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Old 10th December 2011, 11:30 PM   #461
SY is offline SY  United States
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Please take a look at my Figure 3 which clearly shows two Thevenin sources. If it will help, I'll ask Jan for permission to post it, and I'll add numbers and arrows so you can clearly see that there are two of them there.
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Old 10th December 2011, 11:35 PM   #462
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The problem is that the impedances you ascribe to each electrode (plate and cathode) are arrived at by shorting three nodes together to obtain a short circuit current. There is nothing in Thevenin to justify this. He would short only two - the plate and cathode.
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Old 10th December 2011, 11:36 PM   #463
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And you are still ignoring what the current mirror tells you.
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Old 10th December 2011, 11:39 PM   #464
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Well, perhaps I now have an explanation as to why you're confused. In calculating Thevenin source impedances, I used the standard definition of open circuit voltage divided by short circuit current. Since there are two outputs whose loads are specified by the boundary conditions to be equal, and all currents and voltages are referenced to the same point ("ground"), we short each to ground to calculate short circuit currents. So there's two things shorted to ground, not three.
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Old 10th December 2011, 11:39 PM   #465
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Quote:
Originally Posted by Alfred Centauri View Post
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.
lets back up to the two terminal black box.

Click the image to open in full size.

can we all agree that whatever is in the black box has an impedance of 82 ohms?

Now lets reveal a bit of what is in the black box.

Click the image to open in full size.

now we need to short that current source out for accurate results.

Click the image to open in full size.

now we get 20.5K and that is the parallel combo of the 22K plate resistor and the apparent Rp of the DJ8 due to the 22K unbypassed resistor in the cathode. (its late and ian (gingertube) did the math way back.

My question is why do we need to short a device inside the black box to get the proper results?

Thévenin would be sad.
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Old 10th December 2011, 11:43 PM   #466
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Thevenin shorts only two nodes together. They might or might not include ground. Thevenin doesn't care. Thevenin never shorts three together, whether they include ground or not.

It is clear where the confusion lies.

Cite me a three-node Thevenin short.
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Old 10th December 2011, 11:44 PM   #467
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And Thevenin never discussed "boundary conditions". That is pure SY.
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Old 10th December 2011, 11:54 PM   #468
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Dave, I wish you wouldn't ascribe emotions to the dead. They get annoyed.

Second, if you are going to measure impedances in a circuit by inserting a test current into a pair of nodes, you must first set to zero all independent sources in the test circuit.

An independent source is one whose output signal is not influenced by any other signal. An example is a battery, or an oscillator. Your black box contained an independent source whose output was not zero.

Dependent sources are usually called controlled sources. A good example is a triode, whose p-k impdance is controled by its grid -cathode voltage.

Please repeat your experiment with all independent sources set to zero and report back.

I swear I am not making this stuff up. This is basic electrical engineering 101.

A better approach to testing your black box per Thevenin would be as follows:

If there is a voltage between the two nodes you select from your black box, then measure it. Now short the nodes and measure the current that flows. Then divide the voltage by the current. The result is the impedance of the nodes.

If there is no voltage between the nodes that you select to test, then you may connect a signal source between them. The impedance of the nodes is the ratio of the voltage between them to the current thorough the source.

Last edited by CPaul; 11th December 2011 at 12:01 AM. Reason: Another approach
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Old 11th December 2011, 12:39 AM   #469
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Quote:
Originally Posted by CPaul View Post
Thevenin shorts only two nodes together. They might or might not include ground. Thevenin doesn't care. Thevenin never shorts three together, whether they include ground or not.

It is clear where the confusion lies.

Cite me a three-node Thevenin short.
Can you count to two? There are two Thevenin sources in my model. Two. I'll count them for you: One. Two.

So, any progress on a pair of loads that will cause my Thevenin model to give the wrong results?
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Old 11th December 2011, 12:52 AM   #470
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SY, sarcasm ill becomes you and does not advance a rational discussion.

Please review the Thevenin theorem. It talks about two nodes and two nodes only.
You must test the plate to ground impedance by dividing the unshorted p-gnd voltage by the short circuit current that flows when you short only the plate to ground. This is the Thevenin procedure, recounted in many sources.

You must do the same for the cathode with the plate unshorted. Thevenin does not recognize your "boundary conditions". These are your invention.

If you followed the Thevenin procedure, your model would show two very different impedences. But you don't, so it does not.

As I explained, when we can agree on a model, we can test it. Not before.

And you continue to ignore the challenge to your analysis presented by the current mirror circuit. So far, there has been no challenge from you as to its accuracy. What is your justification for continuing to ignore it? I ask you as a person of intellectual integrity to either falsify it, or accept it.
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