Re: The "second pig" ballast: Questions.

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: "Malcolm Watts" <m.j.watts@xxxxxxxxxxxx>

Hi Gerry,

On 8 Feb 2005, at 10:29, Tesla list wrote:

> 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.

The point is that the current can be greatly increased along with
energy storage (E = 0.5LI^2) using the same core.

Malcolm