Optimizing the Design of a Magnetic Circuit Actuator

You are using a sub-optimal core configuration. Use a C-core instead of an E-core and not waste flux in the lower loop that adds very little force to the armature. Interleave the laminations of the lower end of the C-core and the pivot point of the armature to reduce the magnetic gap at the bottom.

A magnetic circuit is analogous to an electric circuit. The driving force is ampere turns instead of voltage. Air gap is analogous to resistance. (Iron has a very low resistance to flux, air a very high resistance). The flux will take the path of least resistance, which is with the least air gap.

There are actually two fields, the B field and the H field. B=μH, where μ is the permeability. The sum (integral) of the H field around the loop is proportional to ampere turns. The B field is the same all the way around. So the result is that the H field (H=B/μ) is much, much smaller in the iron than in the air gap.

To calculate the field from ampere turns, you can assume that the H field is all in the air gap (like the voltage drop across resistance). Once you calculate the H field in the air gap (ampere turns/length), you can calculate the B field.

Since the sum of the H field around the loop is proportional to ampere-turns, the smaller the air gap, the stronger the H field, and likewise the stronger the B field.

For your reading pleasure:

http://www.tdsjp.co.jp/english/techno/tokusei.html

"The objective is to get the highest force/weight ratio"

That being the case, without regard for longevity, maximum current it can withstand to pull in completely before destruction.

How did you decide the length of your windings before you chose the number of laminations?

Seems like your first 70 turns will be about 50mm each in length. The next 70 about 51 mm each....and so on.

Do you have control over where the windings are placed?

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Making your gap even smaller will be beneficial.

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The most area per length will be enclosed by the windings when your depth is equal to the center width.....though since this is force per weight, I think fewer laminations might actually perform better since for the same resistance, you get more turns.

BTW, if you are going to have a representative illustration which needed a straight edge to construct, why not make it proportional?

The winding's are wrapped around a square bobbin of perimeter 74mm. So the length of the wire is equal to Number of turns*74

You're going to need a layer of some kind of insulation between the core and the first layer of wire, to avoid having the corners of the core cut through the wire enamel, and to support the winding while it is being transferred from the winding machine to the core. That will add at least a couple of mm to the length of each turn of the first layer. Each successive layer will be longer by roughly 4 times the wire diameter plus 4 times the corner arc length (which is greater for each successive layer), assuming fully-packed uniform layers.

Don't forget to leave room for the insulating end caps for the bobbin.

Dkwarner is right. You are going to end up with far fewer that 775 turns. Even the first layer of turns on the 74mm perimeter bobbin will require more along the lines of 74.75 mm of wire. Not a huge difference but it adds up quickly.

The space in which the bobbin must fit, end wise, is 24mm. If your wire is 0.32 mm, this means around 70 turns to a layer . You might assume each successive layer requires around 1.5mm more per turn.

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You will get close to 700 turns estimating that way.

Okay, so your winding are on a fixed size bobbin and contribute significantly to the total mass, right? With several hundred ampere turns in the inch available, clearly silicon steel will be in saturation.

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I suspect you will find that given the fixed size of the bobbin on which to wind and therefore the non scaling mass contributed per number of turns of specific wire, that the maximum depth, 16mm of laminations, yielding a square center cross section will yield the greatest force per mass at a given gap size.

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If the windings are massless, it might be a good time to ask what other pertinent attributes of the world of this problem differ meaningfully from those of our real-world reality. A lot of assumptions have been made about the laws governing the problem being similar to those expected in real life.

As somebody already said :

The perimeter of the bobbin may be 74mm, n*74 is valid for the first layer only. The second layer will sit on the first, so the effective perimeter for that layer will be more. So, you need to find out the perimeter of the mean turn, which is 74+half the thickness of the entire winding.

How did you arrive at '...+ half the thickness of the entire winding...'?

It is the mean turn. So half the turns are smaller and the other half are longer than that one. So, we multiply the total number of turns by the length of the mean turn to get the total length of the copper wire in the coil.

"Half the thickness" is more appropriate for a cylindrical coil. For a rectangular coil one needs to add all the four sides as shown in dkwarner's post.

i request you to take a look at that outstanding drawing of dkwarner, which should make this clearer.

Yeah, I get multiplying the number of turns by the mean length of a turn idea, even if that is somewhat putting the whores before the cart, since the number of turns is dependent on how quickly the increasing perimeter uses up the fixed length of conductor.

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By the way, I did look at your excellent illustration and dkwarners. I still do not see, even if the bobbin were round instead of square, how adding 1/2 the total thickness of the stack to the length of a turn in the first layer would equate to the average turn length. You are adding 1/2 the increase in radius to a circumference and trying to obtain an average circumference from that.....

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.....no sir, I don't like it.

Since I am in India, English (UK) is, at best, my second language, and I may have made some faux pas in the 'half-thickness' thing. The OP says that the perimeter of the bobbin is 74mm. So the first layer will have n1*74mm as length of wire in it, n1 being the number of turns in the first layer.. The next layer will sit on this so, its perimeter will be more by two thicknesses of the wire. The next layer will have a still longer perimeter and so on. Let us say, the coil will have a thickness of 12 mm all around. The perimeter of the mean turn will be 74+0.5*12*4 = 98mm..the 4 being the four sides of a rectangular coil. This 98mm, multiplied by the total number of turns, will give the total length of copper wire in the coil, which, when multiplied by the resistivity, will give the total DC resistance of the coil.

