https://www.faulhaber.com/fileadmin/Import/Media/EN_AM1524_FPS.pdf
I am using this motor with a 250:1 planetary gear drive, I am moving a mechanism that requires .3 inch pounds of force. The company says it will produce the required torque at 3.3vdc and only use .26amps, I am having problems with this giving me enough breakaway force to move the mechanism at temperatures of -40*. Right now the motor is set to 1/4 steps because I was under the assumption that it would produce more torque but, I had another individual that said a 256 step it would give me a more significant breakaway torque but at speed it would be less likely to produce the same torque.
I am not all that smart when it comes to this but, I am asking for your advice. The motor will only run 12 seconds from hard stop to hard stop and the accuracy isn't all that important due to the gearbox but I do need a good amount of torque when cold. Would 256 steps work better at cold vs 1/4 steps?
Best Answer
You would need a special oil the gearbox for that temperature. You can refill it by yourself, but you need the specification of the oil and the exact amount of it. And of course you would need to disassemble the gearbox and dump the lubricant, clean the gears and then reassemble and fill the new lubricant.
Also, make sure you use the correct stepper driver.
EDIT:
With increasing microstepping value the torque drops a quite lot and if the value is too great it could happen that the motor can’t produce enough torque to even turn. Usually 1/4, 1/8 or even 1/16 can produce satisfactory smooth movements while still producing enough torque. source
Note that when the current in phase A is at max, the current in phase B is zero when microstepping, meanwhile in full step mode each phase outputs the maximum current. So the output torque is smaller when using microstepping.
Let we have two currents: Phase A and Phase B that are geometrically shifted by 90 degrees. The depicted is a vector sum that has an absolute length of \$\sqrt{2} \cdot I_{phase}\$, this is how full stepping would look like. The position of the vector can only have 4 different positions for one electrical turn.
In microstepping these two phase currents are modulated, so that you get a rotating vector that has many positions. The currents do resemble a sine wave, so imagine these two vectors changing, so that the resultant vector draws a circle. It can be deducted, that a resultant vector has an absolute length that is for the factor \$\dfrac{1}{\sqrt{2}}\$ smaller than the previous vector length of the full step. Therefore the microsteping torque is only 70.7% of the full step torque.
It clearly says that the specified holding torque output is referred when both phases are ON - full stepping.
Your description is missing:
If the supply voltage isn't high enough, then the motor will stall as soon it will start to move. If the environment temperature is always such low, then you could increase the current above the nominal current. With some smart stepper drivers you have the ability even to set the motor current at zero speed, this would be practical for you - example: you set a small current at zero speed, and then a large current for moving. This setup would prevent burning your motor even if you accidentally start without having a motor cooled at -40 C.
Note that for a motor AM1524 0250 (Vn=3.5V, In=0.25A) you would need a supply voltage 12V to 15V, with current capability of at least 0.5A (theoretically). Practically you need more current, since the driver chopper has to be adjusted for 0.25A per phase, in this case the supply current is on its limit. If the torque isn't enough and you need to boost it, then you would need double current. I would start with 12V @1.5A.