I rewound my Axi 2208/20 but need some advice

[StuUK]

Hi All

I smoked my axi a few days ago directly driving a 9x6 prop

I stripped the motor down and the 12 tooth stator was a mess.

It was difficult to tell due to the state of it but it looked as though originally it was double wound with 32-36swg ten turns per tooth.

I had some 32swg(0.28 mm) wire around so I have rewound it double wire 9 turns on each tooth (thats all I could fit on)

As there are 14 poles I wound it AabBCcaABbcC

The motor seems to run fine, but I was wondering if anyone knew how I could calculate the max current this should take??

Cheers

[swatson144]

It seems right up the LBMiller5 alley! he knows that arcane art of engineering. If he doesn't sniff out this thread you should maybe ask PM him and ask If he would prime us. i have some motors I would sure like to calculate a max current on. Lucien posts very informative threads.

[StuUK]

I ve rigged up a shunt ammeter

3 cell lipo, no load, full throttle its pulling appx 1.2 - 1.5 amps

Didnt have any suitable props so I cut down a gws 10x4.7 to 6 inches and slimmed it off a little, then balanced it

At full throttle the motor pulls 15amps with this prop on

I'm getting some mor e props tomorrow so will test it then

I ran the motor at 14 amps on and off for about a minute, it got hot, but not burning hot i.e. you could put your finger on it and leave it there for a few seconds

Does all this sound about right?

[LBMiller5]

Thanks for the compliment Swatson!

The current draw on a motor is proportional to the load extracted from the motor and the speed at which the motor turns. A lot of people do not realize that a motor is also a generator, and when you run a motor, it is actually doing both jobs at the same time.

The maximum current a motor will draw is when the RPM is equal to 0. This is called the "Locked Rotor Current". It is basically the resistance of the wire in the stator poles and nothing else. When a motor turns, it generates a voltage that counter acts the input voltage, placing a smaller load on the motor. When a motor is turning at maximum RPM with no load attached to it, it draws the minimum amount of current.

To illustrate this we can assume a hypothetical motor that has a winding resistance of 0.5 ohms, and has a maximum speed of 10,000 RPMs. We can hook this motor to a 10 volt power supply for a test. With no load applied to this motor lets say we read a speed of 10,000 RPM and measure the input current at 1 amp. This 1 amp is the sum total of all the losses incurred in the motor. It is a combination of the frictional losses in the bearings, the heat losses in the windings, and the hysteresis losses in the stator poles.

If we grab hold of the shaft and stop the motor, the current would go up to about 20 amps, (10 volts divided by 0.5 ohms.) Naturally, there would be some loss in the wire running from the power supply to the motor, but I will neglect these losses to simplify the explanation. With these 2 extremes established, all other conditions between "Locked Rotor Current" and "No Load Current" will lie somewhere between 20 amps and 1 amp.

If the motor has a resistance of 0.5 ohms, why does it only draw 1 amp when running with no load? This is due to the reverse EMF energy that is generated by the motor as it turns. I will calculate this now to show where it comes from.

In any closed electrical circuit, the sum total of all the voltages must add up to zero. In this convention, input voltages are labeled as positive voltages and load voltages are labeled as negative voltages. This way when you add up all the loads, and then add that total to the input, the total sum equals zero.

Back to our motor: At 10,000 RPM the motor draws 1 amp of current with 10 volts supplied. We know that the resistance of the motor is 0.5 Ohms, so the voltage drop across the motor windings is equal to 1 amp times 0.5 ohms which equals 0.5 volts. So what happens to the other 9.5 volts? With no load placed on the motor, the motor is acting as a generator and is producing -9.5 volts at 10,000 RPM. Now when we add the rotor loss of -0.5 volts to the generated voltage of -9.5 volts, we get -10 volts. When this number is added to the input voltage of 10 volts the sum total is equal to zero, and the circuit is complete.

From these measurements, we can calculate that this motor generates 0.95 volts per 1000 RPM. Now that we have established the generator constant for the motor, we can calculate the current draw at any other RPM.

For example, if we put a prop on our motor and measure 5000 RPM with a tach, we can calculate the current draw as follows:

At 5000 RPM, our motor will generate 0.95 x 5 = 4.75 volts. If the motor generates 4.75 volts, then the windings must absorb 10 - 4.75 or 5.25 volts to have the circuit path add up to zero.

If the motor windings are absorbing 5.25 volts, then the current is equal to 5.25 divided by 0.5 or 10.5 amps.

This process can be repeated for a range of voltages to get the calculated current draw for just about any RPM you want. You can set up a spread sheet in Excel with Different RPMs on one side and the proper formula in the cells, and let it calculate the various values. These numbers could be exported into a graph, and you could use it to calculate the current draw at any RPM.

As a side benefit, you can also use these numbers to calculate the power loss in the motor to see if you are overloading it. Looking back at the 3 examples above.

At No-Load Current, we had 0.5 volts dropped across the motor and 1 amp of current. Since power equals Voltage times Current, in the No-Load situation, the motor is dissipating 1/2 of a watt.

At Locked Rotor current, the motor is dropping the full 10 volts across itself and the current was 20 amps. Based on these numbers the Locked Rotor power dissipated is 20 x 10 or 200 Watts. Since 50 watts is about the max per stator in our motors, this little guy would burn up very quickly at this condition.

For our 5000 RPM load condition, the motor is running with 5.25 volts volts across it and the current we calculated was 10.5 amps. Again, taking voltage times current we get 5.25 x 10.5 which equals 55.125 watts. In this condition, the motor is running at a little more than we would like to see, and would probably get a little warm.

That should answer your questions. If you have any other questions or comments, post them here and I will see what I can do to answer them.

Till next time............

Lucien

[swatson144]

sounds like it might be getting hot enough to cook your magnets. A thumb rule is if you can grasp it firmly but don't want too it is nearing 170 f . kind of like a real hot cuffy cup you grab and can manage to set it down to use the handle but you are really in a hurry to get rid of it, and no burn is caused but very uncomfortable. Our magnets start to lose strength at 176 f (80c) unless they are rated M H or SH.

from George the supermagnet man

The suffix indicates the maximum operating temperature of the magnet. Normal neo magnets work well up to 80C, the M is good for 100C, H is good for 120C and SH is good for 150C. All of these have the same Curie temp, which is the temp that totally demagnetizes the magnet, 300C.

So you might want to lower the prop size or gear it. You are probably getting about 150W from what you said assuming 10v, and I'd start thinking more of a 125w max load static which would be reasonably safe flying at speed unless you do a lot of hovering.

[Q]

Great post Lucien!!!!

I should gather up all this info and put it all together. This last post you did will certainly help me get my motor calculation application going.