Which stepper to buy: 68oz 0.9degrees or 70oz 1.8degrees

I've been looking at 3d printing for a while now and finaly decided to take the leap and get into it (Can't wait!!!! ).

I'm looking at building the Rostock: [reprap.org]
The build platform isn't exactly large so I think I'll scale the build platform up from 200mm x 200mm to something around 350mm x 350mm. We'll see when we get there

I'm stuck on a buying decision: I don't know which motor to get.
Here's the ones I'm looking at getting:

#1 [www.sparkfun.com]

Step Angle (degrees) : 0.9
2-Phase
Rated Voltage: 3V
Rated Current: 1.7A/Phase
5mm Diameter Drive Shaft
Holding Torque: 48N.cm
NEMA 17 form factor

Pro's: its cheaper for me, as it will cost me $102NZD for 5 of them with free shipping. 0.9 degrees Its more precise 1.8 .
Con's : its 1.7 amps max, and I'll be using a 2amp driver so it could over heat? less torque. Will I need all that torque? I do plan to print fast

#2 this one: [www.trademe.co.nz]
5x Wantai motor Nema 17
Holding Torque: 4800g-cm(70oz-in)
Phase: 2 (4 wire)
Current / Phase: 2.5A

Pro's: More torque, more speed.
Con's its $145 for all 5 of them. Less resolution than option #1.


So the question is pretty much can I use motor #1 and will it be fine? Or do I need more torque?


Advice is much appreciated.

Edited 3 time(s). Last edit at 11/06/2013 02:50PM by TheTechnicalNoob.

Well the two motors you are looking at both have the same torque to within the round off error in the numbers. Neither one will make a standard Ramps driver board happy. I'd vote for "none of the above". To make the normal driver chips happy, look for a motor down around 0.5A to 1.0A. It also should have a winding resistance that will let you to DC drive it with 9V or less.

I understood its good to have a motor with more amps than your driver, that way the motor is never loaded to its full power meaning you can't overheat the motor right?
The drivers have overheat protection so should be fine I figured..
My power supply is adjustable from 13v to 17 so I'll be running it somewhere in that range.


Why exactly would it not make the driver happy?
Perhaps a better question would have been, how strong motors will I need for my Rostock?

Edited 2 time(s). Last edit at 11/06/2013 09:39PM by TheTechnicalNoob.

No offence uncle_bob but are you sure you are right?
You do know that when the motor is rated at 2.5Amps its not magically going to suck 2.5Amps out of a 2Amp stepper driver. Or is it? Actually its what I'm trying to find out..


EDIT:Ah but that is what the adjustable pot on the driver is for, its for determining how much power it will give the motor. So it should be ok. Should be mmm ^ ^ Advice anyone?

EDIT: I just read on a cnc forum that it will be fine with a motor that has a higher rating than the stepper driver. [www.cnczone.com]

Edited 6 time(s). Last edit at 11/07/2013 12:43PM by TheTechnicalNoob.

The current on the motor is the amount required to deliver the rated holding torque. You do *not* want to exceed the rated current on your drivers. You can always reduce the current into your motors so your drivers do not melt down. That will reduce their holding torque. At some point (half current, maybe less) you will not have enough current for them to accurately step at a reasonable rate. The normal drivers don't like much over 1.2A or so. The max you could do with your motor is 1/2 current. Remember that the holding torque is always greater than output when stepping quickly. Static holding is a "best case" situation.

There is no advantage to a very low voltage / high current motor that poorly matches your supply voltage at the stepping rates we normally use. You are trying to drive a constant sine wave current through an R + L load. The power into the motor is either max or nothing. It's fairly easy to go through the math and see what the result is. (I can give you a pointer to the math if you need it. It goes on for a while).

If you want to do high power stepping at very fast rates, that's something different. You will need high power drivers to get that done. You will also need a machine design that will hang in there with major dynamic loads. You will also need fans on your motors ....

Power supply wise, you would want a supply around 8 volts for these motors rather than 13 to 17V.

So it looks like I'm going to have to find some different motors that are in the range of 0.5 to 1.0 amps.

Could you explain how to calculate what volts to run a x volt rated stepper motor?
That'd be very helpful thanks
Perhaps it would be easier to give me some links to do some further reading instead of typing it all up.

The reason I wanted powerful motors is I want to be able to move the head of the rostock I'm building pretty fast but not too fast. If a smaller motor would do that just fine then I'm happy

To quote Nophead: "We normally use 2.8V motors on 12V." [forums.reprap.org]
So he's running his 2.8v motors on 12v. That contradicts what you said earlier about needing to use 8v. ??? :confused:

Edited 8 time(s). Last edit at 11/07/2013 06:16PM by TheTechnicalNoob.

You *can* try to run a 100A motor that is rated for 0.01V. Your drivers will probably start cutting out at 1.2A so you will have 1% of it's design torque available to do things. If you have a motor that has 100X the needed torque (and it still steps) that's a fine solution. It's also a waste. You have a 12V supply, why put a 3V motor on it?

