[John Byrns]

In article ,
flipper wrote:

On Fri, 08 Aug 2008 22:34:44 -0500, John Byrns
wrote:

Flipper, you seem to be flopping all around, trying to walk both sides
of the street, or perhaps trying to swim both sides of the pier would be
more accurate.

Jokes may be amusing but they're not a valid argument.

It's hard to resist the joke when you're talking to someone who calls
himself "flipper" the "fish". But the point is still valid, you are
trying to have it both ways, which is not physically possible.

The anode and cathode source impedances are either the same or they
aren't the same, you can't have it both ways, you have to pick one or
the other, they can't both be true. Once you have made your choice,
then you can work on trying to justify it.

Of course they can 'both be true' because you're talking about
simplification models, which depend on various assumptions and the
modeling technique chosen.

How does the fact that we are talking about a "simplification model"
make it possible for the anode source impedance of a concertina to be
both equal to the cathode source impedance and different from it at the
same time and in the same "simplification model"?

To illustrate lets get to basics.

THEVENIN’S THEOREM
From the perspective of any load, a linear electric circuit may be
represented by an ideal voltage source in series with an output
impedance or resistance.

NORTON’S THEOREM
From the perspective of any load, a linear electric circuit may be
represented by an ideal current source in parallel with an output
impedance or resistance.

OK, so we have two 'different' models that produce the same results so
which is the 'real' one?

From a modeling perspective they are both the "real one" as you say,
because they both have the same source impedance and equivalent
generator characteristics when viewed from outside the model. When
measuring the source resistance from the outside, as we are doing in the
case of the concertina, it makes no difference which we use inside the
model, there is no way to tell them apart from measurements taken at the
terminals of the model.

Before we get too far from Thevenin and Norton, note that in addition to
a source impedance, they both also include a generator, voltage or
current, which may also have a complex dependence on frequency.

From the physical world perspective neither are because 'ideal'
sources do not exist and it's unlikely the thing we're modeling is
just one resistor. But from a functional perspective, as long as the
underlying assumptions hold, they 'both' are because they both provide
a useful model.

Indeed, the source resistance of the concertina anode or cathode isn't
"just one resistor", they are both complex impedances because the load
connected to the other output is a complex impedance, not just a simple
resistor, that is one of several points I have been trying to make.
Using either the Thevenin or the Norton model doesn't restrict us to
"just one resistor" as the source impedance.

I still don't see how the fact that the two output voltages remain equal
as the frequency is varied, in anyway implies that the anode and cathode
source impedances are equal?

Remember, the 'point' to calculating 'effective impedance' is to
provide something useful for subsequent calculations. So, keeping that
in mind, we're modeling a black box with a voltage input and two
outputs, 180 degrees out of phase, that you say have different
impedances.

Yes, that is correct.

Those black box outputs are connected to a capacitive load so try
calculating the RC time constant. Doesn't work that way? Then what's
the point to the 'effective impedance' calculation?

The point of calculating the effective impedances at the anode and at
the cathode is to allow us to evaluate the effect of unbalanced loads,
in the real world we may not find it convenient to have perfectly
balanced loads, and we may want to evaluate the effects of various
degrees of load imbalance before actually building the circuit.

In the degenerate case of balanced loads the effective RC time constants
at the anode and cathode outputs become equal. There is no real problem
with calculating this single value using the different anode and cathode
source impedances, along with the two generator characteristics.

Or, conversely, observe that the Cs are equal and the roll off is the
same. Simple RC implies the impedances out of the black box are equal
(which answers your question).... but you say they are not.

Correct, the impedances of the two outputs of the black box are not
equal. Everyone seems to agree that an amplifier has different output
impedances from the cathode and anode, yet you and HP expect us to
believe that when we use the amplifier as a concertina the anode and
cathode source impedances suddenly somehow magically become equal! Can
you explain how this transformation takes place?

You handle this conundrum by jumping inside the black box to explain
why the 'effective impedance' doesn't do what RC implies. Well, ok, so
the black box transfer function is a little more complicated, with
internal 'source voltages' changing under load, but that works.

No, the internal source voltages seen at the anode or at the cathode
don't "change under load" they are fixed in the relatively simple model
we are talking about, just as the source impedances are fixed. It is
important to understand that we are measuring the internal source
voltage at one output at a time, and while this source voltage is
independent of the load applied to the terminal we are measuring at, it
is not independent of the impedance connected to the other output
terminal, just as the source impedance isn't. In other words while the
internal source voltage driving the anode output is independent of the
load connected to the anode terminal, it is not independent of the
impedance connected in the cathode circuit, and vice versa.

On the other hand, what's wrong with a simple A, -A transfer function
and equal output impedance model? That gives you the right answer too,
as long as the underlying equal load impedance assumption is valid.

You have partly answered your own question, the "simple A, -A transfer
function and equal output impedance model" is a very limited model that
only works when the two loads are equal. Why even bother with a model
in that case when a simple inspection of the circuit tells us that the
outputs are equal and opposite when the loads are balanced? The two
output impedance model can tell us what happens when the loads are not
balanced or equal.

The fallacy of HP's argument that the anode and cathode output
impedances are equal is that his method doesn't really measure the anode
and cathode source impedances separately, it really only measures one
source impedance, the one between the anode and cathode terminals. HP
then effectively observes that if the single anode to cathode load is
split into two equal halves, with the center junction grounded,
everything will remain balanced. This contrivance on HPs part in no way
proves that the source impedances at the anode and cathode terminals are
equal, although it is obviously useful in calculating the RC time
constant that you are interested in.

--
Regards,

John Byrns

Surf my web pages at, http://fmamradios.com/

[John Byrns]

In article ,
flipper wrote:

On Sat, 09 Aug 2008 12:28:58 -0500, John Byrns
wrote:

But the point is still valid, you are
trying to have it both ways, which is not physically possible.

As Jamie on Mythbusters likes to say "well, there's your problem." The
effective output impedance is not a 'physical' component. It's a model
and, yes, you can "have it both ways," depending on the model you use,
just as you can have a current or voltage source, depending on whether
you use a Thevenin or Norton model.


OK, a model is not the actual physical realization of the concertina,
since you are trying to hide behind the fact that a model is not the
actual physical device let's move on to measuring the actual physical
implementation of the concertina.

Before we do that let me make a few observations about the nature of
models from my perspective. In my world the object of a model is to
duplicate the operation and performance of an actual physical device as
closely as possible, or at least as close as is necessary for the job at
hand. In other words in the case of the concertina the output
impedances of the concertina model should closely approximate those
measured on a physical implementation of the concertina.

From what you have said so far on the subject it is apparent that you
take a completely different approach to modeling, and consider it OK to
build a model that supports a preconceived notion about the device being
modeled, in the case of the concertina that the two output impedances
are equal, rather than actually trying to make the model approximate the
operation of the physical device as closely as possible. My
mathematical model of the concertina was built without any consideration
of what the output impedances might be one way or the other, the output
impedances we only calculated from the model as an after thought without
any preconceived notion of what the results might be.

OK, we have different views on what a model is and how it should be
constructed so let's instead build the real physical thing on a block of
wood and measure the source impedances of the anode and cathode circuits
to see if they are the same or if they are different.

What do you think the results would likely be if we built a physical
concertina, or tapped into one in an existing amp, and measured the
actual source impedances? I suppose the problem with this approach
would be agreeing on a test and measurement protocol to be used in
making the measurements of the source impedances? However if we could
agree on a suitable methodology, measuring the actual physical device
would seem to be a good way to settle the argument once and for all.

Since this equal impedance thing has taken on many of the
characteristics of a myth, perhaps I should submit this problem to the
Mythbusters for a Busted or Confirmed judgment, as my son is a
Mythbusters fanatic.

I think you've missed the rather astonishing, at the time,
breakthrough that, regardless of the circuit complexity (and as long
as the underlying assumptions are met), a linear electric circuit can
be *modeled* as simply a voltage source (or, later, a current source)
and one 'equivalent resistor'.

That is rather astonishing, even today, and I did miss it, probably
because it is false. I believe "one equivalent resistor" is incorrect
and it should actually be an impedance not a resistor. I will have to
look it up.

--
Regards,

John Byrns

Surf my web pages at, http://fmamradios.com/

[Henry Pasternack[_2_]]

John Byrns wrote:
How does the fact that we are talking about a "simplification model"
make it possible for the anode source impedance of a concertina to be
both equal to the cathode source impedance and different from it at the
same time and in the same "simplification model"?

I've been having an email discussion with Flipper and he seems to be a very
friendly and reasonable guy who certainly hasn't done anything to deserve
such sarcastic and generally difficult treatment by you, John.

There's a simple answer to your question. If you're solving a problem with
a number of variables, and if you don't specify all the variables, the
result
may be a family of solutions that depend on the unspecified variable(s).
This
is a perfectly legitimate outcome.

The inverter is a three-port network. If you measure the output impedance
at one port, you still have to specify what you do with the other ports.
Your
definition of output impedance is a measurement where the opposite output
port is left open-circuited. In the test Flipper and I have proposed, the
opposite output port is terminated in an impedance that always matches the
impedance of the test set.

Your test represents a purely unbalanced load while Flipper's represents a
purely balanced load. The unbalanced test tells us more about how the
inverter performs when the following stage is driven to grid current. The
balanced test tells us more about the case where there is no significant
grid
current. In the first case, the two measured impedances are very
different;
in the other they are the same. There is no contradiction between the two
results because the test conditions are different for each.

An infinite number of other results are possible, depending on how you deal
with the two ports. In this way, you could explore the entire state space
defined by the Preisman equations, and only one model is needed.

On the other hand, as Flipper has said, you might ask if there is a way to
simplify the model so that it changes from a general three-port network into
a pair of two-port networks with their inputs in parallel. This
simplification
is possible, if you impose the constraint of equal loading. And when you do
so, it turns out that not only are the two networks independent, but they
are
also identical. If you can accept the equal load constraint, this is a very
practical and economical solution.

The origin of this discussion, which you have insisted over the years in
turning into an argument, was the question of whether or not some kind of
"build-out" resistor or other hack is needed to equalize the impedances
of the split load inverter. From the very start, I always stated as an
explicit assumption that the two loads were equal. Under that assumption,
and using a well-defined model, the two output impedances are the same.
This is so clear as to be indisputable.

(For that reason, I fully expect you to dispute it.)

If you want to use a more complicated model where the output impedances
aren't equal, then by all means, you're welcome to do so! But please don't
insult those of us who realize we can get the answer we want more easily
by simplifying the model. And watch out for a physicist who comes along
and berates you for the simplifications inherent in your own model.

-Henry

[John Byrns]

In article ,
"Henry Pasternack" wrote:

John Byrns wrote:

How does the fact that we are talking about a "simplification model"
make it possible for the anode source impedance of a concertina to be
both equal to the cathode source impedance and different from it at the
same time and in the same "simplification model"?

I've been having an email discussion with Flipper and he seems to be a very
friendly and reasonable guy who certainly hasn't done anything to deserve
such sarcastic and generally difficult treatment by you, John.

There's a simple answer to your question. If you're solving a problem with
a number of variables, and if you don't specify all the variables, the
result
may be a family of solutions that depend on the unspecified variable(s).
This
is a perfectly legitimate outcome.

The inverter is a three-port network. If you measure the output impedance
at one port, you still have to specify what you do with the other ports.
Your
definition of output impedance is a measurement where the opposite output
port is left open-circuited. In the test Flipper and I have proposed, the
opposite output port is terminated in an impedance that always matches the
impedance of the test set.

Your test represents a purely unbalanced load while Flipper's represents a
purely balanced load. The unbalanced test tells us more about how the
inverter performs when the following stage is driven to grid current. The
balanced test tells us more about the case where there is no significant
grid
current. In the first case, the two measured impedances are very
different;
in the other they are the same. There is no contradiction between the two
results because the test conditions are different for each.

An infinite number of other results are possible, depending on how you deal
with the two ports. In this way, you could explore the entire state space
defined by the Preisman equations, and only one model is needed.

On the other hand, as Flipper has said, you might ask if there is a way to
simplify the model so that it changes from a general three-port network into
a pair of two-port networks with their inputs in parallel. This
simplification
is possible, if you impose the constraint of equal loading. And when you do
so, it turns out that not only are the two networks independent, but they
are
also identical. If you can accept the equal load constraint, this is a very
practical and economical solution.

The origin of this discussion, which you have insisted over the years in
turning into an argument, was the question of whether or not some kind of
"build-out" resistor or other hack is needed to equalize the impedances
of the split load inverter. From the very start, I always stated as an
explicit assumption that the two loads were equal. Under that assumption,
and using a well-defined model, the two output impedances are the same.
This is so clear as to be indisputable.

(For that reason, I fully expect you to dispute it.)

If you want to use a more complicated model where the output impedances
aren't equal, then by all means, you're welcome to do so! But please don't
insult those of us who realize we can get the answer we want more easily
by simplifying the model. And watch out for a physicist who comes along
and berates you for the simplifications inherent in your own model.

Hi Henry,

I agree that Flipper is friendly and helpful fellow; unfortunately he is
also a very slippery fish who has tried to muddy the waters with his
contention that different models produce different results, effectively
legitimizing any outcome. Without having given a concrete example, this
is just so much BS. For example I believe you and I are using the exact
same model, yet we have come to different conclusions about what the
anode and cathode source impedances of the concertina are. From my
perspective the model is not the issue here, the issue is what we are
measuring, and more importantly how we choose to interpret the results
of those measurements.

The concertina has three output terminals, first the ground or signal
reference terminal, secondly the anode output terminal, and finally the
cathode output terminal. There are three source impedances that can be
measured between two these three terminals. I measured two of these,
the anode to ground source impedance, and the cathode to ground source
impedance. Your method is effectively measuring the source impedance
between the anode and cathode terminals under the assumption that the
concertina and its loads are balanced.

IIRC from a long ago discussion, what you did was to inject equal and
opposite currents into the anode and cathode terminals, and then measure
the voltages developed each of these terminals and the ground terminal.
Normally you would measure the anode to cathode voltage under this
condition to determine the anode to cathode source impedance. I would
suggest that under the conditions that you assumed, namely that the
concertina and its two loads are completely balanced, that the anode and
cathode voltages that you measured are each precisely one half of the
anode to cathode voltage. Hence what you have measured is the anode to
cathode source impedance divided by two, you have not measured equal
source impedances at the anode and cathode.

Your measurement doesn't in any way imply that the anode and cathode
source impedances of the concertina are equal, all you have done is
measure a value that represent one half of the anode to cathode source
impedance under the condition of equal loads.

Given all that, your measurement method is a very clever and useful way
to calculate the RC time constant number that Flipper wants, given the
assumption of complete balance. Your method is far simpler and easier
to apply in that case than my more general method, although in this day
and age of computer spreadsheets the advantage may be somewhat moot.

As far as the issue of reopening this argument goes, I don't believe
that I was the one that did it. I have not gone back to verify exactly
what was said, but I believe it was someone else, probably either
Flipper or Patrick that brought the subject up again. All I did was
mention in response to a discussion about Morgan Jones, and his status
as a guru, that he had got the build out resistor issue wrong. IIRC I
initially justified this only on the basis that the concertina was
already balanced without the build out resistor and that adding the
build out resistor could only serve to unbalance an already balanced
circuit. I do not remember bringing up the argument about source
impedances, as I remember it I simply based my comment on balance. You
will have to look elsewhere for the person that restarted this argument.

As far as I am concerned the whole argument is now moot, since I now
understand how to correctly interpret your measurements, even if you
fail to understand what it was that you actually measured.

--
Regards,

John Byrns

Surf my web pages at, http://fmamradios.com/

[John Byrns]

In article ,
flipper wrote:

On Sun, 10 Aug 2008 13:32:57 -0500, John Byrns
wrote:

In article ,
flipper wrote:

On Sat, 09 Aug 2008 12:28:58 -0500, John Byrns
wrote:

But the point is still valid, you are
trying to have it both ways, which is not physically possible.

As Jamie on Mythbusters likes to say "well, there's your problem." The
effective output impedance is not a 'physical' component. It's a model
and, yes, you can "have it both ways," depending on the model you use,
just as you can have a current or voltage source, depending on whether
you use a Thevenin or Norton model.

Your wholesale elimination of every point made, but two, is disturbing
because it indicates no argument whatsoever, regardless of it's
meaning or merit, is going to have any effect. You simply 'wash away'
whatever is inconvenient.

OK, a model is not the actual physical realization of the concertina,

It's not just the concertina, A model is a not physical realization.

since you are trying to hide behind the fact that a model is not the
actual physical device

There is no 'hiding'. It's simply what a model 'is' (or is not) and a
fundamental concept of them.

In even the most trivial Thevenin case, it not only 'is not' a
physical realization it is flat impossible to make as ideal voltage
sources simply do not exist. It's an abstraction.

That a model is not a 'real thing', and is developed for a specific
purpose under defined conditions, is neither trivial nor
inconsequential. For example, a Thevenin equivalent is intended to
simulate equivalent output voltage and impedance but it does not
model power dissipation of the original circuit. So any notion that
the abstraction is 'real' can lead to erroneous conclusions.

Neither does the Thevenin equivalent work outside defined conditions,
the most obvious of which being the original circuit operating in it's
linear region.

I submit it should be obvious that if the intended purpose is
different, like simulating power in this case, then a different model
is likely needed.

I further submit that if the defined conditions are different then a
different model is either likely needed or, at least, possible. In
fact, one might intentionally limit conditions in order to create a
simpler model.

Now, all of these models, if done properly, will be valid for the
intended purpose under the defined conditions despite being quite
different.

So which is 'real'?

None of them. They're useful abstractions.

let's move on to measuring the actual physical
implementation of the concertina.

Well, we already have one 'measurement' and that's the observation of
equal roll off, which implies, via the well known RC equation, that
equal output impendences is a valid model.

You have a different model that also satisfies the observation.

At this point it would probably be a good idea for you to explain what
it is you're trying to 'prove', or 'convince' me of, because if it's
that there can be only 'one' model then you've got a serious conflict
with Thevenin and Norton, and the concept of models in general.

Before we do that let me make a few observations about the nature of
models from my perspective. In my world the object of a model is to
duplicate the operation and performance of an actual physical device as
closely as possible, or at least as close as is necessary for the job at
hand. In other words in the case of the concertina the output
impedances of the concertina model should closely approximate those
measured on a physical implementation of the concertina.

Fair enough, except to add that, in this context, a model's purpose is
also to make things simpler and more convenient. I.E., Rather than
having to analyze the entire original circuit each time one wants to
see what 'effect' something has one can use the simpler, more
convenient, model.


From what you have said so far on the subject it is apparent that you
take a completely different approach to modeling, and consider it OK to
build a model that supports a preconceived notion about the device being
modeled, in the case of the concertina that the two output impedances
are equal, rather than actually trying to make the model approximate the
operation of the physical device as closely as possible. My
mathematical model of the concertina was built without any consideration
of what the output impedances might be one way or the other, the output
impedances we only calculated from the model as an after thought without
any preconceived notion of what the results might be.

How in the world do you come up with my contention that both models
are valid as being a "preconceived notion" favoring one or the other?

On the contrary, even a most casual reading of the preceding
discussions shows it is you who 'insists' on a particular outcome.


OK, we have different views on what a model is and how it should be
constructed so let's instead build the real physical thing on a block of
wood and measure the source impedances of the anode and cathode circuits
to see if they are the same or if they are different.

What do you think the results would likely be if we built a physical
concertina, or tapped into one in an existing amp, and measured the
actual source impedances?

As I mentioned, one means of measurement has already been done and the
observation was that the RC time constants appear to be equal. And
since the Cs are, by definition, equal the common conclusion would
likely be that the Rs are also equal.

In fact, the 'commonness' of the conclusion, and the common
understanding of what 'output impedance' means, is what leads people
to the erroneous conclusion, when using the 'different impedance'
model,.that the concertina has unequal roll offs.

This is not particularly surprising because the notion is entirely
consistent with the typical textbook Thevenin equivalent consisting of
a voltage generator and series 'output impedance'. And one of the
typical exercises is slapping a capacitor on the output to observe the
response is a simple Zc, Zo divider because the 'output impedance' is
Zo..

