magnetics design -- 60mJ energy impedance matching

K

Jan 1, 1970
0
I've been posting various out of copyright technical books on scribd,
the one below is old, so old it may not be very useful. It mostly
concentrates on rotating machinery (generators and motors) but has
some stuff on transformers. This is more for steady level AC than
pulsed DC, but I don't have any old stuff on pulse transformers.

http://www.scribd.com/doc/15657613/The-Magnetic-Circuit

J

John KD5YI

Jan 1, 1970
0
In the case of a toroid, I assume Aw has to be computed from N*d^2,
then?

Yes, as follows:

(If you size the wires so that their cross-sectional areas are proportional
to the currents they carry)
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.)

True, but it is easy to integrate it to infinity without much thinking and,
as you say, 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.)

I have never heard that last part before. Wouldn't that mean that the
transformer has to invert the phase of either the current or the voltage but
not both? I don't see how that is possible.

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?

You removed it. It is not there and you have only the capacitor and an
inductor. A free-wheeling resonant circuit with no losses. You are correct,
being a perfect parallel LC, it will ring forever.

Remember that C1 will not discharge via infinite impedance, it will
discharge/charge periodically due to an exchange of energy with the primary
inductance of the transformer.

In a non-perfect circuit, it is probable that, without the load resistor,
the capacitor will hold up long enough for the core to saturate due to
excessive volt*seconds.

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.

No, it makes sense. You wind up with a capacitor in parallel with an
inductor. To simplify matters, we ignored the primary inductance when the
resistor was present because the primary inductance is large with respect to
the load. After all, you are attempting to keep the core out of saturation
for the duration, aren't you? You can keep the inductance in your model when
you analyze it, if you wish.

Both these thought experiments give me serious problems. The real
deal is that the only way energy is dissipated, is in R1.

Yes, that must be true for the circuit you described above. If you don't
have a load, what do you expect? If you leave the 20 ohms in there, you have
that as a dissipator as well. Also, the core will use some of the energy in
the real world.

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 "reflected impedance" didn't work, we would never have had "matching
transformers" for interstage coupling, for speaker output, and a host of
other examples which I cannot think of right now. They are really just
ordinary transformers (with attention given to specialized requirements such
as leakage inductance, iron loss, etc).

I get .0294cm or .000294.

You're probably right. I wasn't trying to actually come up with a solution,
just give an example.

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!

This is not valid. You cannot now assume an arbitrary inductance nor Np
after we calculated the required AwAc with the other chosen parameters
above. Now you must use the calculated Ac and the calculated Aw to compute
your Np and inductance. For simplicity, assume a pot core (that's what I
would use anyway). Assume your core is round and get its circumference from
the area you calculated. Then see how much wire you can put in the square
window. That will give you Np and length. For our imaginary core:

If the window is square in cross section, it is 1.48cm wide and half of that
is .742cm. The core is 2.2cm^2 giving a diameter of 1.674cm. To get the mean
length of a turn, we will add the core diameter to half of the window width.
So, the MLT is 2.416cm. Np is window area divided by wire diameter squared,
Np=.742^2/.0294^2 = 853 turns. And, that many turns will be 2.416*853 = 20.6
meters in length (less than 7 ohms).

I will let you calculate the resulting inductance, if you think it is
needed.

Some notes:

* You will most likely need to recalculate when you settle on a core/window
geometry. Things are not as perfect in a core catalog as they are in a
thought experiment (but, you know that). I don't have much experience with
toroids, but my gut feeling is that the measured/published window area
cannot be as fully utilized as in other geometries. Watch out.

* 300V will probably require at least one layer of mylar tape half way
through the primary winding. I, myself, would probably use two layers so
that no more than 100V would be seen between any two turns. Maybe heavy
insulation on the magnet wire will support that. So, that means that you
might need a little more window (increase AwAc).

* If you settle on ferrite, you will need to re-select based on Bmax. It
will get bigger than our example (by about 3 times).

* I'm getting too old for this.