Here's the math:

Take your steps per mm on the motor. (Marlin firmware config.h)
Take the max rational speed (say 100 mm / sec) (experience)
The steps and the speed will give you the revolutions per second. (math)
Every time you complete 4 steps you go through one cycle of micro stepping drive. (Allegro or TI or Sanyo driver chip data sheet)
You are diving the motor at revs/sec x steps / 4. For a 1.8 degree motor that's 200 steps / 4 = 50 times the revs per second.

The driver chip puts a constant current into the motor (Allegro or TI or Sanyo data sheet). The curet is set by the Vref.
The normal driver boards have a chip rating of 2A maximum (data sheet again).
The normal driver boards have a thermal cut-out / self protect. That cuts in after extended run at about 1.2A (many user's experience).
The normal driver boards are not super duper well designed for thermal performance (50 years of doing this stuff for a living).

From your motor data sheet you can get three basic numbers:

1) The resistance of the winding
2) The inductance of the winding
3) The rated current through the winding

You will find that the rated current and resistance give you the voltage of the winding (E=IR, ohms law).
The reactance of the winding is 2 * pi * frequency * inductance in Henries (EE 201)
The magnitude of the impedance of the winding is square root of reactane squared plus resistance squared (EE201)
The voltage on the winding will be the current times the magnitude of the impedance.

Most of these motors have an inductance in mHy that's about 1.4X the resistance in ohms regardless of what the specific numbers are. That gives them all a common point that the reactance starts to matter.

You can go through the numbers for a number of motors and you get a fairly consistent result, If you have about 30% more voltage than the "voltage rating" on the motor, you can drive it just fine with the typical chips.

Next up is the temperature rating on the motors. They have what's known as a "class B" insulation rating for the most part. That's a 135C rating. Its measured by an odd technique (winding resistance change) so the real inner temperature is a bit higher. When the inside of the motor is 135C, it's outside is pretty warm. That bothers many people. They tend to cut the current back because of it. If you cut the current in half you loose half the torque. You also need half the voltage. Your 3V motor is now a 1.5V motor ....

That's the first layer. There's a lot more, I suspect that's enough for now.

just go for the standard [www.aliexpress.com]

$74.61USD or $90NZD(approx) including shipping

Know to work well on repraps, and no issues getting to New Zealand

Re 0.9 vs 1.8 degrees..

If your running a standard Cartesian (I2, I3 etc) with 1.8 at 1/16 micro stepping using gt2 belts and a 16 tooth pulley = 100/mm ie 10micron resolution
the same thing with .9 you get 5 micron resolution. Ie you will never notice the difference. Especially when you consider the average nozzle, .35mm which extrudes about .52 ie 520 microns.

Now if your building a Delta (which on re reading I see that you are) they have variable resolution across the bed. A 0.9 do apparently help to increase resolution when in its low resolution areas.

NB if you use a 0.9 you have to use twice as many steps/mm so you do lower your max speed compared to a 1.8

Edited 6 time(s). Last edit at 11/08/2013 01:15AM by Dust.

So here's another layer to this - why all the low voltage motors?

We stopped using 12 or 15V for logic supplies a long time ago. Today even 5V for logic is becoming less common. People like to design stuff that uses a single supply. That means 5V or 3.3V in a modern design. If they are running a stepper the low cost design will be one that runs the steppers off of the same supply as the logic. If you are doing direct PWM control it also saves you some money on the electronics.

Volume of manufacture matters when it comes to the cost of something like a motor. There are a lot of surplus motors from various projects out there. They didn’t cost much when they were new and they cost even less as surplus. If you go looking for motors, the low voltage / high current ones are what you will find very quickly. Because of this, they are the ones that everybody talks about. There are probably 100X more 3 and 5V motors out there than anything else. They get mentioned because they are common / low cost, not because they are the best fit for us performance wise.

We are not (for several reasons) running 5V high current drivers on our steppers. We are running drivers that were made for high(er) voltage low(er) current motors. You can make a low voltage motor run with them, but there are significant compromises. The high(er) voltage motors are out there. If you are paying a bundle to ship things in, it’s better to get ones that are a best fit for the drivers.

Thanks for the feed back!

Dust, thats a great find! Thanks! I'm pretty new to all this, I came across your blog as well, very nice. I'm also near christchurch as you mentioned on your blog. Maybe I should visit you and say hi.

Uncle_bob the motors should work (it works for other reprappers right?), the drivers will just fine as long as I have a heatsink on it. Thanks for the explanation.


Aliexpress gives me the following when I try place an order: Sorry, nothing is available to order now. At the bottom of the page its says: This product can't be shipped to the destination selected. Please contact the seller for details.

Dust have you had this before?

