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.

Yes, assuming (and I still need to think about this) the reflected

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

amount of secondary load dissipation in the first PI/2 quadrant (at

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