No, I don't agree with that. You've made an elementary mistake. Like Chris said, we're not making this stuff up as we go.
There are basic, fundamental results that all of circuit theory depends on. I find it remarkable that you're seriously questioning them.
For example, the simulator you are using is based in part on the very circuit fundamentals that you are calling into question!
Don't you see how utterly hopeless it is to use that simulator, which is based on these fundamental results, to question those very results?
Last edited:
(1) When you short each output node to ground, you've connected three nodes together; the two output nodes and ground. This is not in accord with Thevenin's theorem, end of story. To find the Thevenin equivalent circuit between two nodes, you measure the open-circuit voltage between those two nodes and then short those two nodes and only those two nodes together and measure the short-circuit current.
(2) By simply shorting the two output nodes together without also connecting them to ground, you're in accord with Thevenin's theorem and with the "boundary conditions". Now, you can apply Thevenin's theorem and determine the impedance between the output nodes.
Last edited:
How can you look inside the black box to see if there is a independent source to zero?
dave
OK, so you're either saying that it is impossible to have a pair of Thevenin sources (the "counting to two" problem), or "I want to violate the boundary conditions by loading the two outputs differently," in which case you're not dealing with anything I've presented.
My 2 eurocents worth
The open-voltage / shorted current is just a limiting case for determining Zout. You might as well use a load of 10k and 5k and measure different voltages/currents for the two cases to calculate the Zout. In the limiting case you move one load to infinity ('open voltage') and the other to zero ('short circuit current') but it's the same concept.
In the case 10k/5k there's no issue of 'shorting three nodes, not two'. You're not shorting three nodes, your shorting two Thevenin sources. That, in the case of zero load, the shorted nodes happen to be all the same, ground, is just incidental but not specific to the case.
Thevenin indeed never set boundary conditions, but SY is fully entitled to, as it is part of his claim: "IF the loads are the same, THEN the Zout..."
Seems clear to me.
jan
The open-voltage / shorted current is just a limiting case for determining Zout. You might as well use a load of 10k and 5k and measure different voltages/currents for the two cases to calculate the Zout. In the limiting case you move one load to infinity ('open voltage') and the other to zero ('short circuit current') but it's the same concept.
In the case 10k/5k there's no issue of 'shorting three nodes, not two'. You're not shorting three nodes, your shorting two Thevenin sources. That, in the case of zero load, the shorted nodes happen to be all the same, ground, is just incidental but not specific to the case.
Thevenin indeed never set boundary conditions, but SY is fully entitled to, as it is part of his claim: "IF the loads are the same, THEN the Zout..."
Seems clear to me.
jan
janneman, I agree that you can use 5K and 10K loads to test impedanaces - you don't have to use open and short circuits.
But look at Thevenin's theorem. Wikipedia actually has a good presentation. The test is about two nodes and two nodes only. There is no room in the theorem for tests involving three nodes. Thevenin doesn't know and doesn't care about SY's "boundary conditions." Thevenin's theorem applies to any two nodes in any linear circuit. No special cases; no special pleading.
Whenever SY does a test with three nodes, he automatically violates Thevenin.
But look at Thevenin's theorem. Wikipedia actually has a good presentation. The test is about two nodes and two nodes only. There is no room in the theorem for tests involving three nodes. Thevenin doesn't know and doesn't care about SY's "boundary conditions." Thevenin's theorem applies to any two nodes in any linear circuit. No special cases; no special pleading.
Whenever SY does a test with three nodes, he automatically violates Thevenin.
SY, Thevenin's theorem relates to any two nodes in any linear circuit. It does not address pairs of equivalents in a single circuit. Whenever you attempt to do this, you are outside the bounds of Thevenin's theorem.
As Alfred and I have said before, you can have both your boundary conditions and one and one only Thevenin equivalent. Just divide the open circuit difference between the p & k voltages by the current that flows between the p & k when they are shorted together. You get a little less than 2/gm. Both Thevenin and your boundary conditions are satisfied. But the result is a single source - not two sources with impedances of 1/gm.
As Alfred and I have said before, you can have both your boundary conditions and one and one only Thevenin equivalent. Just divide the open circuit difference between the p & k voltages by the current that flows between the p & k when they are shorted together. You get a little less than 2/gm. Both Thevenin and your boundary conditions are satisfied. But the result is a single source - not two sources with impedances of 1/gm.
They can be done simultaneously, if you wish. Nothing prevents that- this isn't quantum mechanics!
Ah, so we've now moved from the "I can't count to two" problem to the "I want to violate the boundary condition" problem. Pick one and stick with it.
If you short one output, you have to short the other. That's the "given." If you can't accept the given, then you have no valid critique of my results given the given.
If you short one output, you have to short the other. That's the "given." If you can't accept the given, then you have no valid critique of my results given the given.
No, SY, that is an incorrect analysis.
I understand your "given" to be that I must always keep the Cathodyne balanced, not that I must short the p & k to ground. Or has the "given" changed?
If I acceed to your demand that I satisfy what I understand to be your "given", an always balanced Cathodyne, then surely it is a small effort on your part to agree to adhere to Thevenin. You can't sacrifice Thevenin on the altar of your boundary condition.
Short the p to the k only. Satisfy your boundary condition, and Thevenin at the same time, and be done with it. Arrive at Vogels result.
I understand your "given" to be that I must always keep the Cathodyne balanced, not that I must short the p & k to ground. Or has the "given" changed?
If I acceed to your demand that I satisfy what I understand to be your "given", an always balanced Cathodyne, then surely it is a small effort on your part to agree to adhere to Thevenin. You can't sacrifice Thevenin on the altar of your boundary condition.
Short the p to the k only. Satisfy your boundary condition, and Thevenin at the same time, and be done with it. Arrive at Vogels result.
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.