Calculating series load resistors

[whateg01]

I have noticed on at least a couple of websites and in a few posts here that there is an error in the way the value of the series load resistor is calculated. It seems that people are being advised to take the rated voltage for the motor to be used (V) and divide it by the current per phase (I) to figure the DC resistance of the windings. This is good information. Then, to subtract the motor voltage from the power supply voltage and divide this by the winding current. This would be great if these were the only devices in the current path. However, there is a voltage drop across the switching device, whether using FETs or bipolars transistors. This must be taken into account, especially when using power supplies that are closer to the rated voltage of the motor! It should also be noted that a 5V power supply will not supply 5V across the windings of a motor when put in series with the switching device.

That's the summary. For those who have the time and want proof...

As an example, let's assume I have a 9VDC p/s capable of supplying the needed current, and use steppers rated for 5V and 1.6A/ph.

My series resistor needs to drop 9V(p/s)-5V(winding)=4V.
Using Ohm's law, we can calculate the DC resistance of our windings. R=E/I, so R=5V/1.6A=3.13ohms. We'll use this later.
Going back to Ohm's law, our series resistor needs to be R=E/I, R=4V/1.6A=2.5ohms.
This resistor would need to be able to handle P=IE, P=1.6A*4V=6.4W.

Right? Wrong.

Let's factor in the FET that I will use in my example.

The datasheet says that my FET will have an on drain-source resistance of 1.76ohms(typical).
Since this is in series with the windings, it will have the same 1.6A of current going through it, so it will drop E=IR, or E=1.6A*1.76ohms=2.82V.
If we add that to the 5V that we want our motor to "see" we get a total voltage drop of 7.82V. Now when we subtract that from our 9V power supply we get 9V-7.82V=1.18V.
Back to Ohm's law again, R=E/I, R=1.18V/1.6A=0.74ohms.
This resistor would need to be able to handle P=IE, P=1.6A*1.18V=1.89W.

If we use the original resistor value we came up with, we would have a total series resistance of 1.76ohms(FET)+3.13ohms(winding)+2.5ohms(resistor)=7.39ohms. With the 9V power supply this results in only, I=E/R, I=9V/7.39ohms=1.2A, or 25% less current than we want! With a bigger difference between the supply and motor voltages this is less of a factor, but should still be taken into consideration.

That's my 2.5 cents.

Feel free to comment.

Dave

[Old_Man]

Dave, you are a right about the FET (switcher) issue, also the same we could and should apply to bipolar switcher, now if you could give me some help I will really apreciate, 10X.
My problems, I have some stepper motors with label that says 12v 1A, well, after I checked the DC resistance with my DVOM I have 22ohms on each winding, I am corect that I need 22V/winding + the FET(or bipolar) issue for 1A per motor winding?
That label is wrong or what??
Regards Gabriel

[whateg01]

Gabriel,

I am by no means an expert on steppers. In fact I am fighting my own problems trying to implement them in my project. I am an electronics tech by trade, and the threads I saw regarding the load resistors caught my eye.

That stated, from what I do know about steppers, are the motors you have of a design that they can be used as either bipolar or unipolar? Or are they a "true" bipolar design? How many wires do they have?

Regards,
Dave

[Old_Man]

Well, the motors I have are 8 wire universal stepper motor, and I measured 22ohms on each winding (4 22ohms winding).
Regards, Gabriel

[AbSat]

Well, the motors I have are 8 wire universal stepper motor, and I measured 22ohms on each winding (4 22ohms winding).
Regards, Gabriel

Than it looks like the label is for bipolar use with two times two windings parallel. That will give you about 1A at 12V / phase.

AbSat.

[whateg01]

Than it looks like the label is for bipolar use with two times two windings parallel. That will give you about 1A at 12V / phase.

AbSat.

Thanks AbSat. That was what I was going to say, but I hadn't had a chance to get back online.

Good to see that somebody is picking up my slack.

Dave

[Old_Man]

Thank you, well, I want to use them as unipolar 6 wire, so I asume 22V will be nominal voltage, if I use the 10X voltage rule I need 220V
What voltage I should use, just to be safe and fast.
Regards Gabriel

[H500]

Unfortunately, your motors require too high of a voltage. You won't be able to get much performance out of them with any of the commonly available drives.

