# Stepper motor question

1) I have a stepper motor in mind :
http://tinyurl.com/2qr4d9
I'm guessing since it has 6 wires that it has 2 coils. But should
that not mean it has 4 phases. Why do they list it as a "2 phase stepper motor"?
2) The motor above is listed as a "bipolar (in series) and unipolar" as well which is kind of confusing. Is it possible to drive it as a bipolar motor (i.e. without using the two center taps)? Is there any downside to driving a unipolar motor as a bipolar motor (or is this question totally ridiculous).
3) Can you suggest a driver chip (IC) i can use to drive the above stepper motor? Would the following driver be ok :
http://www.onsemi.com/pub/Collateral/MC3479-D.PDF
4) Is "load voltage" the voltage rating of the stepper motor? (in this case 4V)
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Hi all, Say I have a stepper motor whose ratings say that it rotates 1.8deg/step. Does that mean 1.8 degs as in Full Step mode or as in Half Step Mode? I mean: Will it rotate at 3.6 deg/step in Full-step mode or will it rotate at 0.9deg/step in Half-Step Mode?
Another question: I know that in order to get a stepper motor rotating u need to give it a sequence of pulses as input. Say I just connect one phase to a battery(that is, only one pulse). Will it just rotate by one step and stop?
Thanx,
Karthik Ravikanti http://students.iiit.net/~karthik_ravikanti
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Karthik Ravikanti wrote:

It should rotate 1.8 degrees per full step or 0.9 degrees per half step.
If you apply a battery to one of the phases, the motor will rotate at most 1 step and possibly not at all. This is dependent on what the mechanical position of the motor is when you apply the battery. If the motor is lined up with the phase that you apply power to, it will not move appreciably. If it's lined up with another phase, it will move to the one you power.
Good Luck, Bob
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This taken from UCN5804 data sheet: The wave-drive format consists of energizing one motor phase at a time in an A-B-C-D (or D-C-B-A) sequence. This excitation mode consumes the least power and assures positional accuracy regardless of any winding inbalance in the motor. Two-phase drive energizes two adjacent phases in each detent position (AB-BC-CD-DA). This sequence mode offers an improved torque-speed product, greater detent torque, and is less susceptible to motor resonance. Half-step excitation alternates between the one-phase and two-phase modes (A-AB-B-BC-C-CD-D-DA), providing an eight-step sequence.
Looking at the description of half-step excitation it seems as if the motor could turn a half step at the transitions. That would depend, however on whether the next phase has enough power to overcome the static friction of the motor and the force of the current phase. Someone with more experience could give you a more definitive answer, although my guess is that a 3.6 degree/step motor might turn 1.8 degrees when half-stepping. Try it and see.
And for the second part of the question, yes.
On 15 Apr 2005 17:19:39 -0700, "Karthik Ravikanti"

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vorange wrote:

Yes. 6 wires indicates two windings and a center tap on each.

2 phases means that the coils are placed in two magnetic locations. You get 4 effective phases by driving each with current in one of two directions.

Bipolar means that the driver circuit applies voltage in either direction across the whole winding (the two halves of the center tapped winding, both halves in series, carrying the same current). Unipolar means that the driving circuit can force current in only one direction, so you get the reversal by applying it to only one half of the center tapped winding at a time, both currents either toward or away from the center tap. The effect is one each half magnetizes the poles in one direction and the other half winding magnetizes the poles in the other direction. The efficiency is lower, because only half of the copper is being used at any time.

An excellent question. It is perfectly reasonable to drive a unipolar motor with a bipolar driver, and it will be more efficient because the whole winding will be in use. Having the current pass through twice as many turns also produces about the same torque as twice the current from a unipolar driver. The down side is that the motor will require 4 times the supply voltage to max out at the same speed, because twice as many turns produce 4 times the inductance. You can give up the current efficiency and recover the full speed by driving only half of the center tapped winding with the bipolar driver, but you will be wasting half of the copper, so the temperature rise will be worse.

Allegromicro makes many:
http://www.allegromicro.com/en/Products/Categories/ICs/motor.asp I would pick a dual bipolar driver with two or three times the series current rating of the motor if you want full torque from the motor. Perhaps the AD3992, because it is available in a DIP package for easy prototyping. http://www.allegromicro.com/en/Products/Part_Numbers/3992/index.asp
> Would the following driver be ok :

It is a bit wimpy, with a current rating of only 0.35 amp, which I would not get close to, for good reliability. It would probably turn the motor but would not get much torque from it.

The chip is rated for up to an 18 volt motor supply, but it puts out about 3 volts less than the supply because of the voltage drops of the power switches. It also has no provision for current regulation like the AD3992 has.
Current regulation allows the supply voltage to be well above the DC rating of the motor, so that there is extra voltage available to overcome the generated voltage as the motor moves faster. Basically, the motor torque is proportional to the coil current, but the voltage it takes to produce that current is proportional to the speed, plus a fixed amount to overcome the winding resistance.
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Thank you John.
As always, you are an encyclopedia of knowledge. You really should write a book on electronics for beginners to intermediate. I predict it would sell like hotcakes.
Just a quick comment :

Yes I noticed that too. I had planned to put transistors on its outputs to boost the current. But the Allegro suggestion is looking good.
Cheers!
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My favorite stepper motor tutorial:
http://www.cs.uiowa.edu/~jones/step/
BRW
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vorange wrote:

unipolar steppers are niffy with the correct controller on it. the CT Fields give you the option of micro stepping. That is if you have the correct controlling electronics. what this does is locks both fields on . You don't get as much holding torque as you would on a single phase position but it's nice where you would have lets say a 100 PPR stepper and emulate a 200 PPR stepper. In some cases, this is used as a soft position index to reduce shock movement between poles. But then again, it depends on the controller.
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Jamie wrote:

(snip)
Micro stepping is used with both bipolar and unipolar drivers, but I have seen it most often with bipolar drivers.
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John Popelish wrote:

It works smoother with unipolar but with less torque. the bipolar gives you more torque but is a little less stable in the micro step. In many cases, more torque is more favorable but does require more current to operate it while in the split phase index. I wrote a little code in an AVR to operate a unipolar motor using PWM to perform soft movement of the rotor. I gotta say, it was an interesting experience. Parameters had to be tweaked to account for the induction sang in the coils. When it was all done how ever, the motor it self was able to turn so smooth between indexes, you couldn't tell it was a stepper motor.
We had a project where we needed smooth movement but be able to lock the rotor once it stopped and index 2 notches . the locking position didn't have to be exact! but the motion to get it there needed to be smooth. it was a toss up between using a stepper or a brushless DC motor.
I guess either one would work, just a matter of how you would want to approach it. And bipolar is more common due to it's simplicity in design.
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