Hood on Negative Feedback
Jack Crenshaw
Hood, author of "Valve and Transistor Audio Amplifiers," has to say:
"For good audio performance, in transformer coupled amplifiers, it is
_NECESSARY_ [emphasis mine] to employ negative feedback (NFB) in the
amplifier design and, in view of the distortion that can arise in the
loudspeaker coupling transformer, it is _ESSENTIAL_ [ditto] that this
component is also included in the NFB loop."
"Simple NFB can reduce the extent of any waveform distortion or noise
introduced by the amplifier block, and can help to make the frequency
response of the system more uniform."
"The distortion of the output signal will be reduced in direct
proportion to as the value of B*A increases."
"Another advantage of NFB is that it can improve the accuracy of
rendition of transient signals, in that any difference between the
input signal, Ei, and that which is fed back, B*Eo, will generate an
error voltage which will be amplified by Ao and subtracted from the
signal output."
"All in all, an amplifier block employing negative feedback, when this
is applied in a properly designed system, will be substantially superior
to that of a similar system without NFB. The fact that claims are
sometimes made that the performance of an amplifier without NFB is
superior to that of one using this technique seems to be due principally
to the lack of design skills of those in respect of whose designs this
claim is made."
Hood goes on to talk about gain margin and phase margin, and the need to
have both. He emphasizes that, in the real world, with complicated
signals and anything but ideal speaker impedance loads, a goodly margin
is necessary. He then goes on to say,
"Understandably, badly designed amplifiers will give unpleasant results,
and these results may worsen when NFB is applied. But the answer is not
to use reduced amounts of NFB, or no NFB at all, but to design them so
thaat they have greater loop stability margins, especially when used
with reactive loads. It is inconvenient that this is a region in which
the overall stability in operation of an amplifier is often in conflict
with the attempts of the designer to achieve the lowest practical level
of total harmonic distortion, when measured with a steady state signal
on a purely resistive load. For this purpose, a large amount of NFB is
often used , and the design of the circuit is then tailored to sustain
the loop gain at a high level up to the frequency at which it must fall
rapidly to reach 0 db before the loop phase approaches -180 degrees.
Circuits designed like this, though stable, often have very small gain
and phase margins."
To paraphrase: NFB can greatly improve the performance of a
well-designed amplifier, particularly if that amp is designed with NFB
in mind, in the first place. It can and will reduce distortion, reduce
hum and noise, increase bandwidth, and _IMPROVE_ _TRANSIENT_ _RESPONSE_,
all roughly in proportion to the amount of feedback.
On the other hand, it can't save a poorly designed amp, and one is
foolish to try to make it do so.
Jack
Wiz
>say:"For good audio performance, in transformer coupled amplifiers, it
>thiscomponent is also included in the NFB loop."
Jack, the below isn't a flame...simply pointing out that the quote from JLH
fails as proof of his thesis on logical grounds alone. You are simply pointing
out that JLH *believes* this to be so, not that it *is* so.
I would agree with the above statement completely if you substitute fo the
words "For good audio performance..." the words "For good performance when
propagating sine waves through the amplifier....."
As the quote goes on, JLH gets farther and farther from statements about which
he can * actually test* for, e.g., you then would need to substitute for
>"Simple NFB can reduce the extent of any waveform distortion or >noise
introduced by the amplifier block, and can help to make the >frequency response
of the system more uniform."
"Simple NFB can reduce the extent of any *repetitive* waveform distortion or
noise introduced by the amplifier block, and can help to make the frequency
response of the system more uniform."
and so on. Your post is far from establishing any positive or negative values
for or against NFB.
The bit about transient response improvement is particularly amusing in light
of later work that shows that the leading edge of transients is the bit that is
*most negatively* affected by NFB.
Lastly, he doesn't address the *good design* of amps without NFB. The proof of
the pudding is in the eating, not in the recipe...it is simply amazing what a
positive difference is made in well-designed amplifiers when the NFB loop is
disabled and the gain adjusted. People who aren't technicians of any stripe
simply listen with dropped jaw. But of course, those who *know better* aren't
likely to experience this, as they won't do the simple mods.
Cheers/Carron
"LABOR SVGIT"
Henry Pasternack
>"Simple NFB can reduce the extent of any *repetitive* waveform distortion
or
>noise introduced by the amplifier block, and can help to make the frequency
>response of the system more uniform."
Not true. Feedback definitely reduces noise, and noise by definition, is
not repetitive. Feedback works regardless of the signal, periodic or
aperiodic.
>The bit about transient response improvement is particularly amusing in
light
>of later work that shows that the leading edge of transients is the bit
that is
>*most negatively* affected by NFB.
Again, not true. Negative feedback improves the bandwidth of an
amplifier,
and therefore its transient response. Feedback amplifiers need not ring.
>Lastly, he doesn't address the *good design* of amps without NFB. The proof
of
>the pudding is in the eating, not in the recipe...it is simply amazing what
a
>positive difference is made in well-designed amplifiers when the NFB loop
is
>disabled and the gain adjusted. People who aren't technicians of any stripe
>simply listen with dropped jaw. But of course, those who *know better*
aren't
>likely to experience this, as they won't do the simple mods.
"Positive difference" is a completely subjective thing. In amps that are
already
extremely linear, adding feedback may take away a subjectively pleasing
"tubey"
warmth and sense of space. The resulting sound will be more accurate, but
less
euphonic to some listeners. In other well-designed amplifers (say, typical
EL34
ultralinear types), removing feedback gives an interesting but ultimately
too colored
sound. In many other cases of supposedly "well-designed" amplifiers,
feedback
has made the amplifier unstable and the result is harshness and grain. I've
heard
this myself on many occasions. Sometimes it's quite subtle, but fixing the
problem is a jaw-dropping revelation.
Feedback is a powerful tool, and a double-edged sword. Most
anti-feedback
arguments are based on technical misunderstandings or careless subjective
generalizations.
-Henry
Jack Crenshaw
> Jack, the below isn't a flame...simply pointing out that the quote from JLH
> fails as proof of his thesis on logical grounds alone. You are simply pointing
> out that JLH *believes* this to be so, not that it *is* so.
Ok, Wiz, I understand that it's not a flame, and to be honest, I didn't really
expect to post such a post in this NG, without drawing some flames. I hasten to
add that my intent was not simply to stir up a hornet's nest, but to introduce some
badly-needed (my opinion) engineering data from an expert, into the discussion.
Let me try to take your comments one at a time, starting with the above.
You say, "simply pointing out that the quote from JLH fails as proof of his thesis
on logical grounds alone. You are simply pointing out that JLH *believes* this to
be so, not that it *is* so."
I suppose, in a sense, what you say is quite true. Most people don't write things
in books because they think they're false. I'm sure that Hood does indeed "believe"
what he says. But there's more to it than that. He believes what he says on the
basis of fifty years of experience with vacuum-tube amplifiers, practical
experience gained at a time when most of us "old-timers" were still learning to
ride our bikes. And I do think you're misrepresenting Hood's position if you
believe that he states his position without backing it up. I quoted what he said
in his book because I thought it captured the gist of his position. I didn't quote
the backup data because (a) to do so, I'd virtually have to quote the book, and (b)
I'd need to copy the equations as well. Suffice it to say that Hood gives
_COMPELLING_ facts to back up his statements. Don't let the fact that I didn't
quote them mislead you into thinking they aren't there. I would urge you to read
the book for yourself.
An analogy might be in order here. If I'm an attorney in court, and I call J.H.
Hood as an expert witness, I don't need to ask him to prove, with equations based
on circuit theory, the conclusions he draws. I only need to prove that he _IS_ an
expert witness. Then I can ask my question, and have him answer it as an expert.
The quotes I gave you are his answer. Is there any question that he is an expert?
By way of background and explanation for the "NFB is essential" statement, Hood
makes the specific point that the output transformer is an imperfect device. He
discusses, at some considerable length, the issues of core magnetization,
saturation, hysteresis effects, the importance of choosing the right core
materials, winding techniques, etc., etc. Because of this, he argues that it's
impossible to get HD down in the 1% level, or below, without having NFB that
includes the output transformer. His book discusses quite a number of attempts
various people have made to get around this step, and the varying degrees of
success they found with them.
> I would agree with the above statement completely if you substitute fo the
> words "For good audio performance..." the words "For good performance when
> propagating sine waves through the amplifier....."
That is such a gross trivialization of the point, and an oversimplification of
Hood's position, that I hardly know where to begin to answer it. It seems
abundantly obvious to me that an expert of 50 years doesn't get to be an expert by
sitting around watching sine waves on a scope, and never bothering to actually
_LISTEN_ to an amplifier. I'm surprised you would seriously consider making such a
suggestion. That position is is definitely _NOT_ Hood's -- in fact, it is the very
antithesis of his position -- and I thought my quotes made that crystal clear. He
is very much aware that both the output transformer and the loudspeaker are
anything but ideal devices. That's the reason for the whole discussion about phase
and gain stability margins. A paragraph which I didn't quote, for sake of brevity,
might help:
"The problem in audio is that the phase shift within an amplifier will be
influenced by the characteristics of the load applied to it, as well as by the
signal frequency, and probably also the amplitude of the signal. Unfortunately,
all of these things are continually varying in any real-life audio system, so any
system without an adequate margin of stability may be flickering into and out of
the regions of near or actual instability for the whole of the time, and this will
lead to the modification of the general waveshape, due to instantaneous changes of
gain, or, at the worst, to the generation of bursts of oscillation, superimposed on
the wanted signal. It is difficult to generate suitable input test signals to
reveal this problem, and it is also hard to simulate in the laboratory appropriate
awkward loads to allow examination of these effects."
Now, tell the truth: Does this _REALLY_ sound to you like someone who is only
concerned with sine waves?
To dispel any last vestiges of doubt that might remain, please indulge me long
enough to let me quote the end of the next paragraph one more time:
"It is inconvenient that this is a region in which
the overall stability in operation of an amplifier is often in conflict
with the attempts of the designer to achieve the lowest practical level
of total harmonic distortion, when measured with a steady state signal
on a purely resistive load. For this purpose, a large amount of NFB is
often used , and the design of the circuit is then tailored to sustain
the loop gain at a high level up to the frequency at which it must fall
rapidly to reach 0 db before the loop phase approaches -180 degrees.
Circuits designed like this, though stable, often have very small gain
and phase margins."
Translation: Hood is decrying the common practice of doing _PRECISELY_ what you
accuse him of: Tuning the amplifier so that it looks good when amplifying sine
waves, without regard to adequate stability margins.
As long as we're on the subject, I might as well bring up here something Hood also
discusses at length in his book, and something I also have mentioned here before:
The use of Nyquist diagrams to tune the open loop response. This is relevant
because a properly designed feedback amp, in both Hood's view and my own, is _NOT_
the same as a properly designed zero-FB amp. You don't just make the amp as
wideband as possible, then add feedback. Neither do you make the amp haphazardly
narrow-band, counting on feedback to save your buns.
The reason is that good stability demands that the open-loop phase shift be modest
before the phase shifts by 180 degrees. Hence the emphasis on Nyquist diagrams.
The best and most effective way to get a good foundation design for a feedback amp
is to design it to be low-distortion and wideband _WITHOUT_ feedback. Then you
deliberately introduce simple, first-order bandwidth-limiting elements. The reason
is that the open-loop response now has a simple, 90-degree shift at each end as the
gain goes to zero. Now you can add feedback around that amp to your heart's
content, and still maintain good stability margin.
> As the quote goes on, JLH gets farther and farther from statements about which
> he can * actually test* for, e.g., you then would need to substitute for
>
> >"Simple NFB can reduce the extent of any waveform distortion or >noise
> introduced by the amplifier block, and can help to make the >frequency response
> of the system more uniform."
>
> "Simple NFB can reduce the extent of any *repetitive* waveform distortion or
> noise introduced by the amplifier block, and can help to make the frequency
> response of the system more uniform."
> and so on. Your post is far from establishing any positive or negative values
> for or against NFB.
Wiz, these statements are not a matter of subjective opinion, and therefore not a
matter open for debate. They are backed up by hard and rigorous mathematics, which
any junior-level student of electronics learns ad nauseum. I didn't reproduce
Hood's equations for reasons already stated, but also because they can be found in
any undergraduate textbook on the subject. They are _NOT_ subjective.
> The bit about transient response improvement is particularly amusing in light
> of later work that shows that the leading edge of transients is the bit that is
> *most negatively* affected by NFB.
I get a little frustrated hearing debated, over and over, the same things EE
students learn and take for granted, early in their careers. The response to a
step function, the ultimate transient, is as familiar to a EE as his favorite soup
bowl. To suggest that there's something about transient response in vacuum tube
amplifiers that's managed to escape the best minds in the business for the last 50
years is a bit much.
If one tries to explain to a layman how a feedback loop works, and how the summing
junction has to "see" a difference at the output before it can call for a
correction, I can see how you could get into the mindset that there is going to be
some kind of spike at the output before the feedback loop has a chance to react. I
defy you, however, to show me that spike on the highest frequency scope you can
beg, borrow, or steal.
I'm a big advocate of square-wave testing, which is about as close to a transient
as anyone could ask for. Anyone who's ever compared the square-wave performance of
an amp with and without NFB should have no doubt as to its ability to improve
transient response.
These days, all circuits that aren't digital use op amps heavily. Op amps
typically have open-loop gains of 100,000 (100 dB) or more. They _MUST_ use NFB to
keep them from hitting the rail; they are useless any other way. For you to
suggest that an amplifier with NFB will have trouble handling transients is like
suggesting that a step function input of 1 volt is going to produce a brief output
of 100,000 volts. It just ain't gonna happen.
> Lastly, he doesn't address the *good design* of amps without NFB.
Gimme a break! He _DOES_ address it!!! I just didn't quote the whole *#@$#% book!
> The proof of the pudding is in the eating, not in the recipe...it is simply
> amazing what a
> positive difference is made in well-designed amplifiers when the NFB loop is
> disabled and the gain adjusted. People who aren't technicians of any stripe
> simply listen with dropped jaw. But of course, those who *know better* aren't
> likely to experience this, as they won't do the simple mods.
Uh, huh. And then there are the people who are certain that speaker cable that
costs $50 per foot makes a big difference, too.
I must confess that I am not a Golden Ear. Never have been. I can't do an A/B test
on two different 1 uf capacitors in a circuit, and tell you that one is more
"rich," while the other is a little "strident." One speaker cable sounds about the
same as any other to me (at least, anything bigger than ordinary zip cord. Come to
think of it, zip cord sounds fine, too). I use Monster Cable stereo connections to
my stereo, gold-plated connectors and all, but to tell the truth I don't sense any
particular greater "fullness" or "detail" compared to my old el cheapo cables.
On the other hand, what I _CAN_ hear is the difference between an amp that's well
designed and one that's not. If, for example, I have an amplifier that's
exhibiting parasitic oscillations, I can hear that. I can't listen and say, "That
amp is oscillating parasitically at 127,000 Hz." But I can say, "That amp sounds
bad."
I could tinker with resistors and capacitors in that amp until I'm blue in the
face, and I'd _NEVER_ be able to hit on the solution, just by ear. The best I
might hope for is, "Now it still sounds bad, but in a different way."
So what I do is I take the amp into my lab, hook it up to scope, drive it with
square waves, and spot the oscillation. I fix it. Then I hook it back up, and
say, "There, that sounds better."
In short, I am not much for subjective, ear tests. There's too much room for ... um
... subjectivity. What I _HAVE_ learned is that if I do things using sound
engineering design, use instruments to trouble-shoot problems, and use a
combination of instruments and design theory to optimize performance, I end up with
a system that sounds good.
I wish, along with Hood, that I had a method that would allow me to test an
amplifier in an environment more like the real world in which it's going to be
used. I don't have that method. If you do, please don't fail to share it. Until
then, I'm going to stick to the things that we engineers have been using
effectively for the last 50 years: Sound theory, good mathematics, and good test
equipment.
Jack
Kevin Olson
impossible to get HD down in the 1% level, or below, without having NFB that
includes the output transformer.
*And those dern musical instruments INSIST on producing HD far in excess of
1% level...
I wish, along with Hood, that I had a method that would allow me to test an
amplifier in an environment more like the real world in which it's going to
be
used. I don't have that method. If you do, please don't fail to share it.
*Errrrahhhh...listen to it?
Until
then, I'm going to stick to the things that we engineers have been using
effectively for the last 50 years: Sound theory, good mathematics, and good
test
equipment.
*Good for you! Bye!
Jack Crenshaw
> Because of this, he argues that it's
> impossible to get HD down in the 1% level, or below, without having NFB that
> includes the output transformer.
>
> *And those dern musical instruments INSIST on producing HD far in excess of
> 1% level...
That is untrue. Musical instruments do not produce distortion. They produce
_HARMONICS_. Big difference. The challenge of a reproduction system is to
reproduce the sound the instruments make, harmonics at all, in precisely the
same proportions and the same phase, even when many other instruments are
producing their own sounds and their own harmonics. The system is supposed to
_REPRODUCE_ the sound of a full orchestra, not embellish it.
> I wish, along with Hood, that I had a method that would allow me to test an
> amplifier in an environment more like the real world in which it's going to
> be
> used. I don't have that method. If you do, please don't fail to share it.
>
> *Errrrahhhh...listen to it?
Doesn't work. As I said in my last post, you can listen to a system and say,
"That one sounds good," or "that one sounds bad." Listening doesn't give you
the slightest _CLUE_ as to what makes the good one sound good, or the bad one
sound bad. Unless you're into blindly copying the good amplifier, hoping that
you managed to somehow capture that elusive "something" without knowing what it
was, you can only keep buying the ones that sound good, and selling the ones
that don't. This may work for Stradivarii, but it's a terrible way to deal with
audio amplifiers.
Things get much more complicated when you consider the fact that one amplifier
might sound good in one room, with one sound source and one speaker system, and
sound terrible in another room, with different transducers. What are you going
to do _THEN_? A/B test the whole universe?
Those of us who design amplifiers, whether for fun or for profit, need something
a little more concrete to hang our hats on.
> Until then, I'm going to stick to the things that we engineers have been using
>
> effectively for the last 50 years: Sound theory, good mathematics, and good
> test equipment.
>
> *Good for you! Bye!
What, exactly, is _THAT_ supposed to mean? Are you going somewhere?
Jack
Ken Gilbert
obviously you have that book. look on page 81. look at fig 5.8. tell
me what it means to you.
it means to ME that if i am to use NFB, i'd better use more than 15dB
of it, so at least i could reap the benefits across ALL orders of THD.
levels less than 15 dB produce GREATER high order distortion as the
feedback factor is increased.
this also tells us there is a good deal of shit going on behind the
scenes.
though i have not had the opportunity to read it myself, i have heard
that someone named mati otala did some seminal work on the study of
NFB, and concluded that the (necessarily) imperfect nature of the loop
itself tended to push harmonic distortion up in order, even as it
reduced THD. thus the amp will measure better, yet could easily sound
worse. one only needs look at steve bench's web page (and the "look"
of distortion) to see how much uglier the upper order distortion looks
when compared to say 2nd.
there is also the possibility of the overall THD of an amplifier to be
LESS than the THD of its constituant stages, when taken one by one.
this is due to the nature of the predominant 2nd harmonic introduced by
your typical triode, and the phase inversion that takes place at each
stage. fire off an email to Jean-Michel Le Cleac'h
(lecl...@cgi.ensmp.fr) and ask him what the THD of his ALL TUBE, ZNFB
amplifier (name: shabda) is. i'm betting that you would be completely
blown away.
i would also like to include this thought-provoking discussion
excerpted from a mailing list i subscribe to. it's rather long, but
good reading. enjoy.
kg
[begin quoted message]
Date: Fri, 29 Sep 2000 04:12:41 +0200 (CEST)
From: Thomas Dunker
To: Remco Stoutjesdijk
cc: "Joe-Net (E-mail)"
Subject: Re: [JN] harmonic distortion patterns filosofies - thoughts?
On Wed, 27 Sep 2000, Remco Stoutjesdijk wrote:
> [ me ]
> Ha! My thoughts exactly! There are not so many people who dare admit
it. For
> one, it's like admitting 'With my trained ears, I like a deteriotated
> (distorted) signal better' and also 'my design can't be all that good,
> because it sounds marvellous' :-)
>
> IMO, this is the one of the reasons why modern amps sound
unpleasing...
> 0.00005% thd...
> [ and only optimised for THD, I meant there ]
>
> I've already tried to make something like this into a research topic
that I
> could explore at the university. For instance, something like
building an
> amp with pots for adjustable H2 and H3 and having people listen to
different
> THD patterns. I think that patterns where the odd harmonics are
larger than
> even (most PP stuff) appear un-natural to the brain. Maybe every next
> harmonic needs to be smaller than the previous one in order to make
sense as
> a natural signal (can you think of any natural signal with another
harmonic
> signature?). The SET mob would agree here. Of course noone backed up
my
> research topic and more importantly, noone wanted to fund it... :(
Hello, just some thoughts:
When we talk about the distortion we measured and the sound we heard,
we
invariably fail to consider a few important things.
First, the sound we hear comes out of the speakers, which has added
harmonics, IMD, dynamic compression, phase distortion etc. etc. and
makes
the relative relevance of simple THD measurements on amps rather
questionable. Speakers and amps, by virtue of their respective
inherent distortion characteristics, can complement each other or make
both show all their worst depending on the accidental amp-speaker
combination.
Second, the distortion we measure at the output of the amp is not the
same as the distortion we would be able to measure in the CURRENT
delivered by the amp unless the amp worked into a true ohmic resistance.
