Well I think your all winners in my book.
Top minds, dueling it out in the streets, yet you managed to be civilized about it.
I think I learned something, though %98 is way over my head.
Top minds, dueling it out in the streets, yet you managed to be civilized about it.
I think I learned something, though %98 is way over my head.
Chris, I'm not sure I follow. The expression for Vp in terms of Ip and Ik will contain two unknowns: the open-circuit plate impedance and the transfer impedance from cathode to plate.
To reject the portion of Vp due to Ik requires that one know the transfer impedance but that is an unknown. To measure it requires that one inject a test current into the cathode output while measuring the open-circuit plate output voltage. But that violates the "boundary conditions" so I don't see how, starting out with no knowledge of the open-circuit output impedances and transfer impedances, one can measure anything but the differential mode impedance without violating the BCs.
So, I figure I must be misreading you. What am I missing?
To reject the portion of Vp due to Ik requires that one know the transfer impedance but that is an unknown. To measure it requires that one inject a test current into the cathode output while measuring the open-circuit plate output voltage. But that violates the "boundary conditions" so I don't see how, starting out with no knowledge of the open-circuit output impedances and transfer impedances, one can measure anything but the differential mode impedance without violating the BCs.
So, I figure I must be misreading you. What am I missing?
OK... you asked for this a while back from me and I thought I already showed it clearly but lets see if I get it right. (They Key is in the phase of the excitation)
Here are the current sources in place and a 1V excitation o the grid. Note the gain is slightly below unity and the phases of the signals are 180 apart.
input set to zero. current sources set to 1ma. (i use this because the voltages can be then directly read as output impedance vs. frequency via ohms law. in the pic below, .082 / .001+ 82 ohms)
It is also important to note that the phases are still 180 apart.
When only use one CCS, I get very different results and the the voltages are suddenly in phase which is not how the circuit works.
When I invert the phase of one of the CCS's I get double the results of a single and both are still in phase.
and finally when I ground the cathode and measure plate to ground I get exactly 2X the plate to ground values.
I don't really care about engineering, theory, nomenclature and the likes. What I care about is how the circuit operates in the manner I intend to use it. I do know that If I built the circuit I and measured the output with a 1V AC excitation on the grid I would get a little less than 1V at both the plate and cathode and furthermore those signals would be in opposite phases.
I also know if I place a pair of 82 ohm resistors from p / k to ground or a single 164 ohm resistor from plate to cathode the output will drop to just under 1/2 a volt and since we know when Rsource = Rload out output becomes half it seems everything works out.
call me an empiricist, but unless you can verify your results in more than one way I am suspect of their validity and since there is no what I know of to inject a 1V signal into the grid of a concertina and not get inverted phases at the output, anything that shows those that happening is a flawed test.
as for my solution to dealing with the consequences of a possible imbalance, I have two ideas... The first is to simple apply the 10X load to source ratio and place a pair of 1K resistors from the grids of the driven tubes to ground. (yes that was a joke...) The other is to simply put a nice differential driver stage immediately after. Lets face it we need some gain anyways so just toss a differential pair after and call it a day.
dave
Hey AC...
I really like this quote. I really think the major flaw in this whole type of discussion is simple communication via a keyboard and if you swapped the keyboard for a face to face with a cocktail napkin or ten I think everyone would learn (or maybe fisticuffs)
I would like to see the same group of people have the discussion on the load each tube sees from a 5K plate to plate PP output transformer since I see a lot of the same arguments happening.
dave
Yes, that is what I have said all along. That much is child's play.
That would be an odd way to find output impedance; to find the source impedance of a Thevenin or Norton equivalent circuit you must set all independent sources to zero... i.e., you don't start with any input voltage.
The trouble is, a phase inverter has at least three ports: anode, cathode and ground. Therefore, there are at least three output impedances we might be interested in:
1: Cathode to ground.
2: Anode to ground.
3: Anode to cathode (the differential output imepdance)
No one is saying that all three of these are the same- demonstrably they are not. The anode output impedance is large, and the cathode output impedance is small. The differential impedance is something else.
