[The Phantom]

On Wed, 20 Aug 2008 22:09:04 -0500, flipper wrote:

On Wed, 20 Aug 2008 17:33:34 -0700, 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.

Interesting. And interesting to compare McFadden's letter to my
original post because I said almost exactly the same thing

He wrote "The circuit behaves as if the output resistance at both
ports is the same {as the source resistance of a cathode follower...]"

And I wrote "...the thing acts as if the source impedances are
identical. I.E. there is no difference in HF roll off (with equal
value grid stoppers)."

I'm want to put forth the notion that it's not necessary to use phrases
like "...behaves as if..." and "...acts as if...". The phrases suggest
that the circuit "behaves as if" some condition were true, but which isn't
really true.

What is in fact true, is that as used in audio, the Concertina has a
differential load (or call it a balanced load; it's the same thing). We
need not say that it "...behaves as if it had a differential load."; it
*does* have a differential (balanced) load.

The outputs, P and K, of the Concertina are exactly equal in amplitude and
180 out of phase; v(P) = -v(K) as long as the loads are identical. This
means that if a load of Z is applied from each of P and K to ground, we may
lift the grounded ends of each Z and connect them together. The junction
of the two Z's will have no voltage present because when -v(t) is present
at P and v(t) is present at K, they cancel at the junction of the two Z's.
That means that the circuit behavior is the same, whether that junction is
grounded or not, and if the junction is *not* grounded, then we have a
purely differential load of 2*Z. This means that connecting identical
loads of Z (one end grounded, of course) to each of P and K is *exactly*
the same as connecting a load of 2*Z *between* P and K; differentially in
other words.

Z may have a capacitive part, and this doesn't alter the previous argument.
If a purely differential load 2*Z is applied, it may be treated as separate
loads of Z at each of P and K, and by the well-known explanation (which I
mention below can be found in RDH4), the current through whatever load is
applied to P will be the same as the current through that same load applied
to K. And therefore the high frequency rolloffs at P and K will be
identical.

Because the Concertina has a vacuum tube as one of its components, its
admittance matrix is not bilateral; the transfer impedance from node i to
node j is not the same as in the reverse direction. This causes the
driving point impedance at P and at K to be different than the half
differential output impedance. Since the intended load on the Concertina
is a differential (balanced) load, if we want to calculate things like
decrease in output voltage due to loading, or high frequency rolloff due to
loading by capacitance, we must use the differential output impedance,
because the load is differential.

However, we may sometimes wish to calculate the effect of unequal
capacitive loads, perhaps due to strays. In that case, the use of the
driving point impedances at P and K is called for.

The differential output impedance could be measured in the real world by
connecting a small LCR meter between P and K:
http://www.mcmconnect.com/tenma/prod...tance%20Meters
This meter should be able to measure resistance with an AC stimulus (1 kHz
typically) instead of a DC stimulus.

The meter will inject a test current from one of its leads while
simultaneously withdrawing the same current from the other lead; in other
words, it injects a current of +i(t) from one lead and a current of -i(t)
from the other lead. This allows a measurement *between* nodes P and K; a
*differential* measurement.

This same method can be used to find the differential impedance between a
pair of nodes in a circuit when doing a mathematical analysis. We simply
inject a current of +1 at one node while simultaneously injecting a current
of -1 at the other node. Using a current of magnitude 1 results in a
voltage appearing at a node, due to the injection of such a current, that
is numerically equal to the impedance at that node.

If we (mathematically) inject a current of +1 into P and simultaneously
inject -1 into K and calculate the voltage appearing at each node due to
those currents, those voltages will be (numerically) the impedances at the
nodes. The differential impedance will be the sum of the individual
impedances appearing at P and K while injecting the +1/-1 differential
currents. The impedance at the node where a current of -1 is injected is
the negative of the voltage appearing there.

That differential impedance can treated as one quantity, or it can be
considered split into a P component and a K component. When the individual
components are used, then capacitive loading can also be split into
separate components. In other words, a 100 pF differential (balanced)
capacitive load can be considered to be separate 50 pF capacitors applied
simultaneously to P and K for the purpose of determining high frequency
behavior. But the effect of the two 50 pF capacitive loads is identical to
the effect of a single 100 pF differential capacitive load.

As it happens, when we inject +1/-1 currents at P and K we find that the
individual impedances at P and K are identical and their sum is the
differential impedance. So, we might say that the impedance at each node,
P and K, which we should use to compute the effect of further
(differential) loading, is the "half differential output impedance". And,
when we are careful to apply any additional loads to P and K
simultaneously, and in identical value, we should realize that we are
really applying a *differential* load.

I recommend using the phrases "differential output impedance" and "driving
point (output) impedance". If we only use "output impedance", we have one
person meaning one thing and another person meaning another thing and
acrimony results. There's no need for this. The circuit can be analyzed
exactly and without resort to waffle-words like "...behaves as though...".
We need not show that the circuit "...behaves as though..."; we can show
how the circuit behaves in fact.

We can do fairly simple mathematical analyses which will show how the
circuit behaves for loads applied to one output, P or K, at a time or for
loads applied to both outputs simultaneously (a differential, or balanced,
load).

I'm going to post over on ABSE, analysis of the circuit using the
admittance technique of Jacob Shekel. His paper has been posted over there
already.


I also made the same 'intuitive' argument with "If, however, you
accept that the concertina is balanced as long as the
two loads are equal then it's an 'of course' the roll off is the same
because the loads are equal, they just vary (equally) with
frequency." (Balanced as long as the loads are equal because anode and
cathode current through the one tube must be one and the same.)

I wonder when this explanation first began to be used. I looked in my RDH3
(printed in 1940) and on page 10 there is a schematic and short description
of the circuit we now call the Concertina. There is no discussion of
impedances at all.

By the time of RDH4 (1952), there is discussion of impedances. RDH4 says,
on page 330:

"The effective output resistance is different for the two output channels,
since P operates with current feedback and K with voltage feedback."