Btw I've changed my mind as I've come across the 3DR-delta and really like the design, which means I'm going to need some spectra 90lb fishing line. I'm planning to scale it up a bit. Dust do you have a few meters I could buy from you?

Edited 7 time(s). Last edit at 11/08/2013 02:06PM by TheTechnicalNoob.

I cant get that error you see.... I've ordered from them before. but not recently.

I don’t have any spectra, Im a firm believer in belts. From what I read on the IRC channel Spectra stretches to much... (ie you need to tighten every 3 months or so)

Both motors in the original post are good. Coil resistance is 1,8 ohms and 1,25 ohms respectively.

The voltage rating on the label of the stepper motor is not the voltage the motor is intended to run at. The whole point is to run the motor at a voltage much higher than labelled, because this makes current rise faster and that means flux rises faster and that means power transferred to rotor is better. If the voltage is too low, and the motor rotates fast, the driver needs to reverse coil polarity before the current reaches the Ipeak set point, so the total power becomes much and much less at speeds. Which happens anyway, meaning torque will go down at speeds, but the point of running voltage much more higher than label, is exactly to help with that.

The voltage rating on a dc motor is the rating the motor is intended to be run at, while the voltage on the stepper is the voltage which together with coil resistance give a rated current that is more like a maximum that we dont actually care about, because the driver takes care to limit that anyway. And the driver ic will burn before the coil wire, so not even second failmode thoughts.

Thanks people! That answers my question. Or at least, I'm satisfied

I ran some tests on my stepper motors. The motors are rated for 2.5 amps for a rated max torque of 4.8kg/cm. My drivers are A4988 compatible 2amp max and I'm running them on their max amps with heatsink and a fan over top of that.

Well you know what. The max torque I get out of them is 2.69kg/cm. I'm not impressed.
Uncle_bob, you told me 2.5 amp motors with 2amps stepper drivers isn't a good idea. Should have listened.
Motors like these would have been better: [ultimachine.com] But I didn't know of the site at the time..

So there you have it people, buy lower amp, higher voltage motors, like good'ol uncle bob told us too

...... or get something that puts out more drive than the little stepper boards. There are a bunch of $15 to $20 driver modules that will work quite well with those motors. Wiring them up is a bit of a mess. I don't know (yet) how well they work. Should know in a few weeks ....

I think you are assuming that if you would put same waveform of voltage/current, you would of got almost double more torque from the other motor. I dont think thats can be the case, because you would need a different law for conservation of energy. Lets keep in mind we talk about same technology, same construction even same form factor, same parameters and no corner cases. If some datasheet says you get @1A the same power you get with another motor @2A, well i suspect that is more marketing than engineering. Or maybe some methodology difference. The flux depends on current. There may be some differences with geometry, materials permittivity, choice of other mats like magnets, but we look past that. Those may count for some, but not so fundamental as double the output. What you get is ultimately what you put in (settings) * some efficiency, and the efficiency part doesnt double without some radical change.

I suspect that its not just you who gets low power from their motors - i would say its everybody. For a 12v&1a setting we start with an ideal 12w if it would be dc, except its not, lots of rise and falls and off times. Also the stepper tech is much worse in efficiency than dc motor - e.g. why steppers dissipate so much heat etc, so while its hard to advance a number for output because its a dynamical parameter, its safe to say its much much lower than the original 12w. So we are talking about just a couple of watts in mechanical domain, its certainly not impressive.

What is different in your case is the bed size, making it almost 40x40cm it means its 4 times heavier than the normal 20x20cm. So the motor needs 4 times more power than the normal printer. I know this first hand because i have this 35*40cm and i run it with nema23/2.5a steppers and my own a3977 drivers which are more powerful and tweakable +double belt mod which increases the power. I thought it would be better, but still just like you, i am still not quite happy with it even in these conditions. It makes quite a noise and its so heavy, plus bed wastes so much heating when printing small parts. It has to be felt in practice to understand all that, although as a concept was initially quite attractive. From this construction i learned that for the best speed and acc settings axes should minimal weight and equal weight (x=y), and only the heaviest matters in acc and speed limitations. Only now i see the benefits to have the x+y together like how is ultimaker and now i understand why makerbot copied that construction after their initial printer which had separate x and y. If i make myself a next one, will be more or less like that, x+y together.

Flux in an electromagnet depends on current, permeability, and number of turns in the coil. More turns with the same current / core and you get more flux. That assumes you don't saturate the core. If you are talking about redesigning a motor for 1 vs 2A by changing the number of turns, the core likely isn't an issue. It's mostly a matter of filling the same winding area with the same amount of copper - divided into smaller strands.

