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#941 |
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diyAudio Moderator
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So what? The loads are connected to ground. That's where we measure them (those of us who actually build and test things) and how the voltages are detected by following stages (if one actually builds amplifiers). The symmetry forces their currents to cancel.
The "equal source impedance" model I presented gives accurate results. The fact that one can model it in other ways doesn't change that- the word "model" is the operant term here.
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If there's a sucker born every minute, where do the rest of them come from? |
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#942 |
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diyAudio Member
Join Date: Nov 2007
Location: Dallas
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Try #908 as-is. It has equal Z and can drive non-linear loads into A2.
Because inductive couplings (at low frequency) prevent DC pumping. The main coupling is capacitive, and not abusing transformer coupling to even the split. Transformer has 2K2 incentive not to get involved and let the caps do the job... Change ratio of cathode feedback attenuation, so its not same as Mu. Observe what non-equal Z's then do with non-linear but equal loads. This test might be valid, because DC pumping is no longer complicating interpretation of the result. Even with local impedance not equal, global feedback path is probably forcing splitter impedance very close to equal. We may have to open the big loop too, to tell any difference. Last edited by kenpeter; 1st October 2012 at 05:04 AM. |
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#943 |
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diyAudio Member
Join Date: Nov 2011
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"So what?" Well, here’s what: You now accept that you are measuring the P - K impedance, not two separate P and K to ground impedances.
The reason that your prediction is successful is that it is that P-K impedance, and not the P and K to ground impedances, that is crucial in determining balanced Cdyne operation. It is wholly appropriate that you now place the name of your model, "equal source impedance" in quotes. Because you clearly haven't measured the impedances of the two sources which you claim to be equal. You’ve only measured the impedance of one. It is crucial, when modeling, to know what is being modeled.
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"It is error only, and not truth, that shrinks from inquiry." - Thomas Paine |
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#944 |
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diyAudio Member
Join Date: Nov 2011
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KenPeter, I'd like to take a look at the 908 circuit and get a better feel for it. I think it'd be best to start with the simpler case, without global feedback, and then add it back in.
Do you have any thoughts about the difference between its operation and that of the earlier circuit with no caps bypassing the transformer? I guess the bandwidth of the newer one would be better. Hope the beer was good. You don't mix that with soldering irons, do you?
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"It is error only, and not truth, that shrinks from inquiry." - Thomas Paine |
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#945 |
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diyAudio Moderator
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In this case, they are equivalent. The symmetry argument is a feature, not a bug.
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If there's a sucker born every minute, where do the rest of them come from? |
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#946 |
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diyAudio Member
Join Date: Nov 2011
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Demonstrably false in general.
I have a three node network, P, K and ground. There is a 100 ohm resistor between P and K, a 10K resistor from P to (G)round and a 20K resistor from K to G. The total impedances Zpg, Zkg and Zpk are unequal, and the sum of any two are unequal to the other. It’s rather strange that you would be making predictions about impedances that we both now agree you were not measuring before, and in fact have yet to even propose, let alone prove, a correct method of testing.
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"It is error only, and not truth, that shrinks from inquiry." - Thomas Paine |
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#948 |
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diyAudio Member
Join Date: Nov 2011
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I interpreted "So what" in 941 to mean that you accept the point that you were only measuring Zpk. I stated that in 943. You had no response to that then. Why has it suddenly become an issue in 946? Are you suddenly claiming that you were measuring P and K to G impedances after all? Do we have to go back to that again?
At least don't go back on what you've already said. And let's not get off on a red hering. I have a three node network, P, K and ground. There is a 100 ohm resistor between P and K, a 10K resistor from P to (G)round and a 20K resistor from K to G. The total impedances Zpg, Zkg and Zpk are unequal, and the sum of any two are unequal to the other. There is no promise of the symmetry you keep calling for. Simply because you have created a model that has symmetry and makes a prediction that is useful does not validate the model. Your argument begs the question – assumes the conclusion. It’s similar to the following: A model states that objects must be viewed through rose-tinted goggles to view the object’s color. The model’s statement that rose is the only color in existence is vindicated by implementing the model. Therefore, the model is correct.
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"It is error only, and not truth, that shrinks from inquiry." - Thomas Paine |
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#950 |
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diyAudio Member
Join Date: Nov 2011
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Your constraints constrain your model – not the circuit. To assume otherwise is voodoo-like: make a model of something, do something to it, conclude the thing that was modeled must be affected. (Put on your rose-tinted goggles.)
You are simply begging the question: assume a model that requires balance. When the model does something useful, assume it correctly defines the circuit. I already demonstrated that your model measured Zpk, not Zpg, not Zpk. You did not even attempt to refute that: your response was, "So what." You have no basis on which to assert anything about something you haven't measured.
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"It is error only, and not truth, that shrinks from inquiry." - Thomas Paine |
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