Speakers, to a greater or lesser extent, always massively distort the
current that they depend on for their own operation due to reactive
impedance and dynamic impedance shifts. A perfectly distortionless amp
measured according to standard amp THD measuring schemes would still
result in distorted current in the speaker. I=V/z. The proportionality
of
voltage and current is immediately lost when a speaker is attached.
The "by the book" THD characteristics of an amp that would always
deliver
the correct CURRENT are anyone's guess. As Kurt and people like Matti
Otala have proved, a "ultra low distortion" amp may very well be
responsible for the highest current distortion. Which brings us to the
third point:
If the amp employs negative voltage feedback (trying to maintain zero
output Z and undistorted voltage amplitude at output) to achieve its
very
low distortion (of voltage), chances are that the disagreement between
what the amp supplies and what the speaker ideally should get becomes
unduly gross and problematic. It is not that NFB doesn't work as
intended,
it is about what you're asking it to do for you. "Give the speaker
undistorted voltage" and you'll get something at least close to that.
Hook a scope to the thing and wow, linear - all very fine in preamps,
line amps and anything not involving reactive loads.
The current in a speaker voice coil translates (with some
considerable linearity problems) into force accelerating the diaphragm
and
consequently the air and that's what we hear. It ought to be obvious
then,
that a speaker can not be more linear than the distortion of the current
driving it permits it to be. Negative voltage feedback makes the amp
entirely blind to the very thing that matters in the speaker, that the
current is not distorted. In this sense, therefore, negative voltage FB
in
an amp driving a real speaker is directly counterproductive in terms of
nonlinear distortion of the signal that energizes the voice coil -
again,
the current. Negative voltage feedback makes sure that any speaker
always
get the "right" voltage, and the wrong current.
Even tubeheads are sometimes very hard to convince that it may not be
the
tubes that are entirely responsible for the sound, but the way in which
tube amps are built - by contrast to most SS amps. You can build tube
amps
that measure - and sound - just like transistor amps if they are doing
the
same things to the signal. I don't claim that the true is equally
possible, but I insist that the most interesting differences is in the
basic philosophy of power amplifier design AS IT RELATES TO WHAT KIND
OF A
LOAD A SPEAKER ACTUALLY IS.
To illustrate the validity of these claims, one could repeat one of
Otala's famous experiments:
One pair of power amps drives a pair of resistive loads. The
voltage signal at their output is stepped down and recorded on tape (or
whatever) or monitored on a reference system or headphones.
A second pair of amps, identical to the first, at same power levels,
drives a pair of any old speakers. The voltage signal at the output of
these amps is similarly stepped down and compared by blind listening
test
to the signal from the first pair of amps.
I have not performed this experiment myself, but the difference is said
to be enormous. One could juggle variables and make similar experiments
with different amps on a reference pair of speakers, record signals and
compare, or different speakers on the same amp, record signals, listen
and
compare. The result would be:
No two identical amps would measure nor sound the same regardless of
whether they drive a resistive load or a speaker.
No two identical amps would produce the same distortion (audible and
measurable) driving considerably different speakers.
Note that this is the first time we listen to the voltage signal at the
output of the amp (which we normally measure and attempt to relate to
the
sound coming out of the speakers), and what we do is listen to the
actual signal that is fed back to the input if the amp employs global
voltage feedback. Now, if the closed loop output signal of an NFB amp
is affected by the presence of a speaker (it can easily be, with back-
EMF
currents on the order of several amperes), and in a detrimentary
fashion,
the NFB is still just doing its job, comparing the input and output
signal. BUT with a speaker attached, distortion coming from outside the
feedback loop (speaker's back-EMF, reactance of crossovers, conceivably
cable capacitances, voltage drop on large currents etc...) is inserted
into the loop at the output of the power amp.
If one repeated the above experiments tapping the current at the output
of the amps rather than the voltage and then amplified it, the
differences
would become grossly accentuated and one would literally be able to hear
the difference between conventionally measured amp distortion and the
actual distortion when the amp is driving the speaker.
In an amp that does not employ global NFB, the amp-speaker interaction
will be restricted to the output stage of the amp. The distortion here
can
be visualized as an ellipse of dynamically varying shape superimposed on
the transfer characteristic of the output stage.
I have some times been speculating about the specialness of triodes in
this regard, since they have a unique current/voltage relationship not
shared by pentodes, which looks interesting when you begin to draw load
ellipses on the plate curves. Particularly, triodes with remarkable
Ip/Vp linearity (well, more linear curves) AND remarkably parallel plate
curves (211, AD1, RE 604) have a particular reputation for unparalleled
"clarity" of sound. I have not looked deep into this, but if anything
makes triodes unique to any other amplification device, it's the Ip/Vp
relationship. These differences may not matter as much if the device is
used for voltage amplification into resistive loads, in which case the
output voltage is very predictable. When you begin to look at current at
various phase angles, as a function of input voltage (to the output
stage)
The situation seems a little different.
Think about it: The output stage of a power amp is the moment of truth
for the signal. This stage drives a load more unpredictable and hostile
than anything else in the whole signal chain. By comparison, pure
voltage
amplification into cables, potmeters or voltage dividers is a breeze.
But you can't cheat your way to "undistorted" driving current for the
speaker.
Unfortunately for us all, some smartass designed
acousto-mechanically undamped resonances into a class of speakers that
would soon dominate the market. The bass reflex speaker. The whole
matter
of damping became a matter of corner cutting. A little dash of
resonance
right THERE makes the speaker play lower frequencies in spite of its
modest size. Only in microphones, phono carts and speakers are
resonances
used on purpose towards some more or less agreed upon end.
Now, the amp's job is twofold: It must have high "damping factor" (in
absurdum, considering VC resistance) - "zero" output Z, to make a bass
reflex speaker work predictably at resonance, AND supply current to the
speaker. It is obvious that you can't have zero ohms output resistance
and undistorted current when you're driving a reactive load. By
demanding
from the speaker that it always shorts out the back-EMF from the speaker
(peaking at resonant frequency), you must have low output impedance.
In other words, a "constant voltage" (Vin prop. to Vout) amp, a reliable
source of distorted current to any real world speaker.
To drive a speaker properly, you need amps and speakers that don't ask
impossible things from each other. If the amp has to try and always
supply
an output current proportional to its input voltage signal (a linear V-
>I
converter with power gain), it can do so more or less successfully with
or
without resorting to NFB, but it can not simultaneously have "zero"
output
impedance. With a reactive load, that's a contradiction of terms.
The speaker therefore, must be designed in agreement with the amp,
which
must be designed in agreement with the inner workings of the speaker,
and
you get a kind of closed loop.
Here is where I get a little upset with people specializing in power
amps
or specializing in speakers but never both. This specialization is the
very problem which complicates all progressive work.
If you want to properly drive a speaker, you first have to have the
proper speaker. With conventional electrodynamic transducers, the first
thing to know is that they are current controlled and not voltage
controlled. That says something about what the amp must be like. It also
says what the amp can NOT be like, and that too must be considered when
designing the speaker. One of the things you can (well, you can...) not
do
is dump massive amounts of back-EMF into the output terminals of the
amp
through crossovers, cables and everything and expect this to damp the
resonance properly. The current thus developed can damp the woofer cone
to
the extent that the current is linearly converted to force for braking.
As if the current was linearly converted to force in the first place...
Now, let massive non-signal currents surge back through the speaker
cables
and crossover and watch the shit hit the fan at the amp's output and IN
the amp.
In reality, for a number of boring reasons, no matter how good your
"damping factor", the voice coil resistance (and thus its temperature)
and
the linearity of the force factor, turns resonant damping into a a
matter
of signal related accident. The more violent the cone excursions, the
more
you need your damping, and the less of it you can expect to get. Measure
impulse response on your bas speakers at 0.1, 1 and 10 amp pulses
respectively and you get the idea.
The damping of the speaker's inherent resonances should be such that
it is independent of signal levels, to preserve the scale of macro- and
microdynamics and a sort of "dynamic coherence" if you will.
Now one will have to argue to what extent the resonances should be
damped. No argument, that a Qt of the whole system (amp-cables-speaker)
if it exceeds 0.5, results in a periodic overshoot on (bass) transients.
I have heard a few dipole bass systems that have this order of damping,
and their articulated and tight bass is addictive. It would be similarly
doable with a sealed enclosure of required size.
A periodic response is information added to the signal -> distortion!
Perhaps difficult to quantify, but extremely easy to hear.
One thing I forgot to mention: The back-EMF-into-low-Z-amp-output type
damping is literally a feedback loop, with a terrifying number of
nonlinearities, phase shifts, instabilities and stuff inside the loop.
It's called back-EMF, because it is something that comes BACK from the
speaker because of complex impedance. The greater the undamped impedance
peak, the greater the back-EMF. It would be better to get rid of the
whole
mess and let the amp control the speaker and not vice versa.
There are so many things that are so easy to forget about power amps
simply because they are used to drive speakers, that I am sure entire
books could be written about it. If we continue to measure our power
amps'
performance in terms of voltage amplification linearity, we can go on
forever looking for clues. If we're measuring the wrong things, and
maybe
even design electronics according to these measurements, we can not know
what we end up with. It's positively not that "you can't measure it,
but I
can hear it" - it's a matter of measuring the right things.
So you measure THD and frequency response into resistive dummy loads
when
the amp is on the bench, and measure frequency response and impedance
(at a single, and low power level) on speakers. The data you get say
absolutely nothing about how the amp and the speaker perform together,
and we should not expect them to.
The mentality has been "We build the amps, they build the speakers".
The consequence is that they had to agree about a few "standards" for
mutual assumptions about each other's projects. Like Ragnar Lian says,
"some idiot took the speaker out of the radio".
Paradoxically, the "high end" is where the division of disciplines is
at
its most total. This is mostly okay except for power amps and speakers
which really should have been built as integrated systems, based on an
enlightened and synergistic approach.
It's the right speakers that can make your amp come alive, and vice
versa. How this all works and all the challenges and potential for
improvement with a unified approach is the last frontier of audio to me.
The only similar field in audio is turntables, which also have complex
action-reaction and mutual influence problems beyond anything seen in
amplification electronics.
When we have the perfect amps and speakers, we still have the
room-speaker interaction to worry about...
The many digressions are necessary - everything IS connected. What is a
system without holistic, synergistic thinking?
And something to think about: What is it about TRIODES...? I don't for
a
second believe it's in the THD spectrum alone....
Thomas
------------------------------------------------------------------------
_/\_ Thomas Dunker \ The Horn Speaker Home Page:
/ \ P.O.Box 2811 \ http://invalid.ed.ntnu.no/~dunker/horns.html
| | 7002 Trondheim \
| | NORWAY \--\ "Those with head above water
\____/ dun...@omegav.ntnu.no \ see only the tip of the iceberg"
|||| phone: (+47)73911068 \ (Gene Dalby)
------------------------------------------------------------------------
[end quoted message]
--
I know that the twelve notes in each octave and the varieties
of rhythm offer me opportunities that all of human genius
will never exhaust. --Igor Stravinski
Sent via Deja.com
http://www.deja.com/
Henry Pasternack
>i would also like to include this thought-provoking discussion
>excerpted from a mailing list i subscribe to. it's rather long, but
>good reading. enjoy.
This is too long for me to respond to at the moment, Ken. But I
have some very serious problems with what Tom is claiming in the
post you quoted.
-Henry
Henry Pasternack
>excerpted from a mailing list i subscribe to.
>
No offense to you, Ken, I always appreciate your support. But
this article by Thomas Dunker is the biggest pile of unmitigated
horseshit I have read on an audio newsgroup in a long, long time.
It's just so maddeningly ill conceived I don't even know if have the
stomach to respond to it. I was arguing this crap six years ago
(probably with Thomas Dunker himself) on the SP mailing list and
got so fed up I quit the list for good.
I think one paragraph ought to be enough to debunk this whole
horrendously moronic missive. Loudspeakers are designed to
work properly when driven by a voltage source, possibly with a
small resistance in series. There is nothing mysterious about
"back EMF". Viewed from the speaker terminals, all of the
various electromechanical things going on in the driver translate
into an electrical rectance (inductance, capacitance, resistance)
seen by the amplifier. There's no other magic going on here.
An amplifier, feedback or not, that is stable and produces low
distortion into a worst-case reactive load will be happy driving a
real-world loudspeaker. An amplifier that is not content to drive
a reactive load (usually a capacitor) will likely have problems
driving speakers in the real world. There is no, I repeat, there
is NO, feedback loop between the coil of the speaker and the
amplifier input or output via "back EMF". The only feedback loop
is between the amplifier's output and input.
All of Thomas Dunker's techno-yammering should be summarily
tossed in the trash can. And if you run into him, please direct him
to this posting.
-Henry
TubeGarden
Waders on? CHECK!
Flamesuit secure? CHECK!
Hello HP!
Let me see if I understand what you are saying:
_____
>Viewed from the speaker terminals, all of the
>various electromechanical things going on in the driver translate
>into an electrical reactance (inductance, capacitance, resistance)
>seen by the amplifier.
So, the properties of the driver change over time, dependant on the previous
signal from the amplifier?
That is, the electrical reactance changes, after some elapsed time, due to the
signal from the amp?
______
>There's no other magic going on here.
Not, unless, perhaps, you are listening to music. :)
_____
>An amplifier, feedback or not, that is stable and produces low
>distortion into a worst-case reactive load will be happy driving a
>real-world loudspeaker.
The Mr. Rogers safe driver amp.
______
>An amplifier that is not content to drive
>a reactive load (usually a capacitor) will likely have problems
>driving speakers in the real world.
Driving Under the Influence of changing electrical reactance ... bad Bad
socially dysfunctional amp.
_____
>There is no, I repeat, there
>is NO, feedback loop between the coil of the speaker and the
>amplifier input or output via "back EMF".
Perhaps not :)
______
Could it be possible that the load on the output stage influences the operating
characteristics of that stage, and changes in the load will change the nature
of the amplification of the now present signal due to the preceding signal
causing changes in the speaker's Z, C and R?
Almost as if the earlier signal fed back, by translating into changes in
electrical reactance, through the speaker terminals, to have an effect on
amplification of the later signal?
Let us not quibble over terminology. "Back EMF" is possibly misleading.
Since the electrical reactance changes are due to the processing, or digesting,
of the earlier signal, let us call this process 'Speaker Farts".
______
To sum up:
"An amplifiers performance may be adversely affected by the stinking speakers".
Happy Ears!
Al B^}
Alan J. Marcy
Phoenix, AZ
PWC/mystic/Earhead
Jack Crenshaw
> jack, couple of points:
>
> obviously you have that book. look on page 81. look at fig 5.8. tell
> me what it means to you.
>
> it means to ME that if i am to use NFB, i'd better use more than 15dB
> of it, so at least i could reap the benefits across ALL orders of THD.
> levels less than 15 dB produce GREATER high order distortion as the
> feedback factor is increased.
It means the same thing to me. Not sure what the conditions were that
Baxandall used, other than the fact that it was a "simple FET amplifier."
Does that mean push-pull? Does it mean class A. It's unclear. Also,
Baxandall's designs tended _NOT_ to use feedback from the speaker winding
(see Figure 6.7). I did see the figure, was as disturbed as much as you by
it, but just don't know what to make of it. In the absence of other data, I
don't think I'm ready to make life decisions based on it.
> though i have not had the opportunity to read it myself, i have heard
> that someone named mati otala did some seminal work on the study of
> NFB, and concluded that the (necessarily) imperfect nature of the loop
> itself tended to push harmonic distortion up in order, even as it
> reduced THD. thus the amp will measure better, yet could easily sound
> worse. one only needs look at steve bench's web page (and the "look"
> of distortion) to see how much uglier the upper order distortion looks
> when compared to say 2nd.
I don't see how NFB can introduce new, higher order distortion, Baxandall or
no. There is one exeption, and that is if we are talking about driving the
amp beyond its capabilities. At high power levels, NFB will try to maintain
flat response by demanding more gain from the amp. If the amp can't
deliver, that's going to cause clipping. Maybe _THAT's_ how Baxandall got
his curves. Otherwise, I just can't see it.
In any case, I defy even the most golden ear to hear 7th harmonic in an amp
that measures 0.1%, _TOTAL_.
Jack
Jack Crenshaw
do agree that one should look at what's going on with a real speaker
connected, and most of us might be both appalled and surprised at what we
see. Not sure exactly what we _DO_ about it.
The writer is quite right that an amp with feedback is basically a
constant-voltage (high damping factor) device. That may well not be what
the speaker needs, although I would argue that high damping has a lot more
chance of being right, with most speakers, than low. Some amps, notably the
Heath W-6M, had a variable damping factor, precisely to allow the adaptation
of the amp to the speaker. This was at a time when bookshelf speakers like
the AR-1 were in vogue. They already had all the damping they needed, and
sounded like they were in a pillow when damped further. Some users went so
far as to insert a resistor between the amp and speaker, thereby putting
even more demands on the power-handling capabilities of the amp than the
horribly inefficient speaker did already.
Anyhow, I do agree that one must give a lot of consideration to the load
presented by the speaker. Why do we still tend to test with a resistive
load? Because it's all the variables we can handle. Otherwise you go nuts
trying to tweak every possible variable.
Jack
Kevin Olson
> Because of this, he argues that it's
> impossible to get HD down in the 1% level, or below, without having NFB
that
> includes the output transformer.
>
> *And those dern musical instruments INSIST on producing HD far in excess
of
> 1% level...
That is untrue. Musical instruments do not produce distortion. They
produce
_HARMONICS_. Big difference.
*Not at 'tall... If you play a musical instrument into a distortion analyzer
it says "tilt...
high distortion". Whether you call it distortion (cause you don't think it
should be there)
or Harmonics (cause you do) is semantics.
The challenge of a reproduction system is to
reproduce the sound the instruments make, harmonics at all, in precisely the
same proportions and the same phase, even when many other instruments are
producing their own sounds and their own harmonics. The system is supposed
to
_REPRODUCE_ the sound of a full orchestra, not embellish it.
*Whoa...you seem to be saying phase matters...there gonna take away your AES
membership if you're not careful! I would basically agree with this but my
point is that
considering the nature of the signals that the amplifier is called on to
err...amplify,
adding .5% of HD might be less harmful overall then some of the measures
taken to
overcome it. But not always. I have heard excellent low feedback amps, high
feedback
amps, SE, push pull and even (gasp) solid state, and bad examples of them
too. That's
why ya gotta listen (after all that IS the intended use of the device...)
> I wish, along with Hood, that I had a method that would allow me to test
an
> amplifier in an environment more like the real world in which it's going
to
> be
> used. I don't have that method. If you do, please don't fail to share
it.
>
> *Errrrahhhh...listen to it?
Doesn't work. As I said in my last post, you can listen to a system and
say,
"That one sounds good," or "that one sounds bad." Listening doesn't give
you
the slightest _CLUE_ as to what makes the good one sound good, or the bad
one
sound bad.
*From the viewpoint of the listener WHO CARES? If it sounds right, its good,
If it
don't it's bad. The designer, of course DOES care why, that's his job.
Here's an
example that may please you. The measurements are the feedback! You listen,
something's wrong. You measure to try and determine in a technical sense
what it
is. You work to improve whatever you find wrong. You listen to see if you
fixed the
problem. If you just blindly work to make needles go lower on test equipment
that's
just another kind of religion.
Unless you're into blindly copying the good amplifier, hoping that
you managed to somehow capture that elusive "something" without knowing what
it
was, you can only keep buying the ones that sound good, and selling the ones
that don't. This may work for Stradivarii, but it's a terrible way to deal
with
audio amplifiers.
*I suspect that is why most audiophiles own so many amps in their lifetime
(hi hi). BTW
are you blindly following Mr. Hood's opinions (just asking)? No, you seem
like a
thoughtful person and I am sure you have thought about them a lot and
established them
to your satisfaction. STILL doesn't mean it's the only was to design an
amp...
Things get much more complicated when you consider the fact that one
amplifier
might sound good in one room, with one sound source and one speaker system,
and
sound terrible in another room, with different transducers. What are you
going
to do _THEN_? A/B test the whole universe?
*Most people care only about their system in their room. I suspect that is
why
there is a whole universe of equipment available, rather than one agreed
upon system.
Those of us who design amplifiers, whether for fun or for profit, need
something
a little more concrete to hang our hats on.
*Sure. And measurements are invaluable to the designer who understands why
he made them. But the phenomenon precedes the measurement. Once upon a time,
I measured audio and video equipment for a major TV network (ok...CBS) in
their
network engineering lab. I rather enjoyed it! But numbers without context
meant
only a certain amount. After all, evaluating the numbers can be subjective,
too!
> Until then, I'm going to stick to the things that we engineers have been
using
>
> effectively for the last 50 years: Sound theory, good mathematics, and
good
> test equipment.
>
> *Good for you! Bye!
What, exactly, is _THAT_ supposed to mean? Are you going somewhere?
*I meant that it's good that you stick to your beliefs. Actually I had to go
watch
my new favorite team, anybody but the Giants errr....the Ravens win their
playoff
game. All my pre-game measurements proved that the Vikings would beat the
Giants, but unfortunately they played the game anyway.
Regards, Kevin
Henry Pasternack
distortion.
Although this is an important issue, it's not really what Dunker was trying
to
explain.
-Henry
TubeGarden
Jack Crenshaw
> Jack said:
> > Because of this, he argues that it's
> > impossible to get HD down in the 1% level, or below, without having NFB
> that
> > includes the output transformer.
