Series vs Parallel vs Half Winding

[jeeper74]

Specs:
Qty 3 - SLO-SYN KML092S-106 - 6 Wire motors
2.4 VDC 6 Amps
Gecko 201 drivers
50 VDC Unregulated PS
Mach III interface on 1.1 ghz Laptop

I need high rpms with torque. Of Course!! I've tried the motors in both Full winding and half winding mode. Full winding with G201 set at 7 Amps has very little torque at 750 RPM. Half Winding with G201 set at 7 amps has considerable more torque at 750 rpm, but extensive heating. With both configs I get sporadic lockups/stalling from 500 rpms to 1440 rpms. I can't find any mechanical binding.

1. What is the best hook up? I've read a post that mentions converting the 6 wire motor into a 8 wire to hook up in Parallel. What does that benifit over full winding or half winding connections?

2. Is a 6 amp current setting correct for half-winding? (This is probably the most commonly asked question. Sorry.)

3. Are steppers more reliable at low RPMS? Would running the stepper at a lower RPM with a finer screw or 1:2 or 1:4 gearing be more reliable than a straight 1:1 at a high RPM?

4. Is resonance and mechanical over powering the only reason a stepper will lock/stall?

I've looked through many posts and have yet to find any definitive answers. Thanks for any help. Roger

[jeffs555]

2)Since they are 6 wire motors, the current spec would be for the unipolar current. Half winding bipolar would use the same current as unipolar so 6 amps should be the max for half winding. For full winding(ie bipolar series), the the current should be even less, a maximum of .707 times the unipolar current or about 4.25 amps. I hope you haven't damaged the motors.

1) Unmodified, the best for a 6-wire is usually half winding. The advantage to converting to 8 wire is that you can run bipolar parallel which will allow higher currents and consequently higher torque. For most motors, the max for bipolar parallel would be 1.414 times the unipolar current. Be very careful if you decide to rewire! You should not remove the rotor from the motor. Mine had just a cap over the wiring connections and removing it left all the bearings in place, so they were easy to rewire. On some motors the cap covering the wiring is also the bearing housing, so you have to be careful with the rotor or you can demagnetize it.

3) Gearing up or using a coarser pitch screw would reduce the motor speed, but would also increase the torque required. Any performance difference would depend your machine and the torque curves of your motors.

Don't know about #4

[jeeper74]

Ok. Quick recap to see if I understand correctly.

Full windings (i.e. using two coils in series and neglecting the center tap wire) - Able to run .707 of current. Why? Inductance?

Half Windings (i.e. using one coil with center tap wire and coil end wire, neglecting the other coil) - Able to run full current - Optimum use of motor without modifications

Modifying to 8 wire - You utilize all 4 coils (Instead of only two in half windings), Able to run 1.414 times current through motor - Most torque possible from these motors - Best Performance

Any corrections??

[jeeper74]

Now to work on my last question.

Speed?? What RPM is resonable for a stepper motor using microstepping with out missing steps. My torque curve for my motor shows a max torque, 640 oz-in, from 0-600 RPM then a sharp, almost linear, decline to 190 oz-in. So with a 5 TPI ball screw I would be optimum at 120 IPM. Any problems with my reasoning??

Is this type of RPM curve typical for most Steppers or are there some whose torque doesn't start deminishing until 1000+ RPMS?

Thanks again. Roger

[jeffs555]

Are you sure about the torque curves? For most motors, at 600 rpm, the torque has dropped to nearly half of the static torque.

DC power disipation(ie heat) is equal to resistance times the square of the current. When you connect two coils in series, the resistance doubles, so to keep DC heating the same, you need to reduce the current by the square root of 2. One divided by the square root of two is 0.707 .

[jeeper74]

Here are the torque curves. My motors are the KML092. Looking at the curves, at 2000 steps/sec(10 rev/sec) the torque is about 520 oz-in. 10 rev/sec x 60 sec/min = 600 rev/min. Right?? Maybe I'm reading it wrong. Let me know. Roger

[Willbird]

I take the chart the same way you do (interesting chart, wish I had one for my Keling 1150 oz/in stepper)

Bill

[ger21]

Originally Posted by jeeper74 Here are the torque curves. My motors are the KML092. Looking at the curves, at 2000 steps/sec(10 rev/sec) the torque is about 520 oz-in. 10 rev/sec x 60 sec/min = 600 rev/min. Right?? Maybe I'm reading it wrong. Let me know. Roger

Those curves are with full step drives (200 steps/rev), and they don't tell you what voltage they're at. Curves can vary greatly with varying voltage, and microstepping drives will probably give a little less torque.

[ger21]

Looked up a similar current model KM 092 stepper at Danaher which had a similar torque curve to yours, with the same SS2000D6 Drive. The curve was at 170Volts!! If the curve you have is also at 170V, expect the torque at 50V to drop off over 3x faster.

[jeeper74]

As long as I'm close to optimum performance, I'm happy. I rewired the motors to run in bipolar parallel and so far so good. The torque is considerably more than bipolar half winding. Thanks for all the help!!

[davidma]

i'm curious where the .707 for series wiring comes in. i thought it was 1/2 but what do i know.

dave

[Mariss Freimanis]

"i'm curious where the .707 for series wiring comes in. i thought it was 1/2 but what do i know."

Going from half-winding to full winding doubles the winding resistance. Both coils are in series now. The current must be reduced to 70.7% of the half winding value to keep the heating due to winding resistance the same as before.

Example: Say your half winding coil resistance is 1 Ohm and the rated current is 1 Amp. Resistive heating will be 1 Watt (Watts = Amps squared times Ohms).

In series the resistance becomes 1 Ohm + 1 Ohm or 2 Ohms. What current will still cause 1 Watt of heating?

Rearrange "Watts = Amps squared times Ohms" to solve for "Amps" and you get Amps = square-root of Watts divided by Ohms. Plug in the numbers:

I = SQRT (W / R) = SQRT (1W / 2 Ohms) = SQRT (0.5) = 0.7071 Amps

Mariss