Reply To: 0.9 vs 1.8 degree steppers?



I’m not trying to cause anyone to get panicked and none of this is meant to come off as confrontational or anything so please don’t take it as such. I’ve nothing but respect for your project.
I’m not saying that the steppers won’t work if you do it that way, I’m just saying that you would not be getting the full benefit of them if you calculate the current in the way that you are saying.
They’ll run fine but you’ll be running them at only part of what they are capable of. These stepper motors are current driven, so if you are not reaching the rated current then you will not get the rated torque. Also stepper motors do always run at full power even when stationary, the motor coils are constantly energised to develop their holding torque.
It’s quite safe to run stepper motors hot also as long as you don’t exceed the rated temperature. My motors when running at the rated spec would be quite hot (as I said, too hot for the PLA brackets).

e.g. My steppers are specified at 2A per phase, 2.8V with a coil resistance of 1.4 ohms.
If I was to use your method and assume that the driver would feed 12V to the motor the I would end up with 2.8Vx2A=5.6W —->> 5.6W/12V=0.47A. —->> 2 steppers =0.94A
That would work but you won’t be getting the power that you think you are getting. The motors won’t receive 12Vx0.47A per coil. They’ll get an initial surge and then the driver will chop the voltage to control the current. You won’t have 5.6W but a fraction of that.

The coil resistance is a constant so if you apply ohms law, 2.8V/1.4ohms=2A. For the driver to achieve the lower current than you have set it to, it must reduce the voltage across the coil, not increase it.
If you were to apply 12V to a 1.4ohm coil then 8.5 amps would flow.

I’ll verify it when I get my drivers but if you want to verify it yourself in the meanwhile all you need are 2 ammeters.
Do it with only 1 stepper and have it hold position. Put an ammeter in series with the output of your power supply and another ammeter in series with 1 motor phase.
Double the motor phase reading because the 2nd phase will have the same. You should find that the current on the supply side is much lower than the total motor current.
That’s because the input voltage is 12V and the voltage applied to the motor will be lower.