magnetics design -- 60mJ energy impedance matching

[Jon Kirwan]

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. 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

 

[Jon Kirwan]

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.

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
25mJ/50ms means .87A, sqrt(.5W/.67ohms). And with thicker wire a 12V
battery system it all easily gets the job done. It's safe, but it's
heavy and I'm trying to make this so convenient you hardly notice it
is there. I expect it to take a minute or two to charge up with the
meager milliamp capability of a CR2032, but that's okay. The main
idea is small and light weight.

Jon

 

[Jon Kirwan]

firing sight

firing site...

Jon

 

[Jon Kirwan]

<snip>

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

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.

These days, I'd probably modify something like this.
$12.00
http://store.sundancesolar.com/midywiledfl.html

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

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

http://www.inertproducts.com/inc/sdetail/1868
It's new!

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

 

[James Arthur]

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.

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
25mJ/50ms means .87A, sqrt(.5W/.67ohms). And with thicker wire a 12V
battery system it all easily gets the job done. It's safe, but it's
heavy and I'm trying to make this so convenient you hardly notice it
is there. I expect it to take a minute or two to charge up with the
meager milliamp capability of a CR2032, but that's okay. The main
idea is small and light weight.

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.

25mJ says you'll need >180V on 1.5uF (at the rocket
site), or 320V for 75mJ at the control site. That
sounds a lot like a disposable camera flash and a 'AA'
cell. Or maybe one of Wenzel's geiger supply circuits.

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

 

[Jon Kirwan]

On Sat, 16 May 2009 21:17:01 GMT, Jon Kirwan

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.
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.

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
25mJ/50ms means .87A, sqrt(.5W/.67ohms). And with thicker wire a 12V
battery system it all easily gets the job done. It's safe, but it's
heavy and I'm trying to make this so convenient you hardly notice it
is there. I expect it to take a minute or two to charge up with the
meager milliamp capability of a CR2032, but that's okay. The main
idea is small and light weight.

Jon

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

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.

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).

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.

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

Like I said, that's not the issue.

25mJ says you'll need >180V on 1.5uF (at the rocket
site), or 320V for 75mJ at the control site.

I'm figuring 75mJ (as I mentioned elsewhere) on the cap. I just keep
300V in mind on the 1.5uF cap. But 320V, perhaps.

That
sounds a lot like a disposable camera flash and a 'AA'
cell. Or maybe one of Wenzel's geiger supply circuits.

I think I can do it with the CR2032. I've played around getting to
300V with a micro doing the calcs for the OFF timing with some
success. I am not comfortable with magnetics enough to design the
standard flash lamp means, with a BJT and a CT transformer driving
with increasing frequency. So I prefer the micro approach for now.

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

100' each way, 20 ohms. The igniter is 2/3 ohm.

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

yes. The specs were in the original post.

Jon

 

[James Arthur]

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

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

James Arthur

 

[Jon Kirwan]

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?

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

 

[Jon Kirwan]

On Sun, 17 May 2009 03:40:27 GMT, Jon Kirwan



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

 

[Jon Kirwan]

On Sun, 17 May 2009 03:40:27 GMT, Jon Kirwan



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

 

[Jon Kirwan]

<snip>
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;
<snip>

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

Jon

 

[Tim Williams]

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.

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'.

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.)

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 = N*At
where At = 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 = V / Xl,
V / Xl = N*At
We don't need to know Xl, but we do know V. How about this, what if
Xl = Xc (since we're implicitly at 'resonance', albeit damped) and we
substitute like so...
V*2*pi*F*C = N*At
But we don't know F. Well, we half do, but that's still no better...
N*At = V*C / sqrt(L*C) = V * sqrt(C/L)
But L = Al*N^2. Hmmm...
N^2*At = 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 =
10AT, C = 1.5uF, Al = 0.5uH/T^2 and V = 300.
N = sqrt(V * sqrt(C / Al) / At) = sqrt(300 * sqrt(1.5e-6 / 0.5e-6) /
10) = 7.2 turns. Let's call it 8 then.

As for checking...
L = Al*N^2 = 32uH
Fo = 23kHz
Ipk = Vpk / sqrt(L/C) = 65A(!)
AT = 65A * 8t = 519, which is a whole **** 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

 

[Jon Kirwan]

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.

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.

[The Laplace solutions I've done before filled a few pages, if I
converted back to the time domain, and the result was expected --
showing both amplitude and phase shift (has both sine and cosine terms
and an exponential in front, memory serving.)]

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'.

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

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?

Yeah. I think I'm with you.

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.

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.

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.

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.

Consider that
Isat = N*At
where At = amp-turns saturation (10 let's say).

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.

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 = V / Xl,
V / Xl = N*At
We don't need to know Xl, but we do know V. How about this, what if
Xl = Xc (since we're implicitly at 'resonance', albeit damped) and we
substitute like so...
V*2*pi*F*C = N*At
But we don't know F. Well, we half do, but that's still no better...
N*At = V*C / sqrt(L*C) = V * sqrt(C/L)
But L = Al*N^2. Hmmm...
N^2*At = 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??)...

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.

Assuming it is, let's try some BS and see what happens. Let At =
10AT, C = 1.5uF, Al = 0.5uH/T^2 and V = 300.
N = sqrt(V * sqrt(C / Al) / At) = sqrt(300 * sqrt(1.5e-6 / 0.5e-6) /
10) = 7.2 turns. Let's call it 8 then.

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

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

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

Thanks,
Jon

 

[Fred Abse]

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.

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

RTFM

 

[Jon Kirwan]

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

RTFM

Indeed I should have.

Jon

 

[Baron]

firing site...

Jon

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

Just Curious Lurking.

 

[amdx]

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

Just Curious Lurking.

Why would anyone launching rockets, limit themselves to 30 gauge wire,
"suspicious minds want to know"

Mike

 

[James Arthur]

James Arthur wrote:

yes. The specs were in the original post.

Jon

Ah, sorry. I didn't realize your spec. was snipped.

ISTM you want to make 0.5W across 0.68 ohms, for 50mS.

To do that you need a pulse transformer that can make
0.6V @ 850mA across your 0.68 ohm ignitor, from a 300V pulse.

To make up for connection losses and such at the ignitor,
we'll take a WAG and say you want V.sec=2V @ 2A, which
allows for 0 to 1.6 ohms of hookup loss. Adjust those
assumptions as you see fit.

That's a 150:1 turns-ratio on the transformer.

I.pri = I.sec/150 = 13mA, which drops just 1/4 volt across
your 20-ohm firing line--transmission losses are minimal.

Critical damping to stop ringing? Why does that matter?
The thing's going to ring regardless of the transformer
design, but your 20 ohm hookup will damp it.

50mS is a half-cycle of 10Hz, which sounds a lot like rusty
old iron territory to my rusty self.

For a physical reality check, you might consider how
large a 220VAC to 2VAC wallwart would be.

If the pulse could be shorter, the core could be a lot
smaller.

HTH,
James Arthur

 

[FunkyPunk FieldEffectTrollsistor]

If the pulse could be shorter, the core could be a lot
smaller.

More quiet as well.

 

[Jon Kirwan]

Why would anyone launching rockets, limit themselves to 30 gauge wire,
"suspicious minds want to know"

Lower cost, ease of packing, flexibility, etc. If I were looking for
anything truly suspicious, it would be 40 gauge or finer. 30 gauge is
pretty obvious and, frankly, I'd like it more obvious by choosing
visible colors for it. Red wire wrap wire, for example.

Jon