Re: calculating inductive factor for c cores

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

Original poster: "Ed Phillips by way of Terry Fritz <twftesla-at-qwest-dot-net>" <evp-at-pacbell-dot-net>

 > > I have some C Cores which I might like to rewind as HV transformers.
 >  >
 >  > How do I calculate the how much power a core can handle.
 >
 > You can get a rough estimate from: 39 * Ae^2  where Ae is the
 > effective pole cross-sectional area. The derivation for this arises
 > from geometrical limitations on transformer design, in particular an
 > acceptable degree of regulation on the loaded secondary winding/s.
 > Although one can build a transformer core with a huge winding window
 > and relatively small pole area, the further the windings are from the
 > core, the poorer the coupling to the core will be and hence
 > regulation also suffers. Additionally, the windings will also suffer
 > from rather poor coupling between each other (not to mention between
 > their own turns).

 > Current throughput is a function of copper area of the winding wire
 > and the number of turns (the winding length) and by proxy, the
 > winding resistance and the ability of the windings to lose power
 > wasted as heat in that resistance. For a given copper mass, the
 > thicker the wire, the fewer the turns, the lower the terminal
 > voltage, the higher the current capability.


 > A simple equation turns
 > this mess into a turns/volt figure for the core and if the primary
 > occupies around 40 - 45% of the total winding window, the magnetizing
 > current will be a low percentage of the reflected load current at the
 > core's VA rating. The equation can be boiled down to:  turns/volt =
 > 7.164/Ae for the 60Hz mains.

	This expression should be burned into the brains of ANYONE thinking
about running a 120V NST on 220V, etc.,etc.  If you try to increase the
voltage across a coil/core very much beyond the design value the
magnetizing current will go out of sight and all of the smoke will come
out of the winding.

Ws