The text usually explains that this is 'why' the Thevenin is so
useful, because the behavior of a complex linear circuit and load can
be reduced to a 'simple' divider equation, and this meaning of 'output
impedance' is pounded home over and over again.

A concertina befuddles this basic notion because it is not a two
terminal device and the outputs interact with each other through the
loads.


I suppose the problem with this approach
would be agreeing on a test and measurement protocol to be used in
making the measurements of the source impedances?

Could be, because all indications, from the general discussion, are
you will insist on whichever means 'confirms' what you have already,
it would seem, decided 'must be' the case.

Might be interesting, though, to see how you intend to measure 'output
impedance' while maintaining equal loads on the concertina, since that
is a defined condition.

However if we could
agree on a suitable methodology, measuring the actual physical device
would seem to be a good way to settle the argument once and for all.

What 'argument' are you trying to resolve?

Since this equal impedance thing has taken on many of the
characteristics of a myth, perhaps I should submit this problem to the
Mythbusters for a Busted or Confirmed judgment, as my son is a
Mythbusters fanatic.

Maybe they could deal with the myth there can be only one model.

Or maybe not, because there's nothing to blow up

I think you've missed the rather astonishing, at the time,
breakthrough that, regardless of the circuit complexity (and as long
as the underlying assumptions are met), a linear electric circuit can
be *modeled* as simply a voltage source (or, later, a current source)
and one 'equivalent resistor'.

That is rather astonishing, even today, and I did miss it, probably
because it is false. I believe "one equivalent resistor" is incorrect
and it should actually be an impedance not a resistor. I will have to
look it up.

I said "at the time" and "at the time" the theorem delft with two
terminal networks composed of voltage sources, current sources, and
resistors. The concept was later generalized to impedances.

But in your rush to quibble with non essentials you miss the point,
which is the concept of modeling and 'equivalent circuits', which are
(simplified) abstractions of the actual circuit.

That was an astonishing thing, that the 'whole circuit', regardless of
how complex, could be reduced to just two things, at least for that
purpose. Granted, none of the things actually exist, and are not
'real', but, shazzam, it's a boatload easier to deal with than the
'real' one.

Flipper, please see my response to Henry Pasternack for further comment.

--
Regards,

John Byrns

Surf my web pages at, http://fmamradios.com/

[Patrick Turner]

John Byrns wrote:

In article ,
flipper wrote:

On Sat, 09 Aug 2008 12:28:58 -0500, John Byrns
wrote:

But the point is still valid, you are
trying to have it both ways, which is not physically possible.

As Jamie on Mythbusters likes to say "well, there's your problem." The
effective output impedance is not a 'physical' component. It's a model
and, yes, you can "have it both ways," depending on the model you use,
just as you can have a current or voltage source, depending on whether
you use a Thevenin or Norton model.

OK, a model is not the actual physical realization of the concertina,
since you are trying to hide behind the fact that a model is not the
actual physical device let's move on to measuring the actual physical
implementation of the concertina.

Before we do that let me make a few observations about the nature of
models from my perspective. In my world the object of a model is to
duplicate the operation and performance of an actual physical device as
closely as possible, or at least as close as is necessary for the job at
hand. In other words in the case of the concertina the output
impedances of the concertina model should closely approximate those
measured on a physical implementation of the concertina.

The best model for the triode is the voltage gene producing an output
voltage of
µ x Vg-k, and Ra is a resistance in series between the low Rout gene and
the anode terminal.

So for a typical CPI with 1/2 6SN7, the model becomes a gene with Vo =
20 x Vg-k, and Ra = 10k at typical 4mA of Iadc.
There is say 20k anode load from anode to a fixed 0V point for ac, 20k
cathode load to 0V at ac.

From this its extremely easy to examine what a change in load makes to
the anode voltage or cathode voltage outputs,
and if the loading change is separate to output, or done equally to each
output.

What everone will find is that the Rout at anode is a lot higher than
the Rout at the cathode, if measure separately
and if loaded separately with the same load change to each.

Really basic stuff.

But everyone will find that while the loads stay equal the Vo stays
balanced, until F goes so high and the C
around th circuit begins to unbalance te loadings, and unbalance the
outputs.

With pentodes, most ppl like to use a constant current generator
shunting Ra
for the model of the tube. But one can use µ x Vgk if you want, but then
you get
large imaginery voltages in the model which confuse those with lame
minds.


From what you have said so far on the subject it is apparent that you
take a completely different approach to modeling, and consider it OK to
build a model that supports a preconceived notion about the device being
modeled, in the case of the concertina that the two output impedances
are equal, rather than actually trying to make the model approximate the
operation of the physical device as closely as possible. My
mathematical model of the concertina was built without any consideration
of what the output impedances might be one way or the other, the output
impedances we only calculated from the model as an after thought without
any preconceived notion of what the results might be.

OK, we have different views on what a model is and how it should be
constructed so let's instead build the real physical thing on a block of
wood and measure the source impedances of the anode and cathode circuits
to see if they are the same or if they are different.

What do you think the results would likely be if we built a physical
concertina, or tapped into one in an existing amp, and measured the
actual source impedances? I suppose the problem with this approach
would be agreeing on a test and measurement protocol to be used in
making the measurements of the source impedances? However if we could
agree on a suitable methodology, measuring the actual physical device
would seem to be a good way to settle the argument once and for all.

Since this equal impedance thing has taken on many of the
characteristics of a myth, perhaps I should submit this problem to the
Mythbusters for a Busted or Confirmed judgment, as my son is a
Mythbusters fanatic.

The CPI has a current output ( high Rout compared to the load) at its
anode,
and voltage output ( low Rout compared to the load) at its cathode,
simple.

Voltage amplitude equality is achieved with local current FB,
and the accuracy of the load match of anode and cathode loads.


I think you've missed the rather astonishing, at the time,
breakthrough that, regardless of the circuit complexity (and as long
as the underlying assumptions are met), a linear electric circuit can
be *modeled* as simply a voltage source (or, later, a current source)
and one 'equivalent resistor'.

That is rather astonishing, even today, and I did miss it, probably
because it is false. I believe "one equivalent resistor" is incorrect
and it should actually be an impedance not a resistor. I will have to
look it up.

The resistor for a load is fine for the tube at AF.
But for above AF, you have to add in the C involved around the whole
circuit,
and then it behaves very differently with impedance loading instead of
just resistive loading.
With CPI, inductance also becomes a factor if F becomes high enough.

Patrick Turner.

--
Regards,

John Byrns

Surf my web pages at, http://fmamradios.com/

[Patrick Turner]

Henry Pasternack wrote:

John Byrns wrote:
How does the fact that we are talking about a "simplification model"
make it possible for the anode source impedance of a concertina to be
both equal to the cathode source impedance and different from it at the
same time and in the same "simplification model"?

I've been having an email discussion with Flipper and he seems to be a very
friendly and reasonable guy who certainly hasn't done anything to deserve
such sarcastic and generally difficult treatment by you, John.

There's a simple answer to your question. If you're solving a problem with
a number of variables, and if you don't specify all the variables, the
result
may be a family of solutions that depend on the unspecified variable(s).
This
is a perfectly legitimate outcome.

The inverter is a three-port network. If you measure the output impedance
at one port, you still have to specify what you do with the other ports.
Your
definition of output impedance is a measurement where the opposite output
port is left open-circuited. In the test Flipper and I have proposed, the
opposite output port is terminated in an impedance that always matches the
impedance of the test set.

Your test represents a purely unbalanced load while Flipper's represents a
purely balanced load. The unbalanced test tells us more about how the
inverter performs when the following stage is driven to grid current. The
balanced test tells us more about the case where there is no significant
grid
current. In the first case, the two measured impedances are very
different;
in the other they are the same. There is no contradiction between the two
results because the test conditions are different for each.

An infinite number of other results are possible, depending on how you deal
with the two ports. In this way, you could explore the entire state space
defined by the Preisman equations, and only one model is needed.

On the other hand, as Flipper has said, you might ask if there is a way to
simplify the model so that it changes from a general three-port network into
a pair of two-port networks with their inputs in parallel. This
simplification
is possible, if you impose the constraint of equal loading. And when you do
so, it turns out that not only are the two networks independent, but they
are
also identical. If you can accept the equal load constraint, this is a very
practical and economical solution.

The origin of this discussion, which you have insisted over the years in
turning into an argument, was the question of whether or not some kind of
"build-out" resistor or other hack is needed to equalize the impedances
of the split load inverter. From the very start, I always stated as an
explicit assumption that the two loads were equal. Under that assumption,
and using a well-defined model, the two output impedances are the same.
This is so clear as to be indisputable.

(For that reason, I fully expect you to dispute it.)

If you want to use a more complicated model where the output impedances
aren't equal, then by all means, you're welcome to do so! But please don't
insult those of us who realize we can get the answer we want more easily
by simplifying the model. And watch out for a physicist who comes along
and berates you for the simplifications inherent in your own model.

-Henry

The apparent paradox which confuses so many about the simple CPI
is the apparent equal Rout at anode or cathode, observed when one
changes the anode and
cathode loads equally.

Since the change of loads causes equal VO to keep appearing, Rout at a
and k must be equal, no?

But they ain't equal at all when measured separately, by changing just
*one* of the TWO loads.

The distinction needs to be remembered about this very simple circuit.

When you add an additional R to the anode to reduce the ohms load, the
voltage sags.
But adding the same R also to the cathode increasess the tube gain with
NFB, and voltage at anode and cathode
become identical if both anode and cathode have the same total RL
connected.
But the addition of the extra RL at anode and cathode *do cause* some
slight
reduction of tube gain and both the anode and cathode voltages fall a
bit, and by the same amount if the load change is equal at a and k.
The *apparent* Rout at and k can be worked out by measuring the VO
change and dividing it by the I change
caused by the additional loading.

I leave you to ponder the math, but its pretty simple.

I doubt I need to work it out, because I know the CPI just works,
and its not a bad way to create two balanced voltages, providing you
don't overload the tube.

The other thing needed to be remembered is that the a and k loads sum to
form the total RL
to determine tube gain without NFB. So 22k plus 22k for 1/2 a 6SN7
make a load of 44k, and the tube gain without FB is 20 x 44k / 54k ( if
µ = 20 and Ra = 10k. )

But with the current FB, tube gain is slightly under two, if one sums
Va&Vk outputs, and divides by Vg.


Patrick Turner.

[Henry Pasternack[_2_]]

"John Byrns" wrote in message
...
As far as I am concerned the whole argument is now moot, since I now
understand how to correctly interpret your measurements, even if you
fail to understand what it was that you actually measured.

The argument isn't about circuits at all, since we agree on the behavior,
but rather it's about your unfriendly insistence over the years that I
somehow don't understand what I'm talking about, that I'm "slippery"
and wrong, and so on. And now poor Flipper is caught up in your
web of suspicion as well. LOL!

It seems to me that Flipper is right, and you are so attached to your
idea of what's "real" that you're completely unwilling/unable to see what
the math is telling you. Namely, given the equal load condition (which
has been a stated assumption since day one), the difference between
the two impedances disappears and they become effectively equal.
Yes, the impedance measured from plate to cathode will then be twice
the individual output impedance. This doesn't refute the conclusion,
but follows logically from it given the symmetry of the output voltages.

I've always agreed with you about the impedances being different when
measured separately. But that's irrelevant because there is no way to
perform the measurement you describe without unbalancing the circuit.
And I agree the argument is moot, but not for the reason you think. It's
moot because you insist on violating the equal load constraint, which is
fundamental to my position . So of course you get different results!

When I think of the word "slippery", I have an image of someone
wriggling out of constraints. That's exactly what you're doing, John!
Anyway, I have a very busy week ahead of me, so that's the last I
have to say on the subject. Hopefully, Flipper will take the time to
follow up to your inevitable inane response to this message.

-Henry

This trait of yours to "wriggle" out of constraints really seems to me to
be the essence of "slipperiness." Like your friend (What's his name?
I seem to have forgotten it.) you

[Henry Pasternack[_2_]]

"Henry Pasternack" wrote in message
...
This trait of yours to "wriggle" out of constraints really seems to me to
be the essence of "slipperiness." Like your friend (What's his name?
I seem to have forgotten it.) you

Please ignore this left-over text, which I didn't realize was still attached
to the end of my message. FYI, I was going to draw a comparison to
the "slippery" behavior of another r.a.t. participant, but thought better
of it. LOL. Maybe we'll hear from him yet.

-Henry

[John Byrns]

In article ,
"Henry Pasternack" wrote:

"John Byrns" wrote in message
...
As far as I am concerned the whole argument is now moot, since I now
understand how to correctly interpret your measurements, even if you
fail to understand what it was that you actually measured.

The argument isn't about circuits at all, since we agree on the behavior,
but rather it's about your unfriendly insistence over the years that I
somehow don't understand what I'm talking about, that I'm "slippery"
and wrong, and so on. And now poor Flipper is caught up in your
web of suspicion as well. LOL!

Well Flipper has no one to blame but himself for trying to attribute our
differences to different models, when as near as I can tell we are using
essentially the same model, differing only on measurement issues and
their interpretation. Flipper seems to feel that a model can reflect
any result you want, while I feel that a model must be constructed to
mimic the performance of the actual circuit, and that if it doesn't it
isn't a valid model.

It seems to me that Flipper is right, and you are so attached to your
idea of what's "real" that you're completely unwilling/unable to see what
the math is telling you.

I am no more attached to my ideas than you are attached to your ideas
which makes you completely unwilling/unable to see what the math is
telling you.

I will have to look back to see what I said when, however my recent use
of the "idea of what's real" was simply a reaction to Flippers statement
that models weren't physical and therefore could be manipulated to show
whatever you wanted them to show, because models weren't "physical" and
hence didn't represent the truth. As a result I suggested that we build
the actual physical, or "real" circuit and measure that, if the problem
was in the models. Flippers use of the Thevenin and Norton equivalent
circuits as examples in this context is complete nonsense because as
black boxes they are indistinguishable from one another by external
electrical measurements, although you can presumably tell which is
inside the black box by measuring the surface temperature of the box
under open and short circuit conditions.

Namely, given the equal load condition (which
has been a stated assumption since day one), the difference between
the two impedances disappears and they become effectively equal.

And here we have the actual slight of hand, you are talking about a
single impedance, the anode to cathode impedance, not two separate
impedances. You have contrived to make it appear that what you are
measuring two separate impedances, when in reality all you are doing is
measuring the anode to cathode impedance in a questionable way that
gives the appearance that you are actually measuring two separate
impedances when only one impedance is actually being measured. The
customary way to measure impedance is to measure the voltage across the
same two terminals that you are injecting the test current into. In
this case that would mean measuring the voltage between the anode and
cathode terminals, yet you measure between one of those terminals and
ground, a procedure that for an impedance measurement appears completely
bogus on the face of it. Can you cite any support for the measurement
procedure you are using, any suggestion that it is a valid impedance
measurement?

Now because of the symmetry constraint on the circuit it is true that
you are measuring the anode to cathode impedance divided by two. The
actual situation under the balanced load constraint is that the anode to
cathode impedance is driving a load impedance of twice that of either of
the two loads alone, since the two loads are in series and under the
balance constraint, no current will flow between ground and the junction
between the two loads, hence the junction of the two loads can be
disconnected from ground and the loads can be considered to be one
single load of twice the value, driven by the singular anode to cathode
source impedance of the concertina tube.

Yes, the impedance measured from plate to cathode will then be twice
the individual output impedance. This doesn't refute the conclusion,
but follows logically from it given the symmetry of the output voltages.

It sure looks like it refutes the conclusion, because you aren't making
a legitimate impedance measurement, but instead are making a bogus
measurement contrived to prove your point. Under the constrain of
balance it is a useful measurement, but it in no way demonstrates that
the anode and cathode source impedances are equal when the circuit is
balanced, all it demonstrates is that when the circuit is balanced the
single anode to cathode source impedance is effectively driving a load
impedance of twice the magnitude of the individual loads. It does not
in any way demonstrate that the cathode to ground and anode to ground
source impedances are equal.

The bottom line is that what you are measuring is not two quantities
representing the cathode impedance and the anode impedance, but is a
single quantity that is essentially the anode to cathode impedance, this
in no way shows the anode impedance to be equal to the cathode impedance.

I've always agreed with you about the impedances being different when
measured separately. But that's irrelevant because there is no way to
perform the measurement you describe without unbalancing the circuit.

That's quite true, but as I believe I demonstrated some years back, if
you also derive the equations for the voltage generator part of the
Thevenin equivalent circuit, for the anode and cathode circuits, you
will see that your objection falls out in the wash and balanced loads
will see balanced voltages even with the unbalanced source impedances.

And I agree the argument is moot, but not for the reason you think. It's
moot because you insist on violating the equal load constraint, which is
fundamental to my position . So of course you get different results!

How do I violate the equal load constraint? I have always maintained,
and at one time demonstrated, how my method handles the balanced load
condition, how is that "violating the equal load condition"? I also
understand how your method takes advantage of the equal load condition
to provide a simplified calculation. Where is the violation?

What I fail to see is how the fact that your calculation, of what is
effectively the anode to cathode impedance, which provides a useful and
simplified result for the equal load condition, in any way implies that
the anode and cathode source impedances are equal? What you don't seem
to understand is that you are not measuring the anode and cathode source
impedances, but that you are measuring only a single impedance, the
impedance between the anode and cathode.

When I think of the word "slippery", I have an image of someone
wriggling out of constraints. That's exactly what you're doing, John!

Tell me again what constraint I am "wriggling" out of, certainly not the
equal load constraint, I have np problem with that?

Anyway, I have a very busy week ahead of me, so that's the last I
have to say on the subject. Hopefully, Flipper will take the time to
follow up to your inevitable inane response to this message.

Perhaps, or perhaps not, but Flipper is plowing a different field
somehow related to different models, not different measurement methods
and interpretations for a single model. If Flipper does respond, it
will be interesting to see where he is going with his different model
theory.

This trait of yours to "wriggle" out of constraints really seems to me to
be the essence of "slipperiness." Like your friend (What's his name?
I seem to have forgotten it.)

I think you must mean Andre, he is doubtlessly thinking you are making a
complete fool of yourself, probably enjoying every minute of it while
making wagers with his friends as to what you will do next.

--
Regards,

John Byrns

Surf my web pages at, http://fmamradios.com/

[Henry Pasternack[_2_]]

"John Byrns" wrote in message
...
[Deleted]

ROTFLMAO!

Enjoy your debate with Flipper, John. :-)

-Henry

[John Byrns]

In article ,
flipper wrote:

On Mon, 11 Aug 2008 09:48:29 -0500, John Byrns
wrote:

Flipper, please see my response to Henry Pasternack for further comment.

Maybe. I don't know. But if I do this may be the last time because I'm
getting a bit tired of putting in a good faith effort only to have you
flush the whole blooming thing down the toilet.

Hi Flipper,

I'm sorry that I have been ignoring your "good faith effort" and have
been "flushing the whole blooming thing down the toilet" for the moment.
I hope to get back to your comments shortly however my primary interest
in this thread is the source impedance of the anode and cathode
terminals of the concertina phase inverter, whether or not they are
equal, and the interpretation of Henry Pasternack's measurements.

As a result of this focus I have been temporarily fast-forwarding over
any mention of models, especially the mention of the word "model" in
connection with the word "different". Why you might ask have I been
doing this? The reason is simple, it is one of four or five simple
facts that I was going to list in my next post to clarify the context of
this discussion before I drop it and move on to other things, like the
things you have been saying, that I have been fast-forwarding over.

So I will start my contextual list with the first of four or five items.

1. Henry and I are using the same model for the concertina circuit.

At least Henry hasn't said otherwise, and from earlier discussions I
believe it is the case. What does this mean, it means that our models
are not different and hence have little relevance to our disagreement.
The rest of the list will hopefully follow.

--
Regards,

John Byrns

Surf my web pages at, http://fmamradios.com/

[Andre Jute[_2_]]

John Byrns wrote:

"Henry Pasternack" wrote:
This trait of yours to "wriggle" out of constraints really seems to me to
be the essence of "slipperiness." Like your friend (What's his name?
I seem to have forgotten it.)

