V475 translation + summing network

[Moby]

This tx current summing looks really great! Does somebody knows ratio of transformer? Ok, primary is 3-4 ohm, what about sec.? Let's say that feedback windings are same like feeding or maybe is possible to skip them in some simpler make-up configuration than neumann. Also, is it possible to go with unbalanced summing or just bal. works :?: Hmm, my other question looks like bit stuppid, but maybe i'm missing something...

 

[[silent:arts]]

the Neumann 90456 transformer = Haufe RK-292.
the ratio is 1:2+2
Haufe RK292 datasheet

 

[Samuel Groner]

the ratio is 1:2+2

Interesting - how do they get a reasonable noise figure? OSI is 8k for the 5534, but source impedance is around 400 ohm for 24 channels - way to low even after step-up...

BTW, can someone explain me the nasty feedback loop of this output-circuit? It looks like the generate negative impedance to cancel the primary resistance.

Samuel

[silent:arts]: did you check your PMs?

 

[Moby]

Ok, 1-2+2 but i'm confused with 90456 datasheet. What is "kuzschlussausgangsscheinwiederstand" :?: Is that ac resistance? And what is "impedanz" 2kohm?
If i calculate ratio 1-2+2 like 4ohmprimary, sec will be 2x16ohm. If that's true, tx is not "so complicated" :green:

 

[Moby]

:roll: Any comments about sec impedance? :roll:

 

[Samuel Groner]

"Wicklungswiderstand" means DC-resistance of the windings.

"Kurzschlussausgangsscheinwiderstand" stays for the impedance (@ 50 kHz in this case) one "sees" at the output if we short the primaries.

Not shure about "Impedance" either; but the 2k pop up in the schema as R1, so it may be related to the special use of this XFMR.

If i calculate ratio 1-2+2 like 4ohmprimary, sec will be 2x16ohm.

I don't know what you want to calculate; if we have i.e. a source impedance of 500 ohms, and use a 1:2 transformer (as here, the second "2" is used for feedback), the following gain stage sees 500*(2^2)=2k, not considering DC-resistance of the windings.

Samuel

 

[Moby]

don't know what you want to calculate

Of course i don't want to calculate DC resistance of TX. That depends from turns and wire diameter... I would like to know AC resistance. I'm not sure did i understood well. Are we talking about AC 3-4 ohm? If that's right, with ratio of 1 to 2, sec will be 16 ohm... Hmm, maybe i'm missing the point.
Or lets say, i'm asking what will "see" resistor summing network? 3 -4 ohm of ac resistance? And "what will drive" ne5534's ? Sec of 16 ohm ?
I'm asking that just to be sure about this "passive-active" summing.
And part of sec. feedback windings...

BTW, can someone explain me the nasty feedback loop of this output-circuit? It looks like the generate negative impedance to cancel the primary resistance

I think that this windings are included in the neg. feedback of ne5534 and that's the way to cancel nice part of distortion generated in transformer. Any comments? :wink:

 

[PRR]

> how do they get a reasonable noise figure? OSI is 8k for the 5534, but source impedance is around 400 ohm for 24 channels - way to low even after step-up...

5534 doesn't suck at 400Ω. Or really 1K6. Self-noise of 5534 is IIRC around 0.7uV. Self-noise of 1K6 is around 0.6uV. And while sum-amps are noise-sensitive, they should not be noise-critical. When you have good amplifiers, your noise figure should be set at the passive transducers (mike capsule), and everything else worked at high enough level that it won't degrade the noise spec. Figure mike-amps run at gain of 40dB, a 16-in mixer has 24dB loss, the level at the summing amp is still 16dB above the level at the mike, so even a somewhat loose noise level won't degrade the mike noise.

In this case, while the bus-amp sets the noise with all faders full-down, at any practical setting of the faders the channel noise will meet or exceed the sum-amp noise. That's part of why this card has all those gain-set options: to optimize the system gain-structure for best dynamic range.

> OSI is 8k for the 5534

But when you can't make the source fit the "perfect" OSI of the amp, you need another number. What is the -range- of resistances where noise is "low"? In BJT amps, this is really the Beta of the input device. Say the 5534 inputs have Beta of 100. The OSI of 8K really is just the midway point between an equivalent input noise-voltage (series) resistance like 800Ω and an input noise-current (shunt) resistance like 80K. If the source is 800Ω or 80K, the noise figure is 3dB. Splitting the difference gives even lower N.F., under 1dB. But in many situations, a 3dB noise figure is not a major contributor to total system noise.

> explain me the nasty feedback loop of this output-circuit? It looks like the generate negative impedance to cancel the primary resistance.

That's one way to look at it.

