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 20:22, Tesla list wrote:

> Original poster: "Gerald  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>
>
> Hi Malcolm,
>
> This I believe.  For a given number of turns N on a gapless toroid, it
> will have an inductance L and(with an applied AC voltage of fixed
> magnitude and freq) a maximum current I.  If the Imax is too small
> then the addition of the airgap will reduce the L and allow the
> current to increase.
>
> I have heard that the addition of the airgap will keep the core from
> saturating and this is the point that I'm not seeing (if it is true)

It isn't true - exactly. What it does do is change the shape of the
BH curve from something squarish to something much more elongated. To
get the core near saturation requires a much higher applied H; i.e.
NI. The core still asymptotically approaches its limit for B.

> Seems like the flux density in the core is a function of the cross
> sectional area. the volts per turn and frequency, and not dependent on
> the presence of an airgap.  Would this be correct??

The airgap has the effect of greatly increasing the effective length
of the core so that for a given applied NI, most of the flux
generated is moved from the core to the gap. Hence the flux density
in the core drops. It helps to think of flux in terms of dots rather
than lines. See additional comments in the original text below:

> Gerry R.
>
> >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.

True for a given applied voltage across the windings.

> > >
> > > 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.

Agree. In a crude resistance - reluctance analogy, where is most of
the power in a circuit dissipated if one sticks a significant
resistance in series with a circuit?

Malcolm

> > >
> > > 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
> >
> >
>
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