# How to choose wire gauge for Brushless DC motor?

I'm planning to build a Brushless DC motor, with inrunner perm mag rotor and stator on the outside. How can i determine the proper copper wire gauge for the stator windings? How much current will the motor pull? At first i thought it was just "Power = Voltage * current", so knowing the power and the voltage i should be able to figure the current. However, when the motor is spinning it generates back EMF, which counters the applied voltage and diminishes the current. At no load, the motor should draw no current for an ideal motor, and only a small current in real life to overcome friction and losses. In any case it would be much lower than the nominal current.

So, for instance, a 1000w motor at 50v should draw 20A nominal, but in most situations it would be less than that? How much is the starting current, when you have to overcome inertia/tire friction?

I know the motor can take higher currents for small amounts of time, before it gets too hot,but how can i determine what will be the "average" current. Is there a rule of thumb that motor designers use?

Also, the amp rating for each AWG seems to vary a lot. Some places list a conservative rating that is used for home wiring. But other places list much higher currents. If the stator is on the outer side and exposed to air, how much current can the copper handle for each AWG?

Any insight will be appreciated, thanks.

## Comments

4 years ago

https://www.instructables.com/id/Make-Your-Own-Mini...

https://www.google.co.uk/#q=DIY+brushless+motor

4 years ago

This is possibly the best motor design instrucable of all time

https://www.instructables.com/id/Make-Your-Own-Mini...

Sorry earlier pointer was off.

4 years ago

An important feature is the kind of wire you will use.. Magnet wire is round bare copper wire with a flexible insulating film so that adjacent stator wire windings do not short circuit. There are grades of films and the make a very big difference in a motor's ability to take temperature Radio_Shack magnet wire has a lo grade film that will char, short and destroy a motor.. While a Polythermaleze wired stator keeps working over 180'C.

Another important feature is forced air cooling through the air gap and over outside air passing through radiating fins to remove the I^2 x R losses..

When you know what kind of magnetic material your stator punching are made of, the ID, stator length, anticipated Speed range, number of phases and poles.. Than you can start playing with wire gauge and turns per pole

A word about fill factor, if you plan to wind your own stator.. Given well formed insulation slot "FishPaper", most first winders will only fill a slot area with 40% copper.

## Heavy Armored Polythermaleze Round Magnet Wire

"HAPT is a multi-purpose film insulation which has a

modified polyester basecoat and a polyamideimide topcoat. This

combination provides a film insulation which has physical toughness,

excellent dielectric properties, and superior chemical resistance to

common solvents and refrigerants. Thermal properties of over 200°C

qualify it for severe thermal overload applications. Compared to nylon

overcoated polyester, the properties most improved are physical

moisture, chemical and thermal resistance. HAPT is an extremely moisture

resistant film insulation coating, suitable for many uses including

open motor, high moisture application and hermetic applications. Other

outstanding properties include thermal life, thermoplastic flow, burnout

and heat shock.

This product is the standard of excellence for

motor repair, maximum result, and minimum reject applications. It is

impressive because it couples the extra advantages of high moisture

resistance with ease of insertion."

Typical ApplicationsAnswer 4 years ago

Sorry I forgot to quote mark the text under the

## Heavy Armored Polythermaleze

"

Answer 4 years ago

I'll check this coating out, thanks!

Answer 4 years ago

Assuming i get a wire with some average coating, what is the max safe temperature that the windings can tolerate?

Answer 4 years ago

As magnet wire can handle higher temperatures, certain manufacturers are making machines with less copper and iron for the same HP. They save $$ you get a lighter motor but it costs more electricity to use because reducing the turns per coil makes a burning hot inefficient machine.

Answer 4 years ago

Yea, the bearings might survive 180'C !!

Answer 4 years ago

You need to know what brand of coating it is by asking the seller.. Sometimes the price and color shade is an good indicator of quality.

Failing to get that information. Wrap ten turns around a bolt and slowly raise a current through the the wire until it smokes and use an IR thermal temp meter.

If plan to run slow, where a fan blade attached to the shaft is ineffective, you will have to limit I^2xR (i squared r) or use an appropriate air fan motor to cool your machine. See the cooling fins on the second pic.

One more thing you need to consider is the appropriate varnish or epoxy that you are going to encapsulate the windings into one solid structure which helps heat transfer out to the iron but more important it prevents micro-wire-rubbing which would eventually scratch the insulating from the copper causing a shorted turns.

Naturally the choice of the varnish / epoxy should be able to survive at or above the temperature of the wire film after curing.

4 years ago

I think I can imagine a motor, or generator, as a resistor in series with a voltage source. The resistor is just the actual physical resistance in the copper windings. The voltage source is there to represent the, what you call, "back EMF", and the magnitude of that voltage is directly proportional to the speed at which the motor is turning. The physical source of this back EMF voltage, is Faraday's Law, N*(dPhi/dt), number of turns multiplied by rate of change of magnetic flux through those turns.

Note that for a DC motor, I can pretend this back EMF voltage is a DC voltage, because there is some mojo happening with a commutator, to periodically flip the winding upside down, or change to another winding, in such a way that this EMF voltage is always in the same direction.

