# magnetics design -- 60mJ energy impedance matching

J

#### Jon Kirwan

Jan 1, 1970
0
Ah, sorry. I didn't realize your spec. was snipped.

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

Yes. .5W * 50ms = 25mJ.
To do that you need a pulse transformer that can make
0.6V @ 850mA across your 0.68 ohm ignitor, from a 300V pulse.

If an average, then SQRT(.5W/.68) is indeed about 850mA. Problem is,
this won't be an average situation. This will be a pulse from a cap
discharge. So the current will likely vary a bit unless I add a lot
of extras to filter and shape it. Which I don't want to do.
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.

Secondary of 2V, roughly, at 2A, roughly. Yes. But these are all
averages and then inflated a bit.

The way I looked at this is to take a slightly different tack at it. I
figured that what I want a turns ratio such that the volts across the
secondary roughly equals the current across the secondary times the
squib resistance. Since V_sec=V_pri/a and I_sec=I_pri*a and finally,
V_squib=I_sec*R_squib, I got a=sqrt(V_pri/I_pri/R_squib), which gives
me about 25:1 as a turns ratio.

I would normally think 150:1 is fine, except for a few problems I can
guess at. I'll get to those.
That's a 150:1 turns-ratio on the transformer.

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

Yes, if the primary current were 13mA. But did you calculate the
primary inductance to achieve that?? We're getting neigh into kH
territory. And I've never seen one of those. Way overdamped, too.

I'd need to buy into copper commodity stocks, first.
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.

Well, I'm actually fine with ringing, as long as enough of the energy
gets across within the required time period. I just used that as a
rough approach. I could loosen up on that. But if it rings and the
primary current is high enough (smaller primary inductance), then I'm
facing losing most of the energy in the wire. Bad news.
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.

Short time periods imply higher currents, since the Joules is a
constant. I'd like to allow simple (idiot proof) wiring, which means
whatever they have on hand (or in a pinch) to attach to the secondary.

I'm looking to keep the peak currents on the secondary in the area of
about 10A. As you already figured, an average of 850mA is required to
get the job done. So 10A is a roughly doable target. Much more than
that and it may create other problems I'd like to avoid.

Jon

J

#### Jon Kirwan

Jan 1, 1970
0

Probably should have written, kHy. Hopefully, it got across.

Jon

T

#### Tim Williams

Jan 1, 1970
0
Starting over with some different assumptions:

Want power in resistor. Want to make resistor look bigger, in
particular, arbitrarily bigger than the transmission line resistance
(maybe by 10 times).
Voltage is higher and current is lower, but how much doesn't matter.

So we'll use an ideal transformer. With infinite inductance, the
resistor is reflected straight through. Offhand, if the line is 20
ohms, we want the squib to look like 200. But it's 0.68 ohms, so we
need a 294:1 impedance ratio or 17:1 turns ratio.

99% of the power is dumped into the load in 2.3 RC, so for t = 50ms,
tau = 22ms and C = 22ms/200 ohm = 109uF. Voltage is V = sqrt(2*E /C)
= sqrt(2*0.05J / 0.0001F) = 32V. So a 100uF, 35V cap will suffice
here. Quite moderate voltage ratings here.

For the transformer not to look like anything, it needs to have an
inductance maybe 10 times the nearest impedance, so a primary
impedance of 2kohms or so (at whatever the operating frequency could
be considered to be) would be prudent. If we ballpark it as 45Hz
average (1/tau), then L = 7H pops out. That's not too horrible.

Peak current from the capacitor is around 32V / 200 ohms = 0.16A. RMS
for most of the pulse is closer to 80mA or so. Small wire is fine.

Now you just need a core with enough winding space and permeability
for 7H worth of ~32AWG wire.

