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.)
In the case of a toroid, I assume Aw has to be computed from N*d^2, then?
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 (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.)
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
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.
I get .0294cm or .000294.
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.)
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.]
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?
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.
(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.
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.
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.
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.
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.
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).
You're probably right. I wasn't trying to actually come up with a solution, just give an example.
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).
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.
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.
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.
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.
And the primary voltage goes to zero since no more energy can be stored in the field, as I gather things.
I'm beginning to see things better. Thanks.
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.
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.
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.
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.
Interesting looking for pot cores on the web. I found one with Ur=3000, a 36x22 pot core, for $2/ea.
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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:
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:
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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.
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.
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.
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.
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.
Now go back and recalculate your required Ac*Aw product based on this much lower flux density.
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.
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?
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:
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.
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?
Self defeating. You need at least 24 AWG wire for that distance. Why are you going so far away? Untested propellants? What size engines are you talking about? (F or larger?)
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).
The former (bobbin) is one piece. Have you ever seen a sewing machine bobbin?
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).
Yes, but let's calculate the single section bobbin first. The winding area is:
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.
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