Q1: How do you measure output impedance of a balanced cathodyne?
Subsidiary question Q1a: How do you measure output impedance?
A1a: By changing the load impedance and seeing how the voltage varies.
Subsidiary question Q1b: How do you maintain a balanced cathodyne?
A1b: By ensuring the same load at both output ports.
A1: By changing both load impedances at the same time and seeing how the voltage varies.
Q2: What do I mean by the output impedance of a balanced cathodyne?
A2: The impedance I would use as the upper part of a potential divider or RC low pass filter in order to calculate the gain or frequency response of a balanced loaded cathodyne.
Q3: Isn't this just half the differential output impedance of a balanced cathodyne?
A3: You might choose to call it that - if you place twice the load on twice the impedance you get the same gain and frequency response. Balance means that the centre-point of the doubled impedance can be grounded, as it is already at ground potential anyway.
Subsidiary question Q1a: How do you measure output impedance?
A1a: By changing the load impedance and seeing how the voltage varies.
Subsidiary question Q1b: How do you maintain a balanced cathodyne?
A1b: By ensuring the same load at both output ports.
A1: By changing both load impedances at the same time and seeing how the voltage varies.
Q2: What do I mean by the output impedance of a balanced cathodyne?
A2: The impedance I would use as the upper part of a potential divider or RC low pass filter in order to calculate the gain or frequency response of a balanced loaded cathodyne.
Q3: Isn't this just half the differential output impedance of a balanced cathodyne?
A3: You might choose to call it that - if you place twice the load on twice the impedance you get the same gain and frequency response. Balance means that the centre-point of the doubled impedance can be grounded, as it is already at ground potential anyway.
DF96:
Q1c: Can the individual output impedances be measured by changing both load impedances at the same time?
A1c: No, unequivocally no. Thevenin's Theorem is clear on this and Chris' LTE circuit unambiguously demonstrates this. There is no rational doubt possible. Whatever it is you measure, it is not the individual output Thevenin impedances.
The outputs of Chris' circuit, by inspection, have vastly different Thevenin impedances and yet, by "ensuring the same load at both output ports", you might incorrectly infer that the outputs have equal Thevenin impedances.
To maintain that it is possible to measure the individual output Thevenin impedances while varying both loads is to dispute one of the most elementary, fundamental results in circuit theory.
Guys, stop trying to fit square pegs into round holes. All the word play in the world isn't going to change reality.
Q1c: Can the individual output impedances be measured by changing both load impedances at the same time?
A1c: No, unequivocally no. Thevenin's Theorem is clear on this and Chris' LTE circuit unambiguously demonstrates this. There is no rational doubt possible. Whatever it is you measure, it is not the individual output Thevenin impedances.
The outputs of Chris' circuit, by inspection, have vastly different Thevenin impedances and yet, by "ensuring the same load at both output ports", you might incorrectly infer that the outputs have equal Thevenin impedances.
To maintain that it is possible to measure the individual output Thevenin impedances while varying both loads is to dispute one of the most elementary, fundamental results in circuit theory.
Guys, stop trying to fit square pegs into round holes. All the word play in the world isn't going to change reality.
You can measure the impedance between the p-k by varying a load connected between p&k, the center point of which is ground. This is why the identical grounded loads in a Cathodyne look like a single (differential) load between the p&k.
I believe I agree with what you say, as long as you are talking about the impedance between the p & k (the differential output impedance of the balanced Cathodyne) and not the impedance from p to gnd or k to gnd.
Your treatment is different from what SY says - that the p to gnd and k to gnd impedances are 1/gm.
I believe I agree with what you say, as long as you are talking about the impedance between the p & k (the differential output impedance of the balanced Cathodyne) and not the impedance from p to gnd or k to gnd.
Your treatment is different from what SY says - that the p to gnd and k to gnd impedances are 1/gm.
I have always been careful to accept that if you measure the output impedances individually then you get different results. That is not in dispute.
As far as I remember, SY's main practical point is that it is incorrect to include a cathode build-out resistor. I think you accept that? If so, we probably are arguing about the meanings of words rather than the design of circuits.
As far as I remember, SY's main practical point is that it is incorrect to include a cathode build-out resistor. I think you accept that? If so, we probably are arguing about the meanings of words rather than the design of circuits.
Merlinb, DF96, just to be clear, I believe that if I start with an unloaded Cathodyne, with equal plate and cathode resistors, I can measure 3 impedances: P-K, P-Gnd and K-Gnd. Assuming an AC grid voltage is applied, I can measure the impedances of each pair of nodes by dividing the open circuit voltage difference between the two nodes by the current that flows between them when I short the two nodes together (of course I would use a large capacitor as an AC short for this test so as not to unbias the triode from normal operation.)
The results of these tests are that Zp-k is a bit less than 2/gm, Zp-gnd is a little less than the plate resistor and much larger than Zk-gnd, which is a bit more than 1/gm. In our Letters to the Editor in Linear Audio, Mr. Vogel and I give identical exact results for these parameters.
SY disagrees. He says Zp-gnd = Zk-gnd = about 1/gm.
This is more than just a difference about the meanings of words.
Certainly, SY has correctly championed the point that you should not add a build-out resistor to the cathode that you don't add to the plate. He recently pointed out that problems due to grid current drawn during overload conditions can be ameliorated if equal value resistors are added as P & K build-outs, and again I agree. This is particularly helpful for the low impedance cathode to limit the current available to charge the grid coupling capacitor. I'm not sure that SY would agree with that last sentence, but it can be easily confirmed that the overload characteristics of the P & K are different.
With the bove explanation, do you still think this is just about the meaning of words?
