Indeed! Although I don't have SY's article, if I had to succinctly summarize what I believe to be SY's point, it would be: when taking the output differentially, use the differential output impedance for time constant calculations.
It's a good point and I don't wish to take away from it. The disagreement I have with him is that he makes the claim that the output node impedances are equal to the differential output impedance (what others have called the "balanced" output impedance). That's just plain and simply incorrect.
As you and I and others here have mentioned, when the loads are equal and the output voltages are equal, the common-mode current (return current) is zero, i.e., the output is taken differentially.
The loads could be connected in series without a ground at their connection and, ideally in theory, the voltage there would be identically zero. So, the ground can be removed and then it's clear you're loading the circuit differentially.
Honestly, I love threads like this. I almost always learn or teach myself something.
As one of my favorite fictional characters said:
Of course the total voltages are equal and opposite. But I think you missed the part about dividing only that portion of the voltage between the nodes that is due to the current flowing through them.
If I apply voltage signal 1 to a linear voltage amplifier input, it appears as an amplified version at the output. If I simultaneously apply current source signal 2 to the output, it produces a second component of the output voltage. If I want to use the current source to measure impedance, I don't divide the total output voltage by the currrent - I divide just that portion of the output due to the current source by that current.
The principle is the same in the example I gave in the post you responded to.
If I apply voltage signal 1 to a linear voltage amplifier input, it appears as an amplified version at the output. If I simultaneously apply current source signal 2 to the output, it produces a second component of the output voltage. If I want to use the current source to measure impedance, I don't divide the total output voltage by the currrent - I divide just that portion of the output due to the current source by that current.
The principle is the same in the example I gave in the post you responded to.
Don't know who Martin A is. To clarify, I agree that for balanced loads, you can model the Cathodyne as a differential driver with impedance slightly less than 2/gm. I have never disagreed with this. This does not imply, and it is incorrect to assert however, that in such balanced cases or even in unbalanced cases, that the plate to ground impedance is identcal to the cathode to ground impedance. It is not.
Two-port analysis with Z parameters is hardly complex and non-intuitive unless you've never seen it before. Once you have, it.just.makes.sense.
Usually, I would use a Norton equivalent representation of the plate circuit since the triode is basically a transconductance device. However, since you were using Thevenin equivalent circuits in your analysis, it dawned on me that describing the phase-splitter as a 2-port (well, really a 2+ port) network would be just the ticket for comparing models.
Regardless, despite our evident irresolvable disagreement, I've enjoyed the debate.
Best,
AC
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CPaul, a suggestion: try quoting a portion of the post you are responding too. That will identify who and which post you are responding to.
Thanks!
Thanks!
MerlinB, you might think so. Instead, he shorted both the cathode and the plate to ground! This is not in accord with Thevenin, of couse, which deals only in two-node equivalent circuits, not three-node ones (plate catthode and ground.)
You are indeed suggesting the correct method, dividing the voltage difference between te plate and cathode by the short circuit current obtained by shorting the plate and cathode. And indeed, you do get about 2/gm. But you cannot conclude from this that the cathode impedance is equal to the plate impedance. You must measure those separately. And when you do, you find that the plate to ground impedance is much higher than the cathode to ground impedance. It may be counter-intuitive, but that is the conclusion that straight forward application of the (two-node) definition of impedance leads you to.
You are indeed suggesting the correct method, dividing the voltage difference between te plate and cathode by the short circuit current obtained by shorting the plate and cathode. And indeed, you do get about 2/gm. But you cannot conclude from this that the cathode impedance is equal to the plate impedance. You must measure those separately. And when you do, you find that the plate to ground impedance is much higher than the cathode to ground impedance. It may be counter-intuitive, but that is the conclusion that straight forward application of the (two-node) definition of impedance leads you to.
Lets look at it another way...
Say for laughs, the *actual* impedances might be different.
But correlated consequence of cathode error arriving at the
plate can make the plate misbehave as-if impedances equal.
The cause of this current is not at the plate, only the effect.
Any illusion of causality at the plate is ignored. It happens
to be correlated to effect only by chance, and only if driven
loads are equal and linear.
If we inject an uncorrelated (maybe different frequency) test
signal back into the plate, we will see the true impedance.
Don't let Theven measure the effect and jump to conclusion
that had anything to do with the cause. Impedance has to
have cause and effect at the same node...
Hate to say it, but I'm leaning toward agreement with the
trolls on this one...
Say for laughs, the *actual* impedances might be different.
But correlated consequence of cathode error arriving at the
plate can make the plate misbehave as-if impedances equal.
The cause of this current is not at the plate, only the effect.
Any illusion of causality at the plate is ignored. It happens
to be correlated to effect only by chance, and only if driven
loads are equal and linear.
If we inject an uncorrelated (maybe different frequency) test
signal back into the plate, we will see the true impedance.
Don't let Theven measure the effect and jump to conclusion
that had anything to do with the cause. Impedance has to
have cause and effect at the same node...
Hate to say it, but I'm leaning toward agreement with the
trolls on this one...
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Your experimental interpretations are at fault. And you seem to forget that Cathodynes can draw grid currents from output tubes - a decidedly unbalanced situation! I agree completely with Vogel's analyses - do you? He talks of the balanced impedance 2/gm and while doing so, he quite correctly never talks about cathode and plate impedances separately. He quite correctly never claims that the plate and cathode impedances are equal, not once in the entire long letter. He also acknowledges that the plate to ground impedance is quite different from and larger than the cathode to ground impedance.
What's not to like about Vogel?
I just don't think he realizes how far apart you and he are.
What's not to like about Vogel?
I just don't think he realizes how far apart you and he are.
I love your ending quote.
