Stepper selection - voltage confusion.

Im getting really confused about stepper motor selection.

I am building a new large printer using the ramps 1.4 and DRV8825 driver from a 12v supply.
I already have a mendelmax using this setup except with standard A4988 drivers and nema17s

It looks like both drivers use the main Ramps 1.4 voltage (12v) as the motor drive.
OK so far, except the nema17s supplied with the mendelmax kit are rated at 3.7v.
I guess that is what the pot power adjustment is all about, finding the right drive power?

Anyway I am trying to select a nema23 size for my new project and have the following choice.

Specifications:
Model No. Voltage Current (phase) Inductance (Phase) Holding
Torque
V A mH Nm
SY57STH51-1008B 9.24 0.7 32.8 1
SY57STH51-3008B 3.1 2.1 3.6 1
SY57STH56-2008B 5.04 1.4 10 1.24
SY57STH56-3008B 3.15 2.1 4.4 1.24
SY57STH76-3008B 4 2.1 6.4 1.85

I thought I was quite good with electronics running 3.7v motors from 12v seems a bit wacky.


(sorry cant get the table formatted properly)

Im thinking the lower one is the best choice.

Any advice welcome.

Edited 1 time(s). Last edit at 11/05/2013 08:24AM by Vince.

I would not go much above 9V, but going below 5V has very little advantage with normal step rates and inductance / resistance numbers. That of course assumes that they don't pull more than the ~ 1.5A the drivers can put out (think about the normal dual motor Z axis) and that they put out enough torque.

Its just ohms law where current = voltage/resistance this is (almost) all. Usually V and A are given, but can deduct directly the resistance and even wattage.

A dc motor has this voltage rating on it and it is the value to run the motor at, because it only has dc in it, no controller, no other current limiter, current stays dc with same polarity. So the coil resistance does the magic thing and naturally limits the current (the current's natural target or the value it aims at, but still). The coil can have any other resistance, and if it would have zero resistance, would put the source in short.

A stepper motor does not use technically dc, the driver input gets dc, but output is say ideally square wave with some frequency reversing the polarity, which is not dc, but actually its more like ac. And the driver also limits the current, so the coil resistance doesn have to. Because of this, the coil has some freedom to have much lower resistance than an equivalent dc motor would. And this is beneficial, because the current rises much faster if its resistance is lower than previously. And the coil doesnt get burned because the driver senses and limits the current before it reaches any dangerous limit.

So both for the dc and stepper that voltage is the voltage given by the coil resistance and an estimation of coil current (for dc is rating, for stepper is more like max). For dc it is a voltage to run the motor at. But for stepper, you are much better if you actually run the motor a a voltage much higher than that.

And the reason we do want the current to rise as fast as possible, its because we run them at frequency, and current needs to rise fast at a good preset value before dropping out again. Its about the power. The flux is proportional to this current, so no current = no power transferred to rotor. And if the current doesnt rise high enough before the driver needs to reverse it, the overal power in the rotor goes down. So this is why the coil resistance needs to be much lower than one of a dc motor. As low as possible. This is the same thing with saying that we run the motor at a much higher voltage than labeled. Because current is v/r, it grows for this relation, if voltage gets higher or resistance gets lower is same thing, the current doesnt care much.

I would choose the ones with lowest coil resistance, and which for 12v have a max current (the labelled one) that is higher than my intention for the setting, because that is how they will run. Also, as a side note, its sort of ac, so instead of resistance would be more correct to talk about impedance instead, sort of R+s*L, but thats just unnecessary complication and we can technically ignore coil impedance for this purpose.

You only need to have a motor that runs at printer speeds. Getting a very low voltage motor with a very high current requirement is not a good idea. Motors with 0.5 to 1.0 A currents are out there cheap. Anything much over that is going to make your drivers a lot less happy.

The label current value is not a "requirement". Its rather a maximum current, e.g. from the coil wire diameter and perhaps something from the motor's magnetics. So the actual driver setting has to be lower than label current, thats the point, not the other way around.

So for example if a motor is labeled 1.2v and 2.4A, that is quite good to run at 12v and 1A driver setting. IMO, much better than 1A label running 1A setting.

If you are after a specific amount of torque and the motor says it supplies that amount of torque, it does it at the rated current. If you put in less current, you get less torque. If you buy a motor that puts out 2X (or 5X) the torque you actually need, then sure, you can put a lot less current through it. You probably pay more for a higher torque motor, it's likely larger and heavier, it may (or may not) be the best design approach.

Derating current is fine. There's nothing wrong with it as part of a rational design. When you do, that also drops the drive voltage required for a given level of performance. Half the current will always require half the voltage ....

My $.02.

A friend of mine bought his printer as a kit. The extruder gear was a printed ABS part and his motor was, IMO, undersized, I don't recal the exact rating. It would get so hot after about 20 minutes of printing that the gear would melt and start to slip on the shaft. I don't recall if he figured it out or if he just put a fan and heat sink on it.

I use oversized 2.5A steppers but I have the problem of keeping my electronics cool so my drivers don't shutdown or lose steps. I think a good compromise would be something in the 1.4 - 1.8A range, won't get hot under constant use and won't tax your drivers as much.

Gyv

There are two very different issues with heat:

1) The heat in the driver as it tries to put the right voltage / current on the motor
2) The heat in the motor it's self.

A given torque motor normally comes in a variety of windings. They all will accept the same amount of energy (volts x amps) to get the job done. Some are higher current / lower voltage for a given amount of energy others are lower current / higher voltage for the exact same level of energy. Any time you cut the energy into the motor in half, it's going to cool off. That's true if it's a 0.5A motor or a 25A motor. It's the volts x amps that counts, not just the amps.

The driver is trying to switch current to deliver it to the motor. The switches are primarily a resistive sort of thing. Their heat will go up by the current squared. Cutting the motor current in half will likely cut the driver heat by a factor of four. The typical driver chips are not heat sunk as well as they might be. Anything past about 1.5A seems to be a bit much for them. In some cases the magic limit is 1.2A. Different suppliers and

At the speeds we run, you can easily get the job done with a motor who's max current rating is more like 0.5A than 2.5A. The driver chips will be much happier. You can still cut the current back (to say 0.25A) and have half the energy into them to keep them cool. You will have more volts into them than into the 2.5A motors, but not more than the normal drivers at normal voltages can supply.

Quote

undersized

What's undersized isn't defined by the printer type, but by the accelerations and speeds you want to achieve. In the case you described I'd first reduce motor current to find out how much the printer really needs. 10% less current mean 20% lower motor temperature (in the range where the motor is substantially above room temperature).

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