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'[PIC]: frying eggs with a ULN2003'
2001\03\01@145342 by Joan Ilari

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I am driving a stepper motor -with 3 windings (4 Ohms each) and
1 common terminal- salvaged from an old disk drive by means of
a 16F84 and a ULN2003. I am feeding the motor with 12V.

The prototype works but after a few seconds working I can fry
eggs in the 2003. I guess is that this phenomenon can be
explained by means of the electric power formula :
 W = V**2/R = 12**2/4 = 36W !?!?

I suppose that the 2003 does not melt because :
1 - I switch off the power supply as soon as the 2003 begins
smelling burned
2 - My self-made power supply cannot deliver 36W

Now the question is : how can I drive my stepper motor without
risking burning my house (and with a as far as possible simple
circuit) ?

Any ideas would be very appreciated.

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2001\03\01@152955 by David VanHorn

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At 08:49 PM 3/1/01 +0100, Joan Ilari wrote:
>I am driving a stepper motor -with 3 windings (4 Ohms each) and
>1 common terminal- salvaged from an old disk drive by means of
>a 16F84 and a ULN2003. I am feeding the motor with 12V.
>
>The prototype works but after a few seconds working I can fry
>eggs in the 2003. I guess is that this phenomenon can be
>explained by means of the electric power formula :
>   W = V**2/R = 12**2/4 = 36W !?!?

The 2003 shouldn't have large voltages across it.
You're asking each driver to sink 3A when it's on. That may be a bit much.


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2001\03\01@153229 by Drew Vassallo

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>I am driving a stepper motor -with 3 windings (4 Ohms each) and
>1 common terminal- salvaged from an old disk drive by means of
>a 16F84 and a ULN2003. I am feeding the motor with 12V.
>The prototype works but after a few seconds working I can fry
>eggs in the 2003. I guess is that this phenomenon can be
>Now the question is : how can I drive my stepper motor without
>risking burning my house (and with a as far as possible simple
>circuit) ?

I'm not an electrical engineer by any means, and I probably shouldn't be
offering any suggestions, BUT, you might try adding some series resistors to
the stepper.  Depends on how fast of a response you need.

The 2003 should be able to drive it just fine, though.  If I remember
correctly, that chip has a heat sink built in... they're meant to be capable
of running pretty hot.

Again, don't take what I'm suggesting as being 100% accurate... just some
ideas for you to look further into.

--Andrew
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2001\03\01@155654 by Diego Sierra

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Hi!

>>I am driving a stepper motor -with 3 windings (4 Ohms each) and
>>1 common terminal- salvaged from an old disk drive by means of
[...]
>>explained by means of the electric power formula :
>>   W = V**2/R = 12**2/4 = 36W !?!?

That´s the whole power you are sinking from the supply. As the 2003 is an open collector transistor, when it is driving the windings the voltage across it is about 0.2 volts, so the power the transistor is disipating is V*I = 0.2*3 = 0.6 W, driving the four windings at the same time is 2,4 W, the remain power up to 36W is disipated by the motor itself.

According to the electrical characteristics of the 2003, it said it can drive up to 30 V and 500mA (4*0.2*0.5=0.4W (!)), hence, 3A for it seems to be too much unless you prefer well cookied eggs :-)

Cheers,
Diego.

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2001\03\01@155928 by Joan Ilari
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David VanHorn wrote :

>  >I am driving a stepper motor -with 3 windings (4 Ohms each) and
>  >1 common terminal- salvaged from an old disk drive by means of
>  >a 16F84 and a ULN2003. I am feeding the motor with 12V.
>  >
>  >The prototype works but after a few seconds working I can fry
>  >eggs in the 2003. I guess is that this phenomenon can be
>  >explained by means of the electric power formula :
>  >   W = V**2/R = 12**2/4 = 36W !?!?
>
>  The 2003 shouldn't have large voltages across it.
>  You're asking each driver to sink 3A when it's on. That may be a
>  bit much.

The problem is that I do not know what is the voltage needed by
the motor. I assumed 12V but perhaps it should be 5V or even less.
I have tried with 5V but the holding torque is greatly reduced.

Joan

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2001\03\01@161240 by Diego Sierra

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01/03/01 22:55:32, Joan Ilari <@spam@joan.ilariKILLspamspamTERRA.ES> wrote:

>The problem is that I do not know what is the voltage needed by
>the motor. I assumed 12V but perhaps it should be 5V or even less.
>I have tried with 5V but the holding torque is greatly reduced.

Could it be drived not by a DC current but by a pulsed one?

I have some stepper motors (TEAC) from old disk drives and they are drived by a DC current, 12V @ 160mA per phase (this is 75 ohms per phase). 3 A sounds a pretty
high current :-)

Cheers,
Diego.

