Why not? The standard definition of Thevenin impedance is open circuit voltage divided by short circuit current. If one sets the boundary condition (and I know that's where I lose you, but boundary conditions are set every time you define a new circuit) that the two nodes be loaded identically, then open them both and measure the voltages, then short them both and measure the currents. In each case, the calculated impedance is the cathode follower impedance- not exactly, it's a bit higher (2/mu, in the limit of low plate resistance and high mu), but close. And when you measure this experimentally, that's exactly what you get, a neat confirmation of the correctness of that model.
Burkhardt Vogel has done a more detailed analysis and comes to the same conclusion- this will be appearing in Linear Audio as well.
Taken one at a time they have different impedances. Taken together, and with an extension of the concept of impedance to cover two ports, they have the same impedance (roughly equal to the cathode impedance when taken singly). These two statements do not conflict because the two ports are not independent of each other.
In this context my definition of impedance is the drop in voltage as you load a port, divided by the current you draw. This is equivalent to your definition, in the sense that it gives the same result.
The practical significance of all this is that you don't need to add build-out resistors to the cathode - in fact this is a bad idea. You just have to ensure that the two loads are the same, under all conditions. The weakness of the cathodyne is that if the two loads are different, perhaps under dynamic conditions, then there is a lack of symmetry between the two ports so P-P balance is lost. The LTP is better in this respect.
In this context my definition of impedance is the drop in voltage as you load a port, divided by the current you draw. This is equivalent to your definition, in the sense that it gives the same result.
The practical significance of all this is that you don't need to add build-out resistors to the cathode - in fact this is a bad idea. You just have to ensure that the two loads are the same, under all conditions. The weakness of the cathodyne is that if the two loads are different, perhaps under dynamic conditions, then there is a lack of symmetry between the two ports so P-P balance is lost. The LTP is better in this respect.
Without having done the test I predict a surprise here. But I love surprises!
Thanks,
Chris
Fortunately, the loads generally seen by phase splitters do not vary dynamically! The LTP is far from immune to load imbalance.
Back to "If the two output ports had different (and independent) impedances then they would drop different amounts of voltage as the load impedance changes." Agreed, but only if there were a third (ground?) node that these outputs could work against and pass different currents through. But there is no such node in a triode as long as the grid carries no current.
Assume for the sake of argument that the P and K have different impedances. They would "try" to cause different voltage drops across identical loads. To do so, they would have to pull different currents. But this can't be; the cathode and plate currents MUST be equal and opposite - there is no place else for the current to go in a triode (with no grid current). So regardless of what the impedances of the P & K might “try” to make the circuit do, the identical P & K currents and loads would force equal and opposite P & K voltages.
The fact that Cathodyne P & K voltages stay equal and opposite implies nothing about P & K impedances; it requires them to be neither equal nor unequal.
If you disagree, please show me the flaw in my logic.
Assume for the sake of argument that the P and K have different impedances. They would "try" to cause different voltage drops across identical loads. To do so, they would have to pull different currents. But this can't be; the cathode and plate currents MUST be equal and opposite - there is no place else for the current to go in a triode (with no grid current). So regardless of what the impedances of the P & K might “try” to make the circuit do, the identical P & K currents and loads would force equal and opposite P & K voltages.
The fact that Cathodyne P & K voltages stay equal and opposite implies nothing about P & K impedances; it requires them to be neither equal nor unequal.
If you disagree, please show me the flaw in my logic.
Let's apply two special cases to your externally forced current test: First we drive both anode and cathode together with out-of-phase (yes, I know it's not really "phase", but.... ) equal amplitude signal currents. What differences between anode and cathode voltages will we observe?
Then we drive both together with in-phase signal currents. What will we observe now? And will this be a different result than when driving each separately?
Thanks,
Chris
To DF46, I agree with all you said, except that I request a clarification. Please define "the concept of impedance to cover two ports". We really need to work with a definition here, not just a concept. Without a definition, it's easy to wrapped around the wheel. Please define.
