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'[EE]: More efficient way to drop voltage.'
2002\07\23@140023 by A.J. Tufgar

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I know this is a bit of a stupid question but oh well.  :)

I've got a 1.2V solenoid running off a 5V supply.  Right now I just drop
the voltage with a resistor but this is wasteful as most of the power
does not go to the solenoid, but is lost as heat in the resistor.

I was looking around the net and saw that a better way to do a voltage
drop is a zener, but isn't the same power wasted through the zener as
the resistor?  Or is less wasted?

I was thinking of using a voltage regulator but this seems like
overkill.

So what's a more efficient way to power the solenoid from the 5V supply?

Thanks,
Aaron

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2002\07\23@140859 by Herbert Graf

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> I know this is a bit of a stupid question but oh well.  :)
>
> I've got a 1.2V solenoid running off a 5V supply.  Right now I just drop
> the voltage with a resistor but this is wasteful as most of the power
> does not go to the solenoid, but is lost as heat in the resistor.
>
> I was looking around the net and saw that a better way to do a voltage
> drop is a zener, but isn't the same power wasted through the zener as
> the resistor?  Or is less wasted?
>
> I was thinking of using a voltage regulator but this seems like
> overkill.
>
> So what's a more efficient way to power the solenoid from the 5V supply?

       ANY linear method (resistor, zener, linear voltage regulator) will waste
the same amount of power. The only way to save power in your case is a
switcher, they can be as high as 95% efficient. Of course the cost can be
quite high, depending on what you've got. What is controlling the solenoid?
Is it possible that the "on" signal can be pulsed instead of just on? TTYL

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2002\07\23@140906 by Brendan Moran

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-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1

A 1.2V solenoid, you say?  Well, it's your lucky day!!  Just use an
LM317, which has a regulation voltage of 1.2V.  Just don't bother
putting a resistor between its common pin and ground.  Tie its common
pin to ground directly.

- --Brendan

- {Original Message removed}

2002\07\23@142206 by Harold M Hallikainen

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       The zener, the resistor, and the linear regulator are all inefficient in
this application. The power dissipated (ignoring the low quiescent power
dissipation of a regulator) is (Vin-Vout)*I where I is the input current
= load current.
       A switching regulator would be the efficient way to do it, or get a
solenoid that runs off the supply you have (fewer parts). You should also
be able to "PWM down" the voltage to the solenoid. Put a diode across the
solenoid, connect one end to +5V, the other end to the drain of an FET.
Drive the FET with PWM output of a PIC with the duty cycle set to 1.2/5
(24%). This requires even fewer parts than the resistor "solution."

Harold


On Tue, 23 Jul 2002 13:59:55 -0400 "A.J. Tufgar"
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2002\07\23@142411 by A.J. Tufgar

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Herbert, an 'F873 is controlling the solenoid.

Right now I use the classic NPN driver circuit and a resistor is series
with the coil the get the 1.2V I want.

I leave the 1.2V on for 30ms then turn that transistor off and turn
another transistor drive circuit on (same thing with a higher value
resistor) that applies .5V to the coil.

I have thought of pulsing it, but I'm not sure it would work.

What would a switcher cost or how would I even begin to design it, I
know some basic principles of switchers but that's about it.

Would it be possible with a switcher to decrease the volatge to .5V
after 30ms?

Thanks,
Aaron

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2002\07\23@144332 by Herbert Graf

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> A 1.2V solenoid, you say?  Well, it's your lucky day!!  Just use an
> LM317, which has a regulation voltage of 1.2V.  Just don't bother
> putting a resistor between its common pin and ground.  Tie its common
> pin to ground directly.

       The poster wants to reduce power waste, I don't see how this solution will
waste less power then a resistor. TTYL

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2002\07\23@145132 by Herbert Graf

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> Herbert, an 'F873 is controlling the solenoid.
>
> Right now I use the classic NPN driver circuit and a resistor is series
> with the coil the get the 1.2V I want.

       Perfect, take the resistor out and replace it with a straight connection...

> I leave the 1.2V on for 30ms then turn that transistor off and turn
> another transistor drive circuit on (same thing with a higher value
> resistor) that applies .5V to the coil.

       Remove the second transistor and it's resistor, you can use the first
transistor for both purposes). Now, use the ccp module (or bit bang if you
are using it for something else) and generate a pulse train set to 24% on
for 30ms.

> I have thought of pulsing it, but I'm not sure it would work.
>
> What would a switcher cost or how would I even begin to design it, I
> know some basic principles of switchers but that's about it.

       In your case you can use what you've got.

> Would it be possible with a switcher to decrease the volatge to .5V
> after 30ms?

       Sure, use the timer in the PIC to time out after 30ms have passed (with
your pulse train set at 24% on) and then simply change the pulse train to be
10% on. I'm not sure what else you are doing with the PIC but if you have
the ccp module and one of the timers free you are good to go, if you have
neither then it takes a little more PIC horsepower, that depends on what
else you've got going on.

       Note though, you should have a diode across the solenoid in the reverse
direction of "normal" current flow, this is to save your transistor. When a
solenoid (or any inductive load) is abruptly shut off (don't worry, I won't
get into the math) it generates what can be a VERY high voltage (certainly
dozens of volts, even more). I once connected a neon bulb across the
terminals of a coil and the bulb flashed every time the coil was shut off,
neon bulbs usually light when the voltage reaches ~ 90V). TTYL

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2002\07\23@150006 by A.J. Tufgar

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Harold,
      I'm still confused about one aspect in your solution.  Did you
mean that I should keep the same solenoid if I PWM the valve or I should
get a new one?

If I do PWM this valve won't I put to much voltage (5V) to the coil,
once the PWM module goes high?  I'm afraid I don't understand the
concept of PWM to change the output voltage.  Does It somehow lower the
voltage to 1.2V?

I do love your solution though thank's alot.

I'll add this information to the post.  My valve works off of 1.2 @
100mA.  How would I PWM this valve off of a 5V supply?  Or how does this
solution work?

Thanks all,
Aaron

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2002\07\23@150833 by A.J. Tufgar

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Herbert,
       Thanks I think this answers my question.  I'm happy as long as I
can keep my valve and use less power.

I would still like understand the concept behind PWM the valve.
(I get it's a simple squarewave freq)  But how does the 5V not kill the
1.2V coil?  And how does the coil stay open during low periods of the
squarewave?

BTW, I do have the reverse diode there for protection.  :)  Although I
do have this question.  I normally pick a diode with a large max voltage
valve (say 200V), but how do I pick the max current valve?

Thanks,
Aaron

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2002\07\23@155634 by Matt Pobursky

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Do a transistor/FET/diode clamp driver to the solenoid from the PIC
pin. PWM the drive to the solenoid. 25% duty cycle = 1.2V, 10% duty
cycle = .5V (for holding current, I assume?). I've done this many many
times driving 5V or 12V solenoids from 24V sources.