Is the amount of copper wire limited ? I assumed it is not. So, there are no mix-up of carts and whatever, IMHO.

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Oops. i posted when doing something else, so my calculation was pathetic. You already have calculated, so i won't belabour this. To each his own

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Yes, there is a limit to the amount of copper. It must fit in an 8mm gap between the legs of the "E". Since he has already used 1.25mm of that gap for the bobbin, that leaves 6.75mm maximum for wire. I used the 0.32mm wire diameter that someone else mentioned, without going back to check the original posts... Theoretically, that would allow 21 layers, but being realistic, I left some insertion room, so 20 layers.

Is 0.32mm just the conductor diameter or conductor and insulation thickness? Also what will be the insulation breakdown voltage?

Ah yes of course. I should have looked at your scale drawing where you have rightly mentioned this very point.

Your worksheet gives what I was trying to say. On a coil thickness of 6.4mm, the mean turn is something between #10 and #11 (not possible physically of course, but a mean turn could be hypothetical), which is half the coil thickness, 3.11mm.

Thank you.

Correct, except that I always thought half of 6.4 was 3.2.

(Off topic) Reiterating what someone else said, your English is excellent! In fact it is not at all rare to find others whose first language is English, but their written use of it is deplorable (much harder to understand than yours).

I'd love to hear you speak, since I usually have great difficulty understanding the English spoken by people from India. It has gotten to the point that, if I get connected to a "help Desk" with such a person, I just hang up, because I know I won't get the technical help I need.

3.2 = approximately 3.11, whatever keeps me making such bloomers ! Sorry.

Well, on the question of pronunciation, i am told that i am easily understood by Americans, but also that i am obviously a non-US person. On my first visit to the USA in 1982, my guide at Cutler-Hammer. Milwaukee told me just to slow down my speech, and i would be understood. Apparently we tend to speak fast (though our government is one of the slowest in the world) and accent a different syllable when compared to an American or an Englishman (who think they have a common language ). i am disappointed that the Indian call-centers are not training their people in US and other accents. They used to be indistinguishable from the genuine article when they started. i will try to spread the word around.

There is also the maximum current limited to 1 amp requirement that will constrain the length in this situation. It will end up around ten layers of turns if well packed and close to the full width is used.

Your English is excellent. So good in fact, that probably shouldn't offer it up as a way to explain a disagreement. It might come across as disengenuous, worded so well.

I'm glad we seem to be on the same page about the winding length calculation, now.

Thank you. Among the many things I have learnt on CR4, some proficiency in English is also there. Regional nuances are difficult to get rid off, so sometimes there are communication disconnects. Sorry.

This thread has been very informative for me. I simply used F=B²*A/2μ₀ when designing EI magnets for contactors (Cutler-Hammer, ABB). Now they use sophisticated FEA programs like CEDRAT, ANSYS, AOS MAGNUM ... i wish i was young enough to work on those ....

I had to check. here is a spreadsheet showing the lengths of all the layers for solid packed winding:

The 20 layers are 6.4mm thick. I'll leave some calculation to others...

So at 12 volts, and assuming 70 turns per layer, resistance should bring current below the 1 amp limit somewhere around the tenth layer.

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The response to this question throws a wrench into the usual 'we don't do homework' response. Apparently we do.

It is amazing the willingness to help that follows a detailed well layer out question with some indication of attempts so far and sticking points.

If he uses the same thickness for the ends of the bobbin as the center, I only see 67 turns fitting per layer. On the other hand, there is almost room for one more turn, which means that if carefully wound, it seems like it might be possible to have the turns interleaved, but since each layer must have the opposite slope to its neighbors, I suspect that is not practical.

"The response to this question throws a wrench into the usual 'we don't do homework' response. Apparently we do." Not really! No one has provided an answer to the objective. We have hopefully provided some guidance, but have not provided a full solution.

"It is amazing the willingness to help that follows a detailed well [laid] out question with some indication of attempts so far and sticking points." Absolutely! Original effort makes all the difference in the world!

I wonder if any production windings either:

loop once through the bobbin for each successive layer, creating a layer of passes under, to layers of all the same slope that can be interleaved;

or cuts the wire at the end of each layer, interleave half the layers in one slope, then the other half in the opposite slope, then solders the appropriate ends for one continuous winding...?

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Seems like even a reduction in total coil thickness of a few percent or similar reduction in conductor length (mass) for a given number of turns might be worth the additional effort for projects where mass/volume need to be as low as possible.

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On second thought, square or rectangular cross section wire would be the better solution.

We may not have provided an exact specific solution, but the guidance is sufficiently explicit and points unambiguously at a specific solution.

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Would be nice if the OP posted the evaluation of the assignment here.

By the way, dkwarners post is even better if you read the text as well as kook at the pictures. It describes NOT adding the four sides, since the straight sections remain constant. Instead the corner radii are what must be added.

