[wakibaki]

Again, are you trying to investigate & understand this or are you trying to display your learning?

Neither jk, I do understand this, better than anybody else who has contributed to the discussion, and I'm trying to make sure that other people aren't misled.

w

[wakibaki]

The cable is 1 metre. The roundtrip time for a 1 metre cable of 0.88 velocity factor is ~7.7nS. The rise time in question is stated to be 3-4pS, but this is absurd, he means nS. The reflection therefore cannot arrive before the transition is complete.

OK John.

I have seen other posts of yours, if you are the same jneutron, you are accessible to reason.

Regardless of whether the rise time is in pS or nS, tell me how the reflection can arrive before the rise time is complete.

This is the recognised issue with short cables. Manufacturers often slow the rise time to avoid EMC issues. See:- http://www.positive-feedback.com/Issue14/spdif.htm. In that case, with a short cable, the reflection can transit the cable in 2 directions and interfere with the rising edge of the pulse. This is what art is talking about, short cables, but he gets it wrong.

Take a look, tell me your understanding of what he means, and let's see if we can get on the same page.

w

[jneutron]

OK John.

I have seen other posts of yours, if you are the same jneutron, you are accessible to reason.

Clearly you've never met my ex..she would not agree with your assertion.

Regardless of whether the rise time is in pS or nS, tell me how the reflection can arrive before the rise time is complete.

One must consider the reflection cause..

1.  If it is passive, such as the ferrite one Pat provided information on, then let's consider what is happening past the ferrite of the input.  From the waveform presented, the reflected signal "issue" is in the 50 to 75 (or so) nano domain of relaxation.  As a consequence, the signal passing on is also within that domain, but it's leading edge will certainly not be clean like the input stimulus.  On the drive side, any reflection which can be bounced by the transmitter can indeed arrive within the horrible rise time caused by this ferrite..it will of course, be subject to the same linear forming so in theory, be less of an issue.

2.  If it is an active node reflection, a node settling time can be either the inductance around the feedback loop, ground bounce, or internal circuitry limitations.  Given such erratic waveforms at an input node, I cannot say what the output will do.  I've seen circuits lose gain when saturation occurs, I've seen circuits invert subsequent signals before recovery (which I had to repeat to believe).  If this is the dominant reflection cause, then a returned bounce will hit the active node while the node is still recovering from the previous slew..  The use of a sufficiently long cable would help.  An attenuator, while losing signal energy, would lessen the return bounce by matching better, and by reducing the input node slew rate as well and reducing the node recovery time.

If one considers only prop delays and reflection coefficients, then a simple timewise analysis would be sufficient.  Over the years, I've been bitten by the simple models.  Nonetheless, one must start with the simple models.

Manufacturers often slow the rise time to avoid EMC issues.

Yep..been there, done that, seen it..

And apparently, they are also known to toss ferrites over high speed signal lines...go figure..

In that case, with a short cable, the reflection can transit the cable in 2 directions and interfere with the rising edge of the pulse.

One may have to consider the stuff further into the input as a result of the slew rate reduction caused by seemingly ridiculous input elements like ferrites or slow dac nodes.  If the signal of interest is slew rate challenged 50 to 75 nSec, it may indeed be necessary to consider reflections along that time frame.

This is what art is talking about, short cables, but he gets it wrong.

I have not been following the "emotional" arguments, so do not know what the specific foible is..  The physics, I do know a tad..


Pat has taken the time to present some nice data and analysis, but I've not followed the specific points of contention between you two or three..

Cheers, John

[jneutron]

I do understand this, better than anybody else who has contributed to the discussion....

30 years ago,even 20, I would have responded in a rather vocal fashion to being lumped as such.

Our strengths, weaknesses, and differences are to be enjoyed, not used as blunt instruments.  On some topics I am quite good, and yet I still learn things from others here.  I find no useful gain to be had by stepping on them.

(yes, on occasion some goofball requires "admonishment", but I consider that to be a perk, not a way of life.)