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2001\03\01@161857 by Dan Michaels

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Joan Alari wrote:
>I am driving a stepper motor -with 3 windings (4 Ohms each) and
>1 common terminal- salvaged from an old disk drive by means of
>a 16F84 and a ULN2003. I am feeding the motor with 12V.
>
>The prototype works but after a few seconds working I can fry
>eggs in the 2003. I guess is that this phenomenon can be
>explained by means of the electric power formula :
>  W = V**2/R = 12**2/4 = 36W !?!?
>


Joan, go to the Allegro site and download the datasheet if you
haven't already done so:

http://www.allegromicro.com/

The 2003 is rated for 600 mA max, so you're running too much current.

Also, the secret to not setting the neighborhood on fire with these
things is to design the circuits so that all the drive transistors
are operating in "saturation" mode when on. Then the power dissipated
by the drivers will be minimal.

In your case - if properly designed - most of the 12v would actually
appear across the stepper windings, and only about 1v would be across
the driver. Then, the dissipation would only be ~ 1v*3A = 3W for "each"
active winding. Note - this is still too much for the 2003 - too
much current.

Also, your stepper motor may not actually be a 12v jobber, and you
may be applying too much voltage to the ckt to start with.

You would have the identical considerations if doing a discrete design
with an NPN inverter ckt. In this case, you guarantee saturation of the
transistor in several ways - make sure that the voltage powering the
ckt matches the motor requirements, use enough base drive to ensure the
NPN is saturated, and use a large wattage transistor.

hope this helps,
- dan michaels
http://www.oricomtech.com
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2001\03\01@161902 by Joan Ilari

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Diego Sierra wrote :

>  >>I am driving a stepper motor -with 3 windings (4 Ohms each) and
>  >>1 common terminal- salvaged from an old disk drive by means of
>  [...]
>  >>explained by means of the electric power formula :
>  >>   W = V**2/R = 12**2/4 = 36W !?!?
>
>  That4s the whole power you are sinking from the supply. As the
>  2003 is an open collector transistor, when it is driving the
>  windings the voltage across it is about 0.2 volts,
>  so the power the transistor is disipating is V*I = 0.2*3 = 0.6
>  W, driving the four windings at the same time is 2,4 W, the
>  remain power up to 36W is disipated by the motor
>  itself.
>
>  According to the electrical characteristics of the 2003, it said
>  it can drive up to 30 V and 500mA (4*0.2*0.5=0.4W (!)), hence,
>  3A for it seems to be too much unless you
>  prefer well cookied eggs :-)

But the frying pan is in the 2003, not in the motor which stays pretty
cool :-)

Joan

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2001\03\01@161910 by Joan Ilari

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Drew Vassallo wrote :

>
>  >I am driving a stepper motor -with 3 windings (4 Ohms each) and
>  >1 common terminal- salvaged from an old disk drive by means of
>  >a 16F84 and a ULN2003. I am feeding the motor with 12V.
>  >The prototype works but after a few seconds working I can fry
>  >eggs in the 2003. I guess is that this phenomenon can be
>  >Now the question is : how can I drive my stepper motor without
>  >risking burning my house (and with a as far as possible simple
>  >circuit) ?
>
>  I'm not an electrical engineer by any means, and I probably shouldn't be
>  offering any suggestions, BUT, you might try adding some series
>  resistors to
>  the stepper.  Depends on how fast of a response you need.

If you are right -> the maximum current in each winding must be
0.5A -> the overall resistance of each coil must be 12V/0.5A =
24 Ohm (aprox). If the resistance of each coil is 4 Ohm this means
that I must add 20 Ohms in each coil ? If I do so, each resistance will
have to dissipate i**2*R = 0.5**2*20 = 5W !!!!!
I am affraid that I will need more space for the 3 resistors than for
the rest of circuit :-(

Joan

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2001\03\01@163318 by David VanHorn

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>
>But the frying pan is in the 2003, not in the motor which stays pretty
>cool :-)

Make sure you really are turning it ON as opposed to on.
Measure the voltage fromcollector to ground. <<1V is normal.
Anything else means it's not ON, and therefore will get very hot.