I think you must mean Andre, he is doubtlessly thinking you are making a
complete fool of yourself, probably enjoying every minute of it while
making wagers with his friends as to what you will do next.

Poor Old Plodnick is too dull and predictable. Betting on which way a
silkworm will turn on a mulberry leaf is more exciting.

Watching silkworms turn...

Andre Jute
Inventor of the "Ridiculus Curse" (vide Harry Potter, see effects on
Poor Old Plodnick)
Visit Jute on Amps at http://members.lycos.co.uk/fiultra/

[John Byrns]

In article ,
flipper wrote:

On Mon, 11 Aug 2008 20:34:40 -0500, John Byrns
wrote:

In article ,
flipper wrote:

On Mon, 11 Aug 2008 09:48:29 -0500, John Byrns
wrote:

Flipper, please see my response to Henry Pasternack for further comment.

Maybe. I don't know. But if I do this may be the last time because I'm
getting a bit tired of putting in a good faith effort only to have you
flush the whole blooming thing down the toilet.

Hi Flipper,

I'm sorry that I have been ignoring your "good faith effort" and have
been "flushing the whole blooming thing down the toilet" for the moment.
I hope to get back to your comments shortly however my primary interest
in this thread is the source impedance of the anode and cathode
terminals of the concertina phase inverter, whether or not they are
equal, and the interpretation of Henry Pasternack's measurements.

I take it then you'll be looking for the 'real', 'true', 'actual',
impedances.

No, that is a game you can play if you are so inclined, I will simply be
looking for the "source" impedance.

Based on what definition of 'output impedance'?

My definition of "source impedance" should be easy to divine given
comments I have recently made on the subject. I am not wedded to that
definition, although it is congruent with common methods used to measure
the value of resistors and the impedance of other components. Do you
have an alternate definition for the source impedance? I think Terman
wrote a book on electrical measurements, it would be interesting to see
what he might have had to said on the subject, unfortunately I don't
have the book in my library.

As a result of this focus I have been temporarily fast-forwarding over
any mention of models, especially the mention of the word "model" in
connection with the word "different". Why you might ask have I been
doing this? The reason is simple, it is one of four or five simple
facts that I was going to list in my next post to clarify the context of
this discussion before I drop it and move on to other things, like the
things you have been saying, that I have been fast-forwarding over.

That's a shame because, judging by the conversation so far, the
concept of modeling and equivalent circuits is precisely where the
problem lies.

Given that Henry and I are both using the same model this conclusion
seems illogical to me, however this is precisely the sort of situation
where it is easy to make a logical error. Can you elaborate on why you
say this "is precisely where the problem lies"?

So I will start my contextual list with the first of four or five items.

1. Henry and I are using the same model for the concertina circuit.

His model has equal output impedances. Yours does not.

No, that is where you have gone off the rails in your thinking. Our
models both have the same source impedances, either equal or not, by
definition since they are the same identical model. The question is a
measurement issue, i.e. who is doing the measurements correctly and who
is doing them incorrectly.

It's patently obvious they're not the same model.

You keep saying that, please explain the differences between our models.
It would be nice if the models were different because that would offer
the potential for resolving the argument without either of us being
wrong.

However the most satisfying solution would be to abandon the model
entirely, along with the questions it raises, and simply build the
physical circuit and measure the resulting source impedances.
Unfortunately it appears that would just take us back to square one,
again raising the crucial question of how to properly measure the source
impedances.

At this point it appears that the only way to make forward progress
would be to agree on a method for measuring source impedance.

At least Henry hasn't said otherwise, and from earlier discussions I
believe it is the case. What does this mean, it means that our models
are not different and hence have little relevance to our disagreement.

Au contraire.

Again you disagree without providing the slightest discussion of why you
disagree. This is becoming a somewhat irritating personality trait of
yours.

If you think our models are different, it should be a trivial matter for
you to simply point out the differences, as you have not attempted this
simple matter, I can only assume that you are playing a game here, if
that is how you feel so be it.

--
Regards,

John Byrns

Surf my web pages at, http://fmamradios.com/

[Patrick Turner]

However the most satisfying solution would be to abandon the model
entirely, along with the questions it raises, and simply build the
physical circuit and measure the resulting source impedances.
Unfortunately it appears that would just take us back to square one,
again raising the crucial question of how to properly measure the source
impedances.


Just building something, observing it, measuring it and deciding what is
there
is trifle to complex and real for some around here. They hafta get orf
their bums, and go
into their workshop, and spend time actually doing something.

But ya don't need to all that. Blind Freddy can see that if you measure
the Rout
at the anode by adjusting the anode load and only the anode load, the
Rout
= Vchange / Ichange, which is how you measure the Rout of anything.

If you repeat the same load adjustment to the cathode load and leave the
anode load alone,
you will find Rout is less then at the anode.

And you'll find the cathode is a voltage source, ie, Rout RL at the
cathode,
and Rout at the anode is a current source, ie, Rout RL at the anode.

Now this is true of any common cathode amplifier with an unbypassed Rk,
and you test the Rout at
anode and cathode separately.

It would be possible to rig up the CPI so that Rout is the same at a and
k,
but it'd need you to reduce anode RL and increase the cathode RL until
the Rout was the same at a and k.
But then the voltages wouldn't be equal at all, so its not a useful
exercize.

The only way to ensure the CPI has equal Rout at a and k would be to
direct couple
a pair of cathode followers to cathode and anode, and then you'd find
the two Routs would be the same
and each would be equal to what you'd get with any CF.


The theorists here who are making a mountain out of a molehill
could draw up the triode and a voltage gene + Ra resistor as a model in
a CPI with equal anode and cathode loads
and prove to themselves how it all works by working out the voltages and
currents and then moving on to
derive the formulas for Rout at the anode, and Rout at the cathode, in
terms of µ, Ra, and RL.


At this point it appears that the only way to make forward progress
would be to agree on a method for measuring source impedance.

I hope I have described a reasonable way to do it.

Measurers will find Rout at cathode a lot less then Rout at anode.

Sorry, but its true.

However, just because the two source resistances are different, it don't
stop the CPI from being a
decent sort of phase inverter, and able to give good enough balance for
most amplifiers
as long as it isn't driven into overload.

I like the CPI as it is in a Williamson better than a Schmitt, or
paraphase
or as in Quad-II, or a transformer.

I am too busy working on tube gear all day long to worry too much about
this
huricane of hot air about CPIs.

Patrick Turner.



At least Henry hasn't said otherwise, and from earlier discussions I
believe it is the case. What does this mean, it means that our models
are not different and hence have little relevance to our disagreement.

Au contraire.

Again you disagree without providing the slightest discussion of why you
disagree. This is becoming a somewhat irritating personality trait of
yours.

If you think our models are different, it should be a trivial matter for
you to simply point out the differences, as you have not attempted this
simple matter, I can only assume that you are playing a game here, if
that is how you feel so be it.

--
Regards,

John Byrns

Surf my web pages at, http://fmamradios.com/

[Andre Jute[_2_]]

On Aug 12, 5:22*pm, Patrick Turner wrote:
Blind Freddy can see that if [...]

On RAT when Pasternack gets involved, the question isn't the truth of
an impedance but whether Blind Freddy is congenitally accident-prone
or blinded himself. Adding a conditional "if" to that volatile mix
merely gives the resident weasels another opportunity to perform their
only party trick: weaseling.

Andre Jute
Ah, the human condition! But that would only apply if the subjects
were human. They're not. They're rats.

[Patrick Turner]

Andre Jute wrote:

On Aug 12, 5:22 pm, Patrick Turner wrote:
Blind Freddy can see that if [...]

On RAT when Pasternack gets involved, the question isn't the truth of
an impedance but whether Blind Freddy is congenitally accident-prone
or blinded himself. Adding a conditional "if" to that volatile mix
merely gives the resident weasels another opportunity to perform their
only party trick: weaseling.

Andre Jute
Ah, the human condition! But that would only apply if the subjects
were human. They're not. They're rats.

But at least we are not settling issues they way they do in Georgia at
the moment.

I'd be nervous lecturing some Georgians about the benefits of Russian
845 triodes.
It'd depend a bit where you were there because some Georgy boys hate
Ruskies, others love 'em....

Its likely that a couple of big bellied heavies with gold fillings would
wander over to me and shove the 845 *sideways*
up my nether regions.....

I'm also reminded of Churchill's comment about democracy,
"Democracy if bloody terrible, until you think about the alternatives"

So let 'em yabba yabba yabba in the dust for a month about the CPI and
models and things, and nothing will be agreed upon,
and that's how it was 8 years ago over the same damn issue of a
concertina phase inverter, CPI.
The indigenous peoples of Oz have won considerable improvements to their
landrights over the last 25 years.
Now the mining companies actually do have to negotiate the terms of land
use when a new mine is planned.
Talk about yabba yabba! goes on for years before anything is settled,
and until the price
and perks offered meet the indigenies' expectations.
Big holes then get dug, and filled in again.

CPI tube talk gets kinda boring afer awhile; talk about the Consumer
Price Index would be less boring,
unless you lived in Zimbabwee, when the price of bread in the morning
rises 300% by late afternoon, and a wheelbarrow is needed to cart the
price for the loaf.
I hear wheelbarrows have risen 1,000% since yesterday.

I always wonder HTF the Zimbies actually get on, not very well unless
you are among the elete.
But although English is widely spoken there we never hear from audio
experimeters from Zimby
arguing the case about concertinas.
No Africans here afaik, and no Indians. Not even anyone from China.
I live in hope things get more evenly representative on the Internet.

Now I am kicking myself that the Oz dollar has declined from near parity
with the US dollar
and because I should have bought a heafty stock of 6550 etc a month ago.

Its only money though ain't it?

Patrick Turner.

[Patrick Turner]

flipper wrote:

On Tue, 12 Aug 2008 16:22:18 GMT, Patrick Turner
wrote:


However the most satisfying solution would be to abandon the model
entirely, along with the questions it raises, and simply build the
physical circuit and measure the resulting source impedances.
Unfortunately it appears that would just take us back to square one,
again raising the crucial question of how to properly measure the source
impedances.


Just building something, observing it, measuring it and deciding what is
there
is trifle to complex and real for some around here. They hafta get orf
their bums, and go
into their workshop, and spend time actually doing something.

You get yourself into trouble with those incessant knee jerk snide
remarks of yours because this matter came up as a result of my doing
exactly that. I BUILT the damn thing and measured it.

So, with all due respect to you and your bums, sit on it.

OK, Point taken, but some here are arm chair wanabes.


But ya don't need to all that. Blind Freddy can see that if you measure
the Rout
at the anode by adjusting the anode load and only the anode load, the
Rout
= Vchange / Ichange, which is how you measure the Rout of anything.

Not when equal loads is inherent to the circuit and a required
condition.

But when one talks about the Rout at one point in a circuit,
what is usually meant is that its taken as is at that point,
and without any load adjustments elsewhere.

The Vchange / I change can be used to measure Rout easily for any amp
output.

In the case of the CPI, it can be done if you specifiy that the load
change
to anode and cathode will be equal.
Then the because the same current flows through cathode load, tube, and
anode load,
the Ichange is equal, and because the TWO loads are kept equal, V change
will also be equal,
so you'd think Rout is the same from both terminals.

It is, but subject to conditions.
Unconditionally, its not.

Unequally loaded it isn't a 'concertina' and it doesn't behave like a
'concertina'. In fact, measuring your way you'd rip the thing up and
throw it out because the outputs would be nothing akin to equal and,
so, of no use as a 'phase inverter'.

Unless you has a pot to lower the higher VO to be equal to the lower VO.

But it wastes gain, so nobody does it.

If you repeat the same load adjustment to the cathode load and leave the
anode load alone,
you will find Rout is less then at the anode.

You're not measuring a concertina.

A concertina is merely a normal single ended common cathode triode amp
stage, but one with
a cathode resistor equal to the anode resistor.

So when measuring a concertina, you are measuring a common cathode
amplifier stage.


And you'll find the cathode is a voltage source, ie, Rout RL at the
cathode,
and Rout at the anode is a current source, ie, Rout RL at the anode.

You're not measuring a concertina.

Now this is true of any common cathode amplifier with an unbypassed Rk,
and you test the Rout at
anode and cathode separately.

That's right, a common cathode amplifier, not a concertina.

As I said, a concertina *is* a common cathode amp......

The word concertina is used so everyone knows the cathode load = anode
load,
hence Va = Vk.


It would be possible to rig up the CPI so that Rout is the same at a and
k,
but it'd need you to reduce anode RL and increase the cathode RL until
the Rout was the same at a and k.
But then the voltages wouldn't be equal at all, so its not a useful
exercize.

Nope, because if one defines 'output impedance' per the Thevenin model
then a concertina already has equal Thevenin 'output impedances' as
long as the loads are equal.

If measured separately, you get what ya get, if measured conditioanlly,
you get something else, a convenient measure.

I have never worked out exactly what the conditional Rout of the
CPI actually is. But its somewhere higher than a cathode folower, and
lower than from
a normal common cathode amp at its anode and with fully bypassed
cathode.

Math experts should have come up with all the releveant figures by now,
but I've never
*needed* to calculate the Rout, so I have never have.


The only way to ensure the CPI has equal Rout at a and k would be to
direct couple
a pair of cathode followers to cathode and anode, and then you'd find
the two Routs would be the same
and each would be equal to what you'd get with any CF.

There is no need because if one defines 'output impedance' per the
Thevenin model then a concertina already has equal Thevenin 'output
impedances' as long as the loads are equal.

"as long as the loads are equal" is the condition needed for equal Rout
at each terminal.

What if they were not in a dynamic working amp?

This becomes a fact when a CPI runs into grid current powering output
tubes.
The grid current affects the anode signal supply far more than the
cathode supply,
and the amp clipping becomes grossly asymetrical, ie, like a dogs'
breakfast,
with gross asymetrical dc offsets. So in guitar amps, 99% use a
balanced amp or LTP to drive the output stage which gives symetrical
clipping and better
recovery, and more even tube wear. Its something worth something in
musicians amps which are
routinely used well into clipping the output stage.



Any measurement with equal loads, such as observing the roll off, will
show equal impedances. And it is trivial to derive mathematically as
well since you have 'one' common current through the equal loads.
Simply factor out the equal external loads leaving equal source
impedances.

The theorists here who are making a mountain out of a molehill
could draw up the triode and a voltage gene + Ra resistor as a model in

You seem to be confusing 'real circuit' with models. In a Thevenin
model there is a ideal voltage source (ideal current source in the
equivalent Norton) and an output impedance. There is no "triode
and..."

The triode model I use is a perfect voltage generator with 3 terminals,
plus a resistance added to be equal to Ra taken to a 4th terminal
being the anode of the model.
One terminal is the cathode, one is a grid, and infinite Rin, and the
other generator output is
NOT the anode, but just the generator output, which is low Rout = 0.0
ohms and the signal output = µ x Vg-k.
From gene output to the anode there is a resistance = Ra.

This model can be used to easily work out whatever signal voltages will
exist at a, k and g for any vacuum tube.
Go to
http://turneraudio.com.au/tube-operation1.html
Take a look at Fig 2 nearly 1/2 way down the page.

I hope I have covered the subject at my website to promote understanding
of the use of modelling to some advanantage to those used to working
things out in a very basic
manner based on current flow and voltage appearances and the use of
Ohm's Law.



a CPI with equal anode and cathode loads
and prove to themselves how it all works by working out the voltages and
currents and then moving on to
derive the formulas for Rout at the anode, and Rout at the cathode, in
terms of µ, Ra, and RL.

That's more akin to what you're doing, except you then ignore that the
supposed 'source impedances' calculated have no apparent utility.or
meaning for an equally loaded concertina.


Separate Rout measurements of Rout at a of k without making oads stay
equal does have limited utility; it tells us exactly what is there.

Meaning for use as a CPI comes when we add a condition of ensuring equal
loads always exist,
and changing anode and cathode loads equally when measuring Rout.




I'd be interested if you can reveal some use for the 'unequal
impedance' calculation because, off hand, I can't think of a single
thing. Doesn't even tell me what happens with unequal loads because it
wasn't calculated with those either.

There is no application I can think of where the fact of unequal Routs
is exploited for a benefit.

Of course RDH4 has a sample CPI where a normal CPI with one triode feeds
a
pair of following CPIs triodes, and in the following pair the
anode and cathode of one CPI are cross coupled with C to the other CPI
cathode and anode,
and the individually tested Routs at a and k become equal.

Its like a kind of CPI overkill to me, and ther's not one amp in which I
have ever seen such a thing.

Maybe it had its uses in scientific circuits of some sort.



At this point it appears that the only way to make forward progress
would be to agree on a method for measuring source impedance.

I hope I have described a reasonable way to do it.

It's reasonable. It just doesn't measure the Thevenin equivalent for a
concertina with equal loads.

Measurers will find Rout at cathode a lot less then Rout at anode.

Measurers who use equal loads will find equal Thevenin impedances.

Measurers who break the concertina with unequal, or no, loads will
find something else.

Sorry, but its true.

Not when one defines output impedance as a Thevenin equivalent output
impedance.

Well, OK, equivalant whatever, Rout is Rout at the end of the day.

However, just because the two source resistances are different, it don't
stop the CPI from being a
decent sort of phase inverter, and able to give good enough balance for
most amplifiers
as long as it isn't driven into overload.

That's because, as long as the loads are equal it has equal Thevenin
source resistances and since the Cs are equal the RC roll off is
equal.

In fact the stray C and Miller C around the circuit can be asymetrical,
and
I have found the anode output sags at lower F than the cathode output.
hence the use of some compensating C across the Rk.
This boosts the anode output as F rises, and HF poles then become the
same at a and k.



Which resolves the century old befuddlement of why the damn thing
works despite those confounding 'different impedances' that had even,
so called, 'experts' shoving 'build out resistors' on the thing to
solve a problem with those confounding 'different impedances' that
didn't exist.

You simply accept that it works, despite the 'different impedances'. I
have shown *why* it works and done so three ways: with the 'one
current through equal loads' derivation and two models, one with your
preferred unequal 'output impedances' (measured with a broken
concertina) and one with the equal Thevenin equivalent output
impedances.

I might have said more stuff than "it just works."

They all work but the purpose behind concepts like 'output impedance'
(the term being derived from Thevenin) is to simplify things and I
submit that the 'unequal impedance' model has caused infinitely more
confusion and problems that it ever resolved, so what is the purpose
of it?

We need to know the unequal Routs are there if the loads become unequal.

But as Vchange / Ichange remain equal if both anode and cathode loads
are kept equal, at AF the CPI make a fine phase inverter.

It also buffers the input triode and balanced amp stage, so for a given
set of tubes the BW is very good.
But its 4 stages for an amp instead of a minimum of two, as in say
Quad-II.
4 triodes intead of two pentodes. We all know how the bean counters
voted.
They voted Williamson out into the Wilderness.

Patrick Turner.


I like the CPI as it is in a Williamson better than a Schmitt, or
paraphase
or as in Quad-II, or a transformer.

I am too busy working on tube gear all day long to worry too much about
this
huricane of hot air about CPIs.

Patrick Turner.

[John Byrns]

In article ,
flipper wrote:

On Tue, 12 Aug 2008 16:22:18 GMT, Patrick Turner
wrote:


However the most satisfying solution would be to abandon the model
entirely, along with the questions it raises, and simply build the
physical circuit and measure the resulting source impedances.
Unfortunately it appears that would just take us back to square one,
again raising the crucial question of how to properly measure the source
impedances.


Just building something, observing it, measuring it and deciding what is
there
is trifle to complex and real for some around here. They hafta get orf
their bums, and go
into their workshop, and spend time actually doing something.

You get yourself into trouble with those incessant knee jerk snide
remarks of yours because this matter came up as a result of my doing
exactly that. I BUILT the damn thing and measured it.

OK, great, that is exactly what I suggested a couple of days ago, you
should have said you had already done it. Of course building the
concertina is a relatively trivial task, the tricky bit is measuring the
concertina's two source impedances.

