>At 11:10 AM 9/6/97 -0800, Rich wrote:
>>
>>
>>This is just another case of someone propagating an 'old wives tale',
>>Harald. Whenever you use two tubes, the output load R is 1/2 as much as
>>it is with one tube, which means that you Need 2x as much C on 10m to
>>achieve the recommended Q of 10 to 15. The same thing applies if you use
>>three tubes.
>
>If you are going to continue holding yourself out as the final arbiter of
>the right and wrong of amplifier design, and deride those with significantly
>more experience than you as "recognized experts,"
Mr. Subich:
RE: Experience: When Charles Thomas Rauch, Junior was in 5th Grade, I
finished the construction of amplifier number-one.
RE: 'Experts'. In the 9/94 issue of QST. [p.72]:
Quoting Mr. Rauch:
".They are not supported by design theory or the experience of recognized
experts in the RF amplifier community that include Fimac, Siemens, ETO,
Henry, and Ameritron. ..."
- Ameritron's amplifier designer is Mr. Rauch. -
> ... at least get your facts
>right. Your comments are only true if one is using two tubes to generate
>twice the output power at a constant plate voltage eg. 77SX vs 77DX).
>
Most folks seem to add a second tube to double the power output..
>1) The design of an output network, be it a simple pi network or a pi-L
> is based simply on the tube optimum load impedence which, in turn, is
> determined by the anode voltage, anode current, and conduction angle
> (class of operation) of the tube(s).
Agreed.
> Thus, for example, the output
>network
> for 2 X 3-500Z would be identical to that of a single 3-1000Z as long as
> the two amplifiers are operating at similar output powers.
Yes, indeed.
>
>2) It is more difficult to achieve the required low "C" for an output
> network tuning capacitor with two tubes than with a single tube simply
> because one must account for the output capacitance of TWO tubes instead
> of one. In the case of 2 X 4CX800A the output capacitance is 22 pF
> (11 pF x 2) plus circuit strays ...
Does this check out? ... ..
- The optimal load R for one tube is 2500V/2(0.8A) = 1600 ohms. To
achieve a Q of 10, the reactance of the tune-C should be 1600/10 = 160
ohms of Xc- - which at 29MHz is 34pF. Since the tube contributes 11pF,
an additional 22pF of tune C would be needed.
The optimal load R for two tubes is 2500V/2(1.6A) = 800 ohms. To achieve
a Q of 10, the reactance of the tune-C needs to be 800/10 = 80 ohms of Xc
- - which at 29MHz is 68pF. Since the tubes contribute 2*11=22pF, an
additional 68 -22 = 44pF of tune C would be needed.
The optimal load R for ten tubes is 2500V/2(8A) = 160 ohms. To achieve a
Q of 10, the reactance of the tune-C would have to be 160/10 = 16 ohms of
Xc - which at 29MHz is 340pF. Since ten tubes contribute 110pF, an
additional 220pF of tune C would be needed.
> where the single 4CX1600B is 12 pF.
okay, and ________________________?
> This capacitance appears in parallel with the plate tuning capacitor
> effectively dominating the minimum circuit capacitance and forcing
> HIGHER network Q on 10 meters (often well above 20) contributing to
> higher circulating currents, and increased losses in the plate tank.
The calculations seem to indicate otherwise, Mr. Subich.
Cheers
Rich---
R. L. Measures, 805-386-3734, AG6K
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