They then give expressions for the "effective output resistance" (what they
are referring to is what nowadays would be called the "driving point
impedance") in formulas 30 and 31, but I'm sure that formula 30 is
incorrect. They continue:

"...but this does not affect the balance at either low or high frequencies
when the total effective impedance of channel P is equal to that of channel
K. The same signal plate current which flows through one impedance Zp also
flows through the other impedance Zk, and if Zp=Zk then the two output
voltages are equal.

As I've shown, when you apply identical load impedances Z to each of P and
K, you are in effect applying a differential load of 2*Z, and the behavior
of the circuit can be analyzed that way. The expression given in formula
34a in RDH4 is what I am now calling the "half differential output
impedance". This is the impedance that can be attributed to each of P and
K for the purpose of determining circuit behavior when identical loads are
simultaneously applied to P and K.


About the only difference is I went on to explain that the 'classic
unequal output impedance model works when you consider different
generator voltages with "If you keep the 'different impedance'
analysis the thing is as cathode impedance drops anode gain increases
in exact proportion to the drop across the 'larger' anode impedance."

And I have been attacked and insulted by Byrns for it.


Btw, I agree with your implied criticism (at least that's how I
interpreted it) that Jones tries to 'hedge' his 'build out' resistor
with the 'problem' of grid drive.

He continues to analyze this as an 'output impedance' issue but, IMO,
that only serves to muddy one's understanding because the actual
behavior is different than a simple 'output impedance' model suggests.
I.E., it isn't that the signal on the 'high impedance' side is
(uniformly) 'reduced', as one might think with a simple impedance
model, it has a large negative spike when the cathode tries to drive
+ve, in addition, of course, to the positive clipping. Due, of course,
to the impedances but not in the manner one would expect for 'Ro into
a load'.

I *have* seen the 'problem' he refers to but I tend to view the issue
as an 'upset' condition with the huge anode (and some on the cathode
in reverse) swing introducing spurious HF components that cause
feedback instability issues, depending on the OPT characteristics and
amount of GNFB.

I mentioned that in an earlier post about grid stoppers on the output
tubes noting that it only takes 'one', on the cathode side, but that I
always put two, equal, grid stoppers.


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 ,
flipper wrote:

On Tue, 19 Aug 2008 18:37:40 -0500, John Byrns
wrote:

Can you give a clear definition of what makes something a
"--Thevenin--- 'output impedance'"?

Yes, as I have repeated so many times I've lost count, a derived
impedance you can plug into RC, or any other simple impedance
calculation, and get a meaningful result with. As in, if I slap a
capacitor on the output of an arbitrary circuit how will the output of
that arbitrary circuit behave? Simple, take the --Thevenin--- output
impedance and use it for the R in the RC calculation. That's what a
--Thevenin--- output impedance *is* and why Thevenin bothered to
develop the model

The values you calculate in the 'unequal output impedance' model don't
work that way. The value derived with the equal output impedance model
does.

Hi Flipper,

Thanks for making a start at defining what a Thevenin model is. Please pardon
my trimming this discussion back to a single item, I would like to try and keep
things more understandable, and less confusing, by discussing only one issue at
a time, starting with the Thevenin model, before moving on to the next issue. I
am clearly a little foggy as to what the restrictions and constraints on a
proper Thevenin model are? IIRC you said that my model, as defined by the
equations I posted, with the typos corrected, is not a Thevenin model, is that
correct?

It is my understanding that the Thevenin model consists of a voltage generator
and an impedance, clearly there are some restrictions and constraints on these
that my model fails to meet? Can you more completely explain the restrictions
and constraints on the Thevenin voltage generator and impedance, and how my
model fails to meet the requirements?

You speak to "the --Thevenin--- output impedance" above, saying that it is a
"simple impedance". I believe the Thevenin model also includes a "
--Thevenin--- voltage generator", can you explain the requirements for the
Thevenin voltage generator?

Finally, I have snipped off much of your post in my attempt to keep things clear
and simple, in the part that I snipped off you mentioned your model for the
"CPI" at one point. Could you post the equations representing "the
--Thevenin--- output impedance" and "the --Thevenin--- voltage generator"
used in your model, as an example of a proper Thevenin model?

--
Regards,

John Byrns

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

[Henry Pasternack[_2_]]

"The Phantom" wrote in message
...
I'm want to put forth the notion that it's not necessary to use phrases
like "...behaves as if..." and "...acts as if...". The phrases suggest
that the circuit "behaves as if" some condition were true, but which isn't
really true.

I disagree. Everything you say about the circuit is really a "...behaves as
if..."
statement if you back off a level or two and consider all of the
abstractions
that go into traditional circuit analysis. So, it it really a question of
choosing
the appropriate level of abstraction for the situation. You, like Byrns,
seem
to be a pretty pedantic fellow. And that's fine, but I have a different
focus.
And it's not for lack of the wherewithal to be pedantic if I wanted.

There is a whole other aspect to the so-called "controversy" on this board
that has nothing at all to do with circuits and everything to do with
abusive
personalities.

Whatever.

-Henry

[John Byrns]

In article ,
The Phantom wrote:

On Wed, 20 Aug 2008 22:09:04 -0500, flipper wrote:

On Wed, 20 Aug 2008 17:33:34 -0700, 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.

Interesting. And interesting to compare McFadden's letter to my
original post because I said almost exactly the same thing

He wrote "The circuit behaves as if the output resistance at both
ports is the same {as the source resistance of a cathode follower...]"

And I wrote "...the thing acts as if the source impedances are
identical. I.E. there is no difference in HF roll off (with equal
value grid stoppers)."

I'm want to put forth the notion that it's not necessary to use phrases
like "...behaves as if..." and "...acts as if...". The phrases suggest
that the circuit "behaves as if" some condition were true, but which isn't
really true.

What is in fact true, is that as used in audio, the Concertina has a
differential load (or call it a balanced load; it's the same thing). We
need not say that it "...behaves as if it had a differential load."; it
*does* have a differential (balanced) load.