What i am saying on short is, apparently the OP is unhappy with the choice, thinking a 1,7A motor would give him approx double power than the 2.5A motor, which is highly unlikely regardless of what datasheets or ebay says. Besides, the holding torque isnt so relevant do dynamic behavior anyway but thats beyond the point. Then i tried to say that power isnt inherent to the motor max current, but otherwise inherent to driver settings. For example take a nema17, nema23, and nema34 motors, if all are used with same driver and same settings, they will give about same power regardless of being different motors. Because output ultimately depends on what is being put in. Beyond this, in some motors can put more current, in some not so much. Therefore while higher current is likely higher torque, still is not enough to change the motor, would need to change the driver, or at least upper driver setting. Also what he could do in the situation is to use double belt aka MadKite belt mod. The bottom problem is the bed is 4x the size and probably same times more weight than normal. Not in that the motors have too high current rating. Imho the latter is probably more fortunate, at least provides a margin if needed.

I have done the same thing as OP, in the link i gave e.g. 4x bed size, and it has its flaws. Has to be felt in practice to understand. Its much heavier, vibrations make more noise, harder to heat up and wastes so much heat when printing small parts - quite inefficient. Beyond this, about motors in this application, ultimately i think this is due to the construction in which x and y being separate, increase of the bed size means bed is heaviest, which limits both axis. Printer designs like ultimaker which have the x and y together in same constriction (if thats correct english word), i think those models scale up in bed size much better because it doesnt add the bed weight to either x or y - those only add the weight of axis increase and that is likely lesser.

The very fundamental problem is that people seem to be looking just at current and not at voltage or resistance (or max torque). Current by its self means absolutely nothing at all. You need the rest of the data. I can *easily* design a NEMA 17 motor that will blow away you 2.5A gizmo when mine is running 0.1A..

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uncle_bob
The very fundamental problem is that people seem to be looking just at current and not at voltage or resistance (or max torque). Current by its self means absolutely nothing at all. You need the rest of the data. I can *easily* design a NEMA 17 motor that will blow away you 2.5A gizmo when mine is running 0.1A..

All circumstances aside. I'll try say what these parameters mean to me. These could mean differently to everyone else, with a margin that is normal, for everyone sees things differently. Voltage/resistance is the factor that gives the final value in the rising edge formula. What is important for the stepper to function as it needs to, is to have that factor several times lower than what ohms law would give, that is for the purpose to have the actual rising edge engaged in its first part, where it rises fast e.g. di/dt is high, and not in the latter part where it rises painfully slow. Resistance with inductance gives the time constant tau at the coeficient of e^(-t/tau) and thats their impact. I think i should care for rise time specifically, simply put, because there is lots of it. Beyond that, after the Ipk has been reached, input voltage isnt what it normally would be. You said those words believing the power relation still holds like in an normal dc load, but i think it doesnt. Actually after the Ipk is reached, the *input* voltage doesnt matter in the way you expect it to, because the driver job is to limit the current. It does that in a what that you can say that either the driver limits the voltage down to fit ohms law for the given coil, or probably bettter you can say the driver chops it (e.g. "chopper" driver). In power switching to get the average is applied some principle named volt*second balance (e.g. webbers). So it doesnt matter what the voltage is at input, the one going through the coil will be dynamically chopped off, and will average to a different level, hence we cant simply multiply the input voltage with the current to get the coil power state. If we want to do that in comparable terms we would need to apply something like volt*second balance. And in case of higher voltage input then duty will be smaller and then you can see it doesnt go the way of increasing input voltage to get proportinally more power. We do say things like that sometimes, but as an abstraction or oversimplification for the purpose of highlighting something else. However increasing the voltage does bring an important benefit, and that is it betters voltage/resistance ratio that is the final value (the natural point would be reached if no intervention) in the curent rise formula so the current increases much faster than it would increase with low voltage, and this way also contributes to improving the average. But ultimately this is very beneficial exactly for the purpose of helping with maintaining torque at speeds.

Maybe some motors may have a "max torque", that is they have a bad behavior at start and low rpm, then they get to a nice torque plateau, then at higher rpm they again behave worse. E.g. like a automobile gas pedal or something. In stepper motors i see nothing similar as a principle. Some things that do that are missing, and the last thing that could do that is rotor inertia but characteristically too low to for the large scale impact, though may be seen its effect on some graphs. One of the things given, holding torque is somthing i dont care for in my printer, because the x and y are horizontal, and may care for if z would had belts, but its on self blocking screws and z motors are off between slices - this may be different with rostocks or milling heads that get a force exerted on head. Instead, what we would certainly care for is in any 3d printer is dynamic torque, that is a dynamic parameter that starts from a detent torque value that is the torque from one full step position to the other i guess like in a solenoid. Also a note about it, normally this should of mentioned methodology or its measuring params, otherwise its not applicable or comparable directly. Instead, this is exactly why they give it as a max, that isnt like peak of a parabola, but instead is just the max start point under best conditions - and after that things only go downhill. And then at speeds, this torque keeps decreasing untill at some speed, the torque tends to zero. And that happens because current cant rise fast enough. So if there is a parameter that goes down, then it goes down more or less depending on its critical conditions, which determines where we end up. We could have a good start and be in trouble mid way, or we could have a lesser start but still be in good shape mid way.