More choices: you could add a CMOS chopper (an H bridge thing, maybe
build it with high-Vgs MOSFETs so you can built it complementary
without worrying about more charge pumps) and have it run (self
excited?) at 1kHz let's say. Just let it buzz down until the cap is
discharged (a couple cycles). A _stock_ 2k:8 ohm audio transformer
would be more than enough on the other end!

For that matter, an audio transformer may already be enough. Worth a
try. Shame on you for not going directly to the junk box and trying
it! ;-)

Tim

J

#### Jon Kirwan

Jan 1, 1970
0
Starting over with some different assumptions:

Want power in resistor. Want to make resistor look bigger, in
particular, arbitrarily bigger than the transmission line resistance
(maybe by 10 times).
Voltage is higher and current is lower, but how much doesn't matter.

So we'll use an ideal transformer. With infinite inductance, the
resistor is reflected straight through. Offhand, if the line is 20
ohms, we want the squib to look like 200. But it's 0.68 ohms, so we
need a 294:1 impedance ratio or 17:1 turns ratio.

99% of the power is dumped into the load in 2.3 RC, so for t = 50ms,
tau = 22ms and C = 22ms/200 ohm = 109uF. Voltage is V = sqrt(2*E / C)
= sqrt(2*0.05J / 0.0001F) = 32V. So a 100uF, 35V cap will suffice
here. Quite moderate voltage ratings here.

For the transformer not to look like anything, it needs to have an
inductance maybe 10 times the nearest impedance, so a primary
impedance of 2kohms or so (at whatever the operating frequency could
be considered to be) would be prudent. If we ballpark it as 45Hz
average (1/tau), then L = 7H pops out. That's not too horrible.

Peak current from the capacitor is around 32V / 200 ohms = 0.16A. RMS
for most of the pulse is closer to 80mA or so. Small wire is fine.

Now you just need a core with enough winding space and permeability
for 7H worth of ~32AWG wire.

More choices: you could add a CMOS chopper (an H bridge thing, maybe
build it with high-Vgs MOSFETs so you can built it complementary
without worrying about more charge pumps) and have it run (self
excited?) at 1kHz let's say. Just let it buzz down until the cap is
discharged (a couple cycles). A _stock_ 2k:8 ohm audio transformer
would be more than enough on the other end!

For that matter, an audio transformer may already be enough. Worth a
try. Shame on you for not going directly to the junk box and trying
it! ;-)

Well, this is my first shot at attempting a magnetics design. And I
enjoy taking the opportunity to learn theory.

Interesting thoughts about H-bridge chopping. Complicates what I'd
imagined as simpler, but what the heck?! I'll play with the idea and
see how it flies. I kind of like it. I'd still like to learn if my
volume calculations are solid, but I can set that aside for another
day, too.

Jon

K

Jan 1, 1970
0
Well, this is my first shot at attempting a magnetics design. And I
enjoy taking the opportunity to learn theory.

I remember being taught the rule of thumb that core cross-sectional
area times winding cross-sectional area was a useful metric for comparing
power transformer designs. But I haven't done much power transformer stuff,
so I haven't practiced much with the formulae.

G

#### [email protected]

Jan 1, 1970
0
Excuse me asking, will the inductance of the 100ft of wire affect the
pulse at the far end ?

Just Curious Lurking.

Thank goodness someone else asked this question.

George Herold

J

#### James Arthur

Jan 1, 1970
0
Jon said:
Yes. .5W * 50ms = 25mJ.

If an average, then SQRT(.5W/.68) is indeed about 850mA. Problem is,
this won't be an average situation. This will be a pulse from a cap
discharge. So the current will likely vary a bit unless I add a lot
of extras to filter and shape it. Which I don't want to do.

You're making this too complicated--25mJ is 25mJ. The thermal
mass surrounding the ignitor integrates it for you. The main
limit I see is if you hit the squib too hard it'll disintegrate
(har) before it can deliver its full thermal payload. How hard
is "too hard"? Dunno.
Secondary of 2V, roughly, at 2A, roughly. Yes. But these are all
averages and then inflated a bit.