Solenoids are pretty forgiving devices, but you may want to AC couple
your drive to the PWM transistor so that if the PIC locks up with PWM
signal "on" you don't cook the solenoid. you may also want to
experiment with the PWM period (my best guess is that 100Hz or so will
work well) and the effect of different duty cycles on pull-in time and
power dissipation.

Matt Pobursky
Maximum Performance Systems

On Tue, 23 Jul 2002 14:22:09 -0400, A.J. Tufgar wrote:
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2002\07\23@163728 by Brent Brown

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> I would still like understand the concept behind PWM the valve.
>  (I get it's a simple squarewave freq)  But how does the 5V not kill
>  the 1.2V coil? And how does the coil stay open during low periods of the
> squarewave?

Two reasons.

First, the limiting factor with a coil is power dispation and power
is proportional to current. Typically it is quite OK to apply more
current for a short period of time as long as the overall average
power dissipation is not exceeded. This is because it is a simple
electromechanical device and it obviously takes time for the wire to
heat up. Same applies to motors & lamps etc.

Second reason is inductance, and this is what applies with the PWM
scenario. You have to the look at it on a pulse by pulse basis. The
current in an inductor can not change instantly. During the first ON
cycle the current starts from zero and begins to ramp up. The current
charge/disharge curves for an inductor are the same as the voltage
charge/discharge curves for a capacitor. Likewise during the OFF
cycle the inductance wants to keep the current flowing (now through
the diode) and it slowly ramps down. Over a number of PWM cycles the
current will reach an average value proportional to the ON/OFF ratio
of the PWM signal. The current will have some ripple on it at the PWM
frequency. Net result is bacsically the same as applying a reduced
volatge. Valves have pretty high inductance so a PWM frequency of a
just few hundred hertz would be good to start with.

> BTW, I do have the reverse diode there for protection.  :)  Although I
> do have this question.  I normally pick a diode with a large max
> voltage valve (say 200V), but how do I pick the max current valve?

Diode needs a current & voltage rating the same as the valve. With
PWM the diode provides a path for the valve current during the OFF
time. Current through the valve can not change instantly as stated
above, so during the OFF time the current begins to ramp down from
the level it was during the ON time.

I hope this helps you understand. There is a quite a bit to grasp,
but a bit of experimenting will help a lot.

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16 English Street, Hamilton, New Zealand
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2002\07\23@165029 by Herbert Graf

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> Herbert,
>         Thanks I think this answers my question.  I'm happy as long as I
> can keep my valve and use less power.
>
> I would still like understand the concept behind PWM the valve.
>  (I get it's a simple squarewave freq)  But how does the 5V not kill the
> 1.2V coil?  And how does the coil stay open during low periods of the
> squarewave?

       What I am proposing only works because you are dealing with a
mechanical/physical system. What is happening is you are pulsing the 5V so
fast that the mechanical solonoid can't possibly react that quickly (think
of a light bulb, it is being "pulsed" at 120 times per second (in countries
with 60Hz grids) yet it doesn't flicker, this is because the bulb can't
possibly heat up and cool down that quickly. Flourescent bulbs on the other
hand CAN react that quickly and DO flicker (only they flicker at a rate so
fast the human eye doesn't notice, for the most part)). The 5V will not kill
the coil since it is there for a relatively short amount of time. I won't go
into the math of how all this works, you can however research this sort of
stuff on the web or at your local library.

> BTW, I do have the reverse diode there for protection.  :)  Although I
> do have this question.  I normally pick a diode with a large max voltage
> valve (say 200V), but how do I pick the max current valve?

       Generally you don't have to worry about it with typical solenoids, a 1A
part is usually sufficient. The current going through the part will be
significant, however it will be of very short duration. In order to properly
figure things out you would need alot more info and more math, and it
certainly is possible to figure out, I just don't think it would be worth it
in this case (unless you are building a million of them?). TTYL

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2002\07\23@174921 by A.J. Tufgar

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It makes so much more sense now.  Thanks to all.  :)
I forgot about how inductors store charge in thier magnetic field.
I'm sure there is alot of Math and as I'm in engineering I'm sure I'm
about to learn alot of it.  :)  But for now this should be sufficent.
Three last questions:  Is there any reason for using a FET or can I use
a run of the mill 2n2222?

If my reg can only supply 200mA at a time will it matter if it
instantaneuosly has to supply more (because of PWM)?
Also is there any reason to PWM LED's?  I also have a 7 seg in the same
project and I thought if there is advantages to PWM it (each segment
turns on for a seven of the time) I might as well.

Thanks again,
Aaron

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2002\07\23@181901 by Harold M Hallikainen

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Aaron,

       The solenoid is an inductor, so it takes time for the current to ramp up
once the voltage is applied (V=L di/dt, so the slope of the current rise
is V/L. Increasing V makes the current ramp up faster while increasing L
makes it ramp up more slowly. If there were no resistance in the
inductor, the current would ramp up for ever, reaching infinite current
in infinite time. The resistance of the inductor limits the current. As
we approach the limit, the ramp gets less steep, giving us the familiar
exponential curve).
       With the suggested circuit, the current in the solenoid ramps up when
the transistor is on, then ramps down through the diode when the
transistor is off. The average current is the same as if you'd applied
1.2V to the solenoid. The frequency of the PWM is normally chosen to be a
little above audible (maybe 30 kHz) so the noise doesn't bother you (but
may bother dogs). In this case, the lower limit to the frequency (other
than it being audible) is probably pretty low since I believe we're
dealing with heating effects instead of core saturation.
       Anyone have any better ideas on choosing the PWM frequency?

Harold

On Tue, 23 Jul 2002 14:59:16 -0400 "A.J. Tufgar"
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2002\07\23@182532 by jdubner

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Aaron,

The solenoid does not require 1.2V -- it requires some amount of
current.  It just happens that in steady state the voltage across it is
1.2V at that current.  But in getting to that steady state, the current
rises at a rate determined by the time constant composed of the L and R
of the solenoid coil plus any series resistance.

So, what to do?  Both efficiency and solenoid activation time can be
improved by keeping the "R" value as low as possible.  Short across the
resistor in your "1.2V driver" and shorten the 30 ms pulse to a length
that allows current to rise to its rated value.  Then leave that driver
off and turn on your ".5V driver" just like your current scheme.  Of
course, efficiency will not be changed in the .5V driver circuit.

Be sure to provide some sort of failsafe means of limiting the current
(fuse, PTC thermistor, etc.) in the event of a malfunction (e.g. software).

--
Joe


On 7/23/02 11:22 A.J. Tufgar wrote:
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2002\07\23@190213 by Russell McMahon

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part 1 1424 bytes content-type:text/plain; (decoded 7bit)

> So what's a more efficient way to power the solenoid from the 5V supply?

An excellent solution, and probably as cheap as you are going to get with a
standalone solution and still save power, is the relay driver proposed on
this list about a year ago OR Roman Black's 2 transistor regulator that was
derived from that.

See archives
No - I'll attach it as it's very good
From Richard Prosser
27 August 2001
re [EE]: re Design Challenge - Low power step down switching regulator.

See long and "interesting" thread in archives if interested where it went
from there.

