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Re: NST Inductance
Terry Fritz wrote:
> The voltage step up of a 15kV NST from 120VAC is 15000/120=125. Since the
> primary to secondary inductance is proportional to this number squared
> (125^2=15625), the primary inductance would be 1300H/15625=0.0832H.
>
> That would give the NST on open load current of 120/(2 x pi x 60 x 0.0832)
> = 3.83 amps. The actual open load is around 0.3 amps.
The high current would be with the output in short circuit, not open.
With the output open you see the equivalent inductance of the primary
inductor alone. This inductance doesn't affect the output, and can't
be calculated by measurements at the output circuit with the primary
connected to the mains.
A convenient model for a lossless transformer is:
1:n
o-----+-----+ +-----Lsec------o
| | |
| ) (
Lprim } {
| ) (
| | |
o-----+-----+ +---------------o
The transformer shown is ideal, with a voltage gain n.
Lprim is the actual inductance of the primary winding. Lsec is the
inductance seen at the output with the input in short-circuit,
or connected to a low-impedance voltage source, and
is what limits the output short-circuit current.
The actual inductance of the secondary winding, measured with
the input circuit open, is:
L2=Lsec+Lprim*n^2.
This circuit is equivalent to a "real" lossless transformer
with primary inductance Lprim, secondary inductance L2, and
coupling coefficient k=sqrt(Lprim*n^2/(Lsec+Lprim*n^2)).
Antonio Carlos M. de Queiroz