Here's the disconnects:

Even if you could put a several thousand volt driver on the motor and run an amp through it no matter what, it's torque wold still drop off. The hysteresis and loss in the magnetic materials used in the motor are the limit here. As the frequency goes up their losses go up as well. All of your added input simply goes to heat. The easy way to look at that is the angle between the voltage and current. You can directly measure what's going on.

In a stepper the torque starts off at the holding torque and pretty much just drops as rpms (or frequency) goes up. Your "dynamic torque" has an upper bound in the static torque. That's what makes static torque a very important parameter on a motor. That's why you can always find it specified.

A motor is a *very* slow thing. Compared to electronics motors simply don't move. Put in a more technical fashion a stepper motor's torque drops off as you try to run it faster. You could get the current up to spec in a nanosecond and the motor would just sit there for a *long* (in nanoseconds) time. You have acceleration and jerk limits on controllers for exactly this reason. You can't take a motor from zero to 10 rpm in zero time. You *slowly* go from static DC to stepping. The only time you ever have high voltage on a stepper is when you are running high RPM's. Except on some very unusual Delta's we just don't run high rpm's.

You can go through the acceleration math and the feed rate math to confirm all of this.

A simple example:

Motor is at rest.
I want to go 10 rpm
I immediately step at 10 rpm
Motor slips

Motor is at rest
I want to go 10 rpm
I slowly accelerate to 10 rpm
Motor does not slip.

In this case "slow" is in electrical terms. I'm accelerating the motor as fast is it can in mechanical terms. On an 200 mm travel at max 200 mm / sec, I spend 1/4 of my time accelerating or slowing down. 250 ms is a *long* time in electrical terms. As the motor starts and finishes it's at low RPM's. Low RPM's mean low drive frequencies. Low drive frequency means that you are completely in the ohms law region.

Edited 1 time(s). Last edit at 12/23/2013 11:22AM by uncle_bob.

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uncle_bob
Here's the disconnects:
Even if you could put a several thousand volt driver on the motor and run an amp through it no matter what, it's torque wold still drop off. The hysteresis and loss in the magnetic materials used in the motor are the limit here. As the frequency goes up their losses go up as well. All of your added input simply goes to heat. The easy way to look at that is the angle between the voltage and current. You can directly measure what's going on.

No, i havent said i would put 100v to a stepper driver, i intuitively though that would of been your idea behind the affirmation about you designing a stepper to run at 0.1A and beat "my gizmo" 2.5A one. Changing the driver input voltage has its benefits, but not like that and not in that sense. If you didnt thought about changing voltage, then tbh dunno what else you thought about when saying that. I think this part is over anyway - if not then feel free to expand.

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In a stepper the torque starts off at the holding torque and pretty much just drops as rpms (or frequency) goes up. Your "dynamic torque" has an upper bound in the static torque. That's what makes static torque a very important parameter on a motor. That's why you can always find it specified.

Imo, the starting point for dynamic behavior is technically the detent torque. The word i believe means jumping distance, and thats what i would be interested in, jumping from one step to the next. The holding torque is about static behavior e.g. "holding" position, i would be somewhat interested for an application where the motor needs to hold weight or otherwise counter forces exerted on the head. And even then still not so much: since any forces on head when one motor holds, can also manifest while is moving, and the moving one is lowest, i guess it means i would still be only interested in detent one just because its lowest value, and i would efectivelly ignore the holding one just coz its higher and thats covered.

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You *slowly* go from static DC to stepping.

Hmm, when the motor stays still, thats not DC. Its still switched mode even then. In case you hear noise, thats where is coming from. If you dont hear noise, then its above 30kHz, hence too high for human ear. Now you see why i said i care for rise times because simply put, there is lots of it. Rise times dont count much when motor holds, but switching does occurs at all times, it has to, as its the only method of control. Without switching, current obviously would rise above setting and would burn something, probably the driver ic. But thats too plain therefore i think you must have been refering to something else by that.

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Low RPM's mean low drive frequencies.

I dunno what you mean by this. It makes me think about vfd as in variable frequency drives which are used for stuff like milling heads, but i dont see the resemblance here, so maybe you mean something else.

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Low drive frequency means that you are completely in the ohms law region.

Again, i am at miss. I think this choice of words may be at least partly my fault. I think i remember saying something like that once, but was in some context of choosing the motor values, and we dont care for ohms law about motor's label values because the driver limits the current anyway, its way we can run it at much higher voltage while not caring for the low resistance coil getting burned. This is why there is switching even when motor stays still, because coils would burn otherwise. I think was wrong thing to say, poor perspective/comparison or at least unfortunate choice of words from my part. We may not care for it and even we may sometimes hide it under the carpet, but we are in fact always under ohms law in some form, even if not about label values. Actually its more than just a law, its the very definition of resistance. So there is no practical region or operating regime that is outside ohms law, if we look down hard enough, ohms law gets its finger in all operating points in one way or another. E.g. in the sense that voltage would eventually rise the current to ohms law for a final value, except the driver senses and interrupts the rise edge at the preset current point. I think you meant something similar along those lines. So again i might of misunderstood you and perhaps you meant something else.