> >
> > *And those dern musical instruments INSIST on producing HD far in excess
> of
> > 1% level...
>
> That is untrue. Musical instruments do not produce distortion. They
> produce
> _HARMONICS_. Big difference.
> *Not at 'tall... If you play a musical instrument into a distortion analyzer
> it says "tilt...
> high distortion". Whether you call it distortion (cause you don't think it
> should be there)
> or Harmonics (cause you do) is semantics.
Kevin, I think we are basically saying the same thing, but I'm not sure it's
only semantics where we differ. Anyone whose ever heard a violin knows that its
output is rich in harmonics. I suppose I could feed a sine wave signal through
a bad enough amplifier, and if I worked at it hard enough, I could make it sound
somewhat like a violin. But the only way I know of to get it right easily, is
to reproduce all the subtleties of the original violin, complete with all its
harmonics, in the right proportion and phase.
> *Whoa...you seem to be saying phase matters...there gonna take away your AES
> membership if you're not careful! I would basically agree with this but my
> point is that considering the nature of the signals that the amplifier is
> called on to
> err...amplify, adding .5% of HD might be less harmful overall then some of the
> measures
> taken to overcome it.
I do think phase matters. I've read all kinds of things that say that our ears
don't seem to be sensitive to phase, and I think that's true if you're sitting,
say, in a concert hall, and the sound source is a point source a long ways
away. On the other hand, our ears are incredibly good at determining the
direction to a source, and that's done mostly by very subtle differences in
phase (or, if you prefer, time delay) between the signals. I'm not prepared to
say that phase doesn't matter. Particularly in a complex signal, where phase
might be useful to separate instruments.
> But not always. I have heard excellent low feedback amps, high
> feedback amps, SE, push pull and even (gasp) solid state, and bad examples of
> them
> too. That's why ya gotta listen (after all that IS the intended use of the
> device...)
No arguments here. I agree 100% that listening is the point, and that if the
conflict comes down to what sounds best vs. what measures best, the "sounds
best" has to win every time. My only problem is how to _FIX_ the amp if it
sounds bad. Suppose you have two amps; one sounds bad, one sounds good. How are
you going to make the bad one sound like the good one? Without some
understanding of _WHY_ the bad one sounds bad, you're flying blind, basically
making changes by trial and error. The point I keep trying to make is that
I've never had much success, trying to improve sound by trial and error. Too
many trials, too much error. The only effective way I've found has been to
_UNDERSTAND_ why the bad one sounds bad (and the good one sounds good), and then
adjust the bad one so it doesn't do that anymore. That's the rationale I used,
awhile back, in the discussion of parasitic oscillation. You will not see it in
frequency response or HD specs. You may not even see it in IM specs. But if
you know that it can occur, and are looking for it, you can spot in in a flash
with an LF square wave test. Once spotted, the fix is a resistor change away.
In short, my suggestion is: Listen to spot a problem; analyze to figure out what
it is.
> *From the viewpoint of the listener WHO CARES? If it sounds right, its good,
> If it don't it's bad. The designer, of course DOES care why, that's his job.
Again, we agree. The listener doesn't care. He's got his good sounding
amplifier. He's happy. The designer is stuck with either reproducing the good
one for the rest of his life, or trying to figure out what went wrong with the
bad one. You can't just change one component in the bad one, and listen to it
again. You will be there for the rest of your life. You need some kind of
systematic approach, and that's where the instruments come in.
> *I suspect that is why most audiophiles own so many amps in their lifetime
> (hi hi). BTW are you blindly following Mr. Hood's opinions (just asking)? No,
> you seem
> like a thoughtful person and I am sure you have thought about them a lot and
> established them to your satisfaction. STILL doesn't mean it's the only was to
> design an
> amp...
I guess I'm leaning towards blindly following Hood's opinions, mainly because
they agree with the ones I had formed independently, before I read his book. In
words of few syllables, Hood must be right because he agrees with me <g>.
> *Most people care only about their system in their room. I suspect that is
> why there is a whole universe of equipment available, rather than one agreed
> upon system.
Stop and think about what you just said for a moment, and I think you will have
to agree that this is completely unworkable. Suppose you just moved to a new
house (as I have), and you want to buy a brand new system (as I have). If each
system sounds different in each room and with each set of transducers, what
scheme do you have for choosing your components. Are you planning to try every
possible permutation and combination? I doubt it.
Jack
Sheldon D. Stokes
wrote:
>Some amps, notably the
>Heath W-6M, had a variable damping factor, precisely to allow the adaptation
>of the amp to the speaker. This was at a time when bookshelf speakers like
>the AR-1 were in vogue.
Lots of amps had current feedback as well, via a cathode resistor
arangement. MY Little heath amps did too. It's supposed to have a
damping factor of one. It sounds awful on my quads, my mission
bookshelf speakers and a set of NHT's. Give me high damping factors
any day.
Sheldon
Jack Crenshaw
> Lots of amps had current feedback as well, via a cathode resistor
> arangement. MY Little heath amps did too. It's supposed to have a
> damping factor of one. It sounds awful on my quads, my mission
> bookshelf speakers and a set of NHT's. Give me high damping factors
> any day.
>
I think that's pretty much the same experience others have had. As someone else
has pointed out, after all is said and done, a speaker really _IS_ a voltage
device, and likes a constant-voltage source.
If you stop and think about it, it's not all that surprising. An electric motor
is also a voltage-dependent device. If you plot the output rotation rate of an
electric motor against the input voltage, you will find an almost perfect
straight line. The reason is that famous back EMF that folks have been talking
about. The friction in an electric motor (and a loudspeaker) is virtually zero.
So the motor speeds up until its back EMF equals the applied voltage. The motor
(and the speaker) has its own internal feedback loop! Any series resistance,
which is what current feedback amounts to, only gets in the way.
Jack
Henry Pasternack
>The motor (and the speaker) has its own internal feedback loop!
Ouch, let's not say that lest we should perpetuate the myth that back
EMF is some kind of feedback mechanism. The motor is really the
mechanical analog of a capacitor. Initially, when you apply a voltage,
the impedance is very low. A high current flows and the motor begins
to spin. The energy flowing into the motor is stored as energy of motion
in the rotor. As time goes on, the current diminishes, the acceleration
goes down, and the rotor approaches a constant speed. Finally, the
open-circuit terminal voltage equals the applied voltage, the current is
zero, and the rotor speed is constant. In the absence of friction, the
terminal voltage would stay constant forever. Put a load across the
terminals, and you would get current back out. In practice, there is
internal loss (friction) and the motor slowly "discharges". There is no
feedback loop here, any more than a capacitor charging is a feedback
loop, or a car accelerating to a top speed is a feedback loop. It's just
a first-order differential solution to Newton's law, f=mA.
The thesis of the anti-feedback zealots as exemplified by Dunker is
this: Given two hypothetical amplifiers having infinitesimal output
impedance, one using feedback and one not, when connected to
identical real-world loads that generate back-EMF and driven with
real-world signals, the terminal voltages will be the same (due to the
essentially zero output impedance), but the current flows will be
substantially different. (As I said, I had this argument several years
ago). Obviously this is an objective technical statement that can be
tested and demonstrated to be false. But for some reason, people
want to cling to the idea of this magical interaction between back
EMF and feedback loops, and the resulting physically impossible
consequences, don't ask me why.
Sorry to seem cross. You're on the good side, Jack, and a lot
more patient than I am.
-Henry
Jack Crenshaw
> Jack Crenshaw wrote in message <3A641541...@earthlink.net>...
> >The motor (and the speaker) has its own internal feedback loop!
>
> Ouch, let's not say that lest we should perpetuate the myth that back
> EMF is some kind of feedback mechanism. The motor is really the
> mechanical analog of a capacitor. Initially, when you apply a voltage,
> the impedance is very low. A high current flows and the motor begins
> to spin. The energy flowing into the motor is stored as energy of motion
> in the rotor. As time goes on, the current diminishes, the acceleration
> goes down, and the rotor approaches a constant speed. Finally, the
> open-circuit terminal voltage equals the applied voltage, the current is
> zero, and the rotor speed is constant. In the absence of friction, the
> terminal voltage would stay constant forever. Put a load across the
> terminals, and you would get current back out. In practice, there is
> internal loss (friction) and the motor slowly "discharges". There is no
> feedback loop here, any more than a capacitor charging is a feedback
> loop, or a car accelerating to a top speed is a feedback loop. It's just
> a first-order differential solution to Newton's law, f=mA.
Ok, I'll buy that, except that it can't be a true analog of a capacitor,
since current
continues to flow. Must be an analog in the voltage realm?
> The thesis of the anti-feedback zealots as exemplified by Dunker is
> this: Given two hypothetical amplifiers having infinitesimal output
> impedance, one using feedback and one not, when connected to
> identical real-world loads that generate back-EMF and driven with
> real-world signals, the terminal voltages will be the same (due to the
> essentially zero output impedance), but the current flows will be
> substantially different. (As I said, I had this argument several years
> ago). Obviously this is an objective technical statement that can be
> tested and demonstrated to be false. But for some reason, people
> want to cling to the idea of this magical interaction between back
> EMF and feedback loops, and the resulting physically impossible
> consequences, don't ask me why.
>
> Sorry to seem cross. You're on the good side, Jack, and a lot
> more patient than I am.
I understand your frustration, and read your earlier comments, so I know
where you're coming from.
Jack
Henry Pasternack
>Ok, I'll buy that, except that it can't be a true analog of a capacitor,
>since current continues to flow. Must be an analog in the voltage
>realm?
I think the analog is correct. The current doesn't go to zero because
some energy has to flow to overcome friction, else the rotor will eventually
slow down and stop. But the steady-state current is definitely much lower
than the starting current. This is why some motors need starting capacitors
to get them going from a dead stop.
The frictional loss is equivalent to a resistor in parallel with the
capacitor.
The winding resistance is equivalent to a series resistance. It slows the
acceleration of the rotor, but doesn't change the final speed, assuming
there are no other losses. A little thought will convince you this is
precisely analgous to the behavior of a capacitor.
-Henry
Corne Janssen
Henry Pasternack wrote:
>
> Jack Crenshaw wrote in message <3A641541...@earthlink.net>...
> >The motor (and the speaker) has its own internal feedback loop!
>
> Ouch, let's not say that lest we should perpetuate the myth that back
> EMF is some kind of feedback mechanism.
But a DC motor hooked to a VOLTAGE source can be descibed as a
regulated system with the back EMF as the negative feedback.
Therefore....
Corné
Corne Janssen
Henry Pasternack wrote:
>
> Jack Crenshaw wrote in message <3A658180...@earthlink.net>...
> >Ok, I'll buy that, except that it can't be a true analog of a capacitor,
> >since current continues to flow. Must be an analog in the voltage
> >realm?
>
> I think the analog is correct. The current doesn't go to zero because
> some energy has to flow to overcome friction, else the rotor will eventually
> slow down and stop. But the steady-state current is definitely much lower
> than the starting current. This is why some motors need starting capacitors
> to get them going from a dead stop.
The starting capacitor is required to produce a phase shift to push
an asychronous AC motor into motion and into a particular direction.
If this isn't done the motor just hums and doesn't move. In this
situation you can start the motor in both directions by turning the
axle by hand in the rotation direction you want.
You can try this by hooking up a 6V AC bike dynamo to a 6V AC source.
Don't try this with large AC motors it's very dangerous.
Corné
henry_pa...@my-deja.com
cj...@oce.nl wrote:
> But a DC motor hooked to a VOLTAGE source can be descibed as a
> regulated system with the back EMF as the negative feedback.
>
> Therefore....
So what you're saying is that when I apply a voltage to the
terminals of a motor, initially the motor looks like a short
circuit and a large current flows. But as the motor spins up,
the back EMF gradually opposes the applied voltage until an
equilibrium is reached where the voltages exactly cancel, the
rotor speed is constant, and the current flow is zero. If a
load is applied to the motor, slowing it down, the back EMF
drops, a voltage drop appears across the windings, current
starts to flow again, and the speed quickly returns to very
nearly its original value. And this is a regulated negative
feedback system because the rotor speed is actively maintained
by comparing the back EMF to the applied voltage and generating
a corrective force that offsets any changes in the speed.
So, I propose to you that a capacitor is also a regulated
negative feedback system. When I apply a voltage to the
terminals of a capacitor, initially the capacitor looks like a
short circuit and a large current flows. But as the capacitor
charges up, the electrostatic potential between the plates
gradually opposes the applied voltage until an equilibrium is
reached where the potentials exactly cancel, the plate voltage
is constant, and the current flow is zero. If a load is applied
across the capacitor, drawing charge from it, the electrostatic
potential goes down, a voltage drop appears across the circuit,
current starts to flow again, and the plate voltage quickly
returns to very nearly its original value. The potential between
the plates is actively maintained by comparing it to the applied
voltage and generating a corrective displacement of charge that
offsets any changes in the potential.
I will go one step further and claim that the same logic can
be used to assert that any passive circuit is a negative feedback
system since, according to Kirkhoff's Law, the current will
"actively" adjust itself so that the net voltage around the loop
is zero.
I appreciate the appeal of your argument, but the premise it's
based on seems to lead to an invalid conclusion. Regardless of all
of this, Thomas Dunker's analysis of loudspeaker and feedback
amplifier behavior defies reality.
-Henry
BTW, yes, you're right about the motor start capacitor.
Corne Janssen
henry_pa...@my-deja.com wrote:
>
> In article <3A65ADCA...@oce.nl>,
> cj...@oce.nl wrote:
> > But a DC motor hooked to a VOLTAGE source can be descibed as a
> > regulated system with the back EMF as the negative feedback.
> >
> > Therefore....
>
> So what you're saying is that when I apply a voltage to the
> terminals of a motor, initially the motor looks like a short
> circuit and a large current flows. But as the motor spins up,
> the back EMF gradually opposes the applied voltage until an
> equilibrium is reached where the voltages exactly cancel, the
> rotor speed is constant, and the current flow is zero. If a
> load is applied to the motor, slowing it down, the back EMF
> drops, a voltage drop appears across the windings, current
> starts to flow again, and the speed quickly returns to very
> nearly its original value. And this is a regulated negative
> feedback system because the rotor speed is actively maintained
It's not actively maintained because the overall (conversion) gain
is less than zero. In fact the speed under load is not maintained
at all because the speed will drop. This is comparable to a
regulated system in which the final error isn't zero. This means
that the intergation action (I part in PID regulated systems) is
zero.
> by comparing the back EMF to the applied voltage and generating
> a corrective force that offsets any changes in the speed.
By putting a constant load on the motor the speed will drop until
((Vsource-Vemf)/Rinternal)*Vsource - (Vsource-Vemf)^2/Rinternal
equals the mechanical power (including friction losses) taken from
he motor. Rinternal is the total internal resistance of motor.
>
> So, I propose to you that a capacitor is also a regulated
> negative feedback system.
I don't want to compare a DC motor to a capacitor because a capacitor
is an energy storage device and a motor is an energy conversion
device which has an inbuild (not always wanted) energy storage.
In fact ironless rotor DC motor were developed to minimize mass
(energy store) in the rotor.
> When I apply a voltage to the
> terminals of a capacitor, initially the capacitor looks like a
> short circuit and a large current flows. But as the capacitor
> charges up, the electrostatic potential between the plates
> gradually opposes the applied voltage until an equilibrium is
> reached where the potentials exactly cancel, the plate voltage
> is constant, and the current flow is zero. If a load is applied
> across the capacitor, drawing charge from it, the electrostatic
> potential goes down, a voltage drop appears across the circuit,
> current starts to flow again,
The big difference here is that the current supplied to the load
comes from the powersupply not from the capacitor. If the
powersupplies internal resistance isn't zero the voltage will
drop. The capacitor will dump part of it's stored energy into the
load until the capacitor voltage has reached the new equilibrium
point.
> and the plate voltage quickly
> returns to very nearly its original value.
> The potential between
> the plates is actively maintained by comparing it to the applied
> voltage and generating a corrective displacement of charge that
> offsets any changes in the potential.
>
> I will go one step further and claim that the same logic can
> be used to assert that any passive circuit is a negative feedback
> system since, according to Kirkhoff's Law, the current will
> "actively" adjust itself so that the net voltage around the loop
> is zero.
>
> I appreciate the appeal of your argument, but the premise it's
> based on seems to lead to an invalid conclusion.
My point here is the a motor/speaker hooked to a voltage source can
be described as a regulated negative feeback system. If the voltage
source is an amplifier with a certain output resistance and feedback
you'll (mathematically) end up with two NFB systems hooked together.
So the back EMF from the speaker will have an effect on the amplfiers
feedback loop.
> Regardless of all
> of this, Thomas Dunker's analysis of loudspeaker and feedback
> amplifier behavior defies reality.
I not familiar with Thomas Dunker's work but I'll look into it.
Corné
Ken Gilbert
> I don't see how NFB can introduce new, higher order distortion,
Baxandall or
> no. There is one exeption, and that is if we are talking about
driving the
> amp beyond its capabilities. At high power levels, NFB will try to
maintain
> flat response by demanding more gain from the amp. If the amp can't
> deliver, that's going to cause clipping. Maybe _THAT's_ how Baxandall
got
> his curves. Otherwise, I just can't see it.
jack,
the way nfb increases higher order distortion products is
multiplicative in nature. say you've got an amp that produces 2nd and
3rd HD. there's a feedback loop in place, so that the output signal
(including 2nd and 3rd HD) is fed back to a point earlier on in the
signal path. now the 2nd and 3rd HD is present earlier on in the
signal path (not just at the output), and thus will ITSELF have some
2nd and 3rd HD added to it... now we've got the 2nd and 3rd HD of 2nd
and 3rd HD, so we've got 2nd, 3rd, 4th, 6th, and 9th order HD present
in the output signal. certainly the 2nd and 3rd HD has been reduced by
the action of the NFB loop, but there are also larger artifacts of
higher order distortion present precisely due to the action of reducing
2nd and 3rd. obviously this will get wrapped yet AGAIN through the NFB
loop, with the same results, only this time even HIGHER in order.
eventually, at some HF breakpoint, you run out of closed loop gain, due
to the requirements of stability... but that doesn't mean you can't
hear those distortion products--it just means that the block of
amplifier enclosed by the loop has unity gain. obviously the lack of
gain at HF will also decrease the ability of the NFB loop to correct
for those very artifacts.
note that this mechanism does not need to drive the amp to clipping at
ALL, which is a whole other failing (imho) of loop NFB amps... music
has a very high peak/average ratio, and i think lots of us are putting
instantaneous voltage peaks through our amps that would very nearly
drive the amp to clipping, if not actually clipping. obviously as soon
as the clip condition occus the loop falls apart, and rarely
gracefully, since a HUGE error voltage is introduced to the NFB node.
> In any case, I defy even the most golden ear to hear 7th harmonic in
an amp
> that measures 0.1%, _TOTAL_.
i think this is where our misunderstanding lies... i claim (and i
certainly stand on the shoulders of many before me) that the higher
order distortions are MUCH more audible than lower order, and that i
while i would agree there's little chance of finding someone who could
discern 0.1% 2nd HD, there's a HELL of a lot more folks out there that
could hear 0.1% 7th HD. as the order increases, the audibility also
increases--some have suggested exponentially. i would agree.
kg
henry_pa...@my-deja.com
cj...@oce.nl wrote:
>It's not actively maintained because the overall (conversion) gain is
>less than zero. In fact the speed under load is not maintained at all
>because the speed will drop. This is comparable to a regulated system
>in which the final error isn't zero. This means that the intergation
>action (I part in PID regulated systems) is zero.
And the "D" part is zero as well. I'll make an assertion, and if
you disagree with it that's fine, but if so we're basically stuck at
an impasse. My assertion is that a feedback system, to be remotely
interesting, has to have some gain in the loop. Clearly a motor is
a purely passive device because all the energy in the system comes
from the input. You're saying there's an implicit proportional
control loop inside the motor. I understand the logic perfectly but
it seems to me to be a degenerate case at best.
>I don't want to compare a DC motor to a capacitor because a capacitor
>is an energy storage device and a motor is an energy conversion device
>which has an inbuild (not always wanted) energy storage. In fact
>ironless rotor DC motor were developed to minimize mass (energy store)
>in the rotor.
The energy in a capacitor is stored in polarization of the dielectric
which, at the atomic level, is a conversion of electrical energy to
mechanical energy. Energy conversion is secondary to this discussion.
The real issue is the mechanism by which internal state is maintained,
regardless of how that state is manifested. You can hook a motor to
a flywheel and use it in place of a capacitor as an energy store and
the analogies still hold. There are even valid applications for this
sort of machine.
>The big difference here is that the current supplied to the load comes
>from the powersupply not from the capacitor. If the power supplies
internal
>resistance isn't zero the voltage will drop. The capacitor will dump
part
>of it's stored energy into the load until the capacitor voltage has
reached
>the new equilibrium point.
Power delivered to the shaft of the motor also comes directly from
the power supply. Yes, there is an energy conversion but I claim the
point is moot. If the power supply has non-zero resistance and a load
is imposed, the motor will dump part of its inertial energy into the
load until the rotor speed has reached the new equilibrium point. A
low-mass rotor is simply equivalent to a low-K dielectric. The fact
that energy conversion occurs doesn't change the basic model. In fact,
I'll argue that it's precisely because people are befuddled by the
electromechanical conversion of energy that they have such trouble
thinking of loudspeakers as ordinary passive electrical loads.
>My point here is the a motor/speaker hooked to a voltage source can
>be described as a regulated negative feeback system. If the voltage
>source is an amplifier with a certain output resistance and feedback
>you'll (mathematically) end up with two NFB systems hooked together.
