Re: Audio Cyclopedia - A highly recommended book

In article <qldra4pkrh3hqvkgjdmff797o26gub2ktl@xxxxxxx>,
The Phantom <phantom@xxxxxxx> wrote:

On Wed, 20 Aug 2008 22:09:04 -0500, flipper <flipper@xxxxxxxx> wrote:

On Wed, 20 Aug 2008 17:33:34 -0700, The Phantom <phantom@xxxxxxx>
wrote:

On Tue, 05 Aug 2008 13:04:52 -0500, John Byrns <byrnsj@xxxxxxxxxxxxx>
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/product/72-960/LCR-%20Capacitance%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/
.

Relevant Pages

  • Re: Gain equations : ref frequency response
    ... A flat frequency response can only exist into a pure resitive load. ... This only applies to circuits with the load in the plate circuit. ... > Graham, if you reallly want to discuss something here, ... > and as long as that stays well below reflected impedance it is ...
    (rec.audio.tubes)
  • Re: Some thoughts relevant to measuring Tx eq src impedance
    ... source impedance of a transmitter with such varying results. ... must therefore be the conjugate of the load. ... The required anode load resistance is the resistance calculated from the ... impedance to small changes in the external 50 ohm load is different. ...
    (rec.radio.amateur.antenna)
  • Re: AC measures 27volts
    ... You are using a low impedance meter ... ... The part where he said "a high impedance meter can load a circuit". ...
    (alt.home.repair)
  • Re: Shorting out a transmission line
    ... >actual numbers for the systems's behavior at those frequencies I can ... Short circuit: a male BNC plug, with a short inserted between center ... check quality of T-line and load, ... creating an abrupt change in the load impedance. ...
    (rec.radio.amateur.antenna)
  • Re: Audio Cyclopedia - A highly recommended book
    ... He wrote "The circuit behaves as if the output resistance at both ... differential load ... the transfer impedance from node i to ... The meter will inject a test current from one of its leads while ...
    (rec.audio.tubes)