>
>This fits well with the topic of people not knowing everything, and
>making mistakes.
>
>> >> is Maxwell's conjugate matching theorem. If it were 100% true,
>> >> amplifiers could be no more than 50% efficient. .
>> >
>> >And the logic behind that statement is???
>
>Rich wrote:
>
>> ? Power transfer is maximal when the resistance of the load equals the
>> internal of the generator. At this point, efficiency is 50%. However,
>> in practice amplifiers are c.60% efficient.
>
>You've made the common mistake of looking at "pictures in books"
>and not reading all the text explaining the pictures.
>
>The rules of the model you are trying to use are:
>
>1.) The model can never be used to explain actions in the source.
>Thevenin was very clear about this. It only describes the box as a
>two terminal black box with a non-dissipative resistance and a
>PERFECT zero impedance voltage source..
>
? I am not trying to use Thevenin's. I am using the max power transfer
theorem, which essentially states that when half of the total P is being
dissipated in the generator, transfer of P to the load is max.
>2.) The source must be linear (by that they mean it has to have the
>same characteristics as power is varied) in order to be modeled
>that way.
>
>3.) The Thevenin model can be FREELY interchanged with a
>Norton model that uses a resistance in parallel with a perfect
>current generator by changing resistance parameters to admittance
>form.
>
>If you do the interchange you will find your 60 percent amplifier
>model made by wrongly applying Thevenin's theorem to a PA
>suddenly becomes less than 50% efficient with the very same
>parameters of operation.
>
? I don't see how doing the Thevenin/Norton interchange is going to
solve the discrepancy. If the T/N interchange is valid, why not do it
twice and we are back to square one?.
>The maximum power transfer theorem clearly states maximum
>power is transferred to the load when the system is conjugately
>matched, and that if a system is conjugately matched at one point
>it is conjugately matched at all points in the linear portion of that
>system. None of this has anything to do with dissipative
>resistances,
? True . It has everything to with dissipative resistances -- i.e., the
dissipative ESR in the generator and the dissipative resistance of the
load. When power in the load is the maximum possible, an equal amount of
power will be dissipated in the ESR in the generator. Thus, the
efficiency is 50% at max. power transfer to the load.
> and only a person who never read the theorems would
>think it did or that it could be used to describe efficiency in a PA.
>
? Yours truly used to teach Thevenin's Theorem and Norton's Theorm to
technicians who worked at the Pacific Misslie Range. I think I see what
your trying to get at, and some of Maxwell's stuff makes sense, however,
this apparently is not good enough for the boys in Newington.
>Breaking two rules of a model is not a good way to illustrate you
>understand the model.
>
>> >It's a bad idea, for a multiband amplifier, to increase unwanted
>> >feedthrough.
>>
>> ? The hat trick is to move the grid's parallel-resonance (where
>> feedback/feedthrough is maximal) away from the anode-resonace, where
>> damped wave VHF ringing takes place whenever the anode current changes.
>
>There are several flaws with that theory.
>
>First, grid resonance is caused by the self-capacitance of the grid
>resonating with the grid lead inductance.
? ok, if by self-capacitance you mean the grid's total capacitance to
other tube elements.
> That self-capacitance is in
>the order of a few pico-farads.
? For an 8877, grid/cathode C is 40pf and grid/anode C is 11pF. Is 51pF
a few pF in your opinion?
>If you add an EQUAL amount of
>capacitance in series with the grid you shift the resonant frequency
>LESS than 50% higher. That would require about 10 pF or so in an
>811A.
>
? close enough
>Adding a few hundred pF barely changes the self-neutralizing
>frequency of the tube, when that parameter is properly measured.
>
? I agree that the resonance shift is relatively small, however, perhaps
it was far enough to decrease feedthough a bit at the anode-resonance.
>Second, adding series reactance INCREASES feedthrough on
>every other frequency.
? Maximal VHF feedthrough/feedback occurs primarily at grid-resonance.
How could moving the grid resonance a bit higher make much difference at
HF?
> Collins (and Bill Orr when he "borrowed" the
>idea from Collins) claim this is to add "negative feedback".
>
>That's pure nonsense, as anyone who calculates the grid voltage
>produced by the capacitive divider formed by the cathode-grid
>impedance and grid-to-chassis impedance would be able to see.
>
? Nonsense indeed, Tom. Everybody makes mistakes.
>> >Especially when the tubes have too much feedback to
>> >be stable without neutralization when the grids are grounded.
>> >
>> ? In the world of VHF, a grid can not be absolutely, positively brought
>> to ground potential. > - later, Tom
>
>True. But they can be neutralized at and near the operating
>frequency range
? However, when the operating freq. is 28MHz and the grid resonants at
80.something MHz, regeneration at 28MHz is virtually impossible.
>and proper negative feedback can be added in the
>cathode.
>
? good point
>A good book is Circuits and Networks by Koehler. Read the
>section on matching theorems.
>
>
- cheers, Tom
Rich...
R. L. Measures, 805-386-3734, AG6K, www.vcnet.com/measures
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