Back driving a CIM

[Nyle]

I think he's asking how much torque it takes to spin a CIM that is not receiving any voltage, which would not be the stall torque. I don't see this anywhere on the datasheet for the CIM, although it might be possible to calculate it from data in there. Unless someone on here has already measured it and has the value available, you'll probably have to just measure it yourself. I'm not sure if connecting it to a speed controller would change the value or not, I believe that would depend on if/how the motor leads are electrically connected when the controller is set to zero. I'd guess that they are not, but I don't know exactly how they work. What do you need this for? For most cases I can think of where you might need it, approximating it as zero would probably not make any significant difference.

[Richard Wallace]

Free speed drag torque for a CIM can be calculated based on data sheet parameters as follows:

Free power at 12V = Free current x 12V = 2.5A x 12V = 30 Watts

Effective resistance = 12V / Stall current = 12V/130A = 0.092 Ohm
Winding loss at free current = 0.092 Ohm x (2.5A)^2 = 0.57 Watt

Drag power at free speed = Free power - Winding loss = 29.4 Watt

Drag torque at free speed = 29.4 Watt / (5300 RPM x pi/30) = 0.053 Newton-meter

I can confirm from measurements in my motor lab that this theoretical figure is pretty close to what we really get, after the motor warms up a bit. Initially the drag (and the free current) are a bit higher because the grease and the brushes are a little stiff when cold.

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Back to the OP: yes it matters a lot if the motor is connected to a speed controller. The calculations above give drag at zero current; when the CIM is connected to a speed controller its induced voltage (back-EMF) will excite the controller and that excitation will cause current to flow, charging up the controller's capacitors and creating additional drag torque. Anyone who has pushed an FRC robot and seen the little LEDs come on knows what I mean.

[cad321]

Relating to the op's question, would disabling 1 motor in a 3 cim gearbox and continuing to power the other 2 cause any damage to either the speed controller on the unpowered cim, or the unpowered cim itself? An example of a practical application for this would be in a 6 cim drive, if the current draw exceeds a certain amount, disable one motor on both gearboxes to lower the overall current draw.