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'[ot]: stepper motor heating up the resistors??'
2002\11\08@054326 by

Hello,

I have a couple of questions about my stepper circuit.

first, the stepper has a resistance of 1.9 ohm/phase (omega/phi)
I'm using 10 ohm resistors, little ones, 1/8 or 1/4 watt,
and 12vdc for the power.
(using a 5804 stepper controller)

These resistors get so hot they melt the plastic proto-board.  I guess
I need more wattage on the resistors, is this correct?  will higher
wattage resistors not get as hot??  And how do you calculate what
watt is required?

Would it be 1.01 amp * 12 vdc for a 12 watt 10 ohm resistor?  This
seems or sounds like something up on the pole....

I wanted to make this motor move, it's late(or early), and I only have
what's in my parts box...so I'm guessing about how to calculate the correct
values of parts required.

ANY help is much appreciated, thanks

Second, the motors seems to jump around occasionally, my program
steps 400 steps in one direction, then reverses and steps 400 steps the
other way. ( this thing is 1.8deg step angle, set on half step)

I'm sending a 10ms pulse then pausing 10ms.  How do you determine or
what's the difference in pulse width using this 5804 controller.  And
what about my pause of only 10ms, any problems with this??

thanks again

--

> I have a couple of questions about my stepper circuit.
>
> first, the stepper has a resistance of 1.9 ohm/phase (omega/phi)
> I'm using 10 ohm resistors, little ones, 1/8 or 1/4 watt,
>  and 12vdc for the power.
> (using a 5804 stepper controller)
>
> These resistors get so hot they melt the plastic proto-board.  I guess
> I need more wattage on the resistors, is this correct?  will higher
> wattage resistors not get as hot??  And how do you calculate what
> watt is required?
>
> Would it be 1.01 amp * 12 vdc for a 12 watt 10 ohm resistor?  This
> seems or sounds like something up on the pole....

Essentially yes.
Power = I^2 x R or V x I or V^2/R.
I will not quote be as ohms law would suggest (V/R) as the currenyt will be
ramping up for part of the time.
I assume that your stepper is rated for this current?

Here total power is not more than V^2/R = 12^2/(1.9+10) = 12.1 watt.
The resistor will dissipate about 10/(10+1.9) th of this or about 10.2
watts.
Even a 10 watt resistor will get very hot at this power level. Note that
this is the power for continuous activation and with a multiphase stepper
any one phase will generally be activated for only part of the time so you
can scale power dissipation per phase downwards accordingly. In some schemes
there is a single resistor used to power all phases and in such cases you
will need to rate it accordingly. Space the resistor off the board somewhat.

A large resistor will dissipate the same energy but will generally run
cooler than a small one as it has a higher radiating area. This sounds like
excessive power to be throwing awy unless you are really really keen on
performance and cannot get it any other way.

> I wanted to make this motor move, it's late(or early), and I only have
> what's in my parts box...so I'm guessing about how to calculate the
correct
> values of parts required.

Correct as above.

> ANY help is much appreciated, thanks
>

> Second, the motors seems to jump around occasionally, my program
> steps 400 steps in one direction, then reverses and steps 400 steps the
> other way. ( this thing is 1.8deg step angle, set on half step)

Whether this will work depends on what your stepper is specified at for the
given load. At 20/ms/step that's 50 steps/second which is well within the
capabilities of many (most?) stepper motors DEPENDING ON LOAD. What speed
you can step at at a given torque/load and a given power level will be
specced by the manufacturer. Is this essentially unloaded or with some load?

RM

--

> 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

> 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?

. 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?

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?

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?

thanks very much
>
> > ANY help is much appreciated, thanks
> >
>
>
> > Second, the motors seems to jump around occasionally, my program
> > steps 400 steps in one direction, then reverses and steps 400 steps the
> > other way. ( this thing is 1.8deg step angle, set on half step)
>
> Whether this will work depends on what your stepper is specified at for
the
> given load. At 20/ms/step that's 50 steps/second which is well within the
> capabilities of many (most?) stepper motors DEPENDING ON LOAD. What speed
> you can step at at a given torque/load and a given power level will be
> specced by the manufacturer. Is this essentially unloaded or with some
>
>
>         RM
this I don't understand at all

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
where torque is on the y axis and frequency(pps) on the x axis?

do you have a quick explanation of these units?
frequency(pps) what does this stand for and is it steps per time
unit -somehow??

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

thanks very much again.

--

If the motor specification is 2.0 amps at a specified voltage, then you will compromise the performance of the motor and the circuitry if you deviate from his.  There are buffer drivers ( pass transistors) that can interface with the chip.  Unless cost and real-estate are a significant you may wish to consider this option.

--

----- Original Message -----
To: <PICLISTMITVMA.MIT.EDU>
Sent: Friday, November 08, 2002 6:39 AM
Subject: Re: [ot]: stepper motor heating up the resistors??

{Quote hidden}

issue?
>
> 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?
>
> I guess 2.0 amps is all you need to worry about.  Then the power
generated,
{Quote hidden}

the
> > > other way. ( this thing is 1.8deg step angle, set on half step)
> >
> > Whether this will work depends on what your stepper is specified at for
> the
> > given load. At 20/ms/step that's 50 steps/second which is well within
the
> > capabilities of many (most?) stepper motors DEPENDING ON LOAD. What
speed
{Quote hidden}

--

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
> 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

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

--

>
> > 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

Why use the UCN5804? Why not sequence direct
from a PIC and use 4 transistors if you only
need crude half stepping?

> This motor is for a sherline mill, a cnc setup, so performance is an issue?

Yes, the sherlines have a lot of static friction
and at 20 turns/inch you will also need decent
motor speeds, probably about 5 revs/second. A
microstepping driver will be a lot better.

> 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?
>
> 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?

Both true, the label rating is usually the
"half copper" rating, ie 2 of the 4 coils
activated, each at 3.2v 2A. That gives your
max heat value, 3.2 x 2 (x2 coils) = 12.8W

If you decide to drive the motor bipolar (in
full copper) mode you need to derate that to
0.707 x amps = 1.4A per coil and about 4.5v
per coil.

You sound to be having difficulties with this
and the UCN5804 are not a cheap chip. Also
like all small stepper chips they are trash
(my view) and only good for disposable printers
for the consumer market.

If you have a nice Sherline metalworking mill
+ motors, worth probably \$1000 why not spend \$75
more and get 3 Linistepper boards from:
http://www.piclist.com/techref/io/stepper/linistep/index.htm

James Newton is offering these stepper kits at
a fraction more than the parts cost and half of
the (small) profit goes to support the piclist. The
design is my own and is RUGGED and will last the
distance, with heavy duty 5 amp transistors doing
the hard work instead of a flimsy DIP chip.

Possibly the best benefit is that it does
microstepping in 1200 and 3600 step modes, MUCH
better smoothness and precision than what you will
get from your 5804 chip. This makes a difference
with resonance and when cutting curves.
-Roman

--
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