Searching \ for '[EE]: Solenoid information' in subject line. ()
Make payments with PayPal - it's fast, free and secure! Help us get a faster server
FAQ page: www.piclist.com/techref/io/motors.htm?key=solenoid
Search entire site for: 'Solenoid information'.

Exact match. Not showing close matches.
PICList Thread
'[EE]: Solenoid information'
2002\10\24@121041 by Dave Dribin

flavicon
face
I'd like to use a small, low power solenoid controlled by a PIC output
in one of my applications.  I'm looking for information on how to
correctly do this, including how to step up 5v or 6v up to 12v for the
solenoid using a switching regulator.  Pointers to books are welcome
(preferred?)  as I'd like to educate myself on this subject.  I'd like
one that has some focus on the "real world" use of solenoids and
switching regulators, not just theory, though.  I'd like to know
specific switching regulators that work well in a low power
environment.

Also, what is a good low-volume source for solenoids?  Digi-key
doesn't carry any (that I could find) and Jameco has a very limited
set.

Thanks!

-Dave

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads


2002\10\24@124335 by Alan B. Pearce

face picon face
>I'd like to use a small, low power solenoid controlled by
>a PIC output in one of my applications.  I'm looking for
>information on how to correctly do this, including how to
>step up 5v or 6v up to 12v for the solenoid using a
>switching regulator.

It sounds to me as though you are going backwards in the way you are
thinking. Where is the power for the PIC coming from? Normally from a
regulator that will be supplied by a voltage greater than 5V. If this is
close to 12V then you can use that to supply the solenoid, and all you will
need will be a transistor to do the voltage level conversion from the PIC to
handle the higher voltage and current of the solenoid.

An even better way of driving the solenoid will be a ULN2003 IC, which
contains 7 suitable drivers. Check the data sheet for this chip at the ON
semiconductor or Texas Instruments web sites. This data sheet also gives
examples of how to connect a solenoid or relay to these chips.

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads


2002\10\24@124747 by Sid Weaver

picon face
In a message dated 10/24/2002 12:11:13 Eastern Daylight Time,
spam_OUTdave-mlTakeThisOuTspamDRIBIN.ORG writes:


{Quote hidden}

Hi, Dave

The 7812 regulator will deliver 12VDC and has a max input of 35VDC.  Use a
ULN2003 running at 5VDC, with one side of the solenoid going to 12VDC and the
other side going to a 2003 output pin.  A high from PIC will close the
solenoid.

Go to parallaxinc.com and search for "solenoids" .  The have 2 or 3
industrial experiments re motors, valves and solenoids.

Would you e-mail me when you get home.  Thanks.

Sid

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads


2002\10\24@125954 by Larry Bradley

flavicon
face
A quick note on using solenoids - the current required to pull in the
solenoid and the current required to hold it in are usually quite different
- the holding current is perhaps 25% of the "pull" current. If power
consumption is a concern, your driver circuit can take this into account.

Larry

At 11:08 AM 10/24/2002 -0500, you wrote:
{Quote hidden}

Larry Bradley
Orleans (Ottawa), Ontario, CANADA

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads


2002\10\24@131045 by Dave Dribin
flavicon
face
On Thu, Oct 24, 2002 at 05:43:34PM +0100, Alan B. Pearce wrote:
> It sounds to me as though you are going backwards in the way you are
> thinking. Where is the power for the PIC coming from? Normally from a
> regulator that will be supplied by a voltage greater than 5V. If this is
> close to 12V then you can use that to supply the solenoid, and all you will
> need will be a transistor to do the voltage level conversion from the PIC to
> handle the higher voltage and current of the solenoid.

I'm trying to build a battery powered application.  I'd like to use
long-lasting lithium batteries.  The problem... lithiums seem to come
only in 3v flavors.  So I was gonna take 2 of those in series to get
6v.  Throw in a diode or two in there to feed 5v to the the PIC and
other digital circuitry.  The next problem would then lie in getting
12v for a solenoid.  This is where I thought a step-up switching
regulator would work.  I've read the data sheet on the LT1930:

 http://www.linear-tech.com/prod/datasheet.html?datasheet=749

It's got a bunch of example circuits that are all different in subtle
ways.  I don't know enough to read them and understand their
differences.

> An even better way of driving the solenoid will be a ULN2003 IC, which
> contains 7 suitable drivers. Check the data sheet for this chip at the ON
> semiconductor or Texas Instruments web sites. This data sheet also gives
> examples of how to connect a solenoid or relay to these chips.

I've got a TD62003AP, which is similar to the ULN2003, I believe.  I
guess I'm not confused on that part of the circuit too much.  I'm more
confused on how to choose a solenoid and estimate their current usage.
The solenoid specs are usually given in terms of a voltage and a
power, like 12VDC, 2W:

 http://www.electromechanicsonline.com/product.asp?pid=17

Can I just use P=IV to calcluate the current going through the
solenoid?  And why should I not pick a 6V solenoid?  That way no
step-up is needed at all.  Are there any gotcha's I'm missing?

