Re: magnetics design -- 60mJ energy impedance matching

On Sun, 17 May 2009 05:19:55 GMT, I wrote:

On Sat, 16 May 2009 21:52:48 -0700, D from BC
<myrealaddress@xxxxxxxxx> wrote:

On Sun, 17 May 2009 03:40:27 GMT, Jon Kirwan
<jonk@xxxxxxxxxxxxxxxxxxx> wrote:

<snip>

Yeah. Something like:

: R2
: ,--------~~~~~~~--/\/\--, ,-------,
: | 20 | | |
: | | | |
: --- C1 )||( \
: --- 1.5u L1 )||( L2 / R1
: | 150m )||( 240u \ .67
: | )||( /
: | / | | |
: | / | | |
: '--o o-~~~~~~~--------' '-------'
:
: 25:1

With 300V sitting on C1 to start. The squiggle characters represent
the 100' of wire length. R2 represents that resistance -- 20 ohms.
The 25:1 is what I guess is about right for critically-damped dumping
of energy. R1 is the squib, itself.

Problem is the core design for the transformer represented by L1/L2.
The rest of the specs and some reasoning for values are in the main
post.

I come up with HUGE volume for the core. I don't like that.

Thanks,
Jon

Oh... signal pulse transfer.
This is not quick for me to figure out.

Yeah. Most of the texts leave this by-the-by -- leaving it for later
on as "non-sinusoidal systems." Which makes me want to just set up
the Laplace transform and do it that way. But that means more double
checking on my end, as I need to be sure and then double sure I got
everything laid out right. So I was hoping for a pragmatic thought
about it.

Run a simulation?

yeah. Been there, done that. LTSpice wise, the design I cooked up
through the approach I listed out works beautifully -- and exactly as
I'd hoped. The ringing is just what I wanted to see. The energy
transfer is beautiful. It's all good.

But then... LTSpice doesn't tell me about B_sat! As far as it's
concerned, sans my adding in the right extras, the world is perfect
and everything is beautiful.

Of course, then I'll go out and buy a nice core, wind it up all neat
and pretty, and find that nothing much gets to the squib. The primary
becomes a dead short when B_sat is hit, and that's that.

I'd begin with figuring out when/if enough energy is transferred
before the core saturates.

( Bsat = UoUrNI/Le )

Well, yeah. I mentioned the equation, already. However, I used an
equation for N and for I and stuffed those in, solving back for
volume. Here's the logic using your terms:

B = U0*Ur*N*I/Le

If you'll forgive my use of L for inductance here (don't confuse this
with your use of Le as a length.)

L = U0*Ur*N^2*Ac/Le

then solving for N gives,

N = sqrt(L*Le/(Ac*U0*Ur))

Since I already know that when the capacitor transfers its potential
energy into the primary's magnetic energy, the current will peak at:

I = V*sqrt(C/L)

(This arrives without Laplace or 2nd order diff eq solutions from a
simple examination of the energy equations for both C and L and
realizing that the energy ping-pongs back and forth, ideally.)

Since there is a secondary, that peak current will be stunted a bit.
But leave it there for now.

Plugging these into your B equation, we get:

B = U0*Ur*N*I/Le
B = U0*Ur* sqrt(L*Le/(Ac*U0*Ur)) *I/Le
B = U0*Ur* sqrt(L*Le/(Ac*U0*Ur)) * V*sqrt(C/L) /Le
B = U0*Ur*V*sqrt(L*Le*C/(Ac*U0*Ur*L))/Le

Now, the 'L' cancels out,

B = U0*Ur*V*sqrt(Le*C/(Ac*U0*Ur))/Le

Folding 1/Le into the sqrt(), gives:

B = U0*Ur*V*sqrt(C/(Ac*U0*Ur*Le))

Folding U0*Ur into the sqrt(), gives:

B = V*sqrt(C*U0*Ur/(Ac*Le))

Now solving for Ac*Le gives:

Ac*Le = [C*V^2] * [U0*Ur/B^2]

Note how neatly L was removed when N and I were replaced out. In
fact, the first factor is directly proportional to the stored energy
on the cap. It's just a small factor, k, different. That value is
fixed by the cap I choose and the voltage impressed on it. So that
baby doesn't change. It's given.

All that's left to play with in figuring the volume, Ac*Le, is U0*Ur
and B. And it looks as though, for iron, that ratio is pretty much
fixed, as well. Transformer iron may have a B_sat that is 10 times
the B_sat of ferrite, for example. But then, guess what? The Ur is
about 100 times lower for ferrite. Oh? Well that means the ratio
didn't really change. Cripes!

Anyway, that's what's bothering me. I'm probably missing something
terribly important. I hope so.

Thanks,
Jon

Okay. It's making more sense, now. I sat down and looked at the B/H
curves for steel and cast iron and played with some delta-B/delta-H
values taken from the curve as finite approximations for Ur. Then I
picked off the central B value, squared it, to get some Ur/B^2 figures
plotted out for various B values through to saturation. The steel
curve I was looking at flattened out at 1.6 Teslas, by the way.

What makes sense now, given the volume equation I derived above and
these figures, is that designers must almost _always_ be designing
near the saturation region; balancing windings with the effective Ur
to get the right balance between copper losses from too many windings
as the core nears saturation and air-like behavior as the field
fringes out into infinity and very few windings at low values of H,
but where the core volume is too big. Somewhere between "few
windings, horribly huge core" and "no core to speak of, but massive
windings for an air core" you get the right in-between thing.

It's making more sense.

Simulating this design region in LTSpice is the next thing to learn
about.

Jon
.