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2001\03\01@163738 by Dan Goddard

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Joan,

If its salvaged from a disk drive, are you sure its a stepper motor? or a
brushless DC motor, they use similar internal wiring but the center tap is
used for sensing the phase on the bdc motor, whereas its driven on the
stepper. The phase for a bdc is 120 degrees I think, and if its the case,
trying to drive that as a stepper motor could cause some interesting load
characteristics to your driver. I've driven high current steppers at 12v
with SOT23's (Zetex FMMT SuperSOT) before without anything getting hot
except the motor (because it was a 5V motor). Good luck and happy hunting :))

Cheers,
Dan

At 08:49 PM 3/1/01 +0100, you wrote:
{Quote hidden}

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2001\03\01@164400 by t F. Touchton

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At the current levels the stepping motor requires (probably 2000oz-in torque or
more) you will need a dedicated drive transistor for each winding.  Also you
will need to beware of the magnetic field collapse as you are stepping.  I have
designed such size stepping motor drivers for an industrial Pratt and Whitney
drilling machine, and I resulted to large heat sinks with forced air cooling.
This was done with bipolars, which at the drive current level (7 amps) had
significant saturation losses.  This was before the age of mosfets, which should
be much more efficient for this application.  You may be able to use your IC as
a predriver.  Beware if it is open collector, for this complicates turning the
switching transistor on rapidly and guaranteeing it is in saturation.

One other thing to watch out about is that this size of motor requires an
acceleration profile which is matched to the intended load.  I learned this the
hard way :-)

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2001\03\01@164832 by t F. Touchton

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part 1 2286 bytes content-type:text/plain; charset=us-asciiOk folks.. before you torture me... I missed the disk drive part and went
straight to the current.....

Unless it is the worlds largest disk drive, the drive current is way out of
line.

Well it did say OLD disk drive... and things were bigger a few years back... : -
)


|--------+----------------------->
|        |          Joan Ilari   |
|        |          <joan.ilari@T|
|        |          ERRA.ES>     |
|        |                       |
|        |          03/01/01     |
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 |       Subject:     Re: [PIC]: frying eggs with a ULN2003                   |
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David VanHorn wrote :

{Quote hidden}

The problem is that I do not know what is the voltage needed by
the motor. I assumed 12V but perhaps it should be 5V or even less.
I have tried with 5V but the holding torque is greatly reduced.

Joan

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part 2 2987 bytes content-type:application/octet-stream; (decode)

part 3 136 bytes
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2001\03\01@165019 by Chris Carr

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> If you are right -> the maximum current in each winding must be
> 0.5A -> the overall resistance of each coil must be 12V/0.5A =
> 24 Ohm (aprox). If the resistance of each coil is 4 Ohm this means
> that I must add 20 Ohms in each coil ? If I do so, each resistance will
> have to dissipate i**2*R = 0.5**2*20 = 5W !!!!!
> I am affraid that I will need more space for the 3 resistors than for
> the rest of circuit :-(
>
> Joan
>
Actually you only need one resistor in the common wire. Also note that you
have 7 drivers available of which you are using only 3 and you are allowed
to parallel the drivers for increased current drive capability. So pair the
drivers, this will give you 1 Amp per coil.

Not exactly, you must derate the current because of thermal dissipation
limitations. Look at the Data Sheets

Regards

Chris

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2001\03\01@170933 by Olin Lathrop

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> I am driving a stepper motor -with 3 windings (4 Ohms each) and
> 1 common terminal- salvaged from an old disk drive by means of
> a 16F84 and a ULN2003. I am feeding the motor with 12V.

What exactly is a ULN2003?  Some switching element, I presume, but a FET,
NPN, PNP, used as emitter follower, common emitter, or what?  It repeatedly
amazes me how people ask questions about a specific part number (other than
PICs) without even a sentence explanation as to what the part is.  Are they
so arrogant to think everyone else must know the part number because they
are using it, or too dumb to realize there are far too many parts for anyone
to know more than a tiny fraction?

> The prototype works but after a few seconds working I can fry
> eggs in the 2003. I guess is that this phenomenon can be
> explained by means of the electric power formula :
>   W = V**2/R = 12**2/4 = 36W !?!?

This would be the power into the 4 ohm load, not the switching element.  If
the 2003 is a saturated bipolar transistor in common emitter configuration,
then the power into it is roughly:

 power = V * I = 500mV * (12V / 4 ohm) = 1.5 watts

This is still enough to get pretty toasty without a heat sink.


