It is valid statement, but valid only and only for this kind of measurement conditions. Reading such scientific conclusions like yours people tend to "compensate" effects of different output impedances loading phase splitter on different resistances, and as the result are getting opposite results. Readers of SY's article even if they don't understand limitations of measurements and their theoretical models will get better results. That's the point. Great article, well done, very useful.
One can (and I did, and experimentally verified) calculate and measure the two Thevenin output impedances under the "equal load" constraints. I also predicted and experimentally verified the disastrous results of treating those two impedances as different.
You can only calculate, simulate, measure the Thevenin impedance looking into one node when you keep the conditions at the other nodes the same.
So, for example, if you inject a small-signal current into the plate node only, you can calculate, simulate, measure the change in voltage there and take the ratio to get the Thevenin impedance looking into that node.
However, if you disturb the cathode circuit as well, you cannot calculate, simulate, or measure that Thevenin impedance due to the interaction between plate and cathode. Whatever it is you get, it is a mixture of the Thevenin impedance and the transfer impedance from cathode to plate.
Let me say this again: Only by varying conditions at one of the nodes can you determine the Thevenin impedance looking into that node.
This is so elementary that I find it hard to believe that two smart guys are being so hard headed about it. I'm beginning to suspect that this disagreement boils down to semantics. We might be on the same wavelength but don't know it because we're using terms differently. I'm using them correctly by the way ;^)
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The single-node impedances depend on the loading at the other node, so they are properties of the circuit+environment, not the circuit alone. I have not thought about this deeply, but I suspect the four impedances you mentioned are not four independent impedances but two pairs: anode and cathode, differential and common-mode, with one pair being a linear combination of the other pair. Balance makes the common-mode impedance drop out of the equations, so the anode and cathode impedances are then both equal to half the differential impedance.Alfred Centauri said:
Quite the opposite- what I have calculated and measured is exactly in accord with Thevenin's theorem. I have treated the cathodyne as a black box, with one input and two outputs- exactly the way it's used. If we model the two outputs as Thevenin sources, as in my figure 3, we find theoretically and experimentally that the Thevenin source impedance of the two outputs is identical (and approximately equal to the cathode follower impedance) as long as the loads are identical.
This seems to be a very difficult concept for a few people, pretty simple, straightforward, and in accord with experiment for the rest of us. I have asked repeatedly for a pair of identical loads (the boundary condition for this model) which will show the model to be incorrect, but to this point, no-one has brought them forward. All that's being said over and over, and no-one disagrees, is that if the boundary conditions (equal loads) do not apply, the model doesn't work. OK, great, no-one disagrees, you can stop repeating that, now let's see those identical loads that disprove the model.
Give me a Nolble Prize, I add one condition to the theorem: when level of disturbance approaches zero. 😉
And the balance condition makes the other impedance, common-mode, irrelevant.Alfred Centauri said:
Harder loads , I didn't mean lower resistances without any capacitance , I mean ( as I say before ) grids of more than two big power valves like the EL34 for example , which their input capacitances increased drasticaly as frequency and drive voltage increases ( dynamic capacitance ) , your test is at 1Khz and with simply low resistances and without any capacitance even static capacitance .this is incorrect. Here is a concertina loaded with 1gigohm compared to one loaded with 1000 ohms which I think you will agree is much "harder". both the plate and the cathode decrease by the same amount and as I pointed out with the math, the output impedance of both is ~82 ohms which is that of a 6DJ8 cathode follower.
It might help to read the article- and my earlier post. I showed results with capacitances much greater than the grid capacitances of any conceivable output tubes, including EL34.
Well SY, since I don't have that Linear Audio article handy, I'll just have to wing it.
Look, your "boundary" constraint of equal and opposite voltages across the equal plate and cathode loads effectively places the entire output circuit in series since there is no net signal current through ground. By constraining the output currents to be opposite and equal, there is no way to measure the impedance of one node to ground. With equal and opposite output currents under any circumstance, there is only one output impedance that can be measured: the equivalent impedance between the output nodes, i.e, the differential-mode output impedance.
The impedance looking into a node is understood to be the impedance to ground from that node in the same way that a node voltage is understood to be referenced to ground.
But and again, your constraint makes it impossible to measure an impedance to ground from either of the output nodes.
Bottom line: you've measured the impedance looking into one node and out the other. Conceptually, you've place an ohmmeter across the output nodes. You need to be able to conceptually place an ohmmeter across one output node and ground. Your constraint won't allow that since that would imply a current into one node and out through ground rather than through the other node.
Gentlemen, it's been fun. But now, I've got to go out and eat with some old friends.
AC
Look, your "boundary" constraint of equal and opposite voltages across the equal plate and cathode loads effectively places the entire output circuit in series since there is no net signal current through ground. By constraining the output currents to be opposite and equal, there is no way to measure the impedance of one node to ground. With equal and opposite output currents under any circumstance, there is only one output impedance that can be measured: the equivalent impedance between the output nodes, i.e, the differential-mode output impedance.
The impedance looking into a node is understood to be the impedance to ground from that node in the same way that a node voltage is understood to be referenced to ground.
But and again, your constraint makes it impossible to measure an impedance to ground from either of the output nodes.
Bottom line: you've measured the impedance looking into one node and out the other. Conceptually, you've place an ohmmeter across the output nodes. You need to be able to conceptually place an ohmmeter across one output node and ground. Your constraint won't allow that since that would imply a current into one node and out through ground rather than through the other node.
Gentlemen, it's been fun. But now, I've got to go out and eat with some old friends.
AC
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That is NOT my boundary constraint, that is the CONSEQUENCE of the boundary constraint with a cathodyne. The boundary constraint is that the LOADS be equal. As a result, the analysis shows, and experiments confirmed, voltages across the loads are equal. If loads are equal and voltages are equal, then inevitably the Thevenin source impedances are equal- unless you're aiming for that Nobel.
As I said, common-mode disappears. You can have a ground connection at the middle of the load, but in the balance condition no current flows through it.
I don't think we are too far apart. Enjoy your meal.
No, if the loads are equal and the voltages are equal and opposite, the load is effectively floating. Since the load is effectively floating, there is just one effective source impedance, the impedance between the output nodes. Don't you see this?
Look, I'll draw something up later and post it. It's easy to see.
Not necessary. You can place 2 voltmeters, each to ground. 😉
By the way, where is my Noble Prize? I've corrected the Theorem in my previous post.