-Dave

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads


2002\10\24@144224 by Olin Lathrop

face picon face
> An even better way of driving the solenoid will be a ULN2003 IC, which
> contains 7 suitable drivers. Check the data sheet for this chip at the ON
> semiconductor or Texas Instruments web sites. This data sheet also gives
> examples of how to connect a solenoid or relay to these chips.

The ULN2003 route is certainly easy, but note that the darlington drivers
will drop more voltage than a single transistor.  I would have the PIC pin
drive the base of an NPN thru a resistor, emitter to ground, collector to
the solenoid coil, other end of solenoid coil to 12V supply.  Don't forget
the backwards diode accross the solenoid coil or the flyback pulse will fry
something.

You said "low power" solenoid.  Let's say you use a 2K ohm base resistor.
That would yield about 2.2mA base current.  A generic small signal TO-92 NPN
transistor (I use 2N4401 in these cases) should give you a beta of at least
100, so counting on 50 is usually safe.  50 times the base current is
107.5mA, which is the maximum solenoid current this circuit will support
with a 2K base resistor.  You will have to work out the exact details for
your case.


*****************************************************************
Embed Inc, embedded system specialists in Littleton Massachusetts
(978) 742-9014, http://www.embedinc.com

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads


2002\10\24@144432 by Olin Lathrop

face picon face
> I'm trying to build a battery powered application.  I'd like to use
> long-lasting lithium batteries.  The problem... lithiums seem to come
> only in 3v flavors.  So I was gonna take 2 of those in series to get
> 6v.  Throw in a diode or two in there to feed 5v to the the PIC and
> other digital circuitry.  The next problem would then lie in getting
> 12v for a solenoid.  This is where I thought a step-up switching
> regulator would work.

Why do you need a 12V solenoid?  Why not chose a 6V solenoid?


*****************************************************************
Embed Inc, embedded system specialists in Littleton Massachusetts
(978) 742-9014, http://www.embedinc.com

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads


2002\10\24@145929 by Mike Singer

picon face
part 1 1704 bytes content-type:text/plain; (decoded 7bit)

Dave Dribin wrote:
> I'm trying to build a battery powered application.
> I'd like to use long-lasting lithium batteries.  The
> problem... lithiums seem to come only in 3v flavors.
> So I was gonna take 2 of those in series to get 6v.
> Throw in a diode or two in there to feed 5v to the
> the PIC and other digital circuitry.  The next problem
> would then lie in getting 12v for a solenoid.  This
> is where I thought a step-up switching regulator
> would work.

  Since you are trying to build a lithium battery powered
application, I suppose you are about using low-power
solenoids. In this case you can use head servos from
crashed PC HDDs. Excellent things.
  You need to measure current over the servo that
could be sufficient to drive your mechanics. Then for this
current you should select appropriate step-up regulator.
trying to choose those having in their datasheets sort of
"Li-Ion 12V Application". For example "LM2704 - Micropower
Step-up DC-DC Converter with 550mA Peak Current Limit"
(___attached .gif___)
   Through shutdown control input you can switch on/of
the servo.
  It's a good idea to use Step-up DC-DC Converter even
if you have 6v servo, as battery voltage is not constant.

> I'd like one that has some focus on the "real world" use
> of solenoids and switching regulators, not just theory,
> though.  I'd like to know specific switching regulators
> that work well in a low power environment.

Circuits in datasheets are "real world" circuits, I think.
Brand chipmakers use to think many times before
placing them in datasheets.

Mike.





part 2 3809 bytes content-type:image/gif; (decode)


part 3 154 bytes
--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads


2002\10\24@160754 by Dave Dribin

flavicon
face
On Thu, Oct 24, 2002 at 02:43:41PM -0400, Olin Lathrop wrote:
> > I'm trying to build a battery powered application.  I'd like to use
> > long-lasting lithium batteries.  The problem... lithiums seem to come
> > only in 3v flavors.  So I was gonna take 2 of those in series to get
> > 6v.  Throw in a diode or two in there to feed 5v to the the PIC and
> > other digital circuitry.  The next problem would then lie in getting
> > 12v for a solenoid.  This is where I thought a step-up switching
> > regulator would work.
>
> Why do you need a 12V solenoid?  Why not chose a 6V solenoid?

Don't need a 12V.  My usual part sources only had 24V and 12V parts
available.  I don't know the pros and cons of using different voltage
solenoids, even if I could find out were to get 'em.  Plus, I was
thinking that even if I go with a 6V or 3V solenoid, I'd still need a
regulator due to the battery voltage variations.

-Dave

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads


2002\10\24@162209 by Dave Dribin

flavicon
face
On Thu, Oct 24, 2002 at 09:59:08PM +0400, Mike Singer wrote:
> > I'd like one that has some focus on the "real world" use
> > of solenoids and switching regulators, not just theory,
> > though.  I'd like to know specific switching regulators
> > that work well in a low power environment.
>
>  Circuits in datasheets are "real world" circuits, I think.
>  Brand chipmakers use to think many times before
>  placing them in datasheets.