It would be helpful, and a positive contribution if you could explain
how you measured the source impedances, as well as how you determined
that they were equal?

So far all we have is people stating that they have measured the
concertina's two output voltages, found them to be equal, and then
forcefully asserting that "obviously if the concertina's two output
voltages are equal, then the two source impedances must also be equal"!
That assertion doesn't prove that the source impedances are necessarily
equal, as can be easily demonstrated by simply providing a counter
example that produces equal output voltages from unequal source
impedances. The bottom line is that a little bit of logic, something
sadly lacking in parts of this thread, tells us that the presence of
equal output voltages says absolutely nothing, one way or the other,
about the source impedances. To prove equal source impedances we need
something more than equal output voltages.

Moving on, later today I will post a response to your previous message
containing the answers to all the questions you say I have failed to
answer. I hope once you have those answers in hand, that you will
reciprocate by answering my question above.

--
Regards,

John Byrns

Surf my web pages at, http://fmamradios.com/

[John Byrns]

In article ,
flipper wrote:

On Tue, 12 Aug 2008 09:31:41 -0500, John Byrns
wrote:


I will simply be
looking for the "source" impedance.

Define it.

The source impedance of an output is the impedance you see, or measure,
looking back into the output.

Since you doggedly insist there can be one and only one model you most
certainly are 'wedded' to 'it', despite refusing to define what 'it'
is.

I did not say there could be "one and only one model", you are putting
words into my mouth again. What I said was that as far as I know, Henry
and I are using the same model, at least we were the last time Henry and
I went around on this subject several years back. I assume that if
Henry has switched models since then, he would speak up and loudly
proclaim the fact. If Henry has changed models I doubt I would have any
problem adapting his new model myself, thereby bringing us back into
congruence.

Did I refuse to define the model? Having not yet defined it in this new
thread is not the same thing as refusing to define it.

Here's the definition. The model of the concertina and its separate
loads consists of six elements. The concertina model proper consists of
a perfect voltage amplifier source and a series resistance representing
the small signal, frequency independent, anode resistance and u of the
concertina's triode, these two components are effectively an internal
Thevenin equivalent generator. In addition to the Thevenin equivalent
generator, two equal value resistors are connected from the Thevenin
equivalent generator's two output terminals to the AC or signal ground.
These two resistors represent the anode and cathode resistors used in
the concertina circuit to provide a path for the DC operating current
required by the triode. A Norton equivalent generator can replace the
Thevenin equivalent generator if desired, the two forms being equivalent.

That accounts for four of the elements in the complete system. The
final two elements in the complete system are two load impedances, Za
and Zk, one connected from the concertina's anode to ground and the
second connected from the concertina's cathode to ground. For the
purposes of this discussion of the balanced concertina Za and Zk are
made equal.

That's all there is to it. Note that this is a small signal model only
and does not take into account frequency dependent effects due to the
triode, such as the grid related capacitances, it does take into account
frequency dependent effects due to the loads Za and Zk.

Finally IIRC you asked me in another post how I measure impedance. I
have used three methods, the first two I have used on physical circuits
in my workshop, and a third theoretical method that I have used only on
mathematical models.

The first impedance measuring method that I use in my workshop is a GR
impedance bridge that I connect across the terminals of the device whose
impedance I wish to measure.

The second method I have used in my workshop is what I will call the
"Turner method" which Patrick described in a recent post to this group.
This method consists of feeding a sine wave signal into the input of the
DUT and measuring the output voltage. A resistor is then connected
across the output, while maintaining the same input voltage, and the
output voltage is again measured, with the added resistor connected.
The source resistance can then be calculated from the two voltages and
the value of the added resistance. Note however that this method only
works when the source impedance is purely resistive, it does not give
the correct answer when the source impedance is a complex impedance as
in the case of the concertina circuit with load.

The third method is what I call the "Pasternack" method after Henry
Pasternack who proposed it several years ago. I have not tried this
method in my workshop, as it would require building some test equipment
I don't have as well as a special test jig to make use of it. I have
however applied this method to mathematical models of various circuits.
This method requires two pieces of virtual or real test equipment. The
first is an AC current generator, and the second is a high impedance AC
voltmeter which can measure the magnitude of a voltage, as well as the
phase angle of the voltage relative to the phase of a reference signal.
In this case the reference signal is the phase of the current generator
output.

HP uses these instruments to measure the source impedances of the
concertina's two outputs by connecting the current generator between the
anode and cathode terminals of the concertina circuit, with the loads
disconnected. HP then uses the high impedance voltmeter to measure the
voltages that appear between the cathode and ground terminals, and
between the anode and ground terminals. From these voltage
measurements, as well as the value of the current produced by the
current generator HP then calculates the anode and cathode source
impedances.

I use the same two pieces of test equipment, however I connect them in
parallel between the two terminals I am measuring the impedance of. I
leave the anode load in place when measuring the cathode source
impedance, removing only the cathode load, and vice versa when measuring
anode source impedance. From the two voltage measurements, magnitude
and phase, and the current generator current setting I can then
calculate the source impedances for the two output terminals. These
source impedances are complex impedances, i.e. they have a reactive
component even for the simple resistive triode model. The complex
source impedance results from the load impedance Z that remains
connected to the output terminal not being measured.

Once I have the two source impedances, I next measure the voltages of
the two Thevenin equivalent voltage generators for the anode and cathode
terminals. I do this by disconnecting the current source from its
parallel connection across the voltmeter, instead feeding a known
voltage into the grid circuit, and then measuring the voltages developed
at the anode and cathode terminals of the concertina, using the same
load protocol as with the source impedance measurements. From these
measurements the Thevenin voltage generator function can be
characterized.

All these measurements must be repeated across the audio band to
completely characterize the source impedances and equivalent voltage
generator characteristics. The resulting source impedances and Thevenin
voltage generator functions can then be used to calculate the
gain/frequency characteristics the unbalanced concertina, as well as the
balanced concertina.

I think that about covers it, if I forgot something you have but to ask
and I will answer if I can.

--
Regards,

John Byrns

Surf my web pages at, http://fmamradios.com/

[Henry Pasternack[_2_]]

"John Byrns" wrote in message
...
It would be helpful, and a positive contribution if you could explain
how you measured the source impedances, as well as how you
determined that they were equal?

IMHO there is a rational and straightforward way to explain this
subject that makes it clear what the equal-impedance model means
and why it's a valid point of view. My compaint against you, Byrns,
is not that you take the position that the impedances are different.
It's clear what you mean and how you get to that point of view.
What does bother me, throughout all these years, is that you are
either unwilling or unable to make the effort to understand the other
point of view, and on that basis characterize me (and now Flipper)
as wrong/slippery/ foolish. I would go so far as to say you have
actively tried to avoid understanding my position because it is
incompatible with your larger belief systems to admit that I may
have a valid point. I say that just so we're clear what the argument
is about.

Anyway, I claim there are two aspects to the definition of output
resistance. The first apsect, an operational definition, says output
resistance is the real part of dv/di, where di is a small current flowing
in the output port. The second aspect, an abstraction, models the
device under test as a voltage source in series with an impedance,
where the output resistance is the real part of that impedance. I
think the first aspect is true by definition, and the second is true by
convention. In any event, I assert this is a reasonable and practical
definition.

The two models in question have the same architecture. It is a three-
port network with one input port and two output ports. The input
port is open-circuited. There are two voltage-controlled voltage
sources in the box, and two series resistors, each connected to one
of the output ports in the expected way. The voltage sources are
characterized by the equation Vout = k * Vin.

The models differ in the values of the components. In my model, R
and k are constant and Vin equals the input terminal voltage. Also,
the two halves are identical, i.e., R1 = R2 and k1=k2. In your model,
R, k, and possibly Vin are variable and depend on the external
components connected to the output ports. One of the features of
my model is that the three-port network can be split into two identical
two-ports with their inputs in parallel; this cannot be done for your
model, since it's not symmetrical and there are cross-dependencies
between the halves. Your model is more general than mine. If you
impose the equal load constraint, your model simplifies and then your
model and mine are identical.

With respect to measurements, we can use a sinewave generator
and an identical pair of test sets each containing an AC voltmeter.
Each voltmeter has a known input resistance. Inside the test set,
a resistor of known value may be connected in parallel with the
voltmeter using a switch. To perform an impedance measurement,
a signal of fixed amplitude is applied to the input port of the device
under test. A test set is connected to an output port and the voltages
are noted with the parallel resistor switched in and out. From these
readings it is possible to calculate the value of the series resistors in
the model of the device under test.

There are two test procedures. The Byrns procedure is:
a) Connect a test set to the plate output port and perform the two
measurements to determined the output resistance.
b) Disconnect the test set from the plate output port and repeat
the measurements with the test set connected to the cathode
output port.

My test procedure is:
a) Connect one test set to the plate output port and one test set to
the cathode output port.
b) Measure the voltages at both ports with the parallel resistors in
both test sets switched out of the circuit.
c) Measure the voltages at both ports with the parallel resistors in
both test sets switched into the circuit.

The result will be: In the Byrns test procedure, the measured output
resistances will differ. In my test procedure, the output resistances
will be equal and significantly lower than either of the two values
obtained in the Bryns test.

Now for interpretation. To the extent that both models and their
corresponding test procedures are documented and understood,
there can be no argument as to their validity. We can only argue
about the relevance and usefulness of the models, and the extent
to which their underlying assumptions match generally accepted
practices and principles.

I believe my model is a good model because it matches the typical
application (equal loads) of the split-load inverter, and it quickly
and economically predicts the HF rolloff behavior of the circuit
(ignoring strays). The notion of constant 'R' and 'k' is intuitive and
consistent with typical use of the Thevenin model. There is nothing
inherently wrong with using two test sets and making the measure-
ments simultaneously.

The Byrns model is more general and has the advantage that it
can handle any combination of unequal loads. The disadvantage
is that the resistances and the voltage source 'k' values are not
constant, but depend on the values of the load impedances. This
variability and interdependence, IMHO, is counter-intuitive to the
notion of a "real" output impedance and to common assumptions
about the the Thevenin/Norton model of output resistance.

This is a subjective call, but I would say the purpose of a model is
to impose a layer of abstraction that simplifies the representation of
a system at that level. I think John's model is of limited value because
it's basically just as complicated as the small-signal circuit model it
replaces. In that sense, our two models are not really comparable
because they represent two different levels of abstraction.

There is some sort of logical contradiction here, else there wouldn't
be this historical "paradox" about this circuit. As an aside, I would
observe again that a perfectly good way to measure output resistance
is to hang a cap across the output and measure the -3dB point. If
you do this to a real split-load inverter with two caps (as is proper
for a balanced circuit), you apparently get equal output resistances.
I agree with Flipper that this is an obvious and quite legitimate test
and interpretation of the results. The difficulty some people have in
accepting it seems to lie mostly in their ingrained notions of what
cathode follower and common cathode amplifier are and how they
"must" behave.

I think the single most important statement Flipper has made in this
discussion is when he pointed out that Byrns is not measuring the
split-load inverter, but is separately measuring a cathode follower
and an unbypassed common-cathode amplifier. To the extent that
'R' and 'k' one one port depend on the load at the other port in
Byrns's model, he is measuring two different circuits with his test,
since connecting the meter to one port or the other changes what's
inside his black box model of the DUT.

I anticipate several of John's objections, but my feeling here is that
I don't have to defend my model as the only viable model (it isn't),
but only to show it's reasonable (it is), contrary to Byrn's tiresome
historical remarks. And that's all I have to say.

I am sure John will have many unsatisfying things to say in reply.
But now I have to get back to work.

-Henry

[John Byrns]

In article ,
"Henry Pasternack" wrote:

I am sure John will have many unsatisfying things to say in reply.
But now I have to get back to work.

Hi Henry,

I was getting really irritated with ³Flipper² and his mantra of ³there
are two different models² when he seemingly wouldn't attempt to ³define²
the two models. His constant reference to Thevenin and Norton Models as
if they explained what he meant by two different models furthered my
confusion. After reading your post I finally realized what he was
probably talking about. All became clear when I realized that I was
being confused because there aren't two different models, there are
actually three different models involved. The first model being the one
I was talking about that mathematically describes the component topology
and values used in the actual concertina circuit implementation. The
second and third models are the two models based on your measurements
and on my measurements. Hopefully I now correctly understand what he
meant, although it doesn't really say much to resolve our disagreement
about whether or not the two source impedances are equal or not.

I was also annoyed when I kept asking ³Flipper² what his basis was for
sayingthat the anode and cathode source impedances were equal and he
would come back with statements something like ³the loads and the
voltages are equal so the source impedances appear to be equal². I have
not looked up one of his exact quotes, but I think they were something
to that effect. I really don't know what he may have been trying to
say, I originally interpreted it as his version of a proof, but after
thinking about it today, it occurred to me that he may have meant the
statement not as a proof, but as a sort of political compromise.
Whether he meant it that way or not, I love it in that new sense, it has
just the right number of weasel words to neatly encompass both your view
and mine without being too definite about what the relation of the two
impedances might be to each other.

There is also a third test procedure in addition to the two you
described. Call this procedure the ³New Byrns² procedure. I would
appreciate your comments on this new procedure.

Note that under the balance assumption, where the two loads and the
voltages across them are equal, no current will flow from the junction
of the two loads to ground. This allows us to disconnect the two loads
from ground while keeping the ends of the two loads connected to each
other.

The third test procedure, the ³New Byrns² procedure then becomes:

a) Remove the two balanced loads and replace then with a single load of
twice the impedance of either of the original two loads, using the
observation above.
b) Connect a single test set between the anode and cathode terminals of
the concertina, and perform the measurements to determine the output
resistance.

This output resistance should be twice the value of the output
resistance measured with your two test set procedure and can be used in
the same way to easily calculate the 3 dB frequency without raising the
prickly issue of whether or not the anode and cathode source impedances
are equal or not.

A question for you, with respect to your test procedure you state that
³In my test procedure, the output resistances will be equal and
significantly lower than either of the two values obtained in the Bryns
test.² You have stated this before, are you sure that your output
resistance is significantly lower than either of the two values obtained
in the original Byrns test? That doesn't seem to meet the common sense
test for reasonableness; however stranger things have proven true, so I
will have to put it on my list of things that need verifying one way or
the other.

--
Regards,

John Byrns

Surf my web pages at, http://fmamradios.com/

[Patrick Turner]

flipper wrote:

On Wed, 13 Aug 2008 10:50:58 GMT, Patrick Turner
wrote:



flipper wrote:

On Tue, 12 Aug 2008 16:22:18 GMT, Patrick Turner
wrote:


However the most satisfying solution would be to abandon the model
entirely, along with the questions it raises, and simply build the
physical circuit and measure the resulting source impedances.
Unfortunately it appears that would just take us back to square one,
again raising the crucial question of how to properly measure the source
impedances.


Just building something, observing it, measuring it and deciding what is
there
is trifle to complex and real for some around here. They hafta get orf
their bums, and go
into their workshop, and spend time actually doing something.

You get yourself into trouble with those incessant knee jerk snide
remarks of yours because this matter came up as a result of my doing
exactly that. I BUILT the damn thing and measured it.

So, with all due respect to you and your bums, sit on it.

OK, Point taken, but some here are arm chair wanabes.

That may be but they're not in the discussion. But, at any rate, I
accept the 'ok'.

But ya don't need to all that. Blind Freddy can see that if you measure
the Rout
at the anode by adjusting the anode load and only the anode load, the
Rout
= Vchange / Ichange, which is how you measure the Rout of anything.

Not when equal loads is inherent to the circuit and a required
condition.

But when one talks about the Rout at one point in a circuit,
what is usually meant is that its taken as is at that point,
and without any load adjustments elsewhere.

I am aware of that but that is precisely the 'problem'. It just
doesn't work with a concertina and if you do it that way you get
essentially useless results.

Well, if you use the classic way to measure the Rout of anything by
changing the load
and recording the Vchange, and I change, Ro can be found, but sure,
because the Rout
at a and k are very different if the measurement is done by applying
only one same load change
to each a and k output, then the answer for Ro is useless.

As I said, you could make the same load change to each of a and k, and
you should measure
equal Vchange and I change, and Ro at a and k will simply be Vchange /
Ichange and the same for a or k.




The Vchange / I change can be used to measure Rout easily for any amp
output.

In the case of the CPI, it can be done if you specifiy that the load
change
to anode and cathode will be equal.
Then the because the same current flows through cathode load, tube, and
anode load,
the Ichange is equal, and because the TWO loads are kept equal, V change
will also be equal,
so you'd think Rout is the same from both terminals.

It is, but subject to conditions.

Bingo... as you said, it *is* (the same).

So why were you chastising folks for saying exactly what you just
said?

Because of the condition of the test for Rout that needs to be
specified.

Conditions apply for the truth to be true.



Unconditionally, its not.

I am aware it's subject to conditions and the conditions have ALWAYS
been stated, from the very first post to the last.

It also happens to be the overwhelmingly common case and folks often
go to great lengths to keep it equal and 'balanced'. Unfortunately, in
the burning quest to do so, the befuddlement of an unequal output
impedance model has screwed up more than a few.

Unequally loaded it isn't a 'concertina' and it doesn't behave like a
'concertina'. In fact, measuring your way you'd rip the thing up and
throw it out because the outputs would be nothing akin to equal and,
so, of no use as a 'phase inverter'.

Unless you has a pot to lower the higher VO to be equal to the lower VO.

But it wastes gain, so nobody does it.

The only reason I can imagine anyone even thinking of such a
contraption is if they were befuddled by the unequal impedance model
and were trying to 'fix' the problem that doesn't exist.

"Whatever" as they say....

If you repeat the same load adjustment to the cathode load and leave the
anode load alone,
you will find Rout is less then at the anode.

You're not measuring a concertina.

A concertina is merely a normal single ended common cathode triode amp
stage, but one with
a cathode resistor equal to the anode resistor.

So when measuring a concertina, you are measuring a common cathode
amplifier stage.

Nope. Because you left out that the concertina has two, not one,
outputs with equal loads, and that is not a trivial condition. As
evidenced by the near 'magical' result of the seemingly disparate
output impedances of the common cathode amp becoming equal. Except, of
course, you don't have two 'output' impedances in the common cathode
amp, there's only one, so there's not really a comparison.

Nothing stopping you from using the cathode as an output from a common
cathode amp
**of any kind** including where you have RLa = RLk.



They just aren't the same thing, which is why measuring it as one
thing doesn't give you the same answer as measuring it when it's the
other.

The concertina to my mind is merely a common cathode stage with a lot of
local current FB from
its unbypassed Rk.

OK, so you take outputs from both a and k.

It doesn't stop the concertina being a common cathode amp.



And you'll find the cathode is a voltage source, ie, Rout RL at the
cathode,
and Rout at the anode is a current source, ie, Rout RL at the anode.

You're not measuring a concertina.

Now this is true of any common cathode amplifier with an unbypassed Rk,
and you test the Rout at
anode and cathode separately.

That's right, a common cathode amplifier, not a concertina.

As I said, a concertina *is* a common cathode amp......

No it is *not*. It looks a hell of a lot like one but a common cathode
amplifier has ONE output and a concertina has TWO, with equal loads.


So what about the number of outputs?

Its still a basic common cathode amp.


Another way to get phase inversion is to simply use a common cathode amp
with
a gain of exactly -1.0, ie, a triode common cathode amp is set up with
an unbypassed Rk
which is slightly lower in value than the RLa, and Vg = -Va.

So you have the input signal from somewhere external, maybe a CD player,
and you create an opposite
phase signal of exactly the same amplitude, and the two phases can then
be applied to a fully
balanced PP power amp with balanced series voltage GNFB. ARC did this
maybe 30 years ago.

The word "Concertina" does not make it illegal to say the concertina is
a common cathode amp.
It is a common cathode amp, but one with special features, implied by
the word concertina.

The "squeeze box" instrument stays put, but the player works the 2
handles in opposite directions.




The word concertina is used so everyone knows the cathode load = anode
load,
hence Va = Vk.

That's hunky dory.

It would be possible to rig up the CPI so that Rout is the same at a and
k,
but it'd need you to reduce anode RL and increase the cathode RL until
the Rout was the same at a and k.
But then the voltages wouldn't be equal at all, so its not a useful
exercize.

