End turns bad?

Buk___

10 kW
Joined
Jul 28, 2017
Messages
750
First. I know this is (probably) wrong, but I want to understand why, so please bear with me :)

The conventional wisdom is that the end turns on a stator are dead weight. A couple of the explanations for this I've seen go:

1) They are outside of the width of the magnets, so do not contribute torque, but have resistance, so still consume power.

2) As they run radially, their current does not cut the magnets fields in a way that contributes (Lorentz) force to the motor's torque, but have res...

3) Others...


But ... it is the field in the teeth that interacts with the magnets, and the windings create the field in the teeth; and if you think of a tooth as a squared off solenoid, each full turn, and all of the length of each turn, is used to calculate the field created by the solenoid.

By that reckoning, end turns contribute their proportion to the tooth's internal field, and are just as valuable and productive as those parts of their coils that run through the slots.

[strike]Discuss?[/strike] By all means, tell me I'm wrong; but please include why :)
 
I'm not certain it's relevant to this specific question, but:

AIUI, a wider stator, with a lower proportion of end turns to tooth-parallel winding, gives more torque for the same power usage than a narrower one, with a higher proportion of end turns. If correct, I think that implies that while end turns aren't "useless", they don't contribute as much as the parallel portion. (This could be incorrect because anohter part of the difference is also more magnet area, and I don't know whihc part makes the difference in torque).
 
You mention Lorentz force. Examine that relationship. See it is a vector product, cross product, or orthogonal. The force is perpendicular to both the magnetic field and charge velocity (or current).

Another clue is Ampere's Law. The contributing current to the magnetic flux in a core is that which passes through the core. Current passing around the outside of the core does not contribute magnetically but only serves to deliver current to the active portion.

You should examine the magnetic circuit. And realize that ampere-"turns" is just a convenience term and the operative quantity is current or ampere-conductors. An ampere-turn being equal to 2 * ampere-conductors. End turns and connection leads are just necessary support for the coil sides occupying space in the air gap.

Regards,

major
 
Dealing with your reply one bit at a time.

major said:
You mention Lorentz force. Examine that relationship. See it is a vector product, cross product, or orthogonal. The force is perpendicular to both the magnetic field and charge velocity (or current).

I'm not sure if that is in agreement or disagreement with the Lorentz Force argument?

(If the torque of the motor was down to the current in the "through-core" portions of the windings interacting with the field of the permanent magnets, then the Lorentz Force due current flow in the half-turn on one side of the tooth, would exactly cancel the force due to the current flowing through the half-turn on the other side of the tooth; for a net zero torque. Except half turn motors and airgap coil motors.

And that's ignoring the 1/r^3 drop-off in the field strength as you move away from the PMs.)

And if that is what you were saying above, you could be clearer.
 
Buk___ said:
Dealing with your reply one bit at a time.

major said:
You mention Lorentz force. Examine that relationship. See it is a vector product, cross product, or orthogonal. The force is perpendicular to both the magnetic field and charge velocity (or current).

I'm not sure if that is in agreement or disagreement with the Lorentz Force argument?
...

Hmm. I thought I was just repeating Lorentz force definition. If you need clearer wording I suggest reading more about it. There are many text books and articles on the web. Perhaps not the best, but easy to find, is https://en.m.wikipedia.org/wiki/Lorentz_force. See section "Force on a current-carrying wire".

My intent is to point you in directions where you can learn about the subject.

And if the coil in the motor has equal flux, in the same direction, cutting both coil sides, then the force from that coil is zero. So by virtue of the commutation and rotor position, and number of slots and poles (magnets), the flux is seldom equal (in the same direction) on both coil sides.

Regards,

major
 
major said:
Another clue is Ampere's Law. The contributing current to the magnetic flux in a core is that which passes through the core. Current passing around the outside of the core does not contribute magnetically but only serves to deliver current to the active portion.

Hm. That is the misunderstanding in a nutshell.