The outputs, P and K, of the Concertina are exactly equal in amplitude and
180 out of phase; v(P) = -v(K) as long as the loads are identical. This
means that if a load of Z is applied from each of P and K to ground, we may
lift the grounded ends of each Z and connect them together. The junction
of the two Z's will have no voltage present because when -v(t) is present
at P and v(t) is present at K, they cancel at the junction of the two Z's.
That means that the circuit behavior is the same, whether that junction is
grounded or not, and if the junction is *not* grounded, then we have a
purely differential load of 2*Z. This means that connecting identical
loads of Z (one end grounded, of course) to each of P and K is *exactly*
the same as connecting a load of 2*Z *between* P and K; differentially in
other words.

Z may have a capacitive part, and this doesn't alter the previous argument.
If a purely differential load 2*Z is applied, it may be treated as separate
loads of Z at each of P and K, and by the well-known explanation (which I
mention below can be found in RDH4), the current through whatever load is
applied to P will be the same as the current through that same load applied
to K. And therefore the high frequency rolloffs at P and K will be
identical.

Because the Concertina has a vacuum tube as one of its components, its
admittance matrix is not bilateral; the transfer impedance from node i to
node j is not the same as in the reverse direction. This causes the
driving point impedance at P and at K to be different than the half
differential output impedance. Since the intended load on the Concertina
is a differential (balanced) load, if we want to calculate things like
decrease in output voltage due to loading, or high frequency rolloff due to
loading by capacitance, we must use the differential output impedance,
because the load is differential.

However, we may sometimes wish to calculate the effect of unequal
capacitive loads, perhaps due to strays. In that case, the use of the
driving point impedances at P and K is called for.

The differential output impedance could be measured in the real world by
connecting a small LCR meter between P and K:
http://www.mcmconnect.com/tenma/prod...tance%20Meters
This meter should be able to measure resistance with an AC stimulus (1 kHz
typically) instead of a DC stimulus.

The meter will inject a test current from one of its leads while
simultaneously withdrawing the same current from the other lead; in other
words, it injects a current of +i(t) from one lead and a current of -i(t)
from the other lead. This allows a measurement *between* nodes P and K; a
*differential* measurement.

This same method can be used to find the differential impedance between a
pair of nodes in a circuit when doing a mathematical analysis. We simply
inject a current of +1 at one node while simultaneously injecting a current
of -1 at the other node. Using a current of magnitude 1 results in a
voltage appearing at a node, due to the injection of such a current, that
is numerically equal to the impedance at that node.

If we (mathematically) inject a current of +1 into P and simultaneously
inject -1 into K and calculate the voltage appearing at each node due to
those currents, those voltages will be (numerically) the impedances at the
nodes. The differential impedance will be the sum of the individual
impedances appearing at P and K while injecting the +1/-1 differential
currents. The impedance at the node where a current of -1 is injected is
the negative of the voltage appearing there.

That differential impedance can treated as one quantity, or it can be
considered split into a P component and a K component. When the individual
components are used, then capacitive loading can also be split into
separate components. In other words, a 100 pF differential (balanced)
capacitive load can be considered to be separate 50 pF capacitors applied
simultaneously to P and K for the purpose of determining high frequency
behavior. But the effect of the two 50 pF capacitive loads is identical to
the effect of a single 100 pF differential capacitive load.

Hi Rodger,

Wouldn't a 100 pF differential capacitive load be the equivalent of separate 200
pF capacitive loads applied simultaneously to P and K, not 50 pF loads?

As it happens, when we inject +1/-1 currents at P and K we find that the
individual impedances at P and K are identical and their sum is the
differential impedance. So, we might say that the impedance at each node,
P and K, which we should use to compute the effect of further
(differential) loading, is the "half differential output impedance". And,
when we are careful to apply any additional loads to P and K
simultaneously, and in identical value, we should realize that we are
really applying a *differential* load.

I recommend using the phrases "differential output impedance" and "driving
point (output) impedance". If we only use "output impedance", we have one
person meaning one thing and another person meaning another thing and
acrimony results. There's no need for this. The circuit can be analyzed
exactly and without resort to waffle-words like "...behaves as though...".
We need not show that the circuit "...behaves as though..."; we can show
how the circuit behaves in fact.

We can do fairly simple mathematical analyses which will show how the
circuit behaves for loads applied to one output, P or K, at a time or for
loads applied to both outputs simultaneously (a differential, or balanced,
load).

I'm going to post over on ABSE, analysis of the circuit using the
admittance technique of Jacob Shekel. His paper has been posted over there
already.


I also made the same 'intuitive' argument with "If, however, you
accept that the concertina is balanced as long as the
two loads are equal then it's an 'of course' the roll off is the same
because the loads are equal, they just vary (equally) with
frequency." (Balanced as long as the loads are equal because anode and
cathode current through the one tube must be one and the same.)

I wonder when this explanation first began to be used. I looked in my RDH3
(printed in 1940) and on page 10 there is a schematic and short description
of the circuit we now call the Concertina. There is no discussion of
impedances at all.

By the time of RDH4 (1952), there is discussion of impedances. RDH4 says,
on page 330:

"The effective output resistance is different for the two output channels,
since P operates with current feedback and K with voltage feedback."

They then give expressions for the "effective output resistance" (what they
are referring to is what nowadays would be called the "driving point
impedance") in formulas 30 and 31, but I'm sure that formula 30 is
incorrect. They continue:

"...but this does not affect the balance at either low or high frequencies
when the total effective impedance of channel P is equal to that of channel
K. The same signal plate current which flows through one impedance Zp also
flows through the other impedance Zk, and if Zp=Zk then the two output
voltages are equal.

As I've shown, when you apply identical load impedances Z to each of P and
K, you are in effect applying a differential load of 2*Z, and the behavior
of the circuit can be analyzed that way. The expression given in formula
34a in RDH4 is what I am now calling the "half differential output
impedance". This is the impedance that can be attributed to each of P and
K for the purpose of determining circuit behavior when identical loads are
simultaneously applied to P and K.


About the only difference is I went on to explain that the 'classic
unequal output impedance model works when you consider different
generator voltages with "If you keep the 'different impedance'
analysis the thing is as cathode impedance drops anode gain increases
in exact proportion to the drop across the 'larger' anode impedance."

And I have been attacked and insulted by Byrns for it.


