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Old 30th November 2011, 11:38 PM   #111
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My opinion is that , in Concertina the output impedance of the anode to the ground is nothing else than the output impedance of an common cathode triode amp , and the output impedance of the cathode to the ground is equal to the output impedance of a cathode follower triod buffer . Of course we have equals output voltages in the two circuit if they aren't connected to anything else , or if they connectet to easy loads , but if we connect them to harder loads and start to drive the Concertina harder , we will see that the voltage of the anode start to fall comparing it with the voltage at the cathode , that happens because they haven't equal impedances , the anode have higher one ..
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Old 30th November 2011, 11:40 PM   #112
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The analogy must be considered for all cases, not just the one you prefer.

Last edited by CPaul; 30th November 2011 at 11:43 PM. Reason: removved last sentence; shouldn't have been there.
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Old 30th November 2011, 11:42 PM   #113
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Let us speak in generalities, but in generalities that are compatible with your demand for impedance tests of a balanced Cathodyne that maintain balance during testing.
I have a circuit with nodes A and B between which I want to determine the impedance Zab. To do so, I need to somehow create a loop that puts current Itest into node A, out of node B, past a point outside the circuit, and returns it to point A. I need to measure the change in voltage deltaV between A and B due to that current and that current only. I can then divide. deltaV by Itest to get Zab. If there is no current flowing through A or B, I can’t measure Zab.

In your circuit, node B is ground and there is no current flowing through it. So you can't measure the impedance between nodes A and B.
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Old 30th November 2011, 11:45 PM   #114
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Agreed, Dmitris, if there is only one harder grounded load that is switched between the Plate and Cathode for comparison.
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Old 30th November 2011, 11:55 PM   #115
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We agreed only in this case ? , but we can not ignore the fact that from the begining the two outputs have different impedances .
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Old 1st December 2011, 12:04 AM   #116
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Agreed. Even with equal loads, anode impedance is much higher than that of the cathode.
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Old 1st December 2011, 12:10 AM   #117
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That exactly what i am saying too .
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Old 1st December 2011, 01:25 AM   #118
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You can't say that. Because doing measurements changing loads separately you disbalance Concertina, so it is not a Concertina anymore. Try to change both loads at once, equally, to keep it balanced, and calculate both resistances separately.

Edit: What is remarkable, chemist SY understands electronics better than the majority on this forum, including many Electrical Engineers. What is so special in chemical education, so it so superior to education of Engineers?
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Last edited by Wavebourn; 1st December 2011 at 01:31 AM.
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Old 1st December 2011, 01:42 AM   #119
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Quote:
Originally Posted by CPaul View Post
As an Electrical Engineer of 30 years, I'm familiar with equivlent circuits.
Excellent. Go back to my figure 3 showing two amplifiers with two outputs.
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Old 1st December 2011, 02:23 AM   #120
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Quote:
Originally Posted by Wavebourn View Post
What is remarkable, chemist SY understands electronics better than the majority on this forum, including many Electrical Engineers. What is so special in chemical education, so it so superior to education of Engineers?
I think what's throwing so many folks is that the concertina has a floating output but its feedback is not symmetrical. If tested with equal and opposite currents driven into its outputs (IOW, as a concertina works) point impedances will measure equal. But if measured with equal but not opposite currents, or just single currents, driven into its outputs point impedances will not measure equal (because its no longer a concertina).

So we have arguments about angels dancing on the heads of pins.

Thanks,
Chris
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