Re: magnetics design -- 60mJ energy impedance matching

"Jon Kirwan" <jonk@xxxxxxxxxxxxxxxxxxx> wrote in message news:dh7815d5eed3mv66s55hsssqtrbff7mo3n@xxxxxxxxxx

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

Yes, as follows:

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).

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

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).



Referring to your earlier schematic:

: R2 T1
: ,--------~~~~~~~--/\/\--, ,-------,
: | 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.

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
(N1/N2)^2, as you'll talk about here. And that capacitive loads
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).


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.


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

.

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