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Stepper Motors. Any Experts?

MadMechanic

Aug 28, 2012
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I've been learning and trying to reverse-engineer the logic, nature and anatomy of a stepper motor.

These motors come with certain specs as far as Vmax, A/phase and so forth. I have learned that a 1.8degree motor is 360/1.8 = 200 steps in the motor.

when trying to calculate desired RPM you use PPS (Pulse Per Second). in this case if we choose a desired 500RPM for the motor to output... the math would show:

(500Rpm / 60sec-min) = 8.33Rps.
Then
(200steps-in-motor-per-rev X (8.33Rps)) = 1,666 PPS.

My question is: If I want 500Rpm out of this motor....and I need 1,666 pulses per second to achieve 500Rpm, how is PPS a function of Voltage, Amperage and resistance? What would be the draw of the motor in and V and A at 500Rpm and 1,666PPS?

:eek:
 

Harald Kapp

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I'm not, as you request, a stepper motor expert. But as far as I can judge PPS is not a function of V,I and R at all. PPS is just a number that tells you how many pulses the motor needs per second to achieve the desired rpm.
What voltage the motor needs, what current it draws and what the equivalent resistance of the motor coil is should be stated in the motor's datasheet. You need to form the pulses according to these specs.
Obviously a small stepper motor for a floopy disk head control (you know floppy disks? :) ) surely has other parameters than the stepper motor for a big robot drive.
 

MadMechanic

Aug 28, 2012
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I'm not, as you request, a stepper motor expert. But as far as I can judge PPS is not a function of V,I and R at all. PPS is just a number that tells you how many pulses the motor needs per second to achieve the desired rpm.
What voltage the motor needs, what current it draws and what the equivalent resistance of the motor coil is should be stated in the motor's datasheet. You need to form the pulses according to these specs.
Obviously a small stepper motor for a floopy disk head control (you know floppy disks? :) ) surely has other parameters than the stepper motor for a big robot drive.

I understand better now. And oh yes I remember the old floppies! The first apples came out when I was in gradeschool. The 5.25-inch versions. I can remember those PC's loading for decades and the disk drive sounding like a meat grinder just to play "The Oregon Trail" game. :cool:

Given that PPS is basically a number corresponding to number of steps, is there another way to calculate rpm given the specifications shown here? (http://www.applied-motion.com/products/stepper-motors/ht17-075) -spec tab at bottom left.

Listed is the V/phase and A/phase and R. Would I assume that if this motor has 200steps and I met the PPS equivalent to the RPM of 200 steps that the motor would only draw the listed voltage shown in the data?
 

BobK

Jan 5, 2010
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=MadMechanic;1495690]Given that PPS is basically a number corresponding to number of steps, is there another way to calculate rpm given the specifications shown here? (http://www.applied-motion.com/products/stepper-motors/ht17-075) -spec tab at bottom left.
QUOTE]
Do you understand that the stepper motor is driven with a sequence of pulses alternating among the different phases? There is no RPM specification for a stepper, becuase the RPM is set by the frequency of the pulses driving it. The current draw is also going to depend on the frequency and shape of the driving pulses as well as the voltage.

The voltage needed when running at high RPMs is much greater than the coil resistance * the current.

Bob
 
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MadMechanic

Aug 28, 2012
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QUOTE]
Do you understand that the stepper motor is driven with a sequence of pulses alternating among the different phases?

Bob[/QUOTE]

True, I understand, but if there are 200 steps in 1 revolution, and we are operating the motor at 1 full step interval continually to provide continuous motion of the shaft at the motors rated voltage, what would be the rpm? I am basically trying to make the distinction between the number of steps in a the motor, and how the number of steps in the motor is related to continuous motion @ the given voltage and amperage rating with no load thus giving me its maximum rpm rating. How is one to know this?
 

BobK

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No, I don't think you do understand. The RPM will depend only on the frequency of the pulses. If you don't apply pulses the RPM is zero. Each pulse (actually sequence of pulses to the different phases) rotates the shaft 1.8 degrees. If you apply 200 pulses per minute the RPM is 1, because the 200 pulses will rotate the shaft by 200 * 1.8 = 360 degrees. If you apply 20000 pulses per minute, the RPM IS 100.

