Paul and Sabrina's Cheap 3 Phase Inverter (AC Controller) with Field Oriented Control

[e*clipse]

Paul:

I was digging through some boxes of stuff I have and found this:

http://www.beisensors.com/pdfs/L25-o...al-encoder.pdf

It's a sooooper high precision (2540 cycles/turn) encoder.

It's got a 1/4" shaft with a flat on it, so it doesn't really meet your motor's requirements, but if we could figure out a way to connect it, it may work.

Does the motor have a shaft out the back?

- E*clipse

[cts_casemod]

Originally Posted by e*clipse

Paul:

I was digging through some boxes of stuff I have and found this:

http://www.beisensors.com/pdfs/L25-o...al-encoder.pdf

It's a sooooper high precision (2540 cycles/turn) encoder.

It's got a 1/4" shaft with a flat on it, so it doesn't really meet your motor's requirements, but if we could figure out a way to connect it, it may work.

Does the motor have a shaft out the back?

- E*clipse

Why don't you guys use the flywheel and a reluctance sensor as an encoder? That would cut a significant portion of the costs.



The input capture peripheral and a free running timer can be used to measure the time between each pulse

[cts_casemod]

Time to use some brain cells folks

I'm designing an algorithm for the induction motor.

Ok, everyone knows that above nameplate speed the V/Hz ratio and the torque start to fall, but exactly by how much I cant find.

I thought about using the HP formula HP = (torque(nm)*Speed(RPM))/5252 but assuming the V/Hz ratio goes down, so does the slip induced current on the rotor, so I would have to increase the slip to have a constant secondary output. Problem is such functions are often not linear.

I cant really find any good source that discusses this on the field weakening region. Any ideas?

[P-hack]

I took a second to look at heubners code, basically you set fweak to the field weakining frequency, it is also the point where max dynamic range is at (i.e. full svpwm). After that you can only increase frequency without increasing v (and still have a sinewave). so field weakening is constant v with increasing hz. I didn't see anything fancy about slip at that point.

[cts_casemod]

Thats just what the literature tells you. But from that you cant calculate the torque.

let me clarify:

Theory says that torque falls by the square of the voltage, so technically at twice the nameplate speed one would have 25% torque as the V/Hz ratio has now fallen to 50%.

Some sources disagrees and state that you'll have half of the torque, since the horsepower is constant. Clearly something is not right here. How can you have the same horsepower if the torque has fallen to 25 and the speed has doubled!? Perhaps one could double the slip, which makes a lot of sense, but this is never mentioned in technical papers.

The idea is to keep the rotor power constant. For that, if the motor is operating at half the V/Hz ratio, one doubles the rotor frequency, by increasing the slip.
The net result would in fact be constant torque, since the primary V/Hz ratio and hence the flux has fallen to 1/2, but the rotor flux is kept constant, but I haven't found any sources mentioning anything about the slip, so for the time being, this is just me putting 2+2 together.

Adding to the problem is how reluctance will play here. Not only the V/Hz ratio is increasing, but the impedance of the motor (XL) is also increasing. More impedance = less current.

[cts_casemod]

I wrote the formulas for all that and calculated the table below

Parameters are:
Motor nominal Voltage: 208VAC
Motor nominal Frequency: 50Hz
Motor rated Slip: 70RPM
Battery cut out voltage: 320VDC



Notice how having a variable slip, with the aim of keeping rotor power constant above nameplate, can attain a larger output torque. There was a topic on DIY car forums where we discussed this 2 years ago.

[e*clipse]

Originally Posted by cts_casemod

Why don't you guys use the flywheel and a reluctance sensor as an encoder? That would cut a significant portion of the costs.



The input capture peripheral and a free running timer can be used to measure the time between each pulse

Funny - that's what I woke up thinking may work! I think it would work great for Paul's induction motor; after all they require slip, so an accurate position isn't necessary - right?

OTOH, accurate position is very helpful for PM motors; and that's the current issue.