I do want to point out that if you choose the Cathodyne P and K as the two nodes you want to test the impedance of, you'd get approximately 1/gm in parallel with RL.
Finally, help out a newbie here. I'm not sure I'm properly responding to the comment I intend to.
I do want to point out that if you choose the Cathodyne P and K as the two nodes you want to test the impedance of, you'd get approximately 1/gm in parallel with RL.
Finally, help out a newbie here. I'm not sure I'm properly responding to the comment I intend to.
Chris Hornbeck, great question!
Driving P and K with ip and ik respectively, equal and oposite ground referenced currents: with no grid excitation, we'd see equal and opposite P and K voltages! Let's not get too excited yet. Now, back to the def of impedance. What part of the voltage at P is due to ip and what to ik? I ask because you must pair only that portion of the voltage change with the current that caused it.
I don't know how to answer that question with a bench test or a sim. I need algebra. When I derive the expressions for Vp and Vk in terms of ip and ik, I can see what part of vp is caused by ip, and I can divide to get its impedance, ignoring the ik contribution. A similar thing can be done with Vk
The great thing about these equations is that you can then see that setting ik to 0 doesn't change the ip contribution to Vp one iota - it stays exactly the same. And so there is no reason to test Cathodyne impedances with the Cathodyne "balanced" with equal and opposite volatges. It just makes you have to solve the problem with algebra, when a simple sim or bench test would have sufficed.
Driving P and K with ip and ik respectively, equal and oposite ground referenced currents: with no grid excitation, we'd see equal and opposite P and K voltages! Let's not get too excited yet. Now, back to the def of impedance. What part of the voltage at P is due to ip and what to ik? I ask because you must pair only that portion of the voltage change with the current that caused it.
I don't know how to answer that question with a bench test or a sim. I need algebra. When I derive the expressions for Vp and Vk in terms of ip and ik, I can see what part of vp is caused by ip, and I can divide to get its impedance, ignoring the ik contribution. A similar thing can be done with Vk
The great thing about these equations is that you can then see that setting ik to 0 doesn't change the ip contribution to Vp one iota - it stays exactly the same. And so there is no reason to test Cathodyne impedances with the Cathodyne "balanced" with equal and opposite volatges. It just makes you have to solve the problem with algebra, when a simple sim or bench test would have sufficed.
To SY on #42, if boundary conditions apply to perfectly balanced Cathodynes, then they must apply to unbalanced ones too. There are just a different set of boundary conditions, right? Please describe the boundary conditions that have to be adhered to for an unbalanced “Cathodyne”, (say, Rp = 2 Rk), and tell me how to respect those boundary conditions when I measure that circuit’s impedances.
Regarding the test you describe, you are testing the impedance between the P & K, not P to ground and K to ground. The proof is that the circuit is balanced, right? So the current is flowing only between P and K, not P and ground or K and ground. Even if you connected the P & K to ground, the circuit is balanced, so P & K currents are equal and opposite, so there is no current left to flow through ground, so it cannot be one of the nodes whose impedances is being measured.
I agree that the impedance you measure in this case is about RL in parallel with 1/gm. But it is not the impedance of anything but the P and K node pair.
Regarding the test you describe, you are testing the impedance between the P & K, not P to ground and K to ground. The proof is that the circuit is balanced, right? So the current is flowing only between P and K, not P and ground or K and ground. Even if you connected the P & K to ground, the circuit is balanced, so P & K currents are equal and opposite, so there is no current left to flow through ground, so it cannot be one of the nodes whose impedances is being measured.
I agree that the impedance you measure in this case is about RL in parallel with 1/gm. But it is not the impedance of anything but the P and K node pair.
But ip = ik. The triode floats with no ground references, ignoring grid current.
We're just counting angels here, methinks.
Thanks,
Chris
I thought you intended that ip = (-)ik.