But there is a 'special case': If we apply identical loads between each output and ground, it is exactly equivalent to connecting those loads in series between anode and cathode, without the ground connection. Under this special circumstance, the "effective" output impedance from anode or cathode can be taken as half the differential output impedance. But these effective figures are merely a contrivance that works for the purposes of finding bandwidth or gain. They are NOT the "anode output impedance" or "cathode output impedance"; those terms are reserved for the true, Thevenin anode-to-ground or cathode-to-ground impedances. It sounds to me that the issue with SY's article is that he uses these special terms too loosely, possibly leading to misunderstanding?
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Thanks Dave. With those steps, you've measure the 6 "open-circuit" parameters of that particular circuit. (Well, almost; technically, you'd have to remove those 470k resistors paralleling the test current sources but close enough!)
Regardless of how you use the circuit and regardless of its intended use, those parameters are what they are within the context in which they are valid: linear, AC analysis.
You mention something about how the circuit is used is all you care about. So sure, if plan to take the outputs differentially, then, in that limited context, what could be wrong with pretending the plate and cathode output impedances are equal, right?
But consider the following as one example of many why you might care about the genuine plate and cathode output impedances.
Every real component generates noise. When we model noise in circuits, each two-terminal circuit element gets a noise current source across it (or equivalently, a noise voltage source in series with it). In a real circuit, there will be noise currents circulating in the plate and cathode output circuits and, in general, they won't be correlated and more, you can't get rid of them. The noise currents will produce noise voltages in proportion to the genuine plate and cathode output impedances which are vastly different.
I think Chris mentioned something earlier about power supply noise and grid current from the following stage. Bottom line, for any circuit non-ideality that results in a common-mode current in addition to the ideal differential-mode load current, you will need to consider the genuine plate and cathode output impedances.
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Alfred, tell me where we diverge in our thinking:
(1) With ip and ik equal and opposite, Vp = -Vk, so we meet the "boundary conditions".
(2) Just using Kirchoff's laws, we can completely solve for Vp and Vk in terms of ip and ik, right? Vp = Vp(ip) + Vp(ik). Vk = Vk(ip) + Vk(ik).
(3) We find the resistor for the Thevenin equivalent of the plate - ground circuit by taking the partial derivative of Vp w.r.t. ip. We do similarly for the cathode - ground circuit.
What am I missing?
(1) With ip and ik equal and opposite, Vp = -Vk, so we meet the "boundary conditions".
(2) Just using Kirchoff's laws, we can completely solve for Vp and Vk in terms of ip and ik, right? Vp = Vp(ip) + Vp(ik). Vk = Vk(ip) + Vk(ik).
(3) We find the resistor for the Thevenin equivalent of the plate - ground circuit by taking the partial derivative of Vp w.r.t. ip. We do similarly for the cathode - ground circuit.
What am I missing?
Scale. If the cathodyne's noise (whether from the power supply, the resistors, or the tube) is significant compared to the 10-50V required at output tube grids, then you have a fundamental problem that needs fixing. If you have noise pickup in the few inches under chassis, ditto.
As I mentioned before, long twin shielded leads from the ImPasse to a balanced-in power amp showed no significant noise pickup. For the far-more-common use (phase splitter inside an amp), this objection doesn't hold much water, it's more analogous to the audiophile obsessing over 0.002% vs 0.001% THD.
Hi Merlin, I am glad that we agree about what you call "child's play." I do not believe SY does.
It is an odd way to find impedance. Normally, I'd do just what you say. But if I want to satisfy the "boundary conditions" demand of SY of having the circuit operate perfectly balanced, that is, Vp = - Vk, then this is the only way I can figure to measure P-gnd and K-gnd impedances.
Agree completely with your 1-2-3. And your subsequent paragraph, except: SY and some others believe that Anode to ground and plate to ground Z's are equal and about 1/gm.
I think your last paragraph hits the nail on the head. I believe that SY believes that what you rightly call a contrivance is real.
