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Re: shunts/Magnetek NST
Original poster: "Malcolm Watts by way of Terry Fritz <twftesla-at-uswest-dot-net>" <m.j.watts-at-massey.ac.nz>
Hi Harvey,
On 18 Apr 01, at 7:09, Tesla list wrote:
> Original poster: "harvey norris by way of Terry Fritz
> <twftesla-at-uswest-dot-net>" <harvich-at-yahoo-dot-com>
<snip>
> 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.
Which shows how small the secondary wire is.
> 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.
Note how small the influence of a significant amount of R is.
> 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. I conclude at this point
> in time that my Jefferson contains no shunts, and as I
> have said it is already current limited by the acting
> impedance of the secondary, based on the 20,000 ohms
> figure that has been measured.
Work the figures. O/C voltage = around 17kV peak. 20 kOhms is going
to allow s/c peak currents of about 850mA pk. That's a lot more than
the transformer is designed to deliver and far more than the wire
will stand on a continuous basis. The leakage inductance in
combination with the shunts effectively does all the current limiting
work. You can already see why you can throw any notion of current
limiting out the window if that inductance is resonated with the
primary cap. Additionally, you can see how such a circuit can ring up
on successive cycles to some huge voltage.
> Terry had also posted figures for other NST's where
> the resistance of this NST secondary had the highest
> figure at 20,000 ohms. Is it possible that only the
> jeffersons function this way without the need for a
> shunt? Any comments on this shunt business, and
> potential errors in my calculations showing that no
> shunt exists on this transformer would certainly bring
> me to a better understanding of this matter. Perhaps I
> have made a silly error, and can stand to be
> corrected.
I repeat - a good transformer of any type has ideally no resistance
at all in its windings. Such resistance is a source of power
dissipation and transformer windings are not designed to be a load
but to transform voltages and currents as efficiently as possible. A
transformer should never be designed so that the resistance of its
windings is a current limiting feature.
On the subject of suitable transformers for TC use: another
type which I frequently use is a CC-core which has its primary and
secondary windings on opposite legs. Physical separation of the
windings makes for a high leakage/limiting inductance regardless of
the presence of the core.
Regards,
malcolm