Re: Phase splitter with capacitance load.

I noticed a couple of errors in Preisman's paper. The last sentence of the
first paragraph on the second page reads:

"...we must use Eqs. (8) and (9) separately rather than use Eq. (11) for
both output voltages."

but it should read:

"...we must use Eqs. (9) and (10) separately rather than use Eq. (11) for
both output voltages."

Also, in Fig. 4, right hand part he has an expression Zg+(m+1)/m*ZL. This
should be Zg+(m-1)/m*ZL as it is in the text.

Re-reading Preisman's paper, I realized what it is that I found misleading.
He makes much of "equivalent circuits", Figs. 2, 3 and 4. But his
"equivalent circuits" are not what would be considered equivalent nowadays
without qualification. He is calling circuits "equivalent" if they have
the same output voltage. But, if you consider two of his equivalent
circuits such as the ones in Fig. 4, you will see that even though they
have the same output voltage, EL, as they stand, if a further load is
placed on them they will not behave the same. This is in contradistinction
to a Thevenin equivalent. A Thevenin equivalent behaves the same as the
circuit it replaces with respect to changing loads; Preisman's equivalent
circuits don't.

The IEEE Dictionary of electrical terms defines impedance, as it applies to
passive networks, like this (not an exact quote):

Inject a current into a node of a network and measure the voltage thus
produced at that node. Divide the voltage by the current. This is the
driving point impedance.

Inject a current into a node of a network and measure the voltage thus
produced at some other node. Divide the voltage by the current and this is
the transfer impedance to the other node.

Consider the following resistor network. It has 3 nodes and 4 resistors.

Network 1

2 ___ 1 ___ 3
.----|___|---.-----|___|---.
| 4 | 4 |
| In |
.-. .-.
| | | |
2| | | |2
'-' '-'
| |
-------------.--------------
|
GND

We'll apply a voltage to node 1 and measure the output voltages at nodes 2
and 3. The ratio of the voltage at node 2 or 3 to the voltage at node 1
will be called the gain to that output node. (Strictly, it should be
called the open circuit voltage transfer ratio.) Since this network is
intended to be driven by a voltage source, we must ground node 1 when
calculating (or measuring) the output impedance at node 2 or 3.

Let's apply 1 volt to node 1 and calculate the voltages at nodes 2 and 3.
The output voltage would be 2/6 or .3333333333 volts. The gain is 1/3 to
both nodes 2 and 3.

Ground node 1 and calculate the resistance at nodes 2 and 3. It's 2||4
ohms, or 1.33333333 ohms. Now let's apply some additional load at nodes 2
and 3, sufficient to reduce the gain to 1/2 its previous value. One would
think that a resistor equal to the output resistance of the network would
do the job. So, connect a 1.33333333 ohm resistor in parallel with each of
the 2 ohm resistors, and calculate the gain to nodes 2 and 3. It's
..166666667, which is indeed half the value without the added resistors.

Now consider this network:

Network 2
___
---------|___|----------
| 4 |
| |
| |
2 | ___ 1 ___ | 3
.-'--|___|---.-----|___|-'-.
| 4 | 4 |
| In |
.-. .-.
| | | |
2| | | |2
'-' '-'
| |
-------------.--------------
|
GND

It's the same as the first one except for the additional 4 ohm resistor
between nodes 2 and 3.

The output resistance at nodes 2 and 3 is 1.066666667 ohms. Don't forget
to ground node 1 when making this calculation.

The voltage gain from node 1 to nodes 2 and 3 is .3333333333.

What additional loads applied to nodes 2 and 3 will cause the gain to those
nodes to be 1/2 of its previous value? One might think that it would
simply be resistors equal to the output resistance. Let's try it. Add a
1.06666667 ohm resistor from node 2 and node 3 to ground, and calculate the
gain to those nodes. The result is a gain of .13761468. But we expected a
gain of .166666667; what went wrong?

The problem is that there is coupling between nodes 2 and 3 due to the 4
ohm resistor we added. If we go back to Network 1 and imagine injecting
some current into node 2 and measuring the voltage produced at node 3
(remember, node 1 must be grounded for these measurements), we can easily
see that there will be none. In other words, the transfer impedance is
zero; there is no coupling between nodes 2 and 3.

Do the same with Network 2; inject 1 amp into node 2 and calculate the
voltage produced at node 3. The result is .266666667 volts, and since 1
amp was injected, the transfer resistance is .266666667 ohms. Does this
help us find the load which will reduce the voltage gain by 1/2? Yes. The
required load is the sum of the driving point impedance and the transfer
impedance. So, connect a 1.3333333 ohm resistor from nodes 2 and 3 to
ground and calculate the voltage gains to nodes 2 and 3. The result is a
gain of .166666667, which is half the gain without the additional load.
(One could have noticed that because of the extreme symmetry of the
circuit, the additional 4 ohm resistor doesn't do much. But it does change
the output resistance at nodes 2 and 3, so from one point of view it's
clear that the same load resistance that we used with Network 1 will also
work with Network 2. But, we also have been told that load resistors equal
to the output resistance should reduce the gain by half. Using the
transfer resistance leads us back to the correct loads.)