Nope, because if one defines 'output impedance' per the Thevenin model
then a concertina already has equal Thevenin 'output impedances' as
long as the loads are equal.

If measured separately, you get what ya get, if measured conditioanlly,
you get something else,

Exactly.

a convenient measure.

Well, one convenience is it measures the thing as it actually is and
used. Another is it produces a useful result, which is THE point to
doing a calculation in the first place.

I have never worked out exactly what the conditional Rout of the
CPI actually is. But its somewhere higher than a cathode folower, and
lower than from
a normal common cathode amp at its anode and with fully bypassed
cathode.

Henry came up with a number lower than a cathode follower and I think
Preisman did as well.

I don't see how it could be lower, but I have never needed to work it
out.

You can have 22k for RLa and RLk for a 6SN7, and have Va = Vk = 2V,
and then change the cap coupled loads from say 150k to 22k, and the Vk
and Va will stay equal,
but the drop in voltage is low, like a CF.

If the following cap coupled loads become too low compared to the dc
carrying RLs,
then the undistorted voltage range is reduced, and you get sudden cut
off distortions.....






Math experts should have come up with all the releveant figures by now,
but I've never
*needed* to calculate the Rout, so I have never have.

That's ok. It doesn't mean someone else doesn't care, though.

Hmm, we have not seen too much care around here and have not seen
anyone say what the Rout actually is for CPI if a&k loads are always the
same.


Just knowing they're equal, never mind the 'value', would have saved a
hell of a lot of confusion over the last century, though.

The only way to ensure the CPI has equal Rout at a and k would be to
direct couple
a pair of cathode followers to cathode and anode, and then you'd find
the two Routs would be the same
and each would be equal to what you'd get with any CF.

There is no need because if one defines 'output impedance' per the
Thevenin model then a concertina already has equal Thevenin 'output
impedances' as long as the loads are equal.

"as long as the loads are equal" is the condition needed for equal Rout
at each terminal.

What if they were not in a dynamic working amp?

Doesn't matter as long as the load impedances are equal,

Well there is a point where the loads suddenly do become unequal, at
grid current
in a following stage.

As one phase goes +, it hits and stalls with Ig, but the other phase
going -
has no grid current.

Guess which output tends to charge up the coupling caps the most.




This becomes a fact when a CPI runs into grid current powering output
tubes.
The grid current affects the anode signal supply far more than the
cathode supply,
and the amp clipping becomes grossly asymetrical, ie, like a dogs'
breakfast,
with gross asymetrical dc offsets.

None of the 'impedance' models are valid when it exits linear
operation and that's a stated condition for a Thevenin output
impedance.

So in guitar amps, 99% use a
balanced amp or LTP to drive the output stage which gives symetrical
clipping and better
recovery, and more even tube wear. Its something worth something in
musicians amps which are
routinely used well into clipping the output stage.

Yes, I know. I've built them. But it's not germane to a discussion of
concertina output impedance.

Any measurement with equal loads, such as observing the roll off, will
show equal impedances. And it is trivial to derive mathematically as
well since you have 'one' common current through the equal loads.
Simply factor out the equal external loads leaving equal source
impedances.

The theorists here who are making a mountain out of a molehill
could draw up the triode and a voltage gene + Ra resistor as a model in

You seem to be confusing 'real circuit' with models. In a Thevenin
model there is a ideal voltage source (ideal current source in the
equivalent Norton) and an output impedance. There is no "triode
and..."

The triode model I use is a perfect voltage generator with 3 terminals,
plus a resistance added to be equal to Ra taken to a 4th terminal
being the anode of the model.
One terminal is the cathode, one is a grid, and infinite Rin, and the
other generator output is
NOT the anode, but just the generator output, which is low Rout = 0.0
ohms and the signal output = µ x Vg-k.
From gene output to the anode there is a resistance = Ra.

This model can be used to easily work out whatever signal voltages will
exist at a, k and g for any vacuum tube.
Go to
http://turneraudio.com.au/tube-operation1.html
Take a look at Fig 2 nearly 1/2 way down the page.

I hope I have covered the subject at my website to promote understanding
of the use of modelling to some advanantage to those used to working
things out in a very basic
manner based on current flow and voltage appearances and the use of
Ohm's Law.

That's fine if one wants to model a triode, and there are good reasons
why one might, but in a classic Thevenin model of output impedance
there is no 'triode' (nor a model of one), or anything else but a
voltage source and a single output impedance. That's it's purpose and
the 'definition' of "output impedance."

a CPI with equal anode and cathode loads
and prove to themselves how it all works by working out the voltages and
currents and then moving on to
derive the formulas for Rout at the anode, and Rout at the cathode, in
terms of µ, Ra, and RL.

That's more akin to what you're doing, except you then ignore that the
supposed 'source impedances' calculated have no apparent utility.or
meaning for an equally loaded concertina.


Separate Rout measurements of Rout at a of k without making oads stay
equal does have limited utility; it tells us exactly what is there.

Maybe, if you define "what.". But 'what' is not "output impedance"
when you have two loaded outputs.

Meaning for use as a CPI comes when we add a condition of ensuring equal
loads always exist,
and changing anode and cathode loads equally when measuring Rout.

That's a different derivation and measurement.

I'd be interested if you can reveal some use for the 'unequal
impedance' calculation because, off hand, I can't think of a single
thing. Doesn't even tell me what happens with unequal loads because it
wasn't calculated with those either.

There is no application I can think of where the fact of unequal Routs
is exploited for a benefit.

Makes me wonder why you're so enthusiastic about a useless
calculation.

Of course RDH4 has a sample CPI where a normal CPI with one triode feeds
a
pair of following CPIs triodes, and in the following pair the
anode and cathode of one CPI are cross coupled with C to the other CPI
cathode and anode,
and the individually tested Routs at a and k become equal.

Its like a kind of CPI overkill to me, and ther's not one amp in which I
have ever seen such a thing.

Maybe it had its uses in scientific circuits of some sort.

I'm not interesting in calculating the output impedance of that.

Too useless for you?

Probably RDH4 has the expressions in math for all the Routs you can ever
think of.

I just don't know them all.

At this point it appears that the only way to make forward progress
would be to agree on a method for measuring source impedance.

I hope I have described a reasonable way to do it.

It's reasonable. It just doesn't measure the Thevenin equivalent for a
concertina with equal loads.

Measurers will find Rout at cathode a lot less then Rout at anode.

Measurers who use equal loads will find equal Thevenin impedances.

Measurers who break the concertina with unequal, or no, loads will
find something else.

Sorry, but its true.

Not when one defines output impedance as a Thevenin equivalent output
impedance.

Well, OK, equivalant whatever, Rout is Rout at the end of the day.

That's the problem, it isn't and you get completely different results
when you 'measure' under different conditions. 'Single sided' you get
one thing, equal loads gives you equal impedances, and unequal loads
gives you a whole range of impedances. 'Rout' is anything but 'Rout'.

However, just because the two source resistances are different, it don't
stop the CPI from being a
decent sort of phase inverter, and able to give good enough balance for
most amplifiers
as long as it isn't driven into overload.

That's because, as long as the loads are equal it has equal Thevenin
source resistances and since the Cs are equal the RC roll off is
equal.

In fact the stray C and Miller C around the circuit can be asymetrical,
and
I have found the anode output sags at lower F than the cathode output.
hence the use of some compensating C across the Rk.
This boosts the anode output as F rises, and HF poles then become the
same at a and k.

Come on Pat, you're grasping at any and everything. Everyone,
regardless of the model, has specifically said local parasitics are
being ignored and we're limiting operation under them.

If you want an ideal square wave at 5kHz in a Williamson,
you would find you'll get the best looking one where you have taken the
trouble to
add a small C across Rk of the CPI.

Lots of ppl do it. I just re-wired a VAC amp, there is the C, and 47pF
across 22k.

Its there to improve the HF performance.


Which resolves the century old befuddlement of why the damn thing
works despite those confounding 'different impedances' that had even,
so called, 'experts' shoving 'build out resistors' on the thing to
solve a problem with those confounding 'different impedances' that
didn't exist.

You simply accept that it works, despite the 'different impedances'. I
have shown *why* it works and done so three ways: with the 'one
current through equal loads' derivation and two models, one with your
preferred unequal 'output impedances' (measured with a broken
concertina) and one with the equal Thevenin equivalent output
impedances.

I might have said more stuff than "it just works."

Well, you usually say 'more stuff'. Whether it's related to the topic
is another question and it's true I haven't heard everything you've
ever said.

But this last time around you didn't say anything about how in the
world different impedances into equal capacitive loads could possibly
come out with an equal roll off. All you said was the unequal
impedances don't stop it from working just fine.




Well, as you said above, the reason is the output impedances *are*
equal, not different.

Well, I suggest you examine a working amp more closely.

You might find the reason why some makers add a C across Rk of a CPI.

Its to slightly boost the anode output which sags before the cathode
output.

Most makers don't bother, like they don't bother to do a whole range of
stuff
because they work down to a price, not up a standard of quality.



They all work but the purpose behind concepts like 'output impedance'
(the term being derived from Thevenin) is to simplify things and I
submit that the 'unequal impedance' model has caused infinitely more
confusion and problems that it ever resolved, so what is the purpose
of it?

We need to know the unequal Routs are there if the loads become unequal.

The Routs you calculate are useless for that as well because it
doesn't take into account unequal loads any more than it does equal
ones.

That derivation only works for ONE output at a time..

But as Vchange / Ichange remain equal if both anode and cathode loads
are kept equal, at AF the CPI make a fine phase inverter.

Right. And the result of a V/I calculation is generally called
"impedance" so equal V/I means equal output impedances.

It also buffers the input triode and balanced amp stage, so for a given
set of tubes the BW is very good.
But its 4 stages for an amp instead of a minimum of two, as in say
Quad-II.
4 triodes intead of two pentodes. We all know how the bean counters
voted.
They voted Williamson out into the Wilderness.

"Bean counters" don't get a vote. They count beans.

See Pat, among other things I've *been* both a designer and a Product
Manager so don't pretend to tell me how 'bean counters' voted or that
they told me 'how many' of this or 'what' I could or could not put in
a product.

Bean counters destroyed most quality in most products.

They are there to *remove quality*

If they'd made cars with stainless steel bodies back in the 1950s,
they'd still be around. But they made cars with thin ordinary sheet
steel that wasn't
meant to last longer than 5 years if you were very lucky.
It rusted away as you watched it in damp weather.
Bean counters created the most important commandment for business:-
Thou Shalt Not Make Anything to Last Long.

Some ******* upset the apple cart and invented a good way of painting
the
steel on the concealed inside areas of a car, so it lasted 15 years.
It was such a saving for the Ordinary Man, and his Struggling Missus,
that cars made which didn't rust got the sales.
Reluctantly, bean counters around the world were embarrassed into
recommending their companies extend the plant to include painting the
hidden surfaces of a car body.

They were all forced into it. They sure didn't like to have to pay more
for production.
They slugged the customer real hard for the improvement of course, more
than they should have,
because they'd point out how much longer something would last, and
instead of buying 3 cars in 15 years you'd only need one.
Boy they had it on easy street, because the marketeering arsols
brainwashed everyone
to feel sick and tired of an old model of anything, so they still
changed cars every 5 years.

Then the funders got in on the act and lent money to anyone for
anything,
and what did a price then mean? Just a repayment.

Many people are useless and making anything, so they end up
becoming an a artist, or an accountant, or doer of something
that is often useless, except to facilitate **** going out the factory
gate
so the company won't go broke.

One person in one company removes quality, or moves production to China,
and sacks all the US workers,
and saves the company having to spend so high in production costs, so
the company can lower its prices and become much more
competitive and profitable. Every other company has to follow after the
first or else they go broke.

Anyway, bean counters ridiculed that Williamson fella, what ****wit they
all said, he'll ruin us all!!

I cannot name a single well known amplifer manufacturer who used
Williamson's OPT design
for a huge run of amps.

Fancy having to make such horribly complex OPTs that take so much
labour! Ridiculous they all howled.

But by 1960, Fancy having to use vacuum tubes!!, send 'em to the tip,
let's get those cool running transistors going.
Then anything that worked in the analog realm became BS, and suddenly,
its all digital everything.

Some aspects of progress are good, like me being able to type this very
very cheap telegram,
and not have to pay 10C a word, after spelling it out across a Post
Office desk to a girl.
So the bean counters have not ruined everything.

Make it cheap, light, smaller; let it run hotter, and harder, and let's
take out the good stuff so it goes up in smoke a week after the warranty
period expires.

And while we are at it, lets have a National trade Conferance to agree
on the same crummy standards,
so that the consumers don't have a choice. After smoke it won't matter
who they go to after changing brands,
its all basically the same ****.

And all the entrepreneurs and accontants and bean counters ansd arsoles
all agreed
and colluded with each other to minimize competition and reduce quality
to
the lowest possible denominator. Nice and cosy. SNAFU.

Its very difficult to avoid the ticky tacky and sameness about many
products of the same price
in this world.

With hi-fi, if you know enough, you can avoid all the junksters, and
spend time
in your workshop and emerge later with something much better than the
junksters ever
would let you have.

But you can't build a motor vehicle of your own design.

Naughty naughty, far too many regulations and licences needed. Ditto
your own aeroplane.
Or any plumbing on your house.

And because more folks would die as a result of such freedoms.

But amps rarely kill, so they are OK, like growing your own tomatoes.

I know who I like to throw rotten tomatoes at.

Patrick Turner.

[Henry Pasternack[_2_]]

"Patrick Turner" wrote in message
...
Henry came up with a number lower than a cathode follower and I think
Preisman did as well.

I don't see how it could be lower, but I have never needed to work it
out.

And yet it is true. In fact, for an example I calculated, the equal-load
effective output impedance was an order of magnitude lower than the
cathode-only output impedance. Read about it he

http://groups.google.com/group/rec.a...5394ad0426aeb1

Consider the case where you infer output resistance by measuring the
-3dB point when driving a capacitive load. With a single capacitor at
the cathode you will get one result. Now, adding an identical capacitor
at the plate will reduce the cathode output impedance because it bypasses
the plate resistance.

-Henry

[Andre Jute[_2_]]

On Aug 14, 7:25*pm, "Henry Pasternack" wrote:
. *Now, adding an identical *capacitor
at the plate

Whyever?

[Patrick Turner]

Henry Pasternack wrote:

"Patrick Turner" wrote in message
...
Henry came up with a number lower than a cathode follower and I think
Preisman did as well.

I don't see how it could be lower, but I have never needed to work it
out.

And yet it is true. In fact, for an example I calculated, the equal-load
effective output impedance was an order of magnitude lower than the
cathode-only output impedance. Read about it he

http://groups.google.com/group/rec.a...5394ad0426aeb1

Consider the case where you infer output resistance by measuring the
-3dB point when driving a capacitive load. With a single capacitor at
the cathode you will get one result. Now, adding an identical capacitor
at the plate will reduce the cathode output impedance because it bypasses
the plate resistance.

-Henry

Thanks for taking us all backwards to 2001 where it wasn't such a
terrible place,
and there was some good figuring outing going on.

From the above reference, you had a triode in CPI with Ra = 1k, µ = 10,
and RLa = RLk = 10k.

( Could be nearly like a EL86 in triode )

From that you reckoned Rout of each output under equal loading
conditions = 82.6 ohms.

OK, now for the triode, cathode follower Rout = 1 / gm , and gm = µ / Ra
= 10 / 1,000 A/V = 10mA/V.

So a CF with such a tube would have Rout = 1 / 0.01 = 100 ohms, slightly
higher than the 82.6 ohms
you calculated with CPI.

If the CF already has a dc carrying cathode load of 10k, it is in
parallel with the tube Rout, but the
difference is so great the 10k makes negligible Rout difference.

So what would happen when you add an anode load?

What if you had a CCS as the anode load?

There would be no change to the tube current.
If the grid goes up 1V, the cathode voltage won't move. All the voltage
movement will be at the anode.
In fact anode voltage will be -10V, for +1V at the grid. Since no
current change occurs in the triode
because of the CCS anode supply, there cannot be any change to cathode
load voltage.
So any external load connected to the cathode would find Rout at the k =
10k, ie, the cathode load R.

So methinks as any load is introduced into the anode circuit of a CF,
the cathode Rout
begins to go higher than a pure CF.

In other words, for the lowest Rout possible from a cathode, you have to
completely omit any anode load.


Don't quote me on this, I'll need to check a circuit out and make tests
of my own with a real
triode in a CPI set up and with tests done by varying both cap coupled
anod and cathode loads
so that the loads stay equal.

Its one thing to post lofty math, but to be really true there has to be
a verifying experiment.


The perception of output resistance of other phase inverters such as the
long tail pair
also needs to be specified for the conditions.

Say you have two triodes each with Ra = 10k such as 6SN7.

When you measure the Rout at just one anode of the pair, you'll get
Rout much higher than Ra, because the cathode of that triode has the
common cathode current sink
resistor, and the cathode input resistance of the other triode as its
unbypassed cathode resistance.

But suppose you load the LTP with an additional load resistance from
anode to anode, or two equal R
with each cap coupled to 0V.
Then you should find Rout = Ra at each anode.


My 2C.

Patrick Turner.

[John Byrns]

In article ,
"Henry Pasternack" wrote:

"Patrick Turner" wrote in message
...
Henry came up with a number lower than a cathode follower and I think
Preisman did as well.

I don't see how it could be lower, but I have never needed to work it
out.

It isn't, you have to keep in mind that you are dealing with the
slippery fish, who in this case is misquoting Henry. Henry's original
claim was that his CPI source impedance was lower than the source
impedance at the cathode of the CPI when measured by the Byrns method,
and that is indeed true, the slippery fish has incorrectly equated the
³Byrns² method CPI cathode source impedance with the source impedance of
a cathode follower in his statement above. They are not the same thing,
the source impedance at the cathode of a CPI is higher than the source
impedance of a cathode follower. The reason for this is the extra
impedance in series with the anode of the CPI, which raises the cathode
source impedance. Hence it makes perfect sense for Henry's CPI source
impedance, which is half of the cathode to anode source impedance, to be
lower than the impedance measured at the cathode of the CPI using the
Byrns method, while still being greater than the source impedance of a
true cathode follower. Henry's response below is correct because he
failed to notice that Flipper had substituted the term ³cathode
follower² for the original ³CPI cathode-only² output. You can see how
this result comes about from the equations defining the operation of the
CPI that are presented in the message that Henry was responding to with
the message he links to below.

And yet it is true. In fact, for an example I calculated, the equal-load
effective output impedance was an order of magnitude lower than the
cathode-only output impedance. Read about it he

http://groups.google.com/group/rec.a...5394ad0426aeb1

Consider the case where you infer output resistance by measuring the
-3dB point when driving a capacitive load. With a single capacitor at
the cathode you will get one result. Now, adding an identical capacitor
at the plate will reduce the cathode output impedance because it bypasses
the plate resistance.

I'm not sure why you are getting into the capacitor business, what you
said is true even without the necessity for adding capacitors, at least
for the component values used in your example referenced above.

I would suggest that anyone interested in this subject read the entire
thread the above referenced message was taken from, it is the definitive
work on the subject. Ideally the thread should be read from point where
the subject changes to the concertina from the original feedback topic,
through to the end of the thread where Henry calls me an ³idiot². At a
minimum the message Henry is responding to, with the above referenced
message, should be carefully read, it can be found here.

http://groups.google.com/group/rec.a...fe562e8a81ded7

Contrary to the warning at the end, I think it turned out that the
equations contained no typos.

There is a lot of other interesting stuff in the thread. There is a
discussion of my anode and cathode transfer functions, and how they
degenerate into a form identical to Henry's when you make the anode and
cathode load impedances equal, as in a well balanced concertina.
Another interesting point is where I ³Aced² what I will call the
³Flipper² challenge. At one point Henry challenged me to use my
equations to tell him the -3 dB point for the example he presented in
the reference above, with 470 pF capacitors added to the anode and
cathode loads. This turned out to be a trivial exercise, as it turns
out my equations are no harder to use to determine the -3 dB point for
the balanced case than his equations are. I have so far only skimmed
over that part of the thread, so I don't remember exactly how I made the
-3 dB calculation at that time, which is not surprising given that I had
forgotten the entire thread until Henry pointed it out above. However
it occurs to me now that if we substitute separate R & C load values
into the transfer function equation in place of the complex ³Z² that I
used, we will get a transfer function equation that explicitly includes
the relevant RC time constant term that Flipper was looking for.