And my point is that a loudspeaker is a passive device that can
and ought to be modeled as nothing other than a passive device. The
conversion of mechanical to electrical energy is no more significant
than the conversion of kinetic to potential energy in an oscillating
spring-mass system. Furthermore, there is no feedback loop because
there is no loop gain and what you call "regulation" is nothing more
than the passive attainment of a normal state of mechanical equilibrium.
>So the back EMF from the speaker will have an effect on the amplfiers
>feedback loop.
Any load will have an effect on the performance of an amplifier.
Reactive loads in particular are difficult to drive and a loudspeaker
is no exception. You can plot the magnitude and phase versus frequency
of the electrical impedance measured at the terminals of a loudspeaker.
Assuming the speaker is linear (which for the purposes of this discusion
is a reasonable constraint), you can synthesize a passive network of
resistors, capacitors, and inductors that will have exactly the same
characteristics.
Arguably, you could build an arbitrarily complex feedback circuit
that would also have the same terminal characteristics. For instance,
you could replace the inductors in your box with gyrators. But this
would be an active circuit and it's patently obvious that a speaker
is not an active device. As much as I respect the intelligence of
your comments, I simply cannot accept your conclusion. Loudspeaker
drivers are not regulated feedback systems.
-Henry
Jack Crenshaw
> the way nfb increases higher order distortion products is
> multiplicative in nature. say you've got an amp that produces 2nd and
> 3rd HD. there's a feedback loop in place, so that the output signal
> (including 2nd and 3rd HD) is fed back to a point earlier on in the
> signal path. now the 2nd and 3rd HD is present earlier on in the
> signal path (not just at the output), and thus will ITSELF have some
> 2nd and 3rd HD added to it... now we've got the 2nd and 3rd HD of 2nd
> and 3rd HD, so we've got 2nd, 3rd, 4th, 6th, and 9th order HD present
> in the output signal.
I'd like very much to see the transfer function that produces that kind of
behavior. I just can't see it.
Jack
Jack Crenshaw
Ok, guys, here's _MY_ take on the generation of higher-order harmonics in
feedback amplifiers. It doesn't have near the fancy "here we go
roundy-round" explanations some have proposed here, but it sure works for
me.
Take an amplifier that has rather poor bandwidth, sans feedback. (I used
to have one of those, made in the 40's. Its Bode plot looked more like
Ayers Rock. Taken from the side.) The same amplitude of signal is going
into the amp, but it's being lost one way or another, because:
1) the coupling caps aren't big enough
2) the output transformer doesn't have enough iron
3) the plate resistors are too big (made so to get gain)
4) etc.
You can get good power out of the amp at midrange (mine was a 6V6, with 10
watts). But if you try to get that power at low or high frequencies, you
simply can't get it; the amp will clip before it can deliver that power.
Ok, now add feedback. The feedback increases the bandwidth, by reducing the
gain at the midrange frequencies. We add extra gain somewhere in the
circuit to make up for the loss, so the end result is to give the amp more
open-loop gain than it had before. That doesn't change the ability of the
amp to perform, just gives us more gain to work with.
Ok, so what happens? In the midrange, the feedback says "we've got plenty
of gain, better cut it back several dB", and it does. At the extremes, the
feedback needs more gain to keep the bandwidth flat, so it reduces the
feedback, which is the same as increasing the gain. The feedback loop is
trying to maintain the output at a constant level, but the amp simply can't
do the job. The output transformer is too small, for instance, and no
amount of feedback is going to change that.
So what happens? The amp clips, and you get odd harmonics in the output.
Simple.
The same amp, without feedback, doesn't clip because it's not striving to
hold a constant gain. It's content to let it droop off.
Sorry for the anthropomorphization, but I think that's the best way to
describe what's happening. The harmonics appear because the amp is
clipping, plain and simple.
Jack
Corne Janssen
>
> In article <3A66B84A...@oce.nl>,
> cj...@oce.nl wrote:
> >It's not actively maintained because the overall (conversion) gain is
> >less than zero.
I should have written:less than one
>In fact the speed under load is not maintained at all
> >because the speed will drop. This is comparable to a regulated system
> >in which the final error isn't zero. This means that the intergation
> >action (I part in PID regulated systems) is zero.
>
> And the "D" part is zero as well. I'll make an assertion, and if
> you disagree with it that's fine, but if so we're basically stuck at
> an impasse. My assertion is that a feedback system, to be remotely
> interesting, has to have some gain in the loop. Clearly a motor is
> a purely passive device because all the energy in the system comes
> from the input.
True the overall gain (conversion from electrical to mechanical power)
is less than one. In descibing the motor as a regulated system the
'gain' in the loop is proportional to 1/Ri. If the internal resistance
is zero the 'gain' would be infinite so the static or final error is
infinitly small (zero).
> You're saying there's an implicit proportional
> control loop inside the motor.
No, I said that a motor hooked to a voltage source can be described
as a regulated system. A motor on its own has no internal feedback.
If you hook the motor to a current source (Ri=infinite) there's
no way the back EMF can influence the current therefore this system
can't be discribed as a regulated system.
> I understand the logic perfectly but
> it seems to me to be a degenerate case at best.
>
> >I don't want to compare a DC motor to a capacitor because a capacitor
> >is an energy storage device and a motor is an energy conversion device
> >which has an inbuild (not always wanted) energy storage. In fact
> >ironless rotor DC motor were developed to minimize mass (energy store)
> >in the rotor.
>
> The energy in a capacitor is stored in polarization of the dielectric
> which, at the atomic level, is a conversion of electrical energy to
> mechanical energy.
Didn't think about that. Good point.
> Energy conversion is secondary to this discussion.
> The real issue is the mechanism by which internal state is maintained,
> regardless of how that state is manifested.
My problem at this point is that I've never tried describing a
capacitor hooked to a voltage source as a regulated system before.
The voltage/speed relation in a motor seem analogous to the
voltage/polarization relation. So substituting the one for the
other should result in discription for a voltage source/capacitor
regulated system.
> You can hook a motor to
> a flywheel and use it in place of a capacitor as an energy store and
> the analogies still hold. There are even valid applications for this
> sort of machine.
>
> >The big difference here is that the current supplied to the load comes
> >from the powersupply not from the capacitor. If the power supplies
> internal
> >resistance isn't zero the voltage will drop. The capacitor will dump
> part
> >of it's stored energy into the load until the capacitor voltage has
> reached
> >the new equilibrium point.
>
> Power delivered to the shaft of the motor also comes directly from
> the power supply. Yes, there is an energy conversion but I claim the
> point is moot. If the power supply has non-zero resistance and a load
> is imposed, the motor will dump part of its inertial energy into the
> load until the rotor speed has reached the new equilibrium point. A
> low-mass rotor is simply equivalent to a low-K dielectric. The fact
> that energy conversion occurs doesn't change the basic model. In fact,
> I'll argue that it's precisely because people are befuddled by the
> electromechanical conversion of energy that they have such trouble
> thinking of loudspeakers as ordinary passive electrical loads.
>
I agree with you that loudspeakers and electrical motors are passive
loads. They can also be described as regulated systems in which the
conversion gain (overall gain in the description) is less than one.
>
> >My point here is the a motor/speaker hooked to a voltage source can
> >be described as a regulated negative feeback system. If the voltage
> >source is an amplifier with a certain output resistance and feedback
> >you'll (mathematically) end up with two NFB systems hooked together.
>
> And my point is that a loudspeaker is a passive device that can
> and ought to be modeled as nothing other than a passive device.
I don't agree with you on that point. I don't have a problem with a
model that uses an imaginary gain stage as long as the overall
response is equal to the passive device it describes.
> The
> conversion of mechanical to electrical energy is no more significant
> than the conversion of kinetic to potential energy in an oscillating
> spring-mass system. Furthermore, there is no feedback loop because
> there is no loop gain and what you call "regulation" is nothing more
> than the passive attainment of a normal state of mechanical equilibrium.
>
> >So the back EMF from the speaker will have an effect on the amplfiers
> >feedback loop.
>
> Any load will have an effect on the performance of an amplifier.
> Reactive loads in particular are difficult to drive and a loudspeaker
> is no exception. You can plot the magnitude and phase versus frequency
> of the electrical impedance measured at the terminals of a loudspeaker.
> Assuming the speaker is linear (which for the purposes of this discusion
> is a reasonable constraint), you can synthesize a passive network of
> resistors, capacitors, and inductors that will have exactly the same
> characteristics.
Yes, and in doing so you are using a 'model' of the real thing.
>
> Arguably, you could build an arbitrarily complex feedback circuit
> that would also have the same terminal characteristics. For instance,
> you could replace the inductors in your box with gyrators. But this
> would be an active circuit and it's patently obvious that a speaker
> is not an active device.
True but if the model acts like the real thing it really doesn't
matter what's in it.
> As much as I respect the intelligence of your comments,
> I simply cannot accept your conclusion. Loudspeaker drivers
> are not regulated feedback systems.
I never said they were but they can be described as regulated
feedback systems.
Corné
Fred Nachbaur
There's a flaw in this iterative way of looking at feedback. If the
amplifier had infinite bandwidth, all harmonics would be treated
equally, and even "harmonics of harmonics" would be reduced by exactly
the same amount. The iterative approach reminds me of the story of the
engineer and the scientist who both went to Hades; their punishment was
that each day they were allowed to traverse 1/3 of the distance towards
a beautiful damsel lolling under a shady tree. The scientist gave up,
sat on his rock and pouted. The engineer enthusiastically took his 1/3
distance for that day. "Stupid engineer," said the scientist, "don't you
realise you'll never get there?" "Stupid scientist," replied the
engineer, "I'll certainly get close enough!"
Back to the amplifier - in real life bandwidth is not unlimited, and the
roll-off causes higher harmonics to be amplified less - resulting in
less feedback - resulting in less amplification. Again we can go round
and round, but remember that it's an instantaneous thing; the signal
does not actually loop the loop, as it were.
> eventually, at some HF breakpoint, you run out of closed loop gain, due
> to the requirements of stability... but that doesn't mean you can't
> hear those distortion products--it just means that the block of
> amplifier enclosed by the loop has unity gain. obviously the lack of
> gain at HF will also decrease the ability of the NFB loop to correct
> for those very artifacts.
I'm not sure what you mean here, when you run out of gain you run out of
gain. Of course it doesn't get fed back, but it isn't even there in the
first place!
> note that this mechanism does not need to drive the amp to clipping at
> ALL, which is a whole other failing (imho) of loop NFB amps... music
> has a very high peak/average ratio, and i think lots of us are putting
> instantaneous voltage peaks through our amps that would very nearly
> drive the amp to clipping, if not actually clipping. obviously as soon
> as the clip condition occus the loop falls apart, and rarely
> gracefully, since a HUGE error voltage is introduced to the NFB node.
True, which I think was Jack's original point.
> > In any case, I defy even the most golden ear to hear 7th harmonic in
> an amp
> > that measures 0.1%, _TOTAL_.
>
> i think this is where our misunderstanding lies... i claim (and i
> certainly stand on the shoulders of many before me) that the higher
> order distortions are MUCH more audible than lower order, and that i
> while i would agree there's little chance of finding someone who could
> discern 0.1% 2nd HD, there's a HELL of a lot more folks out there that
> could hear 0.1% 7th HD. as the order increases, the audibility also
> increases--some have suggested exponentially. i would agree.
I agree also. 0.1% is only down 30 dB. You can most certainly hear a
-30dB 3500 Hz. tone on top of 500 Hz. at 0 dB. It's very easy to prove
this using any decent wav generator/editor such as CoolEdit. Whereas it
is much more difficult to hear the subtlety of 1000 Hz at -30dB on top
of 2000 Hz. at 0 dB.
Fred N
--
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+----------------------------------------------------------+
Fred Nachbaur
Corne Janssen wrote:
>
> > [... previous dialogue snipped]
> >
> > Arguably, you could build an arbitrarily complex feedback circuit
> > that would also have the same terminal characteristics. For instance,
> > you could replace the inductors in your box with gyrators. But this
> > would be an active circuit and it's patently obvious that a speaker
> > is not an active device.
>
> True but if the model acts like the real thing it really doesn't
> matter what's in it.
>
> > As much as I respect the intelligence of your comments,
> > I simply cannot accept your conclusion. Loudspeaker drivers
> > are not regulated feedback systems.
>
> I never said they were but they can be described as regulated
> feedback systems.
>
> Corné
An apparently passive device can indeed act like a regulated feedback
system. I'll repeat an example I posted in another thread -- the use of
caster to regulate steering systems. When a vehicle with castered wheels
is just sitting there, of course nothing happens. But push the vehicle,
and the castered wheels will seek straight-line steering. An error
signal (perhaps changes in friction, presence of small objects, etc.)
will be counteracted by the frictional force of the wheel trying to turn
"sideways", and straight-line steering is restored.
Similarly, a speaker's characteristics can act as such "steering" /when
driven/. Under this condition it is no longer strictly a passive device,
but is rather part of a larger active device including the OPT and
drivers (tubes in this case).
Ken Gilbert
the mutiplicative behavior i just described comes from many sources,
one of which is JL hood's own "art of linear electronics." i shall get
you the page number and isbn when i get back home after work.
if you do a dejanews search for RAT postings by scott frankland
approximately a year ago, i'm sure you'll find more discussions about
the topic.
henry_pa...@my-deja.com
cj...@oce.nl wrote:
> > As much as I respect the intelligence of your comments,
> > I simply cannot accept your conclusion. Loudspeaker drivers
> > are not regulated feedback systems.
>
> I never said they were but they can be described as regulated
> feedback systems.
Well, you've convinced me your model is valid, even if I don't
think it's the best model for this situation. What's more important
is that you've convinced me you know what you're talking about. It
was interesting working through your comments -- thanks for the
feedback (har, har).
henry_pa...@my-deja.com
Fred Nachbaur <fr...@netidea.com> wrote:
> Similarly, a speaker's characteristics can act as such "steering"
> device, but is rather part of a larger active device including the
> OPT and drivers (tubes in this case).
Definitely, if you draw a box around the amplifier AND the speaker,
you can say that the system as a whole is an active feedback system.
But the speaker itself is strictly passive, just as a compensation
network inside the amplifier is strictly passive, even though the
amplifier as a whole is not.
SkrewBall
Feedback helps get rid of ringing,distortion and things like that. it can be
very good,to a certain extent.
A speaker is kind of like an electric motor,in the sense that the more
voltage swing you give it,the louder it gets (the motor spins faster)And
that the more current you feed it,the more force it has to move the cone,and
maintain control (the motor has more torque)
Just a simple analogy..
Personally I think regular old dynamic speakers tend to work better with
more current,as opposed to more voltage.
10 amps at 1 volt tends to sound better to my ears than 10 volts at 1
amp...same amount of power into a given load,just in a slightly different
form. i have noticed this mostly with lower frequencies,and drivers like
woofers,and subwoofers. they sound far more controlled,and less "flabby" and
"punchy" at higher current levels,than at higher voltage levels..
anyways.. its late,and i'm yammering!
Jack Crenshaw wrote:
> OK, all you zero-feedback folks out there. Listen to what John Linsley
> Hood, author of "Valve and Transistor Audio Amplifiers," has to say:
>
> "For good audio performance, in transformer coupled amplifiers, it is
> _NECESSARY_ [emphasis mine] to employ negative feedback (NFB) in the
> amplifier design and, in view of the distortion that can arise in the
> loudspeaker coupling transformer, it is _ESSENTIAL_ [ditto] that this
> component is also included in the NFB loop."
>
> "Simple NFB can reduce the extent of any waveform distortion or noise
> introduced by the amplifier block, and can help to make the frequency
> response of the system more uniform."
>
> "The distortion of the output signal will be reduced in direct
> proportion to as the value of B*A increases."
>
> "Another advantage of NFB is that it can improve the accuracy of
> rendition of transient signals, in that any difference between the
> input signal, Ei, and that which is fed back, B*Eo, will generate an
> error voltage which will be amplified by Ao and subtracted from the
> signal output."
>
> "All in all, an amplifier block employing negative feedback, when this
> is applied in a properly designed system, will be substantially superior
> to that of a similar system without NFB. The fact that claims are
> sometimes made that the performance of an amplifier without NFB is
> superior to that of one using this technique seems to be due principally
> to the lack of design skills of those in respect of whose designs this
> claim is made."
>
> Hood goes on to talk about gain margin and phase margin, and the need to
> have both. He emphasizes that, in the real world, with complicated
> signals and anything but ideal speaker impedance loads, a goodly margin
> is necessary. He then goes on to say,
>
> "Understandably, badly designed amplifiers will give unpleasant results,
> and these results may worsen when NFB is applied. But the answer is not
> to use reduced amounts of NFB, or no NFB at all, but to design them so
> thaat they have greater loop stability margins, especially when used
> with reactive loads. It is inconvenient that this is a region in which
> the overall stability in operation of an amplifier is often in conflict
> with the attempts of the designer to achieve the lowest practical level
> of total harmonic distortion, when measured with a steady state signal
> on a purely resistive load. For this purpose, a large amount of NFB is
> often used , and the design of the circuit is then tailored to sustain
> the loop gain at a high level up to the frequency at which it must fall
> rapidly to reach 0 db before the loop phase approaches -180 degrees.
> Circuits designed like this, though stable, often have very small gain
> and phase margins."
>
> To paraphrase: NFB can greatly improve the performance of a
> well-designed amplifier, particularly if that amp is designed with NFB
> in mind, in the first place. It can and will reduce distortion, reduce
> hum and noise, increase bandwidth, and _IMPROVE_ _TRANSIENT_ _RESPONSE_,
> all roughly in proportion to the amount of feedback.
>
> On the other hand, it can't save a poorly designed amp, and one is
> foolish to try to make it do so.
>
> Jack
Jack Crenshaw
behavior of the subsystem (speaker, motor, steering gear, whatever) cannot be
described as a linear equation, but must be described by a differential
equation. There is plenty of precedent for that POV, but a system described
by an ODE is _NOT_ the same thing as a system with feedback.
Jack
Fred Nachbaur
Jack Crenshaw wrote:
>
> PMFJI, but what you guys are really saying, it seems to me, is that the
> behavior of the subsystem (speaker, motor, steering gear, whatever) cannot be
> described as a linear equation, but must be described by a differential
> equation. There is plenty of precedent for that POV, but a system described
> by an ODE is _NOT_ the same thing as a system with feedback.
>
> Jack
Jack, I think you've nailed it. I think it's - again - a question of
viewpoint and semantics, and harps to the age-old question "if it looks
like a duck and quacks like a duck, is it necessarily a duck?"
Systems with externally added NFB can similarly be described /as a
system/ using diff-eq's, where the "feedbackness" is not viewed as a
unique entity within that set of equations.
So if one's definition of NFB hinges on a clearly distinct mechanism of
introducing these self-correcting attributes, then yes; there is indeed
a distinct difference between, let's say, a resistor added from output
to inverting input and, let's say, a steering gear or a speaker/motor,
etc.
Fred
John Byrns
<eat_...@spam.com> wrote:
> Wiz wrote in message <20010114091620...@ng-fb1.aol.com>...
> >"Simple NFB can reduce the extent of any *repetitive* waveform distortion
> or
> >noise introduced by the amplifier block, and can help to make the frequency
> >response of the system more uniform."
>
> Not true. Feedback definitely reduces noise, and noise by definition, is
> not repetitive. Feedback works regardless of the signal, periodic or
> aperiodic.
Hi Henry,
Is this true, that feedback reduces noise? Didn't I have a lengthy
discussion with Frank Deutschman about this a couple of years ago? If I
remember correctly I took the position you are now taking, that negative
feedback can reduce the noise in an amplifier. Frank, and the other more
sophisticated designers here, argued that this was not true, that feedback
doesn't reduce noise in an amplifier, it just appears to reduce noise
because the gain of the amplifier is reduced, or some such. What gives,
what is the real story, who has the correct story on feedback and noise,
was I mislead?
Regards,
John Byrns
Surf my web pages at, http://www.enteract.com/~jbyrns/index.html
Henry Pasternack
>Is this true, that feedback reduces noise? Didn't I have a lengthy
>discussion with Frank Deutschman about this a couple of years ago? If I
>remember correctly I took the position you are now taking, that negative
>feedback can reduce the noise in an amplifier.
Negative feedback reduces output-referred noise. Deutschman's argument
had to do with noise figure, which is an input-referred parameter. I agree
with
his position.
I will never, ever make the mistake again of having a technical debate
with
you on the newsgroup. It's not worth the pain.
-Henry
Fred Nachbaur
John Byrns wrote:
>
> [...]
>
> Is this true, that feedback reduces noise? Didn't I have a lengthy
> discussion with Frank Deutschman about this a couple of years ago? If I
> remember correctly I took the position you are now taking, that negative
> feedback can reduce the noise in an amplifier. Frank, and the other more
> sophisticated designers here, argued that this was not true, that feedback
> doesn't reduce noise in an amplifier, it just appears to reduce noise
> because the gain of the amplifier is reduced, or some such. What gives,
> what is the real story, who has the correct story on feedback and noise,
> was I mislead?
NFB reduces noise and other artifacts _generated within the loop around
which the NFB is applied_. It will not reduce noise that is applied to
the input - GIGO (garbage in, garbage out) except that the garbage is
less modified by additional garbage.
Sheldon D. Stokes
>NFB reduces noise and other artifacts _generated within the loop around
>which the NFB is applied_. It will not reduce noise that is applied to
>the input - GIGO (garbage in, garbage out) except that the garbage is
>less modified by additional garbage.
