Re: The "second pig" ballast: Questions.

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: "Gerald  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>

Hi Steve,

This is the very subject Ive been trying to learn about recently and I'm a 
little confused as to what you mean by  "it impossible to make an efficient 
ballast no matter what number of turns you use".  Following example will 
use a toroid cause it is easy to wrap your mind around and the math is 
simplier.

If one has a toroid using a core with permability (u), with a cross 
sectional area (A), with number of turns (N), and a magnetic path (length 
Len) (I'm choosing the center of the cross section and following it around 
the toroid for the path) and evaluate:

closed integral {H.dl}= NI      around the path (constant H)

(I=current thru one turn so NI is the amp-turns or total amps enclosed by 
the path)

therefore:

H*Len = NI,   so H= NI/Len,   B=u*NI/Len (B=flux density)

assuming H is constant within the core of the toroid (no air gap) (close 
enough), then BA will be the total flux (phi) flowing around the magnetic 
path.  So the inductance (L) of one turn is:

L = phi/I  = BA/I  =  u * NA / Len

The inductance for N turns (Ln) is:

Ln = u * N^2 * A / Len

Now as the number of turns increases, the inductance goes up squared and 
the ballast current goes down (inverse squared).  So the amp-turns goes 
down as N goes up and B goes down as the turns go up and with enough turns 
the core should not saturate.

If the core did saturate with a given number of turns,  adding an air gap 
somewhere in the magnetic path would reduced both H and B (assuming the 
current didnt change) to keep the core from satuating.  The air gap 
increases the reluctance of the magnetic path.

Since B has to be continous crossing the gap and the permability of the gap 
is uo and the permability of the core is u, then:

Hcore * u = Hgap *uo

therefore Hgap = u/uo  *  Hcore = ur * Hcore  (ur is relative permability)

When the H.dl integral is evaluated, one now gets:

Hcore * (Len - Dgap) + Hgap * Dgap = NI    where Dgap is the gap distance

For high permability core:

Hcore * Len + ur * Hcore * Dgap = NI

Hcore = NI/(LEN + ur*Dgap)

This is what I think I understand to date (please corrent any 
errors).  What I dont understand is this:
Lets say that with a given N that results in a L and a current I (assuming 
no saturation), a gap is introduced to reduce the field intensity and 
corresponding flux density in the core.  If the flux density is reduced 
(for the current I), the inductance goes down.  This would result in the 
current going up and seems to undue the benefit of the air gap.  I must be 
missing something.

Gerry R.


> Original poster: "Steve Conner" <steve.conner@xxxxxxxxxxx>
>
> >The core consists of two separate "spools" of steel (not
> >E's and I's, although the whole core has two "windows"
> >like an "EI" core), and won't come apart in any way
> >which might make winding easy.  Bummer.
>
>
> Bummer is the word. If you can't take the core apart, then you can't
> introduce an air gap to it, so you'll find it impossible to make an
> efficient ballast no matter what number of turns you use.
>
> If it were me I would be thinking of sawing the core in half (after binding
> it in a suitable way so it won't spring apart violently)
>
>
> Steve C.