@Oskar: could you try to compare FW and simply using a higher bus voltage (i.e. 12s) ?

Also, I wonder how these SK3 motors would sound like when playing with such high RPM as 20,000 (near 2000 rad/s in your graph).. Did you already try that ? (10,000 RPM is already quite a lot and generally require a heavy down gearing).

## Field Weakening

### Re: Field Weakening

I've been thinking about this for a while. Say for example that you have a motor that requires 2A of "no load current" at top speed.madcowswe wrote:I can redo these calculations including an iron loss model if you guys are interested?

That current goes towards overcoming mechanical losses in the motor, but also the iron losses. Thinking about what I would design a motor for, I think I'd aim for 30-50% of the losses at top speed (and top torque) to be iron losses. That would also mean that those losses would be very significant really quickly if you go above "rated top speed".

But I haven't done any real measurements.

### Re: Field Weakening

Difficult to tell what it means "top speed" and "top torque" for a given brushless motor, especially RC motors with so poor datasheets.

Top speed can be limited by mechanical reasons (mostly unbalance, bearings, centrifugal forces..) or magnetic losses, in the later case it depends how much extra loss you can tolerate. As for torque, it depends how you cool the motor and how long you want this torque..

Now if iron losses were as high as 30-50% of all losses at high power near continuous rating (say 50V, 40Amps for a ventilated SK6374), this would mean that at no load (low current, near 0 joule losses), there would still be as much as 30-50% of the full load losses (near 200W or so). So this motor would require as much as 80W at no load full speed ? I don't think so. My guess is 4 times less. But iron losses could rise strongly at higher voltages (and scarry RPM).

Top speed can be limited by mechanical reasons (mostly unbalance, bearings, centrifugal forces..) or magnetic losses, in the later case it depends how much extra loss you can tolerate. As for torque, it depends how you cool the motor and how long you want this torque..

Now if iron losses were as high as 30-50% of all losses at high power near continuous rating (say 50V, 40Amps for a ventilated SK6374), this would mean that at no load (low current, near 0 joule losses), there would still be as much as 30-50% of the full load losses (near 200W or so). So this motor would require as much as 80W at no load full speed ? I don't think so. My guess is 4 times less. But iron losses could rise strongly at higher voltages (and scarry RPM).

### Re: Field Weakening

markorman wrote:Did you take into account how much Id current you need to achieve higher speeds? Did you also account in this current when calculating efficiency?

So the current is constrained to be 64A in this case. That is, it's constrained to lie on the blue circle on this plot:

The plot shows Id and Iq, the contours are admissible current values for this bus voltage (22V in this case), for a given rotor speed. The previous plot was plotting the torque and power when the current is the optimal one, subject to these constraints.

### Re: Field Weakening

pf26 wrote:@Oskar: could you try to compare FW and simply using a higher bus voltage (i.e. 12s) ?

Also, I wonder how these SK3 motors would sound like when playing with such high RPM as 20,000 (near 2000 rad/s in your graph).. Did you already try that ? (10,000 RPM is already quite a lot and generally require a heavy down gearing).

Using a higher bus voltage is always more efficient. It would result in a flat torque line, and just a straight line power from zero going linearly up (imagine just continouing the trend from the start of the plot, with the same shape it has before hitting base speed).

The main reason you don't just keep increasing the voltage, at least for electric vehicle applications, is the increased cost of inverters and insulation at higher voltages.

### Re: Field Weakening

rew wrote:I've been thinking about this for a while. Say for example that you have a motor that requires 2A of "no load current" at top speed.madcowswe wrote:I can redo these calculations including an iron loss model if you guys are interested?

That current goes towards overcoming mechanical losses in the motor, but also the iron losses. Thinking about what I would design a motor for, I think I'd aim for 30-50% of the losses at top speed (and top torque) to be iron losses. That would also mean that those losses would be very significant really quickly if you go above "rated top speed".

But I haven't done any real measurements.

So I made some basic calculations that makes a big assumption: the no-load loss is dominated by eddy current loss (a type of iron loss) at high speeds. The eddy current loss power is proportional to the square of the motor speed. As most people know, also the ohmic loss in the windings are proportional to current (and hence torque) squared. So if we want to optimise for lowest overall losses (for a single operating point), we want:

min: P_loss = k1 * I^2 + k2 * w^2, where k1 and k2 are computed from the winding resistance, measured no-load current, etc.