What is the definition of impedance? The change in voltage across a node pair due to and divided by the current that flowed through the node pair. If you have a different def, please advise. We can't get anywhere without an agreed upon def.
If someone insists on applying both ik and ip when we're trying to measure just Zp, I say, "fine." I'll just make sure that I'm dividing the voltage change between plate and ground that was due to ip alone by ip, in accordance with the def of impedance.
Please find the flaw in my definition or logic. Accusing me of playing with angels is not helpful!
What is the definition of impedance? The change in voltage across a node pair due to and divided by the current that flowed through the node pair. If you have a different def, please advise. We can't get anywhere without an agreed upon def.
If someone insists on applying both ik and ip when we're trying to measure just Zp, I say, "fine." I'll just make sure that I'm dividing the voltage change between plate and ground that was due to ip alone by ip, in accordance with the def of impedance.
Please find the flaw in my definition or logic. Accusing me of playing with angels is not helpful!
Well, no, not if the boundary condition is "equal loads." You can't just say, "The consequence of the boundary conditions must be independent of boundary conditions"- that leads to nonsense in any physical model.
Even if you design cathodyne with truly equal impedance split,
provable by driving unequal loads... It matters NOTHING once
you close the global loop. Both impedances are forced equal
anyway, so the entire ordeal of matching them in the open
loop state becomes a silly wasted effort.
provable by driving unequal loads... It matters NOTHING once
you close the global loop. Both impedances are forced equal
anyway, so the entire ordeal of matching them in the open
loop state becomes a silly wasted effort.
Attachments
You say,
You say,
I don't believe that you replied to my reasoning that you are measuring the impedance of the P-K node pair rather than that of the P-Gnd or K-Gnd pairs.
Well, why not? What is so special about a "balanced" circuit that it must adhere to boundary conditions when an unbalanced one need not? Please justify your assertion.
You say,
I don't recall saying anything like this.
I don't believe that you replied to my reasoning that you are measuring the impedance of the P-K node pair rather than that of the P-Gnd or K-Gnd pairs.
Are we going to have this 60 year old concertina splitter debate all over again? OK I'll pitch in to share some of the "heat".
Here (FWIW) is what I posted to a Guitar Amp forum not so long back (letter u used for MU). I hope this stacks up with your analysis Stuart!
For equal anode and cathode loads:
Zout(anode and cathode) = RL.ra/RL(u+2)
At typical values of u (say u>20)_ then u+2 is approximately = u so simplify to
Zout approx = RL.ra/RL.u
Cancel RL from top and bottom lines
Zout approx = ra/u = 1/gm
If the anode load drops significantly:
Zout(cathode) = (RL+ra)/(u+2) + ra/RL
The ra/RL term is insignificant so drop it
Zout(cathode) approx = (RL+ra)/(u+2)
and at typical values of u (where u+2 is approx = u)
Zout(cathode) approx = (RL+ra)/u = RL/u +1/gm
That is the Zout(cathode) increases by approx RL/u
If the cathode load drops significantly:
Zout(anode) = RL x RL(u+1) + RL.ra / RL(u+2)+ra
RL squared x (u+1) is much larger than RL.ra
And RL(u+2) is much larger than ra
So Zout(anode) approx = RLxRL(u+1)/RL(u+2)
At typical values of u (where u+1 is approx = u)
Zout(anode) approx = RL
SUMMARY:
For equal values of anode and cathode loads Zout at both anode and cathode is approximately = 1/gm
As the loads become unbalanced:
Zout(anode) increases from 1/gm toward RL
Zout(cathode) increases from 1/gm by a maximum factor of RL/u
That is why the HIFI boffins have been recommending the use of low mu (and high gm) tubes for concertina splitters when driving anything other than a pure Class A output stage.