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Olin Lathrop, embedded systems consultant in Devens Massachusetts
(978) 772-3129, KILLspamolinspamBeGonespamembedinc.com, http://www.embedinc.com

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2001\03\01@173719 by Olin Lathrop

face picon face
> If you are right -> the maximum current in each winding must be
> 0.5A -> the overall resistance of each coil must be 12V/0.5A =
> 24 Ohm (aprox). If the resistance of each coil is 4 Ohm this means
> that I must add 20 Ohms in each coil ? If I do so, each resistance will
> have to dissipate i**2*R = 0.5**2*20 = 5W !!!!!
> I am affraid that I will need more space for the 3 resistors than for
> the rest of circuit :-(

If it turns out that 12V is too much for the motor, you could drive it via
PWM.  This is much more efficient because the switching elements are either
fully on or fully off at any one time, thereby dissapating very little
power.  At 12V and 3A you can easily find FETs with such low on resistance
that they won't even get warm and won't need a heat sink.  One of my
projects drives a DC motor from 12V up to 4A with four FETs in an H bridge.
The FETs are just soldered onto the board without heat sinks, and the PWM
frequency is 5KHz.  You can run it all day and the FET get slightly warm to
the touch.


*****************************************************************
Olin Lathrop, embedded systems consultant in Devens Massachusetts
(978) 772-3129, @spam@olin@spam@spamspam_OUTembedinc.com, http://www.embedinc.com

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2001\03\01@173730 by Olin Lathrop

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> But the frying pan is in the 2003, not in the motor which stays pretty
> cool :-)

This makes sense given that others have said this 2003 is rated for 600mA.
It can't handle the 3A you are trying to put thru it, and the voltage
accross it is rises as a result, and the voltage accross the motor falls.
Another factor is that the motor probably has much more thermal mass.  This
means it would heat up more slowly that the 2003 at the same power
dissipation.


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2001\03\01@190246 by Dingoblue mail

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A ULN2003 is a open collector bipolar driver.  There are 7 darlingtons
within the package and each can switch 50 volts at 500mA.

There is a total dissipation limit on the package and I suspect this has
been well and truely exceeded.

Rex

{Original Message removed}

2001\03\01@223858 by Joan Ilari

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Dan Goddard wrote :

>  If its salvaged from a disk drive, are you sure its a stepper motor? or a
>  brushless DC motor, they use similar internal wiring but the
>  center tap is
>  used for sensing the phase on the bdc motor, whereas its driven on the
>  stepper. The phase for a bdc is 120 degrees I think, and if its the case,
>  trying to drive that as a stepper motor could cause some interesting load
>  characteristics to your driver. I've driven high current steppers at 12v
>  with SOT23's (Zetex FMMT SuperSOT) before without anything getting hot
>  except the motor (because it was a 5V motor). Good luck and
>  happy hunting :))

If it's not an stepper motor it is very well disguised :-)

The motor does turns clockwise, counterclockwise, I can change its
speed, etc... and I have done all the pic programming assuming
that IT IS a stepper motor. Thus, I understand that there is only
one problem : my ignorance (which, BTW is being incrementally reduced
by means of the comments and suggestions in this list).

I'm going to sleep (here is Spain we said "to discuss the subject
with the pillow") and I hope that tomorrow I will be more enlightened.

Cheers

Joan

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2001\03\02@035652 by Kevin Blain

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Guess that would make me arrogant then.

> What exactly is a ULN2003?

> Are they
> so arrogant to think everyone else must know the part number because they
> are using it, or too dumb to realize there are far too many parts for
anyone
> to know more than a tiny fraction?


For your information, the ULN2003 is one of a family of very commonly used
(certainly in the UK) darlington driver ICs. The family is available in 7 or
8 ways per IC, and is quite useful for stepper motor projects especially.

TIP to original poster: You can parallel up the ULN2003 channels to get more
current, e.g. 2 channels gives approx 1A

4ohms seems a bit low for a 12V stepper motor - in a disk drive anyway.
Running from 5V, you would need about 1A per winding to hold this, so
pairing channels on the 2003 (or 2803 which is the 8 channel version)

Regards, Kevin

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2001\03\02@045534 by wzab

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On Fri, Mar 02, 2001 at 07:49:25AM +0930, Dingoblue mail wrote:
> A ULN2003 is a open collector bipolar driver.  There are 7 darlingtons
> within the package and each can switch 50 volts at 500mA.
>
> There is a total dissipation limit on the package and I suspect this has
> been well and truely exceeded.
>
Is the base current enough, so the transistor gets saturated when switched
on? What is the collector-emitter voltage, when the transistor is on?
If transistor is not saturated, Uce raises and dissipated power raises as
well...
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2001\03\02@052312 by mike

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On Thu, 1 Mar 2001 20:46:15 -0000, you wrote:

>Hi!
>
>>>I am driving a stepper motor -with 3 windings (4 Ohms each) and
>>>1 common terminal- salvaged from an old disk drive by means of
>[...]
>>>explained by means of the electric power formula :
>>>   W = V**2/R = 12**2/4 = 36W !?!?
>
>That´s the whole power you are sinking from the supply. As the 2003 is an open collector transistor, when it is driving the windings the voltage across it is about 0.2 volts,
>so the power the transistor is disipating is V*I = 0.2*3 = 0.6 W
Wrong, the 2003 is a darlington, which will drop at least 0.6V,
therefore it will get VERY hot at 3A.

