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Re: [Amps] Help with HL2200 6 meter conversion

To: "Amps" <amps@contesting.com>
Subject: Re: [Amps] Help with HL2200 6 meter conversion
From: "Gary Myers" <garymyers@powerc.net>
Date: Sat, 22 Jan 2011 11:31:59 -0700
List-post: <amps@contesting.com">mailto:amps@contesting.com>
I believe I'm there. The diameter of the coil was too large I believe (I got it 
from a second article by K8SMC which specified 1 9/16" diameter)... I made it 1 
1/4" and repositioned it slightly. Now I'm able to get about 59% efficiency 
with 1080W out and only about 45W drive. I've not pushed it further as the 
suppressors are still getting too hot. I'm going to try shortening the U loop a 
bit - not sure how to do that but I'll work on it... hopefully that fixes 
it.... the K8SMC article just used something very similar to the original 
suppressors - a 10 ohm with 3 turns of #10... I might try using the originals 
just to see.... can someone tell me what to expect if they aren't working right 
(beyond the resistor dissipating too much energy)? I assume it will be unstable 
but not sure how that shows itself. 

thanks for all your help - also to Richard Solomon who replied privately and 
who has done the same conversion. 

Gary
K9RX

  ----- Original Message ----- 
  From: Gary Myers 
  To: Amps 
  Sent: Saturday, January 22, 2011 8:49 AM
  Subject: Re: [Amps] Help with HL2200 6 meter conversion


  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 n
 ot. 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@aol.com 
      To: garymyers@powerc.net ; amps@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

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