       Russell McMahon

_______________


How about something like this.
The circuit attached is used to drive a 5V(~15mA) relay off a supply
ranging from about 20 to 70V.
I can't guarantee all the component values but they are about right. The
coil is the relay coil and the current is sensed by R7. If R7 became the
load and D1 became a zener at about 5.6V then it should work as a buck
converter. Transistor types are BF422/BF423 or BSR19A/BSR20A for the smt
version. Similarly the IN4148s are actually BAX12 or BAV99.  D2 stabilises
the feedback so that the hysterisis is stable with supply voltage. Input
voltage is limited mainly by transistor & diode ratings.

Richard P

(See attached file: relay_~1.gif)



----------------------------------------------------------------------------
----





part 2 7431 bytes content-type:image/gif; (decode)


part 3 131 bytes
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2002\07\23@190433 by Brent Brown

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On 23 Jul 2002 at 14:35, Harold M Hallikainen wrote:

>  The frequency of the PWM is normally chosen to be a
> little above audible (maybe 30 kHz) so the noise doesn't bother you
> (but may bother dogs). In this case, the lower limit to the frequency
> (other than it being audible) is probably pretty low since I believe
> we're dealing with heating effects instead of core saturation.
>         Anyone have any better ideas on choosing the PWM frequency?

I don't know about "better" but here is another idea. The current
through an inductor is inversely proportional to frequency, that is,
as freq goes up current goes down. This shows up when trying do do
PWM control of an inductive load as a non-linear response.

0% and 100% PWM duty cycle still correspond to 0% and 100% current
respectively, but 50% duty cycle for example may give say 45% load
current when operating at 400Hz but perhaps only 20% load current
when operating at 10kHz.

Get the idea? I am willing to stand corrected if this proves wrong. I
got the idea from playing around with PWM motor control. I found
lower frequency PWM, few hundred Hz, appeared to give greater
starting torque, compared to several kHz, and I think this is one of
the reasons.

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2002\07\23@193340 by Kevinhoward

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In a message dated 23/07/2002 19:01:28 GMT Daylight Time,
EraseMEtufgarajspam_OUTspamTakeThisOuTMUSS.CIS.MCMASTER.CA writes:


{Quote hidden}

Depending on current required-
If you are just experimenting and the application is not critical, you could
use several diodes in series (0.6/0.7v voltage drop per diode)  or  even a
suitable wattage Zener diode - crude but inexpensive!

PWM would be more efficient, I can offer advice on dedicated PWM chips, but I
am sure the Piclisters can solve your problem with a pic!!!!

Regards

Kevin

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2002\07\23@203258 by Mike Singer

picon face
Russell McMahon wrote:
.
.
> An excellent solution, and probably as cheap as
> you are going to get with a standalone solution
> and still save power, is the relay driver proposed
> on this list about a year ago OR Roman Black's
> 2 transistor regulator that was derived from that.
.
.
  I'm not quite sure that using relay or solenoid coil
in PWM is a good idea. If they are intended for
non-pulsed current, then they should be used this
way. (saturation, RF noise, coil reliability etc.)
  There are a lot of very chip DC/DC converters
for this voltages: for CPU on PC motherboard, for
example.

  Mike.

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2002\07\24@031520 by Vasile Surducan

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Tufgar,
There is a methode used "when my grandmother it was a
pretty young girl".

The ideea is very simple and better than any PWM, zenner in series,
resistor in series, etc because all of that are disipating power, and this
power must be dissipated on the solenoid. The solenoid must have two
coils, one with large current for start and another one with
small current only to maintain the solenoid on. There is only one
transistor for command and a self contact which is switching the high
current coil to low current coil after the solenoid was supplied. Is the
best methode for long time
solenoid supply. And can be modified for two transistor command or so.

regards,
Vasile
http://www.geocities.com/vsurducan


On Tue, 23 Jul 2002, A.J. Tufgar wrote:

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2002\07\24@042956 by Michael Rigby-Jones

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It's absolutely fine, really whats the difference between driving a solenoid with AC and PWM?

Mike

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2002\07\24@052328 by Vasile Surducan

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On Wed, 24 Jul 2002, Michael Rigby-Jones wrote:

{Quote hidden}

 None. Both have the same low magnetisation comparing with the same DC
command on the same solenoid. The effective mechanical power will be less
than using a DC command on the same coil. But will work if the
solenoid coil was corectly dimensioned ( number of turns, wire section ),
etc

best, Vasile

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2002\07\24@070527 by Russell McMahon

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> > An excellent solution, and probably as cheap as
> > you are going to get with a standalone solution
> > and still save power, is the relay driver proposed
> > on this list about a year ago OR Roman Black's
> > 2 transistor regulator that was derived from that.
> .
> .
>    I'm not quite sure that using relay or solenoid coil
> in PWM is a good idea. If they are intended for
> non-pulsed current, then they should be used this
> way. (saturation, RF noise, coil reliability etc.)
>    There are a lot of very chip DC/DC converters
> for this voltages: for CPU on PC motherboard, for
> example.

All good thoughts, but this is not PWM in the normal sense although the coil
is indeed turned on/off in a ratio required to achieve the design
current/voltage. It is effectively a "buck regulator" which is about as
unstressed and forgiving a regulation means as is available. Properly
designed the coil voltage is always within the rails and inductor current
flows continually. If desired the current can be designed to vary over quite
a small range around the design point. Coil saturation should no be an issue
if the specified operating current does not saturate the coil. If it does
then any design using the same parameters will do so. It is unlikely that
you could build an IC based switching regulator for a component cost much
cheaper than for this one.
I haven't used this circuit myself but it is broadly conceptually similar to
the one that I posted at about the same time. .



       Russell

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2002\07\24@084552 by Roman Black

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A.J. Tufgar wrote:
>
> I know this is a bit of a stupid question but oh well.  :)
>
> I've got a 1.2V solenoid running off a 5V supply.  Right now I just drop
> the voltage with a resistor but this is wasteful as most of the power
> does not go to the solenoid, but is lost as heat in the resistor.
>
> I was looking around the net and saw that a better way to do a voltage
> drop is a zener, but isn't the same power wasted through the zener as
> the resistor?  Or is less wasted?

Hi A.J, first, anything magnetic (with coils of wire)
like solenoids, motors etc are *current* operated.
More current = more magnetism = works better. :o)

Nobody mentioned diode speed in their posts, so I will
mention it here. If you want efficient PWM in your
solenoid driving, you need a high-speed diode across
the coil. Any high-freq SMPS or PWM needs FAST diodes.

When the transistor is ON, the current through the coil
slowly increases, then when the transistor turns OFF,
the current (and magnetic field) can decay slowly or
fast-ly. :o)

If you have NO coil diode the field decays quickly, giving
a very high peak voltage (bad!). If you have a diode across
the coil, the field will decay SLOWLY, because the decaying
field itself causes current, and that current flows through
the diode, meaning the magnetic field is contained in the
coil and takes some time to decay.

So to operate the solenoid *efficiently* you need the
magnetic field to decay slowly, so you can add a tiny bit
of current into it each pulse from the PWM.

You need a FAST diode, like a 1N4148 (1N914) for solenoids
under 100mA, or something like a 1N4937 good for about 1A
solenoids. You CAN parallel 1N4148's (I can see the piclist
arguments now) to get >100mA, as they have gentle "knee"
and will parallel quite well. :o)

A 1N4007 etc (slow-ass diode) will work ok to reduce the
back-emf (high voltage) produced by the coil when it is
turned off. These slow diodes are perfect for this because
of their effective resistance they are like a diode-resistor
snubber. BUT for efficient PWM into a coil you don't want
to waste heat on the diode so the fast diode is a *must*
especially at PWM fequencies over 5kHz.

I suggest a longer initial turn on, probably 100mS etc,
to ensure nice powerful pull-in of the solenoid, then
after that reduce the PWM to a point where the coil JUST
holds in reliably. You will get *extrememly* efficient
solenoid operation. :o)
-Roman

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2002\07\24@085853 by Roman Black

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A.J. Tufgar wrote:
>
> It makes so much more sense now.  Thanks to all.  :)
> I forgot about how inductors store charge in thier magnetic field.

> If my reg can only supply 200mA at a time will it matter if it
> instantaneuosly has to supply more (because of PWM)?


Hi (again) A.J, i'm not sure your decision to drive
the solenoid coil from the regulated 5v rail is
the best. :o) If you MUST do this, I suggest the 5%
rule, ie a filter that sacrifices 5% of voltage to
give serious decoupling of the inductive coil and
the PIC digital circuit.

Example, if your coil takes 100mA, connect your
regulated 5v rail through a 5.6 ohm resistor, to a
large cap like 1000uF. Then run your solenoid PWM
circuit from that cap. You lose a few percent on the
resistor, but gain much in reliability. :o)
-Roman

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2002\07\24@122225 by Harold M Hallikainen

picon face
On Wed, 24 Jul 2002 07:41:52 +0100 Michael Rigby-Jones
<RemoveMEmrjonesspam_OUTspamKILLspamNORTELNETWORKS.COM> writes:
> >
> It's absolutely fine, really whats the difference between driving a
> solenoid with AC and PWM?
>
> Mike
>


       Also, as the switching frequency goes up (within reason), the ripple
current goes down, so the current through the solenoid becomes almost DC
anyway. It's just that part of the time the current is going through the
FET, and the rest of the time it's going through the diode. But, the
solenoid just sees a relatively constant DC.

Harold


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2002\07\24@122238 by Peter L. Peres

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On Tue, 23 Jul 2002, A.J. Tufgar wrote:

>I know this is a bit of a stupid question but oh well.  :)
>
>I've got a 1.2V solenoid running off a 5V supply.  Right now I just drop
>the voltage with a resistor but this is wasteful as most of the power
>does not go to the solenoid, but is lost as heat in the resistor.
>
>I was looking around the net and saw that a better way to do a voltage
>drop is a zener, but isn't the same power wasted through the zener as
>the resistor?  Or is less wasted?
>
>I was thinking of using a voltage regulator but this seems like
>overkill.
>
>So what's a more efficient way to power the solenoid from the 5V supply?

With a current chopper. Whatever it is your building sounds more and more
dangerous. Clocks, altimeters, solenoids, low power, small ?

Peter

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2002\07\24@123520 by A.J. Tufgar

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Just like to say thanks to all for the suggestions I got.  I've decided
on trying a 16F873 with it's internal PWM module.

Thanks again,
Aaron

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2002\07\26@031415 by Mike Singer

picon face
Vasile Surducan wrote:
> > It's absolutely fine, really whats the difference between
> driving a solenoid with AC and PWM?
> >
>   None. Both have the same low magnetisation comparing
> with the same DC command on the same solenoid. The
> effective mechanical power will be less than using a DC
> command on the same coil. But will work if the solenoid
> coil was corectly dimensioned ( number of turns, wire
> section ), etc

  I'll agree with you, Vasile, that there are no difference
between driving an inductor with AC and, say, with positive
voltage pulses, if you agree with that there are no
difference between driving a capacitor with AC and
with voltage pulses after diode rectifier.

  Mike.

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2002\07\26@032811 by Vasile Surducan

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On Fri, 26 Jul 2002, Mike Singer wrote:

> Vasile Surducan wrote:
> > > It's absolutely fine, really whats the difference between
> > driving a solenoid with AC and PWM?
> > >
> >   None. Both have the same low magnetisation comparing
> > with the same DC command on the same solenoid. The
> > effective mechanical power will be less than using a DC
> > command on the same coil. But will work if the solenoid
> > coil was corectly dimensioned ( number of turns, wire
> > section ), etc
>
>    I'll agree with you, Vasile, that there are no difference
> between driving an inductor with AC and, say, with positive
> voltage pulses, if you agree with that there are no
> difference between driving a capacitor with AC and
> with voltage pulses after diode rectifier.
>
 So, you're not agree...
best, Vasile

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2002\07\26@044556 by Mike Singer

picon face
Harold M Hallikainen wrote
> > It's absolutely fine, really whats the difference between driving a
> > solenoid with AC and PWM?
>
>         Also, as the switching frequency goes up (within
> reason), the ripple current goes down, so the current
> through the solenoid becomes almost DC anyway. It's
> just that part of the time the current is going through the
> FET, and the rest of the time it's going through the diode.
> But, the solenoid just sees a relatively constant DC.
>

  I think, solenoid does not just see a relatively constant
DC, instead, it tries hard to retain the current. The costs
of it's efforts are RF EMI, core heating, mechanical
vibrations, risk of coil isolation damaging due to higher
voltage applied and frequency etc.
  May be I'm wrong, since I didn't experiment with it.
But I wonder: why computer PSU output filters are made
of ferrite, not of steel? These filters "just sees a relatively
constant DC".

  Mike.

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2002\07\26@062116 by Mike Singer

picon face
Russell,
  Only you see this as a "buck regulator". Others think
of it as PWM.
  The difference is that "buck regulator"
applies voltage untill current reached level slightly
higher then desired, then waits untill current dropped
to level slightly less then desired and so on.
  "Buck regulator" frequency is mostly determined
by voltage, inductance and current gate.
  PWM produses more thick pulses when current is
lower then desired and thin pulses when current is
higher then desired.
  Good idea to choose PWM frequency is to choose
one about 2-4 times as "buck regulator" frequency.
  I think, "buck regulator" idea is better in our case,
since it works at lower frequency with our core made
of steel.
  Anyway, may be I've missed something, but didn't
see any hints to above-mentioned current loopback
in all postings of this thread. PWM and "buck regulator"
without (current,voltage) loopback are as US without
US dollars.=(:-)
  Another issue is 20-70v at your "relaydriver.gif".
Are your sure coil isolation, designed for 1.2v could
stand 70v high frequency pulses? Should I cite Olin's
usual remarks to this?
  I think, few bucks on standard (PC motherboard)
DC/DC regulator components are worth to be spent,
to avoid all this headache.

