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Stepper Motors and Ohms LAW

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cannonfodder

New Member
I'm having a bit of a problem running an old 12VDC stepper motor and need some help! The only markings it has are "12V - 3.5W" I've measured the resistance of each winding (it's a 4 phase unipolar) and they each measure at approx 10R.

3.5W/12V = 0.29A... but... 12V/10R = 1.2A!!!!

I know I need to "current limit", but every system I've tried, simply lowers the voltage down to something useless (<3V) in order to keep the current at the correct value.

The only method that sounds feasible is PWM... but I don't want to go down that route unless I have to!

Thanks again for any help!

Cannonfodder
 

Rescyou

New Member
Math

Some possibilites:

This could also mean the whole stepper is rated at 3.5W. So if I'm thinking right that would mean you add up the resistances and divide the voltage by them or: 12v/40ohm = 0.3Amps which is pretty close to your calculation of 0.29Amps

or maybe divide 3.5w by 4 which gives .875 divided by 12v = 0.073Amps

Don't go by my word though, wait for the experts here to assist ya.

Resc.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
I don't see as you need to current limit when feeding a 12V stepper off a 12V supply. Current limiting is usually done when feeding the stepper off a much higher supply rail - so if you were feeding your motor off 48V you would feed the windings through resistors to limit the current - the high voltage would increase it's performance.

To current limit at 12V (if needed), you could simply use constant current sources in place of resistors.
 

cannonfodder

New Member
But surely passing 12V through a 10R coil would try to pull 1.2A. It's the difference between the calculated current flow and the maximum printed on the motor that has me confused. I can't connect the motor to 12V without it getting very very hot (and eventually burning out). Hence the current regulation. I've read all the web articles and found info on all sorts of clever driver ICs but OHMS law still haunts me!

Thanks again... and... keep helping :)
 

Nigel Goodwin

Super Moderator
Most Helpful Member
I've just been looking at specs on stepper motors - they give the winding resistances as well, 12V ones vary as low as 20 ohms (0.6A), but a 5.3W one is listed at 47 ohms, and a 2W one at 120 ohms.

Where did the stepper motor come from?, I've often taken them out of 5.25 inch floppy drives - I've never measured coil resistance, but they work fine on 12V.

Your confusion with ohms law is because you are expecting 12V to the motor, with the current limited - this is impossible - ohms law is nice and simple, and it works - ALWAYS!. If you limit the current to the motor you are simply reducing the voltage to it. I would be inclined to try running it off 5V or so and measure the current it takes, and seeing if it runs warm, particularly when it's locked in a stationary position.
 

cannonfodder

New Member
The motors came from a Mitsubishi RM-101 robot arm... (to be honest... they're still in the robot)... I aquired the arm, but the control and power board is missing. When you examine the base unit, all you find is wires!!!

My new interface circuit is based on the UCN5804 and does indeed drive the motors but the performance is terrible. I assumed that when running from 12V directly, the current and power (not to mention the temperature) seem to be very high.

How "hot" should they run? Is 12V 10R 1.2A OK?
 

Nigel Goodwin

Super Moderator
Most Helpful Member
cannonfodder said:
How "hot" should they run? Is 12V 10R 1.2A OK?
Well 12V at 1.2A is 14.4W, if the motor is rated at 3.5W it looks rather excessive. However, presumably a robot arm has to be able to lift things, so a stepper motor is going to require a fair bit of power to do so.

What voltage are you using with the UCN5804?.
 

cannonfodder

New Member
My first circuit used 12V (straight from the orignal PSU) but I noticed the current draw and heat was high. Since then I've tried just about every voltage.

Not only that... there are 6 motors, assuming that only one phase (in each motor) is ever on at any one time, thats...

1.2A x 6 = 7.2A total (not including the logic and other stuff)

The transformer that's still in the robot base doesn't look to me like it could deliver 7.2A...

I guess I just don't know enough about stepper motors (or at least these ones)

Silly question... but how would you power a 12V Motor with 10R windings with a Phase rating of 3.4W? Am I missing something? :?

Thanks again for your help (It's keeping me from my real job)
 

Nigel Goodwin

Super Moderator
Most Helpful Member
cannonfodder said:
Silly question... but how would you power a 12V Motor with 10R windings with a Phase rating of 3.4W?
You don't! - 10R at 3.4W is only 5.8V. As you've already suggested, you could really do with information from a complete one - a service manual would be even better!.
 

cannonfodder

New Member
Strangely enough... I've already requested any available info from Mitsubishi... They sounded helpfull but I've yet to get anything back from them. I'll post any info I get from them.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
cannonfodder said:
Strangely enough... I've already requested any available info from Mitsubishi... They sounded helpfull but I've yet to get anything back from them. I'll post any info I get from them.
I look forward to hearing how you get on, it sounds like fun anyway!.
 

falleafd

New Member
http://www.cs.uiowa.edu/~jones/step/

I think you should read this tutorial on stepping motors. It's a very important tutorial for beginners.