Cheers, John

[Joseph K]

John,

Just wanted to say hello 
Thanks for the nice words - for me it has been always a delight to discuss with You.
And in lurking mode, You had always been one of the best sources of creative thinking in these forums.. thanks.
So I find especially amusing your humble statement:

On some topics I am quite good,

You bet!

       
 

[jneutron]

So I find especially amusing your humble statement:

Thanks for the kind words.

Humility is a learned entity...learned by having children..

Cheers, John

[pjchappy]

Cool it.


Paul

[sts9fan]

Please keep the topic technical

[brj]

In an effort to keep this rather interesting thread going...

Wakibaki, you're raising some interesting questions and the resulting dialog is providing many of us with useful learning material.  That said, I, at least, have perceived your posting style to be rather agressive and more focused on proving yourself correct than truly promoting member education.  Internet forums are hardly the pinnacle of effective communication mediums, however, and your actual intent may very well be different - that is simply my interpretation of your posts to date.  Unfortunately, the coupling of this perception with your demands for responses and information are likely to hurt your cause.  I'd suggest that a more relaxed and open posting style would engender better responses.

By the way, since you're fairly new to AC, I'd recommend reading the The Lab Guidelines, the AC Posting Guidelines and other Rules.  Also note that each circle has a facilitator, with specific moderation mechanisms.  If you aren't already aware, sts9fan is the facilitator for The Lab.  Likewise, pjchappy is another AC facilitator and "volunteer".

Pat, thanks for starting this and other topics.  You and John (jneutron) are among those very few AC members whose posts I specifically search for in order to expand my knowledge base.  Many thanks, and please keep posting!

[jneutron]

Please keep the topic technical

I will abide by your request.  Sorry for the OT.

Cheers, John

[wakibaki]

brj

Thanks for your post. I understand that often my style is confrontational, but I do make an effort to moderate my tone. I'd like to say in mitigation that I have been seriously ill in the past year with lung cancer which is now thankfully in remission, but I am also suffering from hyperthyroidism, which makes one short tempered. The treatment I had for this condition was unfortunately unsuccessful, so I expect to be given radioactive iodine in the near future to kill off part of my thyroid. Please forgive me if in the meantime I overstep the mark.

I would like to assure everybody that I am more concerned with getting things right than with being right. Obviously this is easy to say but I am capable of changing my position and apologising.

==============================================================

OK, this is the basic setup we are discussing. Sometimes the DAC is removed and the coax is left open circuit to demonstrate the reflection coefficient of +1.



At the scope there is a 50R Tee and a 75R male-female connects the SPDIF transmitter to the Tee, so the Tee connects the transmitter, scope and cable.

Let's look at the picture on the scope.



The yellow arrow on the LHS indicates the rising edge of a pulse. The pulse duration is 7 divisions or 350nS, corresponding to a long SPDIF pulse (@ 44k1). The orange circle surrounds a feature which looks like a thorn, the rising edge of which is ~50nS after the rising edge of the pulse. This feature is the (inductive) reflection from the DAC. The cable is ~5 metres long, and this reflection is as a result of the pulse edge (on the left) having travelled to the DAC and back. This is a roundtrip, there has been one reflection.

There is no question of this being an 'active' reflection, it's shape is entirely characteristic of a passive inductive termination, if it were active one would expect to see a change in level on the RHS when the output has settled. This is anyway entirely irrelevant, what is of concern is the duration.

The second yellow arrow, the one on the right indicates an inverted version of the DAC reflection, which has returned to the transmitting end, been inverted by encountering the partial 50R mismatch at the scope and travelled back to the transmitter and returned finally to be recorded on the scope. This is 2 roundtrips, there have been 3 reflections.

This (main) reflection in no way interferes with either the rising edge or the falling edge of the pulse, which is what we are concerned over, because of its effect on the timing.