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NoobMan

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uncle_bob
Here's the disconnects:
Even if you could put a several thousand volt driver on the motor and run an amp through it no matter what, it's torque wold still drop off. The hysteresis and loss in the magnetic materials used in the motor are the limit here. As the frequency goes up their losses go up as well. All of your added input simply goes to heat. The easy way to look at that is the angle between the voltage and current. You can directly measure what's going on.

No, i havent said i would put 100v to a stepper driver, i intuitively though that would of been your idea behind the affirmation about you designing a stepper to run at 0.1A and beat "my gizmo" 2.5A one. Changing the driver input voltage has its benefits, but not like that and not in that sense. If you didnt thought about changing voltage, then tbh dunno what else you thought about when saying that. I think this part is over anyway - if not then feel free to expand.

The point is that past the point that you can establish the proper sine wave drive signal, there is no advantage to greater driver voltage. That is a very predictable level.

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In a stepper the torque starts off at the holding torque and pretty much just drops as rpms (or frequency) goes up. Your "dynamic torque" has an upper bound in the static torque. That's what makes static torque a very important parameter on a motor. That's why you can always find it specified.

Imo, the starting point for dynamic behavior is technically the detent torque.

The detent torque is simply the torque that will move the shaft with no current through the windings.

The word i believe means jumping distance, and thats what i would be interested in, jumping from one step to the next.

The jumping involved is just the "jump" that you get as the magnets align

The holding torque is about static behavior e.g. "holding" position, i would be somewhat interested for an application where the motor needs to hold weight or otherwise counter forces exerted on the head. And even then still not so much: since any forces on head when one motor holds, can also manifest while is moving, and the moving one is lowest, i guess it means i would still be only interested in detent one just because its lowest value, and i would efectivelly ignore the holding one just coz its higher and thats covered.

Since the holding torque is the number with current through the coils, that's the basis for all the calculations on an operating motor. No current and yes detent torque rules.

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You *slowly* go from static DC to stepping.

Hmm, when the motor stays still, thats not DC. Its still switched mode even then.

When the motor is still there is a constant DC current through the motor. How that current is generated by the driver does not matter. You could use a battery and a resistor to get the same effect.

In case you hear noise, thats where is coming from. If you dont hear noise, then its above 30kHz, hence too high for human ear.

Again, that's just how the driver does it's thing. It has nothing to do with how the motor operates. All it's doing is generating a DC current when the motor is still.

Now you see why i said i care for rise times because simply put, there is lots of it.

They only count if you believe the driver running as an ideal switcher. Due to the Q of the motor coil they are far from an ideal switcher. Since there is no "C" in the system they have some other issues as well. A normal switch configuration has more parts than one of these drivers does.

Rise times dont count much when motor holds, but switching does occurs at all times, it has to, as its the only method of control.

As long as it's DC current the motor will hold. You do not need fast switching to have a DC current.

Without switching, current obviously would rise above setting and would burn something, probably the driver ic. But thats too plain therefore i think you must have been refering to something else by that.

The only thing you need for DC is a PWM pulse width out of the switch. You could be switching at 1KHz with 10us rise times and it would work fine. No need for speed at DC.

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Low RPM's mean low drive frequencies.

I dunno what you mean by this. It makes me think about vfd as in variable frequency drives which are used for stuff like milling heads, but i dont see the resemblance here, so maybe you mean something else.

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Low drive frequency means that you are completely in the ohms law region.

Again, i am at miss. I think this choice of words may be at least partly my fault. I think i remember saying something like that once, but was in some context of choosing the motor values, and we dont care for ohms law about motor's label values because the driver limits the current anyway, its way we can run it at much higher voltage while not caring for the low resistance coil getting burned. This is why there is switching even when motor stays still, because coils would burn otherwise. I think was wrong thing to say, poor perspective/comparison or at least unfortunate choice of words from my part. We may not care for it and even we may sometimes hide it under the carpet, but we are in fact always under ohms law in some form, even if not about label values. Actually its more than just a law, its the very definition of resistance. So there is no practical region or operating regime that is outside ohms law, if we look down hard enough, ohms law gets its finger in all operating points in one way or another. E.g. in the sense that voltage would eventually rise the current to ohms law for a final value, except the driver senses and interrupts the rise edge at the preset current point. I think you meant something similar along those lines. So again i might of misunderstood you and perhaps you meant something else.

