magnetics design -- 60mJ energy impedance matching

What's wrong with a capacitor at the rocket end? Add a SPDT arming switch that shorts the cap before you hook it up to the igniter. Hang a discharge resistor across the cap for extra paranoia.

Connect the cap to the ignitor through some scr-type device, one of those trigistor type things.

At the far end, unshort the leads then apply enough volts to charge the cap and fire the scr thing.

You're more likely to fall into an old well walking back than to have this go off unexpectedly.

John

I want 33% efficiency [75mJ total, per firing, to achieve the 25mJ at the squib], NO energy at the rocket end until the moment of trigger, and two wires only. The benefits need to be light weight, small bulk, and smaller wiring to the firing sight -- but without losing any of what is already expected from practice. This means two wires and no energy at the rocket side until the trigger is pressed. I don't want to give any of that up. Yet I'm thinking about something you drop into your pants pocket, weighs very very little, and does the same task.

The CR2032 has about 200mAh at a mean of 2.8V or about 2000J, which even if I waste 2/3rd of it is just fine. Acheiving 500-600 firings out of one is workable. The main bulk will be the cap, as I see it. But it should meet the pants pocket requirement, all said.

In bog-standard battery systems in common use, the requirement of

Jon

firing site...

Jon

In my too young to buy fireworks days, I used a bicycle dynamo to set off homebrew *cough..*..

These days, I'd probably modify something like this. $12.00

Alternatively, How about a M57 firing device. If it's good enough for claymore mines, should be good enough for rockets.

D from BC myrealaddress(at)comic(dot)com BC, Canada Posted to usenet sci.electronics.design

I want something that is flat, small, and not noticeable in a pants pocket.

By the way, in my fireworks days, I just used a slingshot and let someone light the waterproof fuse before I let go. Or just lit a fuse and ran. I didn't mess around with electrics, at all.

Too big. Should be 1.5cm thick and 3cm x 4cm.

hehe. Still too big.

Now, can you help me with the magnetics? It's driving me crazy the darned thing seems to need so much core volume! I have to be missing something. But it looks independent of L and driven by the energy transfer requirement.

Jon

You can use any old off-the-shelf inductor that can handle your peak charging currents without saturating.

The CR2032 has an ESR of about 22 ohms, so you'll pull about 10mA max, maybe double that peak (isolating the CR2032 from the ripple current with a filter cap).

So you only need a 20mA inductor--trivial.

I wonder about the inductance and resistance of the firing wires: how long are the leads to this igniter- beast?

Isn't your load a low-z thing, a squib or something?

Cheers, James Arthur

As far as the pulsing to charge the cap part, yes. But that's not my problem. That's the easy part. I'm not even worrying about it, at all. It is the impedance matching at the rocket end.

The idea is to have a pocket unit with a CR2032 in it, a charge switch, and a push button. Plus two connections for wire. At the rocket end, 100' away, there is an impedance matching transformer and the squib. That's it.

It's not the circuit in the pocket unit I care about. That's easy. It's the impedance matching transformer at the rocket end of things, on the other end of 100' each way of 30 gauge wire.

I'm figuring on 5mA max. They rate them for pulsed use at 6.5mA or something like that, so that's comfortably under that point.

Like I said, that's not the issue.

I'm figuring 75mJ (as I mentioned elsewhere) on the cap. I just keep

I think I can do it with the CR2032. I've played around getting to

yes. The specs were in the original post.

Jon

Clarification: peak _inductor_ charging current, in case that wasn't clear.

James Arthur

I'm still a newbie with magnetics and I'm half dead from hiking all day.

If I got this right, this is a power transmission loss problem. Like with hydroelectric power.

stepup =====wire===== stepdown

Situation: Little battery, small unit size, lots of wire, low impedance load. Objective: Low system power loss..

Is that it?

And you need the transformer design?

D from BC myrealaddress(at)comic(dot)com BC, Canada Posted to usenet sci.electronics.design

Yeah. Something like:

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.

Run a simulation?

I'd begin with figuring out when/if enough energy is transferred before the core saturates. ( Bsat = UoUrNI/Le )

D from BC myrealaddress(at)comic(dot)com BC, Canada Posted to usenet sci.electronics.design

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.

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.

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

Okay. Maybe not almost always. But in cases like this, perhaps so.