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2001\03\02@061951 by Roman Black

flavicon
face
Joan Ilari wrote:
{Quote hidden}

Hi Joan. The ULN 2003 is NOT suitable for higher current
loads, in fact is sucks for higher current loads.
You should not have more than 0.5A current though each
driver. The drivers have a pathetic 1.5v sat at max 0.5A
load, and the chip has a total package dissipation that
should not exceed 1w, better not to exceed 0.6w by my
preference. ULN2003 series are great for <250mA loads.


First, I suggest you get more acquainted with the currents
in your circuit. For beginners I suggest buying some
1ohm-1watt and 1ohm-5watt resistors, and with any new
device, use the resistor in series. Then you can just
measure volts on the resistor to tell you how much
current things draw. 1amp=1volt. :o)

>From your post you calculated 3A but said your PSU won't
make 3A. You should have measured amps and know for sure!!

The motor you have is not a stepper motor, it is a 3-phase
sync motor from the disc platter (spins the disc). These
are only suitable for low torque, high speeds. It can
be run as a stepper motor but won't have great performance.

You are driving it at too much current in the attempt
at getting enough holding torque. You might be able to do
this if you keep the motor cool enough. It DOES NOT take
3A to run that motor!! The driver chips normally use PWM
to limit the motor current.

My suggestions:

1. if you have to use that motor, get 3 power transistors.
Use low-sat BD203, TIP32, etc. Or mosfets if you are already
happy with them. Then download "jones on steppers" and read
the chapter on current limiting, using chopping.

2. Best option is to get another stepper motor, one more
suited for low speeds and high holding torques. You can get
these for $5 or less from most surplus electronics places
on the net. Or search garage sales for old computer printers
and get the motors from these. You can get two $20 motors
out of a printer that cost you $2. I do it all the time.
:o)
-Roman

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2001\03\02@070230 by Russell McMahon

picon face
> "I've seen things you people wouldn't believe. ULN2003's on
>  fire off the shoulder of Orion. I watched PICs glow in
>  the dark near the Tannhauser gate. All those moments will be
>  lost in time, like tears in rain. Time to die..."
>                                                -Blade Runner- (amended :-)

You've had quite a lot of good advice on the ULN2003.
Here's some practical tips:
If you want to know WHY its getting hot, read on.
If you want to know how you SHOULD be doing it ignore most of the following
and skip to the end (A BETTER WAY) .................
:-)


- Have a look at the total package current capabilities and measure the
ACTUAL current being drawn from the power supply.
- Drive one phase fully on in the configuration you will be using and
measure the DC drop across the driver with a meter. If you have an
oscilloscope you can measure this "saturation" voltage directly.
- Power dissipation is the above voltage and current multiplied together.
Anything over about 500 mW is getting fairly warm and over 1 watt is toasty.

The total package has a rated current of 2.5 amp.
The peak current per driver is 500 mA.
The N (DIP) package has a rated dissipation of 1150 mW and a thermal
resistance of 9.2 mW per degree C.

At 500 mA you can use only 1 driver on average always on at 450 mA
With 4 drivers on you can do 180 mA at 100% on or 4 x 500 mA at 25% duty
cycle.

Each driver has a saturation voltage of 1.5 volts !!!!!!!!!!! when driven at
full rated input current (500 uA).
Paralleling two drivers only reduces this to about 1.2 volts!!!

To get full drive you need the better part of 5 volts on the input.
(Input is 2.7 kohm into a 2 diode drop darlington input)
2.7k x 500 uA + 2 x 0.6v = 2.6 volts

Current gain is about 1000 so you dont need much more drive than the above.

I'd say that the ULN2003 has far too low a rating for what you are trying to
do.

A BETTER WAY .................

Using separate TO126 or TO220 power transistors with a modest heatsink would
work much better.
As normal transistors will not be able to be driven directly by a processor
use of a darlington would be better eg

4 amp rated

   NPN        BD675, BD677,  BD682 etc
   PNP        BD680

5 amp

   NPN    TIP120,/121/122
   PNP    TIP125/126/127

etc

Using a 470 ohm resistor from PIC pin to base will allow direct drive from
PIC with adequate drive (Beta typically 750)


That should be enough to go on with ...



regards


               Russell McMahon


{Quote hidden}

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2001\03\02@090626 by Olin Lathrop

face picon face
> A ULN2003 is a open collector bipolar driver.  There are 7 darlingtons
> within the package and each can switch 50 volts at 500mA.

