>In message <19980510030240.AAB548@[205.231.11.82]>, Rich Measures
><measures@vc.net> writes
>
>big snip
>>
>>You have not explained how lowering R is a parallel L/R suppressor lowers
>>VHF-Q. By your logic, the lowest Q would result when R = 0-ohms
>
>In a parallel circuit, lowering R reduces Q.
>
True, however, there's a bit more to designing a VHF-suppressor than Q.
It seems to me that an optimal suppressor design divides the VHF signal
fairly equally between suppressor R (Rs) and the suppressor L . Thus,
if the reactance of Ls is say100-ohms at the VHF anode-resonance, a 1-ohm
suppressor R (Rs) is not going to divide current as equally as a 100-ohm
Rs.
>If the suppressor has a reasonable path to ground at the matching
>circuit end, then it appears in parallel with whatever the valve looks
>like from anode to ground.
>
.A large if. The sticky wicket is undoubtedly VHF resonances in the
Tune-C that are in the vicinity of the anode-resonance.
. So why not make Rs = 1-ohm in order to reduce the VHF load R on the
anode, and thereby reduce VHF gain?
- later
cheers
Rich...
R. L. Measures, 805-386-3734, AG6K
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