Consider the basic virtual-ground summing amp. The feedback works to keep the amp input at zero volts. When the summing resistors inject current into the summing node, the feedback has to supply a cancelling current to keep the input at zero volts. The resulting output voltage (forced to supply the canceling current in the feedback resistor) is the sum of the inputs.

In this case, the feedback extends into the transformer. It works to keep the flux in the core at zero, or as low as possible. This makes the input winding voltage near-zero. This also makes transformer distortion near-zero, while giving good summing action and transformer isolation.

> "Kurzschlussausgangsscheinwiderstand" stays for the impedance (@ 50 kHz in this case) one "sees" at the output if we short the primaries.

Leakage Inductance plus resistance and some other small details.

> is it possible to go with unbalanced summing or just bal. works?

As in any floating input: ground one side of the transformer and drive the other side.

But that loses the main benefit of this scheme. What is "ground"? Ground is a 6 foot long wire with many branches and many different signals leaking into it. Even if ground is a fat copper bar, it won't have the same voltage at every place. Channel 1 outputs a signal relative to the left end of the ground bar, but the summer references the right end of the bar, after channels 2-16 have dumped their own signals into it. That causes cross-talk in a large console. Also some ground returns are not even dumping clean signal: class AB outputs make big garbage in ground.

With balanced summing, the output of each channel is not referred to some "common point" that is really a dumping-hole, but gives us two terminals with the signal expressed as the difference of those two voltages. The summer takes the sum of the differences of all channels. "Ground" does not matter at all, or only as a minor leakage term (like -50dB).

Unbalanced makes perfect sense in smaller systems. Each channel is much simpler. None of these 4-gang panpots, or dual differential booster-amps after an unbalanced panpot. With careful layout you can run 8- and 16-channel 1-man mixers perfectly well, especially if you do not have a German Engineering Department's obsession with specifications. In rock-n-roll, a little cross-talk (like -50dB) goes completely unnoticed.

 

[Samuel Groner]

In this case, the feedback extends into the transformer. It works to keep the flux in the core at zero, or as low as possible. This makes the input winding voltage near-zero. This also makes transformer distortion near-zero, while giving good summing action and transformer isolation.

Thanks PRR for the explanation. I was asking for the output-circuit, not the make up amp. There it looks like they generate 15 ohms of negative impedance.

But when you can't make the source fit the "perfect" OSI of the amp, you need another number.

I do not claim the non-optimal SI to be a problem, but why didn't they use a higher step-up ratio of the input-transformer? At least it's a German Broadcast product, so why not make it perfect? Of course, higher ratios are more difficult to do well etc. but 1:4 would not be impossible for Haufe.

Samuel

 

[PRR]

> I was asking for the output-circuit,

Ooops.

The THD of a transformer reduces with source impedance. Even with a zero-Z source, the winding resistance allows/causes some THD. Fewer turns of fatter wire does not help: inductance falls as well as resistance. A bigger core helps, but it has to get a lot bigger to help much, and that increases core-loss and stray capacitance. Negating the winding resistance really helps a lot, though you have to be sure the transformer DCR does not change in production, and stabilizing a neg-R loop is hard.

> why didn't they use a higher step-up ratio

Pay me a month of German Engineer's salary and maybe I can work out the reason.

My free opinion is that the feedback depends on very tight coupling, and a higher step-up would couple less well.

Also a radio console may spend much of its life with just a few live inputs: with just one input the source looks like 20K which is actually above the OSI you cite. 1 to 8 inputs sweeps across the lowest-noise area of the 5534 noise curve. At 16 inputs, surely the summed mike noise will overwhelm the mix noise.

There is also a LOT of virtue in using an Existing Part. Especially in broadcast where it is common to have a spare of any possible failed part to minimize downtime. Even if this failure would be board-swapped, it is nice if the central repair depot doesn't need yet another odd part in inventory. Also a new part has a learning curve and production ramp-up. The wise engineer might take a "sub-optimum" part if it was not significantly worse than a "perfect" part, met specs with margin, and was already in production and inventory.

But we'd need to study the Given Specification, including stability and phase-shift, and do a lot of math, to really grok how they came to this end.

 

[Samuel Groner]

Thanks again, as always very helpful. Let me ask another question about the output stage: Why have C3/C7 such high values? Together with the 10k resistor, we get a bandwith of 7 kHz - not great, is it? Or does this feedback loop not determin the bandwith of the hole stage?

Samuel

 

[PRR]

> explain me the nasty feedback loop of this output-circuit? It looks like the generate negative impedance to cancel the primary resistance.

I stupidly replied without looking at the plan. Let's try again.

Here's half the output:

> negative impedance to cancel the primary resistance.

Not at all. Output impedance is essentially zero, a few hundred ohms divided by the excess gain of a 5532 working at unity gain. Under 1 ohm across the audio band, but not negative.

