Re: Stepper current - how bad is underdriving a motor?

Hi,

OK. The .707, 1.41 thing, series and parallel, etc.

Let's setup a hypothetical 8-wire motor with the following ratings:

1A unipolar
1.41A parallel
.707A series

Also, let's say each coil is 1 Ohm.

1) In unipolar operation, you have the motor connected as a 6-wire
motor. 1A thru 1 Ohm is 1W of dissipatiom.

2) You now connect the motor in parallel. Two 1 Ohm coils in parallel
is going to be 1/2 Ohm. How much current can you run thru 1/2 Ohm and
still get 1W dissipation? Well, W = I^2 * R, so solving for I you get
I = square root (W / R), or square root of ( 1 / .5), or square root
of 2, which is 1.414 Amps.

3) Now you connect the motor in series. Two 1-Ohm coils in series is
2 Ohms. How much current can you shove thru 2 Ohms and still keep a
1W dissipation? Same equation as in (2) but now it's 2 Ohms instead
of 1/2. The answer is .707A.

In cases (1), (2) and (3), the motor manufacturer is happy because
his motor won't burn down and by extension, so are you for the same
reason; 1W is dissipated in all 3 cases. This great for full-step
operation.

If you microstep though, things get a little diceier; you now care
how evenly the microsteps are distributed. Enters the concept
of "magnetic saturation".

You can think of iron in a motor as being composed of billions of
little bar magnets called magnetic dipoles. You run some current thru
a coil surrouinding this iron and some of the dipoles will line up
north to south, making the iron an electromagnet. Run yet more
current thru and even more will line up. Eventually, at some current,
all dipoles will be lined up. Increasing the current further will
have no additional effect because 100% of the dipoles have lined up.
This is magnetic saturation.

Most round step motors have about 85% of the dipoles lined up when
the motor is running at its unipolar rating. Going to its parallel
current rating of 141% certainly exceeds the level where all dipoles
are lined up. The result is a non-linear motor having distortion much
the same way an overdriven audio amplifier would be distorted!

So use the motor's unipolar current rating even if it is connected in
parallel, half that amount when series connected for best
microstepping performance.

Mariss


--- In CAD_CAM_EDM_DRO@yahoogroups.com, "jeffalanp <xylotex@h...>"
<xylotex@h...> wrote:

> Hello Tony,
>
> OK, now I've read both 50% and 71.5% from the newsgroup here.
> Still there are some issues that I would like clarified if you will.
>
> John said,
> >840 oz-in,
> >1.45V
> >15.2amps !!
> >It is six wire. So full coil turns it into a 2.90V, 7.6A motor.
> >That's almost perfect. With a 48V supply, I'll have 16.5x

voltage,

> >so I should get good high speed performance, and 7A into a 7.6A
> >motor will deliver nearly full torque.
>
> The book I have says this would turn into a 21.49 Amp, 2.05 Volt
> motor when run in bipolar series. John is appently doubling the
> voltage and halving the current in his calculation above. This

seems

> as if it would be fine if the above motor rating was for bipolar
> parallel, but the rating shown would be unipolar as it is a 6 wire
> motor.
>
> You seem to be in agreement with John that at 7 amps, running the
> motor in bipolar series mode, he is going to get nearly full torque
> rating. How do you figure this? I came up with about 65% full
> rating.
>
> I appreciate any and all comments as I consider this to be not only
> a "newsgroup", but an "eductationgroup" as well (as long as Mom
> doesn't say it's OT).
>
> Thanks,
>
> Jeff
>
>
> --- In CAD_CAM_EDM_DRO@yahoogroups.com, Tony Jeffree <tony@j...>
> wrote:
> > Jeff -
> >
> > As Mariss keeps telling us, derating by 71.5% is not a great idea
> with
> > microstepping, as you tend to get bunching of the microsteps. So
> half the
> > current is probably the right answer here.
> >
> > Regards,
> > Tony
> >
> > At 07:52 23/12/2002 +0000, you wrote:
> > >Hello John,
> > > If the motors are rated at 15.2 Amps UNIPOLAR, then if you

use

> > >them in full-coil BIPOLAR series mode, I beleive you want to

derate

> > >them to 71.5% of the UNIPOLAR rating, not half. This comes to
> 10.87
> > >amps/phase. 7 amps/10.87 gives a 64% of maximum total torque
> > >rating. I would double check this (my evaluation) and get

another

> > >opinion though.
> > >
> > >Jeff
> > >
> > >--- In CAD_CAM_EDM_DRO@yahoogroups.com, "jmkasunich
> > ><jmkasunich@y...>" <jmkasunich@y...> wrote:
> > > > --- In CAD_CAM_EDM_DRO@yahoogroups.com, Tony Jeffree

<tony@j...>

> > > > wrote:
> > > > > At 12:18 21/12/2002 +0000, you wrote:
> > > > > >I just acquired a NEMA42 stepper,
> > > > > >
> > > > > >840 oz-in,
> > > > > >1.45V
> > > > > >15.2amps !!
> > > > > >
> > > > > >I was wondering how much torque I
> > > > > >sould expect out of it when running
> > > > > >on a Gecko at 7A? I do know that
> > > > > >whatever torque I get will stay
> > > > > >there to a rather high speed, thanks
> > > > > >to the very low voltage.
> > > > >
> > > > > You would expect to get approx 1/2 the torque at low
> > > > > speeds, but as you say, the low voltage gives plenty
> > > > > of headroom to get good performance at high speeds.
> > > > >
> > > > > You haven't said whether the motor is 4, 6, or 8 wire.
> > > > > If it is 6 or 8 wire, the alternative is to use it in
> > > > > full coil (6-wire) or series connected (8-wire) at 7A;
> > > > > you should then get close to the rated torque at low
> > > > > speeds.
> > > > >
> > > > >
> > > > > Regards,
> > > > > Tony
> > > >
> > > > Duh! Why didn't I think of that! It is six wire. So
> > > > full coil turns it into a 2.90V, 7.6A motor. That's
> > > > almost perfect. With a 48V supply, I'll have 16.5x
> > > > voltage, so I should get good high speed performance,
> > > > and 7A into a 7.6A motor will deliver nearly full
> > > > torque. Thanks Tony!
> > > >
> > > > John Kasunich
> >
> > Regards,
> > Tony