Btw, I agree with your implied criticism (at least that's how I
interpreted it) that Jones tries to 'hedge' his 'build out' resistor
with the 'problem' of grid drive.

He continues to analyze this as an 'output impedance' issue but, IMO,
that only serves to muddy one's understanding because the actual
behavior is different than a simple 'output impedance' model suggests.
I.E., it isn't that the signal on the 'high impedance' side is
(uniformly) 'reduced', as one might think with a simple impedance
model, it has a large negative spike when the cathode tries to drive
+ve, in addition, of course, to the positive clipping. Due, of course,
to the impedances but not in the manner one would expect for 'Ro into
a load'.

I *have* seen the 'problem' he refers to but I tend to view the issue
as an 'upset' condition with the huge anode (and some on the cathode
in reverse) swing introducing spurious HF components that cause
feedback instability issues, depending on the OPT characteristics and
amount of GNFB.

I mentioned that in an earlier post about grid stoppers on the output
tubes noting that it only takes 'one', on the cathode side, but that I
always put two, equal, grid stoppers.


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.

--
Regards,

John Byrns

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

[The Phantom]

On Thu, 21 Aug 2008 18:38:59 -0500, John Byrns
wrote:
SNIP

Hi Rodger,

Wouldn't a 100 pF differential capacitive load be the equivalent of separate 200
pF capacitive loads applied simultaneously to P and K, not 50 pF loads?

Yes, of course, capacitors in series and all that. It's hard to do such a
long post without a boo-boo or two creeping in. As we all know. :-)

[The Phantom]

On Thu, 21 Aug 2008 19:38:24 -0400, "Henry Pasternack" wrote:

"The Phantom" wrote in message
.. .
I'm want to put forth the notion that it's not necessary to use phrases
like "...behaves as if..." and "...acts as if...". The phrases suggest
that the circuit "behaves as if" some condition were true, but which isn't
really true.

I disagree. Everything you say about the circuit is really a "...behaves as
if..."
statement if you back off a level or two and consider all of the
abstractions
that go into traditional circuit analysis. So, it it really a question of
choosing
the appropriate level of abstraction for the situation. You, like Byrns,
seem
to be a pretty pedantic fellow. And that's fine, but I have a different
focus.

What is that focus?

And it's not for lack of the wherewithal to be pedantic if I wanted.

There is a whole other aspect to the so-called "controversy" on this board
that has nothing at all to do with circuits and everything to do with
abusive
personalities.

It seems to me that the first step toward that kind of abuse is bringing
personalities into the discussion in the first place. I don't see what
personalities have to do with the working of a Concertina phase splitter,
so I don't mention them.


Whatever.

-Henry

[Patrick Turner]

By the time of RDH4 (1952), there is discussion of impedances. RDH4 says,
on page 330:

"The effective output resistance is different for the two output channels,
since P operates with current feedback and K with voltage feedback."

They then give expressions for the "effective output resistance" (what they
are referring to is what nowadays would be called the "driving point
impedance") in formulas 30 and 31, but I'm sure that formula 30 is
incorrect.

Equation 30 is what effective Rp' of the tube is at the anode.
Its quite correct.

The R looking into the anode of the triode is Rp' and with an unbypassed
Rk,
is exactly what they say, Rp + ( [µ + 1]Rk.

The Rout from the anode at the anode if tested separately would be the
above Rpp' figure
in parallel with whatever anode RL you have.

I gave a summary which everyone has ignored in a post......

Subject:
Output resistance of the CPI or "concertina phase inverter"
Date:
Mon, 18 Aug 2008 20:47:31 +1000
From:
Patrick Turner
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.

[Henry Pasternack[_2_]]

"The Phantom" wrote in message
...
You, like Byrns, seem to be a pretty pedantic fellow. And that's fine,
but I have a different focus.

What is that focus?

I wrote a lengthy respose, but decided not to post it. If you would like
to see what I had to say, you are welcome to email me privately.

-Henry

moc.ncr@kcanretsaph

[Andre Jute[_2_]]

On Aug 22, 4:12*am, "Henry Pasternack" wrote:
"The Phantom" wrote in message

...

You, like Byrns, seem to be a pretty pedantic fellow. *And that's fine,
but I have a different focus.

What is that focus?

I wrote a lengthy respose, but decided not to post it. *If you would like
to see what I had to say, you are welcome to email me privately.

-Henry

moc.ncr@kcanretsaph

Yup. Plodnick lost the argument on its merits, so now he whines that
the whole world is against him, he calls them names, and at the same
time offers to conspire behind their backs with anyone stupid enough
to enter into private correspondence with such a discredited and
incompetent plotter.

Par for the course. Plodnick is a slow learner. What else is new.

Andre Jute
Observer

[The Phantom]

On Fri, 22 Aug 2008 02:36:21 GMT, Patrick Turner
wrote:



By the time of RDH4 (1952), there is discussion of impedances. RDH4 says,
on page 330:

"The effective output resistance is different for the two output channels,
since P operates with current feedback and K with voltage feedback."

They then give expressions for the "effective output resistance" (what they
are referring to is what nowadays would be called the "driving point
impedance") in formulas 30 and 31, but I'm sure that formula 30 is
incorrect.

Equation 30 is what effective Rp' of the tube is at the anode.
Its quite correct.

The R looking into the anode of the triode is Rp' and with an unbypassed
Rk,
is exactly what they say, Rp + ( [µ + 1]Rk.

The Rout from the anode at the anode if tested separately would be the
above Rpp' figure
in parallel with whatever anode RL you have.

In my copy of RDH4, on page 330 they have two lines like this:

"Channel P: rp' = (u - 1)Rk where Rk = RL (30)
(from equation 25a, Sect 1.)"

But on page 313, equation 25a says:

"rp' = rp + (u +1)R3 (25a)"

How do they get equation 30 from equation 25a?

[The Phantom]

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.


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.

What you are describing here is the driving point impedance; what flipper
is describing is the differential impedance.


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.

[The Phantom]

1
___ | ___
+-+. ,+-+-|___|--|___|-GND
)|( 50 10k
)|( ___
-. ,+-+-|___|-GND
)|( 1k
)|( ___ ___
+-+' '+-+-|___|.-|___|-GND
50 | 10k
2
Imagine that you had a circuit like this, with the transformer turns ratio
being 1 to 1, and the primary driven with a 1 kHz source having zero ohms
output impedance.