Jon

Nahh, no need for all that. Just wave your hands and say, if it doesn't saturate on the highest peak (which would be every peak for a steady state sine wave with amplitude equal to your capacitor's initial voltage; see below), then it certainly won't saturate ever during the ring-down.

Besides, you never get anything from the L^-1 of an electronic circuit besides exp(tau + j*omega). Laplace can kiss my butt, I'm perfectly happy with a quadratic solution hovering over an 'e'.

This is the same type of assumtion: L's peak energy will of course be lower, but you'd need to solve equations to know how much. It's an extra factor of safety to assume 100%, so who cares about losses or ringdown?

As for the solution, I say, look at the volt-seconds. 'Course, that requires knowing what L is, which requires a core and so on, but still.

Here's a tip-- black (or enamelled green or sometimes blue) high-mu ferrites (mu > 1k) almost always saturate at 10-20At. Experimental fact. A_L varies directly with size, from maybe 100nH/T^2 for fairly small pieces, on up to >3uH/T^2 for big fat toroids and line chokes.

Consider that Isat =3D N*At where At =3D amp-turns saturation (10 let's say). We don't know I, but we know V. (Keep in mind that, since Isat is peak, V will also be in peak volts -- which is good, because so is your capacitor.) Substituting I =3D V / Xl, V / Xl =3D N*At We don't need to know Xl, but we do know V. How about this, what if Xl =3D Xc (since we're implicitly at 'resonance', albeit damped) and we substitute like so... V*2*pi*F*C =3D N*At But we don't know F. Well, we half do, but that's still no better... N*At =3D V*C / sqrt(L*C) =3D V * sqrt(C/L) But L =3D Al*N^2. Hmmm... N^2*At =3D V * sqrt(C / Al) Aha! Finally down to the variables we can know...

Please check this, I'm tired and this doesn't feel right (seems to be turns^3 on the left and turns^1 on the right??)...

Assuming it is, let's try some BS and see what happens. Let At =3D

As for checking... L =3D Al*N^2 =3D 32uH Fo =3D 23kHz Ipk =3D Vpk / sqrt(L/C) =3D 65A(!) AT =3D 65A * 8t =3D 519, which is a whole f*ck of a lot more than the 10 I started with. Ya, I borked the algebra for sure.

Oh well, give it a try. Maybe you can get variables to fall out (or not!) better AND get a useful answer. Have fun.

Tim

Yeah, that's about how I see it. The peak current should occur when the voltage across the inductor reaches zero the first time. There will be additional oscillations (especially with a large turns ratio), but that first one is the larger peak. So yeah, the rest will not be a problem. Never thought otherwise.

Hehe. I actually like Laplace. Some of the pain is dinking around with partial fractions and doing conjugates here and there, after.

Yeah. I think I'm with you.

Well, I took a crack at it to see if I could figure the core volume purely from energy, permeability, and B_sat and I seem to have gotten there. But I just didn't like the answer much. And since I haven't ever tried my hand at these things, I figured I have made some gross mistakes.

I think what I'm gradually getting a clue about (and I may reverse myself there) is that if you are pushing up against a volume problem, then just let it move towards saturation and design for that. As the inductance degrades, di/dt should rise more rapidly than otherwise as it moves towards an air-core behavior. I can live with that, if enough energy gets across.

I've got to think more about the loss side of things. Between the rock of copper losses on one side and the hard place of core losses on the other side, there is some nicer place to be I think.

I haven't dug deeply enough to see the details here. But I suspect that if I sit down and compute things a few times, I'll probably see better what you are saying here. Experiments aside, the theory should support what you are saying, I think. If it doesn't, it needs changing.

I think I don't understand this. I remember reading that Isat is:

Isat = Bsat*Le/(N*U0*Ur)

If those two are equated and I move things around a bit, I get something that looks wrong. But as I said I need to think more about what you wrote earlier. Probably, I'll get more of it then.

Well, I got turned around at the outset. So I need to go back and re-align myself before I agree or criticize. It's late, so I'll leave that for tomorrow.

Hehe. If nothing else, just trying to make sense will help things make sense. Um... I think so, anyway.

Thanks, Jon

There's a hysteretic core inductor model in LTSpice, that accounts for Bs.

RTFM

Thanks.

Indeed I should have.

Jon

Excuse me asking, will the inductance of the 100ft of wire affect the pulse at the far end ?

Just Curious Lurking.