PDA

View Full Version : Stepper Motor Wiring Questions

TRC1
08-15-2003, 01:48 PM
What are the advantages/disadvantages of wiring a 6 wire stepper in half wind mode as opposed to full wind mode? I believe it was WOODKNACK who said he wired his motors with the center tap and 1 leg for each coil. In another thread someone else said to wire them with legs 1 and 2 and ignore the center taps. What are the pros/cons of doing it each way?
I have Vexta steppers (http://www.orientalmotor.co.jp/cgi-bin/WebObjects/UPOMStep.woa/wa/F3?typeNameId=1&modelName=PK268%2D02A&seriesId=2PK&frameSize=56.4)
(PK268-02A) 191 oz-in 4.5 volt 2 A/phase
I hope someone can shed some light on this for me, 'cuz I am really in the dark here!
Thanks!
JD

TRC1
08-15-2003, 02:36 PM
:idea:
OK, I think I may understand this. If using legs 1 and 2 you get full torque at low speed . And if using leg 1 and the center tap (half wind) you get better high speed torque?
JD

ToyMaker
08-15-2003, 03:49 PM
close, but...
torque is related to ampere-turns, number amps through the wire times the numbers of turns of wire.
so with full wind you will get full torque and with half-wind you will get less torque at any given speed.
top speed is related to back emf (the voltage the motor generates when it turns). so with full wind the back emf builds up at lower rpms and limits the top speed. with half-wind the back emf is less at a given speed and the motor can spin faster.

robotic regards,

Tom

Mariss Freimanis
08-15-2003, 04:27 PM
Still no cigar...

You get the same low-speed torque either way. In full coil operation use 1/2 the motor's rated current; in half coil operation use the rated current. Both ways ampere-turns are the same.

At high speeds hal coil operation gives twice the power as full coil at the same power supply voltage.

The reason is inductance. Full coil has 4 times the inductance as half coil.

Inductance has a property called inductive reactance. It is measured in Ohms, but unlike a resistor the value is proportional to frequency (read step rate). Ohms = 2*pi*f*L

Torque is proportional to current and inductive current (I = V/R) is inversely proportional to frequency.

Full coil operation has 4 times the inductance, so it has 1/4 the inductive current at a given speed compared to half coil operation. This current passes thru twice as many turns of wire, so the net impact is 1/2 the torque at high speed (and 1/2 the power).

At low speeds the drive limits current so there is no difference.

Mariss

TRC1
08-15-2003, 10:03 PM
So it sounds like I am better off running my motors at half wind to acheive the higher power at higher speeds?