sypedance:
sy·ped·ance /ˈsīˈpēdns/
n.
1. a ghost like source impedance that haunts, according to believers, the mythical perfectly balanced Cathodyne. Some speculate that sypedance is a curse from the grave cast by Thevenin's departed spirit on those that misapply his theorems.
2. a quantity associated with the sypedor, a chameleon like circuit element that defies all attempts to ascertain what it is.
3. a measure of opposition to any rational argument regarding the true nature of the Cathodyne output impedances.
sy·ped·ance /ˈsīˈpēdns/
n.
1. a ghost like source impedance that haunts, according to believers, the mythical perfectly balanced Cathodyne. Some speculate that sypedance is a curse from the grave cast by Thevenin's departed spirit on those that misapply his theorems.
2. a quantity associated with the sypedor, a chameleon like circuit element that defies all attempts to ascertain what it is.
3. a measure of opposition to any rational argument regarding the true nature of the Cathodyne output impedances.
Am I missing something here?
If we test as you insist and Look at the impedance from net 2 to ground, don't we have to zero the source V?? This makes the current mirror an open circuit so at net 2 the impedance to ground is simply that of the RC combo?
dave
Yes, the load is never part of the source impedance. The source impedance is a property of the source only, the load impedance is a property of the load only.
As you note, the current source is an open circuit so the source impedance for the source in "Net2" is infinite while the load impedance = R||(1/Cs)
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SY, if the currents through identical networks are identical, then they must give rise to identical voltages, which demand identical frequency responses. This is simpler than Thevenin - all we need is Ohm's law to explain this. Nothing more.
The fact that certain of your conclusions happen to be correct does not validate the assertions you made or the logic you employed to arrive at them. If someone said, "A Fairy told me the earth is round," I'd agree with the conclusion, but not with the reasoning used to arrive at it.
As for Thevenin, it is fairly easy to establish that he promulgated theories about two-node equivalent circuits, not the three node ones as you attempt to employ. You have not established a basis for quoting Thhevenin to back you up, unless you can find any reference to him dealing with three-node equivalent circuits.
And I do wish you would face the arguments I put forth instead of changing the subject or simply ignoring them. Find fault with the simple example circuit and reasoning that I offered, proving that equal rise times across identical impedances do not require equal source impedances. Seize the opportunity to prove me wrong! Why would you not?
As for your "counterexample" comment, I have no idea what you are talking about. You might want to state your challenge in its entirety.
I won't run away from your challenge. I trust you won't run away from mine.
The fact that certain of your conclusions happen to be correct does not validate the assertions you made or the logic you employed to arrive at them. If someone said, "A Fairy told me the earth is round," I'd agree with the conclusion, but not with the reasoning used to arrive at it.
As for Thevenin, it is fairly easy to establish that he promulgated theories about two-node equivalent circuits, not the three node ones as you attempt to employ. You have not established a basis for quoting Thhevenin to back you up, unless you can find any reference to him dealing with three-node equivalent circuits.
And I do wish you would face the arguments I put forth instead of changing the subject or simply ignoring them. Find fault with the simple example circuit and reasoning that I offered, proving that equal rise times across identical impedances do not require equal source impedances. Seize the opportunity to prove me wrong! Why would you not?
As for your "counterexample" comment, I have no idea what you are talking about. You might want to state your challenge in its entirety.
I won't run away from your challenge. I trust you won't run away from mine.
Oh great... circuit impedance... another term into the mix to allow folks to wiggle.
I thought you said that the mirror, has an impedance of 1/R. You only get infinite at the mirror if R=0 and if R=0 why include it at all? aside from the fact that the circuit wouldn't work without a finite value of R and a finite voltage V. If you zero both of those to do your test you aren't measuring anything of value. In other words, when the circuit ceases to function as intended, what is the point of analyzing it.
dave
If everytime the elephant standing upon my head rolls its bloodshot
eyes and waves its tentacles, so too does the one in the mirror. But
everytime the one in the mirror sneezes at Dave's lint, the elephant
standing upon my head does something entirely different. Can we
truly say they weigh the same?
eyes and waves its tentacles, so too does the one in the mirror. But
everytime the one in the mirror sneezes at Dave's lint, the elephant
standing upon my head does something entirely different. Can we
truly say they weigh the same?
Exactly. Ohms law is the first order and if it is not satisfied why bother making up excuses to get around it.
This can fairly be rewritten as:
The fact that some of your logic happens to be correct does not validate that the assertions made or the assumptions employed are of any concern to the situation at hand.
dave
I tried this earlier but lets take a parallel case.
Assuming an ideal 5K:8 output transformer, what is the load seen by each tube for pure class A operation?
One group will say it is turns ratio squared and insist it is 1.25K
The other group will say it is 2.5K
what say all of you? and back it up with consistent methodology for analysis of the cathodyne.
dave
Assuming an ideal 5K:8 output transformer, what is the load seen by each tube for pure class A operation?