In a cylindrical air-cored solenoid, the individual fields around the turns of the conductor, combine to produce a larger classic "apple-core" 2D cross section field:[img=https://www.universetoday.com/wp-content/uploads/2009/09/ironcore_solenoid-580x441.jpg][/img]
The iron core serves to intensify that field.

In 3d, that field is onion-skin like, as in the classic images of the Earth's magnetic field:
Earth_Magnetic_Field.png



Note that the fields demonstrate rotational symmetry around the core axis; and all parts of the coils contribute equally to the overall field. There are no sides. Nor does any part of the coil pass through the core, only around.

A BLDC stator tooth, is simply a cylindrical solenoid squashed square or rectangular:
th


All parts -- including all 4 sides -- contribute equally to the combined field the tooth (and its contained core) generate. End turns and all.

It is the field lines emanating from the head of the tooth that interacts with the magnets to produce torque, not the copper conductors, or the individual fields they generate around themselves.
p14654_04_obr04.png


Note how the coils (black-lined quadrilaterals in pale blue) contain almost no magnetic field; it is concentrated within the soft-iron core with its very high permeability relative to the copper {and air) where the coils are.

Bottom line: end turns contribute just as much field strength per unit length as the cross-stator (through-slot) lengths of the coils; to the tooth-core field.

When I first posted this thread, I was trying to do some calculations involving the coils, and kept seeing people say that you should exclude the end-turn copper from field/torque calculations; but it didn't ring true to me, nor match experimental data.

Since then I've done a lot of reading, and reached my conclusion, that despite being wide-spread and strongly believed, I consider it -- the notion that end-turns are dead weight -- wrong.
 
Buk___ said:
First. I know this is (probably) wrong, but I want to understand why, so please bear with me :)
...
[strike]Discuss?[/strike] By all means, tell me I'm wrong; but please include why :)

OK. I told you that you're wrong and why. You choose not to believe me. OK. Good luck.

BTW, I used the term "core" when addressing Ampere's Law because of the motor context. It is actually the magnetic path. So in your solenoid examples the magnetic path completely surrounds the coil, therefore there are no end turns. In the motor FEMM you see only a cross section. Only leakage flux will encompass end turns which, in a decent design, will be very little.

Regards,

major
 
major said:
Hmm. I thought I was just repeating Lorentz force definition. If you need clearer wording I suggest reading more about it. There are many text books and articles on the web. Perhaps not the best, but easy to find, is https://en.m.wikipedia.org/wiki/Lorentz_force. See section "Force on a current-carrying wire".

My intent is to point you in directions where you can learn about the subject.

Hm. I understand Lorentz Force pretty well.

Perhaps you misunderstood my first post when I said "explanations for this I've seen "; and though I was proffering that explanation, rather than just citing it?

(I'm not being rude, just trying to understand what you are telling me.) I cannot make sense of this next para:
major said:
And if the coil in the motor has equal flux, in the same direction, cutting both coil sides, then the force from that coil is zero.

"the coil in the motor": Which coil? There are many coils.
"has equal flux": Equal to what?
"in the same direction": Same direction as what?

major said:
the flux is seldom equal (in the same direction) on both coil sides.

I know it will look like I am being deliberately obtuse, but I'm not.

Which flux?

If you mean the flux from the magnets, then it has almost no affect on the coils. The field from the magnet drops off 1/r[sup]3[/sup]. So, in a typical small geared hub stator, the field from an N43 magnets with 1.35T field at the pole face, has dropped to 0.293T by the time you get to the other side of a 0.5mm airgap. If the teeth have a 7mm head thickness before you get to the first loop of the coil, and the wire is 24awg, then that first coil only experiences only 0.09688T from the magnet and the second turn away from the head only 0.08991T. And it gets less faster thereafter.

If the flux you are talking about it that generated by the same coil as you are discussing, then it is, by definition equal both sides.

If you mean the overall flux density within the tooth that coil surrounds; that is uneven and changing it blends or fights with the magnets in proximity; but the flux lines barely impinge on the coil at all. See the FEA plot above. The magnet's flux impinges the coils a little as it passes by the slot mouths, but not enough to have any great influence on anything.