The voltage and current have nothing to do with it, except that you will have to supply enough current at a high enough voltage at any given RPM and torque.

There will be some maximum RPM after which the motor will not run any faster, but that will depend on the torque required and voltage and current available.

Bob
 

MadMechanic

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Basically what Im trying to figure out is how RPMout is a function of Pulse, Voltage, or current In. If I could calculate this somehow then I would know how much RPMin I need to get V and I out for something like this:


The main reason I need to know RPM-out/RPM-in @ a given voltage via the motor's specs. is because I need to know what pulleys and gears I need for the correct ratio. Unfortunately these motor mfg's dont list rpm, only V, I, R, Mh, per phase. Thus I have been trying to reverse engineer the nature of this and figure out how pulses and frequency relate to input voltage or amperage so that I dont have to buy gears or pulleys 5 times to get the right speed. I think I am just going to buy various motors and experiment.
 

BobK

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So now you link to a video where a motor is being used as a generator. Don't you think it would have been helpful if you had mentioned before that you wanted to use this as a generator, not as a motor?

Bob
 

MadMechanic

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So now you link to a video where a motor is being used as a generator. Don't you think it would have been helpful if you had mentioned before that you wanted to use this as a generator, not as a motor?

Bob

Power in = Torque out as Torque in = power out. Is that helpful? Or do you need to watch this:


If I know what RPM the motor makes at a given voltage, then I know what input rpm I need to get a (near) output voltage +-
 
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(*steve*)

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If I know what RPM the motor makes at a given voltage, then I know what input rpm I need to get a (near) output voltage +-

I'm still not sure what you want to do, but I can tell you that it's not as simple as you seem to think it is.

The output voltage of a motor being used as a generator depends on more factors than you think. The (electrical) load on the output, for example.
 

MadMechanic

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I'm still not sure what you want to do, but I can tell you that it's not as simple as you seem to think it is.

The output voltage of a motor being used as a generator depends on more factors than you think. The (electrical) load on the output, for example.

True and I have calculated all this, I was just trying to estimate my gear ratios by finding a way to calculate the RPM and Voltage relation @ 0.95A draw. I talked with one of the Mfg's of a stepper and now I know which motor I need. I think the easiest way im gonna figure this out before ordering gears is to hook the motor up, rectify it, regulate it and load it to what I need...and then read the Rpm at that moment. This way I can select an appropriate gear before attaching it to my engine.

Originally I was just trying to reverse the math to figure out a ballpark figure of RPM out at a given voltage, so that I might estimate what rpm would provide said voltage.
 

Jotto

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First you must determine the RPM of the motor in question, then you will have to run this motor above synchronous speed. So if you have a 1600 rpm motor, you will have to run about 1700 rpm to generate a output.

My only question is why use a stepper motor, using a dc motor would be cheaper and much easier. If your trying to use a stepper motor as a generator, don't think your going to get the desired result. DC motors are cheap and plenty of different motors, AC is a lot more expensive. Stepper motors are great for indexing position.

Not sure what you wish to accomplish.
 

MadMechanic

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First you must determine the RPM of the motor in question, then you will have to run this motor above synchronous speed. So if you have a 1600 rpm motor, you will have to run about 1700 rpm to generate a output.

My only question is why use a stepper motor, using a dc motor would be cheaper and much easier. If your trying to use a stepper motor as a generator, don't think your going to get the desired result. DC motors are cheap and plenty of different motors, AC is a lot more expensive. Stepper motors are great for indexing position.

Not sure what you wish to accomplish.

True, I will just have to adjust the gear ratio accordingly to match the rpm I need.

The reason I am using a stepper is because steppers generate much more voltage at low rpm compared to normal DC motors. I have experimented with other DC motors, one having a (lower than normal) Kv value of 236Kv, so to get 12V out of it I need to turn it at 2,832 Rpm. That motor has a greater amperage rating than a stepper, but takes more input.

