Reading stepper motor datasheets to get torque and speed

Question

I don't understand how to use the information in a stepper motor datasheet to understand how much torque it can generate. I was going to do a simple board and try to lift a small weight as an example but I realize I need to know how much torque the motor can generate. I'll probably have a little spool out there to hold the string for the weight, which probably also changes how much torque it can deliver.

All I see in this datasheet is "in traction torque" and I'm not sure what that means. I don't see any graphs or other information that relate how much input voltage/power will give me how much torque. What happens if I run it at 5 V instead of 12 V, for instance, or if I current-limit it to 1 mA?

I'm just trying to learn what to look for and how to read the datasheet properly.

Answer 1

@StainlessSteelRat linked to a very good resource for stepper motors at All About Circuits, but I fear he didn't address your questions. I'll go through your question line by line.

I'll probably have a little spool out there to hold the string for the weight so that probably changes how much torque it can deliver too.

First, the sentence above is wrong. The radius of the spool will not change the torque. It will change the weight you can lift, but only because $Fr=T$, where $F$ is the force available for lifting, $T$ is the torque output, and $r$ is the radius of the spool. BTW, this method works fairly well, I have done it myself.

Anyway all I see in this datasheet is "in traction torque", and I'm not sure what that means

Based on the torque curve from All About Circuits, it is probably almost equal to the holding torque. As you see from that curve, torque is fairly constant at the low end of motor speed. All About Circuits also mentions that in stepper applications, the speed of the stepper motor should be gradually accelerated.

What happens if I run it at 5V instead of 12V for instance. Or if I current limited it to 1mA.

If you run stepper motors at a higher voltage (within rated limits), the current, and therefore holding torque, will go up. Another way to look at it is that you can run the motor faster for the same torque. I don't have any equations here, but increased current leads to increased force in electromagnets. If you current limit the solenoid, the holding torque will go down.

One last note: @am304 makes an important note about half/quarter stepping. The torque will reduce for half/quarter stepping because the magnets in the stepper motor are acting in opposing directions, reducing the net torque.

Sorry I don't have any equations, but experiments are fun right?

Answer 2

You are considering it from an electrical engineering perspective. Speed vs torque. It's more a digital motor with a rotor position that can be easily controlled (read head on floppy drives).

It is a brushless motor and its strongest torque (traction or holding) occurs when the coils are energized. They have fractional hp ratings (>34.3 mN·m). To do something in the real world, you typically need some gearing. This one does: 64 5.625° steps.

I extracted the following speed torque curve from Stepper Motors, which is a fairly good resource to understand the theory. with the rearing, I'd expect the holding torque up to 120 Hz (120 revolutions/sec).

So you can do something, but not much. I'd check out Brushless DC (BLDC) motors if you want to do something.

Answer 3

I would say that the information on the datasheet it incomplete, you need at least a torque vs. speed graph for the rated voltage, and whether it is being driven at half steps, quarter steps, etc... You can then scale the torque based on the voltage applied. See for example this datasheet which provides that sort of data.