Pi x total thickness of the layers is what needs to be added to the length of one of the turns in the first layer to obtain the mean turn length.

....of course that also has some assumptions ingrained. One of those assumptions is that all layers have the same number of turn or at least that the variation is symmetrical about the center of the layers. Seeing as how the parameter is constrained by resistance and not the number of turns, it may be difficult to have it come out with a full last layer of turns.

There is no simple way to do this correctly....my advice is to get some software programs for analysis...or do some extensive study, maybe both...

https://www.youtube.com/watch?v=ENDDw-ZCCgA

I disagree. This an elegant problem, that is capable of "hand" calculation requiring no simulation software once one understands all of the critical parameters. (I also suspect the teacher crafting this problem does not grasp all of the nuances I see, but...)

As most of you already know, I will guide but not solve a homework problem. However an elegant problem that will reveals exactly the level of command, on all relevant topics, a student grasps I will "fill in the blanks" as I see them and generate a few more along the way.

What is missing in this analysis is the kinematics required to close the gaps in this spinning magnetic circuit. The amount of torque each half magnetic circuit produces on the rotating bar has not been identified. Often a numeric value for moment of inertia is not needed but that it fits into certain differential equations does matter. IMHO It is best to disregard the orientation of gravity but this can be added later on.

There is one nebulous criteria that bothers me in this problem. What is considered optimal? The least amount of energy to maintain a closed position? Minimizing the speed of closing this solenoid? The minimal amount of coil current to move this mass if motion is orthogonal to gravity?

Generating the differential kinematic equations will allow a student to demonstrate their methodology to solve a fuzzy problem.

Good Luck.

There's spinning? What's spinning? I don't see any spinning....

I would say optimal is greatest force that can be generated within the specified parameters.....

....greatest force per weight within given parameters

Weight or mass? Can we use superconductor at low temp?

OP states 'highest force/weight' in comment #4. I suppose free fall would tilt the scales in favor.

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As far a superconductors, even the so called high temperature variety are used at temperatures I consider pretty low. If can use them at more comfortable temperatures, you won't get any resistance from me.

As I mentioned In #6, expected lifetime for this device is not given.

You're right, spin was a poor choice of words. The original diagram implied a hinge point at the bottom of the diagram. So I guess swing or pivot should've been used.

I agree that the original diagram implied a hinge point, both by the angle of the movable core, and by the dot near the bottom.

If there is a fixed hinge/swivel point, it needs to be centered at or below the bottom edge of the E core, in order for the movable core to close all of the gaps and yet be able to open.

If the dot indicates the center of the hinge/swivel point, then that hinge needs to be spring mounted, or perhaps be a spring, in order for the movable core to open.

To get the force output for the actuator, this is the equation

F = -1/2((N*I)/R)^2 * dR/dx

Where dR/dx = change in reluctance with respect to air gap width

I don't know anything about the orientation due to gravity and stuff like that because I think it's a bit more advanced for this course. I am just a 2nd year engineering student and we haven't learnt how to use software and simulations so I have to hand calculate it

The aim is just to get the maximum force/weight ratio within the constraints listed above

"Coil- the copper windings on the solenoid that provide an electrical element through which a current is passed to generate a magnetic field. During the winding process, precision wound coil follows a prescribed pattern in which each turn is laid precisely beside the previous turn. This allows the maximum amount of copper to be wound in the allotted space. A coil with no specific winding pattern is called a random wound coil."

I thought we didn't do HOMEWORK on this site?

An excellent question, and not at all off topic! I do NOT know the answer for sure, but over the last 60 years, I have seen literally thousands of solenoids, relays, and transformers. I have seen a very few transformers that had windings on more than one magnetic leg of the core. I believe those i have seen (and indeed have on hand) were made for military applications where space was more important than cost. I have never (as far as I recall) seen a solenoid or relay with such windings.

My interpretation: if there is an advantage for multiple windings (which does sound reasonable), it is not worth the cost.

You have stated that you are a second year student. Generally optimization is included in final year. First you should understand what is optimization. Optimization can be obtained in many ways. First you should be able to define optimization with respect to which parameter? It could be with respect to cost, weight, volume so on and so forth. You may have to fix the type of cooling (off course it will be air at your level). If you change the cooling (make it better) then you can put much thinner wire and get more number of turns within the same space but with much higher heat generation. So based on choice of optimization criteria, you should define independent and dependent variables. Then define the constraints ( you have done this part to quite a good extant). Then you have to formulate equations wherein convert all dependent variables into independent variables. Then apply mathematics i.e. differentiating the desired (force in this case) with respect to single independent variable. Equate the result to zero to get the value of independent variable which will give you maximum force. Since there are so many parameters involved (some have non linear characteristic (magnet path which does not have linear B-H relationship), true optimization would not be possible with out writing suitable software. You may have to input magnetic properties and then generate logics suitable. I do not know where you are located, text books on optimization should be available wherever electrical engineering courses are included in institutions.

It's read EI. Quite an unusual design/geometric by the way. Say if I am an investor, what make you think your design is advantage over the usual actuator configuration?