[The Phantom]

I think much of this controversy is due to imprecise terminology.

Imagine that you built a phase inverter consisting of two amplifiers, one with a
gain of +1 and one with a gain of -1. Let the amplifiers have an output
impedance of 100 ohms. They could be modelled as ideal amplifiers with zero
ohms output impedance, in series with discrete 100 ohm resistors.

Now, consider measuring some output impedances. I'm not going to consider
complex impedances initially.

The IEEE dictionary of electronic terms calls the impedance at a single node of
an arbitrary circuit, "driving point impedance".

The driving point impedance at a node is defined as the ratio of an applied
voltage to the current that results. The driving point impedance can be an
"output impedance" or an "imput impedance". All other nodes in the circuit must
be left alone.

If it's an output impedance, it can also be measured by applying a signal to
another node (in audio, probably an "input" node) and measuring the signal at
the output node without any additional load. Then apply a variable load to the
node under consideration and vary it until the voltage at that node is reduced
to 1/2 the unloaded voltage. The value of the additional applied load is the
driving point impedance (which would be commonly called the "output impedance")
at that node.

So, if we were to measure the impedance of each of the amplifiers I mentioned
above, we would get 100 ohms.

What if we applied two 100 ohm resistors, one to each amplifier output, with the
other end of each resistor connected to ground. Now lift the grounded ends of
the resistors and connect them together. If some audio is applied to the inputs
of the two amplifiers, we will find that there is no signal at the junction of
the two 100 ohm resistors, so whether these two resistors are grounded, or
simply connected in series across the outputs of the two amplifiers makes no
difference.

The series connection of the two 100 ohm resistors across the outputs of the two
amplifiers would commonly be referred to as a differential load. Since when the
two resistors were separately grounded, the outputs of the amplifiers were
reduced to half, we took this fact as a measurement of the output impedances of
the amplifiers to be 100 ohms each.

But, when we raised the grounded ends of the resistors off ground and connected
them together, we had a 200 ohm differential load on the amplifiers, and their
outputs were still reduced to half. Apparently the *differential* output
impedance of the amplifier pair is 200 ohms. This makes sense, since the
amplifiers have a 100 ohm output impedance each.

Note that this differential measurement is not a "driving point impedance"
according to the IEEE definition, because we are doing something to more than
one node of the circuit.

Now consider the Concertina phase inverter. Apply a small 1 kHz signal to the
input and apply equal resistive loads to the plate and cathode. Imagine an
ideal pair of ganged potentiometers (wired as rheostats) that track perfectly,
as your load. One end of each pot is grounded. Adjust the pots until the
output at plate and cathode are reduced to 1/2 the unloaded value. The output
voltages at plate and cathode will track perfectly due to the fact that the
current in the plate load is identical to the current in the cathode load; this
has been understood since before RDH4.

If the grounded ends of the potentiometers are lifted from ground and connected
together, this junction will never see any signal, since the signals at the
other ends of each pot are equal and of opposite polarity, and the behavior of
the ciruit will be the same whether that junction is grounded or not.

If the value of the pots which reduced the output to 1/2 half its unloaded value
is R ohms each, when the grounded ends are lifted, the pots will present a
resistance of 2R between plate and cathode. This 2R resistance reduces the
output at plate and cathode to 1/2 its open circuit value, and therefore 2R ohms
is the *differential* output impedance of the Concertina. We can assign R ohms
to each output.

If you use one of those cute LCR meters that look like a handheld DVM, you could
connect it between plate and cathode and directly measure the differential
impedance. And, of course, there are various circuit analysis techniques to
mathematically calculate the differential output impedance.

It is ambiguous to say that the "output impedances" of both plate and cathode
nodes are R ohms when *both* are loaded. In most areas of electionics, the
unqualified term "output (or input) impedance" refers to a driving point
impedance, where the impedance is measured at one node only. Have a look at
almost any opamp data sheet. You'll see differential input impedance and common
mode input impedance specified. Common mode input impedance is measured by
connecting both inputs together and measuring the impedance to that single node.
If two nodes are involved, then it's a *differential* impedance; a
"*differential* output impedance" in the case of the Concertina.