The two amps (on each side) double the current capacity. IC1 drives the load. Voltage at the transformer is equal to input voltage. But load current causes voltage drop in the 15Ω resistor. IC2 matches that voltage on its own 15Ω resistor. So whatever IC1 puts out, IC2 puts out the same. If one 5532 output can drive 30mA, this rig can drive 60mA.

They do it like that because one 5532 really should not see less than 600Ω. But here they wanted more output power. Also they got stuck with a unipolar power supply, and didn't want to use a coupling cap to the transformer. They drive the transformer with a bridge. If the input is 600Ω, each amp sees 300Ω. By doubling each output, each amp sees 600Ω and can supply the current.

If the supply rail is +24V, it can make about 11V RMS in 600Ω, or +23dBm minus transformer loss.

> Why have C3/C7 such high values? Together with the 10k resistor, we get a bandwidth of 7 kHz

Look again. It is unity-gain. Assuming the 5532 input pin is 1MegΩ||22pFd then thegain is 1.01 (10K against 1MegΩ, 2200pF against 22pF). Tenth-dB gain; probably less than that. It has no audio function. They do it because the 5532 input stage can't follow a fast-slewing input signal, and there could be large current into the input junction until the feedback catches up. The 10K limits the current without much effect in the audio band. But if it were just the 10K, then stray capacitance to ground would cause a rising response and an extra pole, degrading MHz stability. The thousand pFd cap swamps stray capacitance.

 

[Samuel Groner]

The two amps (on each side) double the current capacity.

It get's pretty obvious as you write it...

Look again. It is unity-gain.

See above... :grin:

Thanks!

Samuel

 

[Moby]

Thaks PRR, now all looks much nicer. Only transformer impedance is not so clear to me. Is the secondary winding 2x16 ohms? Or something else? :roll:

 

[Samuel Groner]

Only transformer impedance is not so clear to me. Is the secondary winding 2x16 ohms? Or something else?

The summing resistors will "see" the primary DC-resistance plus some leakage garbage, more or less 0 ohms (that's "active" summing).

The 5534s will "see" the summing resistors in parallel (i.e. for 24 inputs 2*5k/24) multiplied by the square of the turns ratio (2^2=4) divided by 2 (the 5534 "share" impedance). So for the 24-channel example we get 833 ohm per 5534.

Hope my math doesn't suck this evening...

Samuel

 

[Moby]

ope my math doesn't suck this evening...

Not at all :green: But, i'm just asking for transformer specs. To be more precise, i woud like to know inductance of sec. windings. Or primary... :roll:

 

[Moby]

Or, let's say that i would like to wound this transformer... For what i'm looking for? If i go for primary 4ohm i will calculate it for around 30mH, and sec. 120mH per winding. Will that work in this circuit, or i'm missing something. :roll:

 

[Moby]

[quote author="rafafredd"]Now I should stop writing alone :grin:[/quote]
It's time for me too, or maybe i'm asking stupid question

 

[PRR]

> maybe i'm asking stupid question

The question is "too smart". The answer is not clear. I'm on very thin ice, but nobody else is jumping in with a better guess.

> If i go for primary 4ohm i will calculate it for around 30mH, and sec. 120mH per winding

I don't think so. The 4Ω is "with feedback". The no-feedback impedance can be much higher.

And I think it has to be higher. If you actually wound for 4Ω (30mH at 20Hz), the signal would probably pass but the noise-gain of the amp would be very high at low frequencies. I think you really want to pick an inductance that is about as high as the signal impedance. Somewhere around 300Ω. In fact I suspect that a common 150:600 transformer is about right, and within a few dB of "perfect". It just needs an extra 600Ω winding for the feedback. (Or have I got the ratios right?)

However, feedback through the core probably can't cancel primary resistance. And if primary DCR is 4Ω, it probably isn't 1H like a common 150Ω winding.

Design for 80Ω primary. The DCR should wind up around 4Ω. This is not at all critical (unless you need an EXACT clone). 5Ω or 10Ω DCR would work too.

I think the inductance should give a secondary impedance over 10K at the lowest frequency where the ear can hear noise. Because of ear response, this can be much higher than 20Hz. Say 400Hz. That gives 4H, which suggests 1H primary. Using 30mH primary 120mH secondary, the noise-corner would be over 10KHz, which does not seem right.

Wind for 150:600:600.

 

[Moby]

Thanks PRR, i expected something like this, but just wanted to be sure. Actually i tried to make something like this few years before, but i never ended. Now, when i saw this topics most of "things" became clear. Only TX real standalone resistance was mystery.
If i someday finish my version of this kind of summing, i will let you know :thumb: