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

Scott,

Less than what? It will be around 75% of a full-step holding torque,
it will be equal to half-step holding torque of the "weak" step. This
however has little meaning in the real world. Here's why:

Assume you have a motor rated at 100 in-oz holding torque with both
phases on (full-step). The motor manufacturer will rate that motor's
dynamic torque (at 5 full steps per second typically) as being 60 to
65 in-oz. Where did the missing 35% to 40% torque go?

It went to resonate the motor as it shakes, rattles and rolls at low
speed. Microstepping invests almost no energy in resonating the
motor, so there is no torque drop from its holding value. Even though
it starts at 75% of the full-step holding torque, it keeps all that
torque and so has the full step drive at a disadvantage (75% versus
65%).

At higher speeds, past the motor's corner frequency, it really
doesn't matter what the current set was. The motor's inductive
reactance limits motor current, not the drive.

Mariss

--- In CAD_CAM_EDM_DRO@yahoogroups.com, "sconkworks <cpscott1@b...>"
<cpscott1@b...> wrote:

> 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