The way I looked at this is to take a slightly different tack at it. I
figured that what I want a turns ratio such that the volts across the
secondary roughly equals the current across the secondary times the
squib resistance.

What you've just said is that V.sec = the squib voltage. They're
directly connected, so no argument there.

Since V_sec=V_pri/a and I_sec=I_pri*a and finally,
V_squib=I_sec*R_squib, I got a=sqrt(V_pri/I_pri/R_squib), which gives
me about 25:1 as a turns ratio.

I would normally think 150:1 is fine, except for a few problems I can
guess at. I'll get to those.

Yes, except....

Yes, if the primary current were 13mA. But did you calculate the
primary inductance to achieve that?? We're getting neigh into kH
territory. And I've never seen one of those. Way overdamped, too.

I'd need to buy into copper commodity stocks, first.

Well, I'm kinda rusty, but ISTM the primary is determined
by the volt*seconds per turn your core allows.

One weber being a volt*second through one turn, and a
Tesla being one weber/m^2,

given a 3x3cm cross section laminated iron core that can
handle 1T,

300V * 50e-3 sec
turns.pri = ------------------ = 16,700 turns
1T * (0.03m)^2

Then choose secondary turns to give the desired
output voltage.

turns.sec = turns.pri/150 = 111 turns.

If you could detonate the squib with a shorter pulse
you could use far fewer turns, plus fatter wire.

Your stated requirements--especially the 10Hz part--demand
a hefty chunk of iron. Sending 400Hz AC down the line
instead would make the transformer _much_ smaller, and
the squib wouldn't care.

For fun, try simulating this chunky beast:

R1
SW1 / 20r
.------/ -----/\/\/\----. .----------.
| C1 _)(_ |
--- 1.5uF 120H _)(_ 5.5mH 0.68r
--- V=320V _)(_ |
| | | ===
=== === ===