A darlington is a particularly bad choice in this instance.  The high on
voltage is a significant portion of the 12V supply voltage.  At the voltages
and currents mentioned here (12V, 3A), FETs are the best choice.


*****************************************************************
Olin Lathrop, embedded systems consultant in Devens Massachusetts
(978) 772-3129, @spam@olinspam_OUTspam.....embedinc.com, http://www.embedinc.com

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2001\03\02@090718 by Olin Lathrop

face picon face
> If it's not an stepper motor it is very well disguised :-)
>
> The motor does turns clockwise, counterclockwise, I can change its
> speed, etc... and I have done all the pic programming assuming
> that IT IS a stepper motor. Thus, I understand that there is only
> one problem : my ignorance (which, BTW is being incrementally reduced
> by means of the comments and suggestions in this list).

Do you have any evidence that this is not a brushless DC motor?  That would
be more likely if it turned the disk drive spindle.  If it moved the heads
then it is more likely a stepper.


*****************************************************************
Olin Lathrop, embedded systems consultant in Devens Massachusetts
(978) 772-3129, olinspamBeGonespamembedinc.com, http://www.embedinc.com

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2001\03\02@104815 by Roman Black

flavicon
face
Russell McMahon wrote:

> A BETTER WAY .................
>
> Using separate TO126 or TO220 power transistors with a modest heatsink would
> work much better.
> As normal transistors will not be able to be driven directly by a processor
> use of a darlington would be better eg


Just to be argumentative, I always use bipolar power
transistors directly driven by PIC pin, when I can.

Try BD203,  TO220 pack, (cost 50c)
= 1A, 0.2v C-E, base current 15mA.
(some TIP series are just as good and cheap)

There are good cheap low-sat bipolars around, you
just know where to look! A darlington will get hot
with 1A continuous @ about 0.9 watt, a good low sat
transistor only 0.2 watt and stome cold. :o)
-Roman

PS. Also with the BD203, I measured 19mV C-E saturation
with a 30mA load! That's almost a relay! :o)

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2001\03\02@185551 by David P. Harris

picon face
Hi-
The disk drive spindle (platter) motors are very flat (and beautiful if you
take them apart).  The stepper motors are short and squat.
D

Olin Lathrop wrote:

{Quote hidden}

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2001\03\02@194242 by Russell McMahon

picon face
> > A ULN2003 is a open collector bipolar driver.  There are 7 darlingtons
> > within the package and each can switch 50 volts at 500mA.
>
> A darlington is a particularly bad choice in this instance.  The high on
> voltage is a significant portion of the 12V supply voltage.  At the
voltages
> and currents mentioned here (12V, 3A), FETs are the best choice.


If using FETs and driving directly from PIC you will need "logic FETs" which
(typically) accept a 5 volt gate level.
A standard FET will need the better part of 10 volts.
You can use extra driver transistors and standard FETs but the extra
complexity is hardly worth while.


A Darlington is OK as long as you use a transistor able to handle the
dissipation and current.
A decent darlington in a TO220 pkg should drop no more than 1 volt (probably
less) at 3 amps. This is a 3 watt dissipation (and probably less as duty
cycle per phase will not be 100% usually) so a fairly minimal heatsink will
suffice.


regards

               Russell McMahon

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2001\03\02@222932 by Russell McMahon

picon face
> Russell McMahon wrote:
>
> > A BETTER WAY .................
> >
> > Using separate TO126 or TO220 power transistors with a modest heatsink
would
> > work much better.
> > As normal transistors will not be able to be driven directly by a
processor
> > use of a darlington would be better eg
>
>
> Just to be argumentative, I always use bipolar power
> transistors directly driven by PIC pin, when I can.

I'll see your argument and raise you a clarification :-)
It's all about Beta (current gain) and how much current you want to switch.
If Beta = eg 100 you need 1 mA of base drive to switch 100 mA of collector
current.

A good everyday small signal transistor such as eg a BC337 has a Beta of
typically 300 and a maximum rated current of 500 mA. This means that you can
drive it with 500/300 = 1.6 mA. Allowing 5 or 10 mA would be wise - a PIC
will handle this easily.

Going to higher currents the sitiuation changes.
You can either buy the utterly superb products from Zetex who shoehorn 3 amp
transistors into an ELine TO92 like package with betas of 100 plus
OR
You can largely buy standard TO220 transistors with Betas of typically 25 to
50 AT HIGH CURRENTS.
Getting good betas at 10's of mA is easy but as soon as you go to currents
of a few amps (as is the case here) beta will usually fall off markedly.

It just happens that I have been Beta testing transistors a few days ago to
see how much bang for the buck I can get. If this was a one off I would use
Zetex parts and be very happy. As it's for a volume task $$$$'s count.