1
___ |
+-+. ,----|___|-
)|( 50
)|( ___
)|(----|___|-GND
)|( 10k
)|( ___
+-+' '----|___|-
50 |
2

Now suppose you measure the impedance at node 1 with respect to ground. It
will be something like 10k ohms. This is a driving point impedance.

Now measure the impedance *between" nodes 1 and 2; it will be something
like 100 ohms. This is a differential impedance.

The differential output impedance is not the same as the driving point
output impedance, as can be easily seen.

It doesn't seem to me that there should be controversy about what the
circuit does if we choose our terminology correctly. It's important to
distinguish between driving point impedance and differential impedance.

When making the differential measurement, half of the differential
impedance is at node 1 and half at node 2.

If we had a circuit like this:

1
___ |
+-+. ,----|___|-
)|( 60
)|( ___
)|(----|___|-GND
)|( 10k
)|( ___
+-+' '----|___|-
40 |
2

we would still have a differential output impedance of about 100 ohms, but
we would no longer have half the differential impedance at each node.

The Concertina is like the first circuit. The differential output
impedance is split, half and half, between P and K (which is rather
fortuitous).

[Patrick Turner]

The Phantom wrote:

On Fri, 22 Aug 2008 02:36:21 GMT, Patrick Turner
wrote:



By the time of RDH4 (1952), there is discussion of impedances. RDH4 says,
on page 330:

"The effective output resistance is different for the two output channels,
since P operates with current feedback and K with voltage feedback."

They then give expressions for the "effective output resistance" (what they
are referring to is what nowadays would be called the "driving point
impedance") in formulas 30 and 31, but I'm sure that formula 30 is
incorrect.

Equation 30 is what effective Rp' of the tube is at the anode.
Its quite correct.

The R looking into the anode of the triode is Rp' and with an unbypassed
Rk,
is exactly what they say, Rp + ( [µ + 1]Rk.

The Rout from the anode at the anode if tested separately would be the
above Rpp' figure
in parallel with whatever anode RL you have.

In my copy of RDH4, on page 330 they have two lines like this:

"Channel P: rp' = (u - 1)Rk where Rk = RL (30)
(from equation 25a, Sect 1.)"

In my RDH4, Pg 330, equation 30 reads...

Channel P rp' = rp + ( µ + 1 )Rk where Rk = RL.
( from eqtn 25A, Sect 1 ).

You have misquoted Equation 30.


But on page 313, equation 25a says:

"rp' = rp + (u +1)R3 (25a)"

How do they get equation 30 from equation 25a?

There is no inconsistency. R3 in eqtn 25a is Rk in eqtn 30.


RDH4 gives what is exactly right for the Rout at the anode for the CPI
if measured
as a separate outlet to power some following stage.

Ditto for the cathode Rout as spelled out in eqtn (31).

Rout at Channel K : rp' = ( rp + RL ) / ( µ + 1 )

If you follow my treatise in a recent post, you'll see what Rout of K
and P
will be where always RL = RK as in a real amp using CPI....

[Patrick Turner]

The Phantom wrote:

On Fri, 22 Aug 2008 02:36:21 GMT, Patrick Turner
wrote:



By the time of RDH4 (1952), there is discussion of impedances. RDH4 says,
on page 330:

"The effective output resistance is different for the two output channels,
since P operates with current feedback and K with voltage feedback."

They then give expressions for the "effective output resistance" (what they
are referring to is what nowadays would be called the "driving point
impedance") in formulas 30 and 31, but I'm sure that formula 30 is
incorrect.

Equation 30 is what effective Rp' of the tube is at the anode.
Its quite correct.

The R looking into the anode of the triode is Rp' and with an unbypassed
Rk,
is exactly what they say, Rp + ( [µ + 1]Rk.

The Rout from the anode at the anode if tested separately would be the
above Rpp' figure
in parallel with whatever anode RL you have.

In my copy of RDH4, on page 330 they have two lines like this:

"Channel P: rp' = (u - 1)Rk where Rk = RL (30)
(from equation 25a, Sect 1.)"

In my RDH4, Pg 330, equation 30 reads...

Channel P rp' = rp + ( µ + 1 )Rk where Rk = RL.
( from eqtn 25A, Sect 1 ).

You have misquoted Equation 30.


But on page 313, equation 25a says:

"rp' = rp + (u +1)R3 (25a)"

How do they get equation 30 from equation 25a?

There is no inconsistency. R3 in eqtn 25a is Rk in eqtn 30.


RDH4 gives what is exactly right for the Rout at the anode for the CPI
if measured
as a separate outlet to power some following stage.

Ditto for the cathode Rout as spelled out in eqtn (31).

Rout at Channel K : rp' = ( rp + RL ) / ( µ + 1 )

If you follow my treatise in a recent post, you'll see what Rout of K
and P
will be where always RL = RK as in a real amp using CPI....

Sorry I forgot to add the post and sign it;

Not one person at r.a.t has disagreed with anything in the post below.
Either its dead right and correct or nobody bothered to read it.

-------------------------------------------------------------------------------------

Subject:
Output resistance of the CPI or "concertina phase inverter"

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 anode 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.

[John Byrns]

In article ,
The Phantom wrote:

On Fri, 22 Aug 2008 02:36:21 GMT, Patrick Turner
wrote:

By the time of RDH4 (1952), there is discussion of impedances. RDH4 says,
on page 330:

"The effective output resistance is different for the two output channels,
since P operates with current feedback and K with voltage feedback."

They then give expressions for the "effective output resistance" (what they
are referring to is what nowadays would be called the "driving point
impedance") in formulas 30 and 31, but I'm sure that formula 30 is
incorrect.

Equation 30 is what effective Rp' of the tube is at the anode.
Its quite correct.

The R looking into the anode of the triode is Rp' and with an unbypassed
Rk,
is exactly what they say, Rp + ( [µ + 1]Rk.