Cheers,
James Arthur
~~~~~~~~~~~~~~~~~~~

Version 4
SHEET 1 880 680
WIRE -48 -48 -288 -48
WIRE -288 -32 -288 -48
WIRE 32 32 0 32
WIRE -288 64 -288 48
WIRE -48 64 -48 -48
WIRE 0 64 0 32
WIRE 32 64 32 32
WIRE -64 112 -128 112
WIRE 48 112 16 112
WIRE 144 112 128 112
WIRE 368 112 256 112
WIRE 368 128 368 112
WIRE -128 144 -128 112
WIRE -128 224 -128 208
WIRE 144 224 144 192
WIRE 256 224 256 192
WIRE 368 224 368 208
FLAG -288 64 0
FLAG 144 224 0
FLAG 256 224 0
FLAG 368 224 0
FLAG -128 224 0
FLAG 32 64 0
SYMBOL ind2 128 96 R0
WINDOW 0 -32 39 Left 0
WINDOW 3 -62 73 Left 0
SYMATTR InstName L1
SYMATTR Value 120H
SYMATTR Type ind
SYMATTR SpiceLine Rser=10
SYMBOL ind2 272 96 M0
WINDOW 0 -33 37 Left 0
WINDOW 3 -65 72 Left 0
SYMATTR InstName L2
SYMATTR Value 5.5mH
SYMATTR Type ind
SYMBOL res 352 112 R0
SYMATTR InstName R1
SYMATTR Value 0.68
SYMBOL voltage -288 -48 R0
WINDOW 0 -4 -25 Left 0
WINDOW 3 24 104 Invisible 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value PULSE(0 5 10mS 10uS 10uS 1S)
SYMBOL res 32 96 M90
WINDOW 0 -39 58 VBottom 0
WINDOW 3 -36 59 VTop 0
SYMATTR InstName R2
SYMATTR Value 20
SYMBOL cap -144 144 R0
SYMATTR InstName C1
SYMATTR Value 1.5µF
SYMBOL sw 32 112 R90
SYMATTR InstName S1
SYMATTR Value ""
SYMATTR SpiceModel MySwitch
TEXT 136 88 Left 0 !K L1 L2 .99
TEXT 104 272 Left 0 !.tran 0 250mS 0 5uS
TEXT -280 312 Left 0 !.model MySwitch SW(Ron=0.1 Roff=10Meg Vt=1 Vh=0
Lser=10n Vser=.6)
TEXT -336 160 Left 0 !.ic V(n002)=320

J

#### Jon Kirwan

Jan 1, 1970
0
Thank goodness someone else asked this question.

I had already considered it. It's in the neighborhood of 80uH each
way. It gets tossed into the model I'm using as leakage inductance.
My fault for not mentioning it, earlier. I apologize for that.

Jon

J

#### Jon Kirwan

Jan 1, 1970
0
You're making this too complicated--25mJ is 25mJ. The thermal
mass surrounding the ignitor integrates it for you.

That's where the dissipation takes place (except for the wiring), yes.
The main
limit I see is if you hit the squib too hard it'll disintegrate
(har) before it can deliver its full thermal payload. How hard
is "too hard"? Dunno.

The all-fire spec says <50ms. The main reasoning for that is its
dissipation into the surrounding rocket fuel material it is near and,
the wiring, the air, etc. Faster gets the necessary temperature rise
for ignition. Really fast is probably okay and it's been my
experience that a direct connection to 300V works just fine, whether a
power source or the 1.5uF cap. So I'm not so worried about fast. In
fact, if it is really fast I might instead just use an 0603 resistor
(or smaller?) instead. Probably would work. Now that I think of
it.... I've got to try it!
What you've just said is that V.sec = the squib voltage. They're
directly connected, so no argument there.
Yeah.

Since V_sec=V_pri/a and I_sec=I_pri*a and finally,

Well, I'm kinda rusty, but ISTM the primary is determined
by the volt*seconds per turn your core allows.

One weber being a volt*second through one turn, and a
Tesla being one weber/m^2,

given a 3x3cm cross section laminated iron core that can
handle 1T,

Egads!! It's the 3cm x 3cm that I'm actually trying to AVOID!
300V * 50e-3 sec
turns.pri = ------------------ = 16,700 turns
1T * (0.03m)^2

Then choose secondary turns to give the desired
output voltage.

turns.sec = turns.pri/150 = 111 turns.

If you could detonate the squib with a shorter pulse
you could use far fewer turns, plus fatter wire.

Your stated requirements--especially the 10Hz part--demand
a hefty chunk of iron. Sending 400Hz AC down the line
instead would make the transformer _much_ smaller, and
the squib wouldn't care.

For fun, try simulating this chunky beast:

R1
SW1 / 20r
.------/ -----/\/\/\----. .----------.
| C1 _)(_ |
--- 1.5uF 120H _)(_ 5.5mH 0.68r
--- V=320V _)(_ |
| | | ===
=== === ===

Instead of the switch, I just use a .IC spice command to set the
initial voltage on C1. But yes, nice curve and low primary current.

I'm just not wanting 3cm x 3cm, kilometers of copper wire, and a 120H
primary! I'm looking for something that can be entirely placed in a
pocket here.... and not noticed, that much. That 120H monstrosity
makes me laugh, just thinking about it!

Another thing crosses my mind... firing a three-engine or dual engine
rocket. In this case, the energy delivery would need to be faster, I
think, than 50ms and sufficiently simultaneous. In that case, I think
