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I am programming a motor controller where I have to specify the run current (coil peak current) and hold current (coil hold current) of the motors. The motor I used is a bipolar stepper motor, with a whopping 1.86Kg/cm of stall torque at 1.5Amp current (0.75A per winding) at 1.8 degree stepping angle.

However, the table in the specification makes me confused: It lists a different current for a different torque.

Here is a link to that motor controller.

How could I determine run current and hold current from this specification?

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  • \$\begingroup\$ Kg/cm is not a unit of torque. I know that datasheet uses that measure, but monkey-see monkey-do is no excuse. Even if they erroneously were using kg as a unit of force, it still doesn't make sense. If they can't even get the basic simple stuff right, what hope is there for the other specs? Also, if the motor fails to produce torque to your expectation, how can you complain? Who's to say the 1 Nmm you get isn't somehow 1.86 kg/cm since the latter is meaningless as torque anyway? \$\endgroup\$ – Olin Lathrop May 1 '12 at 22:53
  • \$\begingroup\$ From your other comment and nonsensical units above I'm starting to suspect maybe you don't actually know what torque is? \$\endgroup\$ – Olin Lathrop May 1 '12 at 22:55
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To determine the current you need, you first have to know the torque you need. That is a function of the external system that only you know. Once you know the torque you look in the table and see what current is required to generate that torque. If you can tolerate a lower torque for holding versus moving, then you can have a lower hold current than run current.

Keep in mind that motion inevitably eats up some torque due to friction, which actually helps for the holding case. The back EMF of the motor will also reduce the apparent voltage applied to the windings as a function of speed. Stepper motors are rather inefficient, so their back EMF is usually not that big. You can measure it yourself by looking at the open circuit voltage when mechanically spinning the motor at a known speed. That's how much less voltage the motor "sees" at that speed than what you apply.

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  • \$\begingroup\$ And that was just holding torque, how can I determine the holding torque. Furthermore, where can I learn the running current. \$\endgroup\$ – Kevin Q May 1 '12 at 20:05
  • \$\begingroup\$ No, that wasn't just for holding torque. As I said, the torques are up to you anyway. We have no way to know what torque you need. Once you know the torque, whether holding or running, you can determine the current as I described. \$\endgroup\$ – Olin Lathrop May 1 '12 at 22:47

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