I recommend a little redundant language for the sake of eliminating any possible
ambiguity.

If an impedance is measured at one node only, with all other nodes left alone,
then it should be called a "driving point impedance" if there is a possibility
that there might be ambiguity, which there certainly is in the case of the
Concertina.

If it's measured betweeen a pair of nodes, then it should be referred to as a
"differential output (or input) impedance".

Since the purpose of the Concertina is to provide a *pair* of signals, out of
phase with one another, the *differential* output voltage is all important.

It's only if the individual loads applied to plate and cathode are identical,
that we can lift their grounds and connect the previously grounded ends of the
loads together, forming a differential load on the Concertina. Obviously, a
differential load will never upset the fact that the output voltages at plate
and cathode are the same, just 180 out of phase.

The differential output impedance of the pair of +1 and -1 gain amplifiers at
the beginning of this post was exactly twice the driving point output
impedances. This is not the case in the Concertina, because the vacuum tube
provides some coupling between plate circuit and cathode circuit.

The driving point impedances at plate and cathode of the Concertina are
different from each other, and different from the differential output impedance,
but this has no bearing on the behavior for differential (balanced) loads. This
difference *does* have bearing on the behavior of the Concertina for
*unbalanced* loads, such as capacitance not present in equal amounts at each
output.

When the normally balanced load presented by the push-pull output stage becomes
unbalanced due to grid conduction, etc., then the circuit performance will be
degraded. I don't speak to that.

So, to summarize: to use the unqualified phrase "output impedance" in connection
with the Concertina is ambiguous. There are 3 impedances of interest in this
circuit. There is the differential impedance *between* plate and cathode, and
there is the driving point impedance at plate and at cathode.

If you want to know the corner frequency caused by equal capacitances at plate
and cathode (two such capacitances can be treated as a differential load), then
the relevant resistance is 1/2 the differential output impedance.

We use 1/2 the differential output impedance since by assigning 1/2 to each
output, we can calculate the rolloff caused by equal capacitances applied to the
individual outputs using that 1/2 differential output resistance.

This resistance is given by formula 34a in RDH4, on page 330 (by the way,
formula 30 on that page seems to be wrong). Apparently, RDH4 got the expression
from the paper referenced by Flipper:

http://www.diybanter.com/attachment....2&d=1213179423

Formula 34a in RDH4 can be derived with a little algebra from the 2nd formula in
the 3rd column of that paper.

The differential output impedance is probably the most important impedance in
connection with the Concertina, but it should not be called simply "the output
impedance"; call it the "half differential output impedance" (at each output)
and avoid amgiguity (and, hopefully, controversy).

[John Byrns]

Hi Rodger,

Great post. Could you elaborate on the IEEE definition of "driving point
impedance"? I don't understand how you can measure any "impedance at a single
node" of a circuit; obviously there is an implied reference node involved in the
measurement. I am curious how the IEEE defines and restricts our choice of this
reference node?

In article ,
The Phantom wrote:

I think much of this controversy is due to imprecise terminology.

Imagine that you built a phase inverter consisting of two amplifiers, one
with a
gain of +1 and one with a gain of -1. Let the amplifiers have an output
impedance of 100 ohms. They could be modelled as ideal amplifiers with zero
ohms output impedance, in series with discrete 100 ohm resistors.

Now, consider measuring some output impedances. I'm not going to consider
complex impedances initially.

The IEEE dictionary of electronic terms calls the impedance at a single node
of
an arbitrary circuit, "driving point impedance".

The driving point impedance at a node is defined as the ratio of an applied
voltage to the current that results. The driving point impedance can be an
"output impedance" or an "imput impedance". All other nodes in the circuit
must
be left alone.

If it's an output impedance, it can also be measured by applying a signal to
another node (in audio, probably an "input" node) and measuring the signal at
the output node without any additional load. Then apply a variable load to
the
node under consideration and vary it until the voltage at that node is
reduced
to 1/2 the unloaded voltage. The value of the additional applied load is the
driving point impedance (which would be commonly called the "output
impedance")
at that node.