Cheers,
John

J

John KD5YI

Jan 1, 1970
0
James Arthur said:
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

Thanks for checking, James. I make lots of misteaks these days, so it is
rewarding to have someone validate my stuff.

I agree with your comments but I don't remember from other posts if he can
tolerate a shorter time. If not, he'll need to re-examine his approach.

Cheers to you, too.
John

J

Jon Kirwan

Jan 1, 1970
0
Yes, as follows:

(If you size the wires so that their cross-sectional areas are proportional
to the currents they carry)

True, but it is easy to integrate it to infinity without much thinking and,
as you say, not much different.

impedance figure is correct, it is then easy to see if my time period
of 50ms is large compared to RC. If so, just assume infinity.
I have never heard that last part before. Wouldn't that mean that the
transformer has to invert the phase of either the current or the voltage but
not both? I don't see how that is possible.

Well, I've been skimming over so much material lately, I feel like
I've gone crazy. This particular piece came from a section on coupled
impedance. Zin = Zp + (wM)^2/(Zs+ZL). The second term is the coupled
impedance term. They write, "Note that since secondary impedances
appear in the denominator, they reflect into the primary with reversed
reactive parts. Thus, capacitance in the secondary circuit looks
inductive to the source and visa versa for inductance."

Since I was skimming and didn't take the "sit down" time to think
about any of this, yet, I just spouted something I'd read without any
real study on my part. My fault. The M term appears to be the
leakage inductance term. If it is tiny, that whole factor goes by-by
and so I'm just wrong. It only comes into play, as I now understand
it, in cases of LOOSELY coupled transformer action!!

Sorry.
You removed it. It is not there and you have only the capacitor and an
inductor. A free-wheeling resonant circuit with no losses. You are correct,
being a perfect parallel LC, it will ring forever.

Hold it here. Assume R1 is very large, then. Let's not completely
remove it, but make it arbitrarily large, yet still finite. It
doesn't dissipate even in this case, so the glib comment of "you
removed it" no longer applies. I'm still keeping it there. Yet
energy isn't dissipated (an arbitrarily small finite amount is, but I
can make it arbitrarily small.) It still makes no sense to consider
such a high resistance reflected by N^2 into the primary circuit. At
least, not in series in the loop replacing Lp. That would imply a
long discharge time.

This is a part I'm struggling with. In the case of a signal assumed
to start at negative infinity in time and proceed to positive
infinity, of a single frequency, I can see the "impedance matching"
method as a useful technique to maximize power transfer -- the classic
case being a speaker. But this is more like a Dirac pulse (not
really), which suddenly starts at time t=0. And I'm not sure how
useful this heuristic is in this case. My gut tells me there is
something important missing. On the other hand, I'd love for it to be
exactly this simple.

I still haven't found the time to sit down with the differentials for
the cap, the transformer parts, and the flux changes (Faraday's law.)
I want a moment when I can clear other pressures before I do that,
which is most of the reason why I haven't, yet.

I hope to __derive__ from fundamentals the idea of the reflection. My
situation is indeed __energy__ transfer and that implies power when
the differential for time is introduced, so I suspect I'll get to the
same place. But I need to see how I got there, I think. Not just
accept an assertion. So I need to spend the time. Oh, well.
Remember that C1 will not discharge via infinite impedance, it will
discharge/charge periodically due to an exchange of energy with the primary
inductance of the transformer.

That has _never_ for a moment left my mind. The way I see it in the
idealized case is that there will be an induced voltage on the
secondary and that this induced voltage with cause a current to flow
through the load. This will dissipate energy from the field. If the
which point the voltage across the primary is zero) is less than the
energy transferred to the primary, then the next quadrant is entered
as the voltage on the cap reverses and secondary load dissipation
continues again. Etc. With a small enough induced voltage on the
secondary, this could go on for a while.
In a non-perfect circuit, it is probable that, without the load resistor,
the capacitor will hold up long enough for the core to saturate due to
excessive volt*seconds.