The results of these tests are that Zp-k is a bit less than 2/gm, Zp-gnd is a little less than the plate resistor and much larger than Zk-gnd, which is a bit more than 1/gm. In our Letters to the Editor in Linear Audio, Mr. Vogel and I give identical exact results for these parameters.
SY disagrees. He says Zp-gnd = Zk-gnd = about 1/gm.
This is more than just a difference about the meanings of words.
Certainly, SY has correctly championed the point that you should not add a build-out resistor to the cathode that you don't add to the plate. He recently pointed out that problems due to grid current drawn during overload conditions can be ameliorated if equal value resistors are added as P & K build-outs, and again I agree. This is particularly helpful for the low impedance cathode to limit the current available to charge the grid coupling capacitor. I'm not sure that SY would agree with that last sentence, but it can be easily confirmed that the overload characteristics of the P & K are different.
With the bove explanation, do you still think this is just about the meaning of words?
He doesn´t do a test with three nodes. He tests two nodes. But why can't he just test two Thevenin sources at the same time? You short TWO thevenin sources, and you MUST do that because the claim says: 'IF the two loads are the same...' so you need to test the two sources at the same time in the same way, infinite load or short circuit.
So you short the two nodes of one source, and the two nodes of the other source. Then you see that two nodes actually are the same so you end up with several nodes (three) connected all together. Well, that may look curious at first sight but it doesn't change anything, thevenin still holds. Why wouldn't it?
OTOH, Chris´ current source circuit in his last LTE does bother me....
jan
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janneman, please take a look at the Thevenin theorem if you are not already familiar with it. A good presentation can be found at Thévenin's theorem - Wikipedia, the free encyclopedia In it, will see first that there is nothing special about ground - it's just another node. Second, you will see that the theorem is about measuring the impedance between a pair of nodes, not a trio of them (ground being one of the three.)
Any attempt to measure impedances by shorting three nodes together lacks theoretical underpinning and violates the Thevenin theorem.
Also, Thevenin will not allow you to place shorts between two pairs of nodes in a circuit. This is because the second short affects the impedance seen at the first, and the first short affects that seen at the second. You wouldn't think of applying a second short somewhere inside a circuit with just one output when you measure its output impedance - that would mess up your result drastically. So why would the mere fact that the circuit has a second output that just happens to be the negative of the first give you dispensation to short the second to ground when you measure the ground-referenced impedance of the first?
I know that there is an intuitive appeal to a number of people (SY chief among them) to subject the circuit to what he calls boundary conditions during testing. But a clear reading of the Thevenin theorem simply does not allow it - UNLESS the "boundary conditions" means that you can connect a single load from the P to the K when you are measuring the Zp-k impedance. That would be a valid two-node Thevenin impedance measurement.
The current source circuit should get you thinking. You'll note I can't get a response to it out of SY.
Any attempt to measure impedances by shorting three nodes together lacks theoretical underpinning and violates the Thevenin theorem.
Also, Thevenin will not allow you to place shorts between two pairs of nodes in a circuit. This is because the second short affects the impedance seen at the first, and the first short affects that seen at the second. You wouldn't think of applying a second short somewhere inside a circuit with just one output when you measure its output impedance - that would mess up your result drastically. So why would the mere fact that the circuit has a second output that just happens to be the negative of the first give you dispensation to short the second to ground when you measure the ground-referenced impedance of the first?
I know that there is an intuitive appeal to a number of people (SY chief among them) to subject the circuit to what he calls boundary conditions during testing. But a clear reading of the Thevenin theorem simply does not allow it - UNLESS the "boundary conditions" means that you can connect a single load from the P to the K when you are measuring the Zp-k impedance. That would be a valid two-node Thevenin impedance measurement.
The current source circuit should get you thinking. You'll note I can't get a response to it out of SY.
I'd like to expand on and amplify Chris' reply to Jan.
If the Thevenin circuits are independent, i.e., represent two separate circuits, then there's no problem with testing both at the same time. They're independent; one doesn't affect the other at all.
But these Thevenin equivalent circuits in question are not independent, i.e., they are coupled.
Let's say that you wanted to find the Thevenin equivalent of some circuit and started by measuring the open-circuit voltage. You are ready to measure the short-circuit current but, before you do, you change the value of one of the resistors in the circuit and then measure the short-circuit current.
I imagine that you would instinctively know that you measured the open-circuit voltage of one circuit and the short-circuit current of another, different circuit and that you couldn't combine the two measurements in any meaningful way.
But look, if both loads are changed at the same time as SY does, the above is precisely what is happening; the change in the cathode load changes the circuit between the anode and ground nodes and vice versa. SY is changing the circuit between measurements. This makes it impossible to determine the actual Thevenin equivalents for the anode and cathode nodes.
Actually, it's quite easy- you apply a signal, open both nodes and measure the voltage at the one you're interested in (or both simultaneously- physics allows you to do that). Then short both to ground and measure the current at the one you're interested in (or both simultaneously- physics allows you to do that). If you honor the boundary conditions, the coupling is irrelevant- the next stage is driven by two equal and opposite sources.
When you do that, honoring the boundary conditions, you find both Thevenin source impedances to be low and equal. Experiment is consistent with that prediction. I'm still waiting for the pair of loads which cause the simple model to make a wrong prediction. The crickets are getting hoarse from all that chirping...
Nope, it's trivially easy.
When you do that, honoring the boundary conditions, you find both Thevenin source impedances to be low and equal. Experiment is consistent with that prediction. I'm still waiting for the pair of loads which cause the simple model to make a wrong prediction. The crickets are getting hoarse from all that chirping...
Nope, it's trivially easy.
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