Re: 8-wire 1/2coil stepper?

Hi,

I think the thing being missed here full-stepping vs. microstepping.

If you have a 5V 1A per phase unipolar rated, 8-wire motor, then you
would run it at 1A unipolar (half-winding) and it would dissipate 5W
(1A squared times 5 ohms).

If you connect the windings in parallel, then the winding resistance
is 2.5 ohms (1/2 what it was in half-winding). You can now increase
the current to 1.41A and still dissipate 5W (1.41A squared times 2.5
ohms = 5W).

What matters here is your torque (from the URL you referenced)
increased 30%. It should have increased 41%; the reason it didn't is
because the iron has magnetically saturated.

If you are full stepping, you don't care. There is no negative
positioning effect from saturating the iron except for a drop in
inductance.

If you are microstepping you do care because saturated iron means a
non-linear motor and the microstep positions will begin to bunch-up
at the full step locations.

An analogy is an audio amplifier. If you overdrive it with a sine
wave, the output will limit or "clip" and produce distortion. If all
you are amplifying is a square wave, then who cares if it clips?

Second point. Take the same motor and drive it at 1.41A half-winding.
You will not be able to tell it apart from the parallel connected one
at 1.41A torque-wise. The only difference is it will dissipate 10W
(1.41A squared times 5 ohms).

Third point. This 30% increase in torque occurs at low speeds where
step motors usually have surplus torque anyway. It does not get you a
single extra oz-in of torque at high speed, where improvement would
be appreciated.

Inductive reactance becomes the bottleneck that limits the winding
current. The drive may be set at 1A or 1.41A, it will make no
difference. 10 mH of inductance will have 62.8 ohms of inductive
reactance at 4,000 full steps per second and will limit winding
current to 0.5A for a power supply voltage of 31.4VDC.

I = 31.4V / 2 pi times 1kHz times 0.01 Henries
4 full steps is one electrical cycle, so 4,000 full steps/sec = 1kHz

Mariss

--- In CAD_CAM_EDM_DRO@y..., "Tim Goldstein" <timg@k...> wrote:
> Tony,
>
> You are missing the difference between a unipolar drive and a
bipolar drive.
> It is not that you are using 1/2 the coil that makes the drives
different.
> Take a look at:
> http://www.wirz.com/stepper/phasesqe.html
> to see that a bipolar will give more torque.
>
> Tim
> [Denver, CO]
>