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phase splitter issue
phase splitter issue
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Old 20th May 2011, 04:32 PM   #31
SY is offline SY  United States
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phase splitter issue
Yeah, I cited Preisman in my article. His analysis strikes me as an alternate but defensible view, albeit far more complicated than my simple Kirchoff approach.
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Old 20th May 2011, 04:59 PM   #32
artosalo is offline artosalo  Finland
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Quote:
Don't underestimate coax capacitance.
No, I do not. I know this phenomenon.
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Old 29th November 2011, 05:17 PM   #33
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Please refer to the Preisman article. Equations (1) and (2) and the surrounding text support Preisman's conclusions that "(cathode impedance) ... is much less than rp, and (the plate impedance) is greater than rp..."

The analysis Mr. Yaniger offers in his Linear Audio 0 article is unfortunately flawed. Plase see my reply to him at the Lnear Audiio Site under Online Resource in the Letters to the Editor section.
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Old 29th November 2011, 05:26 PM   #34
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phase splitter issue
Hi, Chris, nice to see you here!

I would indeed recommend people read your LTE and my response. Ditto Peter's LTE and my response.
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Old 29th November 2011, 05:47 PM   #35
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Hi Stuart, good to make the connection again, despite our differing views!
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Old 29th November 2011, 05:51 PM   #36
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phase splitter issue
For those who don't know him, Chris Paul has written some terrific articles over the years for various Audio Amateur publications. If you're using a mu follower, you probably have Chris to thank for it.

Welcome to the forum!
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Old 29th November 2011, 06:13 PM   #37
DF96 is offline DF96  England
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I think the problem with understanding the cathodyne is that we normally think of output/source impedance as applying to a single output port, but then try to apply the same idea to a device with two output ports which affect each other. Because we don't realise we are making the 'single port' assumption, we don't realise that it doesn't directly apply.

If the two ports really had different and independent impedances then equal loading would give unequal voltages, as has been pointed out. For the cathodyne we either have to insist on equal loads (when we find a low impedance at both ports) or apply the idea of a Thevenin equivalent to a single port at a time. We cannot apply the Thevenin equivalent separately to two inter-dependent ports.
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Old 29th November 2011, 09:39 PM   #38
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You say that if the two ports have different impedances, then equal loading would give unequal voltages. But thatís not really the case. The equal and opposite voltages of the Cathodyne work regardless of whether the plate and cathode impedances are different or identical. This is because of one simple characteristic of a triode: as long as no grid current flows, the current into the plate is always equal (and opposite) to the current into the cathode. As long as the plate and cathode loads are identical, the voltages must be equal and opposite by Ohmís law, regardless of the plate and cathode impedances.
Iím not sure why you say that equal loads on both ports leads to low impedances at both ports. Can you take me through that?
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Old 29th November 2011, 10:41 PM   #39
DF96 is offline DF96  England
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If the two output ports had different (and independent) impedances then they would drop different amounts of voltage as the load impedance changes. This means that only one load impedance could result in equal voltage output, or possibly none at all.

By low impedance at both ports I mean that provided you keep the two load impedances the same you get the same voltage at each, which corresponds to a cathode follower type impedance. For example, a load of 1/gm would give a gain of 0.5 to both ports - implying an output impedance of 1/gm at both ports. Of course, this is stretching the concept of impedance to cover two related ports.

I think the basic problem is that engineers are so used to the concept of impedance that they insist on using it in situations where it is poorly defined or undefined.
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Old 29th November 2011, 10:57 PM   #40
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But they do have different impedances. That is easy to see by simply setting the AC grid voltage to zero, and then inserting an AC current source first into the plate, and then into cathode. The plate voltage is going to be a lot larger, indicating a much higher impedance.

I agree that you can't strtech impedance to deal with two ports at once.

In a discussion like this, it's important to define impedance. I maintain that it is first a characteristic of two nodes. If you cause a current to flow between the two nodes, outside of the circuit under test, then the impedance of the node pair is the change in voltage between the nodes due to and divided by that current.

Do you agree? If not, what is your definition?
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