And the primary voltage goes to zero since no more energy can be
stored in the field, as I gather things.
No, it makes sense. You wind up with a capacitor in parallel with an
inductor. To simplify matters, we ignored the primary inductance when the
resistor was present because the primary inductance is large with respect to
the load. After all, you are attempting to keep the core out of saturation
for the duration, aren't you? You can keep the inductance in your model when
you analyze it, if you wish.

I'm beginning to see things better. Thanks.
Yes, that must be true for the circuit you described above. If you don't
have a load, what do you expect? If you leave the 20 ohms in there, you have
that as a dissipator as well. Also, the core will use some of the energy in
the real world.

Yes, all those things are in mind. I need to be able to quantify all
of them, eventually. For now, it's enough that I have a qualitative
model in mind that accounts for them and will force me to eventually
go through the calculations so that I can see for myself what counts
for what and in what circumstances. Thanks for keeping all the balls
in play and above the table in plain view.
If "reflected impedance" didn't work, we would never have had "matching
transformers" for interstage coupling, for speaker output, and a host of
other examples which I cannot think of right now. They are really just
ordinary transformers (with attention given to specialized requirements such
as leakage inductance, iron loss, etc).

Well, I understand that and I completely accept that reflected
impedance does work for the situations that heuristic was developed
for. I don't _yet_ consider it a law of nature. I imagine it as a
useful deduction to specific circumstances of absolutely generalized
laws of nature. Which means it may... or may not... be appropriate to
my circumstance. In my mind, anyway.

The counter argument is that the heuristic was developed to find a way
of designing to maximize power transfer to a load. And that is pretty
much what I'm struggling with. So ... well, it should apply. But I
need to spend more time on this. There are enough different elements
to make me suspicious and to goose myself to do more work instead of
just assuming.
You're probably right. I wasn't trying to actually come up with a solution,
just give an example.

No problem. I went and looked. It's my job to nail down the numbers.
You were generous enough to point a way and I appreciate that.
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!

This is not valid. You cannot now assume an arbitrary inductance nor Np
after we calculated the required AwAc with the other chosen parameters
above. Now you must use the calculated Ac and the calculated Aw to compute
your Np and inductance. For simplicity, assume a pot core (that's what I
would use anyway). Assume your core is round and get its circumference from
the area you calculated. Then see how much wire you can put in the square
window. That will give you Np and length. For our imaginary core:

If the window is square in cross section, it is 1.48cm wide and half of that
is .742cm. The core is 2.2cm^2 giving a diameter of 1.674cm. To get the mean
length of a turn, we will add the core diameter to half of the window width.
So, the MLT is 2.416cm. Np is window area divided by wire diameter squared,
Np=.742^2/.0294^2 = 853 turns. And, that many turns will be 2.416*853 = 20.6
meters in length (less than 7 ohms).

I will let you calculate the resulting inductance, if you think it is
needed.
<snip>

Thanks, John. I am beginning to feel it is NOT important. Or, at
least, that your experience tells you it's not and I should listen to
that.

I take your point about letting the inductance fall where it may. Once
I know the Ac*Aw, my job is to just find a nearby core for that
product, as you say. The Lm term will be what it is. And I wind the
primary to fill whatever area presents itself. The inductance falls
out, but I shouldn't be focused on that I gather. It's just whatever
Ac*u*Np^2/Lm becomes. It's not a design goal.

That's a way of _seeing_ all this. On the other hand, I may actually
want the behavior of a specific inductance. I know, for example, that
the primary inductance should NOT be larger than some value because if
it is then the transfer of energy into the magnetic field will be too
slow for my needs. So there are boundaries. It's not entirely just
left as an output of other factors. But I think I take your point and
will reflect (sorry, no pun intended at the time) on your comments
more.

I appreciate all you've offered me.

Thanks,
Jon

J

Jon Kirwan

Jan 1, 1970
0
Interesting looking for pot cores on the web. I found one with
Ur=3000, a 36x22 pot core, for 2/ea. http://www.ibsstore.com/scripts/prodView.asp?idproduct=2588 That has about a 2cm^2 Ac, with a minimum of 1.7cm^2. Of course, it's not iron and the Bmax is a little over .3, so things change. Call Bmax .3. To keep under an Ac*Aw of 1.7cm^2, I need 38 gauge wire of 124 micron diameter. This gets me to 1.59cm^2, which is less. From a different resource (unitrode magnetics book), I get Np = Volt-secs/2/Bmax/Ac. This suggests about 2000 turns of the 38 gauge wire. Area would be something a bit larger than 2000*.000124^2, or about 31 mm^2. The above core elsewhere (in a way I can't arrive at looking at the diagrams) says that the winding area for 2 sections leaves 33.9 mm^2 for each section. So that just fits, if my rambling ignorantly through that datasheet means anything. Secondary would be 80 turns of 24 gauge, which is 25.6 mm^2 (no allowances, either.) What isn't clear to me from the datasheet is that they specify these things: Ve = 10700 mm^3 Ie = 53.2 mm Ae = 202 mm^2 Amin = 172 mm^2 core factor (C1) = .264 mm^-1 Probably a zero sized air gap, though I can't find a spec for the 3E2A being offered on that web site, but instead just 3E27 (which does say 0mm air gap.) Bmax seems to be .32 or .315. But the Ac*Aw product isn't listed and it's not clear to me how to arrive at those figures from the datasheet: http://www.ferroxcube.com/prod/assets/p3622.pdf 54 grams! This is new territory for me. Thanks a lot. I like to get some theory down first, then work on the experimental side to make sure that practice comes close to what theory predicts. Right now, I'm struggling with the theory part. I'll get around to building, soon, once I think I have an idea what I'm shooting for and can weigh results. Getting success (good as that may be at times) without knowing why is frustrating. Thanks, Jon J Jon Kirwan Jan 1, 1970 0 That has about a 2cm^2 Ac, with a minimum of 1.7cm^2. I meant Ae. Actually, I'm not sure what Ae specifies. I just have an idea that it is the effective area that the flux passes through. But how it relates, in this case, to Ac and Aw... well, I am not sure. Jon J Jon Kirwan Jan 1, 1970 0 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, this is becoming more interesting! I'm beginning to realize that this is a bad volt-seconds saturation problem, just letting it passively operate. By chop/reverse with an H-bridge, if I understand it correctly, I can keep the volt-seconds into manageable numbers on each side of zero and thereby make this smaller than if I just "let it go," so to speak. Some of the picture is trickling in. Jon J John KD5YI Jan 1, 1970 0 Well, I've been skimming over so much material lately, I feel like I've gone crazy. This particular piece came from a section on coupled impedance. Zin = Zp + (wM)^2/(Zs+ZL). The second term is the coupled impedance term. They write, "Note that since secondary impedances appear in the denominator, they reflect into the primary with reversed reactive parts. Thus, capacitance in the secondary circuit looks inductive to the source and visa versa for inductance." Since I was skimming and didn't take the "sit down" time to think about any of this, yet, I just spouted something I'd read without any real study on my part. My fault. The M term appears to be the leakage inductance term. If it is tiny, that whole factor goes by-by and so I'm just wrong. It only comes into play, as I now understand it, in cases of LOOSELY coupled transformer action!! The M may be mutual inductance. If w = 2*pi*f, then the second term looks like the square of the mutual inductance's reactance divided by the sum of the secondary impedance and load impedance. I think this applies where the inductances are small relative to the load. The relation looks vaguely familiar, but please don't hold my feet to the fire on this. Do you have LTSpice? If so, I encourage you to set up a source, a 10:1 transformer (with K=1) and a secondary capacitor. Make sure that the primary inductance is high enough to make the magnetizing current insignificant (say, 1/1000 the load current). Then look at the voltage/current relationship on the primary. Compare to a secondary resistor. Let me know if the phase relationship is looks like an inductor. Also, if you look at the ratio of voltage to current on the primary, it will look as if the capacitance has been cut to 1/100 the actual value. Now you can play with the winding inductances, coefficient of coupling, leakage inductances, winding resistances, and winding stray capacitances. (Big snip) Happy Thursday, John J John KD5YI Jan 1, 1970 0 What isn't clear to me from the datasheet is that they specify these things: Ae = 202 mm^2 Amin = 172 mm^2 40 years ago, when I was a young and daring lad, I used the Ae (effective area) value. To be safe, use the Amin value. Probably a zero sized air gap, though I can't find a spec for the 3E2A being offered on that web site, but instead just 3E27 (which does say 0mm air gap.) Bmax seems to be .32 or .315. Now go back and recalculate your required Ac*Aw product based on this much lower flux density. But the Ac*Aw product isn't listed and it's not clear to me how to arrive at those figures from the datasheet: http://www.ferroxcube.com/prod/assets/p3622.pdf Bottom of page 4 shows the "former" winding area at around 65mm^2. Multiply this times your Amin. I get about Aw*Ac = 48e3mm^4. However, revising the Aw*Ac product for the lower Bmax indicates that you need about 140mm^4 or about 3 times more core using the 30 gauge wire. Using smaller wire as you seemed willing to do, will reduce the requirement somewhat. By the way, winding 2000 turns on that core will result in about 70 henries. Cheers, John J John KD5YI Jan 1, 1970 0 Jon Kirwan said: I meant Ae. Actually, I'm not sure what Ae specifies. I just have an idea that it is the effective area that the flux passes through. But how it relates, in this case, to Ac and Aw... well, I am not sure. Jon Ac is what I call the core area that you need. That is, it is the area that a manufacturer says you have, not necessarily the physical dimension. Ae, as your guessed, is pretty much the same thing. It probably comes from measurements of the core characteristics which might include fringing. As an example, calculate the center leg of the core area from the drawing of the one you linked. You will see that your calculated number falls between the Ae and the Amin number. I'm not sure what Amin is, but my inclination is to use the less optimistic number. Either that, or use a safer (lower) Bmax. Aw is what I call the available window area. I don't know what comments to make on this as it seems self-explanatory to me. Can you clarify your question? John J Jon Kirwan Jan 1, 1970 0 Ac is what I call the core area that you need. That is, it is the area that a manufacturer says you have, not necessarily the physical dimension. Ae, as your guessed, is pretty much the same thing. It probably comes from measurements of the core characteristics which might include fringing. As an example, calculate the center leg of the core area from the drawing of the one you linked. You will see that your calculated number falls between the Ae and the Amin number. I'm not sure what Amin is, but my inclination is to use the less optimistic number. Either that, or use a safer (lower) Bmax. Aw is what I call the available window area. I don't know what comments to make on this as it seems self-explanatory to me. Can you clarify your question? Thanks. On the last part, about Aw, you earlier mentioned that the "Bottom of page 4 shows the 'former' winding area at around 65mm^2. Multiply this times your Amin." I think you mean this value as Aw. (For the reason that if Amin represents Ac [for all intents and purposes], then you must have meant me to understand this 65mm^2 as Aw.) What I see on page 4 is a table of winding areas: Section Winding Area 1 72.4 mm^2 2 2x33.9 mm^2 3 3x21.0 mm^2 No 65mm^2 there. But I take it you meant 65 as a conservative 72.4. But then, with a primary and a secondary involved isn't it the case that I use the 2-section row and plan a primary with 33.9mm^2 and a secondary with the same? Also, I just plain cannot see how to come up with the 72.4 mm^2 they give for the winding area with one section. Let me walk through my thoughts. There is a "former" that is used for winding. It's designed so that the two halves slide their center regions through the hole in its center and wrap around it. (Keep in mind I have never used one of these, so I'm going from the diagrams and no experience, at all.) Once wound, I just take one half of the core itself, slide the former onto the central pedestal in the middle of the core half, then place the other core half over that and close it up (by a clip?) They designate the outer diameter of the former as 29.6mm (or -.2mm less) and the outer reach of the inner cylinder as 17.9mm (or -.2mm less.) Assuming worst case, I use 29.4mm and 17.9mm -- a difference of 11.5mm. Looking at the diagram on page four for the former, I see a central division (vertically aligned on the page) between the two outer edges, which I take to mean that if I am winding two coils then one goes on one side of that divider and one goes on the other side. In my case, that's the primary and the secondary. Yes? They assume a ..8mm division between two coils and thus (14.4-2*0.8)mm divided by 2, for each. This is 6.4mm. Since I figure 11.5mm x 6.4mm for each coil I get 73.6mm^2. But this is just for one side, not the other. The other side should be the same. So although I get a number perilously close to the 72.4mm^2 they give for one section, it's a value I figure for two sections. Which leaves me thinking I don't know what they mean in the table. Unless the 1 means one secondary and the primary is already assumed. Jon J Jon Kirwan Jan 1, 1970 0 Separate question. The datasheet for that pot core I mentioned earlier suggests the following to me as a "kit": * core, in two pieces * former, one piece * container, one piece * spring, one piece * tag plate, one piece When someone sells something like this and somewhere on the web page it says "UNIT PRICE PER SET" I tend to imagine that I get _all_ the pieces. However, other information on the web site makes me feel a little cautious. This company apparently bought these as surplus items and that suggests