Re: "De-coupling" coefficient?

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Original poster: "Stephen Conner by way of Terry Fritz <teslalist-at-qwest-dot-net>" <steve-at-scopeboy-dot-com>

At 09:44 29/06/03 -0600, you wrote:
>Original poster: "Jolyon Vater Cox by way of Terry Fritz 
><teslalist-at-qwest-dot-net>" <jolyon-at-vatercox.freeserve.co.uk>
>
>Will the equation Vs = k.sqrt(Ls/Lp) hold when the transformer is loaded?

Probably sort of %-9



>What is implication of this (if any) in the example of a current-limited
>transformer (I understand all practical transformers are in fact
>current-limited to some extent by the % "regulation" factor)
>Does k change with loading,

No... The leakage inductance, hence the regulation, is a function of K. 
Shunts in an NST (ferinstance) decrease K because they provide a path for 
the secondary flux that doesn't go through the primary. This parallel 
magnetic circuit looks like an inductor in series with the secondary.

>for if a current-limited transformer delivers
>its max current when the secondary voltage approaches zero volts (i.e. a
>short-circuit condition)
>and the above equation was correct, must not the coupling coefficient be
>less at higher currents to explain the concommitant decrease in secondary
>voltage with loading assuming of course, that resistive losses in the
>winding are discounted?

Again I would say no... The internal EMF in the secondary is always the 
same, it's the volt drop across the leakage inductance that makes the 
output fall. Of course the leakage inductance is a property of the same 
coil that has the "internal EMF", so things get a bit confusing, but I 
think it's OK to think of it as if it were a separate inductor.

Sanity check: If this model is true, then removing shunts from an NST 
should increase the open-circuit voltage of the secondary, as well as the 
short-circuit current, because it increases K. Anyone got any data on this?

Steve C.