  Mike.

Russell McMahon wrote:
{Quote hidden}

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2002\07\26@101546 by Russell McMahon

face
flavicon
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> Russell,
>  Only you see this as a "buck regulator". Others think
> of it as PWM.

I'm not sure why you say that. Recall that this is Richard Prosser's
original circuit . He introduced it as a contestant for conversion from a
relay driver to a buck converter design. The load IS the inductor in his
case but this is the equivalent of coupling a resistive load to the buck
converter inductor.

>  The difference is that "buck regulator"
> applies voltage untill current reached level slightly
> higher then desired, then waits untill current dropped
> to level slightly less then desired and so on.

I agree, that is indeed one way of switching a buck converter. Output
voltage rise is also acceptable. This is in fact the same thing as current x
load impedance = load voltage so they amount to the same thing. That is
exactly how Richard's circuit works. See if you can follow the operation
and, if not, I could write out a description for you, if you were
interested,  to show that it is indeed "Buck action". The sensing here is
voltage based but as noted above, that is the equivalent of current sensing
here. Note that there is NO external driving source here. The oscillation is
determined only by load voltage considerations.

>    Anyway, may be I've missed something, but didn't
> see any hints to above-mentioned current loopback
> in all postings of this thread.

Work out how the circuit switches the load on and off and you'll see that it
is as expected for a buck.

>    Another issue is 20-70v at your "relaydriver.gif".
> Are your sure coil isolation, designed for 1.2v could
> stand 70v high frequency pulses?

No. But that is not what was being suggested. The original question that
started this thread involved running a 1.2v solenoid from 5 volts. (Aaron
said "I've got a 1.2V solenoid running off a 5V supply.") I suggested
Richard's circuit as applicable. The 20 to 70 voltas was what applied in his
original design. The circuit was an example which would need to be
redesigned for 5 volt operation. As Aaron is operating a solenoid and
Richard was driving a relay coil the load requirements are very similar.

I would be certain that any solenoid that I wished to use on 1.2V would
stand 5v breakdown voltage. Any solenoid will have a breakdown voltage of
many times this figure due to minimum insulation standards.

>Should I cite Olin's
> usual remarks to this?

Feel free if you want to :-)

>    I think, few bucks on standard (PC motherboard)
> DC/DC regulator components are worth to be spent,
> to avoid all this headache.

Headache?
Try it. You'll like it. Ask Roman!


               Russell

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2002\07\26@161055 by Michael Rigby-Jones

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>    I think, solenoid does not just see a relatively constant
> DC, instead, it tries hard to retain the current. The costs
> of it's efforts are RF EMI, core heating, mechanical
> vibrations, risk of coil isolation damaging due to higher
> voltage applied and frequency etc.
>    May be I'm wrong, since I didn't experiment with it.
> But I wonder: why computer PSU output filters are made
> of ferrite, not of steel? These filters "just sees a relatively
> constant DC".
>
>    Mike.
>
Because steel cores would be extremely lossy at the high frequencies used in
switching supplies and would probably saturate too easily.

Regards

Mike RJ

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2002\07\26@204104 by Mike Singer

picon face
Yes, I've missed the sense of the phrase (relay_~1.gif):

"The coil is the relay coil and the current is sensed by R7",

may be due to bad command of English and habit to see
this like "The coil _L1_  is..." and to see an external element
(relay coil) apart from a scheme.
Instead of this scheme, I'd rather prefer PIC with ADC,
a pair of transistors and some resistors.
Anyhow, I wouldn't stress 1.2v coil and steel core with 5v
pulses.

Thank you.
   Mike. (What a great is this List!)



Russell McMahon wrote:
{Quote hidden}

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2002\07\26@211225 by lexandre_Guimar=E3es?=

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Hi,

>  Anyhow, I wouldn't stress 1.2v coil and steel core with 5v
> pulses.


   I do not think it would stress the coil if the current is controlled
somehow. It is common practice to drive stepper motors with 10 times their
rated voltage and use a chopper driver to limit the current trough them.

   The circuit based on Richard's circuit that Roman draw as a power supply
works beautifully. I am using it for a while.

Best regards,
Alexandre Guimaraes

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2002\07\27@054041 by Mike Singer

picon face
Alexandre GuimarÇes wrote:
> Hi,
>
> >  Anyhow, I wouldn't stress 1.2v coil and steel
>> core with 5v pulses.
>
>     I do not think it would stress the coil if the current is
> controlled somehow. It is common practice to drive
> stepper motors with 10 times their rated voltage and
> use a chopper driver to limit the current trough them.
>
>     The circuit based on Richard's circuit that Roman
> draw as a power supply works beautifully. I am using
> it for a while.
>

  Could you tell me what voltage drop amplitude should be at current loopback resistor R7 in oder to reliably switch on and off "schmitt trigger"-like circuit. "Schmitt trigger" seems to have a fairly wide voltage gate. Wouldn't this resistor eat most of 5-1.2=3.8V, making this scheme almost equivalent to one resistor.
  Is it hobbyist or profesional "common practice to drive stepper motors with 10 times their rated voltage"?

  Best regards.
  Mike.

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2002\07\27@095138 by Peter L. Peres

picon face
On Fri, 26 Jul 2002, Mike Singer wrote:

>Vasile Surducan wrote:
>> > It's absolutely fine, really whats the difference between
>> driving a solenoid with AC and PWM?
>> >
>>   None. Both have the same low magnetisation comparing
>> with the same DC command on the same solenoid. The
>> effective mechanical power will be less than using a DC
>> command on the same coil. But will work if the solenoid
>> coil was corectly dimensioned ( number of turns, wire
>> section ), etc
>
>   I'll agree with you, Vasile, that there are no difference
>between driving an inductor with AC and, say, with positive
>voltage pulses, if you agree with that there are no
>difference between driving a capacitor with AC and
>with voltage pulses after diode rectifier.

Actually Ukrainean capacitors *are* different. They're Urkainean.

The soenoid has an air gap and that makes it very hard to saturate.
Therefore driving it with pulsed DC is just fine. Always assuming that it
is not Ukrainean, that is.

Peter

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2002\07\27@105504 by Russell McMahon

face
flavicon
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I'll leave the first part of the question to the previous poster.

> Is it hobbyist or profesional "common practice to
>drive stepper motors with 10 times their rated voltage"?

Not usually 10 times unless it is very low power BUT it is certainly very
common to drive stepper motors at significantly above their rated voltages
through a resistor so that the resistor drops the extra voltage when full
design current is reached. This method makes the volrtate supply look like a
crude current source and REDUCES the time constant of the circuit so that
the reqired current is reached more quickly. ir time constant T= L/R so that
the higher the R the lower the time constant. Obviously, adding the R
reduces the peak current that will flow.
For example, imagine that you have a 1.2V stepper that has a design current
of 100 mA. If you run this off 5 volts you need to drop (5-1.2) = 3.8 volts
in the series resistor. So R = V/I = 3.8/0.1 = 38r. In this case most of the
drive energy is wasted but the stepper current rises about 3 times faster.
Using a "real" current source is preferable.

