General Motion Control

Bipolar chopper drive circuit IV: understanding how torque is produced

Author: Bob Parente | May 22, 2012

With this posting we’ll take a look at the torque that is produced by step motor.We now know, through the past several blogs, that we can control the current in the motor’s winding in very small increments by using a bipolar chopper drive. We also know that the current in one phase follows a sine wave and the current in the other phase follows a cosine wave, then we can micro step the motor.

Figure 1: Microstepping cosine wave

The sine, cosine current wave forms are shown in figure one along with a step resolution of four micro steps per full step.

Let’s assume that the windings are energized at the sine equal to zero degree point where the current in phase A is zero and the current in phase B is 1.414 amps. I’m going to assume that our one amp, one millihenry, one ohm can produce 100 oz-in of torque.

Now let’s put a torque wrench on the motor’s shaft and turn it clockwise (CW) and take torque measurements as we force the shaft out of its stable position. We’ll repeat the experiment in the counter clockwise (CCW) direction too.

Figure two below shows the results of our experiment.

Figure 2: Stable torque positions

The motor’s stable “holding torque” position is on step number 1b.

Remember that we haven’t stepped the motor yet, we just energized it.

If we don’t force the shaft out of its stable position, the motor generates no torque.

If it did, it would rotate. Torque, when the motor is stationary, is only generated by the difference between its stable-energized-stator position and the location of the rotor or the shaft position.

The torque that is generated when the shaft is forced away from its stable position follows a sine wave. The peak torque takes place +/- one full step away from the stable position (Positions 4a and 2b.)

We get positive torque if we force the shaft in a CW direction and negative torque if we force it in the CCW direction. In both cases the shaft is trying to get back to its stable position 1b.

If we go beyond the one full step “error” the torque begins to decrease and reaches zero torque when its two full steps away from its stable position. (Steps 3a and 3b.)

These two points are extremely important.

  1. Maximum torque takes place +/- one full step away from the stable holding position.
  2. Zero torque takes place +/- two full steps away from the stable holding position.

Now for a trick question:

Our stable position is 1b. If we move the shaft CW to position 3b and let go of the shaft, will the shaft move CCW back to position 1b?

There are a couple of things that make this question tricky.

The first one is that if the shaft is released exactly on the zero torque value of position 3b, then the shaft would just sit there and not move. No torque, no motion.

The second thing is that if the release position has a slight positive torque then in fact it would return to position 1b. But if it had a slight negative torque, then it would scoot over to position 1c and be as stable in position 1c as it was in 1b. Do you remember from a previous blog posting that there are 50 stable positions for each full step and the motor is happy to be in any one of them?

So the answer to the trick question is: we don’t know.

An important item to note here is that any time the rotor gets more than +/- two full steps out of synchronism with the stator’s commanded position the motor will stall and loose synchronism.

In the next blog posting we’ll continue to take a look at the torque the motor produces and how much each micro step produces and see where this might get you in trouble if you don’t fully understand how it works.

 

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