So, if we were to measure the impedance of each of the amplifiers I mentioned
above, we would get 100 ohms.

What if we applied two 100 ohm resistors, one to each amplifier output, with
the
other end of each resistor connected to ground. Now lift the grounded ends
of
the resistors and connect them together. If some audio is applied to the
inputs
of the two amplifiers, we will find that there is no signal at the junction
of
the two 100 ohm resistors, so whether these two resistors are grounded, or
simply connected in series across the outputs of the two amplifiers makes no
difference.

The series connection of the two 100 ohm resistors across the outputs of the
two
amplifiers would commonly be referred to as a differential load. Since when
the
two resistors were separately grounded, the outputs of the amplifiers were
reduced to half, we took this fact as a measurement of the output impedances
of
the amplifiers to be 100 ohms each.

But, when we raised the grounded ends of the resistors off ground and
connected
them together, we had a 200 ohm differential load on the amplifiers, and
their
outputs were still reduced to half. Apparently the *differential* output
impedance of the amplifier pair is 200 ohms. This makes sense, since the
amplifiers have a 100 ohm output impedance each.

Note that this differential measurement is not a "driving point impedance"
according to the IEEE definition, because we are doing something to more than
one node of the circuit.

Now consider the Concertina phase inverter. Apply a small 1 kHz signal to
the
input and apply equal resistive loads to the plate and cathode. Imagine an
ideal pair of ganged potentiometers (wired as rheostats) that track
perfectly,
as your load. One end of each pot is grounded. Adjust the pots until the
output at plate and cathode are reduced to 1/2 the unloaded value. The
output
voltages at plate and cathode will track perfectly due to the fact that the
current in the plate load is identical to the current in the cathode load;
this
has been understood since before RDH4.

If the grounded ends of the potentiometers are lifted from ground and
connected
together, this junction will never see any signal, since the signals at the
other ends of each pot are equal and of opposite polarity, and the behavior
of
the ciruit will be the same whether that junction is grounded or not.

If the value of the pots which reduced the output to 1/2 half its unloaded
value
is R ohms each, when the grounded ends are lifted, the pots will present a
resistance of 2R between plate and cathode. This 2R resistance reduces the
output at plate and cathode to 1/2 its open circuit value, and therefore 2R
ohms
is the *differential* output impedance of the Concertina. We can assign R
ohms
to each output.

If you use one of those cute LCR meters that look like a handheld DVM, you
could
connect it between plate and cathode and directly measure the differential
impedance. And, of course, there are various circuit analysis techniques to
mathematically calculate the differential output impedance.

It is ambiguous to say that the "output impedances" of both plate and cathode
nodes are R ohms when *both* are loaded. In most areas of electionics, the
unqualified term "output (or input) impedance" refers to a driving point
impedance, where the impedance is measured at one node only. Have a look at
almost any opamp data sheet. You'll see differential input impedance and
common
mode input impedance specified. Common mode input impedance is measured by
connecting both inputs together and measuring the impedance to that single
node.
If two nodes are involved, then it's a *differential* impedance; a
"*differential* output impedance" in the case of the Concertina.

I recommend a little redundant language for the sake of eliminating any
possible
ambiguity.

If an impedance is measured at one node only, with all other nodes left
alone,
then it should be called a "driving point impedance" if there is a
possibility
that there might be ambiguity, which there certainly is in the case of the
Concertina.

If it's measured betweeen a pair of nodes, then it should be referred to as a
"differential output (or input) impedance".

Since the purpose of the Concertina is to provide a *pair* of signals, out of
phase with one another, the *differential* output voltage is all important.

It's only if the individual loads applied to plate and cathode are identical,
that we can lift their grounds and connect the previously grounded ends of
the
loads together, forming a differential load on the Concertina. Obviously, a
differential load will never upset the fact that the output voltages at plate
and cathode are the same, just 180 out of phase.

The differential output impedance of the pair of +1 and -1 gain amplifiers at
the beginning of this post was exactly twice the driving point output
impedances. This is not the case in the Concertina, because the vacuum tube
provides some coupling between plate circuit and cathode circuit.

