Again the point is missed, that currents are equal, and feedback sets the impedance.
The cathode has roughly 1:1 feedback, and the plate has roughly 1/Mu feedback.
If you make the feedbacks equal, the impedances will be equal.
Now, there are at least two ways (and maybe more) to do this:
Isolate the cathode from seeing the direct result of its own current, and feedback a
1/Mu cathode voltage instead. See example in Post #55 if this sounds impossible...
This is kind of neat, because you also gain Mu/2 voltage amplification both nodes.
The other way is simply use a large amount of global feedback, will force both nodes
to a uniformly low impedance. And certainly possible to use both methods at once.
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Regardless which method forces concertina impedance equal, it is pointless unless
you are driving impedances that are unequal, such as when output grids conduct.
Yet fact is that you still cannot drive such outputs!!! Because they pump DC into
the coupling caps, and shove the concertina off its proper operating points.
Why argue about something that makes no sense even once you have achieved it?
You have to buffer the concertina to drive DC without the operating points getting
messed up, so the issue of driving unequal impedance is completely secondary to
that much greater problem of DC pumping.
Once you add buffers, unequal impedance due to grid current is hidden. This is the
obvious soloution to both problems. Concertina with equal impedance split does not
by itself solve anything. Arguing wether a match is real or illusory is never going to
drive output grid current, only cathode or source followers will...
The cathode has roughly 1:1 feedback, and the plate has roughly 1/Mu feedback.
If you make the feedbacks equal, the impedances will be equal.
Now, there are at least two ways (and maybe more) to do this:
Isolate the cathode from seeing the direct result of its own current, and feedback a
1/Mu cathode voltage instead. See example in Post #55 if this sounds impossible...
This is kind of neat, because you also gain Mu/2 voltage amplification both nodes.
The other way is simply use a large amount of global feedback, will force both nodes
to a uniformly low impedance. And certainly possible to use both methods at once.
-------
Regardless which method forces concertina impedance equal, it is pointless unless
you are driving impedances that are unequal, such as when output grids conduct.
Yet fact is that you still cannot drive such outputs!!! Because they pump DC into
the coupling caps, and shove the concertina off its proper operating points.
Why argue about something that makes no sense even once you have achieved it?
You have to buffer the concertina to drive DC without the operating points getting
messed up, so the issue of driving unequal impedance is completely secondary to
that much greater problem of DC pumping.
Once you add buffers, unequal impedance due to grid current is hidden. This is the
obvious soloution to both problems. Concertina with equal impedance split does not
by itself solve anything. Arguing wether a match is real or illusory is never going to
drive output grid current, only cathode or source followers will...
Also true for long tail, see saw, paraphase, and just about every other phase splitter. ANY class AB2 or B2 amp needs followers or a transformer (much worse solution but eliminates the phase splitter) to drive the grids.
That was Morgan Jones's derivation.
I never thought of that way...
DC couple both to the concertina cathode,
AC couple the anode side of the split only.
If we do this, maybe the equal Z split still makes sense?
I might rework #55 using this method for DC instead...
Don't forget about servo from long tail to screen grid; without it the whole directly coupled construction has quite unstable working points. Speaking of results, SY saw them on the screen of his computer. Up to half power 2'nd order only on -80 dB level, when approached clipping a grass grew up. After that measurement I added voltage divider from 1'st anode to the grid of Concertina, for higher anode voltage of the first tube that is as I understood the main contributor of distortions on normal listening levels.
Edit: trick with such coupling of concertia to LTP was also used in one Altec amp, later they gave it up I suppose because such a way it works like APF on very low frequencies.
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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.
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Kindly justify this. What is so special about the characteristic of balance?
Consider the characteristic of voltage amplification. If I short the plate of a triode voltage amplifier to measure its short circuit output current, it is no longer an amplifier. Rather, it is a voltage nullifier. So I can't do this, right, because I'm not measuring/analysing the right circuit?
The purpose to a Concertina is clear. Consider when the Concertina drives an output tube into grid current conduction. The balance is lost. The signal on the output tube grid differs greatly depending on whether it is driven by the single ended high impedance plate or the single ended low impedance cathode.
There is also quite a difference in the Cathodyne plate and cathode PSRR. Much of this is due to the relative proximity of the plate to the supply. But some is due to the difference between the single ended plate and cathode impedances.
There is also quite a difference in the Cathodyne plate and cathode PSRR. Much of this is due to the relative proximity of the plate to the supply. But some is due to the difference between the single ended plate and cathode impedances.
Technically, that's not the output impedance of both that you've measured. As I mentioned before, the output impedance of a node cannot be measured in this way.
Since you have this circuit all set up in the simulator, would you be interested in trying something out? If so, then:
First, zero V3.
Second, Replace R6 and R7 with AC current sources of opposite polarities, i.e., positive current is towards C1 or C2 respectively. Label the source replacing R6 as "Itp" and the sourse replacing R7 as "Itk".
Now, three different AC analysis will be run:
(1) Itp = 1, Itk = 0:
Plot the AC voltage at the "Plate_OC". This gives the magnitude of the impedance looking into the plate node. Call it Zp. It's the output impedance of the plate node.
Plot the AC voltage at "Cathode_OC". This gives the magnitude of the transfer impedance between the plate and the cathode. Call it Zpk.
(2) Itp = 0, Itk = 1:
Plot the AC voltage at "Cathode_OC". This gives the magnitude of the impedance looking into the cathode node. Call it Zk. It's the output impedance of the cathode node.
Plot the AC voltage at "Plate_OC". This gives the magnitude of the transfer impedance between the cathode and plate. Call it Zkp.
I suspect that |Zp - Zkp| = |Zk - Zpk|
(3) Itp = 1, Itk = -1:
The plots of the AC voltages at "Plate_OC" and "Cathode_OC" should be equal to each other I think, and their magnitudes equal to |Zp-Zkp| and |Zk-Zpk|. This is the "differential mode output impedance".
There's a fourth possibility: Itp = 1, Itk = 1 used for finding the "common-mode output impedance" but I don't think that one is important here.
Yes, just like with any other phase splitter. And once again, you're trying to say something that no-one disputes: if the load is not balanced, the source impedances aren't the same. But if the load IS balanced, they are by every measurement means.
Dave, nice sims. But please consider Alfred Centauri's post # 146. If you feel you must keep the circuit balanced, then you must segregate the ip-generated portion of Vp and divide that by ip, rather than dividing the entire Vp which is contributed to by both ip and ik. If you are a calculus person, this means that it is the partial derivative of Vp with respect to ip that determines the plate impedance Zp.
Sure it is. You can get the source impedance by open circuit voltage divided by short circuit current.
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