[Amps] Observations on a Pi-L tank

[Scott Townley]

A rambling set of observations on Pi-L tank design and realization:
Proposed amp is for 17m/12m, single 3-500Z, 650W output.  It fills the gaps...
First come up with design values for the Pi-L from TAP.  Then I built a spreadsheet to design the coils using the Nagaoka formulation rather than the Wheeler approximation (in every ARRL handbook since Wheeler published it, no doubt).  
In playing with the proposed inductor diameter I started wondering what the lower limit was.  With a loaded Q of around 12 then I would need the unloaded Qo to be ~200 to keep the tank losses below 0.5dB (70W).  I could accomplish that with 1/4" tubing or with some existing 2" B&W coilstock, but it’s only 14AWG.  Both will easily give me Qo>200.  So what’s the difference?  I’d rather use the coilstock (less work for me).
Some number-crunching gave inductor losses of 34W and 5.6W (L1 and L2, respectively) for a Qo of 300 (perhaps optimistic).  To get an idea of the temperature rise, I went through a simple DC analogy for the 12m coil:
Total wire in 12m coil (L1)=71" (of 1/4" tubing)
DC resistance of the equivalent wire (2AWG)=973micro-ohms
If it dissipates 34W, then the analogous DC current is sqrt(Pdiss/DCR)=190A!!
A quick look at a wire manufacturers chart (Alpha and Belden both have nice hookup wire charts) implies the wire temperature reaching 100degC.  This inference made from the published current limit for a 2AWG wire with a rated 100C insulation being 190A or thereabouts.  Might be more since tubing has less thermal mass than solid wire.
A similar run for L2, but using 18AWG for the much smaller inductor required gave an equivalent DC current of 21A and a temperature of 125C.
This should be the worst case temperature rise, since at lower frequencies, even though Q decreases approximately as the sqrt(f), the total wire required for the same Xl is roughly ~1/f so equivalent currents will decrease roughly 1/sqrt(f).
Tank inductors get hot!  I can see why Rich mentions forced-air cooling of tank components from time to time.  40W doesn’t sound like much (we don’t force-air cool our house lights, right?).  I’m guessing that the ultimate limit is probably around 180C (solder melting point)
any comments?  Probably less than that given that metal resistance has a positive temperature coefficient.
I always thought the decreasing inductor wire size with decreasing frequency (e.g., on a B&W850 and many others) was more for compactness when realizing the higher inductances involved
is it as much as thermal consideration?
Comments?  Flames?  Further observations?  (already hot enough here as it is
)


Scott Townley NX7U
Gilbert, AZ  DM43di