One group will say it is turns ratio squared and insist it is 1.25K
The other group will say it is 2.5K
what say all of you? and back it up with consistent methodology for analysis of the cathodyne.
dave
Dave, a current mirror has an input and an output. The output current "mirrors" the input current, thus the name. The input resistance of the current mirror in Chris' example is r, the output resistance is infinite.
The 1/R your refer to is not the output impedance, it's the transconductance. The output resistance of an ideal current source is infinite period. That comes right out of the definition of ideal current source. An ideal current source drives the same amount of current through an external circuit regardless of the nature of the external circuit (short circuit up through any finite impedance).
Look guys, anyone reasonably well trained in the art, can take a look at Chris' circuit and, by inspection, tell you that the source impedance in Net1 is r and the source impedance in Net2 is infinite. They can also tell you the the load impedances are identical for the two nets. And, they can also tell you that the two sources create equal voltages across equal loads.
Bottom line: If you find that two different sources create equal voltages across equal loads, you cannot claim that the source impedances must be equal. They may in fact be but you cannot tell one way or the other with the method SY used.
No thanks Dave. The focus is on a single issue: Does the Cathodyne circuit, when perfectly balanced, have equal output ("source") impedances? According to SY, it does. According to my model, it doesn't.
Chris' example proves that SY's methodology is incapable of determining if the source impedances are equal. So, his claim that they are is unjustified.
Thus, since my model and his perform the same when the loads are identical, one way to determine which is correct is to "perturb" the balance slightly and see which model agrees with experiment. Without perturbing the balance, it simply is the case that the (individual) output impedances cannot be determined, only the balanced output impedance.
Chris' example proves that SY's methodology is incapable of determining if the source impedances are equal. So, his claim that they are is unjustified.
Thus, since my model and his perform the same when the loads are identical, one way to determine which is correct is to "perturb" the balance slightly and see which model agrees with experiment. Without perturbing the balance, it simply is the case that the (individual) output impedances cannot be determined, only the balanced output impedance.
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Kirchoff's Law tells you the voltages are the same. The Thevenin model tells you that the source impedances are low and identical. And it doesn't require a complex mixture of Thevenin and Norton sources and multiple branches. BTW, the Norton equivalent of my Thevenin model is equally valid and gives the same (correct) results.
Still, not a hint of a pair of loads that cause the model to fail, just repeated assertion that it somehow will. It's tough to argue against experimental verification, yet...
Still, not a hint of a pair of loads that cause the model to fail, just repeated assertion that it somehow will. It's tough to argue against experimental verification, yet...
IOW, violate the boundary conditions of "equal loads" under which I assert (and have demonstrated) that my simple model gives correct results.
Dave, forgive me for introducing confusion. We can talk about the impedance of the source, the impedance Z of the RC network, and the impedance between Net2 and ground - the impedance of the parallel combination of the source and the network, or the "circuit" impedance if you will. I won't use that term again.
The mirror has a transconductance of 1/r. This is the ratio of the mirror output current to its input voltage. The units are 1/ohms. So you see, it couldn't have an impedance of 1/ohms, the units of which must be ohms. And of course, the impedance of an ideal current source, regardless of its transconductance, is infinite.
Please pose your question again with this understanding.
The mirror has a transconductance of 1/r. This is the ratio of the mirror output current to its input voltage. The units are 1/ohms. So you see, it couldn't have an impedance of 1/ohms, the units of which must be ohms. And of course, the impedance of an ideal current source, regardless of its transconductance, is infinite.
Please pose your question again with this understanding.
SY, you continue to simply ignore what the current mirror says. You can't disprove it, so you simply ignore it. You refuse the challenge.
You are not applying Thevenin. You are applying SY's twist to Thevenin. Thevenin is a two-node theorem. Any citation will show you that. Show me anywhere that Thevenin discussed three-node applications. Or Norton, for that matter. Neither ever did. Only if you could do so could you credibly claim that either of them backs you up. But they didn't. So you can't.
As far as finding a pair of loads that cause the model to fail, why would anyone expend such an effort to disprove a faulty model? Only if the model is logically consistent should an effort be employed to test it. It's as if you're claiming to have a whistle that keeps elephants away and daring us to find any elephants in the vicinity to disprove the efficacy of your whistle!
Let us all try to pay attention to the rules of logic here.
You are not applying Thevenin. You are applying SY's twist to Thevenin. Thevenin is a two-node theorem. Any citation will show you that. Show me anywhere that Thevenin discussed three-node applications. Or Norton, for that matter. Neither ever did. Only if you could do so could you credibly claim that either of them backs you up. But they didn't. So you can't.
As far as finding a pair of loads that cause the model to fail, why would anyone expend such an effort to disprove a faulty model? Only if the model is logically consistent should an effort be employed to test it. It's as if you're claiming to have a whistle that keeps elephants away and daring us to find any elephants in the vicinity to disprove the efficacy of your whistle!
Let us all try to pay attention to the rules of logic here.
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