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over a 1/2 mile of magnet wire in the stator of each Model S motor

over a 1/2 mile of magnet wire in the stator of each Model S motor

What do they mean by "magnetic wire" exactly? I was under the impression that tesla doesn't use rare earth magnets, or any magnets for that matter, in the motor, and the magnetic field is created by current flowing through copper windings both in the stator and rotor. Is there a magnetic core to the windings, or is it just ferrous metal (like iron). I would feel pretty dumb if they actually used magnets, as I have been telling people that tesla AC induction motors don't use rare earth magnets, so no dependency on china for that, and no environmental damage from mining neodymium.

bayoufilter.tx | 18. mars 2014

You are correct; from what I know of Nikola Tesla's induction motor there is no need for permanent magnets. The magnetic fields are induced in the rotor by the (current in the) windings of the stator.

Roamer@AZ USA | 18. mars 2014

You answered your own question. The magnetic wire is wire designed to be wound to create a magnetic field. The wire itself is not magnetic. When properly wound it creates a magnet field when current is induced into the wound magnetic wire.

Roamer@AZ USA | 18. mars 2014

If you have ever looked at the windings on an AC motor the wire used is pretty small gauge so a half mile of it is really not that much actual copper.

NKYTA | 18. mars 2014

@Roamer, about what gauge, do you know? Is it like my house wiring or smaller diameter than that?

Vall | 18. mars 2014

I thought so too, but using the word MAGNET threw me off and made me scratch my head when I saw that post on tesla's facebook page. They could have just said "wire" or "copper wire" or whatever wire they are using, but if the wire itself has no magnetic properties, why would they call it a "magnet wire"?

frankviaje | 18. mars 2014

It is a lot smaller diameter than your house wiring!

AoneOne | 18. mars 2014

Magnet wire often refers to wire with a thin lacquer or similar insulation that maximizes the volume of the coil occupied by the copper. Some magnet wire has a square cross-section to further improve the packing density.

Vall | 18. mars 2014

Thanks, AoneOne!

Good explanation, wasn't familiar with the term, and decided to look for answers here, glad I found them. Didn't even occur to me to look it up on google, the wikipedia article is like the first or second result. Guess that'S what happens when you don't google things...

ye | 18. mars 2014

Why doesn't the wire melt if it's so thin? Doesn't a lot of current flow through it?

Roamer@AZ USA | 18. mars 2014

Ye, when you wind the small wire many times you end up with basically a mega wire that can be manufactured into the shape needed and handle all the stresses involved without breaking.

Out4aDuck | 18. mars 2014

AoneOne is correct about the magnet wire (not magnetic wire). The insulation is very thin so that you can pack a lot of copper windings into a small volume. The copper itself has no magnetic properties. Pretty much anything that you can think of that has copper windings uses magnet wire.

Regarding the wire gauge, it's probably about the same as your household wiring. It does have to carry a lot of current, which is induced into the rotor. The windings are actually shorted so that the current flows in a loop in each pole of the rotor. The current in the rotor windings never leaves the motor.

One of the challenges of designing a high output motor in a small package is how to transfer heat out of the rotor windings. Figure out a better way to do it and you can land a job at Tesla.

ye | 19. mars 2014

Roamer, it sounds like you're talking about stranded wire, which is much more flexible than solid wire because it's made up of lots of thin strands in parallel. The current carried by the wire is split up among all the individual strands, so each one carries only a small fraction of the current. I don't think motor windings work like that. Rather, the winding of a motor is one long, thin wire, through which all the current flows. That's why I asked about it possibly melting from all that current.

Baribrotzer | 19. mars 2014

Well, the stator, from what I understand, runs at a pretty high voltage: The battery pack is charged to 375 V, and the motor probably runs at something pretty close. High voltage gives high power without high current, and doesn't need as thick a wire. And paradoxically, the rotor has an enormous current and almost no voltage - which is why it's essentially a cage of solid copper busbars.

Out4aDuck | 19. mars 2014

300kW divided by 375V is 800 amps. I know that in a Prius, the 200V battery voltage is boosted up to 650V to reduce motor current. Does anybody know what Tesla does?

