Yes, and even when one reduces portion of local feedback seen
by the cathode, such that impedances DO match, an imbalance
of common mode PSRR remains... Plate driven output grid will
receive 2x or 3x the PS ripple of the cathode driven grid...
To PS ripple, its a 3 part voltage divdier. And the node closer
to the rail will always have more ripple, even if all impedance
were a perfect match. Unless both impedance were extremely
low, like nested inside a strong global loop.
That was a good point. And I don't have any ready fix for it
in the open loop condition... Unless you gonna bleed some rip
to the cathode on purpose, that they might cancel in push pull.
But a weird thing about forced equal (but non-zero) impedance
cathodyne, is that any error forced upon one node must have an
equal and opposite reaction at the other end. Not just cathode
to plate, but also from plate to cathode... I'm not quite sure the
consequence of that effect in context of PSRR?
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Dave, zeroing the input to the phase splitter should not, if it's a linear circuit, have any effect on the output Z. Thus, your specification of having identical out-of-phase signals is irrelevant to the question of the output Z.
This is basic. Conceptually, if you reduced the input source to an infinitesimal level in your simulation, you should get the same result as before. Thus, even with zero input, the output Z is what it is.
So, with input sources zeroed and in the AC small-signal domain, we're just looking at a network of impedances and perhaps some controlled sources.
I've attached a circuit with three resistors and three nodes including ground. As you can see there are four different resistance measurements that can be made: two single ended, one differential, and one common-mode.
So, with such a "three wire" box, you can't speak of the output Z, there are 4.
SY, that's simply incorrect. As you write, there are two outputs in the phase-splitter circuit. There isn't a source impedance, there are four. This is because there are two Thevenin circuits needed here.
Looking back into the plate node, there is a Thevenin representation. Looking back into the cathode node, there is a separate Thevenin representation. So, each Thevenin circuit has an associated Thevenin (and generally unequal) impedance. These are the output impedances of the plate and cathode nodes respectively.
There is coupling between the two Thevenin circuits (the transfer impedances).
When you load both nodes equally and observe that the plate and cathode node voltages respond similarly, it's not because the Thevenin impedances are equal but rather, that the difference of the Thevenin and transfer impedances in each circuit are equal. You're measuring the differential-mode output impedance.
Now, I suspect you will claim the the phase splitter isn't used any other way. Fine. But that fact does not mean that those other impedances cease to exist. They exist regardless of how the circuit is used.
And, that fact does not license you or anyone else here to arbitrarily claim that the differential-mode output impedance is in fact the impedance at each output node.
Guys, this is basic EE stuff. It's what I learned years ago from Marshall Leach. It's what I taught when Leach was my thesis advisor. And, it's what I teach now to EE undergrads. All that doesn't make what I say correct in and of itself but heck, one can verify it oneself. Consult a few EE level circuit analysis text books.
If one continues to make these claims after doing that then one is effectively claiming that the fundamentals of EE are wrong.
That would put one decidedly in the crank category.
+1
It is amusing that this argument continues when articles like SY's have been published almost every year since WWII, in one publication or another.
The concept is a simple one:
When finding output bandwidth (or gain) you need to use the 'effective' output impedances, which are identical and quite small, provided the outputs are equally loaded.
When considering immunity to external noise -which is likely to be common mode- you need to use the 'single ended' output impedances, which are very different. i.e., the anode will be much more sensitive to electrostatic noise.
If the outputs are unequally loaded then the effective output impedances begin to approach the single ended values. In the extreme, you would have one output grounded and the circuit reverts to a cathode follower or ordinary gain stage.
It is amusing that this argument continues when articles like SY's have been published almost every year since WWII, in one publication or another.
The concept is a simple one:
When finding output bandwidth (or gain) you need to use the 'effective' output impedances, which are identical and quite small, provided the outputs are equally loaded.
When considering immunity to external noise -which is likely to be common mode- you need to use the 'single ended' output impedances, which are very different. i.e., the anode will be much more sensitive to electrostatic noise.
If the outputs are unequally loaded then the effective output impedances begin to approach the single ended values. In the extreme, you would have one output grounded and the circuit reverts to a cathode follower or ordinary gain stage.
No, there are two. Please refer to figure 3 of my article. If you can suggest a pair of identical loads that will cause those Thevenin resistances to be different, I will do that experiment.
No, there are four output impedances: two single-ended, one differential, one common-mode. The impedance you calculate by using identical loads is the differential-mode output impedance. This is not the Thevenin impedance at either the plate or the cathode.
This is basic stuff, SY. You cannot find the Thevenin impedance at the plate if you make any change to the cathode and vice verse. You can only find the Thevenin impedance at the plate by varying the conditions at the plate. Again, this is basic EE 101 stuff.
That's why probably you see deeper than basic EE 101 stuff.
I suppose, in chemistry that is analytical you always solve equations with multiple variables instead of using tables of recipes, right?
Reread the constraint- equal loads. If you change the load on the plate, you have to change the load on the cathode to meet the constraint. Algebra 101 stuff.
@Chris, good luck finding a new team!
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I've read your constraints SY, but the actual output impedances don't care about your constraints; they exist independently of your constraints.
