Motor current x winding scheme x termination - RC Groups
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Oct 13, 2017, 03:50 PM
Ronaldo Nogueira
ronaldopn's Avatar
Question

Motor current x winding scheme x termination


Hi folks,
I was thinking about how to get more power with less temperature on a brushless motor and a few questions came to mind:
- when a motor is pulling X amps, is this the same current that flows through the wire (or strand) on each tooth?
- if the above is true, then a winding/ termination which leads to less turns/ tooth will be more efficient as it would allow for a thicker wire, am I correct?

I was thinking about choosing winding schemes which lead to less turns and also Y termination to reduce the number of turns to a minimum (as long as I can get the kv I want).
Just wanted to know if there is something I'm missing on this strategy.

Ronaldo
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Oct 13, 2017, 08:21 PM
just look at it smokin'
z-matrix's Avatar
Quote:
Originally Posted by ronaldopn
Hi folks,
I was thinking about how to get more power with less temperature on a brushless motor and a few questions came to mind:
- when a motor is pulling X amps, is this the same current that flows through the wire (or strand) on each tooth?
- if the above is true, then a winding/ termination which leads to less turns/ tooth will be more efficient as it would allow for a thicker wire, am I correct?

I was thinking about choosing winding schemes which lead to less turns and also Y termination to reduce the number of turns to a minimum (as long as I can get the kv I want).
Just wanted to know if there is something I'm missing on this strategy.

Ronaldo
Hello!

It does not work like that, less current is required for more efficiency.
You can parallel windings too, and connect to delta if you want higher Kv, that only means lower voltage and higher current.
2x voltage 0.5x current is the same power.
It is just, mosfets are expensive to handle the current, so going high voltage is an option for high power.
So you should reconsider your strategy, to make the BLDC motor cooler:
-run it with 180 commutation (and sine if required),
-install the strongest magnets available
-rewind and achieve better copper fill (lower resistance)
-decrease the load
-optimize the magnetic path to reduce losses at magnet edges and reduce airgap, if it is an (d)LRK motor you could experiment with stator flx barriers too.
-heatsink/fan the motor
in this order.


Z
Oct 13, 2017, 09:08 PM
Ronaldo Nogueira
ronaldopn's Avatar
Thanks Z, my question was generic but can be also very specific.
Let's say I have the original motor with a dLRK wind with 16T Delta. If i rewind it using dLRK wye I can reduce the turns to 9 which means I would be able to use a much thicker wire than if I would if i had to make 16 turns. Let's say we plan to run it on 12S 80A max power system.
In my very simple view (wouldn't be surprised if I was wrong ), if the same current X amps is flowing on each tooth of both windings then the thicker wire will mean less resistance/ heat and more efficiency.
This is obviously assuming current is the same on all teeth (which I don't know if is true).
Ronaldo
Oct 14, 2017, 06:32 AM
just look at it smokin'
z-matrix's Avatar
Quote:
Originally Posted by ronaldopn
Thanks Z, my question was generic but can be also very specific.
Let's say I have the original motor with a dLRK wind with 16T Delta. If i rewind it using dLRK wye I can reduce the turns to 9 which means I would be able to use a much thicker wire than if I would if i had to make 16 turns. Let's say we plan to run it on 12S 80A max power system.
In my very simple view (wouldn't be surprised if I was wrong ), if the same current X amps is flowing on each tooth of both windings then the thicker wire will mean less resistance/ heat and more efficiency.
This is obviously assuming current is the same on all teeth (which I don't know if is true).
Ronaldo
You need at least 17 turns delta with the same diameter wire to get better efficiency than 16 turn delta.
(Termination does not make any difference in efficiency)
Answering your original question, motor current is 2x for 0.5x voltage @ 0.5 Kv, reason why high voltage is used for high power is the resistive losses, and the cost of high current mosfets.


Z
Last edited by z-matrix; Oct 14, 2017 at 06:51 AM.
Oct 14, 2017, 03:11 PM
Registered User
manuel v's Avatar
Ronaldon.

Con cualquier terminacion obtendras los mismos resultados si igualas el empacado de cobre y la resistencia.
With any termination you will get the same results if you equate copper packing and resistance.

https://aeromodelismoelectrico.blogs...rmination.html
Oct 15, 2017, 02:15 PM
Ronaldo Nogueira
ronaldopn's Avatar
Quote:
Originally Posted by manuel v
Ronaldon.

