2013-04-24 06:21 AM
Hi,
I'm exploring the EVAL6470H with my stepper motor.according to the motor electrical parameters, the INT_SPEED should be around 2000 steps/s. The problem is that the L6470 has a maximal limitation of 976 steps/s for this parameter.So if I want my application to work at 6000 steps/s, I cannot get a steady current while accelerating to that point (I can, though, ''trick'' the parameters so I would get the wanted current at 6000 steps/s).Why do you limit the INT_SPEED parameter so low, and do you have a workaround solution for me, or is it impossible to use this motion control with my motor?Thank you. #l64702013-04-29 02:49 AM
In this case you can set the INT_SPEED parameter to the maximum value.
If you need more details to the voltage mode, you can consult the specific application note .RegardsEnrico2013-04-29 03:24 AM
That is what I do, but since this is not the real INT_SPEED of my motor, I can't get a steady current in the acceleration/deceleration phases, and I have to calculate myself a workaround ''final slope'' that will give me the desired current at the desired speed (for every wanted speed, a different ''final slope''!).
2013-04-30 04:35 AM
I see.
Can you post the parameters of your application and motor (target current, supply voltage, Rm, Lm, ke, etc)?Enrico2013-04-30 07:46 AM
Sure.
target current - 1Asupply voltage - 24VRm - 6.2 OhmLm - 1.88 mHke - 0.009 V/HzThis gives an INT_SPEED of ~2100 step/s, but I'm limited to ~1000 step/s.If I want to run at 6000 step/s for example, I can calculate a workaround final slope that will give the wanted current at that speed (by intersecting the graphs), and I get a steady current up to 1000 step/s, but from 1000 to 6000 step/s I get an overcurrent since the final slope starts too early.Thank you.2013-05-06 02:03 AM
Ok,
I made some calculations.First of all also using the ''ideal'' compensation curve the device reaches the 100% of duty cycle before the 6000 step/s speed (at about 4600 step/s). That meas that above this speed the voltage sinewave applied to the motor is not further increased.Try this configuration and let me know how it work:kval_acc = kval_dec = kval_run = 0x42st_slp = 0x06fn_slp_acc = fn_slp_dec = 0x0bint_spd = 0x3fffEnrico2013-08-29 02:33 AM
Hi Enrico,
Sorry for the late response.You are right, and I do encounter a decreasing phase current starting from around 5000 steps/sec due to lack of BEMF compensation (driver is maxed out).I understand there's nothing I can do to resolve this issue, except for using higher motor voltage (which I don't want to do currently).Will the L6472 be able to drive my motor with the same specifications? or would it also be unable to reach the target phase current at high speed?Thanks,Ran.2013-09-04 07:45 AM
Hi Ran,
The different control mode implemented in the L6472 could give you a bit of extra current (and torque). However the limitation is due to the supply voltage, so it is still present. Enrico2013-09-12 02:00 AM
Hi Enrico,
I tried to play a little more with the settings of the L6470 (using the SPIN eval software) to see if I can get better results at high speed, and I found out that if I change the Step Mode to Full Step I do achieve the desired phase current at high speed.However, if I use anything other than Full Step while setting Full-Step Speed to a value much lower than the intended speed, I do not get the wanted phase current (although it does change to step mode).This is a bit unexpected, as I would assume going to full step after passing the Full-Step Speed would behave the same as using Full Step mode from the start.This finding also show that electronically it is possible to achieve the wanted phase current at my desired speed, it's just a driver issue.I would appreciate your feedback on my finding.Thanks,Ran.2013-09-22 02:00 AM
Hi Ran,
When the device switches between microstep and full-step using the Full Step Speed threshold the amplitude of the voltage squarewave (Full-step) il limited in order to avoid discontinuities (see respective section in the device datasheet). So your finds are correct: the automatic full-step gives a lower voltage (and then a lower current) that the normal full-step. Enrico