, as recommended would give a primary DCR of 1000r.
The secondary DCR is given as 1.3k, so w/o Zobel termination, the reflected impedance, assuming that the DA output/line source is 100r, would be 2.4k (1k+1.3k+100r).

So 4 of these in series would be 9.6k; yes?

Correct

For the overall bus Z,

Doesn't mean a thing.

I would also have to account for the reflected Z of the bus xfmr, and probably in series with others, right?

No.

If the input xfrms need to see a 8k load on the secondary, could I connect a 50k resistor in parallel/shunt to bring the bus resistance to 8K?

No. Each secondary will need to have its own Zobel; if you had only one for the string, each secondary would "see" 3 secondaries + whatever Zobel and "bus" xfmr.

Then, would placing a single 1.2nF capacitor in series with this loop meet the capacitance requirements for loading each xfmr? (8k in series with 1.2 nF)

No

The bus xfmr will be connected to the inverting/non-inverting inputs of a discrete differential amp  (neve 10852)  for each bus and I'm wondering how to work out what impedance that amp would prefer to see, and what it would look like as a load.

You'd better forget about the "bus" xfmr and connect the string directly to a non-inverting stage. Most low-noise solid-state circuitry is perfectly happy with a source Z of 10k.

The bus xfmr is a 600r center-tapped deal with 2 coils on each primary/secondary at 150r, which can be wired in parallel for 150r, or in series for 600r

It would definitely be a serious mismatch with the string impedance.

If the amp stage is OK with an source (from the bus) impedance of 8-10k, then should be fine, yes?

Correct

BTW, what's the problem with vertical x-talk, since each signal coming into the box seems to be used for that and only that?

Hmmm...I guess the concern is whether input signal from ch1 would affect the primary loading  & signal transfer from the other channels, as PRR seemed to imply (or as i misunderstood). The point is to mix these signals, after all. It seems unlikely that the signals on the bus would counteract the xfmr action/input on each channel, to the point where it would cause any relevant problems. could the induced voltages from the other secondaries on the bus interfere? It seems that they are additive in this case, yes?

There would be a problem if one or several inputs had to be mixed on the same bus.

 

[quantyk]

OK, let me see if I can get some of this straight...

I'm sure all the electrical wizards here are having a chuckle as I try to figure this probably obvious stuff out...

By "overall bus Z", I meant the load each input secondary on the string sees, from the other elements on the string.

As JR said: "When the secondaries are in series, only the common load across them all gets coupled back to the primaries. This load and sharing is impacted by differences between the windings and perhaps signals in each."

Note: each input xfmr is identical in terms of windings. (signals will be different, except for the john cage performance of 4'12")

So each input xfmr would 'see' the other 3 secondaries, the primary of the bus xfmr and any other elements in the string in series.

As Abbey said: "Each secondary will need to have its own Zobel; if you had only one for the string, each secondary would "see" 3 secondaries + whatever Zobel and "bus" xfmr."

So without Zobel, each input xfmr would see 7.8K from the rest of the string (3 x 2.4k) + DCR of bus xfmr of about 600r, and reflect this to the primary for an input impedance of about 10K.

If I still want use the pesky bus xfmr to get to the other side (a la pedestrian chicken), it would see a source Z from the bus of the sum of all Z's on the string, so 9.6K, with DCR about 10.2k

Since the xfmr itself, while it has a DCR for each winding, largely reflects impedance based on the square of the turns ratio, how does the mismatching between the 9.6k of the bus and the 600r DCR of the xfmr primary work? (or not work?)

Wouldn't the bus xfmr just reflect the 10.2k to the secondary at whatever ratio I chose to use? either 1:1 (about 10k) or 4:1? ( about 3k)?

If I wanted to manage impedance and voltage level mismatches between the bus and amp stages, isn't that what xfmrs are used for?

Seems like between a little insertion loss and saturation, if the summed voltages got rowdy, the bus xfmr would help keep the amp stage from getting too riled up...

