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Old 20th May 2011, 11:30 AM   #21
DF96 is offline DF96  England
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Quote:
Originally Posted by artosalo
The results are as follows:
Your measured results are exactly what theory predicts. They show that, taken in isolation, the cathode has lower impedance. Taken together (i.e. load both outputs equally), equal load gives equal voltage. Taken in isolation, the impedance at the anode is equal to the anode resistor because the valve anode resistance is huge due to the large amount of cathode degeneration.
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Old 20th May 2011, 11:51 AM   #22
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For interest, following the concertina, if a Williamson diff is used with a small amount of regenerative feedback from the split cathode resistor to the grids with a CCS to ground, the diff driver offers perfect capacitive balance for the concertina, proof is the near cancelling of the 2nd harmonic see trace....this only occurs with near matched tubes but illustrates how excellent this combination is.
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Old 20th May 2011, 11:56 AM   #23
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Quote:
Quote:
Then, if the load impedance is too low - but equal at both outputs - the output voltages can be different.
This is entirely incorrect.
This is already clear and agreed.

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What frequency are you measuring at ?
1 kHz. The reason is low (maybe 2...3 Mohms) input impedance of HP, not the capacitance I guess.
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Old 20th May 2011, 12:05 PM   #24
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Quote:
Originally Posted by richwalters View Post
For interest, following the concertina, if a Williamson diff is used with a small amount of regenerative feedback from the split cathode resistor to the grids with a CCS to ground, the diff driver offers perfect capacitive balance for the concertina, proof is the near cancelling of the 2nd harmonic see trace....this only occurs with near matched tubes but illustrates how excellent this combination is.
This is also true for p-p output stages. One point on which there have been some errors in standard texts is the issue of class A versus class AB/B p-p output stages- it makes no difference, the load presented to the cathodyne is still equal.
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Old 20th May 2011, 12:26 PM   #25
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Quote:
Your measured results are exactly what theory predicts. They show that, taken in isolation, the cathode has lower impedance. Taken together (i.e. load both outputs equally), equal load gives equal voltage. Taken in isolation, the impedance at the anode is equal to the anode resistor because the valve anode resistance is huge due to the large amount of cathode degeneration.
This is easy to understand and agree.


Quote:
If the outputs are loaded equally (and that is indeed the case with the cathodyne in situ), the source impedances at plate an cathode are identical and low (~1/gm).
I see that it is now demonstrated that equal loads will give equal output voltages and unequal loads will give unequal output voltages.
Therefore the actual output impedances can not be equal, even if they seem equal in case of equal loads.

If the output impedance at anode were 1/gm, then the small change at anode load would not cause essential difference at output voltage.

According to Langford-Smith (Radiotron Designers Handbook):

At anode: Rout = (-1)Rk where Rk = Ra
At cathode: Rout = (rp+Ra)/(+1)

Summary: Equal loads will give equal output voltages and unequal loads will give unequal output voltages.

This proves that output impedances are NOT equal.
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Old 20th May 2011, 12:50 PM   #26
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Quote:
Originally Posted by artosalo View Post
I see that it is now demonstrated that equal loads will give equal output voltages and unequal loads will give unequal output voltages.
Therefore the actual output impedances can not be equal, even if they seem equal in case of equal loads.
Well, if the loads are equal, the output impedances HAVE to be equal, else the voltages would not be. With equal loads, the output impedances are not only equal, they are low. This can be seen by calculation of open circuit voltage divided by short circuit current (keeping loads equal), the usual definition of source impedance. This can also be seen by injecting equal (but opposite, to keep loads equal) test currents into the cathode and plate. And it can be verified by putting equal capacitances on each output and comparing rise times.

Please see my analysis and experimental verification in Linear Audio Vol 0.
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Old 20th May 2011, 01:42 PM   #27
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I see the explanation given at Radiotron Designers Handbook (4th edition) more natural and convincing:

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The effective output resistance is different for the two output channels, since P (plate output) operates with current feedback and K (cathode output) with voltage feedback...... but this does not affect the balance at either low or high frequencies when the total effective impedance of channel P is equal to that of channel K. The same signal plate current which flows trough one impedance Zp also flows through the other impedance Zk, and if Zp = Zk then the two output voltages are equal.
See 7.2 (ii) (B) page 330
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Old 20th May 2011, 01:51 PM   #28
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You might find it more convincing, but my analysis is backed up by experimental verification.

We do learn new things as time goes on.
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Old 20th May 2011, 02:53 PM   #29
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Don't underestimate coax capacitance. One is Easily fooled. Like RF in cables is able to create standing wave reflections that fight the tube capacitance. Anode in particular.
A test: A fast rise & fall 10Khz square wave with a dual input scope is a good check. One can also do the Lissajous test which is even more revealing. Do identical length coax connections to concertina upper and lower o/ps with exactly same type of coax and length, connect to scope Ch1& Ch2 same amplitude settings to get trace mid screen; and phase invert one channel and adjust shift to coincide.....one will find the upper concertina slews a bit more, which in other terms = slight phaseshift.
However, Sine wave amplitude is equal.
Some tubes may oscillate sensing capacitance, a 600R resistor in each O/p usually cures this.
I think <I've got this correct....disconnect one of the leads and everything becomes glaringly obvious !

richy
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Old 20th May 2011, 04:27 PM   #30
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Found it.......sorry SY....I uncovered this from my University notes. Everyone should read it.
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