Instructables

Step 6: Actually Winding the Motor

If you've never wound a motor before, the diagrams of LRK windings are probably pretty meaningless. This is a time when you need to learn the nomenclature of motor hobbyists.

An example would be the dLRK winding:

AabBCcaABbcC

or the classic LRK winding,

A-b-C-a-B-c

What? Did you just sing the alphabet song or something? Kind of. The three phases of the motor are referred to in this case as A, B, and C.

A capital letter indicates one winding chirality, a lower case means the other. For instance, if A is designated "make a loop of wire in the clockwise direction", then a means "wind the loop of wire in the counterclockwise direction". And a dash or space means an unwound tooth.

The general convention is capital letter equals clockwise loop, lower case equals counterclockwise loop. But, what is more important is consistency. If you do it one way, stick with it. 

So what does the above string of gibberish mean? Starting at any tooth (mark this as your index!), begin making loops of wire around it according to the designation. To wind two teeth Aa style, wind one of them clockwise, and the other counterclockwise (or vice versa - keep track of this.)

There is no "right method" to obtain clean windings, but the last thing you want to do is just bundle wires around the tooth with reckless abandon. For large motors, use latex gloves to ease hand abrasion and a wooden dowel to wrap wire around for extra leverage.

Unfortunately, I don't currently have any pictures of video of me winding a motor. This might change in the near future to save a thousand words of explanation.

Perhaps one of the most valuable resources available is the Combination Table. Input your number of stator teeth ("nuten") and your number of magnets ("pole") and it will automatically generate the correct winding pattern! The above table was generated by one of the Crazy German R/C Airplane Dudes, who seem to be the source of all technological advancement in the model motor scene.

Single Layered, Multi Layered

You may find that you can't get the N number you want by only winding one layer of wires on the stator. Simple solution: Keep winding and make a second layer.

Two to three layer windings are generally the limit of heating & cooling unevenness for small motors, and the Rm gets ridiculous as well. As more layers are added, the end turn effect will become more and more of a factor.

If you find yourself having to wind many layers, perhaps switching down a size of wire will alleviate that.

How many turns (N) do I need? 

The other killer question of small motor design. Given other motor parameters, you can backsolve easily for the minimum N needed to achieve a certain design goal, usually torque. Accounting for losses and assumptions, N should be above this number by a comfortable margin explained shortly.

Example (Updated 3/28/2012 to correct the math which has been wrong for over 2 years! I keep meaning to fix it, then never getting around to it. Ultimately, enough of you called me out on it, so congrats. Here's the fixed math using also the new torque constant factor m).

Let's say that I want to design a motor inside a 12cm (0.12m) wheel that will let me climb a 10% grade (or about 5.5 degrees inclination) at velocity v = 5 m/s (about 11mph), and I weigh m = 65kg. The force of gravity F pulling me back down the hill is

F = m * g * sin 5.5° = 61N, or thereabouts. I

I want to climb the hill at 5 m/s. Mechanical power is torque * rotational speed, but it is also linear force * linear velocity.

Thus Pm = 61 * 5 = 305 W

Seems reasonable, right? Assume the motor is a perfect transducer (it's definitely not). The electrical power required is also 305 watts.

Assume my battery is 28 volts, so i = 305 W / 28 V = 10.9A

To exert a linear force of 61N at a radius of 0.06m (wheel radius), the torque T is 3.66Nm.

Two variables, T and i, have now been established. The motor is a 12-tooth, 3 phase motor, so m is 4 (there are four teeth per phase). You can now reduce the equation to

T / (4 * m * i) =  N * L * R * B

R is ultimately limited by the size of my magnet rotor and inner diameter of my tire - a topic which is forthcoming. Let's say that my wheel choice has forced a  maximum stator diameter of 70mm, and the motor can't be more than 30mm wide to fit in my vehicle. 

B is my magnetic field strength. Let's assume it is 1 Tesla for now - we will see soon that this is not a bad guess if your motor magnets are reasonably thick.

T / (4 * m * i * L * R) = N

Let's see what this comes out to.

3.66 / (4 * 4 * 10.9  * 0.03 * 0.035) = 19.98 = N

This value is a reasonable first approximation for the number of turns per tooth you need.  Since hundredths-precision turn fractions aren't possible, take the closest integer: 20.

Fiddle Factors and Hand Waves

Every nonideality and inefficiency in the world will work to make your motor faster (read: less torquey) than what the number of turns alone would indicate. Therefore, it is good sense to consider this as the absolute minimum number of turns per tooth. The torque constant value derived from using NIBLR is generally 20 to 33% too high for average fractional-slot, permanent-magnet motors like the type we are considering.

Remember also that motors are not perfect transducers. The average efficiency of a decent BLDC motor is somewhere around 90%. So, if I want to perform this hillclimb at maximum efficiency, that's much different than attempting it at maximum power output. The efficiency of a motor at maximum power output is always less than 50%. This is something to be well aware of - if you are using this 'target output force' method to design your turn count, then you should take the speed to be somewhere close to your anticipated cruising speed. This makes sure that, if anything, you overdesign the motor for torque as nonidealities only take it away from you.

The above motor example is the motor for Project RazEr. In actuality, RazEr's motor has 25 turns per tooth - overspecified by roughly 25%.

To wrap up, R and L are mechanical constraints dictated by your vehicle's mechanical parts while m, B, N, and i are electromagnetic constraints dictated by your choice of magnets, wire, and coil layout.

 
Remove these adsRemove these ads by Signing Up
Ottoclav1 year ago
Im an electrician, and house wiring is done in 14, 12 and 10 gauges mostly. Winding a motor in 18 gauge must be a chore! But im sure chris farley would say, " It builds dexterity!"
renhit3 years ago
Can somebody explain how 61N is calculated? what values are considered for m & g?

F = m * g * sin 5.5° = 61N

and below equation how did u get 12.7?
3.66 / (4 * 12.7 * 0.03 * 0.07) = 37 = N

should it not be 10.9 amps as calculated above?

61N = 61 Newtons. m is 65kg as listed above, it is the mass of the rider + the scooter.  g is the gravitational acceleration constant, or 9.81 meters per (second squared)
F = 65kg * 9.81m/s^2 * sin(5.5) = 61.11 kg m/s^2 = 61.11 N
(Force = Mass * Acceleration.  Since this is on a hill with an angle from the horizontal plane of 5.5 degrees, take sin(5.5 degrees) to get the component of the force pushing back down the hill)

I'm guessing that it was either adding a margin of safety to account for other errors in the system, or he simply grabbed the wrong number.
DG4ever4 years ago
It's a really good work and explanation!

I'm planning to build such an electric scooter too!!
But I want to have more power:

Lets say I want drive 10m/s with 1500W(yes it seems oversized) and I use 24V:

1500W / 10m/s = 150N

The radius of my wheel is 0.075m and of my stator 0.0465m the width is 0.026m

150N*0.075 = 11.25Nm and 1500W/24V = 62.5A

-> 11.25/(4*62.5*0.026*0.0465) = 37.2*1.5 = 56

My stator has got 18 teeth, so I have to wind 10 windings per tooth!!

My question is, is that possible and which magnet wire do I have to use to handle the 62.5 A?
megapix4 years ago
Thanks for including the torque calculation. Wasn't sure where to get even a ballpark number for required torque, and I'd sure like to know if my design was off by a factor of 100 before I started winding coils.