Re: A Ballast Mystery

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

Original poster: "Sean Taylor by way of Terry Fritz <twftesla-at-qwest-dot-net>" <taylorss-at-rose-hulman.edu>

This isn't too tough to explain - it's just circuit theory stuff.

With the welder leads open and the pig horns open, you just have the
welder's inductance (a constant between items 2 &3 in your list) and the
inductance of the LV side of the pig, call those Lwelder and Lpigopen,
respectively.  The current that flows is then V/X, which equals 240 /
(2*Pi*f*(Lwelder + Lpigopen)), where X is the reactance of the load.  The
magnitude of the reactance is 2*Pi*f*L, or approximately 377*L for 60 Hz
line frequency.  With the HV side of the pig shorted, the short circuit is
seen almost as a complete short circuit on the LV side, as the transformer
should do (there are some losses, but not much in a pig, esp. if it has
copper windings).  The current with the leads shorted is then V/X, where X
is now just 2*Pi*f*Lwelder, so the reactance has decreased considerably, and
the current will then go up.

For ballasts in general, I think that they are often thought of as "magical"
current limiting devices - i.e. the current will NEVER go above x amount.
The current that a ballast will limit to greatly depends on the load . . .
an inductive ballast with a capacitive load can draw more current than the
capacitive load alone.  This is the same thing about an NST being current
limited (ie 30 ma short circuit current) but being able to draw more current
with a capacitive load (TC tank circuit) than the short circuit current.

Sean