Re: Unipolar as bipolar.

Alan,

It is counter-intuitive, but here's the reason:

A step motor reaches its rated power at the corner frequency. The
corner frequency is where the motor's inductive reactance begins to
limit current and not the drive.

Beyond that speed, inductive current (I = V / 2 pi f L) drops rapidly
with speed and I squared R losses drop even more rapidly.

The main effect of using only 50% of the copper is the winding
resistance is twice as high. This "extra" resistance develops a
voltage drop that subtracts from the voltage the motor "sees".

This voltage drop becomes insignificant (<3%) because of the
diminishing inductive current at higher speeds.

Since motor output power is proportional to supply voltage, this 3%
voltage drop results in a 3% power drop.

At low speeds there is no performance penalty at all because torque
is proportional to current, and it is the drive that sets the current.

Dyno tests bear this out.

Example: a 3V, 4A unipolar motor would be the same as a 1.5V, 4A
parallel wired 8-wire motor. The difference is 1.5V. If this motor
were run at 60VDC, the first motor would see 57V (60-3) while the 2nd
motor would see 58.5V (60-1.5). The ratio would be 57/58.5 or 97.4%
the power of a parallel wired motor.

Mariss

--- In CAD_CAM_EDM_DRO@egroups.com, Alan Marconett KM6VV <KM6VV@a...>
wrote:

> Thanks Mariss,
>
> I wasn't aware that only using 1/2 the winding would be that close

to a

> parallel connection. The old Slo-Syn tables I have don't look that
> good. More like 20%, as I recall. Still, you waste 1/2 the

windings!

>
> Alan