Sorry. I'm rambling. Thinking out loud really, trying to understand what it is you are telling me.
 
major said:
Buk___ said:
First. I know this is (probably) wrong, but I want to understand why, so please bear with me :)
...
[strike]Discuss?[/strike] By all means, tell me I'm wrong; but please include why :)

OK. I told you that you're wrong and why. You choose not to believe me.

Hm. It''d be hard to "believe you", when it was (and is) unclear that you were telling me I was wrong; much less any explanation as to why I'm wrong.

You misquote wikipedia;then suggest I read it; and then conclude I'm wrong because...

Oh well. Time to shut up and take my own council it seems. No point in constructing clear, logical, well documented discussion points; if this is the response.
 
Buk___ said:
You misquote wikipedia;then suggest I read it; and then conclude I'm wrong because...

I didn't misquote wikipedia. The only reason I went there was to get a further resource for you so you could possibly understand what I was talking about.
 
Buk___ said:
major said:
And if the coil in the motor has equal flux, in the same direction, cutting both coil sides, then the force from that coil is zero.

"the coil in the motor": Which coil? There are many coils.

Any coil.

Buk___ said:
major said:
And if the coil in the motor has equal flux, in the same direction, cutting both coil sides, then the force from that coil is zero.

"has equal flux": Equal to what?

Equal flux in the same direction cutting both coil sides. Coil side #1 has the same magnitude and direction of flux as side #2. A coil in the motor has two active coil sides which are the conductors parallel to the axis of rotation.


Buk___ said:
"in the same direction": Same direction as what?

In the same direction of each other.
 
major said:
Buk___ said:
If you mean the flux from the magnets, then it has almost no affect on the coils.

I wonder where the generated voltage comes from.

If you replaced the soft iron core with a (say) plastic former, would much voltage be induced in the coils by rotating the magnets 5mm or 7mm or 10mm away from them? (That's the top coil(s); the rest are successively further away.)

Given that the field strength that far from the magnets is (typically) <7% of the field strength at the pole face.

Without the core in place to guide the flux down past the coils, the magnets barely affect them. (Directly.)

The magnets induce a field/flow across the airgap, through the tooth, and around through the adjacent magnets and backiron, back to its own back-face.

It is the field flowing through the tooth in close proximity to the coil turns that induces the bemf.
 
major said:
Buk___ said:
major said:
And if the coil in the motor has equal flux, in the same direction, cutting both coil sides, then the force from that coil is zero.

"the coil in the motor": Which coil? There are many coils.

Any coil.

Buk___ said:
major said:
And if the coil in the motor has equal flux, in the same direction, cutting both coil sides, then the force from that coil is zero.

"has equal flux": Equal to what?

Equal flux in the same direction cutting both coil sides. Coil side #1 has the same magnitude and direction of flux as side #2. A coil in the motor has two active coil sides which are the conductors parallel to the axis of rotation.


Buk___ said:
"in the same direction": Same direction as what?

In the same direction of each other.
Sorry again. I read your words over and over; but I cannot visualise what you are trying to describe. Maybe a picture or two?
 
Buk___ said:
major said:
Buk___ said:
major said:
And if the coil in the motor has equal flux, in the same direction, cutting both coil sides, then the force from that coil is zero.

"the coil in the motor": Which coil? There are many coils.

Any coil.

Buk___ said:
major said:
And if the coil in the motor has equal flux, in the same direction, cutting both coil sides, then the force from that coil is zero.

"has equal flux": Equal to what?

Equal flux in the same direction cutting both coil sides. Coil side #1 has the same magnitude and direction of flux as side #2. A coil in the motor has two active coil sides which are the conductors parallel to the axis of rotation.


Buk___ said:
"in the same direction": Same direction as what?

In the same direction of each other.
Sorry again. I read your words over and over; but I cannot visualise what you are trying to describe. Maybe a picture or two?

Here you go. Imagine the motor shaft being way off to the right.

Motor_coils_corrected.jpg
 
Buk___ said:
major said:
Buk___ said:
If you mean the flux from the magnets, then it has almost no affect on the coils.

