Snide remark. Not informative or enlightening.
The fact that you can draw two in one figure is irrelevent to the fact that Thevenin's Theorem makes no allowance for you to test them in that way.
Here is what it does allow one to do: Measure the equivalent impedance of any pair of nodes in any linear circuit. If a voltage is present between them, measure and divide it by the current that flows between them when they are shorted. The result is the impedance of the node pair.
When you do this one at a time for the P-gnd, K-gnd and P-K node pairs, you find that Zp-gnd is typically a little less than the resistance of the plate resistor, Zpg is typically a little less than 2/gm, and Zk-gnd is somewhat more than 1/gm.
The fact that you can draw two in one figure is irrelevent to the fact that Thevenin's Theorem makes no allowance for you to test them in that way.
Here is what it does allow one to do: Measure the equivalent impedance of any pair of nodes in any linear circuit. If a voltage is present between them, measure and divide it by the current that flows between them when they are shorted. The result is the impedance of the node pair.
When you do this one at a time for the P-gnd, K-gnd and P-K node pairs, you find that Zp-gnd is typically a little less than the resistance of the plate resistor, Zpg is typically a little less than 2/gm, and Zk-gnd is somewhat more than 1/gm.
Apparently two circuits gets you confused. I'm sorry, I can't help you with that.
To spare any more injuries to your tender feelings, I won't contribute any more until you suggest a pair of loads for which my model makes an incorrect prediction.
To spare any more injuries to your tender feelings, I won't contribute any more until you suggest a pair of loads for which my model makes an incorrect prediction.
More snide remarks. Not informative or enlightening.
Here is what Thevenin allows one to do: Measure the equivalent impedance of any pair of nodes in any linear circuit. If a voltage is present between them, measure and divide it by the current that flows between them when they are shorted. The result is the impedance of the node pair.
It is clear that you have no effective rejoinder to that simple previous paragraph. Otherwise you would offer it.
And you have already agreed in post 575 that the lack of a pair of loads that falsify your model only proves it is not self-contradictory. When I then pointed out in post 580 that you have appointed your model the sole arbiter of whether the model gets it right, and that you are therefore engaging in circular reasoning, you again had no effective rejoinder.
So why would I try find you two loads that don't contradict your model? So I would fail and show again only that your model doesn't seem to contradict itself? What would be the point? Been there, done that.
I could say that your sudden concern for my sensibilities is touching, but I recognize sarcasm when I see it - again, uninformative and unenlightening.
But I do invite all to form their own conclusions about why you are now choosing to fall silent.
Here is what Thevenin allows one to do: Measure the equivalent impedance of any pair of nodes in any linear circuit. If a voltage is present between them, measure and divide it by the current that flows between them when they are shorted. The result is the impedance of the node pair.
It is clear that you have no effective rejoinder to that simple previous paragraph. Otherwise you would offer it.
And you have already agreed in post 575 that the lack of a pair of loads that falsify your model only proves it is not self-contradictory. When I then pointed out in post 580 that you have appointed your model the sole arbiter of whether the model gets it right, and that you are therefore engaging in circular reasoning, you again had no effective rejoinder.
So why would I try find you two loads that don't contradict your model? So I would fail and show again only that your model doesn't seem to contradict itself? What would be the point? Been there, done that.
I could say that your sudden concern for my sensibilities is touching, but I recognize sarcasm when I see it - again, uninformative and unenlightening.
But I do invite all to form their own conclusions about why you are now choosing to fall silent.
Hey CPaul and Alfred Centauri,
instead of debating Sy endlessly, why not show us something you have built in terms of the cathodyne phase spitter and explain in easy language the virtues of your creations........
that way we gain more insights...........theory is fine, but practice is more fun, wouldn't you agree?😉
instead of debating Sy endlessly, why not show us something you have built in terms of the cathodyne phase spitter and explain in easy language the virtues of your creations........
that way we gain more insights...........theory is fine, but practice is more fun, wouldn't you agree?😉
every year around the holidays I curl up with a good book next to the (f)ire. This year it was one of my favorites. Dynamical Analogies by Harry O.
analyze this.
my vision of the split load is a seesaw with a pair of 80 pound 5 year old kids on either side. take one off and all balance is lost.