I am using PNP transistors which are generally slightly inferior to NPN but
the same general lessons apply.
With a forced Beta of about 30 (Collector current divided by base current =
30) I get saturations around 0.2 volts using a TIP30 (BD640 equivalent).
With a forced Beta of 100 saturation varies from not much more than this to
as much as 5 volts !!!! :-)
The difference is in the static beta of each transistor. Selecting onntest a
transistor with a "meter beta" (ie measured using a standard meter beta test
facility) of 50 will need a forced beta of 10 or 20 at 1 amp. A transistor
with a meter-beta of 100+ will be OK at 20 or 30 forced beta.
YMMV (nay, YMWV).
With a max PIC current of 20 mA and a forced Beta of even 30 the best
current you can supp[ort is 600 mA! (20*30)
To support 3 amps you ned a beta of 3000/20 = 150.
Reach for the Zetex transistors quickly !!!!!!!!

The BD203 you mention (slightly higher current rating than a TIP41) has a
specified minimum Beta of 30 at 3 amps.

A FET can be the cure for this.
to get a Vsat of 0.2 volts at 3amps you need an Rdson of 0.2/3 = 0.066 ohms.
This is easily achieved at quite low cost. (Under $US1 in 100's for a 14 amp
part with an Rdson of less than this).
To run this from a PIC pin you will need a logic FET.

A Darlington has beta's of typically 200 to 1000 at 3 amps.
With 20 mA drive this gives 20 * 200 = 4 amps at a beta of 200.

{Quote hidden}

The 3 is a good transistor but is still only as good as its sepcification.
Measure it at 3 amps Ic  tell us what you get.
Also do it at different base drive levels.
Try it with a PIC driving the base directly.
Even with NO resistor twixt base and PIC I reckon you won't get much more
than 1 amp collector curreent at any sort of reasonable saturation voltage.


regards


               Russell McMahon

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2001\03\03@041111 by Jinx

face picon face
From: Roman Black

> Just to be argumentative, I always use
> bipolar power transistors directly driven by PIC pin, when I can

In the case of HDD spindle motors though, to the best of my
recollection, all of the older ones I've ever seen/dismantled have
used discrete SMT FETs

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2001\03\03@045935 by Peter L. Peres

picon face
Reduce the PSU to 4V or use series resistors for each phase of 8 ohms each
(and several watts). The ULN should only switch 0.5A (max) per phase. You
need to use something else for 3A. Also your motor sounds like it's the
spindle motor and that's impedance is definitely not 4 ohms at its working
speed (of 7200 rpm or so). It should be driven with a three phase driver
not with a unipolar driver like the ULN. There are no 36W motors in any
disk drive I know of after about 1995 (which does not mean much). The real
spec of your motor is likely around 8W electrical (at nominal speed). At
low speed it may require forced cooling.

Peter

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2001\03\03@073445 by Roman Black

flavicon
face
Jinx wrote:
>
> From: Roman Black
>
> > Just to be argumentative, I always use
> > bipolar power transistors directly driven by PIC pin, when I can
>
> In the case of HDD spindle motors though, to the best of my
> recollection, all of the older ones I've ever seen/dismantled have
> used discrete SMT FETs
>


This is usually because the engineer is trying
to get max bang for the buck with a tiny cheap
PWM solution.

This doesn't mean it's the best or most reliable,
especially for a hobby use!! The original post
gave the impression that Joan was cooking a simple
darlington driver chip, probably not the best to
suggest tiny fet PWM options... :o)
-Roman

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2001\03\03@093514 by miked

flavicon
face
> To get full drive you need the better part of 5 volts on the input.
> (Input is 2.7 kohm into a 2 diode drop darlington input)
> 2.7k x 500 uA + 2 x 0.6v = 2.6 volts
I roasted some once trying to drive them with the parallel port of a laptop
which didn't have full TTL level drivers.

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2001\03\03@150626 by Olin Lathrop

face picon face
part 1 2378 bytes content-type:text/plain; (decoded 7bit)

> If using FETs and driving directly from PIC you will need "logic FETs"
which
> (typically) accept a 5 volt gate level.

That may be a good solution, although they can be more expensive and have a
higher on resistance.

> A standard FET will need the better part of 10 volts.
> You can use extra driver transistors and standard FETs but the extra
> complexity is hardly worth while.

There are FET driver ICs just for this purpose.  The choice may come down to
a FET driver chip or a heat sink.

