The thing is though that the steppers won’t be getting a constant 12v(or whatever your input voltage is). You’re not meant to do the ohms law calculations in this situation, that’s the stepper driver’s job. You set the current that you want it to aim for and it will chop the voltage at high speed to try and maintain that current.
For example take your motor with a coil current spec of 1.5A at 4.2V. That means your motor’s coil resistance is 2.8 ohms. Also assuming a Pololu DRV8825 with .1ohm sense resistors.
If you set up VRef to 0.75V then you’ll get 1.5A max. At the time that 1.5A is flowing, 4.2V will be applied across the motors coil (V=IxR) (4.2=1.5×2.8)
If you set VRef to .5V then you’ll get 1A max. When the 1A is flowing, 2.8V will be applied across the coil. (2.8=1×2.8).
The motor won’t actually get fed the full input voltage constantly (12V in this case) as that would damage it. (I=V/R) (4.29=12/2.8)
That would end up with over 4 amps flowing and that motor would be toast pretty quick.
You’re right about the motor being hot with the VRef turned up. At it’s rated current a stepper motor can be quite hot. I guess in this situation it’s all about finding the balance.
All that said, that’s just the theory on how to properly adjust the current limit. We don’t necessarily need the motors running at their full current/torque either. We just need enough torque to mill at the feed rates we want without skipping steps or having issues.
- This reply was modified 7 months, 1 week ago by C.