Your constraints mean only that you've limited yourself to measuring one of the four output impedances that exist for this circuit. Why would you do that?
Moreover, you don't even have to change the loads at all to find the output impedances associated with this circuit.
Your constraints mean only that you've limited yourself to measuring one of the four output impedances that exist for this circuit. Why would you do that?
Moreover, you don't even have to change the loads at all to find the output impedances associated with this circuit.
Because he is already limited by conditions to measure properties of Concertina phase splitter. Only and only when loads are equal it is concertina. If they are unequal it is an amplification stage with resistors in cathode and anode.
Edit: as I mentioned before, another term is Split-Load Phase Splitter. That means it is a Phase Splitter when load is split in half. If you add some load to anode, or cathode only, it is no more split in half. That means, your measurements are destructive, and can't be used to measure parameters that you alter by your measurements.
Suppose, in order to measure temperature of body of an anymal you are killing it by measurement tools, then measure temperature of body of killed anymal. It can be measured, but can't be attributed to live anymal. The same way, when you kill Concertina by measurement, your results will be valid for dead concertina only, but say nothing about alive one.
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Bit of a distraction: Pic of an early crude lab "still" as I worked on. Yes your'e right....Lewis dot notations and molecular model understanding, all done in the head-with the aid of a slide rule to quantify.. So in comparison, diagnostic (electron)ics becomes a by-product. In essence the mind deals similiarily with both, so electrons & magnetics become a breeze, it needent be a "black" science.
Hence in my 200W amp design I diverted from the traditional circuit design thoughts and instead designed around the normalised transient response analogy.
Yes I use a concertina. I like it but am aware that it really behaves as see-saw, that is unbalance one half output and the properties tip. One avoids the undesirable after-effects by Williamson buffering with a fixed voltage CCS in the common cathode and using low capacitance video pentodes (conf as a triode). That effectively wraps the main problems up.
The last problem is that the anode side of the concertina on a Bode plot shows a HF pole that drops steeper than the cathode side. As the anode isn't really part of the 100% concertina grid/cathode feedback structure and proof of this is to put two perfectly compensated x10 scope probes one on the anode and the other on cathode and pump through a fast square wave and compare. The anode side will show a slew compared to the cathode as tube capacitances begin to interact.
It's not uncommon to see oscillations esp on some high ft tubes when termination becomes misbalanced. The ECC88 will oscillate given the chance.
richy
Hence in my 200W amp design I diverted from the traditional circuit design thoughts and instead designed around the normalised transient response analogy.
Yes I use a concertina. I like it but am aware that it really behaves as see-saw, that is unbalance one half output and the properties tip. One avoids the undesirable after-effects by Williamson buffering with a fixed voltage CCS in the common cathode and using low capacitance video pentodes (conf as a triode). That effectively wraps the main problems up.
The last problem is that the anode side of the concertina on a Bode plot shows a HF pole that drops steeper than the cathode side. As the anode isn't really part of the 100% concertina grid/cathode feedback structure and proof of this is to put two perfectly compensated x10 scope probes one on the anode and the other on cathode and pump through a fast square wave and compare. The anode side will show a slew compared to the cathode as tube capacitances begin to interact.
It's not uncommon to see oscillations esp on some high ft tubes when termination becomes misbalanced. The ECC88 will oscillate given the chance.
richy
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That may be the case Dave and, in fact, I pointed that out early on.
My point all along is that what is being measured (by insisting on balanced loads) is an output impedance, not the output node impedances.
But, as someone else has pointed out, there is reason to care about the actual output node impedances, e.g., noise currents.
Nonetheless, when the constraints are followed and the output impedances are measured or calculated, the Thevein source Zs are the same and low (~2/gm). So far, no-one has proposed any equal loading which cause the source impedances to be different or anything other than (approximately) the cathode follower source impedance. And also as shown experimentally in the article, if you treat the impedances as different (as was done in several texts), the results are disastrous.
Regardless of what you want to call it, it is a circuit with two outputs referenced to ground. Thus, regardless of what you want to call it and regardless of the loading constraints you wish to impose on it, there are four output impedances associated with it. This is fact that, no matter how hard you may try to ignore it or protest it, will not go away.
Looking into the plate node, there is a single-ended impedance. Period. Looking into the cathode node, there is a single-ended impedance. Period. They are different, they can be calculated, they can be simulated, they can measured and they do not depend on whether you have balanced loads or not. They exist regardless of how the amplifier is used in practice.
Balance the loads to your heart's content. But, if you then asked the question: what would the change in the plate voltage be if I infinitesimally changed the current into (or out of) the plate node? The answer is assuredly not going to be found by multiplying that current by the impedance you claim that node impedance to be. Rather, it will be the product of that current and the actual node impedance. This is the very definition of the node impedance: the ratio of the (AC) change in voltage to the (AC) change in current due to an infinitesimal change in the current into the node - all other factors held constant.
Simply insisting that the circuit must be operated under certain conditions does not magically transform a differential-mode impedance into a node impedance.
I more than insist, I use it in certain conditions, getting desired results: please see schematic attached to one of my previous posts.
Here it is, again:
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