I wonder where the generated voltage comes from.

If you replaced the soft iron core with a (say) plastic former, would much voltage be induced in the coils by rotating the magnets 5mm or 7mm or 10mm away from them? (That's the top coil(s); the rest are successively further away.)

Given that the field strength that far from the magnets is (typically) <7% of the field strength at the pole face.

Without the core in place to guide the flux down past the coils, the magnets barely affect them. (Directly.)

The magnets induce a field/flow across the airgap, through the tooth, and around through the adjacent magnets and backiron, back to its own back-face.

It is the field flowing through the tooth in close proximity to the coil turns that induces the bemf.

So you say that the magnet flux has almost no effect on the coil regarding torque production but has full effect on the coil regarding generation of voltage. You are mistaken.

You've developed some strange ideas about motor theory. I suggest you let go of those and seek education. It appears that task is more than I can provide. Perhaps other members can help you. Go back to the basics.

Good luck,

major
 
major said:
So you say that the magnet flux has almost no effect on the coil regarding torque production but has full effect on the coil regarding generation of voltage.

That is not what I said. Not even vaguely. I restate it.

The magnet's flux has very little direct affect on the coils. For torque product or (re)generation. They -- the coils -- are simply too far away from the magnets, and mostly wrapped around by highly permeable core iron, to feel much of an effect.

Ie. Using
BlockMagnetFluxDensityFallOff.jpg

And a 21.25 x 1.75 x 2.85 Q100 magnet spec., and assuming N45 1.35T, an air gap of 0.5mm, 0.511mm (24awg) wire, and a 7mm "hammerhead" on the tooth; the centre of the first coil is 7.2555mm away.

Update: corrected small math error. Instead of 7.7% for the first coil, its down to 4.3%.

Plug the numbers in and you get: Br:1.35 z:7.2555 => B:0.0589403517046224 / 1.35 * 100 = 4.3659519781201777777777777777778%

These are the number for the face of the tooth (just this side of the 0.5mm airgap, and then the first 10 turns going away from airgap:
Code:
Br:1.35 z:0.5 => B:0.269963211293125
Br:1.35 z:7.2555 => B:0.0589403517046224
Br:1.35 z:7.7555 => B:0.0532897132616767
Br:1.35 z:8.2555 => B:0.0483009356367416
Br:1.35 z:8.7555 => B:0.0438835830102462
Br:1.35 z:9.2555 => B:0.039961122181606
Br:1.35 z:9.7555 => B:0.0364686433155437
Br:1.35 z:10.2555 => B:0.0333509348247649
Br:1.35 z:10.7555 => B:0.0305608805520384
Br:1.35 z:11.2555 => B:0.02805813507232
Br:1.35 z:11.7555 => B:0.0258080320853783

That tenth turn away would experience less than 2% of the magnet's (direct) flux -- if it weren't shielded by the tooth's head.

The flux that affects the coils, is that running round the magnetic circuit. Ie. through the tooth core.

Torque is generated by the flux lines flowing out of the tooth face interacting with the magnets.

Lorentz force plays little or no part.

(Except in coreless/airgap motors like axial flux and thingap; and high-school science toys like the wire loop brushed motor you pictured above.

major said:
You are mistaken.

So you keep saying, but still offering little by way of cogent explanation for your position.

major said:
Go back to the basics

I did.
 
Guys got me confused now....


But I understand the original question and have asked it myself.

First off, for the purposes of this discussion, assume an iron core. In the case of a coreless motor, it is intuitive that the end turns won't contribute to torque.

In the case of an iron core, the flux is a function of amp-turns. It doesn't matter whether the copper is in the slot or on the end, it just needs to loop around the core to make a turn. So in that sense, the end turns are contributing to the flux. There's not many ways I can think of that you could make a turn around an iron core without some copper in the ends. If the copper goes over to the next tooth instead of around, then you have less than a turn and the flux will be less.
 
