# RE: ballasting problems

```Original poster: "Basura, Brian by way of Terry Fritz <twftesla-at-uswest-dot-net>" <brian.basura-at-unistudios-dot-com>

Mark,

Just a comment that no one has mentioned -

IF the secondary is open you can hook a pig directly to the mains without a
ballast and nothing will happen. It will just set there consuming a small
amount of magnetizing current. As soon as you draw an arc off the HV
terminals though you've created a short circuit on the secondary side. This
short is reflected on the primary side of the transformer and blows the
breaker.

> I was attempting to calculate the proper ballast for The Geek Group's
Pig.  We
> wound the core (he posted the size in "The Mighty Inductor of Doom")
according
> to my calcs (actually used more turns for a Safety Factor).  The result:
I've
> NEVER seen a circuit breaker pop that hard!  Wow!  There was enough
resistance,
> or enough inductance to prevent the circuit breaker from tripping until
we drew
> an arc from the pig outputs.

See Bert's excellent post on calculating windings for ballasts....

Regards,
Brian B.

-----Original Message-----
From: 	Tesla list [mailto:tesla-at-pupman-dot-com]
Sent:	Friday, May 11, 2001 7:09 AM
To:	tesla-at-pupman-dot-com
Subject:	ballasting problems

Original poster: "Mark Broker by way of Terry Fritz <twftesla-at-uswest-dot-net>"
<broker-at-uwplatt.edu>

I was attempting to calculate the proper ballast for The Geek Group's Pig.  We
wound the core (he posted the size in "The Mighty Inductor of Doom") according
to my calcs (actually used more turns for a Safety Factor).  The result:  I've
NEVER seen a circuit breaker pop that hard!  Wow!  There was enough resistance,
or enough inductance to prevent the circuit breaker from tripping until we drew
an arc from the pig outputs.

Anyway, I used XL=w*L to determine the inductance needed.  Then I used the
standard inductance equation L=N^2*A*mu/len modified such that len =
N*wire_diameter.  A is the area of the core, from wire center to wire center.
It's approximate.  mu is 4*pi*10^-7 * 5000 (approximate mu for silicon steel).

So, the results show that I need a whopping ONE TURN to ballast 10kVA (5.8mH)!
Woah, something's amiss - I've seen the 200 pound inductors with what looks
like 10000 turns that some list members use....   We lack a Henry meter to
verify its inductance. (plenty of RC meters - even one that measures the hfe of
transistors! but no L meters)

So, all my calcs are in an Excel sheet, that I posted on my "web site".  It's
macro and virus free.
<http://mbroker.tripod-dot-com/ballast.xls>http://mbroker.tripod-dot-com/ballast.xls
(No, I don't have an actual home page - just an "online cache" so I can easily
share files).

Please, tell me where I messed up.

Thanks
Mark Broker
G-3 Geek #1019

```