Motion System Design, Stepper Motor Sizing

Sizing: Ringing and inertia mismatch

Author: Bob Parente | June 9, 2015

I’m sorry if you Googled “ringing” because of a hearing issue and end up on this blog. You should Google “tinnitus” instead or search the WebMD site.

Now where were we? Oh yea, in our last posting we were talking about load-to-rotor inertia matching. We had established the following load-to-rotor inertia ratios:

  • 1:1 load-to-rotor inertia as an ideal ratio.
  • Greater than 1:1 and up to 5:1, with the load inertia being greater than the rotor’s inertia, as good.
  • Greater than 5:1 and up to 10:1 as okay.
  • Greater than 10:1 try to avoid.

1:1 – snappiest performer

A 1:1 load-to-rotor inertia is the snappiest performer when compared to the other ratios. It has the best move times and the ringing at the end of the move is minimal.

Figure 1: Inertia matching and ringing
Figure 1: Inertia matching and ringing

Ringing defined

Ringing is an unwanted back-and-forth motion when the motor stops. If you back up to some of the earlier blog postings you’ll see that we described the holding torque of a stepper motor to be sinusoidal in shape. If you missed that, and I can’t imagine why you would have, go back and read it or to simplify things here I’m going to say it’s linear and looks like an upper case “V” The bottom point of the V is a fully energized stepper motor’s stable holding position. When the motor is at rest and no outside forces are interfering with its position the motor generates no torque.

If it did, the motor shaft would spin. Physically twist the shaft by hand Clockwise (CW) and the torque that the stepper generates is represented by the right side of the V. A slight twist CW results in a torque value that is part way up the right side of the V. Twist it more in the CW direction and the torque value rise even higher. The stepper resists the outside twisting and tries to twist the load back to the bottom of the V. Twist the shaft counter clockwise (CCW) and the torque runs up the left side of the V and again the stepper tries to twist the load back to the stable state at the bottom of the V. Got it?

Now physically twist an unloaded motor’s shaft CW until its one full step out of position. The value of the torque will be represented by the upper right of the V.  Let go of the shaft and the motor “springs” back to its stable position, but overshoots the stable position by some value and runs part way up the left side of the V. The torque on the left side of the V then forces it back the other way. This “ringing” goes on back and forth between the CW and CCW sides of the V until all the energy is dissipated and the motor settles into its stable position at the bottom of the V.

I don’t know of any applications that have a motor operating with no load, so let’s add some load inertia. Let’s assume we have a 1:1 load-to-rotor inertia and we do the same experiment that we just did with no load. Do you think the motor will ring for a longer period of time than it did with a no load? How about a 5:1 or a 10:1 load-to-rotor inertia? I think you can see fairly easily that a higher load-to-rotor inertia ratio is going to have a larger effect on the magnitude and duration of the ringing

More next time.

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