to me that they may have been reboxed per whatever stocking methods the original buyer may have had and they may be selling pieces separately or never received all the other components in their own purchase. I've written them for details after talking with someone on the phone, today. Hopefully, I'll get more info tomorrow from them, directly. But the question remains... does one need to be vigilant in buying things like this? Or can one just 'assume' and go buy with confidence that a kit is a kit is a kit most of the time when looking around for pot cores? Do people break these things up and sell the bits and pieces as separate items? Jon J John KD5YI Jan 1, 1970 0 On the last part, about Aw, you earlier mentioned that the "Bottom of page 4 shows the 'former' winding area at around 65mm^2. Multiply this times your Amin." I think you mean this value as Aw. Yes. My Aw is their "Winding Area". (For the reason that if Amin represents Ac [for all intents and purposes], then you must have meant me to understand this 65mm^2 as Aw.) What I see on page 4 is a table of winding areas: Section Winding Area 1 72.4 mm^2 2 2x33.9 mm^2 3 3x21.0 mm^2 No 65mm^2 there. But I take it you meant 65 as a conservative 72.4. But then, with a primary and a secondary involved isn't it the case that I use the 2-section row and plan a primary with 33.9mm^2 and a secondary with the same? Yes, I meant the 65 to be a rough average of the three areas shown above. A 1-section bobbin (former) has a single 72.4 winding section, a 2-section bobbin has two winding sections at 33.9 each and a 3-section bobbin has three at 21 each. These total winding areas are 72.4, 67.8, and 63. The average total is 67.7 mm^2. I just glanced at the numbers and mentally estimated about 65. There are two common approaches to winding a two-winding transformer. The first is to use a single-section bobbin and wind the primary on first, insulate, then wind the secondary over the top of it. The other approach is to wind the primary on one of the two halves of a two-section bobbin and wind the secondary on the other half. There are benefits and penalties with either way (as is always the case, don't you know). Also, I just plain cannot see how to come up with the 72.4 mm^2 they give for the winding area with one section. Let me walk through my thoughts. There is a "former" that is used for winding. It's designed so that the two halves slide their center regions through the hole in its center and wrap around it. The former (bobbin) is one piece. Have you ever seen a sewing machine bobbin? (Keep in mind I have never used one of these, so I'm going from the diagrams and no experience, at all.) Once wound, I just take one half of the core itself, slide the former onto the central pedestal in the middle of the core half, then place the other core half over that and close it up (by a clip?) Except for my comment above, I think that sounds right. I never used the clip. I always ran a nylon screw through the center hole and used a nylon nut to secure the two core halves (with the enclosed bobbin). They designate the outer diameter of the former as 29.6mm (or -.2mm less) and the outer reach of the inner cylinder as 17.9mm (or -.2mm less.) Assuming worst case, I use 29.4mm and 17.9mm -- a difference of 11.5mm. Looking at the diagram on page four for the former, I see a central division (vertically aligned on the page) between the two outer edges, which I take to mean that if I am winding two coils then one goes on one side of that divider and one goes on the other side. In my case, that's the primary and the secondary. Yes? Yes, but let's calculate the single section bobbin first. The winding area is: (14.4-.8-.8)*((29.4-17.9)/2) = 73.6 (close enough) They assume a .8mm division between two coils and thus (14.4-2*0.8)mm divided by 2, for each. This is 6.4mm. Since I figure 11.5mm x 6.4mm for each coil I get 73.6mm^2. But this is just for one side, not the other. The other side should be the same. So although I get a number perilously close to the 72.4mm^2 they give for one section, it's a value I figure for two sections. Which leaves me thinking I don't know what they mean in the table. Unless the 1 means one secondary and the primary is already assumed. Jon Jon, I think you are expecting too much from the numbers, drawing, dimensions, specs. It sounds like you expect everything to match if you just make the right calculation. Let me show you why you should not expect such exactness: From the drawing, the winding width is 14.4-.2-.8-.8 = 12.6 Now look below the drawing at the column labeled "Minimum Winding Width". It says 12.5. Oops! Where did that silly .1 millimeter go? Also, now take an additional .8 from the 12.6 for a two-section bobbin and the winding width of each section should be (12.6-.8)/2 = 5.9 whereas the column says 5.8. Now you are missing .2mm (.1mm per half). Don't try to pin down these numbers, just use the catalog values. Remember that fudge factor I mentioned? You're going to have much more inexactness than .1 mm. For example, I can almost guarantee that you will not get the wire you calculate on the bobbin unless you layer wind it (and maybe not even then). If you scramble wind it, you will have a