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magnetics design -- 60mJ energy impedance matching

J

Jon Kirwan

Jan 1, 1970
0
I assume that you are going to dump the charge from a capacitor into a
100 ft transmission line with essentially a 0.6 ohm load at the far
end. A voltage step change. What is the waveform going to look like at
the load ?

Same question re-worded.

Well, a long line with a load on the other side of a transformer is my
hope. It doesn't work without something like that, otherwise. All
the energy winds up in the long wiring if just directly connected.

The waveform isn't crucial. The discussion I had with those who make
and sell the squibs is that the current at the squib cannot exceed
40A, as there may be some problems elsewhere (bonding connections,
clipping to the leads, etc.) before the tip gets a chance to heat. The
pyro material has very poor thermal conductivity, so it heats up
quickly and, so long as one isn't too slow about it, will ignite.

Jon
 
J

Jon Kirwan

Jan 1, 1970
0
It's a bit better than a rule of thumb. It comes from the following:

V = N*d(phi)/d(t) so that V*t = N*phi (volt*seconds). But phi = B*Ac making
V*t = N*B*Ac.

B is in Tesla, Ac is core area in square meters, N is turns of wire.

A given core has a core area Ac and a window area Aw. The number of turns
you can put on that core is a function of Aw and the wire diameter (d) so
that N = Aw/(d*d). The primary uses half of this and the secondary uses the
other half. So you have Np = Aw/(2*d*d). Wire doesn't always stack perfectly
and you might use wire with heavier insulation and you might wind the wire
loosely, and so on. So, there is a fudge factor associated with the number
of turns you can put on a core. Try .8 for estimation purposes so that Np =
.8*Aw/(2*d*d).

So now you have V*t = .8*Aw*B*Ac/(2*d*d). Insert the maximum B from the core
curves or data, insert your wire diameter, calculate your V*t, and you will
have the AwAc product.

John, I thought I'd comment on this thread one more time.

Thanks for the input. I've had some more time to consider the details
and your point that "it's a bit better than a rule of thumb" is
nail-on, forgiving the pun. ;)

You explained where the AcAw product arrives, well. I've come at it
from different angles, arriving at similar places. So it feels a lot
better to me. Here's one angle I tried:

mmf = N I B = U0 Ur H
H = mmf / l
dH = (N / l) dI dB = U0 Ur dH

so,

dB = (N / l) U0 Ur dI

introducing time,

dB/dt = [(N / l ) U0 Ur] dI/dt

But dI/dt = V/L = (V l) / (Ac U0 Ur N^2)

So,

dB/dt = [(N / l ) U0 Ur] [(V l) / (Ac U0 Ur N^2)]
= V / (Ac N)

Lots of stuff canceling out, which then rearranged gives me this,

Ac N = (V dt) / dB

Using finite differences to replace the infinitesimals gets me the
same place. Units work out, as well. (I gather that Tesla is just
Joules-seconds/Coulombs-m^2.)

Things seem to make more sense. mmf is equivalent to volts, but
driving a hypothetical magnetic fluid through a hypothetical magnetic
reluctance. Since electrons moving across a linear magnetic field
have a force vector acting on them which is always perpendicular to
their motion vector, they spiral around. But this also means that
forcing electric field sources (electrons) to spiral around, generates
a linear magnetic force field with what amounts to a + and - end. That
acts on vacuum, air, or whatever, to induce a "current," so to speak.
H is like a local field strength (similar to volts/meter); B is tied
to H, but includes the effective permeability; etc.

I'm still working at the concepts. The main thing is that the window
area thrust into the AcAw product makes a great deal more sense, now.
I see how and why and at least some times when not to go there. There
is another factor, current density in copper, that comes into play at
times. And I can see the replacement of dt with 1/f in equations, as
well. So it is slowly coming together.

I still need to see how to develop mmf=N*I as a consequence, I assume,
of the spherical integrals in Maxwell. But that's for another day.

Jon
 
J

Jon Kirwan

Jan 1, 1970
0

Yes, they give it (in SI units) as mass*time^-2*current^-1, which
yields what I wrote above (using the derived-SI unit of Coulombs,
rather than the primary unit of Amps.)