Voltage only induces in the copper wires which are parallel to the rotation axis. In the endturns there happens nothing in terms of that.
If we think about laminations it makes sense. They are AXIALLY stacked to avoid eddy currents, which tells that voltage is just induced axially and not radially or the direction of the endturns.

So much about back emf Voltage.
In terms of torque, i believe they also do nothing, but i am not sure.
 
Major is undoubtedly correct, but I guess the reason why is not intuitive and you need a understanding the underlying principles of how the motor functions from first principles.

I think the analogy of a solenoid is misleading.
 
Punx0r said:
Major is undoubtedly correct, but I guess the reason why is not intuitive and you need a understanding the underlying principles of how the motor functions from first principles.

Apart from siding with/showing support for major, you don't add anything.

No explanation of why you think he is correct. Or what parts of my supporting arguments for my opinion are wrong?

Punx0r said:
I think the analogy of a solenoid is misleading.

In what way?

Is the coil wrapped around a tooth not a solenoid?

Why would it -- a coil around a tooth -- behave differently to the (well-analysed and documented) basis of all induction-based electro-magnetic machines?
 
madin88 said:
Voltage only induces in the copper wires which are parallel to the rotation axis. In the endturns there happens nothing in terms of that.

Have you ever seen the classic demonstration of oscillating a bar magnet in and out of an air coil to generate electricity? Essentially, one of these:
Linear_induction_flashlight.jpg


Square off the coil, and magnets; still functions the same.

Now put a soft iron core inside the coil and move the magnet towards and away from the end of that core. The field in the core changes, and that indices current in the coils. That's a tooth (minus the back iron and adjacent teeth to complete the circuit and improve the flow).
 
Buk___ said:
Punx0r said:
Major is undoubtedly correct, but I guess the reason why is not intuitive and you need a understanding the underlying principles of how the motor functions from first principles.

Apart from siding with/showing support for major, you don't add anything.

No explanation of why you think he is correct. Or what parts of my supporting arguments for my opinion are wrong?

Punx0r said:
I think the analogy of a solenoid is misleading.

In what way?

Is the coil wrapped around a tooth not a solenoid?

Why would it -- a coil around a tooth -- behave differently to the (well-analysed and documented) basis of all induction-based electro-magnetic machines?

Major is correct because he succinctly explained that end turns do not produce stator flux that is orthogonal to the rotor flux and therefore do not contribute anything towards torque production. The solenoid analogy is misleading because the act of wrapping a turn around an iron core does not magically turn the flux lines generated by the end turns through 90 degrees to align with the flux lines of the two sections of the turn that are in fact orthogonal to the rotor flux.
 
Wow what a lot of assuming going on here! Anyone ever get a magnetic field indicating plastic sheet from kjmagnetics and LOOK at the magnetic field outside the end turns? Off the top of my head I would say the core side of the end turn gets into the core so it adds (magnetic flux lines radiate from the core outward and couple the flux to the pole face to the PM) but the air side goes into the cover material which is ... Aluminum? Then it induces Eddy currents or if magnetic material then it couples back into the core a little and mates with the end turn magnetic field of opposite polarity of the adjacent poles/end turns. Keep them short with longer radial slots and you reduce the ratio of end turn to slot turns and have a more efficient motor.
 
kiwifiat said:
end turns do not produce stator flux that is orthogonal to the rotor flux and therefore do not contribute anything towards torque production.

The coils do not interact with the magnets to produce torque. They are too far away.

The end turns -- as with the through-slot runs -- induce flux into the tooth core, and it is that flux emanating from the face of the tooth that interacts with the magnets to produce torque.

If BLDCs relied upon Lorentz force for torque, they would produce very little, because the closest turns only receive <10% of the magnets field strength; and those at the bottom of the tooth <0.0001%. (Check the math above or do your own!)

Beside which, the current in the half of a turn on one side of a tooth runs in the opposite direction to that flowing in the other half of that same turn. Same current but 180° opposite direction, same field strength (more or less) same direction; the Lorentz forces produced by the two sides of each and every turn cancel each other out!

It ain't guesswork, assumption or speculation. It's simply fact.
 
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