Soooo, the idea is to take the encoder off the induction motor he has and make it work on the PM motor he just got. The reluctance encoder could be set up for the induction motor. I think it's a great set of experiments.

Paul- if you'd like to try this, I'll machine an adapter for the different motor shaft diameter.

- E*clipse

[cts_casemod]

Originally Posted by e*clipse

Funny - that's what I woke up thinking may work! I think it would work great for Paul's induction motor; after all they require slip, so an accurate position isn't necessary - right?

OTOH, accurate position is very helpful for PM motors; and that's the current issue.

Soooo, the idea is to take the encoder off the induction motor he has and make it work on the PM motor he just got. The reluctance encoder could be set up for the induction motor. I think it's a great set of experiments.

Paul- if you'd like to try this, I'll machine an adapter for the different motor shaft diameter.

- E*clipse

For PM, I believe If one were to weld two teeth of the crankshaft together the longer pulse could be used as zero point to synchronize the timer. Thats similar to what conventional vehicle engines use (petrol/gas).

From there just count the number of pulses to find the angle.

If possible, a IR or hall sensor will avoid magnetization noise, while reading down to very low speeds.

For me it was not an option since I didn't had space to install the encoder.

[e*clipse]

Originally Posted by cts_casemod

Thats just what the literature tells you. But from that you cant calculate the torque.

let me clarify:

Theory says that torque falls by the square of the voltage, so technically at twice the nameplate speed one would have 25% torque as the V/Hz ratio has now fallen to 50%.

Some sources disagrees and state that you'll have half of the torque, since the horsepower is constant. Clearly something is not right here. How can you have the same horsepower if the torque has fallen to 25 and the speed has doubled!? Perhaps one could double the slip, which makes a lot of sense, but this is never mentioned in technical papers.

The idea is to keep the rotor power constant. For that, if the motor is operating at half the V/Hz ratio, one doubles the rotor frequency, by increasing the slip.
The net result would in fact be constant torque, since the primary V/Hz ratio and hence the flux has fallen to 1/2, but the rotor flux is kept constant, but I haven't found any sources mentioning anything about the slip, so for the time being, this is just me putting 2+2 together.

Adding to the problem is how reluctance will play here. Not only the V/Hz ratio is increasing, but the impedance of the motor (XL) is also increasing. More impedance = less current.

I have a similar question relating the output power of a motor, related to the bus voltage. The reason I'm wondering about this is because I would like to know how the motor's performance is compromised - and not just top speed - if the bus voltage is reduced. This will be a significant help in my battery calculations.

Theory says P = IV. Great, so if we hold current constant, we should get a simple linear reduction in power if we linearly reduce voltage - right? Where this seems to fall apart is exactly where it should be textbook easy. If you're lucky, the motor supplier or controller supplier will provide a torque - speed curve based on a voltage. They may even provide a power - speed curve as well.

For my motor, I'm relying on the data produced by ORNL's testing of the 2010 Prius. For the Prius motor, the torque output is pretty much linearly ( in fact about 1:1 ) related to current. In the region before the base speed, torque output is constant. BEMF has little to do with this because it's almost negligable at low speed - about 10V or 20V at 500rpm. Additionally, they did all these tests at the same switching frequency (5kHz) and used their own inverter so they knew inverter output wouldn't affect the data.
Motor stator resistance is very low ( about 0.7 Ohms L-N ) and the inductance is fairly low at about 1mH. ( Ld and Lq are different - I can't find exact numbers) For all the tests, efficiency was in the 90% range.

Now for the strange part:
At 650Vbus, the torque output is 200Nm from 0 to about 2000 rpm.
At 500Vbus, the torque output is 150Nm from 0 to about 2000 rpm.
At 225Vbus, the torque output is 120Nm from 0 to about 1000 rpm.

This is not linear. Any suggestions about what is going on?

-E*clipse

[P-hack]

@cs, not really sure what the problem is there I guess. More slip is more torque, up to the breakdown torque. Slip above or below rated slip has worse power factor. Is all that taken into consideration in your tables?