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

Mariss,

Perhaps you could clarify something for me.

In your examples of unipolar and series, if I follow your advice and
use half the unipolar current in series, do I have the same or less
torque created?

Thanks
Scott

--- In CAD_CAM_EDM_DRO@yahoogroups.com, "Mariss Freimanis
<mariss92705@y...>" <mariss92705@y...> wrote:

> 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