The Rout from the anode at the anode if tested separately would be the
above Rpp' figure
in parallel with whatever anode RL you have.

In my copy of RDH4, on page 330 they have two lines like this:

"Channel P: rp' = (u - 1)Rk where Rk = RL (30)
(from equation 25a, Sect 1.)"

But on page 313, equation 25a says:

"rp' = rp + (u +1)R3 (25a)"

How do they get equation 30 from equation 25a?

Seems straight forward to me, on page 330 of my copy of the RDH equation 30 for
"Channel P" is shown as:

rp' = rp + u * Rk

Equation 25a on page 313 of my copy of the RDH is shown as:

rp' = rp + u * R3

These two equations appear virtually identical to me. The problem I see is that
while they are the same, they differ from equation 5 in the Byrns report which I
believe to be the correct version, and is as follows:

rp' = rp + (u + 1) * Zk

--
Regards,

John Byrns

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

[The Phantom]

The reason for wanting to know the Concertina "output impedance" is to be
able calculate the effect of further loading on the frequency response.

If "equal output impedance" model is used, then how will the effect of
unequal loading be calculated? This is a very real concern, because the
stray capacitances at the plate and cathode will not likely be identical.

And, of course, a model which only deals with the unequal driving point
impedances will not properly give the effect of the balanced loading which
will dominate the loads (we hope).

It's not necessary to use only one of these. Better to do a detailed
analysis using Shekel's method such as the one I posted a week or so ago.

From this analysis, expressions for the gain from grid to plate and to
cathode are derived. These expressions can then be used to calculate the
frequency responses from grid to plate and cathode regardless of whether
the loads are balanced or not; the correct result will be obtained for all
loads.

Using the following symbols:

ra = plate resistance of the tube

gm = transconductance of the tube

Zp = additional load applied to plate; include all loads in this,
including such things as stray capacitances.

Zk = additional load applied to cathode.

Then the grid to plate gain is given by:

ra*gm*Zp
- -------------------
(ra*gm+1)*Zk+Zp+ra

The grid to cathode gain is given by:

ra*gm*Zk
-------------------
(ra*gm+1)*Zk+Zp+ra

If Zp = Zk, then these expressions are identical, which suggests that the
impedances at plate and cathode are equal before additional identical loads
are applied. This is quite true, because as I've explained elsewhere in
this thread, applying identical loads simultaneously to plate and cathode
is exactly equivalent to applying a differential (balanced) load to the
Concertina, and we would expect that would not upset the equality of gains
at plate and cathode.

Knowing the output impedances allows us to calculate frequency response by
treating the outputs as sources with some impedance and then using a
voltage divider formula to calculate the effect of loads. If the loads are
perfectly balanced, then the appropriate impedance to use for these
calculations is what I've called the "half differential output impedance".
But what if the loads aren't perfectly balanced? Then I suppose we could
separate the load into a balanced component and an unbalanced component,
solve separately and use superposition to get the final result.

It's much easier to just use the gain expressions above and not worry about
output impedances. You get a guaranteed correct result for frequency
response no matter whether the loads are perfectly balanced, or if there is
a slight unbalanced component.

I've posted an analysis and some plots of the 3 Concertina output
impedances (2 driving point and 1 differential) versus frequency. There's
also a plot of plate and cathode frequency response with a slightly
unbalanced capacitive load.

[The Phantom]

On Wed, 27 Aug 2008 13:15:55 -0500, John Byrns
wrote:

In article ,
The Phantom wrote:

On Fri, 22 Aug 2008 02:36:21 GMT, Patrick Turner
wrote:

By the time of RDH4 (1952), there is discussion of impedances. RDH4 says,
on page 330:

"The effective output resistance is different for the two output channels,
since P operates with current feedback and K with voltage feedback."

They then give expressions for the "effective output resistance" (what they
are referring to is what nowadays would be called the "driving point
impedance") in formulas 30 and 31, but I'm sure that formula 30 is
incorrect.

Equation 30 is what effective Rp' of the tube is at the anode.
Its quite correct.

The R looking into the anode of the triode is Rp' and with an unbypassed
Rk,
is exactly what they say, Rp + ( [µ + 1]Rk.

The Rout from the anode at the anode if tested separately would be the
above Rpp' figure
in parallel with whatever anode RL you have.

In my copy of RDH4, on page 330 they have two lines like this:

"Channel P: rp' = (u - 1)Rk where Rk = RL (30)
(from equation 25a, Sect 1.)"

But on page 313, equation 25a says:

"rp' = rp + (u +1)R3 (25a)"

How do they get equation 30 from equation 25a?

Seems straight forward to me, on page 330 of my copy of the RDH equation 30 for
"Channel P" is shown as:

rp' = rp + u * Rk

My copy of RDH4 has:

rp' = (u - 1)Rk for equation 30

Is it really the case that your RDH is different from mine?

There is a downloadable copy available on the web, and it is the same as my
printed copy. What copyright/printing date is your copy?


Equation 25a on page 313 of my copy of the RDH is shown as:

rp' = rp + u * R3

My copy of RDH4 on page 313 for equation 25a has:

rp' = rp + (u + 1)R3

but you are saying that your copy has (and I've cut and pasted from just
above):

rp' = rp + u * R3

They're not the same. How can this be?


These two equations appear virtually identical to me. The problem I see is that
while they are the same, they differ from equation 5 in the Byrns report which I
believe to be the correct version, and is as follows:

rp' = rp + (u + 1) * Zk

[The Phantom]

On Wed, 27 Aug 2008 20:48:59 -0500, flipper wrote:

On Tue, 26 Aug 2008 11:37:26 -0700, The Phantom
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.


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.

What you are describing here is the driving point impedance; what flipper
is describing is the differential impedance.

No, what I described is a Thevenin style 'two' output impedance model.

Do you mean that there are two output ports, each having the same output
impedance, and the output voltage at each port is the same magnitude as the
other, but the opposite polarity?



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.

[Patrick Turner]

The Phantom wrote:

The reason for wanting to know the Concertina "output impedance" is to be
able calculate the effect of further loading on the frequency response.

If "equal output impedance" model is used, then how will the effect of
unequal loading be calculated? This is a very real concern, because the
stray capacitances at the plate and cathode will not likely be identical.