Roamer@AZ USA | 19. mars 2014

Ye. You may be right. I have to admit I have never taken apart a 400 HP AC elec motor. Something special has to be going on to get that out of a motor that size.

mallynb | 19. mars 2014

"Magnet wire" is an industry term for single strand insulated copper wire used to wind coils for electromagnets. Everything from doorbells to motors. Insulation can be one of various forms of varnish depending on the application. The Tesla motors are 3-phase alternating current(AC)induction motors powered by 3-phase inverters that change the DC battery voltage to variable frequency AC. The speed of an induction motor depends on the frequency of the applied voltage.

ye | 19. mars 2014

Baribrotzer said: "High voltage gives high power without high current, and doesn't need as thick a wire. And paradoxically, the rotor has an enormous current and almost no voltage - which is why it's essentially a cage of solid copper busbars."

Now wait just a minute here. If a very low voltage results in enormous current, why won't a high voltage result in even more current?

(I bet you're going to spout some nonsense about back-emf and whatnot, but you can't fool me. I know the truth. Motors are magical, and that's all there is to it.)

Vall | 19. mars 2014

low voltage and enormous current gives you the same power as enormous voltage and low current. Actually less, because dissipated power goes up with the square of the current. Or?

Rheumboy | 19. mars 2014

So is this thing called a COPPERMAGNOSTATOROTORMOTOR? Please help!

david.baird | 19. mars 2014

There's an induction motor open-sectioned in the showroom here in Brussels (I imagine all Telsa showrooms have one??), I thought the wire looked pretty thick, but maybe it's tightly stranded.

I'll have a closer look one of these days...

david.baird | 19. mars 2014

There's an induction motor open-sectioned in the showroom here in Brussels (I imagine all Telsa showrooms have one??), I thought the wire looked pretty thick, but maybe it's tightly stranded.

I'll have a closer look one of these days...

TeslaTap.com | 19. mars 2014

The inverter can deliver up to 1000 amps to the motor. This is from Tesla's 2010 10K filing (but this might be referring to the Roadster motor, which is similar). 400v * 1000 amps = 400 kW of power! This doesn't mean it can sustain this continuously, and more likely its for a very brief period when starting to move, fully loaded on a steep hill.

If the wire length is 1/2 mile, I doubt it is as thick as 12-14 gauge house wiring - I'm sure it is a bit thinner. Liquid cooling helps prevent the wires from melting under high currents. Tesla also monitors the temperature, and would limit the current if the motor gets too hot.

Leeo | 19. mars 2014

High voltage and low current means less heat too. Right?

ye | 19. mars 2014

Vall said: "low voltage and enormous current gives you the same power as enormous voltage and low current."

Yes, but how can you get low current if the voltage is high? Why doesn't the high voltage push a lot of current through the wire? If the wire were made of some high-resistance material, that would explain it, but here the rotor and stator windings are both copper. So how do you get low voltage and high current in one, but high voltage and low current in the other?

ye | 19. mars 2014

mrrjm said: "High voltage and low current means less heat too. Right?"

Yes, that's correct.

Baribrotzer | 20. mars 2014

@ ye: "Yes, but how can you get low current if the voltage is high? Why doesn't the high voltage push a lot of current through the wire? If the wire were made of some high-resistance material, that would explain it, but here the rotor and stator windings are both copper. So how do you get low voltage and high current in one, but high voltage and low current in the other?"

Because the stator wire is half-a-mile long, and has a narrow cross-section. Even copper has some resistance, if only a little, and half-a-mile of thin copper wire could add up to quite a bit. Whereas the copper cage that functions as the rotor windings is quite short and quite thick, so it has very little resistance and very little voltage drop.

ye | 20. mars 2014

No, it's trickier than that. The current in the stator windings is less than what Ohm's law says it should be. If the low current were caused solely by the high resistance of the winding, all of the input power would be dissipated as heat in the winding and none would be left to turn the motor.

ye | 20. mars 2014

TeslaTap said: "The inverter can deliver up to 1000 amps to the motor. ... If the wire length is 1/2 mile, I doubt it is as thick as 12-14 gauge house wiring - I'm sure it is a bit thinner."

Are we sure that the wire is 1/2 mile long? Vall says so in the thread title, but doesn't say where that number comes from.

If it really is correct, and if the motor is run at 400 volts, the wire needs to be at least 7 gauge. Otherwise, it would have too much resistance to let 1000 amps flow.

jkn | 20. mars 2014

3 phase current, so there is only 1/6 mile of wire for each phase.

mrrjm said: "High voltage and low current means less heat too. Right?"

