[AMPS] Another arc question

[Rich Measures]

>Steve Thompson wrote:
>>In message <19980510030240.AAB548@[205.231.11.82]>, Rich Measures
>><measures@vc.net> writes
>>
>>big snip
>
>>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 

I'd call this necessary inductance because it is necessary if one is to 
conduct current from the tank to the anode.  

>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).

The equivalent Impedance of the suppressor does not connect to 
chassis-gnd, instead it connects in series with C-Tune, which connects to 
chassis-gnd.  The rub is that C-Tune is other than a low-Z near its 
resonance points.  
>
>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,

However, by working the parallel L/R circuit as an Admittance problem, 
only one transformation to an equivalent Impedance (series L/R circuit) 
is needed.  

>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. 

... and also on the Tune-C resonances.   //  The way that the Y/Z 
transformation takes place is cast in bronze.  The amount of susceptance 
and the amount of conductance are the variables

>That's
>why the optimum suppressor configuration always depends on the amplifier
>layout. 

How it depends on layout, you seem to have left out.  .  It is my opinion 
that a suppressor design should roughly divide the anode circuit's VHF 
current.  Otherwise, the suppressor's stagger-tuning effect is 
compromised.  
>
>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:

However, at 100 MHz, the 200-ohm/200nH suppressor produced an 
equivalent-Z resistance-component (which Wes designated "Rp") of c. 56 
ohms compared to the 100nH/109-ohm suppressor's 166 ohms.  .  .  Since 
VHF voltage gain is essentially MU x Rp, the 200-ohm/200nH suppressor 
would produce a lesser amount of VHF gain.  The trade-off of using such a 
suppressor is that with a 3-500Z, the 200-ohm resistor will dissipate 
over 40w at 28mHz.  

>1. Low-to-zero shunt R - shunts out the whole suppressor.

Lowering the shunt R decreases the equivalent-Z of the suppressor, which 
increases VHF voltage gain.  

>2. High-to-infinite shunt R - leaves only the inductor active, so it
>shifts the resonance but doesn't suppress it.
>
Since the electromagnetic fields generated by the suppressor L and the 
suppressor R are at right-angles, why would there not be two VHF 
resonances near each other?  

>Therefore there's an optimum value of R, somewhere between those two
>extremes, and it will depend very much on the amplifier layout. 
>
... and how does one calculate this?

>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". 
>
It is my opinion that the design principally hinges on the dissipation 
ability and the intrinsic L of the parasitic suppressor's R.  
High-dissipation and low L are necessary, however, this combination is 
difficult to find.  {ref. March, 1989 *QST*, p.25}
-  later

cheers
Rich...

R. L. Measures, 805-386-3734, AG6K   


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