It is an odd way to find impedance. Normally, I'd do just what you say. But if I want to satisfy the "boundary conditions" demand of SY of having the circuit operate perfectly balanced, that is, Vp = - Vk, then this is the only way I can figure to measure P-gnd and K-gnd impedances.
Agree completely with your 1-2-3. And your subsequent paragraph, except: SY and some others believe that Anode to ground and plate to ground Z's are equal and about 1/gm.
I think your last paragraph hits the nail on the head. I believe that SY believes that what you rightly call a contrivance is real.
Chris, I see that we're thinking in two different modes here. Of course you are correct, that analytically, the individual output impedances can be calculated under the BCs ("boundary conditions") as long as you know the functions you're taking the partial derivative of, right?
However, if you don't know these (say, you're given SY's black box), you have to measure them. But the functions can't be measured without operating the circuit outside the BCs.
So, it seems to me, the only impedance that can be measured while holding to the BCs is the differential output impedance which is a linear combination of the open-circuit output impedance and the transfer impedance.
Am I still missing something?
However, if you don't know these (say, you're given SY's black box), you have to measure them. But the functions can't be measured without operating the circuit outside the BCs.
So, it seems to me, the only impedance that can be measured while holding to the BCs is the differential output impedance which is a linear combination of the open-circuit output impedance and the transfer impedance.
Am I still missing something?
And not one has yet shown any experimental data contradicting the Thevenin model shown in Figure 3 without trying one way or another to unbalance the load.
If we can't agree on how to model, it is nonsensical to present data in support or refutation of one.
Dave, I've been thinking about your simulation with the current sources and one of your comments and something occurred to me.
In case I haven't been clear, the current sources on the output are what are called test sources. Connecting a test current source to an output and measuring the resulting voltage due to the test current is, conceptually, equivalent to connecting an (AC in this case) ohm-meter from ground to one of the outputs.
That's the reason I asked you to put the test sources in there. It's one way to measure input or output impedances.
When you activate just one, you're measuring the output impedance of one output and the transfer impedance to the other.
When you activate both such that they act as a single source, you're measuring the differential output impedance. It's conceptually like placing the ohm-meter across the outputs.
Anyhow, that's the purpose of the test sources. It's a common technique analytically, in simulation, and in actual circuits.
It's also possible to do this with test voltage sources. If we were trying to find the admittance parameters (the short-circuit parameters), we'd have used test voltage sources.
In case I haven't been clear, the current sources on the output are what are called test sources. Connecting a test current source to an output and measuring the resulting voltage due to the test current is, conceptually, equivalent to connecting an (AC in this case) ohm-meter from ground to one of the outputs.
That's the reason I asked you to put the test sources in there. It's one way to measure input or output impedances.
When you activate just one, you're measuring the output impedance of one output and the transfer impedance to the other.
When you activate both such that they act as a single source, you're measuring the differential output impedance. It's conceptually like placing the ohm-meter across the outputs.
Anyhow, that's the purpose of the test sources. It's a common technique analytically, in simulation, and in actual circuits.
It's also possible to do this with test voltage sources. If we were trying to find the admittance parameters (the short-circuit parameters), we'd have used test voltage sources.
Unfortunately, amplifiers aren't built from algebra. 😀 The mechanistics of WHY my simple model works that Alfred and Burkhardt have done is interesting but doesn't change the results.
SY and others reading, two different models cannot both be correct. Since my model and SY's give identical results when the "boundary conditions" are met, we need a "razor" to determine the correct model. This razor is, I submit, the question: which model gives the correct results when the "boundary conditions" are not met even if just by an infinitesimal amount?
You decide. It's your mind, your choice.
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Yes, I believe one thing that we can all agree on is that "amplifiers aren't built from algebra." Thank you for that.
We have to celebrate consensus where we can find it!
We have to celebrate consensus where we can find it!
Of course they can, happens all the time. For example, Schroedinger's equation versus Heisenberg matrices- they give the same answers, but depending on the problem, one may be simpler and more convenient than the other. Likewise, if I'm computing the path of a thrown baseball, I'm not going to worry about relativistic corrections or multibody solutions for internal molecular vibrations.