But, we calculated an output resistance of 1.06666667 ohms; why did we need
loads of 1.3333333 ohms to reduce the gain by half? With Network 1, the
required load was equal to the calculated output resistance. Does this
mean that the output resistance of the network changes to 1.33333333 ohms
when we add two loads instead of just one?

No. The problem is that the classical voltage divider formula only works
for one output at a time. In fact, if you add a 1.066666667 ohm resistor
from node 2 to ground (but not from node 3 at the same time) and calculate
the gain to node 2, it will be .166666667, half the previous value. The
1.33333333 ohm value is not an output impedance; it is the sum (it would be
the difference if one of the gains had a negative sign) of the output
impedance and the transfer impedance. It is a number which will give you
the voltage gain when used where you shouldn't use it, in the voltage
divider formula, which is only guaranteed valid for one output at a time.
We can see that the voltage divider formula, using the actual output
impedance, is not ALWAYS valid when used for more than one output at a
time. Where it is invalid is when the network has non-zero transfer
impedances. In that case we can derive a another number (which is not the
output impedance) which will work in the voltage divider formula for two
outputs at a time. But this does not mean that the number is the output
impedance of the network. And, by the way, the only time one such number
will work for two outputs simultaneously is when the network has a high
degree of symmetry, like Network 2.

Network theory teaches that passive networks like these are completely
characterized by their driving point impedances (this is what would
normally be meant by the term output impedance) and their transfer
impedances.

In a thread back in 2001, Henry Pasternack did some example calculations
for the concertina phase splitter:
-------------------------------------------------------------------------
I suggest we plug in some values and form a practical example:

Zp = Zk = 10K
rp = 1K
u = 10

These are extraordinary values for a tube, but within reason, and easy
to push through the calculator.

For the test, we will set the source voltage to 1.0V and we will use
a test load of 990K Ohms. Note that the value of 990K in parallel with
Zp = Zk = 10K is 9.9K. We will measure the no-load voltage first, and
then observe what happens at each output when we apply the test load
first to one output, then to the other, then to both. Finally, we will
use the resistor divider equation to compute the apparent source
impedances.

Because the load resistor is so high, I'm carrying the calculations
out to many decimal places so as not to lose track of the differences.
I will call the case where the load resistor is on the leg under test
only the "proximal" case. With the resistor on the opposite leg only,
I will call it the "distal" case. With a load on both legs, I will
call it the "dual" case. To compute the apparent source impedances,
I will use the equation:

Zo = Rtest * (1 - r) / r

where 'r' is the test output divided by the no-load output

Here are the results:

Cathode circuit:

1) No load: Vout = 0.826446281
2) Proximal: Vout = 0.825688073 Zout = 909.1 Ohms
3) Distal: Vout = 0.827129860
4) Dual: Vout = 0.826377296 Zout = 82.64 Ohms

Plate circuit:

1) No load: Vout = 0.826446281
2) Proximal: Vout = 0.818858561 Zout = 9174 Ohms
3) Distal: Vout = 0.834028357
4) Dual: Vout = 0.826377296 Zout = 82.64 Ohms
---------------------------------------------------------------------

For these values:

Zp = Zk = 10K
rp = 1K
u = 10

the impedance matrix for the circuit without the added 990k loads, which
has the driving point impedances on the main diagonal and the transfer
impedances off diagonal, is:

[ 909.09091 826.44628 ]
[ 9090.9091 9173.5537 ]

Notice that the difference (difference, not sum, because of the sign change
at the plate) of the cathode driving point impedance and transfer impedance
is 909.09091 - 826.44628 = 82.64463 and the difference of the plate driving
point impedance and transfer impedance is 9173.5537 - 9090.9091 = 82.6446.
This is the "magic" number, which when used "illegally" in the voltage
divider formula applied to both outputs at once, will in fact give the
right gains. If the circuit had not been symmetrical, there would have
been two "magic" numbers.

This could be called an "apparent" output impedance, but it isn't a driving
point impedance at all. It is the difference of a driving point impedance
and a transfer impedance. It also happens to be half the impedance that
would be measured between plate and cathode. It's also the number that
would be given by an application of formula (34a) on page 330 of RDH4.

Loading both outputs of the phase splitter at once doesn't change its
driving point (output) impedances (before the loads are applied, of
course). The circuit behavior after the loads have been applied can be
calculated using the output impedances as calculated individually, one at a
time. But, of course, the calculation must also include the transfer
impedances, since they are not zero. A matrix solution of the circuit does
this automatically, and is guaranteed to give the right answers. It's very
easy to add parasitic capacitances. I recommend it.

The driving point impedances (output impedances) before any loads are
applied are plate impedance = 9173.5537 ohms and cathode impedance =
909.0909 ohms, as seen on the main diagonal of the impedance matrix.

.