Flipper has given me a lot of guff about Thevenin and Norton; it is
interesting to read the old thread, where I referred to the source
impedances I had calculated as ³Thevenin equivalent source impedances².
Jack Crenshaw complained about my use of the ³T² word, saying I was
merely using it to impress people by ³name dropping², and that in this
group it would be more appropriate for me lose the ³T² word and simply
refer to the ³source impedance². To accommodate Jack I dropped the ³T²
word from my posts, however I soon had Henry on my case asking why I
didn't discuss the ³Thevenin source impedance² suggesting it was because
I didn't understand Thevenin. One can but wonder why Jack didn't take
Henry to task for name-dropping too?

That old thread it is a good read and contains much other material of
interest, I plan to carefully read the entire thread, having quickly
scanned parts of it last night. I have to say thank you to Henry for
pointing out this wonderful old thread.

--
Regards,

John Byrns

Surf my web pages at, http://fmamradios.com/

[Patrick Turner]

John Byrns wrote:

In article ,
"Henry Pasternack" wrote:

"Patrick Turner" wrote in message
...
Henry came up with a number lower than a cathode follower and I think
Preisman did as well.

I don't see how it could be lower, but I have never needed to work it
out.

It isn't, you have to keep in mind that you are dealing with the
slippery fish, who in this case is misquoting Henry. Henry's original
claim was that his CPI source impedance was lower than the source
impedance at the cathode of the CPI when measured by the Byrns method,
and that is indeed true, the slippery fish has incorrectly equated the
³Byrns² method CPI cathode source impedance with the source impedance of
a cathode follower in his statement above.


Gee, what slippery fish? Ah Flipper?

I forgot what the Byrns method is....

But as I said in another post, the CPI aRout = kRout if loads are
constant,
and each is above Rout of a CF, 1/gm.

As you add in an anode load to a CF, the cathode Rout rises, and when
the anode load = a CCS,
the cathode Rout = the cathode load R only.


They are not the same thing,
the source impedance at the cathode of a CPI is higher than the source
impedance of a cathode follower.

That's what I would think.

The reason for this is the extra
impedance in series with the anode of the CPI, which raises the cathode
source impedance. Hence it makes perfect sense for Henry's CPI source
impedance, which is half of the cathode to anode source impedance, to be
lower than the impedance measured at the cathode of the CPI using the
Byrns method, while still being greater than the source impedance of a
true cathode follower. Henry's response below is correct because he
failed to notice that Flipper had substituted the term ³cathode
follower² for the original ³CPI cathode-only² output. You can see how
this result comes about from the equations defining the operation of the
CPI that are presented in the message that Henry was responding to with
the message he links to below.

And yet it is true. In fact, for an example I calculated, the equal-load
effective output impedance was an order of magnitude lower than the
cathode-only output impedance. Read about it he

http://groups.google.com/group/rec.a...5394ad0426aeb1

Consider the case where you infer output resistance by measuring the
-3dB point when driving a capacitive load. With a single capacitor at
the cathode you will get one result. Now, adding an identical capacitor
at the plate will reduce the cathode output impedance because it bypasses
the plate resistance.

I'm not sure why you are getting into the capacitor business, what you
said is true even without the necessity for adding capacitors, at least
for the component values used in your example referenced above.

I would suggest that anyone interested in this subject read the entire
thread the above referenced message was taken from, it is the definitive
work on the subject. Ideally the thread should be read from point where
the subject changes to the concertina from the original feedback topic,
through to the end of the thread where Henry calls me an ³idiot². At a
minimum the message Henry is responding to, with the above referenced
message, should be carefully read, it can be found here.

http://groups.google.com/group/rec.a...fe562e8a81ded7

Contrary to the warning at the end, I think it turned out that the
equations contained no typos.

There is a lot of other interesting stuff in the thread. There is a
discussion of my anode and cathode transfer functions, and how they
degenerate into a form identical to Henry's when you make the anode and
cathode load impedances equal, as in a well balanced concertina.
Another interesting point is where I ³Aced² what I will call the
³Flipper² challenge. At one point Henry challenged me to use my
equations to tell him the -3 dB point for the example he presented in
the reference above, with 470 pF capacitors added to the anode and
cathode loads. This turned out to be a trivial exercise, as it turns
out my equations are no harder to use to determine the -3 dB point for
the balanced case than his equations are. I have so far only skimmed
over that part of the thread, so I don't remember exactly how I made the
-3 dB calculation at that time, which is not surprising given that I had
forgotten the entire thread until Henry pointed it out above. However
it occurs to me now that if we substitute separate R & C load values
into the transfer function equation in place of the complex ³Z² that I
used, we will get a transfer function equation that explicitly includes
the relevant RC time constant term that Flipper was looking for.

OK.

But my observations are that the anode output tends to sag before the
cathode output
and it affects the square wave whose shape can be made more symetrical
at top and bottom
by "tuning" with a small C across Rk.
Most makers just don't bother, because artifacts produced by not having
the C are extra 2H at above say 40kHz,
and totally unimportant.

Flipper has given me a lot of guff about Thevenin and Norton; it is
interesting to read the old thread, where I referred to the source
impedances I had calculated as ³Thevenin equivalent source impedances².
Jack Crenshaw complained about my use of the ³T² word, saying I was
merely using it to impress people by ³name dropping², and that in this
group it would be more appropriate for me lose the ³T² word and simply
refer to the ³source impedance². To accommodate Jack I dropped the ³T²
word from my posts, however I soon had Henry on my case asking why I
didn't discuss the ³Thevenin source impedance² suggesting it was because
I didn't understand Thevenin. One can but wonder why Jack didn't take
Henry to task for name-dropping too?

That old thread it is a good read and contains much other material of
interest, I plan to carefully read the entire thread, having quickly
scanned parts of it last night. I have to say thank you to Henry for
pointing out this wonderful old thread.

One day I'll work out a formula for the Rout of a CPI at its a and k,
but I ain't in any hurry. I just work out what I want, a good working
point for the triode involved,
then test it all well and if it works well that's it, done like sunday
dinner.

Patrick Turner.


--
Regards,

John Byrns

Surf my web pages at, http://fmamradios.com/

[Patrick Turner]

flipper wrote:

On Fri, 15 Aug 2008 13:01:41 -0500, John Byrns
wrote:

In article ,
"Henry Pasternack" wrote:

"Patrick Turner" wrote in message
...
Henry came up with a number lower than a cathode follower and I think
Preisman did as well.

I don't see how it could be lower, but I have never needed to work it
out.

It isn't, you have to keep in mind that you are dealing with the
slippery fish, who in this case is misquoting Henry.

I've tried to stay cordial but if hurling insults is your opinion of
the 'good way' to make a case then let me know because, in the spirit
of collegial accommodation, I'll gladly hurl a few grenades your way
if it'll make you happy.

Henry's original
claim was that his CPI source impedance was lower than the source
impedance at the cathode of the CPI when measured by the Byrns method,
and that is indeed true, the slippery fish has incorrectly equated the
³Byrns² method CPI cathode source impedance with the source impedance of
a cathode follower in his statement above.

I'm glad to see you finally agree with Henry.

They are not the same thing,
the source impedance at the cathode of a CPI is higher than the source
impedance of a cathode follower.

And just who said a cathode follower can't have a resistor in the
anode circuit?

Well, the guy who said the lowest Rout is only possible if there is no
resistor in the anode circuit of a CF.

When one selects to use a CF, one does so to get Rout as low as
possible.

No need to waste a resistor.


Any rational person would presume I meant the circuit as it is because
that's the circuit under discussion. And they'd be correct.

Lord only knows how in the world you come up with the whole cloth
fantasy I somehow meant a different circuit with the anode load
removed.

And if you wondered what I meant then why didn't you ask, instead of
accusing me of intentional deceit?

Or, if you thought I goofed then why didn't you 'help' me by
explaining it, instead of accusing me of intentional deceit?

What is it with you that everything gets interpreted as some kind of
nefarious plot?

People are people around here, suspicious, devious even, wacky, even
interested in vacuum tubes of all things
to be interested in. Crazy. Just human nature.



The reason for this is the extra
impedance in series with the anode of the CPI, which raises the cathode
source impedance. Hence it makes perfect sense for Henry's CPI source
impedance, which is half of the cathode to anode source impedance, to be
lower than the impedance measured at the cathode of the CPI using the
Byrns method, while still being greater than the source impedance of a
true cathode follower.

Oh, so now it's a 'true' cathode follower. As opposed to what? A
'false' cathode follower?

Henry's response below is correct because he
failed to notice that Flipper had substituted the term ³cathode
follower² for the original ³CPI cathode-only² output.

Because Henry knows I mean the ³CPI cathode-only² output when I say
cathode follower.

I thought that too, but maybe JB didn't, and who meant what gets a
little confusing afer awhile.

But I thought HP said that in ab CPI, aRout = kRout where loads are
always equal and
each Rout CF Rout if the triode was used as a pure CF with no anode
load.

It can easily be checked by setting up a 6SN7 triode in a typical CPI
circuit, say 22k a and k RLdc loads,
then have 1uF coupling caps to place additional 22k equal cap coupled
RLs to both a and k
so that the output voltage can be measured with 44k total tube load, or
22k total tube load.
To convert to CF, the anode cap is taken to 0V, and the VO with 22k is
again noted,
and then with another 22k cap coupled to 0V.

Better to know by finding out rather than shooting the breeze endlessly
here.



You can see how
this result comes about from the equations defining the operation of the
CPI that are presented in the message that Henry was responding to with
the message he links to below.

And yet it is true. In fact, for an example I calculated, the equal-load
effective output impedance was an order of magnitude lower than the
cathode-only output impedance. Read about it he

http://groups.google.com/group/rec.a...5394ad0426aeb1

Consider the case where you infer output resistance by measuring the
-3dB point when driving a capacitive load. With a single capacitor at
the cathode you will get one result. Now, adding an identical capacitor
at the plate will reduce the cathode output impedance because it bypasses
the plate resistance.

I'm not sure why you are getting into the capacitor business,

Because it's a valid means of measuring the circuit despite your
confusion about it.

what you
said is true even without the necessity for adding capacitors, at least
for the component values used in your example referenced above.

I would suggest that anyone interested in this subject read the entire
thread the above referenced message was taken from, it is the definitive
work on the subject. Ideally the thread should be read from point where
the subject changes to the concertina from the original feedback topic,
through to the end of the thread where Henry calls me an ³idiot². At a
minimum the message Henry is responding to, with the above referenced
message, should be carefully read, it can be found here.

http://groups.google.com/group/rec.a...fe562e8a81ded7

Contrary to the warning at the end, I think it turned out that the
equations contained no typos.

There is a lot of other interesting stuff in the thread. There is a
discussion of my anode and cathode transfer functions, and how they
degenerate into a form identical to Henry's when you make the anode and
cathode load impedances equal, as in a well balanced concertina.
Another interesting point is where I ³Aced² what I will call the
³Flipper² challenge.

Oh? And just what is the ³Flipper² challenge?

At one point Henry challenged me to use my
equations to tell him the -3 dB point for the example he presented in
the reference above, with 470 pF capacitors added to the anode and
cathode loads. This turned out to be a trivial exercise, as it turns
out my equations are no harder to use to determine the -3 dB point for
the balanced case than his equations are. I have so far only skimmed
over that part of the thread, so I don't remember exactly how I made the
-3 dB calculation at that time, which is not surprising given that I had
forgotten the entire thread until Henry pointed it out above. However
it occurs to me now that if we substitute separate R & C load values
into the transfer function equation in place of the complex ³Z² that I
used, we will get a transfer function equation that explicitly includes
the relevant RC time constant term that Flipper was looking for.

No one has ever disputed that both models work and I said so, along
with the explanation of why, in my very first post.

Flipper has given me a lot of guff about Thevenin and Norton;

What "guff?"

Stuff to read. Reading material. Information.

Fine old guys Thevie and Norty.

One's brain falls into their lines of thought intuitively after
farnarkling around with tube circuits for long enough
and you don't even think about their theorems consciously as one plods
through a design.

Do I worry about the speed of light when I watch a sunrise?



it is
interesting to read the old thread, where I referred to the source
impedances I had calculated as ³Thevenin equivalent source impedances².
Jack Crenshaw complained about my use of the ³T² word, saying I was
merely using it to impress people by ³name dropping², and that in this
group it would be more appropriate for me lose the ³T² word and simply
refer to the ³source impedance².

As I said earlier, it's always risky to take a third part description
of someone else's comments and I'm doubly reluctant because my
experience with you on this topic is your recollection of what I said
bears almost no resemblance.

But, judging from your claim I gave you "guff," I'd guess he suspected
you either didn't fully get the meaning of a Thevenin output impedance
or, in any case, that using the term was putting the cart before the
horse because what the "output impedance" 'is' was the subject of
discussion.

I.E. Simply calculating something doesn't automatically make it a
--Thevenin--- 'output impedance'.

To accommodate Jack I dropped the ³T²
word from my posts, however I soon had Henry on my case asking why I
didn't discuss the ³Thevenin source impedance² suggesting it was because
I didn't understand Thevenin. One can but wonder why Jack didn't take
Henry to task for name-dropping too?

Because the whole point is that one model's 'output impedance' doesn't
behave like a Thevenin model and the other's does. Specifically, the
values derived in the unequal 'output impedance' model cannot be used
like Thevenin output impedances while the value of the equal output
impedance model can.

That old thread it is a good read and contains much other material of
interest, I plan to carefully read the entire thread, having quickly
scanned parts of it last night. I have to say thank you to Henry for
pointing out this wonderful old thread.

I've read it and have no problem with either Henry's model or yours.
You're the one who's bent out of shape over there being two of them.

Gee, isn't there anyone straight around here, or are they somewhat
queer?

Life is rather too short to be too serious forever.

Patrick Turner.

[Patrick Turner]

flipper wrote:

On Thu, 14 Aug 2008 16:30:51 GMT, Patrick Turner
wrote:



flipper wrote:

On Wed, 13 Aug 2008 10:50:58 GMT, Patrick Turner
wrote:



flipper wrote:

On Tue, 12 Aug 2008 16:22:18 GMT, Patrick Turner
wrote:


However the most satisfying solution would be to abandon the model
entirely, along with the questions it raises, and simply build the
physical circuit and measure the resulting source impedances.
Unfortunately it appears that would just take us back to square one,
again raising the crucial question of how to properly measure the source
impedances.


Just building something, observing it, measuring it and deciding what is
there
is trifle to complex and real for some around here. They hafta get orf
their bums, and go
into their workshop, and spend time actually doing something.

You get yourself into trouble with those incessant knee jerk snide
remarks of yours because this matter came up as a result of my doing
exactly that. I BUILT the damn thing and measured it.

So, with all due respect to you and your bums, sit on it.

OK, Point taken, but some here are arm chair wanabes.

That may be but they're not in the discussion. But, at any rate, I
accept the 'ok'.

But ya don't need to all that. Blind Freddy can see that if you measure
the Rout
at the anode by adjusting the anode load and only the anode load, the
Rout
= Vchange / Ichange, which is how you measure the Rout of anything.

Not when equal loads is inherent to the circuit and a required
condition.

But when one talks about the Rout at one point in a circuit,
what is usually meant is that its taken as is at that point,
and without any load adjustments elsewhere.

I am aware of that but that is precisely the 'problem'. It just
doesn't work with a concertina and if you do it that way you get
essentially useless results.

Well, if you use the classic way to measure the Rout of anything by
changing the load
and recording the Vchange, and I change, Ro

'An' Ro can be found. Whether it's useful or not is another matter and
with an equally loaded concertina the 'classic' method you describe
produces not only useless results but misleads people into the
'classic' misconception it's unbalanced.

The CPI is unbalanced if used to obtain one output voltage only.
But we force it into balance by insisting on equal loads at a and k.
Everyone knows this. As long as the test gear has very high Rin,
we will measure equal VO at a and k.


can be found, but sure,
because the Rout
at a and k are very different if the measurement is done by applying
only one same load change
to each a and k output, then the answer for Ro is useless.

So what's the point of it?

To understand how it behaves. Human curiosity.


As I said, you could make the same load change to each of a and k, and
you should measure
equal Vchange and I change, and Ro at a and k will simply be Vchange /
Ichange and the same for a or k.

Which is a different Ro than the ones calculated in the other case.

So 'Ro' is not 'Ro' and the question becomes, which is useful?

Now I am confused.

Ro, Ro, Ro your boat along, and hope the pretty girls see you Ro-ing
down the river.

'Tis sunday here, I refuse to be serious.



The Vchange / I change can be used to measure Rout easily for any amp
output.

In the case of the CPI, it can be done if you specifiy that the load
change
to anode and cathode will be equal.
Then the because the same current flows through cathode load, tube, and
anode load,
the Ichange is equal, and because the TWO loads are kept equal, V change
will also be equal,
so you'd think Rout is the same from both terminals.

It is, but subject to conditions.

Bingo... as you said, it *is* (the same).

So why were you chastising folks for saying exactly what you just
said?

Because of the condition of the test for Rout that needs to be
specified.

Conditions apply for the truth to be true.

The conditions have always been specified, over and over and over and
over and over and over and over and over and over and over....

So your 'explanation' is invalid.

Really?



Unconditionally, its not.

I am aware it's subject to conditions and the conditions have ALWAYS
been stated, from the very first post to the last.

It also happens to be the overwhelmingly common case and folks often
go to great lengths to keep it equal and 'balanced'. Unfortunately, in
the burning quest to do so, the befuddlement of an unequal output
impedance model has screwed up more than a few.

Unequally loaded it isn't a 'concertina' and it doesn't behave like a
'concertina'. In fact, measuring your way you'd rip the thing up and
throw it out because the outputs would be nothing akin to equal and,
so, of no use as a 'phase inverter'.

Unless you has a pot to lower the higher VO to be equal to the lower VO.

But it wastes gain, so nobody does it.

The only reason I can imagine anyone even thinking of such a
contraption is if they were befuddled by the unequal impedance model
and were trying to 'fix' the problem that doesn't exist.

"Whatever" as they say....

If you repeat the same load adjustment to the cathode load and leave the
anode load alone,
you will find Rout is less then at the anode.

You're not measuring a concertina.

A concertina is merely a normal single ended common cathode triode amp
stage, but one with
a cathode resistor equal to the anode resistor.

So when measuring a concertina, you are measuring a common cathode
amplifier stage.

Nope. Because you left out that the concertina has two, not one,
outputs with equal loads, and that is not a trivial condition. As
evidenced by the near 'magical' result of the seemingly disparate
output impedances of the common cathode amp becoming equal. Except, of
course, you don't have two 'output' impedances in the common cathode
amp, there's only one, so there's not really a comparison.

Nothing stopping you from using the cathode as an output from a common
cathode amp
**of any kind** including where you have RLa = RLk.

No one said otherwise.

They just aren't the same thing, which is why measuring it as one
thing doesn't give you the same answer as measuring it when it's the
other.

The concertina to my mind is merely a common cathode stage with a lot of
local current FB from
its unbypassed Rk.

OK, so you take outputs from both a and k.

It doesn't stop the concertina being a common cathode amp.

Nothing stops you from changing any topology but, if you do, it isn't
the same topology.

And the proof to the pudding here is the different values for 'Ro'
you, yourself. know one gets if you use it one way vs the other.

Frankly, Pat, you're tilting at windmills and just being obstinate
because *you* agree the output impedances are equal under the defined
conditions, which have *always* been stated. You have no complaint
despite complaining for nothing more than the sake of complaining.

But that's what everyone else seemed to be doing around here.

Its the style, the agony, that hooks me right in.

Peer group pressure at work.

I must try to resist it.