I may be forced to re-consider the idea of energy storage at the
remote site (which doesn't injur the existing desire to avoid that in
the single-engine case) and to use flash-lamps for simultaneity and
_safety_. It's very highly unlikely they will be triggered without a
fairly high voltage, which I can generate over a single conductor wire
and ground stakes at each end, I think. I'll need to consider how to
avoid static charges that can only develop very, very small currents,
just to be dead sure, but that is simple enough. More to play with.

Jon

J

#### Jon Kirwan

Jan 1, 1970
0
I remember being taught the rule of thumb that core cross-sectional
area times winding cross-sectional area was a useful metric for comparing
power transformer designs. But I haven't done much power transformer stuff,
so I haven't practiced much with the formulae.

Yet another thing I need to go think about. On first blush, I read
your use of "winding cross-sectional area" as being not too different
from "core cross-sectional area" in the case of toroids that I was
thinking about. But do you mean the cross-section of the wire in the
first phrase? That doesn't sound right, but I have to ask. Because
otherwise, if I'm winding around a torus, the winding cross section is
about the same as the core cross section.

Can you clarify what was being taught?

Jon

J

#### James Arthur

Jan 1, 1970
0
Jon said:
That's where the dissipation takes place (except for the wiring), yes.

The all-fire spec says <50ms. The main reasoning for that is its
dissipation into the surrounding rocket fuel material it is near and,
the wiring, the air, etc. Faster gets the necessary temperature rise
for ignition. Really fast is probably okay and it's been my
experience that a direct connection to 300V works just fine, whether a
power source or the 1.5uF cap. So I'm not so worried about fast. In
fact, if it is really fast I might instead just use an 0603 resistor
(or smaller?) instead. Probably would work. Now that I think of
it.... I've got to try it!

That sounds promising, and like fun.

Egads!! It's the 3cm x 3cm that I'm actually trying to AVOID!

Of course, but, meanwhile, your specs demand it. Hopefully I've
illustrated that, and why.

Instead of the switch, I just use a .IC spice command to set the
initial voltage on C1. But yes, nice curve and low primary current.

I'm just not wanting 3cm x 3cm, kilometers of copper wire, and a 120H
primary! I'm looking for something that can be entirely placed in a
pocket here.... and not noticed, that much. That 120H monstrosity
makes me laugh, just thinking about it!

You actually could do it with about a fist-full of iron and copper;
it's not a ridiculous approach, it's just way bigger than your goal.
Another thing crosses my mind... firing a three-engine or dual engine
rocket. In this case, the energy delivery would need to be faster, I
think, than 50ms and sufficiently simultaneous. In that case, I think
I may be forced to re-consider the idea of energy storage at the
remote site (which doesn't injur the existing desire to avoid that in
the single-engine case) and to use flash-lamps for simultaneity and
_safety_. It's very highly unlikely they will be triggered without a
fairly high voltage, which I can generate over a single conductor wire
and ground stakes at each end, I think. I'll need to consider how to
avoid static charges that can only develop very, very small currents,
just to be dead sure, but that is simple enough. More to play with.

Jon

Right. You either need to store energy at the rocket, safely
somehow, or go to a.c. transmission down the firing wires. A
300V, 400Hz squarewave, transformed down to 1.5VAC, direct into
the squib, for example.

The no-local-storage philosophy is excellent and preferred, I
think.

Fun project.

Cheers,
James Arthur

J

#### James Arthur

Jan 1, 1970
0
Jon said:
Yet another thing I need to go think about. On first blush, I read
your use of "winding cross-sectional area" as being not too different
from "core cross-sectional area" in the case of toroids that I was
thinking about. But do you mean the cross-section of the wire in the
first phrase? That doesn't sound right, but I have to ask. Because
otherwise, if I'm winding around a torus, the winding cross section is
about the same as the core cross section.

Can you clarify what was being taught?

Jon