If this were a short pulse, half the length of the one shown, the reflection still would not interfere with the pulse edges, and it is quite obvious to the eye, that even were the cable 10 metres long (the maximum in the SPDIF specification) the short pulse would still not have its edges interfered with by the pulse. It's about 10nS for a 1 metre roundtrip, with a comparatively slow cable of 0.66 velocity factor.

The reflection you see here (the big shape) has been reflected only once, you see it returning to the transmitter. In order for it to return to the DAC, it must encounter a mismatch at the transmitter, and when it returns to the DAC it will be reduced in magnitude, perhaps not as much as the small reflection you see at the arrow on the right, but perhaps more so, since if the transmit termination is good, there will be no reflection at all.

Interference with pulse edges by reflections are similarly unlikely in 48k setups with a 10 metre cable. Even at 96k, a 3 metre cable will be unaffected by reflections.

I consider it unfair and self-indulgent to inflict suggestions about RF attenuators on an unsuspecting public without taking the time to make these facts clear. How many people are there out there running 16/44k1 on a 2 metre cable going 'doesn't it sound better with the attenuators'. Don't tell me it won't happen.

w

[jneutron]

Thanks for the schematic Joeseph..

With this in mind, consider what the output will look like if R2 is inductive.

1.  The input node will not remain at zero volts if a high slew is forced in.  The output of the monolithic amp will reflect the input error times some gain, and the R2 current will rise at some rate defined by the inductance.  The net result will be an output spike that will eventually drop to the circuit gain times the input. Reduce the inductance of R2, this spike will reduce.

2.  At the onset, the amp input node is not zero because of R2 inductance..now, you have an amp with an input that is too large...there is a distinct possibility that the amp will behave badly as a result.  Yes, the output will go in the correct direction initially, but once it has reached the final value, will it stop or overshoot the mark?  If internal stages saturate as a result of overdrive on the input node, who knows.

3.  Knowing the amp has been "severely" overdriven as a result of high di/dt into R1, it may need some time to recover, time beyond it's normal bandwidth.

4.  Any reflection off the input node will have a rapid rise, as pat provided pics of.  If the driver "box" reflects that back to the receiver, the input node will immediately pass the spike through to the output...exactly as it did the initial.

Where will that second spike be in relation to the initial waveform?

Inquiring minds want to know...

Cheers, John

ps..wakibaki...I'll discuss this tomorrow..gotta go for now..

[wakibaki]

With this in mind, consider what the output will look like if R2 is inductive.

Fascinating John,

What is your estimate on a percentage basis of the likelihood of encountering such an arrangement in a DAC?

w

[wakibaki]

1.  The input node will not remain at zero volts if a high slew is forced in.  The output of the monolithic amp will reflect the input error times some gain, and the R2 current will rise at some rate defined by the inductance.  The net result will be an output spike that will eventually drop to the circuit gain times the input. Reduce the inductance of R2, this spike will reduce.

Yeah, the circuit (with R2 inductor) is an inverting differentiator, differentiate a square wave you get a dirac delta function. (Why would you?) You might, just, conceivably put a ferrite bead in there to kill oscillation, but it'll have a low Q, you'll have to keep the inductance tiny in relation to the resistance to keep the output waveshape, which'll mean no significant effect on the input match, and anyway there are so many better solutions. If it's a real inductor with some Q, the resistance will be tiny so the gain will be -R2/R1, so if R1 is any size at all the gain will be less than 1. Maybe .01? Less? This just isn't useful.

2.  At the onset, the amp input node is not zero because of R2 inductance..now, you have an amp with an input that is too large...there is a distinct possibility that the amp will behave badly as a result.  Yes, the output will go in the correct direction initially, but once it has reached the final value, will it stop or overshoot the mark?  If internal stages saturate as a result of overdrive on the input node, who knows.

3.  Knowing the amp has been "severely" overdriven as a result of high di/dt into R1, it may need some time to recover, time beyond it's normal bandwidth.

