overcoming holding torque

Also shown below are one set of 5 graphs during a position move. The x-axis is in milliseconds. I see that IqRefA maxes out at 15.0.

The second set of 3 graphs shows the motor response when overcoming the holding torque position.

In both cases the current was measured / verified with a scope at power supply.

I don't quite understand the theory of vectored current drive.. Does this by chance suggest that there is actually 15 Amps through one of the motor phases (reaching the peak setting) and other current components in other phases - that altogether sum up out to the roughly 3.8Amps that we're seeing at the power supply draw?

Bernard Pottner said:

Question 1: We've found that when motor is holding zero, or idle position,  - it seems to max out at approx 3.8A current draw (this was verified with a scope). What is triggering this limit?

#define USER_MOTOR_MAX_CURRENT

sets the limit on the input to the Iq_Ref_A.  Try increasing the value you are using and see if you get correspondingly more torque.

Yes, increasing #define USER_MOTOR_MAX_CURRENT allows Iq_Ref_A to go higher. This allowed more current to flow, and correspondingly more torque.


But I am puzzled as to what the USER_MOTOR_MAX_CURRENT should be now..

The motor rated current is 15 amps. But as the testing shows, setting #define USER_MOTOR_MAX_CURRENT to 15.0 limits actual current draw from power supply to 3.8A.

I am guessing this 3.8A maximum is now well below the motor rated current.

Should #define USER_MOTOR_MAX_CURRENT be set to the motor max current?

Does it make sense then that the actual current never goes above 3.8 Amps?

Thanks for your help.

see if this thread helps
e2e.ti.com/.../1441564

Bernard,

Everything you have posted seems correct. If your motor is 32 oz-in / A, and you are providing 15 A into the motor it will produce 480 oz-in (or 3.4 Nm) of torque. The current drawn from the power supply will be less than the current supplied to the motor. The power supply will be feeding the caps on the DC bus, and depending on how you are measuring the current draw on the power supply it could be averaged over a long period. If you really wanted to confirm the current running into the motor, you should use a current probe on one of the motor phase lines.

Thank you all for the input. Your time is much appreciated!

I understand what you are saying, but I'm having trouble grasping my head around what is actually happening..

Dave, your enthusiasm in "teaching old motors new tricks" is a gift. I've been working with the sim files you created to gain a better understanding. Some more effort is required there on my part. I was disappointed that I couldn't find Adam's follow-up talk to that seminar..

Back to my lack of understanding. At the moment I'm trying to rationalize what the motor manufacturers' current rating means.

For example, if it is 15 amps. Would that be continuous.
And if so, then my first ignorant thought is, why can't the FOC put out 15 amps continuous.

And in the case for example, where the motor is holding a position;
    - say the rotor is 90 degrees out of phase with respect to winding A
    - then I might expect to see a full 15 amps on winding A
    - No BEMF, no sinusoidal switching

But then I started thinking, how would the inverter actually inject 15 amps, my guess;
    - by applying a PWM duty cycle that results in 15 amps being measured through shunt resistor
    - which would be a pretty small value, since no BEMF, current only limited by phase resistance
    - so within one PWM period, a small fraction of time, I am getting 15 amps, majority of time, inverter switches OFF
    - so yes, average current, and PSU current signficantly smaller

This kind of makes me wonder;
    - if the PWM frequency was doubled, I might expect 2 short pulses of 15 amps in equivalent time period, effectively twice the power

I'm sure there is something I'm missing here, back to the basics of motor theory.. I'll try to get a better understanding to get these questions resolved.

But for now, thanks again for your input.

Hi Bernard,

I'm glad you liked the "Teaching Old Motors New Tricks" seminar.

With regard to motor current rating, different manufacturers do it in different ways.  Perhaps the most common rating for AC motors is RMS current.  But when we say 15 amps in our FOC code, we are talking about 15 amps peak.  That means that you may or may not see 15 amps on any motor winding when you are standing still, depending on the electrical angle of the current waveform.  But it is true that if you set the max current limit in InstaSPIN-FOC to be 15 amps, then the sinusoidal peak of the phase current waveforms will be 15 amps (assuming of course that your power supply can deliver the voltage and current necessary to support it).

To look at the current waveforms, I don't recommend looking at the shunt signals, as they are discontinuous representations of the phase currents.  Instead, use a current probe or something which allows an in-line measurement of phase current.  Also, don't forget that under normal operation, within one PWM period, the inverter never "switches OFF".  Assuming no dead-time, whenever one transistor for a given phase turns OFF, another transistor for that phase turns ON.  And another effect which complicates the analysis is that current can flow in the reverse direction through a transistor, whether it is turned ON or OFF, due to the back-body diodes.  As a result, currents can recirculate within phases inside of the inverter which never show up as part of the bus current.

I have a SPICE simulation which shows all of these effects, which may help you in understanding what is going on.  It runs on LTSpice, which is a FREE simulation tool.  I can send you the simulation if you are interested.

Best Regards,

Dave

 

Great! it's starting to make sense - thanks for all the input.

For now I have plenty of simulation files to help clarify. Found on your "Simulation Power" blog at e2e.ti.com/.../simulation-power

As well some dropbox items mentioned in the thread "Controlling Low Side FETs (DRV8301-69m-kit)" at e2e.ti.com/.../1628199

All the best.