You have an R-L circuit. You generate a sine wave current to drive it. All of the switching stuff is just an odd way to generate a sine wave current. You would get exactly the same motor performance if you drove it with a stereo amplifier, a couple of resistors, and signals out of your PC. The switching stuff just confuses things because it applies only to the "amplifier" that's driving the coils.

You are quite right about the differences of detent and holding torques. See, it appears to be hard to admit, but its not that hard, and the fact itself has a great deal of benefits, in other words, its well worth saying. Its a feat of doing stuff that is worth doing, saying that or doing this is equal to learning. And saying "you are right" is as consequence, a feat of ppls that were able to learn from others. Without this, i would be abnoxious, pretending to be always right, trying to play with words to back up weird stuff, and as consequence not being able to improve. Getting out of a state of mind like that is worth anything, any effort. Ofc, this is my opinion.

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uncle_bob
When the motor is still there is a constant DC current through the motor. How that current is generated by the driver does not matter. You could use a battery and a resistor to get the same effect.

First, the current is not generated by the driver at all. The driver is not a current source. The driver just plays with the voltage by its mosfets. The current is "generated" or better said, it grows and decays inside the inductor itself. And that current is anything but "constant". Its switched mode, e.g. pulsating or else, otherwise not constant.

We have 2 waveforms to study, voltage and current. The driver puts the voltage across the inductor, on and off, so the voltage waveform is a simple square wave. The current in inductor by its element law, is defined to be the integral of the voltage by time, so when the voltage starts to be on, then the current starts from zero, and has a rising edge. This rising edge initially grows fast, then tends to grow very slowly. This is because the integral definition makes the grows to be the new addition versus the old one, so after some time it grows very slow because each new portion of addition is relatively smaller in comparison to the bigger "past" integral. In the end this current would tangentially reach the point of ohms law for the voltage and coil resistance, but much before that, the driver will interrupt the voltage. After voltage interrupts the inductor current has a falling edge, that has a different shape but it goes to zero as the voltage stays off. So this is what makes the current sort of speaking.

We can not use a battery and resistor to "get same effect". If we ignore the resistor divider action and that resistor losses, then we are left with a current source which we cant use for inductor, because we would need a voltage source. So what we could actually use is a battery with just a switch instead of resistor. But well it turns out this is what we are actually using, a voltage source that is driver input voltage (e.g. battery), and driver mosfet (which is a switch).

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uncle_bob
They only count if you believe the driver running as an ideal switcher. Due to the Q of the motor coil they are far from an ideal switcher. Since there is no "C" in the system they have some other issues as well. A normal switch configuration has more parts than one of these drivers does.

A switch is simply only one part: a switch is a mosfet or bjt or igbt, etc. Even a simple diode is a switch. Switches dont have more parts, what they do have are quadrants, and quadrants just show on which direction they conduct or block voltage or current. The quadrants help the designer to establish how a switch is implemented, e.g. why we would choose a diode specifically for that spot - see "switch realization". Other parts associated with switches may be things like snubbers, but those are not technically part of the switch itself, just support components or stuff to deal with imperfections of the switch or the load.

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uncle_bob
You have an R-L circuit. You generate a sine wave current to drive it. All of the switching stuff is just an odd way to generate a sine wave current. You would get exactly the same motor performance if you drove it with a stereo amplifier, a couple of resistors, and signals out of your PC. The switching stuff just confuses things because it applies only to the "amplifier" that's driving the coils.

So again, its not a current sinusoid that is generated eslewhere than forced into the coil. Current just grows in the coil as result of playing with voltage, and that is the elemental law of the inducutor. Voltage across inductor builds the flux, to which the current is proportional. Now if you understand that, you will see previous statements differently and, even if most ppls dont see it, some other parts.

Textually, talking of sinusoid currents here is interesting choice of words. We do not take a current sinusoid and put it in the coil - its not how it is, and anyway pure sinusoids are not that easy to come by with h-bridge. There are two perspectives here, one large scale, one small scale. Its ok for large scale to say that under the circumstances like microstepping the current levels will average and one would appear to be like a sinusoid, even if motor positions are discrete levels (e.g. positions are not continuous). This is because to microstep, the driver needs to change the discrete levels from one step to the next, and incidentally this is what a sinusoid is by definition, a smooth transition from one position to its reverse. So even if our driver levels are discrete (not continuous) we can talk about sinusoid in large scale context. In large scale, it appears to be a sinusoid, or actually a level stepped sinusoid, even if once you zoom in enough, then its something else. But to say in the small scale perspective, that you take a sinusoid and put it in the coil to drive the motor - thats a little different. To see this, you can query for example what changes to that sinusoid when the motor is in full step mode.

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NoobMan
So again, its not a current sinusoid that is generated eslewhere than forced into the coil. Current just grows in the coil as result of playing with voltage, and that is the elemental law of the inducutor. Voltage across inductor builds the flux, to which the current is proportional. Now if you understand that, you will see previous statements differently and, even if most ppls dont see it, some other parts.