bulge in the middle and you won't be able to get the bobbin into the core. John J John KD5YI Jan 1, 1970 0 Jon Kirwan said: Separate question. The datasheet for that pot core I mentioned earlier suggests the following to me as a "kit": * core, in two pieces * former, one piece * container, one piece * spring, one piece * tag plate, one piece When someone sells something like this and somewhere on the web page it says "UNIT PRICE PER SET" I tend to imagine that I get _all_ the pieces. Don't imagine anything. One of the core halves is pretty useless, so they come in pairs which some people call a "set". I bet they mean you pay that price for the "set" of two matching core halves. However, I could be wrong. Wait for their reply. John J Jon Kirwan Jan 1, 1970 0 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 Self defeating. Well, I've got it working already with a CR2032 __without__ the transformer. The problem I wanted to wrestle with was the 100' away part. You need at least 24 AWG wire for that distance. What I'd like to do is allow most anything to work. Some people _like_ the idea of having their wire small enough to put into one of those fishing line disks for easy packing around in the pocket. I'd like to support that, if possible. Obviously, if it isn't possible at all, then ... well ... I'll still try to do the impossible. Why are you going so far away? Distance gives you a lot of things. One obvious one is ease of tracking and giving a decent baseline for angle measurements for height, if someone wants to try that. Crowd safety could be another (the ability to _see_ someone approaching and to then have the _time_ to get to them before they reach the rocket site can be a help and clearances aid this.) To be honest, it may be even further. Untested propellants? I've done a LOT of that in my life. Had quite a number of explosions, too, due to things like hairline cracks in the fuel. I can't even remember how many bags of sand I've hid behind! And good thing. What size engines are you talking about? (F or larger?) This would be mostly for casual users (and kids) and this is a hobby project, not a commercial one. Wire is cheap, safety is worth a lot. And I've got to put a lot more thought into details, yet. That will take some practice with this to learn what is good and what is bad about what I start out with. Eventually, I'll get enough of the details right enough that I'll feel comfortable passing it on. Jon P.S. For me? I'd be using picric acid and potassium chlorate and much worse, if I can lay hands on it, and all of it in a steel cylinder with a turned steel engine nozzle I design and make. So for me, lots and lots of distance is a _very_ good thing. But I'm using the flash lamp method for anything I do, not this device. I get precision timing of multiple energy pulses, that way. No current, just a nice 6kV trigger on a wire. J Jon Kirwan Jan 1, 1970 0 <snip> Everyone seems to be thinking voltage transformers when current transformers would be more appropriate. Bigger wire and much lower turns counts. And yes, a 50 to 75 cycle 1 kHz pulse would solve many problems. One of the FIRST things I did was to look up current transformers. Not because I know what I'm doing -- it's plain enough I don't -- but because of the word "pulse" that often is included with the phrase "current transformer." I was looking for pulse stuff and sure enough that came up right away. However, I quickly realized from reading: http://en.wikipedia.org/wiki/Transformer_types#Pulse_transformers That they were designed for something else and, due to my own lack of knowledge (ignorance), I also knew I wasn't competent to decide if they would fit the application. So I dropped the thought and went "back to the books," again. Yes. I'm really beginning to see those volt-second Webery things in every waking moment of my life, now. Damnable things. I am learning to like __micro__-volt-second stuff. Tiny bits of volt-seconds are cool. Mega-volt-seconds are uncool. Jon J Jon Kirwan Jan 1, 1970 0 Jon Kirwan wrote: On Sat, 16 May 2009 15:43:27 -0700, John Larkin 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 I only skimmed it. I remember your initial calc's for the power pulse core with Yikes! Alternative approach? Yeah. I'm getting a small clue, gradually. Since I figured on using a micro anyway, chopping is a doable approach. More parts, more expense, etc. But one cannot argue with results, either. I'm still going to give a shot to the massive P36/22 pot core (54 grams!) and see what I can dish out with that bastard. But then I expect I'm going to go back to the H-bridge choppy way of doing it and use a tiny little bump of a transformer on the other end. The universe can be cruel. I want a free lunch, once in a while. Jon J Jon Kirwan Jan 1, 1970 0 Don't imagine anything. One of the core halves is pretty useless, so they come in pairs which some people call a "set". I bet they mean you pay that price for the "set" of two matching core halves. However, I could be wrong. Wait for their reply. Okay, thanks. I liked your idea of a nylon bolt and nut. I can go there.2 for 54 grams of mass is pretty good, I figure, so I won't
look a gift horse in the mouth. And thanks for the time.