With a "real current source" you may well apply the full drive voltage to
the stepper until the current has ramped to where you want it and then
reduce it to hold the cureent steadyish. (A "real" current source starts off
with infinite terminal voltage but obviously we don't wanty those in our
circuits :-).

While an overvoltage factor of 10 times would be asking for great trouble in
circuits rated at any decent voltage (say 20 volts rated),. when the rated
running voltage is very low it is extremely probable that the votltage that
it will withstand is much higher. This is because eg insulation that can
withstand say 12 volts is extremely thin and other considerations mean that
it usually much thicker than what is required for electrical purposes. For
instance, enamel or similar on wire to withstand only 12 volts would be so
thin that it would easily be damaged during winding unless one used a very
special and probably expensive insulating material.

I would imagine that the minimum withstand voltage of any coil would be 10's
of volta sand typically 100 to huindreds (and can be thousands in special
cases.)  It would always be wise to check with the manufacturer or the spec
sheets ;-)



       Russell McMahon

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2002\07\27@105629 by Alexandre Guimaraes

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Hi,

>    Is it hobbyist or profesional "common practice to
> drive stepper motors with 10 times their rated voltage"?

   All professional chopper drivers for steppers use much higher voltages
to drive the motors. One example is the gecko drivers that are very well
known by people that make CNC machinery. Another example is any reasonably
fast dot matrix printers head stepper. If you look at any epson dot marix
printer schematics you will find 40 to 58 volts to drive 5 volts steppers !!
It is "professional common practice"  for sure.
   It is not my specialty but if you need I can try to dig some printer
drivers schematics that I have lying around somewhere. They get quite fancy
at using high voltages with current control and in the way they "get away"
the kickback pulses of the motors. Some of them even use the pulses to
recharge the power supply capacitors ! You get much better torque and speed
from the motors by driving them with higher voltages. The limit is the
breakdown voltage of the coils. That should be much higher than 10 times the
rated voltage for almost any coil.

Best regards,
Alexandre Guimaraes

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2002\07\27@115310 by Roman Black

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Russell McMahon wrote:

> > Is it hobbyist or profesional "common practice to
> >drive stepper motors with 10 times their rated voltage"?
>
> Not usually 10 times unless it is very low power BUT it is certainly very
> common to drive stepper motors at significantly above their rated voltages
> through a resistor so that the resistor drops the extra voltage when full
> design current is reached. This method makes the volrtate supply look like a
> crude current source and REDUCES the time constant of the circuit so that
> the reqired current is reached more quickly. ir time constant T= L/R so that
> the higher the R the lower the time constant. Obviously, adding the R
> reduces the peak current that will flow.


One system that works *really* well to drive steppers
at very high speeds and any beginner can do, is to use
the common unipolar driver (ie 4 transistors, driven
in 2-phase on "wave" mode) but supply it from a simple
constant current source and higher voltage supply.

For instance I have one circuit with a 5v stepper
motor run unipolar mode (4 transistors) from a 24v dc
supply via a 1.6A constant current source on a decent
heatsink. I get really great high speed performance
from it. :o)

Engineers can argue for weeks about bipolar vs unipolar
and chopper constant current vs linear CC but the fact
is you can get really good high speed power with the
simplest of unipolar drivers.

The constant current transistor on the heatsink can
be as simple as one transistor, one zener and 2 resistors.
Nothing to be afraid of for beginners. :o)

Also unipolar linear CC drivers don't need the diodes
that chopper CC circuits do as they don't get the hv
spikes, they don't need h-bridges, and motors driven in
unipolar mode have slightly less holding torque than
bipolar driven motors but MUCH better torque at higher
speeds due to having only half the effective inductance.

With 4 low-sat transistors being almost free these days
I have given up building bipolar drivers, with the
clumsy discrete h-bridges or 4v sat (lousy) h-bridge
chips. In fact I have some unipolar CC microstepping
boards due back from the board house any day now.
:o)
-Roman

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2002\07\27@134207 by Katinka Mills

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On Sat, 27 Jul 2002 23:23, Roman wrote:

<elsnipo>

> One system that works *really* well to drive steppers
> at very high speeds and any beginner can do, is to use
> the common unipolar driver (ie 4 transistors, driven
> in 2-phase on "wave" mode) but supply it from a simple
> constant current source and higher voltage supply.

<elsnipo>

Hi Roman

Any chance of a rough schematic for the constant current source, I was
thinking of using a few LM317K,s but this would get a bit expensive (although
nowhere near a PBL / PBM chip set)

Regards,

Kat.
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2002\07\27@134337 by John Marshall

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part 1 3093 bytes content-type:text/plain; charset=us-ascii (decoded 7bit)

What do you define as " really good high speed" ? I am about to do an
update to a scanner design that currently operates at 4 ips @ 200 lpi.
Right now my limiting factor in speed is the stepper. It's a little 6
wire tin can job, with at 25:1 gear head. (Yes, it really does scan at 4
ips, and that's 12" wide, 24bit color. This is not a 59.95 scanner)

       Thanks,
       John Marshall



Roman Black wrote:
{Quote hidden}


part 2 382 bytes content-type:text/x-vcard; charset=us-ascii; name=john.vcf
(decoded 7bit)

begin:vcard
n:Marshall;John
tel;cell:817-917-7190
tel;fax:817-275-8938
tel;work:817-275-1311
x-mozilla-html:FALSE
url:niecor.com
org:Nova Instrument & Engineering
version:2.1
email;internet:JohnSTOPspamspamspam_OUTAFBEngineering.com
title:Sr. Design Engineer
adr;quoted-printable:;;1817 East Division St.=0D=0A;Arlington;TX;76011;USA
fn:John Marshall
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part 3 144 bytes
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2002\07\27@161559 by Peter L. Peres

picon face
>> One system that works *really* well to drive steppers
>> at very high speeds and any beginner can do, is to use
>> the common unipolar driver (ie 4 transistors, driven
>> in 2-phase on "wave" mode) but supply it from a simple
>> constant current source and higher voltage supply.

And burn power like crazy. The easiest current limiter is a resistor. By
Roman's standards, running a 12V 1A/phase stepper with a 220V directly
rectified supply and a small cooking ring (~300W) as limiting resistor is
ok, using 3A 400V transistors (which have horrible Vcesat and require
radiators).

By my standards driving with resistors is acceptable if running at most
50% effcient (power supply = 2x rated coil supply and resistor equals coil
R).

A chopped current drive is much much much better and can be achieved using
an opamp to sense the current at the bottom of the transistor switch. This
is used to turn on and off a high side switch that powers that phase (or
pair of phases). The easiest steering scheme uses schmitt trigger on the
read current. The coils do the rest (to provide for nice switching).
Kickback diodes are mandatory.

Normally a 1.5A,30V supply driver (45W input) for 12V/2..3Amp/ph motors
requires no heatsinks if designed correctly. The schmitt chopper together
with the kickback diodes and the motor coils forms a buck converter that
explains the discrepancy in the amps above.