Not so simple. If you double voltage and keep motor power same, you need to double length of wire and cut current to half. This seems to cut resistive loses to half, because P=I*I*R. But you have to make wire thinner so that longer wire fits into same space. This again doubles resistance, so resistive loses don't change.

The inverter can deliver up to 1000 amps to the motor. There are rather large resistive losses in the battery. So with low power it can give 400 V, but with 1000 A much less. I tried to estimate resistance from Panasonic doc, but Teslas cells must have lower resistance than Panasonics.

Battery voltage also drops while discharged. Half empty (Panasonic) battery gives about 20% lower voltage than full. So what is 0 to 60 time with half empty battery?

ye | 20. mars 2014

jkn said: "If you double voltage and keep motor power same, you need to double length of wire and cut current to half."

I don't understand. Why do you need to double the length of the wire?

carlk | 20. mars 2014

Current and number of coil turns determine strength of the magnetic field generated. You can increase the magnetic field by either increase the current or coil turns (wire length) or both. Either one requires increase of voltage from the ohm's law I=V/R. Remember longer the wire higher the R.

ye | 20. mars 2014

carlk said: "Current and number of coil turns determine strength of the magnetic field generated. You can increase the magnetic field by either increase the current or coil turns (wire length) or both."

Yes, I agree with this.

carlk continued: "Either one requires increase of voltage from the ohm's law I=V/R."

Ohm's law is not the only factor here. For a given voltage, the resistance of the winding places an upper limit on how much current can flow. But the actual current is almost always less than this. If the rotor can turn freely, for example, the stator current is tiny, compared to what Ohm's law would predict. Electric fields in the stator windings, produced by the spinning rotor, counteract the applied voltage, reducing the current.

GaryREM.va.us | 20. mars 2014

Remember this is an AC motor not DC! (That inverter converts DC into AC).

So your simple DC analyses don't apply since you are not considering the impedance of the motor elements.

ye | 20. mars 2014

GaryREM, I think we might be saying the same thing in different words. What do you think?

TeslaTap.com | 20. mars 2014

I should also point out the 1000 amp number is likely a peak number, never continuous. Also it was stated without a voltage, so we don't know what voltage is being applied when it's at 1000 amps.

Most of the wire charts you see are rated for continuous current at a rated voltage (and typically a specific maximum length). These are designed around two objectives - so the wire does not get too hot and melt the insulation AND the length is not so long as to have too large of a voltage drop. These are likely secondary issues related to motor windings, since you may not care about voltage drop, and heat can be measured and controlled.

There is no way to easily translate the one fact of using 1000 amps with an unknown duration and unknown voltage to a suitable wire size. Tesla's engineers clearly know these values, but I don't believe they are public.

I took an electrical motor engineering class years ago, and for me it was one of the most complex engineering classes I had. I really appreciate how difficult quality motor design is. I don't profess to be any kind of expert in this field, and most what knowledge I had is long gone :(

carlk | 20. mars 2014

Yes you guys are saying the same thing and you're being right too. All you need to do is to replace R with Z (impedance) for Ohm's law to take care of the inductive part of the equation. Incidentally, like the resistance, the inductance part of impedance is also proportional to wire length, or number of coil turns (at any fixed frequencies) and requires higher voltage when there are more "magnet wires".

Wow I can't believe I can put my high school physics to use now.

ye | 20. mars 2014

Just to clarify, in deriving my 7 gauge figure, I wasn't even considering safety issues. The resistance of 1/6 mile of 7 gauge copper wire is 0.44 ohms. If you put 400 volts across that, you'll get 910 amps. It doesn't matter how good your cooling system is; more current than that simply won't flow.

Of course, if the voltage is less than 400 volts, the current will be even less.

I agree that motors are complicated. I know enough to know that there's an awful lot that I don't know.

Baribrotzer | 20. mars 2014

@ ye -

Carlk raised an important point: Impedance. Because of the stator's inductance, the voltage and current will be out of phase, and may be considerably out of phase. And I believe that only the components of the voltage and current that are IN phase - those components caused by the resistive component of the impedance - give rise to heat that must be dissipated. The inductive component flows over to the rotor, and much of it appears as useful work.

Car t man | 20. mars 2014

Those are all peak peak powers. Bursts really..