Unless your requirement is very light, it's really a lost cause. Automation direct has "real" motors for $30-$40.

The same goes for drives. Any that need limiting resistors are good for learning only.

[pminmo]

Whateg01
Your correct, the only thing I would take exception to is
"The datasheet says that my FET will have an on drain-source resistance of 1.76ohms(typical)."
If your using a MOSFET with 1.76 ohms typical on resistance, you need to change them. You should be looking a 40 to 50 milliohm MOSFETS, then the drop is insignificant unless your running huge currents.

Bipolar transistors do need to be taken into account. However to make more than a rule of thumb drop of .5 to 1.2V depending on current, you would have to measure the actual performance. The comes the issue of getting a high power resistor in that value and wattage.

[whateg01]

Phil,

You are right about these FETs having a very high resistance. I used them because of their very good availability. (I had them in my drawer.) I have since changed to a lower impedance FET.

I am still having problems though with my motors. I can't seem to get enough speed out of them. And I don't think I am getting all of the usable torque out of them.

Dave

[pminmo]

What ratings are your motors? What is your power supply voltage? I assume you must be using a unipolar setup?

[whateg01]

:frown:

I have a set of 100 oz.in motors. I am using them in a unipolar configuration. I have been able to spin them, but the max I was able to get was about 500 RPM. And there was almost no torque at that speed. I was able to grab the shaft with my bare fingers and they just started skipping.

Unfortunately, I haven't gotten a board put together yet to drive 2 coils at once, so I know I may get up to 40% more torque when I do that, but I need more torque and I need about 25% more speed to get the IPM to do job. I will be using this to carry my oxy/acetylene torch, so there won't be much lateral loading, but I need to maintain a decent speed to get a desirable cut.

The motors are rated for 5.2V 1.4A. These are the seemingly popular HP3 printer motors. I have used an O-scope and a series resistor to adjust my variable p/s to get 1.4A, but I don't remember the voltage I had at the time.

My setup is a 5' piece of 1/2-13 allthread moving a 8-10 lb carriage on linear bearings using skate bearings. The carriage moves easily with my pinky, so there is little resistance there. Just momentum. I intend to change to 10tpi acme, but this was locally available at 7:30pm and I know that this will decrease the mechanical advantage that the motors have. From what I have read, I will probably get that back in efficiency, though. Actually, I will have to change it because the threads on this piece of allthread are not very good. (when the allthread rotates, the nut wobbles!) I have eliminated the binding by very loosely attaching the nut to the carriage for my x-axis. I can turn the allthread very freely by hand, but the motor can only manage about 4 IPM, or about 50 rpm. I wasn't expecting 500 IPM, but with the success others have had using this motor to drive routers at up to 60 IPM, I figured that with no lateral load, I should have not trouble obtaining similar results.

Obviously, I need to get more torque. Like I mentioned, I know that I will get more when I drive 2 coils at the same time, but I have my doubts that it will be sufficient, even then. I have read that bipolar motors have more torque. I have also read that motors that can be configured either way (in this case 6-wires) are not capable of that kind of torque. How can I maximize the output from these (or for that matter, any) motors?

Other than having to synchronize the motors and run them in opposite directions, are there any drawbacks to using two motors per shaft, one at each end? The reason I ask that is because when I get ready to assemble my Y-axis, I will have substantially more mass to move. It seems that I can get 2 smaller motors for much less than 1 bigger motor. I don't mind the added complexity, to a degree. Or do I just have unrealistic expectations from these motors? If these will not work for my application, how big of a motor do I need?

I appreciate any advice you can give me. I am sure I probably left out some essential detail that would allow you to assist me, but let me know what other information you need and I'll provide it, as well as I can.

Thanks,
Dave

BTW, I am hoping to have the parts to build one of the boards on your website. I am kind of thinking of the "discrete" board at http://pminmo.com/discrete/discrete.htm . I intend to do a surface mount board, though, to eliminate all the drilling. I'll send you a copy of the artwork when I do, if that's okay with you.