Con cualquier terminacion obtendras los mismos resultados si igualas el empacado de cobre y la resistencia.
With any termination you will get the same results if you equate copper packing and resistance.

https://aeromodelismoelectrico.blogs...rmination.html
Saludos Manuel,
This is what I don't understand. If I have a winding with more turns (thus longer length), wouldn't it have a higher resistance than a shorter wire (less turns) with thicker diameter?
Oct 15, 2017, 08:24 PM
Registered User
manuel v's Avatar
como en que ejemplo?
Oct 16, 2017, 06:43 AM
Ronaldo Nogueira
ronaldopn's Avatar
Como este amigo
I ordered this Sk3 6364 for a new project and it should arrive in a week.
The motor specs say it has 16T (probably delta) for 213kv. I want to rewind it for the same kv but more power/ less heat.
On TC5 you can see my options (yellow cells). As far as I understand if I rewind dLRK wye I'll need only 9 turns instead of 16, 17, 18 or 30 for LRK delta. This means I will be able to use a much thicker wire or maybe 2 wires to fill the same slot which translates into less resistance. Also, 9 turns mean a shorter wire which translates into even lower resistance.
Obviously I'm assuming the same current flows through each tooth but I don't know if it is true.
Ronaldo
Oct 16, 2017, 01:04 PM
Registered User
manuel v's Avatar
in dlrk if you use the same% of copper packing, for the same Kv, the resistance will be the same.
In LRK, you will occupy less wire therefore the resistance between phase will be smaller for the same Kv compared to DLRK.
I thought you could fit an example with your motor using TC7.2. but I lack several data.
First the dimensions of the stator slot.
Second the length of the wire for a given Kv.

You can test the TC7.2 options yourself and feed the wire length and resistance according to the gauge used.
It is not possible to adjust the 100% the packaging, by the used gauges. the same with the kv. but if you make approximations, you will see very similar numbers.


What I can not determine is if under these circumstances the thickness of the wire equivalent to the final circuit will be the same in both delta and Star. In spite of being the same resistance.
In fact, I never thought of it.

en dlrk si usas el mismo % de enpacado de cobre, para el mismo Kv, la resistencia sera igual.
En LRK, ocuparas menos alambre por lo tanto la resistencia entre fase sera menor para el mismo Kv comparado con DLRK.
Pense que te podria cuadrar un ejemplo con tu motor usando el TC7.2. pero me faltan varios datos.
Primero las dimensiones del slot del estator.
Segundo la longitud del alammbre para un Kv determinado.

Tu mismo puedes probar el el TC7.2 las opciones y alimentando la longitud del alambre y la resistencia de acuerdo al calibre usado.
No se puede ajustar al 100% el empacado, por los calibres usados. lo mismo con el kv. pero si haces aproximaciones, veras numeros muy parecidos.
Lo que no puedo determinar es si bajo estas circunstancias, el grosor del alambre equivalente al circuito final sera el mismo tanto en delta como en Star. A pesar de ser la misma resistencia.
De hecho, nunca se me habia ocurrido
Oct 22, 2017, 07:01 PM
Registered User
manuel v's Avatar
then I have the measures of slot of the stator of the Hex 750. that is of 28.2mm2

which are rewound with:

Delta. 750Kv.
18T 21AWG, 80 In, equals to 8.538mm2 * 2, for 60.53% packed, .060mohms phase.

Star 742Kv.
10.5T, 18AWG, equals 9.50 mm2 * 2, 67.8% packed, .056mohms phase.

Star 760Kv.
10.25T, 18AWG, equals 9.33 mm2 * 2, 66.19% packed, .055mohms phase.

Star 742Kv.
10.5T, 19AWG, equals 7.76 mm2 * 2, 55.3% packed, .071mohms phase.

Star 760Kv.
10.25T, 19AWG, equals 7.57 mm2 * 2, 53.7% pounded, .069mohms phase.

Then: if we equalize the Kv, and the copper packing, the resistance will be almost the same, either in Delta or star. so the equivalent thickness of both will be the same and will have the same capacity to handle the amperage.


Regarding resistance, the best advantage is obtained with an LRK winding, the same in delta or star.

Delta. 756v. LRK
34T 21AWG, 78.6 in, equals to 8.065 mm2 * 2, for 57.16% packed, .0557mohms per phase.

Delta. 756v. LRK
34T 20AWG, 78.6 in, equals to 10.01 mm2 * 2, for 70.99% packed, .0441mohms per phase.

Note.
I do not know if anyone can do this packing.
but you can safely do one with 20.5 awg for a resistance of about .050mhoms.
Last edited by manuel v; Oct 22, 2017 at 10:39 PM.
Oct 22, 2017, 07:27 PM
Jack
jackerbes's Avatar
When we have a running motor there are rising and falling current levels in each phase and on each stator arm as the motor runs so I would say that we are seeing the total current that all of the arms drawing.

This page has some nice images of the phases in a running motor as seen on an oscilloscope.

http://www.aerodesign.de/peter/2001/...DY-BL_eng.html

Jack


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