And, of course, a model which only deals with the unequal driving point
impedances will not properly give the effect of the balanced loading which
will dominate the loads (we hope).

It's not necessary to use only one of these. Better to do a detailed
analysis using Shekel's method such as the one I posted a week or so ago.

From this analysis, expressions for the gain from grid to plate and to
cathode are derived. These expressions can then be used to calculate the
frequency responses from grid to plate and cathode regardless of whether
the loads are balanced or not; the correct result will be obtained for all
loads.

Using the following symbols:

ra = plate resistance of the tube

gm = transconductance of the tube

Zp = additional load applied to plate; include all loads in this,
including such things as stray capacitances.

Zk = additional load applied to cathode.

Then the grid to plate gain is given by:

ra*gm*Zp
- -------------------
(ra*gm+1)*Zk+Zp+ra

The grid to cathode gain is given by:

ra*gm*Zk
-------------------
(ra*gm+1)*Zk+Zp+ra

If Zp = Zk, then these expressions are identical, which suggests that the
impedances at plate and cathode are equal before additional identical loads
are applied. This is quite true, because as I've explained elsewhere in
this thread, applying identical loads simultaneously to plate and cathode
is exactly equivalent to applying a differential (balanced) load to the
Concertina, and we would expect that would not upset the equality of gains
at plate and cathode.

Knowing the output impedances allows us to calculate frequency response by
treating the outputs as sources with some impedance and then using a
voltage divider formula to calculate the effect of loads. If the loads are
perfectly balanced, then the appropriate impedance to use for these
calculations is what I've called the "half differential output impedance".
But what if the loads aren't perfectly balanced? Then I suppose we could
separate the load into a balanced component and an unbalanced component,
solve separately and use superposition to get the final result.

It's much easier to just use the gain expressions above and not worry about
output impedances. You get a guaranteed correct result for frequency
response no matter whether the loads are perfectly balanced, or if there is
a slight unbalanced component.

I've posted an analysis and some plots of the 3 Concertina output
impedances (2 driving point and 1 differential) versus frequency. There's
also a plot of plate and cathode frequency response with a slightly
unbalanced capacitive load.

I guess all this deserves some sort of Concertina Medal for the
most serious attempt to mathematically define what happens above say
20kHz to a CPI
as a result of capacitance effects due to strays, Miller, and loading
from following
tube Miller C etc.

I just observe the roll offs one gets, after getting the CPI
to work as well as i want it to at 1 kHz.
Whatever will be will be.

If a CPI drives an output stage, as in the case of a Dynaco ST70,
then the Miller C "seen by" the CPI varies with output load because gain
changes
with output load, and the load change is actually dynamically changing
all the time
because F changes, and speaker Z changes with F.
Miller C reduces with a C load at the output, so tubes should drive ESL
speakers well.
Because speaker loads are not so consistent and simple as a pure
resistance dummy load,
I know I could work it all out mathematically if I tried, but the real
situation does not make me want to try.

In a Williamson, the balanced drive amp buffers the CPI from
both output tube Miller C and the onset of grid I of the output tubes.
Meanwhile the CPI buffers the VI input triode from the Miller C of the
balanced amp.

Last time I built a Williamson with 6CG7 throughout, and measured F pole
at the output tube grids the pole was above 250kHz, and above anything
else I have ever tried.

The Williamson line up is a fast circuit.
Somewhat too fast perhaps, because Willy amps tend to be HF unstable
with a pure 0.1uF to 0.47uF load, so HF gain has to be reduced with a
shelving network
to reduce HF gain above 30kHz.

The CPI output resistance below 20kHz is a simple thing to understand if
you make one assumption,
and that is that RL at a and k will always be equal, and all tests to
measure Rout at a or k are to be made
with RL at a and k always remaining equal, so hence Rout at a = Rout at
k.



Patrick Turner.

[John Byrns]

In article ,
The Phantom wrote:

On Wed, 27 Aug 2008 13:15:55 -0500, John Byrns
wrote:

In article ,
The Phantom wrote:

On Fri, 22 Aug 2008 02:36:21 GMT, Patrick Turner
wrote:

By the time of RDH4 (1952), there is discussion of impedances. RDH4
says,
on page 330:

"The effective output resistance is different for the two output
channels,
since P operates with current feedback and K with voltage feedback."

They then give expressions for the "effective output resistance" (what
they
are referring to is what nowadays would be called the "driving point
impedance") in formulas 30 and 31, but I'm sure that formula 30 is
incorrect.

Equation 30 is what effective Rp' of the tube is at the anode.
Its quite correct.

The R looking into the anode of the triode is Rp' and with an unbypassed
Rk,
is exactly what they say, Rp + ( [µ + 1]Rk.

The Rout from the anode at the anode if tested separately would be the
above Rpp' figure
in parallel with whatever anode RL you have.

In my copy of RDH4, on page 330 they have two lines like this:

"Channel P: rp' = (u - 1)Rk where Rk = RL (30)
(from equation 25a, Sect 1.)"

But on page 313, equation 25a says:

"rp' = rp + (u +1)R3 (25a)"

How do they get equation 30 from equation 25a?

Seems straight forward to me, on page 330 of my copy of the RDH equation 30
for
"Channel P" is shown as:

rp' = rp + u * Rk

My copy of RDH4 has:

rp' = (u - 1)Rk for equation 30

Is it really the case that your RDH is different from mine?

Yes, it is really true if your copy doesn't show equation 30 on page 330 as:

"rp' = rp + uRk"

There is a downloadable copy available on the web, and it is the same as my
printed copy. What copyright/printing date is your copy?

The copyright notice in my copy does not specify a date, various dates are
mentioned elsewhere including 1952, 1953, and February 1954. The latter being
the date it was reproduced in the USA for RCA Victor.

I can find none of the usual numbers indicating what printing my copy might be.

Equation 25a on page 313 of my copy of the RDH is shown as:

rp' = rp + u * R3

My copy of RDH4 on page 313 for equation 25a has:

rp' = rp + (u + 1)R3

but you are saying that your copy has (and I've cut and pasted from just
above):

rp' = rp + u * R3

It appears exactly as:

"rp' = rp + uR3"

They're not the same. How can this be?