Good luck..

I think the best way to measure stepping motor current is that, you apply 12VDC to your motor, and then measure the current on a winding. Try it.


By the way, you have to distinguish the 2 pairs of wires 1a, 1b and 2a, 2b by apply a AC supply. For example, your motor is 12VDC, you may apply a 4 or 5 VAC to the centre wire and one of other wires. Measure the voltage between the center wire and 3 others. you will see one of the voltages is 4 or 5 VAC, and two seem to be zero. The wire from which you get 4 or 5 VAC and the first wire is a couple. Two other is a couple.

I tell you this, because you said you use a old motor and I'm sure it's difficult to find the datasheet.
 

nics

New Member
Please help with stepper circuit

Hi there,

I need help here. I am a sort of a self taught elctronic circuit designer and are still doing things that works but I am not sure why :D

My problem is the following: I have a mineba astrosyn 23LM-C309-10 miniangle stepper. On the motor it is specified that the motor is a unipolar 4-phase motor that runs on 6V/phase, 0.85A/phase with winding resistance of 7.1Ohm.

What baffles me is the following questions I have:
1) Does this mean it is a 24V motor, in other words do I have to supply the motor with 24V.
2) In the datasheet it says you have to put a 24.7 Ohm resistor in serie with both commons and supply 48V. How does this work, why not supply 6V straight without any series resistor?
3)How many amps maximum does this motor draw?
4)I am using a UCN5804B to drive the stepper and therefore needs to know what is the most eficient way to connect the motor to the UCN5804 which can handle a max of 1.25A. Maybe 24v into each common with a 12.35Ohm resistor in serie with each?

I do have the datasheet in PDF format if you want to have a look at it!

Regards

Scratchbuild
 

motion

New Member
1) Does this mean it is a 24V motor, in other words do I have to supply the motor with 24V.
No, you don't have to use 24V. You may use any voltage up to its insulating rating. The higher the voltage the faster the motor can run before the motor loses all torque. However, you can only apply a maximum of 0.85Amps/phase. This means that at 24V you have to insert a resistor in series to limit the current. This is called an L/R drive. This is very inefficient. A chopper (PWM) drive applies a pulsed voltage to limit the current.

2) In the datasheet it says you have to put a 24.7 ohm resistor in serie with both commons and supply 48V. How does this work, why not supply 6V straight without any series resistor?
A 6V supply will not give good speed performance.

3)How many amps maximum does this motor draw?
0.85Amp/phase. Normally a maximum of two phases are turned ON at a time. An R/L drive will consume a maximum of 1.7Amps at any voltage. A chopper (PWM) drive will have reduced current draw at higher voltage.

4)I am using a UCN5804B to drive the stepper and therefore needs to know what is the most eficient way to connect the motor to the UCN5804 which can handle a max of 1.25A. Maybe 24v into each common with a 12.35Ohm resistor in serie with each?
Yes. Although you should consider using the chopper drives from the same manufacturer:

http://www.allegromicro.com/ic/motor.asp#unipolarstepper
 

nics

New Member
Hi there,

Thank you for the reply.

I went back and made some calculations and it does not look good. :cry: So what it boiled down to was that I could reduce the schematic to the attached image. This however means if I want to run three motors I must have a 24V 9A power supply to supply power to the motors. Or should it be a 72V power supply?

Even worse, if i want to add a fourth motor I may need a 12A supply?

Does my logic make sense or do I loose myself along the way?

I had a look at the PWM drivers and would look at them in the future, but at the moment I do have the UCN5804's which I would like to use for my first try. So if I make my calculations correctly I do understand why the PWM drivers may be better.

You said it is an RL circuit but I replaced the L of the motor with an R in my deduced schematic. I sthi allright?

regards

nic
 

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motion

New Member
I went back and made some calculations and it does not look good. So what it boiled down to was that I could reduce the schematic to the attached image. This however means if I want to run three motors I must have a 24V 9A power supply to supply power to the motors. Or should it be a 72V power supply?
It's 1.7 amps per motor (0.85amps x 2-phases). Therefore it's 1.7x3=5.1 amps for 3 motors. 24V power supply may be good enough. How fast do you plan to run the motors?

You said it is an RL circuit but I replaced the L of the motor with an R in my deduced schematic. I sthi allright?
The "R" is the sum of series resistor and the winding resistance. The "L" represents the winding inductance which reduces the current while the motor is running. This is why you need to raise the voltage.

This brings up another way to limit the current using unipolar drivers. You can use 6V supply to supply power while the motors are stopped and momentarily apply a higher 24V during each step. Since it is just momentary, you don't need to use resistors. See the ff. link on current limiting:

http://www.cs.uiowa.edu/~jones/step/current.html#boost

BTW, "L/R" or "L divided by R" is the time constant of the circuit. It determines how fast the current rises when the motor steps. The unit in msec or usec should be smaller than the time between steps.
 
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