IF the noise generated in the circuit is sufficiently wide-band, the
NFB can actually amplify the noise due to the phase shift through the
loop and the feedback at the very high frequencies being positive.
This should be taken care of in the filtering within the feedback
loop, and my example above is what I'd call a bad design. But
nonetheless, it is a case of feedback increasing noise; or should that
be filed under a stability issue?
Sheldon
John Byrns
<eat_...@spam.com> wrote:
> John Byrns wrote in message ...
> >Is this true, that feedback reduces noise? Didn't I have a lengthy
> >discussion with Frank Deutschman about this a couple of years ago? If I
> >remember correctly I took the position you are now taking, that negative
> >feedback can reduce the noise in an amplifier.
>
> Negative feedback reduces output-referred noise. Deutschman's argument
> had to do with noise figure, which is an input-referred parameter. I agree
> with
> his position.
Hi Henry,
Yes, you have jogged my memory, Frank was talking noise figure, but he was
calling it just plain "noise", which was what resulted in all the
confusion. If I am remembering correctly though, after all that was
straightened out, he still wouldn't give up the idea that feedback could
reduce what you are calling the "output-referred noise". I agree with
your position on this issue of feedback and noise, I am not sure however
that your position is really in agreement with Frank's, as I don't believe
he ever came around to the idea that feedback could reduce the
"output-referred noise".
Sheldon just made an interesting comment, pointing out that negative
feedback could possibly increase the out of band noise. Without thinking
much about it, I would think that there would be serious stability
problems before this became much of a factor. Feedback and stability are
right up your alley, any comments on Sheldon's post?
> I will never, ever make the mistake again of having a technical debate
> with
> you on the newsgroup. It's not worth the pain.
Well you did bring it on yourself, but I don't think it would be wise to
dredge all those issues up again.
Fred Nachbaur
Every generality has exceptions. (Itself a generality, which produces an
interesting paradox...) Going a step further, I'd say that the example
you bring up would be indeed a stability issue; in such a network noise
is likely to be the least of one's worries, since it would very probably
be a dandy oscillator.
Jack Crenshaw
your statement as it stands -- that feedback can increase noise if the
open-loop plant is very wideband -- the statement is very misleading.
The open-loop plant should never _BE_ so wideband that its bandwidth exceeds
that of the feedback loop. It stands to reason that feedback can do nothing
to fix noise that's outside its own range to measure. All of feedback
theory is based on the idea that the feedback loop compares output to input
and corrects for any difference. The feedback can't work if the comparator
circuit can't keep up with the open-loop amp.
To avoid this, all properly designed feedback amps are set up so that the
feedback loop is much wider in bandwidth than the open-loop plant. If
necessary, trimming RC circuits should be used to make sure this is the
case.
If this is done properly, the feedback can _NEVER_ increase noise or
distortion since it is always in control.
Jack
Jack
henry_pa...@my-deja.com
jby...@enteract.com (John Byrns) wrote:
> [Drivel]
You're still an idiot, John.
Jack Crenshaw
> In article <jbyrns-0402...@207-229-151-249.d.enteract.com>,
> jby...@enteract.com (John Byrns) wrote:
> > [Drivel]
>
> You're still an idiot, John.
>
Gee, Henry, what brought _THAT_ on?
Jack
John Byrns
<jcr...@earthlink.net> wrote:
Hi Jack,
You are better off not asking, you really don't want to know.
Henry Pasternack
>Gee, Henry, what brought _THAT_ on?
John and I had a disagreement several years ago. I've tried every which
way to resolve the matter with him, but to no avail. It has become a fixed
feature of my universe that there is this bubble of irrationality in him
that I
will never be able to do anything about. Rather than arguing about it (for
I have thoroughly defended myself in public on several occasions), every
time he brings the matter up I just simply call him an idiot and leave it at
that.
-Henry
John Byrns
<eat_...@spam.com> wrote:
Hi Henry,
Human nature is an interesting thing, you call me an "idiot" because I
pointed out the error of your ways when you made a serious technical
blunder, which was all the more glaring because it was perhaps the only
time you made such an egregious mistake. On the other hand, you don't
mention the times that I have arguably actually been an "idiot".
FT
but you both seem quite adept at acting like one.
John Byrns wrote:
> You are better off not asking, you really don't want to know.
That's right; even those of us who already know really don't want to
know!
If you keep this up, I'll have to invite our old friend Andre Ju...
naw, forget I even thought such a thing :-(.
Dig a hole in the back yard, place the hatchet in said hole, and be
done with it.
Please!
Fred
Henry Pasternack
Quit being an idiot, John.
-Henry
John Byrns
<eat_...@spam.com> wrote:
Well Henry, since you seem to get a kick out of calling me an "idiot", I
thought I better check back through "deja-news" to see what you had to say
about me while I was gone. I found that I was wrong when I said that your
mistake on the optimum triode load impedance, was probably the only time
you made a mistake here. I found the following post of yours from 23
November 2000:
Sheldon D. Stokes wrote in message
<6qkm1t0655tg17n1d...@4ax.com>...
>But the split load is not without problems, it's got different
>impedances on each leg. There's no free lunch.
Ouch! Not true!
If DejaNews worked, you could see the lengthy discussion (I guess John
Byrns was involved, so as I recall it was an argument) dating from mid-1999
in which I explained why the statement above is a myth.
-Henry
I'm not sure why you dragged my name in to that post, as we were probably
both on the same side of the argument. It is one of my hot buttons, so I
have posted on it numerous times, including at least once with complete
mathematical formulae.
You are not quite correct though, when you say "Not true!" Those who
point out that the source impedance's of the two outputs from the
concertina are different, have got it right. Where they go wrong is in
assuming that this implies an imbalance between the two outputs, which is
wrong, at least to a first approximation. Their error is easily seen by
noting that if you ignore the grid to plate capacitance, and grid to
cathode capacitance, then the same current must flow in the plate circuit
as flows in the cathode circuit. This means that if the load impedance's,
including the cathode and plate resistors, are balanced plate and cathode,
then the two output signals will be balanced. I ran some spice
simulations way back when to check the effects of the grid capacitance's,
and found that they didn't really cause much of any problem until well
above the audio range.
The down side with the concertina is that the cathode and plate loads are
not instantaneously balanced when it directly drives the grids of the
power tubes, and is pushed to clipping. The instantaneous unbalance
caused by the grid current causes some very unusual distortion to be
generated during clipping. Not only is there the effect of the different
source impedance's, but when grid current flows in the tube connected to
the cathode of the concertina, the gain on the plate side actually
increases instantaneously, creating excess drive to the tube connected to
the plate of the concertina. It is best to use some sort of buffer stage
following the concertina, as Williamson did.
Henry Pasternack
>[Deleted]
I am extremely familiar with the principle of operation of the
split-load phase inverter. As I said, I have no interest in being
dragged into a technical discussion with you.
-Henry
John Byrns
<eat_...@spam.com> wrote:
Doesn't sound that way, if you believe as your earlier post implies, that
the source impedance's of the two outputs are the same. Note for other
readers, the fact that the cathode and plate outputs have different source
impedance's does not imply that the concertina has a balance problem, for
reasons previously mentioned.
Henry Pasternack
>[Deleted]
John, it's not my fault that you haven't the intellectual subtlety to
follow
a simple and elegant argument. If you haven't got it by now, why the hell
should I bother to explain it again? All you're trying to do is provoke me.
Now piss off.
-Henry
John Byrns
<eat_...@spam.com> wrote:
What simple and elegant argument? I haven't seen one from you, can I find
it on "deja-news"? If I can't, it's not a matter of subtlety, but rather
of not having seen your simple and elegant argument at all. I don't
remember your weighing in on this argument in the past millennium, and I
only argued in the past that there was not balance problem with the
concertina. I am now claiming that while there is no balance problem, it
is true as they say that the source impedance's are different. I would be
interested in seeing your simple and elegant argument to the contrary.
But I don't expect to see it, not because you don't want to argue with me,
but rather because you don't actually have a valid argument, simple and
elegant or not, showing that the source impedance of the concertina
looking into the cathode is the same as that looking into the plate.
Ken Gilbert
> elegant or not, showing that the source impedance of the concertina
> looking into the cathode is the same as that looking into the plate.
john, believe it or not, it is.
this is a common misconception about the concertina, or split-load
inverter. so long as the impedances attached to the plate and cathode
nodes remain even (and assuming you have well-matched load resistors),
the balance will be about the best there is, at least among non-tranny
phase splitters. the only time this topology runs into problems is
when the impedance on one of the nodes (plate or cathode) is NOT equal
to the other. a typical cause for this might be running into grid
current in the output tubes.
as long as the loads are equal, the impedance "seen" by one load is
reflected back through the tube to the other. you will not see ANY
difference in output impedance, so long as the load impedance is equal,
as stated above.
you can see why williamson's design was so effective, since he used
another stage (a differential driver) to isolate the split-load from
the output grids, thereby assuring that the split load would always be
working into a balanced impedance.
for an elegant analysis:
Notes on the Cathodyne Phase Splitter
by Albert Preisman, Audio, April, 1960:
http://www.aikenamps.com/cathodyne.pdf
kg
--
I know that the twelve notes in each octave and the varieties
of rhythm offer me opportunities that all of human genius
will never exhaust. --Igor Stravinski
John Byrns
<ride...@ride.ri.net> wrote:
> > elegant or not, showing that the source impedance of the concertina
> > looking into the cathode is the same as that looking into the plate.
>
> john, believe it or not, it is.
No it isn't, and I defy you to show any evidence or proof that the source
impedance of the plate and cathode outputs of the concertina are the
same. It is not a common misconception, it is in fact the true facts, any
text book will show you. You need to learn to listen louder, and not run
of head first in defense of Henry. All I said was the source impedance's
are not the same, I did not say that the concertina has a balance
problem. In fact I was one of the first, probably even the first to argue
in this group that the concertina was nearly perfectly balanced, at least
as long as the loads are balanced. I agree with most everything you say
below, except your claim that the cathode and plate have the same source
impedance. For the more advanced student there is also the problem that
the concertina has with nonlinear loads, like grids being driven into
clipping. Listen up, we are mostly in agreement, stop and think about
what I have said, as I am sure Henry is doing at this very moment.
Regards,
John Byrns
Ken Gilbert
> > for an elegant analysis:
> >
> > Notes on the Cathodyne Phase Splitter
> > by Albert Preisman, Audio, April, 1960:
> > http://www.aikenamps.com/cathodyne.pdf
that's all.
KG
John Byrns
<ride...@ride.ri.net> wrote:
> HEY, LISTEN...
Kenny, Kenny, what are we going to do with you? I said you need to listen
louder, not start shouting. I suggest you pay attention to Henry, he
knows when he is wrong and should keep his mouth shut. You are dead wrong
on this issue. Now pay attention, my claims are two, which you can check
by looking back at my earlier postings.
1. The source impedance of the plate and cathode outputs of the concertina
phase splitter are not the same.
2. The concertina phase splitter has very good balance, unaffected by the
different source impedance's of the plate and cathode circuits.
> > > for an elegant analysis:
> > >
> > > Notes on the Cathodyne Phase Splitter
> > > by Albert Preisman, Audio, April, 1960:
> > > http://www.aikenamps.com/cathodyne.pdf
>
> that's all.
Not quite.
Thanks for the reference, I checked the library and read that article, as
well as Albert Preisman's article it referenced in the RCA Review. Since
Albert Preisman worked for "The General", he must be a good guy.
Mr. Preisman's Audio article, "Notes on the Cathodyne Phase-Splitter"
supports my claims exactly. Equations (1) & (2) in the article support my
claim, #1 above, that the source impedance's of the plate and cathode
circuits are different. Equations (4) & (5) are all you need to
understand why the concertina is nearly perfectly balanced. Unfortunately
Albert should have ended the article with equation (11) at the top of the
third column. The rest of the article just serves to confuse the issue by
attempting to use a different argument to explain why the concertina
remains balanced, for those who don't understand the elegant implications
of equations (5) & (6). His basic point in the remainder of the article
is that the Voltage across the load on a Thevenin equivalent generator
will remain constant as you change the source impedance of the generator,
if you also make the appropriate compensating changes in the magnitude of
the Voltage generator in the Thevenin equivalent generator circuit. I
think this alternate explanation just serves to confuse the unwary, when
equations (5) & (6) say it all. He should have spent more time explaining
the direct implications of those two equations, rather than writing the
last 67% of the article.
The ball is in your court, and please don't shout, it doesn't really
advance your argument.
Regards,
John Byrns
Ken Gilbert
towards the end of the article (p.23), this appears:
"In the case of the phase-splitter circuit just analyzed, we can
conclude that so long as the two load impedances are maintained equal
at all frequencies of interest, no concern need be felt about the
differences in apparent source impedance."
so what is your major concern about the differences in apparent source
impedance?
you yourself state:
>1. The source impedance of the plate and cathode outputs of the
concertina
>phase splitter are not the same.
>2. The concertina phase splitter has very good balance, unaffected by
the
>different source impedance's of the plate and cathode circuits.
if the impedance is different, but it doesn't matter, what in god's
name is this all about? apparently, it matters to YOU that the
impedances are different... and as far as i can tell it matters ONLY to
you.
so what's the deal? do you have a bone to pick? an old scab to
scratch at? have you decided that this issue shall be your last
stand? is all of this strife worth it? what the hell do you want from
all of this, anyway--some kind of public eating-of-crow? i don't think
this is in the honest pursuit of practical knowledge, since the
practical ramifications are so cut and dry... you keep the damned load
impedances equal and you forget about it.
as far as i'm concerned, i agree with the author's opinion of 41 years
ago--the unbalance of source impedances is a non-issue. if you want to
belabor the point and therefore treat it as an issue, fine, but don't
expect everyone else to hinge on your every word.
if the ball is indeed "in my court" as you stated, then consider me
walking away from the game... suffice it to say i don't see the point
in competing.
ken
John Byrns
<ride...@ride.ri.net> wrote:
> john, i've got to ask you, what is the point of all this?
That's simple Ken, Henry Pasternack has made several posts recently in
which he calls me an "idiot", so Friday I did a "deja-news" search to see
what he might have said about me while I was away from rec.audio.tubes. I
found a post that Henry had made on 22 November last, which while somewhat
ambiguous in its meaning, seemed to be saying that the poster he was
responding to was wrong in saying that the cathode and plate outputs of
the concertina had different impedance's. He then went on to attach my
name to the discussion, with a slightly negative connotation as if to
imply that I had been on the losing side of some earlier discussion about
the concertina phase splitter circuit. The point is simply to set the
record straight as to where I stand on the operation of the concertina
phase splitter, and indicate that the truth is not necessarily what Henry
would have us believe. The Audio magazine article by Albert Preisman,
that you pointed out, is exactly congruent with my claims, and is a
complete vindication of what I have said about the concertina, Henry's
comments not withstanding.
> towards the end of the article (p.23), this appears:
>
> "In the case of the phase-splitter circuit just analyzed, we can
> conclude that so long as the two load impedances are maintained equal
> at all frequencies of interest, no concern need be felt about the
> differences in apparent source impedance."
That is exactly the point I have been trying to make, read my posts of the
last couple day's.
> so what is your major concern about the differences in apparent source
> impedance?
I have no concern about the differing source impedance's, that is my whole
point, no one should be concerned about the source impedance differences,
it doesn't affect the balance, as long as the loads are balanced.
> you yourself state:
>
> >1. The source impedance of the plate and cathode outputs of the
> concertina
> >phase splitter are not the same.
>
> >2. The concertina phase splitter has very good balance, unaffected by
> the
> >different source impedance's of the plate and cathode circuits.
>
> if the impedance is different, but it doesn't matter, what in god's
> name is this all about? apparently, it matters to YOU that the
> impedances are different... and as far as i can tell it matters ONLY to
> you.
See above, where I say it is about setting the record straight.
> so what's the deal? do you have a bone to pick? an old scab to
> scratch at? have you decided that this issue shall be your last
> stand? is all of this strife worth it? what the hell do you want from
> all of this, anyway--some kind of public eating-of-crow? i don't think
> this is in the honest pursuit of practical knowledge, since the
> practical ramifications are so cut and dry... you keep the damned load
> impedances equal and you forget about it.
Again what I want to do is set the record straight, and I think I have
done that with your, and Mr. Preisman's help. I would disagree that there
are no practical ramifications, as this discussion started several years
ago after a popular writer published a book pointing out the differing
source impedance's and suggesting that this supposed "problem" could be
solved by adding a series resistor to the cathode output. After the
publication of the book there were several amplifier designs published in
magazine articles, which used the series resister idea to "fix" the
problem, unwittingly breaking a circuit that didn't wasn't broken in the
first place. The differing source impedance's of the two concertina
outputs is still widely viewed as a "problem" by many audiophiles.
> as far as i'm concerned, i agree with the author's opinion of 41 years
> ago--the unbalance of source impedances is a non-issue. if you want to
> belabor the point and therefore treat it as an issue, fine, but don't
> expect everyone else to hinge on your every word.
I agree it's not an issue, but contemporary writers have made it into an
issue, and the modern audiophile needs to be made aware of the old truth.
> if the ball is indeed "in my court" as you stated, then consider me
> walking away from the game... suffice it to say i don't see the point
> in competing.
The point is made, the game is over, everyone is walking away.
Henry Pasternack
news:966jm1$a6g$1...@nnrp1.deja.com...
> as far as i'm concerned, i agree with the author's opinion of 41 years
> ago--the unbalance of source impedances is a non-issue. if you want to
> belabor the point and therefore treat it as an issue, fine, but don't
> expect everyone else to hinge on your every word.
If the output impedances were different, we would expect different
frequency responses when the two outputs were loaded with equal
capacitors. Since this doesn't happen (ignoring second-order effects),
we can say the output impedances are the same. For this to be true,
both load impedances must be the same. The reason is the the load
on one leg affects the output impedance of the other. Obviously, if
you vary one load impedance while holding the other constant, or
otherwise treat the two outputs asymmetrically, you will observe
different responses. But again this violates the constraint that the
two loads on the split-load inverter must be equal.
There's absolutely no reason for all the fuss.
-Henry
John Byrns
"Henry Pasternack" <eat....@spam.com> wrote:
> If the output impedances were different, we would expect different
> frequency responses when the two outputs were loaded with equal
> capacitors. Since this doesn't happen (ignoring second-order effects),
> we can say the output impedances are the same. For this to be true,
> both load impedances must be the same. The reason is the the load
> on one leg affects the output impedance of the other. Obviously, if
> you vary one load impedance while holding the other constant, or
> otherwise treat the two outputs asymmetrically, you will observe
> different responses. But again this violates the constraint that the
> two loads on the split-load inverter must be equal.
Hi Henry,
That's an appealingly simple argument, but it puts you at odds with the
article that Ken pointed out by Albert Preisman in the April 1960 Issue of
Audio Magazine. There are several possible explanations for the
discrepancy.
1. It could be an April Fools joke, the article is in the April issue!
2. Just because something is in print doesn't make it correct.
3. You have overlooked an important variable, the open circuit Voltage
response. Even if the source impedance's are different, it is possible
for the frequency response to be the same, if the open circuit Voltage
responses at the cathode and plate are also different in such a way that
they compensate for the effects of the differing source impedance's.
Measuring the source impedance is only half the job, you have to also
measure the open circuit Voltages vs. frequency at both the plate and the
cathode. When measuring either the plate or cathode source impedance and
open circuit Voltage response, the load impedance must be disconnected
from the output being measured, while the load impedance is left in place
on the opposite output.
> There's absolutely no reason for all the fuss.
It all depends on whether you believe in science or not.
Henry Pasternack
news:jbyrns-1302...@207-229-151-181.d.enteract.com...
> [Deleted]
Sorry, John, you're grasping at straws.
End discussion.
-Henry
John Byrns
"Henry Pasternack" <eat....@spam.com> wrote:
OK, you clearly plan to ignore the Audio Magazine article Ken pointed out,
probably because it doesn't follow the gospel by Pasternack, and you can't
refute it. Who am I to argue with a computer scientist, if Audio
Magazine and Albert Preisman don't even command any respect from you.
FT
Pasternack, Protagonist.
Here's John's testimony: <snipped wasted words> "duh."
Here's Henry's testimony: <snipped wasted words> "double duh."
And now the Moral Court Judge's Decision:
John, YOU LOSE!!!
Henry, YOU LOSE!!!
$5000 to the audience from each of you! That's the Moral Court's
Decision!
Afterwords, the protagonists are interviewed outside the Moral
Courtroom.
What do you have to say about all this, John? "Um, Duh."
What do you have to say about all this, Henry? "Um, Double Duh."
Although he has not been asked for a response, John comes back to
the interviewer and proclaims, "Duh Double Duh."
Well there you have it, folks. Another exciting day in Moral Court.
Too bad everyone, especially our audience, lost this time. Duh!
Fred
Jack Crenshaw
> In article <9641rd$2rl$1...@slb1.atl.mindspring.net>, "Henry Pasternack"
> What simple and elegant argument? I haven't seen one from you, can I find
> it on "deja-news"? If I can't, it's not a matter of subtlety, but rather
> of not having seen your simple and elegant argument at all. I don't
> remember your weighing in on this argument in the past millennium, and I
> only argued in the past that there was not balance problem with the
> concertina. I am now claiming that while there is no balance problem, it
> is true as they say that the source impedance's are different.
