Re: Driving LEDs with a battery pack

On Mon, 13 Jul 2009 18:05:36 +1200, greg <greg@xxxxxxxxxxxxxxxxxxxxx>
wrote:

Jon Kirwan wrote:

Sorry, I should have added something else. The effective inductance
goes very close to zero if there is no place to put energy. This can
allow an I that you'd imagine as worthy of some energy storage when,
in fact, the L just went effectively close to nil.

Yes, but as long as other things are equal, I'd
expect a higher permeability core to give you a
higher inductance, as long as the core doesn't
saturate.

Yes. Higher inductance means "SLOWER." But it doesn't mean more
energy per unit time. In fact, the inductance cancels out of the
equations in calculating power. The great thing about higher perm is
fewer windings. However... they are often wasteful of energy at
lower frequencies (for example, #75 ferrite is horribly lossy even
down at 1MHz with more than 20 times the loss as #64 ferrite at 1MHz,
while #75 has max permeability of 5000 and #64 is only 375.) Also,
the lower perm stuff seems to be non-conductive while the higher perm
seems to be pretty much a semi-conductor, at least. Still, some of
the higher perm stuff is okay, loss-wise. But it seems most of it is
intended for RFI shielding, not transformers. But I'm still reading
and learning, so there probably are some things I'm missing still.

Maybe saturation is the issue, though? That is,
although the inductance is higher, the current
can't go very high before the core saturates,
dropping the inductance to zero.

I think I take your point here. Higher mu means the same B will be
reached a earlier, I_peak wise, if keeping the inductance fixed and
the magnetic loop length fixed. B = mu*H = mu*N*I/l_m, but L is
proportional to N^2 and mu... so I guess B is reached by the square
root of the ratio of mu's. A mu=5000 material would saturate with 10
times lower I_peak than with a mu=50 material, assuming L is constant
and so is the magnetic loop length (same physical core shape?)

I'm studying the references you gave me now. They
do seem to say fairly explicitly that the energy
is stored in the gaps. I'll have to think about
that some more.

Yes. I had to sit and think about that, too. You mentioned just in
the prior paragraph the "dropping the inductance to zero." But that
doesn't actually happen. What happens is that it goes down to what
amounts to an air/vacuum core. It still has some inductance, but not
much. That's because when the core itself saturates, it's done with
and over for the core. So if you keep trying to store more energy the
field lines expand out into space around the core and use the
air/vacuum there to store energy, just like an air core would. So
there is still room and that means there is still inductance. Just
not much, because all those windings are now just packing energy into
the vast regions of air around it once the core is saturated up.

If you look closely at some of the better saturation curves, you will
see that there is still a _slight_ slope after saturation. That
slight slope is the emerging air core inductance.

It can't be as simple as "gap good, iron bad",
because then the more gap you have, the more
energy you could store, so you'd be best off
with no core at all. That doesn't appear to be
the case, so there must be some opposing effect
at work.

I think it is called N. Too many windings are required. Think about
the reluctance, which is inversely related to the permeability. High
perm materials have low reluctance. Low perm, like air especially,
have high reluctance. Net reluctance is affected by inserting an air
gap -- in fact, the air gap may so completely dominate things if it is
big that the other core material doesn't matter and you might as well
be winding an entirely air core from the start and just not bother
with the ferrite, powdered iron, etc.

Think of a series resistance circuit where you have some aluminum you
can run out some length of and the some carbon of the same diameter to
make a combined loop. What is the total resistance? Well, if it is
all aluminum and no carbon at all, pretty low. But if you start
packing in much of a length of carbon instead of aluminum, it won't
take much before the total resistance is pretty much based only on the
carbon and the aluminum that is left doesn't really matter, anymore.
It's like that.

Unless the gap is VERY small (smaller than I have ever seen), it
pretty much takes over. Permeability of the rest of the core doesn't
matter anymore and the calculation of inductance depends on the mu for
air and the magnetic length of the air gap (no longer the magnetic
loop length for the rest of the core.) You really need SMALL air gaps
in order to achieve any reasonable value for the resulting L. You
wrap the windings around the air gap itself to help contain its
fringing into the wild world and you use the otherwise useless core
with its high perm material primarily to give the flux a tight narrow
conduit around from one end of the air gap to the other end of the air
gap, without fringing all over hell and gone. Read chapter 5 from
that site.

Jon
.

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