I almost always go back to SI units to do a dimensional analysis check
on any expression I see to make sure the units work out. If they
don't, usually it means there is a constant whose units I didn't apply
correctly, I misunderstood the units of the variables involved, or
else there are hidden constants with units the author didn't include
but which points out my own need to go track it down. There is
another possibility, of course, which is that the author didn't know
what they were quoting well. Which means setting that aside and
looking for better advice.

I'm more of a 'counter' type person. I prefer thinking in terms of
objects I can count, like electrons into Coulomb units, than in terms
of combined units like Amps, which SI prefers because of our ability
to measure, right now. And I keep in mind a few things that also make
sense to me from classical mechanics, like angular momementum which is
easily derived as a necessary consequence of assumed Euclidean space
and linear time, so Joule-seconds are meaningful to me for that reason
and for keeping one idea about electron spin in mind.

So Volts become Joules/Coulomb to me, which is easy to understand from
accelerating electons across a pair of charged plates in a vacuum.
Complete sense there. Ohms are in Joule-seconds/Coulomb^2, Farads are
in Coulomb^2/Joule, and Henries are in Joules-second^2/Coulomb^2.
Clearly, then, the multiplication of Henries and Farads yields
seconds^2, which must be square-rooted to get seconds out. Etc.

I am in the process, now, of going back to understanding Maxwell,
conduction current, displacement current and dielectrics, E fields, H
fields, Poynting vectors, which if I'm guessing right moves me towards
a closer understanding of near-field and far-field, as well (out of
phase nearby moving towards in-phase further out.) I need to factor
in ideas on back-to-back electric charge motion in a loop which from a
distance appears to be no motion of charge at all, quantum
fluctuations (1/2*k*T in each of 3 dimensions), etc. I've never taken
it on and I can see I need/want to. I'd like to get to the point
where I can derive mmf=N*I from a more fundamental understanding.

Jon
 
J

Jasen Betts

Jan 1, 1970
0
I'm considering a low power design for a rocket launcher. It will use
an MSP430 and a CR2032 as the power source for everything. (I need to
test this, but I'm hoping to pull up to 5mA, for about 100,000 pulses,
to reach a necessary 60mJ charge on a 1.5uF cap.) I'll adjust the OFF
times accordingly to reduce the time-to-charge to a minimum (shorter
and shorter as charge is added. ON time for each pulse is fixed by
the 3V, the inductor, and the peak current I can reach.

I want this to be absolutely safe at the rocket end. In other words,
no energy stored there.
Where am I going wrong?

you want a transformer that doesn't store energy.
there's no such thing.

I'd consider using a buck converter.
 
J

Jon Kirwan

Jan 1, 1970
0
you want a transformer that doesn't store energy.

If you are referring to my comment about "no energy stored there" I
meant I don't want an energy source at the rocket end. No batteries,
etc.
there's no such thing.

If you are referring to the transformer idea itself and not to that
comment, then transformers aren't designed with an eye to energy
storage. Flybacks are. But not transformers. So far as I'm aware,
anyway.
I'd consider using a buck converter.

That would likely be unworkable in this application. At least,
although I have built one before I still don't have a clear idea what
to consider here that would work for the intended purpose.

Jon
 
N

Nobody

Jan 1, 1970
0
you want a transformer that doesn't store energy.

No he doesn't.

A transformer is capable of storing energy, but until you put some energy
into it, it isn't actually storing energy.

He's asking for something where the energy stays in his pocket until he
is at a safe distance from the rocket, at which point the energy is sent
down the wire to the rocket.
 
J

Jasen Betts

Jan 1, 1970
0
No he doesn't.

A transformer is capable of storing energy, but until you put some energy
into it, it isn't actually storing energy.

He's asking for something where the energy stays in his pocket until he
is at a safe distance from the rocket, at which point the energy is sent
down the wire to the rocket.

someone suggested a capacitor and sidac(or other thyristor) at the rocket end, but that
was no good for some reason...
 
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