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I wanted to ask a relatively simple question:

Currently I am using a 3 phase AC motor that has a Gearbox torque of 5.1 Nm. A 16 teeth gear is attached to the shaft and this is connected to a 72 teeth gear using a gear belt.

So straightforward calculation would be:

Gear Ratio = 4.5 : 1
Output torque = 5.1 x 4.5 = 22.95 Nm ~ 23Nm

I need to replace this motor with a stepper motor with a holding torque of 2.7Nm.

Am I right in thinking that when the stepper motor is unloaded, each step (1.8deg) gives a torque of 2.7 Nm?

If that's the case, using the current gear ratio, would the output torque be 2.7 Nm x 4.5 = 12.15 Nm?

Secondly, if I was to use microstepping (32), does this calculation make sense?

1.8 step angle/ 32 microsteps = 0.05625 deg

Does each microstep angle output 2.7 Nm of torque or 2.7Nm / 32 = 0.084 Nm of torque?

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  • \$\begingroup\$ Why do you have no design specs for load , acceleration, v and x ? Microstep is also reduced torque.! \$\endgroup\$ – Tony Stewart EE75 Nov 19 '20 at 15:46
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Am I right in thinking that when the stepper motor is unloaded, each step (1.8deg) gives a torque of 2.7 Nm?

No. When the motor is unloaded gives exactly zero torque. It would however output a torque of 2.7 Nm when loaded with the same - third Newton's law of motion.

If that's the case, using the current gear ratio, would the output torque be 2.7 Nm x 4.5 = 12.15 Nm?

Yes. Usually the torque is declared for full step operation and usually it's a holding torque, which is much higher compared to the torque at nominal speed, where the AC motor has its nominal torque declared at nominal speed. You have also to include the efficiency of the transmission. The belt transmission has almost 0.9 (90%) efficiency if high quality bearings are used.

Secondly, if I was to use microstepping (32), does this calculation make sense? 1.8 step angle/ 32 microsteps = 0.05625 deg

It doesn't make sense at all. With the micro stepping the torque will decrease for a factor of 0.707 compared with the torque produced in full step mode.

P.S. :

Almost every hobby engineer neglects the datasheet of maximum permitted radial force of the motor, when doing belt transmission. With a small diameter pulley, this is exceeded very soon.

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  • \$\begingroup\$ Hi Marko thanks for the reply. Can you please explain the first comment? when loaded with the same what? WRT the last comment, I wanted to know what the output torque would be if I was using microstepping of 32. hence the calculation in the question. I know that with a gear ratio of 4.5 and microstepping of 32, for a motor with step angle of 1.8deg, the load would turn by 0.0125 deg. (1.8/32 uSteps = 0.05625 deg, 0.05625 deg/4.5 gear ratio = 0.0125 deg) for each 0.0125 deg, what would be the output torque ? thnx \$\endgroup\$ – LabMat Nov 19 '20 at 15:42
  • \$\begingroup\$ Motor torque x 4.5 x efficiency x 0.707 (microstepping) \$\endgroup\$ – Marko Buršič Nov 19 '20 at 15:44
  • \$\begingroup\$ Where do you get the 0.707 from? \$\endgroup\$ – LabMat Nov 19 '20 at 15:50
  • \$\begingroup\$ electronics.stackexchange.com/questions/528591/… \$\endgroup\$ – Marko Buršič Nov 19 '20 at 16:23

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