Since this is a least squares problem, it has a very nice solution: make your gearing (aka trade torque for speed) such that P_eddy = P_ohmic; i.e. k1 * I^2 = k2 * w^2.

I stuck these calculations in this spreadsheet, that tells you what the optimal gearing is, given some motor parameters: https://docs.google.com/spreadsheets/d/ ... 2106166316

Of course this is a single operating point, and as you guys pointed out, this is far from the realistic case. Turns out that you can use the RMS value of the torque profile over your entire cycle, and same thing with speed.

OK, concrete example:

Log the current (hence torque) and speed from a normal cycle, compute their RMS values. Compute their ratio, say you get 30 Ncm/krpm. The spreasheet spits out an optimal gearing of 20 Ncm/krpm for your motor. Therefore you should actually gear your motor with slightly more reduction, so that it runs faster and uses less torque.

I hope that makes sense. I whipped these calculations up last night, so there may be some mistake, but I think the concept should be okay.

### Re: Field Weakening

I made some lab measurements on SK6374 149kV and Turnigy 4114 320kV motors:

At no load, the total power from the dc Power supply (including VESCs losses) varies nearly linearly with RPM. I just subtracted the VESC idle power with FETs gates driven at 20kHz (1.8Watts only, including 0.9W for idle VESC)

-> Around 0.015W/RPM for SK6374 up to 7500RPM (slightly more in FOC than BLDC mode), so 110Watts at 7500RPM (I was wrong !)

-> Around 0.0019W/RPM for 4114 up to 5500RPM -> slightly above 10W loss at 5500RPM no load

I also measured the motor current on one phase, using ACS758, and it is very near to what BLDCtools shows (within +/-5%) - Contrary to the battery current displayed by BLDCtools (generally much lower than reality, especially at low duty cycles).

For SK6374, 30Amps in each motor phase -> 50.5Watts losses (rotor locked), so around 18.7 mOhm per phase assuming all 3 are supplied with the nominal current at all time (BLDC tools says 28.7mOhms but I don't know if it is meant to be exactly)

Finally, this should make around 160Watts losses at 50V/30Amps, or 89% efficiency (assuming the iron losses do not increase more with RPM and current applied together than separately). Going for highest RPM is not always better: 40V/40Amps has also around 89% efficiency..

Turnigy 4114 has less iron losses. In this case, higher RPMs are recommanded (but limited here to 4800RPM from the manufacturer).

FOC has a slightly higher power consumption at no load high speeds than BLDC. This is probably due to non optimal control when playing with low motor currents that create large measurement errors on VESC shunts. BLDC uses bemf, so no such issue, especially at higher RPMs.

At no load, the total power from the dc Power supply (including VESCs losses) varies nearly linearly with RPM. I just subtracted the VESC idle power with FETs gates driven at 20kHz (1.8Watts only, including 0.9W for idle VESC)

-> Around 0.015W/RPM for SK6374 up to 7500RPM (slightly more in FOC than BLDC mode), so 110Watts at 7500RPM (I was wrong !)

-> Around 0.0019W/RPM for 4114 up to 5500RPM -> slightly above 10W loss at 5500RPM no load

I also measured the motor current on one phase, using ACS758, and it is very near to what BLDCtools shows (within +/-5%) - Contrary to the battery current displayed by BLDCtools (generally much lower than reality, especially at low duty cycles).

For SK6374, 30Amps in each motor phase -> 50.5Watts losses (rotor locked), so around 18.7 mOhm per phase assuming all 3 are supplied with the nominal current at all time (BLDC tools says 28.7mOhms but I don't know if it is meant to be exactly)

Finally, this should make around 160Watts losses at 50V/30Amps, or 89% efficiency (assuming the iron losses do not increase more with RPM and current applied together than separately). Going for highest RPM is not always better: 40V/40Amps has also around 89% efficiency..

Turnigy 4114 has less iron losses. In this case, higher RPMs are recommanded (but limited here to 4800RPM from the manufacturer).

FOC has a slightly higher power consumption at no load high speeds than BLDC. This is probably due to non optimal control when playing with low motor currents that create large measurement errors on VESC shunts. BLDC uses bemf, so no such issue, especially at higher RPMs.

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