Steve Thompson wrote:
>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.
>
>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.
>
Actually the suppressor appears in series with the stray inductance that
is causing the VHF resonance, and it's that [combination] that appears
in parallel across the valve/tube and its output capacitance (plus
strays).
What matters is the damping resistance across the whole VHF-resonant
circuit, as seen by the anode of the tube. Unfortunately it takes at
least two series<->parallel transformations to get to that point,
starting from the suppressor itself. The way that the R in the
suppressor is transformed into parallel R across the tube will depend on
the L in the suppressor, and also on the external circuit strays. That's
why the optimum suppressor configuration always depends on the amplifier
layout.
It's easy to see that there must be an optimum value of R. Either very
low or very high shunt R will give poor parasitic suppression:
1. Low-to-zero shunt R - shunts out the whole suppressor.
2. High-to-infinite shunt R - leaves only the inductor active, so it
shifts the resonance but doesn't suppress it.
Therefore there's an optimum value of R, somewhere between those two
extremes, and it will depend very much on the amplifier layout.
At the high side of the optimum, reducing R will improve suppression;
and at the low side of the optimum, increasing R will improve
suppression - but both of those are partial answers. There is no
universal "easy answer".
73 from Ian G3SEK Editor, 'The VHF/UHF DX Book'
'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.demon.co.uk/g3sek
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