Gee, I hate to interrupt you guys' pissing contest, with which you both seem to be
having so much
fun. But just for purposes of making conversation, let's say that the two source
impedances _ARE_
different. Why does this matter to you?
By definition, the source impedance of any circuit, passive or active, is a
measure of the way current delivered changes, when the load presented to it
changes. Unless you're planning some kind of new circuit in which you plan to put
pots in the grid circuits of the output tubes, I'm wondering why you care.
Jack
Jack Crenshaw
Ken: the only time this topology runs into problems is
when the impedance on one of the nodes (plate or cathode) is NOT equal
to the other. a typical cause for this might be running into grid
current in the output tubes.
John: For the more advanced student there is also the problem that the
concertina has
with nonlinear loads, like grids being driven into clipping.
Apparently, Ken _IS_ one of those advanced students. So why are you arguing
with him?
Jack
Jack Crenshaw
Jack
Henry Pasternack
news:3A8C9707...@earthlink.net...
> Excellent!!! Some folks need to take up a new hobby.
If only you knew, Jack. Remember, no good deed goes unpunished.
-Henry
Tubie
> frequency responses when the two outputs were loaded with equal
> capacitors. Since this doesn't happen (ignoring second-order effects),
> we can say the output impedances are the same. For this to be true,
> both load impedances must be the same. The reason is the the load
> on one leg affects the output impedance of the other. Obviously, if
> you vary one load impedance while holding the other constant, or
> otherwise treat the two outputs asymmetrically, you will observe
> different responses. But again this violates the constraint that the
> two loads on the split-load inverter must be equal.
>
path is an AC bypass to the cathode resostor and it affects the plate
current (or the cathode current, which is the same).
Now the load bypass at the plate does not change much to the cathode
path. I just imagine a load equal to cathode and plate resistor. The
effective load halfes, so the current almost doubles due to the
low cathode impedance.
Ha, I see, with same load on plate it will again have the original
AC swing due to double current but half impedance to drive.
So as a result I see load variances simultaneous affecting the signal
keeping full symmetry.
But the output impedance at the plate from the circuit analysis
point of view still isn't the same as the cathode path.
Thankx for teaching me,
Michael
Henry Pasternack
news:96m49t$jr4$1...@unlisys.unlisys.net...
> But the output impedance at the plate from the circuit analysis
> point of view still isn't the same as the cathode path.
>
> Thankx for teaching me,
I don't have the article John and Ken were talking about
earlier, so I derived the formulas for the two output impedances
from scratch. I assumed the input was at AC ground and that
the resistors in the plate and cathode legs were identical. I then
calculated the output voltage when a test current was fed into
the inputs, one at a time.
Here are the results (want to check my calculations?):
Rk = R * (R + rp) / (2 * R + rp + gm * R * rp)
Rp = (R + gm * R * rp) / (2 + gm * rp)
where:
Rk = cathode output impedance
Rp = plate output impedance
R = resistor in series with plate or cathode
rp = small-signal plate resistance
gm = transconductance
Plugging typical values into these formulas:
R = 22K
rp = 5k
gm = 5000 umho
Rk = 992 Ohms
Rp = 21.2K Ohms
Ordinarily, I wouldn't bother with the math. Instead, I'd observe
that the mu of the tube, gm * rp, is 25. The plate circuit resistance
is the sum of R + rp. Looking into the cathode, this is transformed
down by a factor of mu + 1. So, (22 K + 5K) / (25 + 1) = 1038 Ohms.
Now, we have to put that in parallel with the 22K cathode resistor.
We compute (1038 * 22K) / (1038 + 22K) = 992 Ohms.
Similarly, looking into the plate, I'd transform the cathode series
resistance up by the same factor and add the plate resistance.
22K * (25 + 1) + 5K = 577K. Put that in parallel with the 22K
plate resistor: (577K * 22K) / (577K + 22K) = 21.2K.
Hoo, boy, we're cookin' with gas.
Now this is all fine and dandy and seems to vindicate Mr. Byrns
completely save for one crucial point that completely turns the
whole analysis upside-down.
As I clearly stated on several occasions, my point of view is that
strict symmetry must be observed in dealing with this circuit. This
means not only that the load impedances must be the same, but
also if you are going to inject a current into one output to test the
impedance, you have to inject an equal and opposite current into
the other output at the same time.
If this condition is maintained, it is immediately obvious from
the symmetry of the circuit that the respective voltages induced
in the two outputs are the same. And therefore, since impedance
equals change in voltage divided by change in current, the
inescapable conclusion, under the conditions of my test, it that
the output impedances are identical.
For the intellectually challenged, here is the quick summary.
Two different test conditions, two different results. Is that so
damned hard to comprehend?
If the two outputs are loaded with equal capacitances, the
voltages will remain equal and opposite at all frequencies,
regardless of the absolute level at any particular frequency.
I say again, a little thought reveals this would not be the case
were the output impedances different, even if the gains were
adjusted to be equal at midband.
You might want to ask whether or not the condition of
symmetry is a valid one. I would respond that since the
whole point of the exercise is to invalidate the assersion that
the presumed difference of output impedances unbalances
the split-load phase inverter at high frequencies, it only makes
sense to impose that condition. Otherwise it is obvious from
the get-go that the thing will be out of balance and there is
no need to go to the trouble of analyzing the circuit in the
first place.
Duh and double duh.
I will implore Mr. Byrns not to further test my patience by
replying with one of his characteristically stupefying remarks.
Have a nice day.
-Henry
John Byrns
"Henry Pasternack" <eat....@spam.com> wrote in message
<oYzj6.13743$Nj5.9...@bgtnsc07-news.ops.worldnet.att.net>
[Snip]
> Now this is all fine and dandy and seems to vindicate Mr. Byrns
> completely save for one crucial point that completely turns the
> whole analysis upside-down.
Ignoring for a moment that the above "analysis" is only the beginning of
the calculations necessary to demonstrate how the concertina phase
splitter works, it is questionable if your "crucial point" is really
even valid.
> As I clearly stated on several occasions, my point of view is that
> strict symmetry must be observed in dealing with this circuit.
We are in agreement on this point. Prior to reading the article in
Audio that Ken pointed out, by Albert Preisman, I didn't even feel a
need for an explanation from the point of view of the differing source
impedance's, how the Concertina could possibly be balanced. It seemed
obvious to me that neglecting the effects of grid capacitance, the
concertina is perfectly balanced, if its two loads are perfectly
balanced, by virtue of the fact that the plate and cathode currents must
of necessity be equal. I guess that simple explanation couldn't
overcome the "problem" of the differing source impedance's, for some
people, so Mr. Preisman wrote his article. Henry, if you would like to
read the article, I will scan it and email you a copy if you would like.
> This
> means not only that the load impedances must be the same, but
> also if you are going to inject a current into one output to test the
> impedance, you have to inject an equal and opposite current into
> the other output at the same time.
This is where we differ, I don't think this is the standard definition
of source impedance, and how you measure it. I don't believe it is
valid to inject currents into nodes other than the one that is being
measured, of course there is also a complimentary current in the return
path. But to simultaneously inject a current into a second node is
invalid as far as I know. Do you have any references that indicate this
is a valid procedure? It would be valid to make a measurement between
the cathode and plate, with the current injected into the plate
returning via the cathode node, but that would be measuring the source
impedance between the cathode and plate terminals, which is not exactly
how the concertina is actually used.
> If this condition is maintained, it is immediately obvious from
> the symmetry of the circuit that the respective voltages induced
> in the two outputs are the same.
OK, I'm not sure if I followed that, but if I did it sounds OK.
> And therefore, since impedance
> equals change in voltage divided by change in current, the
> inescapable conclusion, under the conditions of my test, it that
> the output impedances are identical.
I don't see how this follows, your test is connecting to three points at
once, the plate, the ground reference, and the cathode. As far as I
know you are only allowed to inject current at two points when measuring
source impedance, the node being measured, and the reference node, you
are injecting a current at a third node which invalidates the
measurements.
> For the intellectually challenged, here is the quick summary.
> Two different test conditions, two different results. Is that so
> damned hard to comprehend?
For the intellectually challenged, here is the quick summary. Two
different test conditions, two different results, one test condition is
valid, the other is not. Is that so hard to comprehend?
> If the two outputs are loaded with equal capacitances, the
> voltages will remain equal and opposite at all frequencies,
> regardless of the absolute level at any particular frequency.
Fine so far.
> I say again, a little thought reveals this would not be the case
> were the output impedances different, even if the gains were
> adjusted to be equal at midband.
I disagree with this. If you read Preisman's article you will
understand that the concertina remains balanced even with differing
source impedance's because just as the source impedance's are different,
so are the Voltage gains from the grid to the plate and the cathode.
The two effects work to cancel each other out.
> You might want to ask whether or not the condition of
> symmetry is a valid one. I would respond that since the
> whole point of the exercise is to invalidate the assersion that
> the presumed difference of output impedances unbalances
> the split-load phase inverter at high frequencies, it only makes
> sense to impose that condition. Otherwise it is obvious from
> the get-go that the thing will be out of balance and there is
> no need to go to the trouble of analyzing the circuit in the
> first place.
The condition of symmetry of the two loads is not only valid, it is
required for the concertina to be balanced. I would also point out that
if you read the recent audiophile literature, it is not just the high
frequency issue that bothers people, but also the low frequencies.
There have been articles that state that the differing source
impedance's cause the low frequency time constants of the plate to grid
coupling capacitors to be different, even when the two coupling
capacitors are of equal value. This is not really the case, just as
with the high frequencies.
> Duh and double duh.
Indeed.
> I will implore Mr. Byrns not to further test my patience by
> replying with one of his characteristically stupefying remarks.
I trust these remarks have been stupefying enough to severely try your
patience.
> Have a nice day.
OK, you too.
There are a few more things I would like to mention. Calculating the
source impedance's is only about 25% of the problem of analyzing the
concertina phase splitter. Besides calculating the source impedance at
both the plate and cathode, it is also necessary to calculate the gains
at those two points. Once the plate and cathode source impedance's and
gains are in hand we can create Voltage dividers between the source
impedance and the load impedance for both the plate and cathode, and
then combine that with the gain function for both the plate and cathode
to get the two transfer functions for the plate and cathode outputs. It
is important to note that when measuring the source impedance and gain
at either the plate or cathode node, the load impedance must be
disconnected from that node, but the load impedance must be left in
place on the opposite node. The plate and cathode resistors can either
be considered part of the load, or they can be considered part of the
concertina circuit proper, and left in place for the measurements. The
coupling capacitors to the grids of the following tubes, the grid
resistors of those tubes, stray capacitance's, and miller capacitance's
associated with the following grids are all part of the load impedance.
If there is interest in the complete analysis, I should have time in
about three weeks to post the whole thing. It is a lot of boring math,
and I will skip it and spare you all the drudgery, unless there is some
interest expressed.
Henry Pasternack
if it is to be balanced and work properly. We agree that if that is so,
the currents flowing in and out of the two outputs will be equal and
opposite. Under that condition the ground returns cancel and there
is effectively zero net ground current in the output circuit.
Any test that breaks this symmetry breaks the phase inverter.
The conclusions of that test may be valid, but not to the operation
of this phase inverter in its typical application.
Your test mandates breaking the symmetry:
"It is important to note that when measuring the source impedance
and gain at either the plate or cathode node, the load impedance
must be disconnected from that node, but the load impedance
must be left in place on the opposite node."
Since your test destroys the inverter's balance, It follows that the
measured output impedances will differ. My own calculations, done
to prove I am not missing the point, show a ratio of twenty-to-one with
typical component values.
You are trying to solve the wrong problem. The phase inverter is
not intended to be used with one output disconnected. The outputs
are not independent. The entire premise of your experiment is wrong
because you are not taking into account the interconnectedness of
the two halves of the circuit. You are trying to test the circuit under
conditions in which it is guaranteed to be broken. You totally fail to
grasp how wrong results arise from flawed premises. You miss the
forest for the trees.
Your test would be valid if you built two split-load phase inverters,
and used the cathode circuit of one to drive one output tube, and
the plate circuit of the other to drive the other tube. But then it
wouldn't be a split-load phase inverter, would it?
I got back into this discussion because a nice gentleman asked
me to explain my thinking to him. I've now done that to the best of
my ability. You are completely free to respond however you like.
If that means spending three weeks scribbling down ten or twenty
pages of math, please be my guest.
I would prefer if you hid in a hole. I seriously doubt anyone else
is vaguely interested in this anal-retentive discussion. I know I'm
not.
You are in error, Nomad. Exercise your prime directive.
Sterilize.
SSSTTTEEERRRRIIIIILLIIIIIZZEEEEE.
-Henry
John Byrns
"Henry Pasternack" <eat....@spam.com> wrote:
> We agree that the split-load phase inverter has to have equal loads
> if it is to be balanced and work properly. We agree that if that is so,
> the currents flowing in and out of the two outputs will be equal and
> opposite. Under that condition the ground returns cancel and there
> is effectively zero net ground current in the output circuit.
Yes we agree on that. It is also another way look at the balance issue,
although it feels like a complex variation on the simple observation
that the same current flows in both the cathode and plate circuits,
therefore if the loads are balanced, the output Voltages must also be
balanced. The problem is that you can't measure the output impedance's
of the plate and cathode circuits that way. If you disconnect the
ground returns, you are simply measuring the single source impedance
between the plate and cathode terminals.
> Any test that breaks this symmetry breaks the phase inverter.
> The conclusions of that test may be valid, but not to the operation
> of this phase inverter in its typical application.
In the typical application of the concertina, the ground returns of the
loads are not broken, hence it is your test that is broken and not a
valid representation of the typical operation of the concertina.
> Your test mandates breaking the symmetry:
>
> "It is important to note that when measuring the source impedance
> and gain at either the plate or cathode node, the load impedance
> must be disconnected from that node, but the load impedance
> must be left in place on the opposite node."
>
> Since your test destroys the inverter's balance, It follows that the
> measured output impedances will differ. My own calculations, done
> to prove I am not missing the point, show a ratio of twenty-to-one with
> typical component values.
My test does not destroy the inverter's balance, it just separates the
transfer function for each output into it's component parts, the source
impedance's, and the gains. It is common engineering methodology. By
the way it is not really my test, it is Albert Preisman's test. I
originally only observed that the two output impedance's were different,
yet the concertina was balanced with balanced loads. I did not try to
reconcile the unbalanced source impedance's of the concertina with the
balance of the total circuit, as that balance was obvious to me simply
by looking at the constraints on the currents. You said the output
impedance's were equal and that you had an elegant proof. I asked for
the proof, and Ken posted the reference to Preisman's article as proof.
As it turned out Preisman says the output impedance's are different, not
the same, and his whole thing is to show how the concertina can be
balanced in the face of differing source impedance's.
> You are trying to solve the wrong problem. The phase inverter is
> not intended to be used with one output disconnected. The outputs
> are not independent. The entire premise of your experiment is wrong
> because you are not taking into account the interconnectedness of
> the two halves of the circuit. You are trying to test the circuit under
> conditions in which it is guaranteed to be broken. You totally fail to
> grasp how wrong results arise from flawed premises. You miss the
> forest for the trees.
Sure I am taking the "interconnectedness" of the two halves of the
circuit into account. I only measure the internal source impedance's,
and internal generator characteristics with one load disconnected. The
final computations connect both loads back up and the final transfer
function is derived from the source impedance, the internal generator
characteristics, the load impedance's, and the Voltage divider formed
between the source impedance's and the loads. Everything is taken into
account, the result of all the computations is that the concertina is
balanced in spite of the unequal source impedance's. I didn't really
know or care about this complex argument until Ken brought it to my
attention. A forest is made up of trees, if you put all the trees
together you will have a forest.
> Your test would be valid if you built two split-load phase inverters,
> and used the cathode circuit of one to drive one output tube, and
> the plate circuit of the other to drive the other tube. But then it
> wouldn't be a split-load phase inverter, would it?
That would only work if you had four equal loads, one for each plate and
grid circuit. At that point what difference does it make which cathode
drives one output tube grid and which plate drives the other? Why not
just throw out one tube and drive the grids from the plate and cathode
of one tube.
> I got back into this discussion because a nice gentleman asked
> me to explain my thinking to him. I've now done that to the best of
> my ability. You are completely free to respond however you like.
> If that means spending three weeks scribbling down ten or twenty
> pages of math, please be my guest.
I did not say I was going to spend three weeks scribbling down ten or
twenty pages of math, I said I would post it in about three weeks, when
I would have time to organize and clean it up. I also said I would only
bother typing it up if there was any interest in it, and while I didn't
say it, I specifically meant interest from someone other than you. I
exclude you because I suspect that you have already gone through all the
math, and that you understand Albert Preisman's argument, and are just
trying to obscure the issue to save face. It is really ironic that Ken
jumped in to defend you, but the paper he served up turned out to really
be explaining how balance was still possible even with different source
impedance's driving the two outputs.
> I would prefer if you hid in a hole. I seriously doubt anyone else
> is vaguely interested in this anal-retentive discussion. I know I'm
> not.
>
> You are in error, Nomad. Exercise your prime directive.
>
> Sterilize.
>
> SSSTTTEEERRRRIIIIILLIIIIIZZEEEEE.
I am not sure what that is all about, but if you are equating me to this
"Nomad", it is not me that is in error, it is you that are in error. If
the "Nomad" were to exercise its prime directive it would have to
"Sterilize" you.
RD
<eat....@spam.com> wrote:
> I would prefer if you hid in a hole. I seriously doubt anyone else
>is vaguely interested in this anal-retentive discussion. I know I'm
>not.
A friend of mine used to play a Concertina. He got lots of negative
feedback. The more he got, the worse it sounded. There's a lesson in
that. :o}
RD
Steve O'Neill
John Byrns wrote in message ...
>>You are in error, Nomad. Exercise your prime directive.
>>
>> Sterilize.
>>
>> SSSTTTEEERRRRIIIIILLIIIIIZZEEEEE.
>
>I am not sure what that is all about, but if you are equating me to this
>"Nomad", it is not me that is in error, it is you that are in error. If
>the "Nomad" were to exercise its prime directive it would have to
>"Sterilize" you.
NOMAD's prime directive: seek out new life forms and sterilize all those not
PERFECT. NOMAD mistakes James T. Kirk the captain for James Roy Kirk "The
Creator". Thus NOMAD is not perfect. Therefore NOMAD must
SSSTTTEEERRRRIIIIILLIIIIIZZEEEEE . Bye-bye NOMAD. Thank God for recursive
Vulcan logic else we all be STERILIZED.
Steve
Tubie
check it out.
Seems that I developed a kill-file filter in my head after reading
this NG for a while, It's skipping the childish crossfights between
posters.
Michael
"Steve O'Neill" <steve....@worldnet.att.net> wrote in message
news:K1Ij6.3544$TD1.2...@bgtnsc05-news.ops.worldnet.att.net...
Jack Crenshaw
I, too, used to worry about differences in output impedance, but it only
takes a few minutes of thought to convince oneself that, as everyone in
this discussion seems to agree, the split-load phase inverter absolutely
must deliver the same voltage to both outputs, since the same current goes
through matched resistors.
After a little more thought, I realize that the source of the confusion is
this: The term, "output impedance" is a little misleading when it comes
to circuits that have more than one output. It's really a term intended to
be used with single-output devices.
The general idea of an output impedance is that you can replace the device
in question with an equivalent circuit represented by an ideal voltage
amplifier, followed by a source resistance. The output impedance is
really a measure of how "stiff" the circuit is; i.e., how much the output
voltage falls off as you draw current from it.
You can do this, for example, with a cathode follower. In this case, the
cathode follower is very stiff, meaning that, because of the cathode
feedback, the output voltage hardly changes at all as you change the
load. The same is true, in fact, for entire (feedback) power amplifiers.
The feedback tends to keep the output voltage constant regardless of load,
so its output impedance is much lower than the 4-16 ohms of the load.
Hence the term, "Damping factor."
In the case of the split-load inverter, you have _TWO_ output terminals,
and the voltage presented to either output reflects changes in the load on
either. How, in a case like that, does one use the term "output
impedance" at all? What equivalent circuit of ideal voltage source and
ideal source resistance, can you build that correctly reflects the actual
behavior?
The impedances that Henry (and John, and others) have worked out would
give the correct output impedance at one terminal, _IF_ the other terminal
were not connected to a load at all <!>.
I guess it's my opinion that one can't even come up with an equivalent
circuit with two outputs that behave in the same way as the split-load
inverter. Therefore, any computation of output impedance is moot. It's
pointless to calculate an equivalent source impedance for an equivalent
circuit that doesn't exist.
Jack
Henry Pasternack
interesting development (and one that totally supports my position)
I will make another reply to John's erroneous comments.
> If you disconnect the ground returns, you are simply measuring the
> single source impedance between the plate and cathode terminals.
No, because the voltage you look for is the single-ended voltage,
not the voltage between the two terminals.
> In the typical application of the concertina, the ground returns of the
> loads are not broken, hence it is your test that is broken and not a
> valid representation of the typical operation of the concertina.
It doesn't matter whether or not the ground returns are broken.
Since the currents flowing in and out of the two outputs are equal
and opposite, the net ground current is zero.
But none of this is really related to the heart of the matter.
> As it turned out Preisman says the output impedance's are different,
> not the same, and his whole thing is to show how the concertina can
> be balanced in the face of differing source impedance's.
This, my friend, is where you take your break with reality. Let's see
what Preisman himself says. He begins by observing that in the case
where the two loads are equal, the source impedances (no apostrophe)
are the same:
"[Fig. 2]
If we now impose the phase-splitter
condition that Zk = Zl = Z, we obtain:
Elk = Elp
= u * eg * Z / [rp + (1 + u) * Z + Z]
...