The driving point impedances at plate and cathode of the Concertina are
different from each other, and different from the differential output
impedance,
but this has no bearing on the behavior for differential (balanced) loads.
This
difference *does* have bearing on the behavior of the Concertina for
*unbalanced* loads, such as capacitance not present in equal amounts at each
output.

When the normally balanced load presented by the push-pull output stage
becomes
unbalanced due to grid conduction, etc., then the circuit performance will be
degraded. I don't speak to that.

So, to summarize: to use the unqualified phrase "output impedance" in
connection
with the Concertina is ambiguous. There are 3 impedances of interest in this
circuit. There is the differential impedance *between* plate and cathode,
and
there is the driving point impedance at plate and at cathode.

If you want to know the corner frequency caused by equal capacitances at
plate
and cathode (two such capacitances can be treated as a differential load),
then
the relevant resistance is 1/2 the differential output impedance.

We use 1/2 the differential output impedance since by assigning 1/2 to each
output, we can calculate the rolloff caused by equal capacitances applied to
the
individual outputs using that 1/2 differential output resistance.

This resistance is given by formula 34a in RDH4, on page 330 (by the way,
formula 30 on that page seems to be wrong). Apparently, RDH4 got the
expression
from the paper referenced by Flipper:

http://www.diybanter.com/attachment....2&d=1213179423

Formula 34a in RDH4 can be derived with a little algebra from the 2nd formula
in
the 3rd column of that paper.

The differential output impedance is probably the most important impedance in
connection with the Concertina, but it should not be called simply "the
output
impedance"; call it the "half differential output impedance" (at each output)
and avoid amgiguity (and, hopefully, controversy).

--
Regards,

John Byrns

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

[The Phantom]

On Wed, 20 Aug 2008 13:12:29 -0500, John Byrns
wrote:

Hi Rodger,

Great post. Could you elaborate on the IEEE definition of "driving point
impedance"? I don't understand how you can measure any "impedance at a single
node" of a circuit; obviously there is an implied reference node involved in the
measurement. I am curious how the IEEE defines and restricts our choice of this
reference node?

The reference node is indeed implied, but it's up to the person making the
measurement to pick an appropriate reference node. All measurements and
stimuli are with respect to this reference node. The IEEE doesn't say
anything about restrictions.

The driving point impedance is defined like this: inject a current into a
single node (call it node A) with respect to a "reference" node. Measure
the voltage at node A resulting from the injection of current. Divide the
resulting voltage by the stimulus current. This is the driving point
impedance.

There is also transfer impedance (and the related parameter, transfer
conductance, or transconductance), which is the ratio of the voltage at
another node with respect to the current injected at the first node.

Differential impedance would be measured (or calculated) by injecting a
current, i, into node A while simultaneously extracting the same current
from node B (this is what would happen if a small handheld LCR meter were
connected between node A and node B). Measure the voltages induced at node
A, v(A), and node B, v(B), with respect to the reference node. Calculate
(v(A)-v(B))/i; this is the differential impedance between node A and node
B.

[The Phantom]

On Wed, 20 Aug 2008 13:12:29 -0500, John Byrns wrote:

Hi Rodger,

Great post. Could you elaborate on the IEEE definition of "driving point
impedance"? I don't understand how you can measure any "impedance at a single
node" of a circuit; obviously there is an implied reference node involved in the
measurement. I am curious how the IEEE defines and restricts our choice of this
reference node?

I've posted a second paper on indefinite admittance matrices over on ABSE.

[John Byrns]

In article ,
The Phantom wrote:

On Wed, 20 Aug 2008 13:12:29 -0500, John Byrns wrote:

Hi Rodger,

Great post. Could you elaborate on the IEEE definition of "driving point
impedance"? I don't understand how you can measure any "impedance at a
single
node" of a circuit; obviously there is an implied reference node involved in
the
measurement. I am curious how the IEEE defines and restricts our choice of
this
reference node?

I've posted a second paper on indefinite admittance matrices over on ABSE.

See my comment on ABSE in the "Audio Cyclopedia" thread.

--
Regards,

John Byrns

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