Re: What's happened to this NG????

In article <76idnae3VciGzvnanZ2dnUVZ_t2inZ2d@xxxxxxx>,
"News Client" <newsclient@xxxxxxxxxxxxxxxxxxx> wrote:

"Jack Crenshaw" <jcrens@xxxxxxxxxxxxx> wrote in message
news:13m8g09kokutp63@xxxxxxxxxxxxxxxxxxxxx
Hmm. I guess I never thought of that. Ordinarily, and strictly from an
aesthetic POV, I'd be violently against adding resistance
to _RAISE_ the output impedance of a naturally low-impedance ckt. But if
the alternative is a huge imbalance as the signal
drivers the output grids into conduction, I guess I can see the point.

I'd much more prefer a driver stage between them, as in the Heath W-5M, so
the output impendance of the p.i. stage doesn't matter
so much.

There is no imbalance. Or at least not the way you are thinking. A
build-out resistor, as some designers use, is completely the wrong
solution to the problem that may not even need to be solved.

If the two outputs of the split-load inverter were independent, and if
the output impedances were different, then the magnitude gains and
frequency responses from the input to the two outputs would be
different when driving matched loads. To a close approximation, this
is not true. The gains and frequency responses are the same. This
is obvious when you consider, neglecting strays and grid current, that
the plate and cathode currents are exactly equal.

Balanced operation of the split-load inverter depends on equal loading
of the outputs. Any time you do something unequal to one output, you
end up with unbalanced behavior. That includes trying to measure the
output impedance one output at a time. The reason for this is that the
outputs are not isolated and independent. Once you realize that there
is strong coupling between the outputs, the seeming paradox of the
split-load inverter is resolved.

Morgan Jones, for what it's worth, discusses this in detail in "Valve
Amplifier". He has a separate analysis of the output impedance for
equal and unequal loading. At the end of the section titled "Output
resistance with both terminals equally loaded (Class A1 loading)"
he writes, "Because Zk = Za, the frequency response at each output
is forced to be the same, so the output resistances must also be
equal, and rout(k) = rout(a)."

It is worth noting that the first edition of "Morgan Jones" advocates
using a build out resistor in the cathode circuit to equalize the source
impedances, in the later editions he changes his tune. I have often
wondered how he came up with the build out resistor idea he advocated in
the first edition, or was it a Wireless World article where he first
presented the build out resistor idea?


Regards,

John Byrns

--
Surf my web pages at, http://fmamradios.com/
.

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