Note, then, that from this viewpoint, we
have the same apparent generated volt-
age, u * eg, acting for either output, and
the same apparent source impedance
[rp + (1 + u ) * Z], rather than unequal
source impedance as given by Eqs. (1)
and (2)."
This result is extremely significant and you just gloss right over it.
It says that if we treat the tube as a voltage generator with a series
resistance, then from the point of view of either output, provided the
loads are the same, both the generator voltage and the series
resistance are the same. This, of course, is exactly what I have
been saying all along.
He then goes on to show how to transform the expression for the
cathode impedance so that it corresponds to the lower value of the
original equations. But he notes:
"The equivalent circuit is shown in Fig.
3. It holds whether Zk is equal to Zl,
as in the case of the phase-splitter, or
Zk is not equal to Zl...
In the special case where Zk = Zl, the
circuit of Fig. 3 becomes exactly equiva-
lent to that of Fig. 2, so that we can say
in this case that the internal source im-
pedance for the plate output terminal
Ap of Fig. 1 is also as low provided we
also accept a lower apparent generated
voltage."
Preisman reiterates that under the constraint of equal load impedances,
it is valid to assume the same low source impedance for both the plate
and cathode legs, provided the generator voltages are adjusted properly.
He goes on:
"When Zl = Zk, the individual impedances
lose their separate identities, as do also
Eqs. (9) and (10), whereupon we can
regard either output terminal as having a
higher or lower source impedance, provided
we also adjust the apparent generated
voltages to correspondingly higher or
lower values. It is only when the permit
Zk and Zl to be unequal that we must use
Eqs. (8) and (9) separately rather than
use Eq. (11) for both output voltages."
In other words, it is only when the load impedances differ that we
must treat the source impedances as being unequal. This completely
validates what I have been saying all along. I am astonished that you
can read Preisman's paper and come to any other conclusion. The
only explanation is that you don't fully understand the paper.
> Sure I am taking the "interconnectedness" of the two halves of the
> circuit into account. I only measure the internal source impedance's,
> and internal generator characteristics with one load disconnected.
> The final computations connect both loads back up and the final transfer
> function is derived from the source impedance, the internal generator
> characteristics, the load impedance's, and the Voltage divider formed
> between the source impedance's and the loads. Everything is taken into
> account, the result of all the computations is that the concertina is
> balanced in spite of the unequal source impedance's.
And this is precisely what you cannot do, because, as Preisman
himself notes:
"We see, therefore, that the paradox
is resolved if we take into account not
only the change in source impedance
but also the change in source-generated
voltage. One can compensate for the
other, but only in the case where Zl = Zk.
Otherwise, a variation in either imped-
ance causes the opposite effect upon
the output voltage of the other terminal."
Translating, if you disconnect one load impedance while performing
your test, you will induce a change in the apparent generator voltage
that will dramatically change the source impedance measured at the
other terminal. Yes, when you put the circuit back together, it is
evident that the phase splitter is balanced. If you have to take the
phase splitter apart to test it, then you are not measuring the same
circuit that actually exists inside your amplifier. You are measuring
the impedances of a taken-apart phase inverter. You are violating the
constraint that the impedances in both legs of the split-load phase
inverter must be the same.
Preisman doesn't go all the way in resolving the paradox. In his
explanation of Eq. (12), he notes that when the loads are the same, it
is equally valid to presume either the low or the high value of source
impedance, provided the generator voltage is adjusted appropriately.
Since it's obvious the generator voltage can't have two values at the
same time, something remains to be explained.
The definition of impedance is the derivative of voltage with respect to
current. To measure impedance, you need to induce a change in
current and measure the resulting change in voltage at some node in
a circuit. This can be done under zero-signal conditions by injecting
a test current directly (which is how an ohmmeter works, so don't give
me any crap about my test methodology being wrong). Otherwise
it can be done indirectly by driving the circuit with a signal and varying
the load resistance. Either way, the result is a change in output current
and voltage. Let's assume for the sake of clarity that we use the latter
method for our analysis.
If we perform the test on one output at a time as you propose, we
change one load resistor while keeping the other one the same. In
this case, Preisman states that the separate equations for source
impedance must be used. It follows that we will measure very
different source impedances in the two legs.
On the other hand, it is a prerequisite for the split-load phase
inverter that the load impedances must be equal. If we make the
measurement as before, but simultaneously vary the value of the
opposite load resistor to keep the circuit in balance, we will get a
completely different result. The reason is that changing the value
of the opposite load resistor changes the apparent generator voltage
during the course of the test. Normally this would invalidate the test
but in this particular instance it is a prerequisite if proper results are
to be obtained. This is because by definition we have to have equal
loads when dealing with the split-load phase inverter (I assert for
the dozenth time).
We can now explain Preisman's paradoxical assertion that either
Eq. (11) or (12) is valid for the case of Zk = Zl. In the limit, for
arbitrarily small values of dR, di, and dv, we can say that the source
impedance is measured at a single value of load impedance. The
first equation represents the case where the load impedances vary
simultaneously during the test while the second equation represents
the case where one impedance changes and the opposite impedance
is fixed.
We have two different test cases and two different results. Which
one is valid? Clearly if we impose the equal load criterion, the second
test is the proper one and the values we measure will be identical on
either output.
Granting some consideration to your point of view, Preisman shows
that it is valid to presume the lower or higher source impedance (as
the case may be) and adjust the implicit generator voltage accordingly
to fit the measured results. This is not something that would ever be
apparent from an external measurement but if it gives you pleasure to
think of the circuit in these terms, please feel free to do so.
What I would ask of you is to quit insisting that I have it all wrong,
thereby forcing me to delve exhaustingly into progressively deeper
and deeper levels of detail, but never giving me the satisfaction of
having a resulution to the disagreement. If history is any guide, I
will be rewarded for my efforts by your enduring accusations. Kindly
do not blame me for your inability to comprehend a subtle line of
argument. I have been most accommodating in laying this out for
you in excruciatingly painstaking detail.
And that, Jack Crenshaw, if you're still with me, is why I say no
good deed goes unpunished on rec.audio.tubes.
-Henry
Henry Pasternack
interesting development (and one that totally supports my position)
I will make another reply to John's erroneous comments.
> If you disconnect the ground returns, you are simply measuring the
> single source impedance between the plate and cathode terminals.
No, because the voltage you look for is the single-ended voltage,
not the voltage between the two terminals.
> In the typical application of the concertina, the ground returns of the
> loads are not broken, hence it is your test that is broken and not a
> valid representation of the typical operation of the concertina.
It doesn't matter whether or not the ground returns are broken.
Since the currents flowing in and out of the two outputs are equal
and opposite, the net ground current is zero.
But none of this is really related to the heart of the matter.
> As it turned out Preisman says the output impedance's are different,
> not the same, and his whole thing is to show how the concertina can
> be balanced in the face of differing source impedance's.
This, my friend, is where you take your break with reality. Let's see
"[Fig. 2]
...
He goes on:
> Sure I am taking the "interconnectedness" of the two halves of the
> circuit into account. I only measure the internal source impedance's,
> and internal generator characteristics with one load disconnected.
> The final computations connect both loads back up and the final transfer
> function is derived from the source impedance, the internal generator
> characteristics, the load impedance's, and the Voltage divider formed
> between the source impedance's and the loads. Everything is taken into
> account, the result of all the computations is that the concertina is
> balanced in spite of the unequal source impedance's.
And this is precisely what you cannot do, because, as Preisman
himself notes:
> I exclude you because I suspect that you have already gone through
> all the math, and that you understand Albert Preisman's argument,
> and are just trying to obscure the issue to save face.
This is utterly preposterous, and it is precisely because you always
say this that I think you are a damned lunatic.
Henry Pasternack
> The impedances that Henry (and John, and others) have worked out would
> give the correct output impedance at one terminal, _IF_ the other terminal
> were not connected to a load at all <!>.
Or, specifically, to the fixed resistor that is required by the circuit
to
establish its operating point. You are absolutely correct, and it is for
just this reason that I question John's attempt to analyze the inverter
as two separate half-circuits.
> I guess it's my opinion that one can't even come up with an equivalent
> circuit with two outputs that behave in the same way as the split-load
> inverter. Therefore, any computation of output impedance is moot. It's
> pointless to calculate an equivalent source impedance for an equivalent
> circuit that doesn't exist.
Quite true. Since the two outputs interact so intimately, it's futile to
try to build a model using two independent outputs. What happens
to the load on one output has a big effect on the behavior of the other
output.
What we can do is model the tube as a floating generator having a
single amplitude and series resistance. This is exactly what the
Preisman paper proposes and it makes perfect sense. Under this
condition, if and only if the load impedances are the same, we can
establish a virtual ground in the center, split the generator in two,
and assign equal half-amplitudes and half-impedances to each side.
The concept is elegantly obvious. I hope you will forgive me for
my irritation with Mr. Burns for telling me there is something wrong
with me for proposing it.
-Henry
John Byrns
"Henry Pasternack" <eat....@spam.com> wrote:
> "Jack Crenshaw" <jcr...@earthlink.net> wrote in message
> news:3A8FAA2F...@earthlink.net...
> > The impedances that Henry (and John, and others) have worked out would
> > give the correct output impedance at one terminal, _IF_ the other terminal
> > were not connected to a load at all <!>.
>
> Or, specifically, to the fixed resistor that is required by the circuit
> to
> establish its operating point. You are absolutely correct, and it is for
> just this reason that I question John's attempt to analyze the inverter
> as two separate half-circuits.
>
> > I guess it's my opinion that one can't even come up with an equivalent
> > circuit with two outputs that behave in the same way as the split-load
> > inverter. Therefore, any computation of output impedance is moot. It's
> > pointless to calculate an equivalent source impedance for an equivalent
> > circuit that doesn't exist.
>
> Quite true. Since the two outputs interact so intimately, it's futile to
> try to build a model using two independent outputs. What happens
> to the load on one output has a big effect on the behavior of the other
> output.
Not true at all, see the analysis below, which analyzes the plate and
cathode outputs separately. You need to review your notes from EE102,
Introductory Circuit Theory.
> What we can do is model the tube as a floating generator having a
> single amplitude and series resistance. This is exactly what the
> Preisman paper proposes and it makes perfect sense. Under this
> condition, if and only if the load impedances are the same, we can
> establish a virtual ground in the center, split the generator in two,
> and assign equal half-amplitudes and half-impedances to each side.
It is only one of the things Preisman discusses in his paper, it is not
the actual thesis he presents however. I indicted in a response yesterday
that is a valid way to look at the balance question, I consider it a
complex variation on the equal current argument, but it in no way implies
the two source impedance's are equal.
> The concept is elegantly obvious. I hope you will forgive me for
> my irritation with Mr. Burns for telling me there is something wrong
> with me for proposing it.
Yes as I said in an earlier response, it is a valid explanation of the
balance issue. The reason I say you are wrong for proposing it is because
you are trying to travel both sides of the street at once. While it is a
valid analysis of the balance issue, it in no way represents the way the
concertina is actually used in an amplifier. In a real amplifier the
center if the load is not connected to a virtual ground, it is connected
to the real ground. You claimed my analysis is wrong for including a
ground, so should I be irritated with you? That is where my analysis
proceeds from, two separate, and not necessarily equal loads with a real
ground, as in a real amplifier. My analysis provides the actual output
Voltages for unbalanced loads as well as the balanced case. My, actually
Preisman's analysis, is the general case.
A few comments on Preisman's article. Preisman clearly states in his
article that the source impedance of the plate and cathode outputs of the
concertina are different. Nowhere in the article do I see any statement
that they are equal. When he first discusses the fact that the outputs
are balanced in spite of the differing source impedance's, he says the
"regulation" of both outputs are the same, not the source impedance's are
the same. As he gets into the main part of his thesis, he unfortunately
uses the term "apparent source impedance", which is ambiguous in its
meaning, and can lead to confusion with the true source impedance.
OK Henry, you have goaded me into prematurely publishing a rough draft of
my analysis of the concertina. I don't really consider this to be my
analysis, I consider it to be Preisman's work, but since you take
exception with my interpretation of what Preisman is saying in his paper,
I will call it my own for the purposes of discussion, but I give full
credit to Preisman. I have modified my analysis slightly to align it more
closely with the way Preisman presents it in his paper. Mainly I have
subsumed the plate and cathode resistors into the plate and cathode
loads. Equations 1, 2, & 3 for the cathode equivalent circuit come
directly from Preisman's figure 3. Equations 5, 6, & 7 are what I will
call figure 3.5, which Preisman didn't include, and are the equations for
the plate equivalent circuit. Notice that the generator impedance's and
Voltages are different in the cathode and plate circuits, yet if the load
impedance's are balanced, the output Voltages will also be balanced.
Notice that this analysis is completely general, and can be used to
calculate the two output Voltages even under the condition where the plate
and cathode loads are not equal.
The equivalent generator impedance at the cathode of the concertina is:
(1) Zgk = (rp + Zp) / (u + 1)
The equivalent generator Voltage at the cathode of the concertina is:
(2) egk = es * u / (u + 1)
Combining 1 & 2 with Zk the load Voltage at the cathode of the concertina is:
(3) ELk = es * (u / (u + 1)) * Zk / ((rp + Zp) / (u + 1) + Zk)
Rearranging equation 3 we get:
(4) ELk = es * u * Zk / (rp + (u + 1) * Zk + Zp)
The equivalent generator impedance at the plate of the concertina is as follows:
(5) Zgp = rp + (u + 1) * Zk
The equivalent generator Voltage at the plate of the concertina is as follows:
(6) egp = es * u
Combining 5 & 6 with Zp the load Voltage at the cathode of the concertina is:
(7) ELp = es * u * Zp / (rp + (u + 1) * Zk + Zp)
Notice that equation 4 for the Voltage across the cathode load is
identical with equation 7 for the Voltage across the plate load, with the
exception Zk in the numerator of eq 4, and Zp in the numerator of equation
7. If the loads on the concertina, Zk and Zp, are equal, then the output
Voltages at the plate and cathode are equal. This is true even though the
source impedance at the cathode terminal, given by equation 1, and the
source impedance at the plate, given by equation 5, are no where near
equal.
The meaning of the symbols in the above equations is as follows:
es is the source Voltage at the input of the concertina circuit
rp is the plate resistance of the tube (valve) in the concertina circuit
u is the amplification factor of the tube (valve) in the concertina circuit
Zp is the load on the concertina plate, including the plate resister
Zk is the load on the concertina cathode, including the cathode resister
Zgp is the generator source impedance at the plate terminal of the concertina
Zgk is the generator source impedance at the cathode terminal of the concertina
egp is the generator source Voltage at the plate terminal of the concertina
egk is the generator source Voltage at the cathode terminal of the concertina
ELp is the load Voltage at the plate terminal of the concertina
ELk is the load Voltage at the cathode terminal of the concertina
OK Henry, have at it, tell me where I screwed up by treating the cathode
and plate circuits as if they were independent? Keep in mind that this is
a quickly typed rough draft, and may contain many typos.
Henry Pasternack
> > OK Henry, have at it, tell me where I screwed up by treating the cathode
> and plate circuits as if they were independent?
Your calculations are correct and I will now use them to prove my
point. I suggest we plug in some values and form a practical example:
Zp = Zk = 10K
rp = 1K
u = 10
These are extraordinary values for a tube, but within reason, and easy
to push through the calculator.
For the test, we will set the source voltage to 1.0V and we will use
a test load of 990K Ohms. Note that the value of 990K in parallel with
Zp = Zk = 10K is 9.9K. We will measure the no-load voltage first, and
then observe what happens at each output when we apply the test load
first to one output, then to the other, then to both. Finally, we will
use the resistor divider equation to compute the apparent source
impedances.
Because the load resistor is so high, I'm carrying the calculations
out to many decimal places so as not to lose track of the differences.
I will call the case where the load resistor is on the leg under test
only the "proximal" case. With the resistor on the opposite leg only,
I will call it the "distal" case. With a load on both legs, I will
call it the "dual" case. To compute the apparent source impedances,
I will use the equation:
Zo = Rtest * (1 - r) / r
where 'r' is the test output divided by the no-load output
Here are the results:
Cathode circuit:
1) No load: Vout = 0.826446281
2) Proximal: Vout = 0.825688073 Zout = 909.1 Ohms
3) Distal: Vout = 0.827129860
4) Dual: Vout = 0.826377296 Zout = 82.64 Ohms
Plate circuit:
1) No load: Vout = 0.826446281
2) Proximal: Vout = 0.818858561 Zout = 9174 Ohms
3) Distal: Vout = 0.834028357
4) Dual: Vout = 0.826377296 Zout = 82.64 Ohms
I've omitted the impedance calculation for the "distal" case since
it makes no practical sense.
Now for some interpretation. Under no-load conditions, the two
output voltages are equal as expected. In the "proximal" test cases,
which are the ones where you do the measurement conventionally, one
output at a time, the source impedances differ by a wide margin as
you and I both have predicted previously.
The "distal" data show us what happens to one output while the
other output is under test. Applying the test load on one side
actually causes the output of the other leg to rise slightly, as
noted by Preisman. This illustrates the extent to which the test
throws the inverter out of balance.
Finally, the "dual" data reveal two important facts. The first,
of course, is that both impedances are equal. The second is that
the apparent source impedance is lower than the value in either of
the single-ended cases. This is due to the compensating effect of
the increased generator voltage when the opposite load impedances
are reduced.
I have always stated as a condition to my argument that the load
impedances must be the same. I claim that your test methodology is
inappropriate because it breaks this condition as demonstrated by
the data. I propose that symmetry can be preserved and meaningful
results obtained by performing the test on both outputs at the same
time. Using your own formulas, I prove that the source impedances
observed using this method are equal.
With respect to your other comments, I'm surprised by your claim
that Preisman says nowhere in his article that the output impedances
are equal. In my previous post I quoted several passages straight
from the article where he says just that. It is apparent to me that
you are still misinterpreting the article (in part because, for all
the fuss that's been made over it, I don't think the article does
a very good job of explaining the "paradox").
Regarding the question of virtual ground, the fact that there is
a ground connection in the stage following the inverter does not
mean that any net current from the inverter's output flows in that
leg. This is true provided the inverter is truly balanced. You
can visualize this by connecting a 1:1 voltage divider between the
two outputs and measuring the signal voltage between the center tap
and ground; it will be zero and the circuit will behave the same
whether or not the tap is grounded.
I now brace myself for your next shocking revelation.
-Henry
John Byrns
"Henry Pasternack" <eat....@spam.com> wrote:
> "John Byrns" <jby...@enteract.com> wrote in message
> news:jbyrns-1802...@207-229-151-194.d.enteract.com...
> > > OK Henry, have at it, tell me where I screwed up by treating the cathode
> > and plate circuits as if they were independent?
>
> Your calculations are correct and I will now use them to prove my
> point.
I don't think you really mean to say my calculations are correct, because
my calculations include formulas for the cathode and plate source
impedance's, if you accepted that, the argument would be over, so clearly
since the argument is not over, you don't really mean to say my
calculations are correct.
> I suggest we plug in some values and form a practical example:
>
> Zp = Zk = 10K
> rp = 1K
> u = 10
>
> These are extraordinary values for a tube, but within reason, and easy
> to push through the calculator.
>
> For the test, we will set the source voltage to 1.0V and we will use
> a test load of 990K Ohms. Note that the value of 990K in parallel with
> Zp = Zk = 10K is 9.9K. We will measure the no-load voltage first, and
> then observe what happens at each output when we apply the test load
> first to one output, then to the other, then to both. Finally, we will
> use the resistor divider equation to compute the apparent source
> impedances.
>
> Because the load resistor is so high, I'm carrying the calculations
> out to many decimal places so as not to lose track of the differences.
> I will call the case where the load resistor is on the leg under test
> only the "proximal" case. With the resistor on the opposite leg only,
> I will call it the "distal" case. With a load on both legs, I will
> call it the "dual" case. To compute the apparent source impedances,
> I will use the equation:
>
> Zo = Rtest * (1 - r) / r
>
> where 'r' is the test output divided by the no-load output
OK, good, let's have at it!
> Here are the results:
>
> Cathode circuit:
>
> 1) No load: Vout = 0.826446281
> 2) Proximal: Vout = 0.825688073 Zout = 909.1 Ohms
> 3) Distal: Vout = 0.827129860
> 4) Dual: Vout = 0.826377296 Zout = 82.64 Ohms
>
> Plate circuit:
>
> 1) No load: Vout = 0.826446281
> 2) Proximal: Vout = 0.818858561 Zout = 9174 Ohms
> 3) Distal: Vout = 0.834028357
> 4) Dual: Vout = 0.826377296 Zout = 82.64 Ohms
I have not checked your Vout values, I am assuming them to be correct.
The "Proximal" source impedance test also seems to give the correct source
impedance for the "No load" condition. You have made an error in applying
my test for the "Dual" or loaded case. Remember that I said that when you
test the cathode source impedance, you must have the load on the plate
connected, and vice versa for the plate source impedance test.
The corrected results are as follows:
Cathode circuit:
1) No load: Vout = 0.826446281
2) Proximal: Vout = 0.825688073 Zout = 909.1 Ohms
3) Distal: Vout = 0.827129860
4) Dual: Vout = 0.826377296 Zout = 901.57 Ohms
Plate circuit:
1) No load: Vout = 0.826446281
2) Proximal: Vout = 0.818858561 Zout = 9174 Ohms
3) Distal: Vout = 0.834028357
4) Dual: Vout = 0.826377296 Zout = 9165.97 Ohms
These corrected "Dual" Zout test values are the true source impedance's
for the conditions you state, and they are not equal. These "Dual" Zout
values are identical to the values that result from applying my formulas
for Zgk and Zgp to your circuit parameters.