That's what my example calculated--the max flux density for
a given core.

Flux density =
total flux / cross sectional area

I took it a little farther & calculated turns per winding.

Cheers,
James Arthur

K

Jan 1, 1970
0
Pardon my lack of clarity.

While the core area sets the amount of flux that can be linked
by any one turn of wire, the cross-section through which one can
thread wires (however many) is also important.

The cross-sectional area of a single wire controls the copper
loss and sets limits on current levels. BUT the cross-section
available through which to send all the wires in the winding
sets a limit on the transformer design realted to the power
handling capacity.

For a given winding window, you can put a few large diameter
wires (low voltage, high current) or lots of small diameter
wires (high voltage, low current) or anything in between.
But the available window limits the Volt-Amp product of
the winding.

That is as fundamental a limit on the design as the limit
imposed by core cross-section.

Is this any clearer? If not, ask away, and I will endeavour
to pull my explanations out of the mud.

J

#### John KD5YI

Jan 1, 1970
0
I remember being taught the rule of thumb that core cross-sectional
area times winding cross-sectional area was a useful metric for comparing
power transformer designs. But I haven't done much power transformer
stuff,
so I haven't practiced much with the formulae.

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.

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

The .67 ohm load reflects to the primary as 25*25*.67 or 418 ohms. Adding in
the 20 ohms shown gives 438 ohms load on the capacitor. The V*t will be the
integral of the usual time constant which I calculate to be 300*R*C = .197.
If the core is iron, B can be up to maybe 10kGauss or 1 tesla. I think you
said the wire is 30 AWG so the diameter is about .00028 meters in diameter.

Unless I've done something wrong, this give an AwAc product of 38.6e-9. So,
your core should have about 2 square centimeters of iron area with an equal
amount of window area depending on your choice of geometries.

It's been many years since I've gone through this, so please cut me some
slack if I've made any glaring mistakes. Anyway, you get the idea.

Cheers,
John

J

#### Jon Kirwan

Jan 1, 1970
0
Pardon my lack of clarity.

While the core area sets the amount of flux that can be linked
by any one turn of wire, the cross-section through which one can
thread wires (however many) is also important.

The cross-sectional area of a single wire controls the copper
loss and sets limits on current levels. BUT the cross-section
available through which to send all the wires in the winding
sets a limit on the transformer design realted to the power
handling capacity.

For a given winding window, you can put a few large diameter
wires (low voltage, high current) or lots of small diameter
wires (high voltage, low current) or anything in between.
But the available window limits the Volt-Amp product of
the winding.

That is as fundamental a limit on the design as the limit
imposed by core cross-section.

Is this any clearer? If not, ask away, and I will endeavour
to pull my explanations out of the mud.

Oh, no. I have a hard time imagining anything clearer! It's so
clear, in fact, that I think I actually understand some more of what

Thanks very much,
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.

The .67 ohm load reflects to the primary as 25*25*.67 or 418 ohms. Adding in
the 20 ohms shown gives 438 ohms load on the capacitor. The V*t will be the
integral of the usual time constant which I calculate to be 300*R*C = .197.
If the core is iron, B can be up to maybe 10kGauss or 1 tesla. I think you
said the wire is 30 AWG so the diameter is about .00028 meters in diameter.

Unless I've done something wrong, this give an AwAc product of 38.6e-9. So,
your core should have about 2 square centimeters of iron area with an equal
amount of window area depending on your choice of geometries.

It's been many years since I've gone through this, so please cut me some
slack if I've made any glaring mistakes. Anyway, you get the idea.

Thanks very much for the thoughts. I need to go through these,
together with the other materials I have around (Unitrode magnetics
book and some basic teaching materials, mostly, and zero experience)
and make sure I follow well. (I'm definitely NOT going back to
