J

#### Jon Kirwan

- Jan 1, 1970

- 0

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. A circuit will be clipped to the ignitor, but

no source of energy present at that end -- nothing that can even

remotely place things at risk as an absolute guarantee is important

until the remote end is wired up 100' away. I had considered crazy

ideas such as using a flash tube with 300V across it and requiring the

4-5kV trigger to come from the remote end, but that still means the

energy is present at the remote end and that if there is a short of

some kind it's possible for an accident to take place. So scratch

that idea.

The ignitors have an all-fire requirement of the delivery of 0.5W in

50ms to what amounts to a .68 ohm resistor. This is 25mJ in 50ms. The

wire will be 30 gauge with 100' out and 100' back, so about 20 ohms or

so in the wire. Finer wire would be a convenience, but the ohms go up

fast. 40 gauge is an ohm/ft, so that would be 10 times the

resistance.

So it appears its important to keep the current __low__ going out.

I've found that 300V across a 1.5uF cap does the job nicely, directly

connected. (I don't want to waste energy, either, which is a part of

why I'd prefer something on the 1.5uF side and not something larger.)

However, with 20 ohms of 100' out and back 30 gauge wire, only about

2mJ gets to the squib and nothing happens, at all. (The rest goes

into the wire. Not good.)

Impedance matching is suggested. So I started looking at transformers

I might design and make (toroid windings.) Not being very experienced

in this, I think I need some help considering the details.

Considering that I'd like a critically damped delivery of the energy

(not a lot of ringing), the a=N1/N2 ration works out to the following:

a*V_sec = V_pri

I_sec = a*I_pri

R_squib * I_sec should be approximately equal to V_sec, for

the critically damped case (rough guess)

inputs,

C = 1.5uF

V_pri = 300V

I_pri <= 5% of 300V across 20 ohms or about 0.75A

This yields:

a = sqrt(V_pri/I_pri/R_squib), or about 25:1 turns ratio.

The inductance is proportional to N^2, so 625:1 inductance ratio.

The 0.75A limitation sets up another thing. Roughly speaking, for the

critically damped case, I estimate about 78-80% of the energy will get

transferred in the first time-constant. [I'm going from intuition

here, taking sqrt(63%).] Since I=V*sqrt(C/L) when all the energy has

been transferred to the inductor and since I'm guessing 80% of that in

the first time period, I get L= .64*C*V^2/I^2 or about 150mH. This

means a secondary of about 240uH.

Okay. If I haven't already gotten into trouble, now I am. The issue

is the relationship of B_sat and permeability. Computing B is very

new to me. I note that for ferrites, should that be the way to go, I

should limit myself to a B_sat of about 200mT.

I gather that:

permeability * N B_sat 0.2T

-------------------- <= ------ = ----- = 0.266666

magnetic loop length I_peak 0.75A

But inductance is (permeability*N^2*area/(magnetic loop length)) and

that is set to 150mH.

From some playing around of the expressions, I finally arrive at

something like this:

area*(loop length) = permeability * 2.4

(The 2.4 constant comes from N*area must be greater than .5625 and I

selected .6 as a rounded value. That gets squared into .36 and taken

in ratio with the inductance, .15, for reasons I can explain but am

skipping right now. It's in the above equations, though.)

Picking a low permeability of about 200 for ferrite and using the u_0

of 4*pi*1e-7, I get a volume of 600e-6 m^3. But using transformer

iron for about 10 times the Tesla as a limit, I figure the constant is

100 times smaller:

area*(loop length) = permeability * 0.024

However, permeability is probably 100 times greater, too. So this

isn't being very helpful.

I'm in the same position, again.

So then I took the B=permeability*N*I/(magnetic length), substituted

in N=sqrt(L*magnetic length/(permeability*area)) [derived from basic L

equation), plugged in I=80% of V*sqrt(C/L), and wound up with:

volume = 80%^2 * V^2 * C * permeability / B^2

= [0.64 * V^2 * C] * [permeability/B^2]

Which confirms the general idea that I need low permeability with high

B_sat. However, since B_sat is about 10 times higher for iron than

for ferrite and since permeability is on the order of 100 times

higher, it's hard to get much of a win. Same position either way,

except that I can use fewer turns with the iron core and accept the

eddy currents.

The volume required seems roughly set by the first factor, which is

frankly proportional to energy. That's basically the 60mJ, plus or

minus a small constant factor difference. And I can't change that.

That is required by the application, itself. It's a given. So that

leaves me with finding low mu, high B_sat, to get the volume down.

Am I thinking about this right? With some rough figures, I'm coming

up with a cube more than 8cm on a side!!

Am I facing finding a toroid with 600e-6 m^3?? Setting the cross

section r to be 1/3 of the torus radius, I get 6.5cm radius to the

midline and a cross section radius of almost 2.2cm.

Yikes!

Where am I going wrong?

Jon