That is not how an inductor works. Flux is generated from the current. Even in an inductor, voltage generates the current. It does not just grow by itself.

An inductor only generates current "by itself" when you remove the external voltage, and the flux collapses.

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We very much do generate a sinusoidal current to drive a stepper motor. The stepper is being operated as s synchronous AC motor when it's in micro stepping mode. The data sheets for the stepper chips show this very clearly.

At high and constant speed and if you disregard the constant current chopper drive
it might look like a sinus curve. In all other cases such as slow speed and/or with frequent starts
and stops the current will always show the telltale signs of the chopper working. Nothing
sinusoidal in this document for instance: http://www.ti.com/lit/an/snva029b/snva029b.pdf

Anything less than top speed will have distinct flat current levels that jump to another level
at each step. Not a sinus at all unfortunately, it all discrete steps. The actual current also has
contributions from the fact that the rotor (and it's permanent magnets) moving can both take energy
from the magnetic fields or return energy to the windings. Most if not all data sheets show the ideal
case with an unloaded motor, get an oscilloscope and check what happens in the real (and ugly) world.

In the world of printers, we (hopefully) smoothly accelerate and decelerate the stepper. That means we have a varying frequency sine wave that's the drive in the high speed case that this all came out of. The discussion is revolving around a high rpm rather than a low RPM use case. In the low RPM case you do get steps in the current, but you are in what we are calling an "ohms law" region.

Ohms law never plays a direct role in deciding the current through the motor.
The chopper circuits always limits the current to the set level.

The drivers in printers are mixed analogue/digital/mechanical control systems with feedback.
Ohms law is never used to describe such systems.

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Dala.horse
Ohms law never plays a direct role in deciding the current through the motor. The chopper circuits always limits the current to the set level.

If we dig deep enough, i guess something technically correct is something just a little bit different:
"Ohms law never plays a direct role in deciding the peak current through the motor. The chopper circuits always limits the current to the peak set level."

Why it matters, because if the driver would just leave things go naturally, then the current would eventually settle at the value of I(final)=V/R. This means the V/R is the final point in the *rise time formula*. The rise time formula is tought i believe in first year of any EE class or something like that. Ok, so the rise time formula can be in several forms, here i would put it like Fcurrent(t) = initial value * e^-t/tau + final value * (1 - e^-t/tau) , and we can set the initial value to zero then only the second part remains. So the Fcurrent(time) = V/R * (1 - e^-time/tau), where tau is the time constant ofc. This shows current as continuous function of time. This is the function which will tell us current=.... for any point of time, and in reverse will tell how many nanoseconds it takes for the current to grow to 1A or 2A or w/e we set the peak or limit. Put this in an math software like wxmaxima or even a spreadsheet, and see how much time it takes to get to 1A with different values of V/R. So V/R matters that way, because a high value of V/R will get at 1A in short time of nanoseconds, while a small value of V/R will need much much more time to get to same set point. And we are interested to get to 1A as fast as possible <=> e.g. to energize the coils as fast as possible.

What is unrelated to V/R, is the peak, that is simply something we manually set. Ofc that controlling the peak means we sort of control the average, this is the purpose. But to say that ohms law does not apply or that there is a region that is "outside" of ohms law - anybody with an elementary course in electronics can not say that. Imagine that there was a time when ppls only knew about current and voltage. And then a guy came and introduced a totally knew concept, it named it resistance and he said this resistance = V/I. At first didnt belived him, but ultimately ppls accepted this new concept of "resistance". So ohms law is not just a law, its much more, its the very definition of resistance. Without ohms law, we count not even measure the resistance of the coil, we could not have resistance as a concept. Probably we could not do anything and we could not understand anything without the concept of resistance and hence without ohms law. So its not a question if ohms law is there or not, the question is only how deep you need to dig for it, coz at the lowest level its always there, one way or another.

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Dala.horse
The drivers in printers are mixed analogue/digital/mechanical control systems with feedback. Ohms law is never used to describe such systems.

The fact that is not used to describe something, does not mean ohms law is not there. It just means the explanation is one abstraction layer above.
I guess its like that because the ones reading the subject are supposed to already know and understand things like ohms law, rise time formulas, inductor element law, and things like that, which again are the basics of ee. Then it does suffice to say we get an average current, and we deal with this average current and thats all. And there is no need to get too deep into the matter of exactly how that average current is obtained, which is a separate matter to itself, not specific to a certain system, but rather something too generic to be repeated over and over. Talking about same thing at different abstraction levels is confusing, and sometimes a thing that is true at some level turns out to be false/different at another level. Or sometimes its just semantics. But anyway the abstraction process is a tool we need today, because how things grow in complexity. Anyhow, its wise to just be aware these abstractions exist and sometimes they can have their "traps" or contexts or at very least their semantics.