The changes when loads become unbalanced (as they do when driving a Class AB push pull output) actually sounds lovely in a Guitar Amp and we deliberately emphasize its "warts" by using a High Mu, Low gm tube. 6SL7 driving a pair of 6V6 in Class AB is a Guitar Power Amp to die for BUT IT AIN'T HIFI!!!
Spice Modelling I have seen CONFIRMS that Zout from Anode and Cathode is equal when load impedances are equal.
Priessman is often sited as the definitive article on Cathodyne BUT this article preceded it by 9 years.
Jones. G. E. (1951). Analysis of Split-Load Phase Inverter. Audio Engineering, 35
(12), pp16+40-41.
Stuart, I will try to get hold of your most recent article on the Cathodyne. I recall that when we discussed this elsewhere about 2 or more years ago your comment was "In the words of Hitchhiker's Guide, "Reality was at Fault"".
Cheers,
Ian
Here (FWIW) is what I posted to a Guitar Amp forum not so long back (letter u used for MU). I hope this stacks up with your analysis Stuart!
For equal anode and cathode loads:
Zout(anode and cathode) = RL.ra/RL(u+2)
At typical values of u (say u>20)_ then u+2 is approximately = u so simplify to
Zout approx = RL.ra/RL.u
Cancel RL from top and bottom lines
Zout approx = ra/u = 1/gm
If the anode load drops significantly:
Zout(cathode) = (RL+ra)/(u+2) + ra/RL
The ra/RL term is insignificant so drop it
Zout(cathode) approx = (RL+ra)/(u+2)
and at typical values of u (where u+2 is approx = u)
Zout(cathode) approx = (RL+ra)/u = RL/u +1/gm
That is the Zout(cathode) increases by approx RL/u
If the cathode load drops significantly:
Zout(anode) = RL x RL(u+1) + RL.ra / RL(u+2)+ra
RL squared x (u+1) is much larger than RL.ra
And RL(u+2) is much larger than ra
So Zout(anode) approx = RLxRL(u+1)/RL(u+2)
At typical values of u (where u+1 is approx = u)
Zout(anode) approx = RL
SUMMARY:
For equal values of anode and cathode loads Zout at both anode and cathode is approximately = 1/gm
As the loads become unbalanced:
Zout(anode) increases from 1/gm toward RL
Zout(cathode) increases from 1/gm by a maximum factor of RL/u
That is why the HIFI boffins have been recommending the use of low mu (and high gm) tubes for concertina splitters when driving anything other than a pure Class A output stage.
The changes when loads become unbalanced (as they do when driving a Class AB push pull output) actually sounds lovely in a Guitar Amp and we deliberately emphasize its "warts" by using a High Mu, Low gm tube. 6SL7 driving a pair of 6V6 in Class AB is a Guitar Power Amp to die for BUT IT AIN'T HIFI!!!
Spice Modelling I have seen CONFIRMS that Zout from Anode and Cathode is equal when load impedances are equal.
Priessman is often sited as the definitive article on Cathodyne BUT this article preceded it by 9 years.
Jones. G. E. (1951). Analysis of Split-Load Phase Inverter. Audio Engineering, 35
(12), pp16+40-41.
Stuart, I will try to get hold of your most recent article on the Cathodyne. I recall that when we discussed this elsewhere about 2 or more years ago your comment was "In the words of Hitchhiker's Guide, "Reality was at Fault"".
Cheers,
Ian
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They actually do vary drammatically when grid currents of output tubes can't be anymore neglected in respect to currents through resistors in anode and cathode of Concertina. That's why (particularly) my Pyramid that we measured a little bit in your laboratory contained LTP after Concertina.
Well, when I loaded each node with equal caps to ground, the rise times and amplitudes changed versus unloaded, and changed equally.
QED.
The distortion that this causes is alone the reason that tubes with grid current need cathode or source followers to drive them. But most tubes that draw grid current (not counting defective tubes!) are the sort that need a HV drive, like DHTs, where a cathodyne is unsuitable for driving them directly due to swing requirements.
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