I mean, mathematically what I want to do, is come up with a simple expression for this back EMF voltage, like

Vmotor = k*omega

where omega is angular speed in rad/s. Moreover, all those varibles, {Vmotor, k, omega} are non-negative.

It might look like I have just pulled "k" out of the land of make believe, and to some extent that is true. Although I imagine k really could be derived from Faraday's Law, from the number of turns in each winding, the strength of the permanent magnets, and the geometry that couples changing shaft angle to changing flux. Or k could be found empirically from an existing motor.

I call the resistance in the windings: Rmotor.

Next I attach a constant voltage source, called Vbattery, to make a complete circuit.

The current that flows is simply:

I = (Vbattery-Vmotor)/Rmotor

In words, Vbattery and Vmotor are kind of pushing against each other, through this resistor Rmotor.

Recall that Vmotor is a function of omega, the speed the motor is turning.

I can imagine there is a speed at which,

k*omega = Vmotor = Vbattery, and thus I=0,

and I think this is your, what you call, "no load" condition for an ideal motor.

For those cases with current I greater than zero, it looks to me like the battery is supplying power.

P = I*Vbattery = I*(I*Rmotor + Vmotor) = I^2*Rmotor + I*Vmotor

Part of that power goes to heating the windings, I^2*Rmotor. The other part, I*Vmotor, I will naively assume that is electrical power being converted to mechanical power.

It is also possible to imagine the motor turning faster than its "no load" speed, which is saying the same thing as

k*omega = Vmotor > Vbattery

and for this case, I = (Vbattery-Vmotor)/Rmotor < 0. I is less than zero, or in other words, flowing the other direction. For this case, the motor is apparently supplying power, acting as a generator, charging the battery, and putting some power, I^2*Rmotor, into heating the windings too.

Anyway, I think is a pretty good model for a place to start, since it connects back EMF voltage and motor speed (i.e. using Vmotor=k*omega), and it explains where the power goes (ie., I*Vmotor is electrical power converted to mechanical, I^2*Rmotor is power lost to heating the windings)

Answer 4 years ago

Thanks for the reply. So in this case, Vmotor determines the current, given Vbattery and Rmotor which are constant, right?

Your expression for Vmotor is

V motor=k*omega

Omega is the angular speed, so how can i calculate or estimate "k"?

Answer 4 years ago

Yeah. Vbattery and Rmotor are constants.

Regarding Vmotor, the back EMF voltage induced in the winding, this comes from Faraday's Law.

Faraday's Law for a coil says the induced voltage on that coil is simply:

N*(dPhi/dt), where N is number of turns and (dPhi/dt) is the time rate of change of magnetic flux through the coil.

The amount of magnetic flux through a winding, it changes as the angle of the shaft changes.

If I were to plot this flux through a winding, Phi, as a function of shaft angle, theta, it would be some periodic function.

So Phi = Phi(theta)

When I take the derivative of Phi, with repect to time, I get:

(dPhi/dt) = (dPhi/dtheta)*(dtheta/dt)

from the chain rule, and (dtheta/dt) = omega

So, omega is in there, for sure. The rest depends on how Phi is changing as shaft angle changes.

One sort of easy approximation, is to say the amount of flux through the winding is constantly increasing for half a cycle, then constantly decreasing for the other half of the cycle. When you plot this, it looks like a triangle wave, bouncing back and forth between a maximum of Phi = Phimax, and a minimum of Phi = -Phimax.

So just looking at the part where Phi is increasing, the derivative wrt theta, is change in Phi, divide by change in theta,

(dPhi/dtheta) = (Phimax -(-Phimax))/pi = 2*Phimax/pi = (2/pi)*Phimax

Now I can substitute this back into previously written expressions, to get:

Vmotor = N*(dPhi/dt)

= N*(dPhi/dtheta)*omega

= N*(2/pi)*Phimax*omega

So I guess, k = N*(2/pi)*Phimax

where N is number of turns in the winding, and Phimax is the maximum value of the magnetic flux through the winding, and the minimum value is just -Phimax.

This would all go better with some pictures. So I'm going to up some pictures of pages, 448-453, from a book I found on my bookshelf, SJ Chapman's, "Electric Machinery Fundamentals. 3rd Ed."

Info from Google(r) Books page for this book:

Answer 4 years ago

Excellent readable pics

Answer 4 years ago

'k' comes from interaction between the motor magnet field strength and your motor windings. Its not readily calculable, unless you've designed the motor from first principles, and besides, as I said, you design on the basic either of I=V/R and maximum acceptable motor winding temperature - bearing in mind that THAT is in the centre of the motor windings, and inaccessible.

4 years ago

You can't rate the wire on the basis of nominal, only on maximum, for safety. The maximum current that can flow is at switch on or stall, when RPM=0, and the terminal voltage equation V= IR + E, where E is back EMF is zero. So V=IR, I=V/R

And yes, its a scary big number, for all but trivial motors. Don't be surprised if it can be 100A or more. Of course, your SOURCE impedance willl limit the maximum current, but if its a half-decent SLA, that can easily give you 100A. So size wires appropriately,

Answer 4 years ago

Wouldn't the controller, like a PWM, limit the current at low speeds?

Answer 4 years ago

No. Not without specialist circuitry.