It's a real world circuit not an ideal differentiator, what do you expect? You're going to get a useless output. You made R2 an inductor.

4.  Any reflection off the input node will have a rapid rise, as pat provided pics of.  If the driver "box" reflects that back to the receiver, the input node will immediately pass the spike through to the output...exactly as it did the initial.

Yeah, if the driver "box" reflects that back to the receiver. If the amplifier has recovered from being "severely" overdriven. So what? We're talking about the reflected wave artifact and the edges of the incident wave, not the output. And anyway, it'll still be one roundtrip time behind. Since they're very narrow pulses, they'll only coincide if the roundtrip time is one bit time (long pulse or short pulse, depending).

Where will that second spike be in relation to the initial waveform?

In exactly the same relation any reflection having made a round trip in that length of cable would be. The reflection is generated when the incident wave reaches the input. It's not as though having an inductor in the feedback loop delays it (or magically shunts it forward in time). It returns one roundtrip time after the edge that caused it. The decay is extremely quick because the output pulse duration is short, the disturbance on the input is brief, it lasts the propagation delay of the amp plus the pulse width.

w

[jneutron]

Fascinating John,

What is your estimate on a percentage basis of the likelihood of encountering such an arrangement in a DAC?
w

Hmm..for the linked example to diyhifi...posted by joseph K.....100%

here it is, the response of a video amp, with feedback, when "pinged" with a TDR.
That is, the same way like Pat has done it here with that Dac.
(For those who don't have access to the Diyhifi site - which is a pity  )
It's a little bit different, because this TDR shot contains also the effect of the saturated amplifier.
Without that it would "relax" back all the way to the baseline.

What is your estimate of the percentage of DACs being pinged by a TDR setup..the one you've analyzed the waveforms of?

Cheers, John

[jneutron]

Thanks for your post. I understand that often my style is confrontational, but I do make an effort to moderate my tone.

More effort would be appreciated..thank you.  Your understanding of t-lines is better than most, it would be a pity if your knowledge were lost or rejected as a result of your attitude.

Obviously this is easy to say but I am capable of changing my position and apologising.

Noted.  It is however, easier to not have to apologize.

OK, this is the basic setup we are discussing. Sometimes the DAC is removed and the coax is left open circuit to demonstrate the reflection coefficient of +1.
The yellow arrow on the LHS indicates the rising edge of a pulse. The pulse duration is 7 divisions or 350nS, corresponding to a long SPDIF pulse (@ 44k1). The orange circle surrounds a feature which looks like a thorn, the rising edge of which is ~50nS after the rising edge of the pulse. This feature is the (inductive) reflection from the DAC. The cable is ~5 metres long, and this reflection is as a result of the pulse edge (on the left) having travelled to the DAC and back. This is a roundtrip, there has been one reflection.

There is no question of this being an 'active' reflection, it's shape is entirely characteristic of a passive inductive termination, if it were active one would expect to see a change in level on the RHS when the output has settled.

Good.  You at least concur with my initial statement that this is a passive reflection, not an active one. 

However, an active node reflection will also have a different downslope to the reflection, consistent with the active circuit attempting to bring the node (and consequent node error voltage) back to zero.

This is anyway entirely irrelevant, what is of concern is the duration.

No.  As you so nicely pointed out (almost accurately), the output signal of a driven active circuit will not be a simple linear response to the input.  It is not an inverting differentiator.  It will be an inverted square wave output for the low frequency components, but the leading edge of the wave will have a spike consistent with the circuit attempting to drive the input node through the inductor. This is not a dirac function...

The spikes pass through without delay.  A returned reflection will do exactly that.  Remember your calc of a 10 nSec double transit with a 3 to 4 rise??   Will the active circuit have settled in 10 nSec?

While your t-line analysis of the passive setup is good, remember...an active termination will not act the same..

