Re: shunts/Magnetek NST

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

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

> > Does anybody have a photo of the transformer
> > "shunts" that can be removed for
> > greater power?
> >
> > Thanks
> >
> > --jeff
> Yes I would like to see those also. My basic question
> for the list is this, do ALL NST's have these shunts?
> Recently I tried to measure the inductance of the
> secondary of my 15,000 volt,30 ma Magnetek Jefferson
> NST's with a Wavetek LCR meter. The Highest inductance
> reading scale is 200 Henry, and it shows a overrated
> value on the highest scale reading, meaning the
> inductance is too high for the meter to measure. It
> does however make a reading of 20,000 ohms on the
> resistive reading.
>     Discouraged I decided to calculate the value of
> inductance by finding the impedance that the secondary
> will have at 60 hz and delivering its rated 30 ma as a
> shorted secondary. At 15,000 volts and delivering the
> 30 ma across the short, the acting impedance of that
> secondary would be by Ohms Law extended to AC where
> V=IR then becomes V= IZ :  15000(v)=.03(A)*Z where Z
> then =15,000/.03= 500,000 ohms. Using the equation
> Z(impedance)=sq rt [X(L)^2 +R^2]= 500,000=sq rt
> [X(L)^2+ (20,000)^2] where then squaring both sides
> yeilds
> 2.5 *10^11= X(L)^2+ 4*10^8 showing that X(L)^2
> =2.496*10^11 and therefore the sq rt of that value at
> ~499,600 ohms as the inductive reactance figure X(L)
> Since X(L)= 6.28 Freq(L), we can then divide the
> 499,600 by 6.28(60) to obtaing the L value as ~1326
> Henry.
> 
> So by the above calculations we can see that the 1326
> henry secondary with 20,000 ohms resistance will
> permit a current of only ~ 30 ma if short circuited.
> This rated current that is delivered is already
> current limited by the acting impedance of that
> secondary. This is why I cannot understand what, why,
> and where the shunt exists.

	The shunts serve to increase the leakage inductance (the value you
calculated IS correct, and the leakage reactance IS 500 k).  Without the
shunt the maximum current would probably be 15000/40000 = 0.375 amps. 
That assumes that the effective resistance of the primary and secondary
are about equal.  In this case the reflected resistance of the secondary
is 20000 divided by the square of the ratio 120/15000.

	Note that if you connected a "matched" capacitor (about 0.005 ufd in
this case) to the secondary it would series resonate with the leakage
inductance and, in principle, the current would be limited by the
resistance to the 0.375 amps calculated above.  It nothing broke down
and the core didn't saturate the resulting voltage would be 0.375 x
(15000/0.03), or around 187 kV!!!  Of course, the core would saturate
and the windings would probably short.....

Ed