Re: More power, fewer turns of primary?

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "by way of Terry Fritz <teslalist-at-qwest-dot-net>" <Mddeming-at-aol-dot-com>

In a message dated 4/9/03 7:51:00 PM Eastern Daylight Time, 
tesla-at-pupman-dot-com writes:



>Original poster: "Adam Britt by way of Terry Fritz <teslalist-at-qwest-dot-net>" 
><beans45601-at-sbcglobal-dot-net>
>
>
>Is it a general rule that the more power you put in, the fewer turns on the
>primary you have to use? I ask because I am running on a 12/60 NST now and
>I had to add 5 extra turns (and they hang off the side) to my primary, and
>I want to make sure I won't have to add any more (I am getting another
>12/60 NST soon, for 12/120).
>
>Thanks
>
>Adam


Hi Adam,
It is a question of electrical balance. Yes, a larger NST means you can use 
a larger tank cap. A larger tank cap means a smaller primary coil IF you 
are keeping the frequency the same. For a fixed frequency, L(pri) x C(pri) 
= const. However, a smaller primary coil means larger I^2R losses in the 
primary circuit. Lowering the frequency means increasing the L or C on the 
secondary side. But this means increasing the top load, the secondary size, 
or both, since L(sec) x C(sec) = const. for a given frequency. Too small a 
primary can result in instantaneous currents tearing up your cap, severely 
shortening its life. Just where the break-even point is for primary size vs 
losses depends on the fine details of your coil. It's not a trivial question.

Good Luck,
Matt D.