AoneOne | 21. mars 2014

One more thing: as a 3-phase motor, I'd expect that there are three separate stator windings, each taking the full voltage at one-third of the current and with one-third of the wire length. That substantially reduces the resistive losses.

GaryREM.va.us | 21. mars 2014

@ye

You are stating numbers as if this were a DC problem. It is not.

First, the motor is AC.

Second, the impedance (more accurately the reactance since the motor presents itself as a combination of inductances and resistances) is going to be a higher value than the simple wire resistance.

Therefore your simple DC calculations have no meaning here.

Bubba2000 | 21. mars 2014

Tesla must be getting the voltage they need by adding a certain number of batteries in series. The higher the voltage, less amps, lower the transmission losses and thinner wires needed but more insulation. I imagine the inverters have optimum voltages to operate. Then there is motor design.

Anybody can put a package like this together. However, I have never seen anything so compact, efficient with this kind of HP with flat torque curve. All works perfectly.

ye | 22. mars 2014

Baribrotzer said: "And I believe that only the components of the voltage and current that are IN phase - those components caused by the resistive component of the impedance - give rise to heat that must be dissipated. The inductive component flows over to the rotor, and much of it appears as useful work."

I'm not sure, but I think that, to the extent that electrical energy gets converted by the motor to other forms, whether heat in the winding resistance or motion of the car, the motor appears resistive to the power supply. The inductance of the motor, on the other hand, causes it to draw power from the power supply during part of each AC cycle and immediately return it to the power supply during the remainder of the cycle.

ye | 22. mars 2014

AoneOne said: "as a 3-phase motor, I'd expect that there are three separate stator windings, each taking the full voltage at one-third of the current and with one-third of the wire length. That substantially reduces the resistive losses."

Yes, thanks. I think you're right. I remembered to divide the wire length by three but forgot to divide the current by three.

Redoing the calculation for the minimum wire size of the motor windings yields a value of 11 gauge rather than 7 gauge.

ye | 22. mars 2014

GaryREM said: "You are stating numbers as if this were a DC problem. It is not. ... the motor is AC. ... Therefore your simple DC calculations have no meaning here."

I'm not sure what you're disagreeing with. My calculation gives (what I believe to be) the smallest wire gauge that could possibly carry 1000 amps at 400 volts. Do you think that the inductance of the motor could somehow allow a smaller wire to carry that amount of current at that voltage?

david.baird | 22. mars 2014

I took a look this morning in the Tesla showroom. The wire looks about the same diameter as 13amp household wiring.

Here's a photo (maybe...):

david.baird | 22. mars 2014

Bugger... Another try...

ye | 22. mars 2014

So, here I am, enjoying my theoretical discussion, and then someone has to come along and try to confuse me with actual real-world data. :(

Just kidding. Thanks for the picture. Do you know how thick the 13-amp household wiring is that you're referring to? In AWG or square millimeters or however it's measured where you are. (You're not in the US, right?)

david.baird | 22. mars 2014

Hey @ye, I would say it's 1millimetre in diameter - that's about 1/25 of an inch in old-world money :-)

Of course this is a visual guess, but I'm confident +/- 25%

jkn | 22. mars 2014

ye,

It is a 4-pole motor? That means 2 coils/phase? Those coils could be connected parallel?

battery out: 355 kW, 355 V, 1000 A
inverter out: 345 kW, 345 V peak, 244 VAC, effective 3*471 A AC (peak 667 A)
motor output: 310 kW

I know that 3*471 > 1000. Current from battery goes trough a coil inside inverter. When current from battery is switched off, energy in the coil pulls more current through diode.

gauge 7 -> 1.634 Ohm/km -> 0.219 Ohm/coil

How much heat?

coils in series:
Voltage drop: 471 A * 0.219 Ohm = 103 V / coil
heat: 471 * 471 * 0.219 = 49 kW, 6 coils total 291 kW ... Too much!

Coils parallel:
Voltage drop: 471 A / 2 * 0.219 Ohm = 52 V
heat: 471 * 471 / 4 * 0.219 = 12 kW, 6 coils total 73 kW ... Too much!
But correct number of 0s. Perhaps coil is thicker. Perhaps I made a mistake. I'm not going to find it today. So good luck!

Induction motor is roughly resistive load. It does not cause large phase shifts.

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