Since the goal of the seesaw is fun, how would you console the fat kid with his *** in the sand?
now use thevenin to look at the same situation and analyze how it works.
what next?? a bumblebee cannot fly?
dave
analyze this.
my vision of the split load is a seesaw with a pair of 80 pound 5 year old kids on either side. take one off and all balance is lost.
Since the goal of the seesaw is fun, how would you console the fat kid with his *** in the sand?
now use thevenin to look at the same situation and analyze how it works.
what next?? a bumblebee cannot fly?
dave
Dave, I agree. When the mean ol' grid current bully knocks the first kid off his seat, the remaining one develops a sudden and keen interest in how an unbalanced see-saw works.
Thevenin to the rescue.
Thevenin to the rescue.
Exactly! And that is also what the claim says. 'IF the two load are the same...'.
And nobody is shorting three nodes. The node pairs of each source are shorted together, and it is just a particularity of this particular circuit that the sources have a node in common so you end up with three nodes connected together. But that has nothing to do with the concept we're discussing.
jan
grid current doesn't knock the kid off... Thevenin is the bully. in order to find the weight of either of the kids by Thevenin you need to remove the other (zero independent sources) at this point you are just being a stick in the mud since you interrupt a working model to measure the obvious.
dave
Jan, let's try to disentangle the ideas you've conflated here.
(1) If you wish to find the Thevenin equivalent circuit of either of the outputs (to ground), you change the load on that output only, i.e., you must not change the load at the other output. Do you agree?
(2) If you wish to find the Thevenin equivalent circuit between the outputs, you change the load across the outputs. Do you agree?
(3) If you restrict the testing such that the "boundary constraints" are always met, you cannot do the tests in (1). Do you agree?
(4) If you cannot perform the tests in (1), you cannot find the Thevenin equivalent circuit between either output and ground. Do you agree?
(5) Assuming (1) through (4), SY's claim that his model is confirmed by experiment is false. Only the far weaker claim that his Thevenin model is not ruled out by his experiment is justified.
The bottom line: SY's experiment confirms that, when operated according to the "boundary constraints", the differential output impedance is 2/g_m.
His choice to split that impedance equally across two independent Thevenin circuits is arbitrary and not confirmed by his experimental results. The reason is simple: to verify that arbitrary choice requires that he perform experiments in accord with (1). That's a basic, elementary fact of Thevenin's Theorem. No amount of word play can change that basic fact.
It's time to face it guys. The fat lady has sung.
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Trying to understand here, Dave... I thought the kids were the loads, not the independent sources.
But to tell you the truth, arguing by analogy is confusing. I'd prefer to discuss any circuit issues directly if it's OK with you.
But to tell you the truth, arguing by analogy is confusing. I'd prefer to discuss any circuit issues directly if it's OK with you.
they are both loads and sources to each other. replace one of them with a 80 pound weight and again no fun.
in the case of the cathodyne, when measuring the plate you need an equal but inverted signal in the cathode otherwise you are measuring the out of balanced characteristics when we are interested in the balanced characteristics
we all know as you shift from the balanced to unbalanced situations, the anode Zout becomes the load resistor and the cathode that of a bit more the ideal cathode follower.
how do you apply thevenin to the specific case of the balanced situation when the first thing you insist on doing is throwing the circuit out of balance to measure?
dave
BINGO! You are so very close to the heart of this issue Dave.
The answer is simple, in the balanced situation, the only impedance of concern is the (as in singular) differential output impedance.
To put it another way, differential output currents do not "see", so not "observe", do not "interact" with the individual output impedances to ground.
Why? It's easy: there is no current through the ground return path. That's the essential nature of a differential current.
Let's summarize.
(1) The individual output impedances are "invisible" when the circuit is balanced.
(2) Only unbalanced currents "see" these individual output impedances.
(3) Testing that occurs under balanced conditions cannot reveal the individual output impedances.
(4) Any claim to have confirmed the individual output impedances under balanced conditions is bogus. It's not.even.wrong.
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