SY has not been trained as an engineer, and that may explain some of this. In his article, he somehow develops the differential drive impedance by shorting cathode to plate - and ground! - to get the short circuit current. Some sort of three-terminal "Thevenin" equivalent, I guess. Later, he drives a square wave into a Cathodyne grid. The Cathodyne has identical plate and cathode R-C loads. He correctly observes equal rise times across the loads, and concludes that that means that the plate and cathode source impedances must be identical!
Of course it is easy to construct a circuit with identical loads and rise times but with differing source impedances - just drive an R-C network from a square wave voltage source into a unity gain current mirror input, and plop an identical network on the mirror output. The identical networks see identical currents, develop identical voltages and therefore have identical rise times - but the source impedances driving them are very different indeed!
I can't make him see it.
SY has not been trained as an engineer, and that may explain some of this. In his article, he somehow develops the differential drive impedance by shorting cathode to plate - and ground! - to get the short circuit current. Some sort of three-terminal "Thevenin" equivalent, I guess. Later, he drives a square wave into a Cathodyne grid. The Cathodyne has identical plate and cathode R-C loads. He correctly observes equal rise times across the loads, and concludes that that means that the plate and cathode source impedances must be identical!
Of course it is easy to construct a circuit with identical loads and rise times but with differing source impedances - just drive an R-C network from a square wave voltage source into a unity gain current mirror input, and plop an identical network on the mirror output. The identical networks see identical currents, develop identical voltages and therefore have identical rise times - but the source impedances driving them are very different indeed!
I can't make him see it.
A posts-per-minute meter for diyaudio?
CPaul was away from the thread for a while, and many bytes have flown under the bridge since. But this did make me wonder if a ratemeter might be an appropriate addition to the forum (only half-kidding). Hopping into these threads can be a bit like experiencing the Autobahn for the first time in a Ford Pinto.
And keep your eye on Paul Schimel's part two piece on tubes versus transistors via Electronic Design, which an editor (i.e. not Paul) elected to title Transistors, Tubes Sound The Same... I suspect it will set new records for posts per minute for ED 😱
CPaul was away from the thread for a while, and many bytes have flown under the bridge since. But this did make me wonder if a ratemeter might be an appropriate addition to the forum (only half-kidding). Hopping into these threads can be a bit like experiencing the Autobahn for the first time in a Ford Pinto.
And keep your eye on Paul Schimel's part two piece on tubes versus transistors via Electronic Design, which an editor (i.e. not Paul) elected to title Transistors, Tubes Sound The Same... I suspect it will set new records for posts per minute for ED 😱
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I may not be trained as an engineer (though I've trained many engineers), but I can count to two. The equal loads constraint for a two output device means that you measure (in a conceptual sense) two short circuit currents. And two open circuit voltages.
Not if you're trying to find a Thevenin equivalent. Thevenin dealt in two-node equavalent circuits. Not three or more. I'd be very interested if you can find a reference to a three-node Thevenin equivalent circuit.
You probably forgot about my Noble Prize correction of Thevenin theorem. For math higher than preschool it needs to be a little bit corrected.
Of course you didn't say that but nonetheless, that is precisely your claim. It's what I and others have tried unsuccessfully to reveal to you. You measured the differential output impedance, a single number, and the only impedance that can be measured given the "boundary conditions" , and then claim it to be the output impedance of either output.
In a response to Peter van Willenswaard's LTE, you write:
In fact you found the, as in singular, source impedance to be low. Under the so-called "boundary conditions", and this is trivial to show, the common-mode current vanishes and the loads are effectively a single load connected differentially across the output nodes. In this case, the only output impedance that can be measured is, of course, the differential output resistance.
To reveal the genuine output impedance of each output, you must, you must have a common-mode current, i.e., the test current must enter one node and return through ground.
This is so basic, so fundamental. Under the "boundary conditions", the genuine output impedance of each output simply cannot be measured.
Of course, we've been round and round on this and I don't harbor any hope at all that I'll convince you of the error of your ways but I do hope that other readers of this thread will think this thing through on their own and, not taking my word or anyone else's, work it out and come to their own conclusion.
Thanks! I will be happy to accept the Prize from your hands! 😉
By the way, why don't you want to redraw the schematic, putting both resistors in series, under cahthode, for convenience?
I just started doing that but gave up: I found KCAD very inconvenient...
Attachments
Actually, I think there may be a way to measure it, even under the restrictive "boundary conditions" that SY demands.
Imagine a Cathodyne with identical loads and no grid excitation. Equal and opposite ground-referenced AC current sources ip and ik drive the plate and cathode respectively. The circuit is perfectly balanced.
Now derive expressions for Vp and Vk. Each will contain contributions from ip and ik.
However, the definition of impedance is the change in the voltage between nodes A and B due to the current flowing into A and out of B, divided by that current.
Accordingly, before we divide Vp, we must reject that portion of its voltage that is due to ik, which does NOT flow directly into the plate and out of ground. Only ip does. When we divide Vp(ip) by ip and Vk(ik) by ik, we get Zp and Zk, where Zp >> Zk.
Imagine a Cathodyne with identical loads and no grid excitation. Equal and opposite ground-referenced AC current sources ip and ik drive the plate and cathode respectively. The circuit is perfectly balanced.
Now derive expressions for Vp and Vk. Each will contain contributions from ip and ik.
However, the definition of impedance is the change in the voltage between nodes A and B due to the current flowing into A and out of B, divided by that current.
Accordingly, before we divide Vp, we must reject that portion of its voltage that is due to ik, which does NOT flow directly into the plate and out of ground. Only ip does. When we divide Vp(ip) by ip and Vk(ik) by ik, we get Zp and Zk, where Zp >> Zk.