>Rich said:
>
>>The I^2 x R heating of the grid at 144MHz is probably minimal because
>>gold is a fairly good electrical conductor, and there are over 100
>>grid-bars to share the current burden of 24Arms.
>
>well, the skin depth at 144 is .216 mil (i.e. .000216 inches). The gold flash
is
>probably a few microinches, so there's the base grid material to consider,
>and its resistance - that's going to dominate.
The gold plating is pretty thick judging by the quantity of meltballs I
find. RE: Figure 24 on my Web site. . My guess on the frequency that
causes gold evaporation is above 700MHz.
>What's the diameter of the grid bars?
>
Looks like maybe 0.8mm x 0.2mm. The gaps are c. 1mm. There are 108
bars.
>Now consider the current distribution even at LF. The capacity per unit area
is
>pretty constant, but all the current has to come in at one point - the
>bottom - so you would expect maximum heating there.
>
Gold is too good of a conductor, but with a gaphite grid, perhaps,
somewhat. .
>I guess at the end of the day, the question is 'What is the effect of RF grid
>current on the heating of the grid?' With the currents we're talking
>about, is it negligible? So far, I'm not convinced. It's interesting to look
at what
>happens in a 4CX1000 at 144MHz: that will have (in grounded cathode) an RF
>grid current of 3.26 amps, yet it has a zero watt dissipation rating. So if
the
>grid has a 0.1ohm resistance, there's one watt of dissipation.....
>
1W is the rated grid-diss for a 4cx1500B, which has a virtually identical
grid, so the "0W" rating is questionable.
- R. L. Measures, a.k.a. Rich..., 805.386.3734,AG6K,
www.vcnet.com/measures.
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