From what i've researched, I should be able to get the same or more voltage by turning a stepper motor at 200 to 300 rpm and wiring the 2 phase output across a voltage doubler circuit.

The project objective is to obtain enough DC power after rectification and regulation, to meet the 5V 950Ma rating of standard USB power for charging USB devices. The other reason for trying to use a stepper is the scale. My power plant is a 1/2" bore micro steam engine with a maximum rpm of around 500. I want to be able to obtain unconventional dc power from fire and water to charge devices and batteries. :D The best I can figure the engine is probably capable of 40 watts output. I want to use a stepper for the lower rpm so I dont lose much torque due to increasing the speed of the output gear.

So in the end, I basically started this thread to see if I there was a way to calculate Rpm through the math and given values of the motor. I know it varies with voltage and at this point I think my best bet is to attach the stepper to another DC motor on a PoT and slowly increase speed until I read the value I need across the rectified stepper output circuit.
 

Timescope

Aug 30, 2012
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Hi MadMechanic,

Stepper motor drive circuits are quite complex and use dedicated ics to control the current and voltage waveforms. In your case, I would suggest that you download the datasheet of the ic that controls the stepper motor you are reverse engineering.The datasheets of L6219, PBL3717A and TEA3718 stepper motor drivers also have detailed information.
You can build this 74194 circuit to experiment.
http://forum.sibnet.ru/index.php?act=Attach&type=post&id=186973
Useful tutorials can be found at
http://homepage.cs.uiowa.edu/~jones/step/index.html
http://www.stepperworld.com/

Timescope
 
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duke37

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As I understand it, MadMechanic is intending to use a stepper motor as an alternator. Experiments would be needed to determine the voltage/speed constant. There will be no synchronous speed considerations.

The current output and voltage sag with load could be esimated from winding resistance.

There will be loss in the rectifiers, this may be more or less than the rubbing brushes of a DC motor. Two phase full wave rectification will give a smoother than normal single phase output.

It will be interesting to see the outcome.
 

MadMechanic

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Thanks for the linkage and comments guys. I am pretty sure this is going to work, I will just have to continue testing. I've become obsessed with the project and with winter coming on I should have a bit of time to iron it out.....Today I started thinking about the exhaust of the steam engine and the concept of a coiled exhaust system coupled to the hot side of a thermoelectric generator plate.

This way I can harness the heat lost as the steam condenses back to liquid and convert it to extra energy! the question is... how do I incorporate a lower thermoelectric current back into the output power of the alternator? It would be a lower voltage than the alternator but would it be wired in series or parallel? The quest continues. seems like I keep adding more and more to it, but the end result should be pretty cool. I am hoping to fit the entire unit into a .50 caliber steel ammo box. I will post a good video up to youtube when I complete this.

Hopefully it will charge the iPhone and other devices with no problem. I am going to modify the resistance on the signal wires of the USB out so that the iPhone will only only draw 500mA instead of the 1A the proprietary apple charger pulls.
 

Timescope

Aug 30, 2012
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I just ran a stepper motor type 17PM-M024-G1 with my electric drill. I got a sine wave of about 80v pk-pk off-load at 166 Hz frequency from one winding (yl-wh). With a 68 ohm load it produced about 30v pk-pk.

Timescope
 

Merlin3189

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...sine wave of about 80v pk-pk off-load .... With a 68 ohm load it produced about 30v pk-pk.
Timescope
Which would imply an internal resistance of about 113 Ohm.

As Dave says, the basic generator model is, an emf proportional to speed and an internal resistance in series. And the only realistic way to determine the emf/speed ratio is to measure it as you did here.
What you also found is that the internal resistance is higher than you would expect for an alternator, because the stepper motor is designed (among other things) to provide precise control of motion rather than efficient conversion of energy.
So while you can use them as alternators, they may not be well regulated or efficient.
 

BobK

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I am not an expert on alternators, but wouldn't the voltage drop Timescope saw be more due to back EMF when current is flowing in the coils rather than resistive?

Bob
 
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