Jon

J

Jon Kirwan

Jan 1, 1970
0
...

I have the impression you are thinking of the reflected impedance as
being in series with the primary, rather than in parallel, as model at
(eg) <http://www.butlerwinding.com/elelectronic-transformer/current/>
(near end of page) shows.

Yes, I was thinking that and it doesn't makes sense what I was
thinking. Your point _does_ make sense to me. I thought that way
because a diagram I saw __replaced__ the primary with the reflected
impedance. I had considered the thought of it being in parallel, but
because of the diagram I saw I set my own thoughts aside. I had since
come to the conclusion that had to be in parallel, reflecting on the
quandary posed by my own hypotheticals and the very real fact of
others' experience, resolving that was the only choice that answered
both things simultaneously and figured on showing it to myself, later
on.
...

I have the impression you want to transfer most of the energy in a
large initial pulse. Couldn't you use a small xfmr, ringing at a
few kHz, to transfer the energy?

Yes. And I don't mean to say I fully apprehend all the details, but
the problem I see here is that if the primary is "ringing" around a
lot, I'm going to lose a lot of energy to the wiring out to the
primary. I liked the idea of ringing because it can be used to cut
down on the volt-seconds (so I imagine, anyway) requirement... but I
expect to lose out to wiring losses, then.

(Regarding the volt seconds: the first positive part wouldn't be
entirely compensated by the next negative going part, but that would
be more than the following positive going part... etc., but it's
better overall [I think] than going for the critically damped case
where the hog's share is all on the first part.)

Ringing may be an approach with higher primary inductance (lower
currents?) Now that you are making me think, the key is to minimize
this:

Integral from 0 to 50ms of: (Vc(t)-V_primaryinductance(t))^2/R*dt

(Vr(t) of the wiring is the difference: Vc(t)-V_primaryinductance(t).)

I'll think a little on that.

Jon

B

Baron

Jan 1, 1970
0
Jon said:
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

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

Same question re-worded.

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