[Amps] Help with HL2200 6 meter conversion

[Gary Myers]

Gerald,

duh! Of course its I sqrd R! I guess I was thinking that less current would flow through it if the impedance of the U part were similar. Someone else told me they did this same conversion and indeed found the inductance of the U too high - they reduced it - so I'll give that a try. 

One concern I have is that I didn't follow the layout as they did it.... I placed the plate tune cap (which by the way is not a 100pf max - it is a 9-45pf cap) closer to the tubes using the available mounting point - this meant that the tank coil had to go over the caps. Not sure if that matters or not - I am a bit concerned that if I tune it up with the perforated cover installed I get about 350W out at a pretty bad efficiency.... if I remove that cover (the HV Short on these amps never did work properly - and is not in this case as well) then I get 550W out with about 800W in! So obviously the tank coil is being effected by the proximity of the perforated top. The coil dimensions I used, because its near impossible to wind 1/4" copper on a 1.25" form per the article, was 1 9/16" which makes it closer to the top (as well as the caps).... my question is this: will it effect the tank at all to have this coil closer to the caps? Not sure if the field will effect the caps or not. Can't quite wrap my head around that one. I'm a EE with pretty good field experience (use to do testing for EMI issues on commercial high power products for CE test conformance) but again not sure what happens with the field around this tank coil. 

g.
    ----- Original Message ----- 
    From: TexasRF at aol.com 
    To: garymyers at powerc.net ; amps at contesting.com 
    Sent: Saturday, January 22, 2011 4:58 AM
    Subject: Re: [Amps] Help with HL2200 6 meter conversion


    Hi Gary, it sounds like you are close to success on this project. 

    On the issue of efficiency, it will be the better at higher drive levels for a couple of reasons. One, as drive is lowered, the plate current swing is reduced and that pushes the class of operation more in the direction of class A and away from class B. And two, lower plate current swing means higher plate load impedance when tuned for maximum power. Higher plate load impedance in turn increases the loaded Q which is accompanied by higher tank losses.

    As the parasitic suppressors are in series with the rf current flow, heating HAS to be a function of current squared times resistance. So an increase of resistance will cause an increase in dissipation. If the shunt inductance is made smaller, more of the series rf current will flow through the inductor and that of course forces the resistor current to decrease. Since the dissipation is related to current squared, a 30% current reduction in the resistors will yield a 50% dissipation reduction.

    73,
    Gerald K5GW