And you'll find the cathode is a voltage source, ie, Rout RL at the
cathode,
and Rout at the anode is a current source, ie, Rout RL at the anode.

You're not measuring a concertina.

Now this is true of any common cathode amplifier with an unbypassed Rk,
and you test the Rout at
anode and cathode separately.

That's right, a common cathode amplifier, not a concertina.

As I said, a concertina *is* a common cathode amp......

No it is *not*. It looks a hell of a lot like one but a common cathode
amplifier has ONE output and a concertina has TWO, with equal loads.


So what about the number of outputs?

Its still a basic common cathode amp.

You know better than to argue 'just one thing' doesn't make any
difference and, in a conventional amp, all it takes is 'just one
resistor' from output to cathode to 'change Ro'.

Go ahead, a GNFB amp both with and without the 'one thing' and then
ask me "so what?:

Another way to get phase inversion is to simply use a common cathode amp
with
a gain of exactly -1.0, ie, a triode common cathode amp is set up with
an unbypassed Rk
which is slightly lower in value than the RLa, and Vg = -Va.

So you have the input signal from somewhere external, maybe a CD player,
and you create an opposite
phase signal of exactly the same amplitude, and the two phases can then
be applied to a fully
balanced PP power amp with balanced series voltage GNFB. ARC did this
maybe 30 years ago.

The word "Concertina" does not make it illegal to say the concertina is
a common cathode amp.
It is a common cathode amp, but one with special features, implied by
the word concertina.

See? You can't even make the argument without saying "special" and
invoking a different 'name' for it.

To use your own admonition, 'build one' and 'make measurements'. And
if you measure it the way it is *used*, with balanced loads, you'll
find equal output impedances.

The "squeeze box" instrument stays put, but the player works the 2
handles in opposite directions.




The word concertina is used so everyone knows the cathode load = anode
load,
hence Va = Vk.

That's hunky dory.

It would be possible to rig up the CPI so that Rout is the same at a and
k,
but it'd need you to reduce anode RL and increase the cathode RL until
the Rout was the same at a and k.
But then the voltages wouldn't be equal at all, so its not a useful
exercize.

Nope, because if one defines 'output impedance' per the Thevenin model
then a concertina already has equal Thevenin 'output impedances' as
long as the loads are equal.

If measured separately, you get what ya get, if measured conditioanlly,
you get something else,

Exactly.

a convenient measure.

Well, one convenience is it measures the thing as it actually is and
used. Another is it produces a useful result, which is THE point to
doing a calculation in the first place.

I have never worked out exactly what the conditional Rout of the
CPI actually is. But its somewhere higher than a cathode folower, and
lower than from
a normal common cathode amp at its anode and with fully bypassed
cathode.

Henry came up with a number lower than a cathode follower and I think
Preisman did as well.

I don't see how it could be lower, but I have never needed to work it
out.

You can have 22k for RLa and RLk for a 6SN7, and have Va = Vk = 2V,
and then change the cap coupled loads from say 150k to 22k, and the Vk
and Va will stay equal,
but the drop in voltage is low, like a CF.

If the following cap coupled loads become too low compared to the dc
carrying RLs,
then the undistorted voltage range is reduced, and you get sudden cut
off distortions.....

I haven't calculated it either. All I said is Henry did.

I don't find it difficult to imagine, though, because the same
mechanism that reduces anode output impedance, load 'feedback', is at
work on the cathode.

Seemed to me it wouldn't be 'much' but Henry said it was a fair
amount, which is why I remembered it.

Math experts should have come up with all the releveant figures by now,
but I've never
*needed* to calculate the Rout, so I have never have.

That's ok. It doesn't mean someone else doesn't care, though.

Hmm, we have not seen too much care around here and have not seen
anyone say what the Rout actually is for CPI if a&k loads are always the
same.

Well, you were criticizing without bothering to know what was said and
by who, then, because I explained in my very first post why I 'cared'.

The root and soul of the matter could be expressed in math so that
less time is wasted on endless discussions and more time is available
for soldering and
being useful.

If I seem to be aloof its because ppl are yardi yardi ya all day and
I don't have time to concentrate on each atom of information, and each
tiny grain
of upset emotion.



Just knowing they're equal, never mind the 'value', would have saved a
hell of a lot of confusion over the last century, though.

The only way to ensure the CPI has equal Rout at a and k would be to
direct couple
a pair of cathode followers to cathode and anode, and then you'd find
the two Routs would be the same
and each would be equal to what you'd get with any CF.

There is no need because if one defines 'output impedance' per the
Thevenin model then a concertina already has equal Thevenin 'output
impedances' as long as the loads are equal.

"as long as the loads are equal" is the condition needed for equal Rout
at each terminal.

What if they were not in a dynamic working amp?

Doesn't matter as long as the load impedances are equal,

Well there is a point where the loads suddenly do become unequal, at
grid current
in a following stage.

As one phase goes +, it hits and stalls with Ig, but the other phase
going -
has no grid current.

Guess which output tends to charge up the coupling caps the most.

None of the models work if the circuit leaves linear operation and
THAT is another stated condition.

It's also a stated condition for the Thevenin model and what you call
'Ro'.

This becomes a fact when a CPI runs into grid current powering output
tubes.
The grid current affects the anode signal supply far more than the
cathode supply,
and the amp clipping becomes grossly asymetrical, ie, like a dogs'
breakfast,
with gross asymetrical dc offsets.

None of the 'impedance' models are valid when it exits linear
operation and that's a stated condition for a Thevenin output
impedance.

So in guitar amps, 99% use a
balanced amp or LTP to drive the output stage which gives symetrical
clipping and better
recovery, and more even tube wear. Its something worth something in
musicians amps which are
routinely used well into clipping the output stage.

Yes, I know. I've built them. But it's not germane to a discussion of
concertina output impedance.

Any measurement with equal loads, such as observing the roll off, will
show equal impedances. And it is trivial to derive mathematically as
well since you have 'one' common current through the equal loads.
Simply factor out the equal external loads leaving equal source
impedances.

The theorists here who are making a mountain out of a molehill
could draw up the triode and a voltage gene + Ra resistor as a model in

You seem to be confusing 'real circuit' with models. In a Thevenin
model there is a ideal voltage source (ideal current source in the
equivalent Norton) and an output impedance. There is no "triode
and..."

The triode model I use is a perfect voltage generator with 3 terminals,
plus a resistance added to be equal to Ra taken to a 4th terminal
being the anode of the model.
One terminal is the cathode, one is a grid, and infinite Rin, and the
other generator output is
NOT the anode, but just the generator output, which is low Rout = 0.0
ohms and the signal output = µ x Vg-k.
From gene output to the anode there is a resistance = Ra.

This model can be used to easily work out whatever signal voltages will
exist at a, k and g for any vacuum tube.
Go to
http://turneraudio.com.au/tube-operation1.html
Take a look at Fig 2 nearly 1/2 way down the page.

I hope I have covered the subject at my website to promote understanding
of the use of modelling to some advanantage to those used to working
things out in a very basic
manner based on current flow and voltage appearances and the use of
Ohm's Law.

That's fine if one wants to model a triode, and there are good reasons
why one might, but in a classic Thevenin model of output impedance
there is no 'triode' (nor a model of one), or anything else but a
voltage source and a single output impedance. That's it's purpose and
the 'definition' of "output impedance."

a CPI with equal anode and cathode loads
and prove to themselves how it all works by working out the voltages and
currents and then moving on to
derive the formulas for Rout at the anode, and Rout at the cathode, in
terms of µ, Ra, and RL.

That's more akin to what you're doing, except you then ignore that the
supposed 'source impedances' calculated have no apparent utility.or
meaning for an equally loaded concertina.


Separate Rout measurements of Rout at a of k without making oads stay
equal does have limited utility; it tells us exactly what is there.

Maybe, if you define "what.". But 'what' is not "output impedance"
when you have two loaded outputs.

Meaning for use as a CPI comes when we add a condition of ensuring equal
loads always exist,
and changing anode and cathode loads equally when measuring Rout.

That's a different derivation and measurement.

I'd be interested if you can reveal some use for the 'unequal
impedance' calculation because, off hand, I can't think of a single
thing. Doesn't even tell me what happens with unequal loads because it
wasn't calculated with those either.

There is no application I can think of where the fact of unequal Routs
is exploited for a benefit.

Makes me wonder why you're so enthusiastic about a useless
calculation.

Of course RDH4 has a sample CPI where a normal CPI with one triode feeds
a
pair of following CPIs triodes, and in the following pair the
anode and cathode of one CPI are cross coupled with C to the other CPI
cathode and anode,
and the individually tested Routs at a and k become equal.

Its like a kind of CPI overkill to me, and ther's not one amp in which I
have ever seen such a thing.

Maybe it had its uses in scientific circuits of some sort.

I'm not interesting in calculating the output impedance of that.

Too useless for you?

Probably RDH4 has the expressions in math for all the Routs you can ever
think of.

I just don't know them all.

I wasn't building an amp with that. I was building one with a
Concertina.

At this point it appears that the only way to make forward progress
would be to agree on a method for measuring source impedance.

I hope I have described a reasonable way to do it.

It's reasonable. It just doesn't measure the Thevenin equivalent for a
concertina with equal loads.

Measurers will find Rout at cathode a lot less then Rout at anode.

Measurers who use equal loads will find equal Thevenin impedances.

Measurers who break the concertina with unequal, or no, loads will
find something else.

Sorry, but its true.

Not when one defines output impedance as a Thevenin equivalent output
impedance.

Well, OK, equivalant whatever, Rout is Rout at the end of the day.

That's the problem, it isn't and you get completely different results
when you 'measure' under different conditions. 'Single sided' you get
one thing, equal loads gives you equal impedances, and unequal loads
gives you a whole range of impedances. 'Rout' is anything but 'Rout'.

However, just because the two source resistances are different, it don't
stop the CPI from being a
decent sort of phase inverter, and able to give good enough balance for
most amplifiers
as long as it isn't driven into overload.

That's because, as long as the loads are equal it has equal Thevenin
source resistances and since the Cs are equal the RC roll off is
equal.

In fact the stray C and Miller C around the circuit can be asymetrical,
and
I have found the anode output sags at lower F than the cathode output.
hence the use of some compensating C across the Rk.
This boosts the anode output as F rises, and HF poles then become the
same at a and k.

Come on Pat, you're grasping at any and everything. Everyone,
regardless of the model, has specifically said local parasitics are
being ignored and we're limiting operation under them.

If you want an ideal square wave at 5kHz in a Williamson,
you would find you'll get the best looking one where you have taken the
trouble to
add a small C across Rk of the CPI.

Lots of ppl do it. I just re-wired a VAC amp, there is the C, and 47pF
across 22k.

Its there to improve the HF performance.

That's great.

Still has nothing to do with the Ro discussion.

I've given a simple method to measure the a and k Routs.

C affects CPI, and I sure don't have the math on how it does,
but AF amp HF response is affected by C in the tube circuits.



Which resolves the century old befuddlement of why the damn thing
works despite those confounding 'different impedances' that had even,
so called, 'experts' shoving 'build out resistors' on the thing to
solve a problem with those confounding 'different impedances' that
didn't exist.

You simply accept that it works, despite the 'different impedances'. I
have shown *why* it works and done so three ways: with the 'one
current through equal loads' derivation and two models, one with your
preferred unequal 'output impedances' (measured with a broken
concertina) and one with the equal Thevenin equivalent output
impedances.

I might have said more stuff than "it just works."

Well, you usually say 'more stuff'. Whether it's related to the topic
is another question and it's true I haven't heard everything you've
ever said.

But this last time around you didn't say anything about how in the
world different impedances into equal capacitive loads could possibly
come out with an equal roll off. All you said was the unequal
impedances don't stop it from working just fine.




Well, as you said above, the reason is the output impedances *are*
equal, not different.

Well, I suggest you examine a working amp more closely.

I already did. They're equal.

To be equal they depend on equal loading.

At HF the C affects the CPI adversely.

HF roll off poles occurs at different F unless trimming C are placed.


And you said so as well, so I suggest you argue with yourself about
it.

You might find the reason why some makers add a C across Rk of a CPI.

Its to slightly boost the anode output which sags before the cathode
output.

Most makers don't bother, like they don't bother to do a whole range of
stuff
because they work down to a price, not up a standard of quality.

That's nice. It still has nothing to do with the Ro discussion.

!

They all work but the purpose behind concepts like 'output impedance'
(the term being derived from Thevenin) is to simplify things and I
submit that the 'unequal impedance' model has caused infinitely more
confusion and problems that it ever resolved, so what is the purpose
of it?

We need to know the unequal Routs are there if the loads become unequal.

The Routs you calculate are useless for that as well because it
doesn't take into account unequal loads any more than it does equal
ones.

That derivation only works for ONE output at a time..

But as Vchange / Ichange remain equal if both anode and cathode loads
are kept equal, at AF the CPI make a fine phase inverter.

Right. And the result of a V/I calculation is generally called
"impedance" so equal V/I means equal output impedances.

It also buffers the input triode and balanced amp stage, so for a given
set of tubes the BW is very good.
But its 4 stages for an amp instead of a minimum of two, as in say
Quad-II.
4 triodes intead of two pentodes. We all know how the bean counters
voted.
They voted Williamson out into the Wilderness.

"Bean counters" don't get a vote. They count beans.

See Pat, among other things I've *been* both a designer and a Product
Manager so don't pretend to tell me how 'bean counters' voted or that
they told me 'how many' of this or 'what' I could or could not put in
a product.

Bean counters destroyed most quality in most products.

They are there to *remove quality*

You can babble that nonsense till you're blue in the face but I've
been there, done that, got the T-Shirt, know what the hell I'm talking
about.

Bean counters count beans. That's why they're called bean counters.

Ah, you got a T-Shirt!

Wow, I never knew that. Fantabulous news.

Bean counters also count dollars, and screws, and nuts, and tubes and
sockets,
and millimetres. Acording to them, all such things they count must come
to low totals.


snip of Pat showing he doesn't understand business

I undertsand it very well. Business doesn't understand me.

If everyone was like me there'd be no bean counters.
No need. Ppl would make things adequately well without these parasitic
*******s.
Natural competion would still function and be accepted, but bean
counters bring unatural competition into
every thing they touch, and they symbolise the greed of the world.
Bean counters = efficiency overdose.

Business likes to get everything it can made in China because some bean
counter
said its far cheaper than employing YOU.

As a result, all manner of cheap gadget arrive by boat, and everyone
marvels and gets sucked in.
But oops, the greenhouse effect is ruining the Dream, but western
BUSINESS people
have no conscience troubles with Chinese working like slaves on $2 per
day,
and what's more, the western business ppl won't even give the Chinese an
extra $1 per day
to clean up their emmisions, and all the environmental damage they are
doing in China and around the world.

Things could have been worse, and the standards of living could have
remained at 1955 standards,
and with the really high CO2 emissions of that era, where americans
drove cars with appalling thirst for gas.
But reducing CO2 a lot is offset by population increase, and the fact
that 4 billion ppl
now want equal lifestyles to YOU.

So in 100 years there won't be much rain forest, everything minable will
have been mined,
there won't be any lions, tigers, elephants left, the world's oceans
will be fouled,
food will be hard to grow, and I guess everyone will fight it out over
what's left
to fight over, and the temperature will be +5C, and the waters will be
flooding major cities,
and we won't feel like re-building them all so soon.
There will be lots of mosquitoes though.

Sorry to be pessimistic, but I see a lot of doom and gloom ahead because
of business.

Business simply represents human nature.

When things get real bad, people won't be able to breed so easy.

But here in Oz, much of generation Y is postponing family life
indefinately,
because they see the writing on the wall, and they see a house costs 1/5
a million at least,
and they can't get job security. Their parents live far too long so they
never
see their inheritance until they are old as well.

Maybe in 100 years we'll have learnt to genetically modify ourselves to
cope a bit better,
like removing the gene for depression for starters. And to breathe in
C02, and breathe out 02.
And to eat garbage. The garbage pile from humans is getting big, but its
nothing much compared
to what it'll be in 100 years. So the best thing the bean counters can
achieve is to
force engineers to design people to eat it. Cheaper than growing new
food.
Same old solutional thinking.
Doctors will need to addapt to keeping genetically altered humans alive
at the right price of course.

Don't worry, Be happy, doom is a hundred years away, and you and I will
be long gone. Not our problem.
Somebody eles's!


Patrick Turner.

[John Byrns]

In article ,
flipper wrote:

On Fri, 15 Aug 2008 13:01:41 -0500, John Byrns
wrote:

In article ,
"Henry Pasternack" wrote:

"Patrick Turner" wrote in message
...
Henry came up with a number lower than a cathode follower and I think
Preisman did as well.

I don't see how it could be lower, but I have never needed to work it
out.

It isn't, you have to keep in mind that you are dealing with the
slippery fish, who in this case is misquoting Henry.

I've tried to stay cordial but if hurling insults is your opinion of
the 'good way' to make a case then let me know because, in the spirit
of collegial accommodation, I'll gladly hurl a few grenades your way
if it'll make you happy.

Sorry to be so long in getting back to you but I was away at the beach
for three days and it has taken me all day to recover from this little
"vacation".

Your silly fish name, and your anonymous posting habit, make it hard to
respect you. The "empty calories" you have been posting in this thread
don't help much. Your constant references to Thevenin and Norton in
attempts to criticize my posts without any indication of their actual
relevance to my posts is a constant irritant. You also spoke of
checking your 13FD7 amplifier, yet I never saw the results of any
impedance measurements posted.

Throw some grenades my way if it will make you feel better, I don't
mind, it seems unlikely you will ever be able to equal the gang during
the "Wars", who even went so far as to hurl insults at my wife.

I am making a modest effort at improving my attitude towards Henry, but
don't forget that he has called me far more names during and since the
"War" than I have ever called him.

And yet it is true. In fact, for an example I calculated, the equal-load
effective output impedance was an order of magnitude lower than the
cathode-only output impedance. Read about it he

http://groups.google.com/group/rec.a...5394ad0426aeb1

Consider the case where you infer output resistance by measuring the
-3dB point when driving a capacitive load. With a single capacitor at
the cathode you will get one result. Now, adding an identical capacitor
at the plate will reduce the cathode output impedance because it bypasses
the plate resistance.

I'm not sure why you are getting into the capacitor business,

Because it's a valid means of measuring the circuit despite your
confusion about it.

You missed my point, my point was that the capacitive load was
unnecessary to demonstrate his point which was already illustrated in
the post he referenced, without the need for a capacitive load.

There is a lot of other interesting stuff in the thread. There is a
discussion of my anode and cathode transfer functions, and how they
degenerate into a form identical to Henry's when you make the anode and
cathode load impedances equal, as in a well balanced concertina.
Another interesting point is where I ³Aced² what I will call the
³Flipper² challenge.

Oh? And just what is the ³Flipper² challenge?

The "Flipper" challenge is my new name for the "Pasternack" challenge of
2001. Back in 2001 Henry challenged me to calculate the -3 dB point of
a CPI using my CPI equations, while you aske3d me a week or so ago how
the "RC time constant" of the CPI could be calculated with my CPI
equations. It is pretty much the same challenge.

I.E. Simply calculating something doesn't automatically make it a
--Thevenin--- 'output impedance'.

Can you give a clear definition of what makes something a
"--Thevenin--- 'output impedance'"?

I've read it and have no problem with either Henry's model or yours.
You're the one who's bent out of shape over there being two of them.

I have two problems, first it isn't obvious to me that Henry's method is
a valid source impedance measurement, what he is measuring is the source
impedance between the cathode and anode driving a balanced load, and
then assuming that half that source impedance is attributable to the
cathode and half to the plate. It isn't obvious to me that that is a
valid assumption.

The second problem I have is that if there are "two of them" how do we
know that there aren't really three or four or more "models" for the
source impedance? Can we choose any arbitrary value for the source
impedance that we want to?

--
Regards,

John Byrns

Surf my web pages at, http://fmamradios.com/

[John Byrns]

In article ,
Patrick Turner wrote:

John Byrns wrote:

In article ,
"Henry Pasternack" wrote:

"Patrick Turner" wrote in message
...
Henry came up with a number lower than a cathode follower and I think
Preisman did as well.