I consider it unfair and self-indulgent to inflict suggestions about RF attenuators on an unsuspecting public without taking the time to make these facts clear. How many people are there out there running 16/44k1 on a 2 metre cable going 'doesn't it sound better with the attenuators'. Don't tell me it won't happen.

w

I would agree in principle with your assessment..however, you have in fact done the same.  You have considered only one very specific analysis of a very specific system (with ferrite beads no less) and generalized it to the entire universe.

Reality can be somewhat different.  For you to not consider this is how you say, "unfair and self indulgent".

BTW..your verbage on T-line is very good.  Thanks for the discussion.. I rarely see this level.

Cheers, John

[jneutron]

Yeah, the circuit (with R2 inductor) is an inverting differentiator, differentiate a square wave you get a dirac delta function.

Uh, no.  If a low slew square is input, there will be a low slew inverted square output.  Once the input slew exceeds the ability of the active circuit to keep the input node at zero, the output will start to demonstrate a leading edge spike and a trailoff.  Build a circuit and look at the output to confirm this..

(Why would you?) You might, just, conceivably put a ferrite bead in there to kill oscillation, but it'll have a low Q, you'll have to keep the inductance tiny in relation to the resistance to keep the output waveshape, which'll mean no significant effect on the input match,

What oscillations?

and anyway there are so many better solutions. If it's a real inductor with some Q, the resistance will be tiny so the gain will be -R2/R1, so if R1 is any size at all the gain will be less than 1. Maybe .01? Less? This just isn't useful.

Please remain focussed.  The discussion is about a real world circuit with a real world resistor, and fabricated on a real world PC board.  The inductance being discussed is not a passive element added to the circuit, it is the parasitic inductance of the loop path the current has to take to zero the input node.

It's a real world circuit not an ideal differentiator, what do you expect? You're going to get a useless output. You made R2 an inductor.

Um...you called it (incorrectly) an ideal differentiator, I didn't.  And I did not make R2 an inductor.  It has a parasitic inductance, and the PC board does as well..

Yeah, if the driver "box" reflects that back to the receiver. If the amplifier has recovered from being "severely" overdriven. So what? We're talking about the reflected wave artifact and the edges of the incident wave, not the output.

It has been clear from the jump that you are considering everything but the output.  Unfortunately, everybody you are arguing with is more worried about the output, as that is the entire purpose of the equipment.  Not too many people build this equipment to look at pretty waveforms...perhaps you do (I know I do)   

It is important that you understand more than what you've analyzed..perhaps you should go back and re-read all my posts..you have the talent to understand what I've posted, you just need to take the time..

ps..I realize that last statement appears condescending, but I assure you that it was not meant to be.  I apologize for it looking that way, it is not my intent.

Cheers, John

[wakibaki]

Hmm..for the linked example to diyhifi...posted by joseph K.....100%

100% of one example.

And where is the spike followed by the trailoff? The picture shows a resistive response moderated by the amplifiers response time, not an inductive response.

Are these your words?

Pat's scope is clearly more consistent with an inductive termination, whereas your scope doesn't have the exponential decay.

w

[wakibaki]

The only reason that you see that hump in JosephK's picture is because it's done on a TDR. TDRs have a very fast rise time, that's how they achieve their resolution. The TDR's rise time exceeds even that of the video amplifier, which has a respectable slew rate. Consequently the voltage at the input node rises faster than the amplifier can compensate by feeding back through R2.

If, however, the video amplifier were hit by the ~6nS typical rise time of an HCMOS chip, this hump would be barely visible, if at all, because the amplifier would be able to maintain the voltage at the inverting input close to 0V. N'est ce pas?

Trivially, the stray inductance in the feedback loop is insignificant

w

Oui, c'est vrai.

[jneutron]

100% of one example.

And where is the spike followed by the trailoff? The picture shows a resistive response moderated by the amplifiers response time, not an inductive response.

Please read the posts very carefully.  Go back, and do so slowly.

Now, look at the pictures.  They are all from the input side.  None are from the side after the amp.

You are missing far too much of the detail.  Slow down.

Cheers, John