Maxwell equations, closed but arbitrary 3D surface integrals for
magnetic fields, electric fields from point source electrons, etc. At
least not today!) But one point is that I'm NOT necessarily using 30
gauge for winding. That's what I'm using to get the 100' out to the
rocket site. The primary winding wire can be adjusted as appropriate.
Same for the secondary. I just hated the volume results I was
getting.

Anyway, much appreciated. I _will_ take advantage of the generous
offer here and see where it takes me.

Jon

J

#### James Arthur

Jan 1, 1970
0
John said:
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.

Looks right to me John. I got the same result (earlier post) and
solved for Np:

V * t
turns.pri = -------
B * Ac

Jon spec'd a 50mS pulse, which forced a big core, lots of turns,
and a squawk of protest from Jon!

If we change the spec to permit the same energy in 1/100th the
time, it takes a lot less copper and iron.

For the 2 cm^2 cross-section core you mention below,

300V * 500e-6 sec
turns.pri = ------------------ = 750 turns
1T * 200e-6 m^2

I'll leave the core window & wire gauge calcs to Jon.

Cheers,
James Arthur

J

#### Jon Kirwan

Jan 1, 1970
0
Looks right to me John. I got the same result (earlier post) and
solved for Np:

V * t
turns.pri = -------
B * Ac

Jon spec'd a 50mS pulse, which forced a big core, lots of turns,
and a squawk of protest from Jon!

If we change the spec to permit the same energy in 1/100th the
time, it takes a lot less copper and iron.

For the 2 cm^2 cross-section core you mention below,

300V * 500e-6 sec
turns.pri = ------------------ = 750 turns
1T * 200e-6 m^2

I'll leave the core window & wire gauge calcs to Jon.

I actually appreciate every moment of these contributions. This is
going to help me a great deal in interpreting some "terse" docs I
have. I will get some time later on today to actually sit down for a
moment and reflect some more and what's been added here.

But my initial salvo (which means it is probably wrong, but I've yet
to see someone tear apart the development of the Ac*Lm equation I
produced) showed that core __volume__ was independent of inductance.
Instead, it appeared to be purely based on the energy storage of the
cap and the mu_0*mu_r/B^2 ratio of the core material. Naturally, I
accept that my development is dead wrong. In fact, I HOPE it is. But
I still wonder which part of what I did was in error.

When I get a moment later on to review this and think a little more,
perhaps it will arrive to me. I expect it will. But for now, I am
still both mystified and interested in the implications of that
equation for Ac*Lm.

Can anyone develop a different equation (parameterized differently)
for _that_ product? (I'll be trying later on, but if anyone wants to
take a shot at telling me something new in the meantime, I will
appreciate it a great deal.)

Thanks,
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.

In the case of a toroid, I assume Aw has to be computed from N*d^2,
then?
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.

The .67 ohm load reflects to the primary as 25*25*.67 or 418 ohms. Adding in
the 20 ohms shown gives 438 ohms load on the capacitor. The V*t will be the
integral of the usual time constant which I calculate to be 300*R*C = .197.

Yes, I get about the same figure. The integral out to infinity is
V*R*C, but 1e-34 or something silly like that less for shorter periods
on the order of the 50ms I care about. (In other words, not much
different.)

However, I've got another problem with this "reflect" concept. I've
seen it in books, obviously, and I'm aware of arguments about turns
ratios reflecting the secondary load to the primary according to
reflect back as inductive and visa versa, due to the presence of the
impedance in the denominator (memory serving.)

But here's the "problem" I have.

Ignore leaking inductance for now (which I know is at least 160uH.)
Also, let's assume away R2 for now. Call it zero ohms. Now we've got
a clean slate to work with for a moment. It's just C1, T1, and R1.

Okay. So let's establish an agreed upon value for N1 and N2 for my
25:1 ratio. Call it N1=250, N2=10, for A=250/10=25. Turns ration is
25:1. A^2 is 625. Now, according to this reflection theory, R1 gets
reflected into the primary circuit as A^2*R1 or 625*R1. As R1 rises,
therefore, so does 625*R1 as "seen" by the primary circuit.