Regarding Zobels, the input xfmr self-resonance is at 80khz, and I would like a flat frequency response at this stage, without passing any unnecessary frequencies on to the mix bus or to the mix amp. As long as I'm still on the drawing board, I'd like to figure that bit out too.

As Abbey said: " The Zobel is there to compensate for the increase of impedance due to the leakage inductance (which appears in series with the reflected impedance."

To clarify: does the Zobel network's impedance appear in series with the xfmr's reflected impedance on the string, or did you mean that the increased impedance caused by leakage inductance appears in series with the reflected impedance? (and the zobel, by placing 8k in parallel with this increased impedance, compensates?)

Or put another way, is the equivalent series resistance for each string 'node' (secondary refl. Z1 + Zobel network Z2) found by adding Z1+Z2 (series), or by (Z1*Z2/Z1+Z2) (parallel)

This would change how I would calculate the string/common load, etc. and would prompt even more questions.

thanks again for putting up with all my questions, and I hope this is a helpful process to anyone else interested in this stuff.

I'm hoping to identify the right components I'll need to build this, so I guess I'm asking 4 times, to understand 3 times, to measure twice and cut once...

cheers,

AK

 

[abbey road d enfer]

So without Zobel, each input xfmr would see 7.8K from the rest of the string (3 x 2.4k) + DCR of bus xfmr of about 600r, and reflect this to the primary for an input impedance of about 10K.

No; each input xfmr would see 7.8K from the rest of the string (3 x 2.4k) + the primary impedance of the bus xfmr, which depends on the load at the secondary + the primary DCR.

If I still want use the pesky bus xfmr to get to the other side (a la pedestrian chicken), it would see a source Z from the bus of the sum of all Z's on the string, so 9.6K, with DCR about 10.2k

You can't use a 600ohm xfmr here, you need at least a 10k type.

Since the xfmr itself, while it has a DCR for each winding, largely reflects impedance based on the square of the turns ratio, how does the mismatching between the 9.6k of the bus and the 600r DCR of the xfmr primary work? (or not work?)

The match (or mismatch) is between the source impedance and the primary impedance of the bus xfmr, which comprises the CDR but also the reflected secondary load.

Wouldn't the bus xfmr just reflect the 10.2k to the secondary at whatever ratio I chose to use?

If the inductance of the xfmr is too low, it becomes dominant as a parasitic element in parallels with a perfect xfmr. A 600r xfmr cannot really reflect 10k. This is frequency-dependant, but I won't go further this way because obviously you want to run but you haven't learnt to walk on that subject.

either 1:1 (about 10k) or 4:1? ( about 3k)?

Reflected impedance depends on square turns, so a 2:1 would reflect 2.5k and a 4:1 would reflect 600r.

If I wanted to manage impedance and voltage level mismatches between the bus and amp stages, isn't that what xfmrs are used for?

What's the nominal impedance of your make-up amp? You need a xfmr with a 10k primary and whatever secondary matches it.

To clarify: does the Zobel network's impedance appear in series with the xfmr's reflected impedance on the string, or did you mean that the increased impedance caused by leakage inductance appears in series with the reflected impedance? (and the zobel, by placing 8k in parallel with this increased impedance, compensates?)

The Zobel, which appears in parallels with the secondary, is meant to linearize the frequency response AND the impedance. If you want to include it in your sim, you also need to include the complete parasitic behaviour of your xfmrs. The alternative is to consider that the xfmrs are "perfect" and forget the Zobels. That's what Zobels are for.

 

[quantyk]

Thanks for clarifying some of these points, and will continue the research & experimentation.

 

[quantyk]

Still experimenting here, although the parameters have shifted somewhat, as i've been more focused on walking. Have collected a few different pairs of transformers, and more recently focusing on getting the make-up amp working - started a separate thread on Neve 10852.

Still curious as to how transformers would interact when stacked- assuming that I use xfos with similar DCR & inductances, is there any hope for an even frequency response?