[AMPS] Re:

[Jon Ogden]

>This means the phase shift through the suppressor is 45 degrees.  Why is 
>this optimal?  Since our concern is what load impedance the anode sees, 
>would it not be more sensible to make the overall VHF current flow from 
>the plate to the entire output network 45 degrees, assuming this 45 
>degrees is an optimum of some sort?

Why is our concern what load impedance the anode sees?  IMHO, that load 
wil vary quite a bit over frequency.  IMHO the load impedance is 
important only in so far as resonances are concerned.  I also don't think 
that phase shift as you state is the important consideration here either.


>>> .A large if.    The sticky wicket is undoubtedly VHF resonances in 
>the 
>>>Tune-C that are in the vicinity of the anode-resonance.  
>Precisely.  This is why the supressor pales to insignificance in most 
>designs.  
Arlen, you are thinking in terms of the supressor "abosorbing" the VHF 
energy.  A 100 Ohm resistor in series with a 2000 Ohm anode will do 
nothing to absorb anything.  Rather, from what I can conclude the whole 
function of the supressor is to reduce the Q of the VHF resonance to the 
point where an oscillation is less likely to take place.

>If my matching network has some sort of high impedance 
>resonance (which you allude to by claiming that a low impedance is a big 
>"if"), and since this network, at the intended frequency is several 
>kilohms, then at VHF must be several kilohms plus several more, then 
>what difference would putting a suppressor in series with it of a 
>hundred ohms or so make? 

How can you guarantee a high impedance resonance in a typical amplifier 
circuit?

Quoting again from Solid State Radio Engineering page 38:

"The impedance or admittance of an RLC circuit is a complicated function 
of frequency, and normally has both a resistive (real) and a reactive 
(imaginary) component.  For some circuits the reactive part vanishes at 
one or more frequencies; this condition (pure real impedance and 
admittance) is called resonance......"

So what you are saying is that whatever VHF load your anode sees at 
resonance is a high resistance.  Ok, if you can guarantee this, then your 
idea of no supressor will work.  However, please show me the highly 
resistive components in the tank circuit of a typical amplifier circuit?  
There aren't any as far as I can see.  If one takes Rich's definition 
(and I believe it to be correct) for the VHF resonant circuit to be 
everything from the anode to the tune C, then the only high impedance 
looking part of the circuit would be the switched inductor in the output 
tuning circuit.  And depending on it's resonances, it's impedance at VHF 
will vary as well.

Additionally, the only real way to have full abosrption of energy in an 
RF circuit is to MATCH impedance.  If my tube has an output impedance of 
2000 Ohms plus some reactance at VHF, then you would need to match 
exactly into that in order to get full absorption.  Otherwise you would 
only get partial power absorption in your load. And since the reactive 
impedance of the anode will vary with frequency, there is no way to 
guarantee you are entirely absorbing any unwanted signal.

>and since this network, at the intended frequency is several 
>kilohms, then at VHF must be several kilohms plus several more, then 
>what difference would putting a suppressor in series with it of a 
>hundred ohms or so make?

Huh?  How can you guarantee that a tube having one impedance at HF will 
have a higher impedance at VHF?  I would say it's likely but one can't be 
sure.  Again, the supressor is NOT there to absorb VHF energy. 


>  The only time I've ever put a suppressor in an 
>anode is when my output network did look like a shunt to ground at VHF.  
>If it looked high impedance, then I found suppressors totally unneeded.  
>And in most cases, I do measure such high impedances, and use no 
>supressors.  

Again, how can you guarantee this?  An open circuit is a high impedance 
indeed.  However, if at VHF resonance, your output network looks like an 
open circuit, you won't get abosroption of your VHF signal.  You will get 
total and full reflection.  An open circuit will do exactly the same 
thing as a short circuit: reflect all the energy.  The phase is just 180 
degrees out of phase from the short.

And again, how can you guarantee what a "high impedance" is?

The whole idea of the supressor is to DETUNE the VHF resonant circuit, to 
screw up its Q, etc.  It certainly won't absorb it.


73,

Jon
KE9NA


--------------------------------------------------------------------------
Jon Ogden

jono@webspun.com
www.qsl.net/ke9na

"A life lived in fear is a life half lived."


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