So what happens when I clip R1 out of the circuit? Obviously, R1 is
effectively infinite now. (Close to it, yes?) Then, clearly
according to this theory, C1 should discharge via an infinite
impedance, reflected to the primary by T1. Which means C1 will
suddenly take a "long time" to discharge. But this isn't the case at
all. Instead, since no _energy_ is lost (we are assuming perfect
cases here and there are no longer any dissipative processes, the
C1/L1 circuit should "ring" forever without loss at a frequency
related by a constant and sqrt(L1*C1). Where is this reflected,
infinite R1 now?

Now imagine it another way. Imagine that I change N2 from 10 down to
1. This should, by all rights, mean that A=250/1=250 and that
A^2=62500. Whatever R1 is, the reflected impedance is 62500*R1.
Obviously, higher still. Suppose I took this to an extreme: 0
windings for N2! Obviously, infinite impedance is reflected back into
the primary. Again, nonsensical.

Both these thought experiments give me serious problems. The real
deal is that the only way energy is dissipated, is in R1. How much?
Well, the instantaneous value of (V(t)^2/R)*dt should represent the
tiny amount of energy dissipated. However, this is also certainly
limited by the energy available in the magnetic field (d/dt of
1/2L*I^2 or L*I*dI.) And starting at zero energy, it would be limited
by volume*H(t)*dB or I(t)*V(t)*dt. I think, anyway. Another key is
that things in nature arrange themselves such that minimum energy
occurs, which is why a voltage is induced in the first place in the
secondary -- to generate a counter flux to minimize the energy state
at all times.

So I have a qualitative problem with the "reflected impedance" in my
case. We aren't talking about a fixed frequency and a driven source
of _power_ here. It's a charged cap. It seems to me everything has
to be analyzed using instantaneous values to develop the picture
accurately.
If the core is iron, B can be up to maybe 10kGauss or 1 tesla. I think you
said the wire is 30 AWG so the diameter is about .00028 meters in diameter.

I get .0294cm or .000294.
Unless I've done something wrong, this give an AwAc product of 38.6e-9.

With my revised wire size I get 42.6e-9. Adding in something extra
for ineffeciency on the wire insulation, etc., let's just call it
48.4e-9 (because it makes a nice sqrt() result of 220e-6.)
So,
your core should have about 2 square centimeters of iron area with an equal
amount of window area depending on your choice of geometries.

Okay. I think I computed more like 1.4cm to 1.44cm on a side with the
two earlier figures (not the 50e-9 one.) [sqrt() of your figure,
assuming equal areas. Then sqrt() again to get square sides.]
It's been many years since I've gone through this, so please cut me some
slack if I've made any glaring mistakes. Anyway, you get the idea.

Well, let's go further. All this has done is get us a proposed core
area and wire area as a thin slice view. Now we need the length.

I did some dB/dH computations on a B/H curve for steel that topped out
at 1.6 Teslas. A good max is 1 Telsa for that material. I chose the
region from .4T to .8T as a clean slope upwards for my computation
around the .6T center and came up with Ur of about 19000. I assume
there are higher figures. But I'm going with Ur=20000 for now. U0 is
4e-7*pi (SI stuff.) So u=0.025, roughly. Let's just go with the L1 =
150mHy, for now. So we can figure N^2/Lm = .15/(.025*220e-6). This
is about 27300 for N^2/Lm. Um... Assume N1 for the primary is that
250 figure I mentioned before. This works out to 2.3m!! What?!?! Not
happing in my universe!

The problem is that the inductance is so small. BIG is GOOD, it
seems. But I have to actually wind stuff, too. So let's assume just
3 windings for N2. N1 is 75. Now we are down to 0.21 meters, 21cm,
over 8 inches. Yikes!

Criminently! One winding on the secondary??? Well, at least that is
only 2.3cm in length. An inch or so. But really?

Something feels VERY WRONG here.

I'll rework this with .2T for ferrites and look at a lower Ur.

Anyway, any thoughts on the impedance reflection thought experiments,
in the meantime?

Jon

J

#### Jon Kirwan

Jan 1, 1970
0
I should add that with so few windings on L1, the distribution of Ac
verses Aw isn't 50:50, most likely. But that's just another item to

Jon

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