[AMPS] Another arc question

Rich Measures measures@vc.net
Mon, 11 May 98 00:12:40 -0800


>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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