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Re: Danger, and I don't understand why.
Tesla List wrote:
>
> Original Poster: Parpp807-at-aol-dot-com
>
> In a message dated 8/17/99 12:25:09 AM Central Daylight Time,
> tesla-at-pupman-dot-com writes:
>
> << Interesting speculations! However, NST's are actually wound with many
> turns of fine copper magnet wire, an this alone accounts for the
> comparatively large resistance of an NST secondary. The current limiting
> behavior in an NST actually comes primarily from the addition of
> magnetic "shunts". These bypass a portion of magnetic flux that would
> otherwise link the primary and secondary windings. This results in a
> substantial increase in the transformer's leakage inductance - and its
> external behavior is similar to connecting an ideal transformer in
> series with a very large inductor.
>
> <SNIP>
>
> -- Bert --
>
>
>
> >>
> Bert,
>
> That's an excellent explanation of how a transformer is current limited. Why
> don't the
> textbook writers use so simple an analogy of a "perfect transformer" in
> series with
> an inductance. I have never dissected a NST. Physically, where are the
> magnetic shunts
> placed in the xformer? Is the flux shorted out and converted into Eddy
> currents? How is
> this calculated?
>
> Thanks for the eddyfication.
> Ralph Zekelman
Ralph,
Thanks for the kind words! For this technique to work, the primary and
secondary need to be on seperate "legs" of the main core. Magnetic
shunts are placed between the primary and secondary windings, and their
presence reduces the coupling coefficient between the windings while
increasing the leakage inductance. This technique is used not only in
NST's, but also in ferroresonant/constant voltage transformers, many
adjustable-current AC welders, and microwave oven transformers.
While most of the primary's magnetic flux is carried by the main core, a
portion (which in an NST may be around 5%) is diverted through the shunt
plates, which are made of the same transformer steel as the main core,
but have appreciably less cross sectional area. Because some of the flux
is merely being diverted, there's little additional overall loss
introduced when using the technique, and eddy current losses are
virtually unchanged. The coupling coefficient is appreciably less
(~0.95) than it would be if the shunt plates were not present.
The ASCII art diagram (use a fixed-font for proper viewing) should help
you visualize the relationships of the windings, core, and shunts.
--------------------------------------
| Main Core |
| |
| |---------| |----------| |
| | || | | || | |
----------- || ----------- || ------------
| | || | | || | |
| | || | | || | |
| Sec1 | || | Pri | || | Sec2 |
| | || | | || | |
| | || | | || | |
----------- || ----------- || ------------
| | || | | || | |
| |---------| |----------| |
| ^ ^ |
| | | |
--------------------------------------
| |
Shunt Shunt
Plates Plates
While NST manufacturers may use various design approaches to accomplish
this function, the basic principles are all thge same.
A transformer with primary inductance Lp, secondary inductance Ls and a
coupling coefficient k can be modelled as an ideal transformer in series
with inductors:
Lp*(1-k) "ideal" Ls*(1-k)
-----OOOOOOO------ -------OOOOOOO-----
(Leakage) O ||| O (Leakage)
O ||| O
O ||| O
k*Lp O ||| O k*Ls
O ||| O
O ||| O
-------------------- -------------------
<-- M -->
As you can see, as we reduce k by adding shunts, the leakage inductance
increases and the the short-circuit current is limited by the increased
leakage inductances in the primary and secondary. Through a series of
measurements, you can measure Lp, Ls, and the mutual inductance, M.
Since k = M/(sqrt(Lp*Ls), you can then solve for k to characterize the
NST. Hope this helped!
Safe coilin' to you!
-- Bert --