> A Darlington is OK as long as you use a transistor able to handle the
> dissipation and current.
> A decent darlington in a TO220 pkg should drop no more than 1 volt
(probably
> less) at 3 amps. This is a 3 watt dissipation (and probably less as duty
> cycle per phase will not be 100% usually) so a fairly minimal heatsink
will
> suffice.

If you want to stick with the simplicity of driving bipolar transistors,
there are still better choices than darlingtons.  I would use a regular NPN
power transistor and drive it with another smaller NPN cascaded.  This is
similar to the darlington configuration, except the power transistor can now
go to saturation.  This reduces its on voltage and thereby reduces its power
dissipation.  See the attached GIF file.

Assuming a low gain of 10 for the power transistor (Q2) at 3A, it
would require 300mA of base drive.  There are lots of NPN transistors that
can do 300mA with a minimum gain of 50 (being very conservative here).  R2
is chosen to provide 300mA into the base of Q2 when the PIC output is high.
The voltage accross R2 is 5 volts minus the Q1 E-B drop, minuse the Q2 E-B
drop.  Assuming 700mV for the E-B drops, then R2 = (5 - .7 - .7) / .3 = 3.6V
/ .3A = 12 ohms.  The max power dissipated by R2 is V**2 / R = 3.6**2 / 12 =
1.1 watts, so use a 2W resistor.  The purpose of R1 is to make sure Q2 stays
off despite any leakage and to decrease the turn off time.  R1 = 300 ohms
causes it to only draw 2.3mA maximum, which is small enough to ignore for
the purpose of chosing R2.

Since Q2 is now being driven into saturation, its on voltage will be much
lower than in a darlington configuration.  Assuming its saturation voltage
at 3A is 400mV, its power dissipation when on is 3A * 400mV = 1.2 watts.
This would only require a modest heat sink.


part 2 10704 bytes content-type:image/gif; (decode)


part 3 347 bytes

*****************************************************************
Olin Lathrop, embedded systems consultant in Devens Massachusetts
(978) 772-3129, olinspam_OUTspam@spam@embedinc.com, http://www.embedinc.com

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2001\03\03@165014 by Bill Westfield

face picon face
Well, I've been wanting to drive spindle motors with digitial logic (normally
they get three-phase sine waves, right?)  And I'm pretty happy to see an
existance proof that this works, even if it did happen due to misconceptions.

Has anybody tried driving motors like this with things like TTL bus drivers?
At 64mA current capability, it seems like this would be close to being able
to make a motor move, at least in limitted torque situations...

BillW

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2001\03\03@172006 by Joan Ilari

flavicon
face
Bill Westfield wrote :

>  Well, I've been wanting to drive spindle motors with digitial
>  logic (normally
>  they get three-phase sine waves, right?)  And I'm pretty happy to see an
>  existance proof that this works, even if it did happen due to
>  misconceptions.
>
>  Has anybody tried driving motors like this with things like TTL
>  bus drivers?
>  At 64mA current capability, it seems like this would be close to
>  being able
>  to make a motor move, at least in limitted torque situations...

Yes it moves, but with this current, the motor is not able to go from
one step to the following one. It only slightly moves. Torque is almost
null.

OOPS !!! Sorry !! I did it again ! I didn't mean step ! I mean position !
No, I didn't mean position ! I mean ... ;-)

Joan

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2001\03\03@173654 by Joan Ilari

flavicon
face
I got an off-list e-mail from Joe Colquitt with the reference
to a page where a circuit and a program is described to drive
(without feedback) a spindle motor with a Pic. I include it
because I think it can be of interest :

http://home.clear.net.nz/pages/joecolquitt/0hdd.html

Thanks, Joe !

--------------------------------------------------------------
    Joan Ilari                 spamBeGonejoan.ilari@spam@spamterra.es
    Barcelona                  Voice:  +34 93 431 96 39
    Spain

"I've seen things you people wouldn't believe. Attack ships on
fire off the shoulder of Orion. I watched C-beams glitter in
the dark near the Tannhauser gate. All those moments will be
lost in time, like tears in rain. Time to die..."
                                              -Blade Runner-
---------------------------------------------------------------

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2001\03\04@131201 by rottosen

flavicon
face
Not recommended, but I once drove a small stepper DIRECTLY from a
PIC16C84. The motor was spec'ed for 5 volts and 35 ohms per phase. I
drove each of the 4 phases from 2 PIC pins. I also put a resistor in
series with the motor windings but I don't remember the value. There was
just enough torque to drive a pointer.

Note that this was a classic case of PIC abuse! The PIC survived but is
marked with a big X scratched into it to make sure I never use it for
anything real.


-- Rich

p.s.  When a friend asked where the protections diodes were I just
grinned. Of course, they are in the PIC.




William Chops Westfield wrote:
{Quote hidden}

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