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.
lets back up to the two terminal black box.
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.
now we need to short that current source out for accurate results.
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.
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.
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.
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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?
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.
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.
Chris, Alfred may well have been flirting with the line here. But if you are concerned about personal attacks, you may be interested that in post 469, SY questioned whether I am able to count to two.
What do you think?
Personally, I get pretty frustrated by some of what gets written here. But if I posted what I really thought, I'd be banned in a New York minute.
Personal stuff gets in the way of trying to find the truth. I like to think that we're all engaged in trying to find the truth. But maybe I'm just a Pollyanna.
What do you think?
Personally, I get pretty frustrated by some of what gets written here. But if I posted what I really thought, I'd be banned in a New York minute.
Personal stuff gets in the way of trying to find the truth. I like to think that we're all engaged in trying to find the truth. But maybe I'm just a Pollyanna.
I suspect Thévenin' would be happy.
which dead guy gave you permission to look inside the black box?
dave
My dead guy tells me that if I can't look inside the box, and I don't first measure between the nodes to see if there is a voltage there, I can't test by just connecting a test source across the nodes.
So I make sure to first measure the voltage. The test method I choose is dictated by the presence or absence of a voltage at the frequency I want to test.
So I make sure to first measure the voltage. The test method I choose is dictated by the presence or absence of a voltage at the frequency I want to test.
Nearly 200 posts ago, I stated that there was a LOT of debate but NO agreements....
Has this been resolved yet now we're up to 474 Posts on the subject...
After all, There's Only Three Parts in the circuit Not inc the PSU!
BTW, Outta curiosity, How much in V P-P in each phase is the maximum would it be possible from a 6SN7 in Concertina running 320V +B....??
Has this been resolved yet now we're up to 474 Posts on the subject...
After all, There's Only Three Parts in the circuit Not inc the PSU!
BTW, Outta curiosity, How much in V P-P in each phase is the maximum would it be possible from a 6SN7 in Concertina running 320V +B....??
OK then, I don't look. Good rule. So I inspect the voltage between the two nodes of interest.
If there is a component at the frequency I'm interested in, I record it. Then I short the nodes and record the short circuit current at the frequency of interest. I divide the recorded voltage by the recorded current to get the impedance.
If there is no component at the frequency of interest, I connect a signal source at the frequency of interest to the two nodes. The ratio of the voltage across to the current through the source at the frequency of interest is the impedance.
If there is a component at the frequency I'm interested in, I record it. Then I short the nodes and record the short circuit current at the frequency of interest. I divide the recorded voltage by the recorded current to get the impedance.
If there is no component at the frequency of interest, I connect a signal source at the frequency of interest to the two nodes. The ratio of the voltage across to the current through the source at the frequency of interest is the impedance.
Well, Alastair, we're doing our best. Care to join in?
The answer to your question is, unfortunately, "It depends." Since the voltage gain from the grid to plate is roughly -1, and from grid to cathode is roughly +1, one limit is how far the stage driving the Concertina grid can swing.
A second limit is the DC bias voltage between the plate and cathode. Ignoring limitations on the stage driving the Concertina grid, maximum symmetrical p-p swing would occur if the bias were Ebb/2. Then the cathode could swing from 0V to Ebb/2 while the plate went from Ebb to Ebb/2. However, the electrodes wouldn't get to Ebb/2, because you'd probably be drawing Concertina grid current at that point. But this gives you a rough idea of where to start in the analysis.
The answer to your question is, unfortunately, "It depends." Since the voltage gain from the grid to plate is roughly -1, and from grid to cathode is roughly +1, one limit is how far the stage driving the Concertina grid can swing.
A second limit is the DC bias voltage between the plate and cathode. Ignoring limitations on the stage driving the Concertina grid, maximum symmetrical p-p swing would occur if the bias were Ebb/2. Then the cathode could swing from 0V to Ebb/2 while the plate went from Ebb to Ebb/2. However, the electrodes wouldn't get to Ebb/2, because you'd probably be drawing Concertina grid current at that point. But this gives you a rough idea of where to start in the analysis.
Dave, he's being cute. He means that it's not just wrong, it's very wrong. And it's wrong for all the reasons we've been saying. SY does not apply Thevenin correctly. He shorts three nodes together (plate, cathode and ground) to measure his short circuit current. Thevenin only deals with and shorts two nodes.
I checked the Wikipedia reference for Thevenin. It's pretty good, and pretty short. I recommend you check it out if you are not familiar with the Thevenin theorem. You'll see it only deals with two nodes in a circuit.
I checked the Wikipedia reference for Thevenin. It's pretty good, and pretty short. I recommend you check it out if you are not familiar with the Thevenin theorem. You'll see it only deals with two nodes in a circuit.
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