I don't see how it could be lower, but I have never needed to work it
out.

It isn't, you have to keep in mind that you are dealing with the
slippery fish, who in this case is misquoting Henry. Henry's original
claim was that his CPI source impedance was lower than the source
impedance at the cathode of the CPI when measured by the Byrns method,
and that is indeed true, the slippery fish has incorrectly equated the
³Byrns² method CPI cathode source impedance with the source impedance of
a cathode follower in his statement above.

Gee, what slippery fish? Ah Flipper?

I forgot what the Byrns method is....

It's the equations I developed to define the elemental "CPI" which are presented
a few paragraphs further down.

But as I said in another post, the CPI aRout = kRout if loads are
constant,
and each is above Rout of a CF, 1/gm.

Isn't Rout for a CF equal to rp/(u+1)? IIRC Henry's Rout for both the anode and
cathode terminals of the CPI is rp/(u+2), which is always lower than Rout for a
CF using the same tube.

[Snip]

One day I'll work out a formula for the Rout of a CPI at its a and k,
but I ain't in any hurry. I just work out what I want, a good working
point for the triode involved,
then test it all well and if it works well that's it, done like sunday
dinner.

No need to bother, the dinner has already been cooked and is on the table below.

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)

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

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 nowhere near equal.

The RC time constant determining the -3 dB rolloff point when Zk equals Zp can
be calculated from the pole location in equation 7, by replacing Zk and Zp with
the parallel combination of Rl and Cl, giving TC = ((rp/(u +2))||Rl) * Cl.

--
Regards,

John Byrns

Surf my web pages at, http://fmamradios.com/

[Patrick Turner]

John Byrns wrote:

In article ,
Patrick Turner wrote:

John Byrns wrote:

In article ,
"Henry Pasternack" wrote:

"Patrick Turner" wrote in message
...
Henry came up with a number lower than a cathode follower and I think
Preisman did as well.

I don't see how it could be lower, but I have never needed to work it
out.

It isn't, you have to keep in mind that you are dealing with the
slippery fish, who in this case is misquoting Henry. Henry's original
claim was that his CPI source impedance was lower than the source
impedance at the cathode of the CPI when measured by the Byrns method,
and that is indeed true, the slippery fish has incorrectly equated the
³Byrns² method CPI cathode source impedance with the source impedance of
a cathode follower in his statement above.

Gee, what slippery fish? Ah Flipper?

I forgot what the Byrns method is....

It's the equations I developed to define the elemental "CPI" which are presented
a few paragraphs further down.

But as I said in another post, the CPI aRout = kRout if loads are
constant,
and each is above Rout of a CF, 1/gm.

Isn't Rout for a CF equal to rp/(u+1)? IIRC Henry's Rout for both the anode and
cathode terminals of the CPI is rp/(u+2), which is always lower than Rout for a
CF using the same tube.

[Snip]

One day I'll work out a formula for the Rout of a CPI at its a and k,
but I ain't in any hurry. I just work out what I want, a good working
point for the triode involved,
then test it all well and if it works well that's it, done like sunday
dinner.

No need to bother, the dinner has already been cooked and is on the table below.

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)

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

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 nowhere near equal.

The RC time constant determining the -3 dB rolloff point when Zk equals Zp can
be calculated from the pole location in equation 7, by replacing Zk and Zp with
the parallel combination of Rl and Cl, giving TC = ((rp/(u +2))||Rl) * Cl.

--
Regards,

John Byrns

Surf my web pages at, http://fmamradios.com/

I'll have to chew slowly through all the above and see if it compares
whith what I said in a
new post topic name some days ago to which nobody replied.....


Subject:
Output resistance of the CPI or "concertina phase inverter"
Date:
Mon, 18 Aug 2008 10:47:27 GMT
From:
Patrick Turner
Organization:
Comindico Australia - reports relating to abuse should be sent
to
Newsgroups:
rec.audio.tubes




Much has been said about the CPI last week, and many must be wondering
what the F... does all that hot air mean.

When let's consider a 1/2 6SN7 used in a typical CPI case in an amp.

Let us assume that at the chosen Ia for the triode, µ = 20, and Ra =
10k.

Let us examine the gains and output resistance of the two outputs
assuming that the
load at anode and cathode will always remain equal up to 32 kHz, so we
don't have to worry about
capaictance effects.

Suppose total RL for the triode = 40k, so cathode RL = 20k = anode RL =
20k.

Internal open loop gain OLG of the tube, A, = 20 x 40k / ( 10k + 40k ) =
16.0.

So if +1V appears across Vgk, there will be -8V at anode, and +8V at
cathode, and
to get this you must apply +8V + +1V to the grid, so total
closed loop gain, CLG = A' = -16/+9 = -1.778 total, or -0.889 if the
gain is between grid and anode only,
or +0.889 if measured between grid and cathode.

So for +9V VOLTAGE CHANGE applied to the grid, you get -8V at a, and +8
at k and Ia change = 8V / 20k = 0.4mA.

Now let's say we change the load conditions so IaQ and EaQ remains the
same,
but RL total = 20k, so RL at a and k each equal 10k.

CLG = 20 x 20 / (10 + 20) = 13.333.

So for +1V change across Vgk, there must be -/+ 6.666V change at anode
and cathode respectively.
A' between grid and anode = -6.666 / ( 6.666 + 1 ) = -6.666 / 7.666 =
-0.8695.
And +0.8695 gain between grid and cathode.

So let's apply the same +9V change to the grid as we did when RL = 20k +
20k, and we'll get
Va = -7.82V, and Vk = +7.82V.

Current flow = 7.82V / 10k = 0.782mA

So, what is the output resistance of the two outputs in our circuit?

For one thing, while ever RLa = RLk, the two remain equal.

But we can work out from the two load states what is the Rout for each
with 9V applied to either load state of 20k or 10k at a and k.

With 20k+20k, Io = 0.4mA, 10k+10k Io = 0.782mA, so current change
between
the two loadings = the difference = 0.382mA.

The Va or Vk change is from 8.000V to 7.82V, which gives a V change =
0.174 V.

Rout = Vchange / Ichange = 0.174V / 0.382mA = 0.455k = 455 ohms.



What else can we say? We could say Rout measured between anone and
cathode = 2 x 455 ohms because
there are two oppositely moving voltage changes so Ra-k = 910 ohms.

Now what if the same tube under identical Ia and Ea conditions was set
up as a cathode follower?
Asume there was a CCS cathode current sink, Rout = 1/gm, and gm = µ / Ra
so Rout of the CF = Ra / µ = 10k / 20 = 500 ohms.

The CPI can be said to have Rout at each of the two terminals slightly
below that of a pure CF.

So a CF has an Rout slightly higher than either Rout of th CPI with
resistance loading on which it
depends to work.

But the CPI Rout expressed as between a and k is 910 ohms, and well
above the CF case.


There is of course an easier way to calculate the Rout a to k for the
CPI.

If the loads always remain the same for a and k, its as if the CPI
had transformer coupled loadings between a to 0V and k and 0V,
and then it is said to operate with voltage NFB instead of current NFB.

The effective Ra, or Rout, with the FB where 1/2 the total Vo a to k is
fed back
becomes simply Ra' = Ra / ( 1 + [µ x ß] ) .

In the case of the 1/2 6SN7, we get Ra' = 10k / ( 1 + [ 20 x 0.5 ] ),
= 10k / 11 = 909 ohms, which remarkably enough, is very close
to what I so laboriously calculated above.

Measurements should confirm all the above, providing the µ and Ra
of the tube are exactly as I stated above for all tests.

One could have two transformer windings, one for a and the other for k,
to accomodate the difference of a and k operating Vdc, and have a
tertiary winding to alter the loading.
Rout will be the same as calculated as long as the tranny has a large
amount of Lp.

Because of the variations of Ra and gm due to Ia variations, and thus
slight changes to µ as a result of µ = gm x Ra,
there isn't any need to measure such things as long as the circuit works
as required in an amplifier.

We don't need to count photons during a sunrise, or calculate what
allowance must be made for mist
or clouds, if its a beautiful sunrise, all is well.

Clear enough for the nit pickers here?

Patrick Turner.

[The Phantom]

On Tue, 19 Aug 2008 18:47:39 -0500, John Byrns
wrote:

In article ,
Patrick Turner wrote:

John Byrns wrote:

In article ,
"Henry Pasternack" wrote:

"Patrick Turner" wrote in message
...
Henry came up with a number lower than a cathode follower and I think
Preisman did as well.

I don't see how it could be lower, but I have never needed to work it
out.

It isn't, you have to keep in mind that you are dealing with the
slippery fish, who in this case is misquoting Henry. Henry's original
claim was that his CPI source impedance was lower than the source
impedance at the cathode of the CPI when measured by the Byrns method,
and that is indeed true, the slippery fish has incorrectly equated the
³Byrns² method CPI cathode source impedance with the source impedance of
a cathode follower in his statement above.

Gee, what slippery fish? Ah Flipper?

I forgot what the Byrns method is....

It's the equations I developed to define the elemental "CPI" which are presented
a few paragraphs further down.

But as I said in another post, the CPI aRout = kRout if loads are
constant,
and each is above Rout of a CF, 1/gm.

Isn't Rout for a CF equal to rp/(u+1)? IIRC Henry's Rout for both the anode and
cathode terminals of the CPI is rp/(u+2), which is always lower than Rout for a
CF using the same tube.

[Snip]

One day I'll work out a formula for the Rout of a CPI at its a and k,
but I ain't in any hurry. I just work out what I want, a good working
point for the triode involved,
then test it all well and if it works well that's it, done like sunday
dinner.

No need to bother, the dinner has already been cooked and is on the table below.

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:
typo here; should be: ^^^^plate^^^^

There should probably be a minus sign in here somewhe
(7) ELp = es * u * Zp / (rp + (u + 1) * Zk + Zp)

So it should be:
(7) ELp = -es * u * Zp / (rp + (u + 1) * Zk + Zp)


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

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 nowhere near equal.

The RC time constant determining the -3 dB rolloff point when Zk equals Zp can
be calculated from the pole location in equation 7, by replacing Zk and Zp with
the parallel combination of Rl and Cl, giving TC = ((rp/(u +2))||Rl) * Cl.

[The Phantom]

On Tue, 05 Aug 2008 13:04:52 -0500, John Byrns
wrote:

SNIP

Could you elaborate on what the goof was?

The goof was not a typo or an error in a formula, his idea of placing a
build out resistor in the cathode of a concertina phase inverter to
equalize the source impedances of the plate and cathode circuits was
simply a goofy idea. It was a "bright" idea intended to fix an imagined
problem that didn't actually exist, that instead created a real problem.

I saw the magazine article

I'm posting the page from the magazine article where he shows a schematic
"...featuring--possibly for the first time--cathode build-out resistor in
the driver stage.", and several letters that followed.

It's over on ABSE.

and the first edition of his book where he
presented this goofy idea, I have never seen the second edition of his
book to see how he extricated himself from the predicament he created
for himself, although I have been told by people that have seen the
second edition that he did somehow extricate himself.

[John Byrns]

In article ,
The Phantom wrote:

On Tue, 19 Aug 2008 18:47:39 -0500, John Byrns
wrote:

In article ,
Patrick Turner wrote:

John Byrns wrote:

In article ,
"Henry Pasternack" wrote:

"Patrick Turner" wrote in message
...
Henry came up with a number lower than a cathode follower and I
think
Preisman did as well.

I don't see how it could be lower, but I have never needed to work
it
out.

It isn't, you have to keep in mind that you are dealing with the
slippery fish, who in this case is misquoting Henry. Henry's original
claim was that his CPI source impedance was lower than the source
impedance at the cathode of the CPI when measured by the Byrns method,
and that is indeed true, the slippery fish has incorrectly equated the
³Byrns² method CPI cathode source impedance with the source impedance of
a cathode follower in his statement above.

Gee, what slippery fish? Ah Flipper?

I forgot what the Byrns method is....

It's the equations I developed to define the elemental "CPI" which are
presented
a few paragraphs further down.

But as I said in another post, the CPI aRout = kRout if loads are
constant,
and each is above Rout of a CF, 1/gm.

Isn't Rout for a CF equal to rp/(u+1)? IIRC Henry's Rout for both the anode
and
cathode terminals of the CPI is rp/(u+2), which is always lower than Rout
for a
CF using the same tube.

[Snip]

One day I'll work out a formula for the Rout of a CPI at its a and k,
but I ain't in any hurry. I just work out what I want, a good working
point for the triode involved,
then test it all well and if it works well that's it, done like sunday
dinner.

No need to bother, the dinner has already been cooked and is on the table
below.

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:
typo here; should be: ^^^^plate^^^^

There should probably be a minus sign in here somewhe
(7) ELp = es * u * Zp / (rp + (u + 1) * Zk + Zp)

So it should be:
(7) ELp = -es * u * Zp / (rp + (u + 1) * Zk + Zp)


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

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 nowhere near equal.

The RC time constant determining the -3 dB rolloff point when Zk equals Zp
can
be calculated from the pole location in equation 7, by replacing Zk and Zp
with
the parallel combination of Rl and Cl, giving TC = ((rp/(u +2))||Rl) * Cl.

Hi Rodger,

Thanks for pointing out these errors, you are clearly far and away the Ace
proofreader in the group. I am absolutely amazed that neither Henry nor Flipper
noticed these errors as they both had approved the equations without noting the
errors. I assume they missed the minus sign error because it wasn't
particularly relevant to the argument at hand.

I guess the discovery of these errors vindicates my original warning when I
first posted these equations, that they ³may contain many typos.²

You did miss the error in equation 6 which should probably also contain a minus
sign as in equation 7. The omission of the minus sign in equation 6 was
obviously the root cause of the missing minus sign in equation 7, which is the
result of combining equations 5 & 6.

--
Regards,

John Byrns

Surf my web pages at, http://fmamradios.com/

[John Byrns]

In article ,
The Phantom wrote:

On Tue, 05 Aug 2008 13:04:52 -0500, John Byrns
wrote:

SNIP

Could you elaborate on what the goof was?

The goof was not a typo or an error in a formula, his idea of placing a
build out resistor in the cathode of a concertina phase inverter to
equalize the source impedances of the plate and cathode circuits was
simply a goofy idea. It was a "bright" idea intended to fix an imagined
problem that didn't actually exist, that instead created a real problem.

I saw the magazine article

I'm posting the page from the magazine article where he shows a schematic
"...featuring--possibly for the first time--cathode build-out resistor in
the driver stage.", and several letters that followed.

It's over on ABSE.

Just as a point of interest a large number of US residents, myself included, no
longer have access to these binary groups as a result of the New York AGs
actions.

--
Regards,

John Byrns

Surf my web pages at, http://fmamradios.com/

[The Phantom]

On Wed, 20 Aug 2008 21:56:17 -0500, John Byrns
wrote:

In article ,
The Phantom wrote:

On Tue, 19 Aug 2008 18:47:39 -0500, John Byrns
wrote:

In article ,
Patrick Turner wrote:

John Byrns wrote:

In article ,
"Henry Pasternack" wrote:

"Patrick Turner" wrote in message
...
Henry came up with a number lower than a cathode follower and I
think
Preisman did as well.

I don't see how it could be lower, but I have never needed to work
it
out.

It isn't, you have to keep in mind that you are dealing with the
slippery fish, who in this case is misquoting Henry. Henry's original
claim was that his CPI source impedance was lower than the source
impedance at the cathode of the CPI when measured by the Byrns method,
and that is indeed true, the slippery fish has incorrectly equated the
³Byrns² method CPI cathode source impedance with the source impedance of
a cathode follower in his statement above.

Gee, what slippery fish? Ah Flipper?

I forgot what the Byrns method is....

It's the equations I developed to define the elemental "CPI" which are
presented
a few paragraphs further down.

But as I said in another post, the CPI aRout = kRout if loads are
constant,
and each is above Rout of a CF, 1/gm.

Isn't Rout for a CF equal to rp/(u+1)? IIRC Henry's Rout for both the anode
and
cathode terminals of the CPI is rp/(u+2), which is always lower than Rout
for a
CF using the same tube.

[Snip]

One day I'll work out a formula for the Rout of a CPI at its a and k,
but I ain't in any hurry. I just work out what I want, a good working
point for the triode involved,
then test it all well and if it works well that's it, done like sunday
dinner.

No need to bother, the dinner has already been cooked and is on the table
below.

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:
typo here; should be: ^^^^plate^^^^

There should probably be a minus sign in here somewhe
(7) ELp = es * u * Zp / (rp + (u + 1) * Zk + Zp)

So it should be:
(7) ELp = -es * u * Zp / (rp + (u + 1) * Zk + Zp)


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

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 nowhere near equal.

The RC time constant determining the -3 dB rolloff point when Zk equals Zp
can
be calculated from the pole location in equation 7, by replacing Zk and Zp
with
the parallel combination of Rl and Cl, giving TC = ((rp/(u +2))||Rl) * Cl.

Hi Rodger,

Thanks for pointing out these errors, you are clearly far and away the Ace
proofreader in the group. I am absolutely amazed that neither Henry nor Flipper
noticed these errors as they both had approved the equations without noting the
errors. I assume they missed the minus sign error because it wasn't
particularly relevant to the argument at hand.

I guess the discovery of these errors vindicates my original warning when I
first posted these equations, that they ³may contain many typos.²

You did miss the error in equation 6 which should probably also contain a minus
sign as in equation 7.

I didn't miss it. I figured I would give you one freebie. :-)

The omission of the minus sign in equation 6 was
obviously the root cause of the missing minus sign in equation 7, which is the
result of combining equations 5 & 6.

Yep, I saw that, but I thought I would just point it out at the final
expression.

[The Phantom]

On Wed, 20 Aug 2008 22:25:44 -0500, John Byrns
wrote:

In article ,
The Phantom wrote:

On Tue, 05 Aug 2008 13:04:52 -0500, John Byrns
wrote:

SNIP

Could you elaborate on what the goof was?

The goof was not a typo or an error in a formula, his idea of placing a
build out resistor in the cathode of a concertina phase inverter to
equalize the source impedances of the plate and cathode circuits was
simply a goofy idea. It was a "bright" idea intended to fix an imagined
problem that didn't actually exist, that instead created a real problem.

I saw the magazine article

I'm posting the page from the magazine article where he shows a schematic
"...featuring--possibly for the first time--cathode build-out resistor in
the driver stage.", and several letters that followed.

It's over on ABSE.

Just as a point of interest a large number of US residents, myself included, no
longer have access to these binary groups as a result of the New York AGs
actions.

Yes, that is a problem. If anybody wants it, including you, post an email
address and I'll email it to you.

In your case, John, you would probably like to have the Shekel paper, too.

I see what looks like a good email address in the header of your posts.
Shall I send you these items at that email address?

[John Byrns]

In article ,
The Phantom wrote:

On Wed, 20 Aug 2008 22:25:44 -0500, John Byrns
wrote:

In article ,
The Phantom wrote:

On Tue, 05 Aug 2008 13:04:52 -0500, John Byrns
wrote:

SNIP

Could you elaborate on what the goof was?

The goof was not a typo or an error in a formula, his idea of placing a
build out resistor in the cathode of a concertina phase inverter to
equalize the source impedances of the plate and cathode circuits was
simply a goofy idea. It was a "bright" idea intended to fix an imagined
problem that didn't actually exist, that instead created a real problem.

I saw the magazine article

I'm posting the page from the magazine article where he shows a schematic
"...featuring--possibly for the first time--cathode build-out resistor in
the driver stage.", and several letters that followed.

It's over on ABSE.

Just as a point of interest a large number of US residents, myself included,
no
longer have access to these binary groups as a result of the New York AGs
actions.

Yes, that is a problem. If anybody wants it, including you, post an email
address and I'll email it to you.

In your case, John, you would probably like to have the Shekel paper, too.

I see what looks like a good email address in the header of your posts.
Shall I send you these items at that email address?

Yes, the email address should be good, please send the Morgan Jones stuff, and
the Shekel paper too. Thanks.

--
Regards,

John Byrns

Surf my web pages at, http://fmamradios.com/