> With respect to your other comments, I'm surprised by your claim
> that Preisman says nowhere in his article that the output impedances
> are equal. In my previous post I quoted several passages straight
> from the article where he says just that.
I was just asking for an indication of where in his article Preisman
claimed the cathode and plate source impedance's are the same. Thanks for
pointing out your earlier posts, which unfortunately I haven't read yet.
I will check out whatever quotes you have included and comment after I
have read them.
> It is apparent to me that
> you are still misinterpreting the article (in part because, for all
> the fuss that's been made over it, I don't think the article does
> a very good job of explaining the "paradox").
I agree that Preisman's article doesn't do much to resolve the "paradox",
I think I indicated something along that line in an earlier response to
one of Ken's posts. I may very well be misinterpreting Preisman's
article, but that is really neither here nor there, as I have no
obligation to defend his position, I am only trying to defend my position,
and if Preisman doesn't agree with it, then it makes sense I am not going
to defend him. My arguments stand on their own if they aren't the same as
Preisman's. In any case Preisman's work, whether or not I have
misinterpreted it, has been very helpful to me in formulating my own
position.
> Regarding the question of virtual ground, the fact that there is
> a ground connection in the stage following the inverter does not
> mean that any net current from the inverter's output flows in that
> leg. This is true provided the inverter is truly balanced. You
> can visualize this by connecting a 1:1 voltage divider between the
> two outputs and measuring the signal voltage between the center tap
> and ground; it will be zero and the circuit will behave the same
> whether or not the tap is grounded.
I agree that the virtual ground is valid if the concertina is balanced,
but by the same token, when the circuit is balanced, it is also perfectly
legitimate to have the center of the load grounded, as in my calculations,
and in most real amplifiers. I only disagree with your claim that the
virtual ground is the only valid "connection".
> I now brace myself for your next shocking revelation.
Well I have just made my revelations, I hope you were braced well enough.
I will check the quotes you had provided where Preisman said the cathode
and plate source impedance's are equal, and get back to you. Before
posting my previous post, I had read through the entire article again
specifically looking to see if he said they were equal anywhere, I guess I
missed the places where he stated that the cathode and plate source
impedance's are the same. In any case he is wrong if he said that.
Henry Pasternack
> [Deleted]
I am totally fucking flabbergasted, John. Where the hell do
you get those numbers from? Are you off your rocker, or doing
this on purpose just to annoy me?
Cripes!
I should have known better. Have a nice life.
-Henry
Henry Pasternack
> Cripes!
Let me add a couple more comments now that I have recovered
from the shock of reading your latest response. I certainly
did mean to say your calculations are correct, and I am quite
confident of the results I reported earlier. In case you were
confused (is there any question of that?), I want to make clear
what my proposed test setup was.
The 10K plate and cathode resistors, Zk and Zp, are presumed
to be built into the phase inverter. The 990K test loads are
placed from the outputs to ground, either singly or in tandem.
When the test loads are applied, the values of Zk and Zp become
effectively 9.9K, representing the parallel combination of 10K
and 990K. In the "proximal" and "distal" cases, therefore, one
of the two combined load impedances will be 10K and the other
will be 9.9K. In the "dual" cases, both load impedances will
be 9.9K.
The resistor divider equation is used to determine the
source resistance. The ratio of the voltage under load to
the no-load voltage is computed and plugged into the equation
I specified. It's easy enough to see, even in your "corrected"
results, that the values, and hence the ratios, are the same
for both "dual" cases. Therefore the resistor divider equation
must yield the same source impedance for either circuit. I
can't imagine a calculation that would yield the results you
have posted and I don't feel like puzzling out the answer.
I am literally blown away by my inability to get any kind
of intelligent response out of you. Under normal circumstances
I am quite successful at conveying complicated technical ideas
to other people. As I am completely assured of the validity
of my assumptions and of the manner in which I have stated
them, I am left once again with the inescapable conclusion
that you are, well, fucked in the head.
As it is evident that no effort, no matter how strenuous,
is likely to succeed in bringing you to grips with reality,
I am forced to conclude this discussion for good.
Have a life.
-Henry
John Byrns
"Henry Pasternack" <eat....@spam.com> wrote:
Calm down Henry. I have stated several times where I get the numbers,
including in my previous post, while you don't agree with my methodology,
I thought you understood what it was. To remind you again, I said in my
last post, "Remember that I said that when you test the cathode source
impedance, you must have the load on the plate connected, and vice versa
for the plate source impedance test."
What this means, is for example when testing the source impedance of the
cathode terminal, the load on the plate must be connected, while the load
on the cathode is disconnected, to measure the open circuit Voltage, and
then the load is connected to get the Voltage under load, and the "ratio"
"r", used in your test methodology to calculate the source impedance.
Since the 10 k plate and cathode resistors were considered part of the
concertina circuit for the calculation of your Vout values, they are both
left in place for all tests. when testing at the cathode the 990,000 Ohm
test load is left connected to the plate circuit, and the cathode Voltage
is measured both with and without the 990,000 Ohm test load connected to
the cathode terminal, to get the "ratio" "r". The source impedance in the
plate circuit is measured just the same way, with the two roles reversed.
What this boils down to in your test methodology is calculating the
"ratio" "r" using the "Distal" and "Dual" values for Vout. The source
impedance's calculated in this manner, using your Vout values, agree
exactly with the source impedance's calculated with my equations 1, and 5,
for Zgk, and Zgp.
The beauty of this approach is that it can be used to easily calculate the
Voltage vs. frequency response at both the cathode and plate terminals of
the concertina, even when the loads on the two outputs are unbalanced.
Your methodology using the "No Load", and "Dual", values of Vout to
calculate "r", gives the value of one half of the source impedance between
the cathode terminal and the plate terminal. This is fine for
demonstrating balance, but is more difficult to make use of when the loads
are not balanced. Try connecting a 1 nF capacitor across the cathode
resistor, and see which method most easily gives the Voltage vs. frequency
response at each output terminal. My equations 4, and 7 provide the
values of ELk and ELp vs. frequency directly, without further
manipulation.
John Byrns
"Henry Pasternack" <eat....@spam.com> wrote:
> "Henry Pasternack" <eat....@spam.com> wrote in message
> news:fZ0k6.4483$TD1.3...@bgtnsc05-news.ops.worldnet.att.net...
> > Cripes!
>
> Let me add a couple more comments now that I have recovered
> from the shock of reading your latest response. I certainly
> did mean to say your calculations are correct, and I am quite
> confident of the results I reported earlier. In case you were
> confused (is there any question of that?), I want to make clear
> what my proposed test setup was.
I am having trouble reconciling your statement that my "calculations are
correct" when those calculations are for the source impedance's Zgp and
Zgk, which you claim are wrong? There is nothing wrong with your results
for demonstrating balance, but they are a convoluted way to calculate the
Voltage response vs. frequency when the loads are different. I understood
perfectly well what your test setup was the first time you explained it,
and I understand where your numbers come from, I just don't think they are
of much practical use.
> The 10K plate and cathode resistors, Zk and Zp, are presumed
> to be built into the phase inverter. The 990K test loads are
> placed from the outputs to ground, either singly or in tandem.
> When the test loads are applied, the values of Zk and Zp become
> effectively 9.9K, representing the parallel combination of 10K
> and 990K. In the "proximal" and "distal" cases, therefore, one
> of the two combined load impedances will be 10K and the other
> will be 9.9K. In the "dual" cases, both load impedances will
> be 9.9K.
>
> The resistor divider equation is used to determine the
> source resistance. The ratio of the voltage under load to
> the no-load voltage is computed and plugged into the equation
> I specified. It's easy enough to see, even in your "corrected"
> results, that the values, and hence the ratios, are the same
> for both "dual" cases. Therefore the resistor divider equation
> must yield the same source impedance for either circuit. I
> can't imagine a calculation that would yield the results you
> have posted and I don't feel like puzzling out the answer.
It's very simple, I have stated my methodology four or five times already,
go back and read my earlier posts again. Your calculation does not give
the actual source impedance's, instead giving a value of one half the
source impedance measured between the plate and cathode terminals.
> I am literally blown away by my inability to get any kind
> of intelligent response out of you. Under normal circumstances
> I am quite successful at conveying complicated technical ideas
> to other people. As I am completely assured of the validity
> of my assumptions and of the manner in which I have stated
> them, I am left once again with the inescapable conclusion
> that you are, well, fucked in the head.
I will ignore these insults, I thought my response was very intelligent.
> As it is evident that no effort, no matter how strenuous,
> is likely to succeed in bringing you to grips with reality,
> I am forced to conclude this discussion for good.
The problem is that you assume me to be wrong without even trying to
understand the meaning of my equations. I have taken the trouble to read
and understand your method, just because I don't think it is as generally
useful as mine, and doesn't give the true source impedance's, doesn't mean
I don't understand what you are doing. I was even able to use your
measured Vout values to calculate the same source impedance's as given
directly by my equations 1 and 5 for Zgk and Zgp.
> Have a life.
OK, Thanks.
Henry Pasternack
of balance and then bringing it back into balance by applying
the single-ended load. To get the right results, the circuit
must be balanced whenever measurements are made. No wonder
you get the wrong results.
The difference in our numbers directly from the fact that
my test is the right one and yours is wrong.
I suspect you know this is true and are trying to obscure
the issue to save face.
Silly boy.
-Henry
John Byrns
"Henry Pasternack" <eat....@spam.com> wrote:
> You are measuring the no-load voltages with the circuit out
> of balance and then bringing it back into balance by applying
> the single-ended load.
I am calculating/measuring both the source impedance and the open circuit
Voltage with the circuit out of balance, it is not necessary to bring the
circuit into balance to make the calculations/measurements. I suggest you
review the textbook used in EE102, or EE201, or whatever the course number
was for Introductory Circuit Theory at the university you went to.
> To get the right results, the circuit
> must be balanced whenever measurements are made. No wonder
> you get the wrong results.
Why do you say the circuit must be balanced to make the source impedance
and open circuit Voltage calculations? You have provided no reason beyond
vigorous assertion. I am not sure what you mean when you say I get the
wrong results? The output Voltages calculated with my equations are
correct, I think you even stated that yourself, so if the output Voltages
are correct, how could my results be wrong? Not only are the output
Voltages equal to yours for the balanced load case, which I assume means
they are correct, but my equations will also give the correct output
Voltages when the two loads on the concertina are not balanced, something
your theory doesn't easily accomplish. You have yet to give an example
that demonstrates wrong Voltages being calculated by my equations, can you
supply an example where my equations 4 and 7 for ELk and ELp give an
answer that is wrong? Do you take exception with Preisman's figure 3? My
equations 1, 2, 3, and 4 are taken directly from Preisman's figure 3, did
Preisman also get it wrong?
> The difference in our numbers directly from the fact that
> my test is the right one and yours is wrong.
>
> I suspect you know this is true and are trying to obscure
> the issue to save face.
>
> Silly boy.
What ever you say Henry.
Arny Krueger
> "Henry Pasternack" <eat....@spam.com> wrote in message
> <oYzj6.13743$Nj5.9...@bgtnsc07-news.ops.worldnet.att.net>
>
> > This
> > means not only that the load impedances must be the same, but
> > also if you are going to inject a current into one output to
test the
> > impedance, you have to inject an equal and opposite current into
> > the other output at the same time.
> This is where we differ, I don't think this is the standard
definition
> of source impedance, and how you measure it.
The standard definition of source impedance usually involves the
presumption of linearity and just one relevant output terminal.
These presumptions aren't necessarily correct for any phase inverter,
particularly one using a somewhat nonlinear active device.
> I don't believe it is
> valid to inject currents into nodes other than the one that is
being
> measured, of course there is also a complimentary current in the
return
> path.
Its a simple rule of thumb that test conditions should resemble the
conditions of actual operation. The "standard definition" of output
impedance applies far better to something like a highly linear mono
amplifier that has exactly one output.
But a phase inverter has two outputs and in typical use they have
equal and opposite signals on them.
>But to simultaneously inject a current into a second node is
invalid as far as I know.
IMO it is not invalid in the case of a phase inverter.
> Do you have any references that indicate this is a valid
procedure?
At this point, just 35 years of experience and common sense such as
mine is.
>It would be valid to make a measurement between
> the cathode and plate, with the current injected into the plate
> returning via the cathode node, but that would be measuring the
source
> impedance between the cathode and plate terminals, which is not
exactly
> how the concertina is actually used.
Here is a "novel" idea. Why not base your measurements on actual
operation?
Take a working amp. Carefully and precisely measure the AC voltage at
first one phase slitter output, and then the other.
Then apply a load composed of a large capacitor in series with a
resistor that is about 10 times larger than one of the two resistors
across which the signal is developed, to each output or the phase
splitter (or make a pair and do both at the same time). Other end(s)
of the network(s) go to signal ground. Again carefully and precisely
measure the voltage.
Now, do the usual calculations to determine the actual source
impedances at those points in the circuit.
If there is global feedback, I suspect you will find that the source
impedance at those points in the circuit are far lower than you
might otherwise expect.
If you want "open loop" measurements for an amp with loop feedback,
you will have to disable the feedback and deal with the fact that the
amp's voltage gain is greatly increased.
Henry Pasternack
The whole point of this exercise was to explain why the frequency
responses of the two outputs of the inverter are equal in spite of
the apparent impedance imbalance. It's common knowledge that the
-3dB frequency can be found using the simple one-over-two-pi-R-C
formula. Why don't you calculate the frequency response assuming
the outputs are loaded down with, say, equal 470pF capacitors? And
then show how the impedance values you calculated relate to the
answer.
-Henry
Patrick Turner
I must say I'm a trifle amused by the disscussion on the concertina phase
splitter.
Most of the answers to questions about the concertina are answered for you in
the Radiotron Designer's Handbook , 4th
edition , from 1955 , put together by the brilliant Fritz Langford-Smith.
Taken individually , the two outputs from the concertina have very different
Ro's yet both outputs yield the same voltage but
at 180 degrees phase difference.
However at frequencies above say 5 Kc the miller effect and uneqal stray C
begins to unbalance the concertina outputs.
In a classic Williamson circuit this doesn't matter too much but if you want
to equalise the two outputs up to say 300Kc then
try connecting a small C of about 15 PF across the concertina's cathode
resistor. A variable trim cap would do then examine
all with a scope probe until all remains balanced up to a higher frequency.
But some concertinas are used to directly drive output tubes. All is well
until overload is reached and the onset of grid current begins in the output
tube grids.
At this point, the output tube grid being driven positive presents a much
lower load than the other tube which is being driven negative. The voltage
output from the concertina plate circuit becomes much less and more distorted
than the voltage from the cathode circuit. The resulting direct voltage
charges across the two coupling caps become very dissimilar if the overload
continues long enough. I've always thought this behaviour and the resulting
paralysis after severe overload and recovery behaviour to be much worse than a
well done long tail pair which behaves much more symetrically during overload
or an occasional clipping signal. Some folk like concertinas because the
overload behaviour generates more 2nd harmonic distortion thus mimicking their
beloved single ended sound. Who am I to argue if that's their preference? I
can well understand why musicians prefer lots of low order distortion as the
amp can be as important as the Guitar to get just the desired effect.
Well that's all I reckon for now from down under,
Keep the home triodes burning,
Patrick.
http://www.turneraudio.com.au
Henry Pasternack
news:%Duk6.470$JF7.38...@newssvr16.news.prodigy.com...
> Here is a "novel" idea. Why not base your measurements on actual
> operation?
Excellent idea.
>
> Take a working amp. Carefully and precisely measure the AC voltage at
> first one phase slitter output, and then the other.
>
> Then apply a load composed of a large capacitor in series with a
> resistor that is about 10 times larger than one of the two resistors
> across which the signal is developed, to each output or the phase
> splitter (or make a pair and do both at the same time). Other end(s)
> of the network(s) go to signal ground. Again carefully and precisely
> measure the voltage.
>
> Now, do the usual calculations to determine the actual source
> impedances at those points in the circuit.
Bear in mind that you will get very different results if you do
the tests individually as opposed to simultaneously. It is the
(equal) values you get when both tests are done at the same time
that you can use to make meaningful predictions about the effect
of loading on the stage.
I agree with Patrick Turner that in practice the inverter will
not remain balanced at high frequencies due to the effect of stray
capacitance. I've measured this myself in my own amplifiers.
-Henry
Arny Krueger
> "Arny Krueger" <ar...@pop3free.com> wrote in message
> news:%Duk6.470$JF7.38...@newssvr16.news.prodigy.com...
> > Here is a "novel" idea. Why not base your measurements on actual
> > operation?
> Excellent idea.
>
> > Take a working amp. Carefully and precisely measure the AC
voltage at
> > first one phase slitter output, and then the other.
> > Then apply a load composed of a large capacitor in series with a
> > resistor that is about 10 times larger than one of the two
resistors
> > across which the signal is developed, to each output or the
phase
> > splitter (or make a pair and do both at the same time). Other
end(s)
> > of the network(s) go to signal ground. Again carefully and
precisely
> > measure the voltage.
> > Now, do the usual calculations to determine the actual source
> > impedances at those points in the circuit.
> Bear in mind that you will get very different results if you do
> the tests individually as opposed to simultaneously.
How "very" different it would be should be somewhat dependent on how
large the load resistor is. The larger it is, the harder it would be
to have accurate results, but the less likely there would be a big
difference between individual versus simultaneous tests.
One nice thing is that even if you measure one side at a time, both
sides are swinging at least 90% of the usual operational voltage
range.
>It is the
> (equal) values you get when both tests are done at the same time
> that you can use to make meaningful predictions about the effect
> of loading on the stage.
> I agree with Patrick Turner that in practice the inverter will
> not remain balanced at high frequencies due to the effect of stray
> capacitance. I've measured this myself in my own amplifiers.
Seems like a likely outcome. The open loop measurements probably
would show this the best.
The consequence of imbalance would most likely be somewhat more even
order distortion at high frequencies due to the loss of loop gain in
one half of the output /driver stage combination.
John Byrns
"Henry Pasternack" <eat....@spam.com> wrote:
> Sorry, John, but you're just digging deeper.
How so? You have already stated on at least two occasions that my
"calculations are correct", referring to the 7 equations I posted. That
would seem to settle the issue as far as I am concerned.
> The whole point of this exercise was to explain why the frequency
> responses of the two outputs of the inverter are equal in spite of
> the apparent impedance imbalance.
Not so, I think we have been in agreement from the bigining that excepting
effects due to grid capacitance to the plate and cathode, the concertina
is perfectly balanced when its loads are balanced. This is obvious from
the simple observation that the same current must flow in both the plate
and cathode circuits, again ignoring grid capacitance effects. The whole
point of this "exercise" was to show that the plate and cathode source
impedance's of the concertina are not the same, yet the output Voltages
are balanced in spite of the differing source impedance's..
> It's common knowledge that the
> -3dB frequency can be found using the simple one-over-two-pi-R-C
> formula. Why don't you calculate the frequency response assuming
> the outputs are loaded down with, say, equal 470pF capacitors? And
> then show how the impedance values you calculated relate to the
> answer.
Henry, I have already posted the relationship between my calculated source
impedance values and the frequency response at the plate and cathode
terminals. The relationship is contained in the equations I posted
earlier. The functions for the response at the cathode and the plate,
equations 4 & 7 respectively, are derived from the source impedance's
given by equations 1 & 5, combined with the generator Voltages given by
equations 2 & 6, and the Voltage divider formed with the load
impedance's. The beauty of this approach is that it will give the correct
frequency response at the plate and cathode terminals when the loads are
unbalanced, as well as when they are balanced. For the case where the
loads are balanced, Zk = Zp = ZL, equations 4 and 7 become identical and
have this form:
ELk = ELp = es * u * ZL / (rp + (u + 2) * ZL
I will post the -3 dB frequency predicted by my equation using the circuit
parameters you posted earlier, and with the addition of 470 pF capacitors
across both the plate and cathode loads, just as soon as you catch up with
the discussion and post your equation for the balanced load case.
John Byrns
"Henry Pasternack" <eat....@spam.com> wrote:
> "Arny Krueger" <ar...@pop3free.com> wrote in message
> news:%Duk6.470$JF7.38...@newssvr16.news.prodigy.com...
On my return the rec.audio.tubes a couple of weeks ago I observed that the
group had taken on something of the flavor of rec.audio.high-end. The
apperance of Arny Krueger confirms that observation as to the present
direction of the group.
The problem is not stray capacitance, which can be balanced out, but
rather as I posted a week or so ago is due to the grid to plate, and grid
to cathode capacitance of the tube in a real concertina circuit, which
results in some high frequency imbalance. The problem is not serious
though, and Patrick Turner's characterization that it starts at 5 kHz
makes it sound much worse than it really is. I posted some numbers on
this effect a couple of years ago. Perhaps once Google gets the old Deja
database back on line, we can unearth those old posts. Another good point
that I posted on earlier in this thread, which Patrick has reiterated, is
the distortion problem the concertina has when the following grids are
driven into conduction, dynamically unbalancing the loads. I have
suggested on several occasions that this problem can be minimized by using
a buffer stage between the concertina outputs and the power tube grids. I
hadn't thought of the point Patrick made about guitar people actually
liking the distortion effect generated by the concertina driving the
following grids into conduction.