There are far more experienced stepper people here but I'm awake (barely -
1:30am in NZ :-) ) so I'll give a quick guide.

> > I assume that your stepper is rated for this current?
> it's rated for 2.0 amps, but I needed to limt it to 1.25 for the
> 5804 controller

That will work.
Ultimately you are going to want to drive it at 100% if you are using it to
do real work. You said "Sherline Mill" - is that as a cutter driver or table
mover?

> > Even a 10 watt resistor will get very hot at this power level.
> how can I make it cooler, and use this controller chip??  is there
> a way?

If you dissipate a given power you have to get rid of it somehow.
Several 10 watt resistors mounted well clear of board will get rid of 10
watts and only get "very hot".
If using two in series each should be half of desired resistance.
If using two in parallel each shouyld be twice target value.

You can run on less voltage and dissipate less power BUT the motor response
will not be as good. Steppers "like" to be run from a constamnt current
drive ad this is approcimated by the higher volatge and resistive feed.

> . This sounds like
> > excessive power to be throwing awy unless you are really really keen on
> > performance and cannot get it any other way.
> This motor is for a sherline mill, a cnc setup, so performance is an
issue?

OK - you probably need whatever power you can get then.

> I don't really understand what voltage is OK or what voltage should be
used
> for this motor.  On the list of specs, it says 3.2V.  What is this? a
> minimum?

MAXIMUM!
But you are getting lesw sthan that across the motor at present.
You could run off 3.2v with no resistor at all but performance would be
degraded as time constant would be poor. (Save that for another
lesson/time). .


> I guess 2.0 amps is all you need to worry about.  Then the power
generated,
> or heat to be dissapated away somehow?  is this true?

Essentially.



> the load right now is a piece of tape, so I can see the shaft move.
> (it is continuously running though)
>
> looking at the spec sheet, I see a 'motor torque curve' chart for a
> given driver/current/exciting mode/inertial load
> where torque is on the y axis and frequency(pps) on the x axis?
> do you have a quick explanation of these units?

Torque = twisting power = product of force and radius.
a given force requires more torque to apply it if it is applied at a larger
radius.

Power is torque x speed.

Stepper will drive with a given torque under various loads and drive
conditions.
Inertial load is load when motor is standing still. This can be quite low as
the motor has to step while held in place by this load from a "standing
start".

> frequency(pps) what does this stand for and is it steps per time
> unit -somehow??

pps = pulse per second.


> and torque (kgf-cm) - this isn't in my memory banks either...

twisting powerv as above.
Product of force to be applied x radius it is applied at.
Hold 10 pounds at arns length for 1 minute.
Note how hard this is.
Now hold 210 pounds horizontally on end of 6 foot stick.
Note how impossible this feels.
Torque increases with distance of leverage arm.

Here the force is in kilograms-force (kgf) and radius or torque arm in
centimetres (cm)
1 kgf = 2.2pounds
1cm - 0.4 inches close enough.
SO 1 kgf.cm = 0.88 pound-inch.
eg if your motor can produce 10 kgf.cm torque = 8.8 pound inch it can apply
a force of 8.8 lbf at a radius of 1 oinch or 88 lbf at a radiuis of 0.1 inch
etc or 0.88 lbf at 10 inch radius etc
.

If you have no load your stepper should step at 50 pps with ease - see spec
sheet.
How do you know it is missing steps sometimes?

        Russell

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