Peter

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2002\07\28@004010 by Mike Singer

picon face
  Thanks to all who participated in the thread on
feeding an inductive device with pulsed DC current.
The thread was very usefull to me.

  Mike.

To Peter L. Peres:
How do your comments connect to message you
replied to? Personally I like Ukrainian theme jokes,
as well as other nations themes. (You know what
nation I mean in the first place)
My reply to Vasile's postings was about that
feeding a solenoid with _AC_ you can't reach values
of current, when driving with same width _DC pulses_,
due to inductance's ability to "accumulate" currents
from different DC pulses.
No any hint in my comments to Vasile's posting,
that I hate to drive it with pulsed DC.
If you suffer of some sort of incontinence of
postings, no problem, get you thoughts in separate,
properly tagged messages. We all will have fun.

Peter L. Peres wrote:
{Quote hidden}

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2002\07\28@004630 by Russell McMahon

face
flavicon
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part 1 1428 bytes content-type:text/plain; (decoded 7bit)

> > One system that works *really* well to drive steppers
> > at very high speeds and any beginner can do, is to use
> > the common unipolar driver (ie 4 transistors, driven
> > in 2-phase on "wave" mode) but supply it from a simple
> > constant current source and higher voltage supply.

> Any chance of a rough schematic for the constant current source, I was
> thinking of using a few LM317K,s but this would get a bit expensive
(although
> nowhere near a PBL / PBM chip set)

As long as you want currents in the 1 amp range (a big IF :-) ) the LM317 is
actually a very cost effective current source compared with most
alternatives and it has some advantages that are hard to match. One resistor
"programs" it. It is in a TO220 package with reasonable dissipation (also
smaller pkgs available) and it has internal thermal and short circuit
protection which is hard to achieve as cost effectively in other ways.  The
voltage rating may be a limitation as it must withstand full driving voltage
when the load is first applied.

In NZ the LM317T (TO220 package) costs about $NZ0.60 = about $US0.27 each in
tube quantities. Getting an equivalently rated transistor fir less is not
impossible but it has no thermal protection.


A simple current source schematic is attached
I = Vzener-0.6)/R1 approximately


       Russell McMahon



part 2 1857 bytes content-type:image/gif; (decode)


part 3 131 bytes
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2002\07\28@072025 by Roman Black

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Katinka Mills wrote:
>
> On Sat, 27 Jul 2002 23:23, Roman wrote:
>
> <elsnipo>
>
> > One system that works *really* well to drive steppers
> > at very high speeds and any beginner can do, is to use
> > the common unipolar driver (ie 4 transistors, driven
> > in 2-phase on "wave" mode) but supply it from a simple
> > constant current source and higher voltage supply.
>
> <elsnipo>
>
> Hi Roman
>
> Any chance of a rough schematic for the constant current source, I was
> thinking of using a few LM317K,s but this would get a bit expensive (although
> nowhere near a PBL / PBM chip set)


Hi Kat, no need for more than one LM317 is there?
You only need one constant current source and the
4 fets or transistors to switch the coils.

The simple one transistor circuit Russell posted
is what I use for most simple apps like this. :o)
-Roman

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2002\07\28@072856 by Katinka Mills

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On Sun, 28 Jul 2002 19:16, you wrote:
{Quote hidden}

Only need one per motor, but one of my designs has 18 steppers in it :o)

Regards,

Kat.

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2002\07\28@073313 by Roman Black

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John Marshall wrote:
>
> What do you define as " really good high speed" ?

My definition of "really good high speed" is my CNC
X/Y axis drivers, old (cheap) motors 5.1v 1A (which you
can get anywhere) currently doing 10 RPS, throwing a
2kg payload, with a toothed belt and 56mm diam pulley.
Accelerates to 10 RPS in half a revolution, travel
speed is 56mm x 10 = 560mm/sec = 22 IPS with a 2kg
load. It will do much higher top speeds than that, but
the G-forces currently throw the whole machine across
the table on accel and decel, which is the limiting
factor. Very nice for a $5 motor and driver built from
junk. :o)

> I am about to do an
> update to a scanner design that currently operates at 4 ips @ 200 lpi.
> Right now my limiting factor in speed is the stepper. It's a little 6
> wire tin can job, with at 25:1 gear head. (Yes, it really does scan at 4
> ips, and that's 12" wide, 24bit color. This is not a 59.95 scanner)

A tin-can motor will normally do about 10 RPS +
depending on it's voltage/current ratio and what
voltage you use in your CC circuit. Can you give
more details about the motor spec? These 7.5' motors
are a bit fussy regarding their primary mechanical
resonance.
-Roman

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2002\07\28@073741 by Roman Black

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Katinka Mills wrote:

> Only need one per motor, but one of my designs has 18 steppers in it :o)

Sounds interesting. There is a simple way to do
CC linear driving without using the CC source at the
top. My new driver does it like this with the 4 cheap
transistors doing both the phase switching AND the
linear CC control. What current/speed etc needs do
your motors have, and if you don't mind me asking
what type of project needs 18 steppers? A motorised
solar system maybe?? ;o)
-Roman

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2002\07\28@074802 by David Duffy

flavicon
face
>Katinka Mills wrote:
>Only need one per motor, but one of my designs has 18 steppers in it :o)

Roman wrote:
>Sounds interesting. There is a simple way to do
>CC linear driving without using the CC source at the
>top. My new driver does it like this with the 4 cheap
>transistors doing both the phase switching AND the
>linear CC control. What current/speed etc needs do
>your motors have, and if you don't mind me asking
>what type of project needs 18 steppers? A motorised
>solar system maybe?? ;o)

More likely "lighting related" with Kat? Do I smell a DMX fixture here?
If it is, it's going to be a doozy!
Regards...

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2002\07\28@075217 by Katinka Mills

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On Sun, 28 Jul 2002 19:35, you wrote:
> Katinka Mills wrote:
> > Only need one per motor, but one of my designs has 18 steppers in it :o)
>
> Sounds interesting. There is a simple way to do
> CC linear driving without using the CC source at the
> top. My new driver does it like this with the 4 cheap
> transistors doing both the phase switching AND the
> linear CC control. What current/speed etc needs do
> your motors have, and if you don't mind me asking
> what type of project needs 18 steppers? A motorised
> solar system maybe?? ;o)
> -Roman

Smart light for disco's :o) I am desiging one atm, it is a challange and a
half, DMX 512 in (250Kbps) then move 18 steppers (1 for pan, 1 for tilt, 1
for focus, 1 for dimmer, 1 for strobe, 3 for colour 4 for gobos (shapped
effects) 4 to rotate each gobo on it wheel, 1 for prism effects (to make 1
-12 coppies of image)

Now I know why pulsar and the other big boys have multiple 8051's, it is fun
updating lots of steppers at once. when I find some decent steppers I will be
laughing, the dynsin (IIRC) ones I got from a Martin 1220 are the best so
far, the ones out of 5 1/4" floppies are next to useless in this project,
they just do not appear to step reliably at high speed.