I guess it was one of the wonders of modern printing technology circa 1954.

--
Regards,

John Byrns

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

[The Phantom]

There is a downloadable RDH4 at:

http://headfonz.rutgers.edu/RDH4/

I checked chapter 7, and it appears to be exactly like my printed copy.

If you want a version to compare, there it is.

If I look carefully, it's apparent that the typeface of equation 30 in my copy
is different than that of equation 31.

Changes were made somewhere along the way.

Does your copy have an equation 34a?

It appears to have been derived from the paper by George Jones that is available
at http://www.diybanter.com/attachment....2&d=1213179423

[The Phantom]

On Wed, 27 Aug 2008 01:37:29 GMT, Patrick Turner
wrote:



The Phantom wrote:

On Fri, 22 Aug 2008 02:36:21 GMT, Patrick Turner
wrote:



By the time of RDH4 (1952), there is discussion of impedances. RDH4 says,
on page 330:

"The effective output resistance is different for the two output channels,
since P operates with current feedback and K with voltage feedback."

They then give expressions for the "effective output resistance" (what they
are referring to is what nowadays would be called the "driving point
impedance") in formulas 30 and 31, but I'm sure that formula 30 is
incorrect.

Equation 30 is what effective Rp' of the tube is at the anode.
Its quite correct.

The R looking into the anode of the triode is Rp' and with an unbypassed
Rk,
is exactly what they say, Rp + ( [µ + 1]Rk.

The Rout from the anode at the anode if tested separately would be the
above Rpp' figure
in parallel with whatever anode RL you have.

In my copy of RDH4, on page 330 they have two lines like this:

"Channel P: rp' = (u - 1)Rk where Rk = RL (30)
(from equation 25a, Sect 1.)"

In my RDH4, Pg 330, equation 30 reads...

Channel P rp' = rp + ( µ + 1 )Rk where Rk = RL.
( from eqtn 25A, Sect 1 ).

You have misquoted Equation 30.

In another post, John Byrns says that his copy has:

rp' = rp + u * Rk

for equation 30.

My copy of RDH4 has:

rp' = (u - 1)Rk where Rk = RL

which is the same as the downloadable copy of RDH4 at
http://headfonz.rutgers.edu/RDH4/ has.

and you (Patrick) say that yours has:

rp' = rp + ( µ + 1 )Rk where Rk = RL

It would seem that there are at least 3 different expressions for equation
30 in various editions/printings of RDH4.

I wonder what other differences there are?




But on page 313, equation 25a says:

"rp' = rp + (u +1)R3 (25a)"

How do they get equation 30 from equation 25a?

There is no inconsistency. R3 in eqtn 25a is Rk in eqtn 30.


RDH4 gives what is exactly right for the Rout at the anode for the CPI
if measured
as a separate outlet to power some following stage.

Ditto for the cathode Rout as spelled out in eqtn (31).

Rout at Channel K : rp' = ( rp + RL ) / ( µ + 1 )

If you follow my treatise in a recent post, you'll see what Rout of K
and P
will be where always RL = RK as in a real amp using CPI....

[John Byrns]

In article ,
The Phantom wrote:

On Wed, 27 Aug 2008 01:37:29 GMT, Patrick Turner
wrote:

The Phantom wrote:

On Fri, 22 Aug 2008 02:36:21 GMT, Patrick Turner
wrote:

By the time of RDH4 (1952), there is discussion of impedances. RDH4
says,
on page 330:

"The effective output resistance is different for the two output
channels,
since P operates with current feedback and K with voltage feedback."

They then give expressions for the "effective output resistance" (what
they
are referring to is what nowadays would be called the "driving point
impedance") in formulas 30 and 31, but I'm sure that formula 30 is
incorrect.

Equation 30 is what effective Rp' of the tube is at the anode.
Its quite correct.

The R looking into the anode of the triode is Rp' and with an unbypassed
Rk,
is exactly what they say, Rp + ( [µ + 1]Rk.

The Rout from the anode at the anode if tested separately would be the
above Rpp' figure
in parallel with whatever anode RL you have.

In my copy of RDH4, on page 330 they have two lines like this:

"Channel P: rp' = (u - 1)Rk where Rk = RL (30)
(from equation 25a, Sect 1.)"

In my RDH4, Pg 330, equation 30 reads...

Channel P rp' = rp + ( µ + 1 )Rk where Rk = RL.
( from eqtn 25A, Sect 1 ).

You have misquoted Equation 30.

In another post, John Byrns says that his copy has:

rp' = rp + u * Rk

for equation 30.

My copy of RDH4 has:

rp' = (u - 1)Rk where Rk = RL

which is the same as the downloadable copy of RDH4 at
http://headfonz.rutgers.edu/RDH4/ has.

and you (Patrick) say that yours has:

rp' = rp + ( µ + 1 )Rk where Rk = RL

It would seem that there are at least 3 different expressions for equation
30 in various editions/printings of RDH4.

I wonder what other differences there are?

That was exactly the thought that came to my mind too.

--
Regards,

John Byrns

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

[Spike[_2_]]

Henry,

Good to see you are alive and kicking.

Pop me and email.

Spike

Henry Pasternack wrote:
"The Phantom" wrote in message
...
You, like Byrns, seem to be a pretty pedantic fellow. And that's fine,
but I have a different focus.
What is that focus?

I wrote a lengthy respose, but decided not to post it. If you would like
to see what I had to say, you are welcome to email me privately.

-Henry

moc.ncr@kcanretsaph

[Spike[_2_]]

Testing 1, 2 3.

Spike wrote:
Henry,

Good to see you are alive and kicking.

Pop me and email.

Spike

Henry Pasternack wrote:
"The Phantom" wrote in message
...
You, like Byrns, seem to be a pretty pedantic fellow. And that's fine,
but I have a different focus.
What is that focus?

I wrote a lengthy respose, but decided not to post it. If you would like
to see what I had to say, you are welcome to email me privately.

-Henry

moc.ncr@kcanretsaph