Oh and to make this all fun, each stepper also has a home opto aswell, need to
know where 0 is when spinning a wheel.

When I finally finish this all I will put the whole thing online (in the next
10 years as it is not a paid job ;o) The case will be the old Martin 1220 as
it was free (thanks to a nightclub who thought they could cover the fans as
they let light out ;o) worked for an hour then the light caught fire (1200
HMI globes tend to run hotter than most cable insulation can stand ;o)

Regards,

Kat.

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2002\07\28@075426 by Katinka Mills

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On Sun, 28 Jul 2002 20:06, you wrote:
> >Katinka Mills wrote:
> >Only need one per motor, but one of my designs has 18 steppers in it :o)
>
> Roman wrote:
> >Sounds interesting. There is a simple way to do
> >CC linear driving without using the CC source at the
> >top. My new driver does it like this with the 4 cheap
> >transistors doing both the phase switching AND the
> >linear CC control. What current/speed etc needs do
> >your motors have, and if you don't mind me asking
> >what type of project needs 18 steppers? A motorised
> >solar system maybe?? ;o)
>
> More likely "lighting related" with Kat? Do I smell a DMX fixture here?
> If it is, it's going to be a doozy!
> Regards...


Lol you got it dave, If I could find someone to rebuild my NEC 500mw tube I
would put that in with a few galvos and make one kewl effect ;o)

BTW any one know of any good sorces for laser tubes in AU ?

Regards,

Kat.

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2002\07\28@080253 by Chris Wheeler

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From: "Katinka Mills" <spamBeGonekatinkaSTOPspamspamEraseMEMAGESTOWER.COM>
> Lol you got it dave, If I could find someone to rebuild my NEC 500mw tube
I
> would put that in with a few galvos and make one kewl effect ;o)
>
> BTW any one know of any good sorces for laser tubes in AU ?

Oatley electronics used to sell them. Now mostly laser LEDs.

http://www.oatleyelectronics.com/

Worth emailing them, they may be able to direct you to a supplier.

--
CW

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2002\07\28@081744 by Roman Black

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Peter L. Peres wrote:
>
> >> One system that works *really* well to drive steppers
> >> at very high speeds and any beginner can do, is to use
> >> the common unipolar driver (ie 4 transistors, driven
> >> in 2-phase on "wave" mode) but supply it from a simple
> >> constant current source and higher voltage supply.
>
> And burn power like crazy. The easiest current limiter is a resistor. By
> Roman's standards, running a 12V 1A/phase stepper with a 220V directly
> rectified supply and a small cooking ring (~300W) as limiting resistor is
> ok, using 3A 400V transistors (which have horrible Vcesat and require
> radiators).

Sigh... By MY standards, using more than about 5x
the motor rated voltage is ridiculous and will give
almost no gains. In most cases with modern motors using
more than 40v is never needed. So you're an expert on
my standards huh? Since when? A resistor current limiter
won't give anywhere near the performance of a good CC
driver, you SHOULD know that.

> By my standards driving with resistors is acceptable if running at most
> 50% effcient (power supply = 2x rated coil supply and resistor equals coil
> R).

I never said to use resistors, and I *don't* use
resistors. Read my posts before you criticise, i'm
growing tired of correcting you because you have lousy
comprehension.

> A chopped current drive is much much much better and can be achieved using
> an opamp to sense the current at the bottom of the transistor switch. This
> is used to turn on and off a high side switch that powers that phase (or
> pair of phases). The easiest steering scheme uses schmitt trigger on the
> read current. The coils do the rest (to provide for nice switching).
> Kickback diodes are mandatory.

No it is NOT "much much much" better! Your attitude is
typical of semi-educated stepper "experts" who have read
some datasheets and manufacturers design notes, or done
"steppers 101" through some institution in *recent* times.
Chopper drivers DO NOT offer the same quality of performance
of a good linear constant current system. They will give
OK results at low speeds, and the ONE way they are better
than linear drivers is the reduced power needs.

They have significant problems with magnetic ripple,
recirculating currents, needing to switch between fast/slow
decay modes, etc etc. Chopper designs are a real tradeoff
of performance, where the only real winner is total power
dissipation at the cost of many performance issues.

Have you built a microstepping driver good for 200,000
microsteps per second? My new linear board will do that.
How do you propose to control the exact current of 200,000
individual steps/sec with a chopper driven PWM freq of
20,000Hz and 5% magnetic ripple? You only have ONE PWM
PULSE for TEN actual steps?

In most cases the linear driver can be operated at 33%
current in "hold" mode, with equivalent dissipation to
a chopper driver (good choppers run about 70%). But when
the motor is turning, especially at higher speeds the linear
dissipation is reduced by large amounts and a good linear
driver will run almost cold when the motor is *really*
pushing some loads. When the load stops you put it in back
to low power mode.

Chopper drivers are ok for low performance needs, where
you want to use one cheap chip to drive a motor at low
speeds and low rates of acceleration. They cannot compare
to the massive accelerations you can get with linear
drivers especially where the chopper PWM freq starts to
approach 25% of the step freq, which for a half-step
motor at 400 hsteps/rev equals 12.5 RPS. Long before that
you hit the major resonances, both mechanical and electrical,
and if they are near any harmonic of your chopper freq
then god help you.

If you are only interested in a couple of RPS, chopper
drivers are ok. If you don't want to get through the
first major resonances than choppers are ok. If you need
max holding torque at very low speeds choppers are great.
If you are afraid of heatsinks (seems you are) choppers
are great. If you need high accelerations and max
power/stability through resonant speeds choppers suck.

Just because the chip makers all want to push sales of
their one-chip chopper chips you shouldn't believe that
it is the ultimate way to drive a motor, it simply isn't.
-Roman

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2002\07\28@144611 by Mike Singer

picon face
Roman,
Peter's suggestions do not seem to me as ones
by"semi-educated stepper "experts"" in this case.
I admit, you have practical experience with
steppers, I have almost not. Nevertheless, it is not
absolutely clear for me why linear drivers can not
be succesfully replaced by some digital scheme.
Look, for example, at
---------------------
www.electronicstalk.com/news/pof/pof100.html
The TA3020 is the first 1bit digital amplifier IC using the
patented digital power processing technology (DPPT). This new
audio IC provides a breakthrough level of power and fidelity in a
highly integrated single package. The 2 x 350W Class-TT digital
amplifier is designed for the consumer electronics market. DPPT
incorporates a 1bit sigma-delta digital to analogue convertor
with a sampling frequency of 50MHz together with proprietary
signal processing algorithms to provide uncompromised fidelity
and unsurpassed power efficiency.
----------------------------

Or take regulated high-frequency PWM PSU, which
Brendan Moran experiment with now. I admit, stepper
motor coils can't be used as high-frequency supply
chokes. But nobody said they should.
You just replace linear driver with high speed regulated
DC/DC converter for energy saving purpose. Usually
they work at 300-400 khz, if I'm not mistaken.

  Mike.

Roman Black wrote:
{Quote hidden}

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