My main problem with the data sheet circuits is that they don't
explain them.  Take the LM2704 data sheet.  Figure 6 and Figure 7
differ only by 1 capacitor, yet they seem to have completely different
applications.  One takes an input of 2.5V to 4.5V whereas the other
takes an input of only 5V.  One can deliver 40ma while the other
180ma.  I don't really understand what the circuits are doing enough
to not only use them, but design one that fits my needs better than
one of their canned examples.  I figure there has to be a text book or
something that explains this stuff.

-Dave

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads


2002\10\24@173244 by William Chops Westfield

face picon face
> I'm trying to build a battery powered application.  I'd like to use
> long-lasting lithium batteries.  The problem... lithiums seem to come
> only in 3v flavors.  So I was gonna take 2 of those in series to get
> 6v.  Throw in a diode or two in there to feed 5v to the the PIC and
> other digital circuitry.  The next problem would then lie in getting
> 12v for a solenoid.

Solenoids, in general, are NOT "low power" devices, and I think you will
have a lot of trouble trying to drive any of the common types of solenoids
off of 2 lithium coin cells, no matter how clever you make hypothetical
voltage conversion circuitry.  (A typical coin cell will put about 40-100mA
into a short circuit.  That's not enough for most solenoids (whose power
consumption is typically ~10W, or nearly an AMP at 12V) , and the available
CURRENT will go down as you boost voltage.

When an application that traditionally uses solenoids is converted to low
power, you start to see things like motor-operated valves rather than
solenoid operated valves.  This way you do not consume power except when
changing state of the valve, and you can trade off current consumption
agains operating speed (but I think it'll STILL be a tough job to get
something that works well with a couple of coin cells!)

BillW

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads


2002\10\24@175323 by Bob Blick

face picon face
> > I'm trying to build a battery powered application.  I'd like to use
> > long-lasting lithium batteries.  The problem... lithiums seem to come
> > only in 3v flavors.  So I was gonna take 2 of those in series to get
> > 6v.  Throw in a diode or two in there to feed 5v to the the PIC and
> > other digital circuitry.  The next problem would then lie in getting
> > 12v for a solenoid.

"Long-lasting batteries" and "solenoids" in the same project? I don't
think so.

Look at how hose-end water timers (the type that screw onto a faucet) deal
with the problem:

First, do not start with more voltage than the processor uses. Typically
these timers will run on 3 or 4.5 volts. Any voltage regulator, no matter
how expensive, will use some current, typically more than a pic running at
32kHz.

Second, use a motor and geartrain instead of a solenoid. Motors not only
draw less current than solenoids, but you can shut them off, unlike a
solenoid that must be powered continuously to stay activated.

Third, how about alkaline batteries? Shelf life is not going to be your
problem, so why use lithum?

It may be that none of this applies to you, but your project hasn't really
been described so don't blame me if I didn't read your mind.

Cheerful regards,

Bob

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads


2002\10\24@180651 by David Minkler

flavicon
face
Dave,

An excellent primer to switching power supplies can be found at
http://www-s.ti.com/sc/techlit/slup067.pdf .  From the early '80s
Unitrode offered a series of annual (or semi-annual) seminars on various
topics related to swiching power supplies and their application.  When
TI purchased Unitrode they continued the seminar series and have put
most of the seminar topics from all of the seminars on their site.  Hope
this aids in understanding.

Regards,

Dave Minkler

Dave Dribin wrote:
{Quote hidden}

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads


2002\10\24@182540 by Dave Dribin

flavicon
face
On Thu, Oct 24, 2002 at 02:50:53PM -0700, Bob Blick wrote:
> "Long-lasting batteries" and "solenoids" in the same project? I don't
> think so.

Hmm... perhaps I am approaching this incorrectly.

> Look at how hose-end water timers (the type that screw onto a faucet) deal
> with the problem:
>
> First, do not start with more voltage than the processor uses. Typically
> these timers will run on 3 or 4.5 volts. Any voltage regulator, no matter
> how expensive, will use some current, typically more than a pic running at
> 32kHz.

Ok, but I'm only gonna be running the solenoid for a very short period
of time (see below).

> Second, use a motor and geartrain instead of a solenoid. Motors not only
> draw less current than solenoids, but you can shut them off, unlike a
> solenoid that must be powered continuously to stay activated.
>
> Third, how about alkaline batteries? Shelf life is not going to be your
> problem, so why use lithum?

I would like to approach on the order of years without changing the
battery on this stand-alone unit, if possible.  Maybe I'm incorrect
here, but I was under the impression that lithium batteries had far
more capacity (mAh) than conventional alkaline.  And rather than
target the coin lithiums, there are some AA sized lithiums that I was
looking at.  I think you can get a couple of these wrapped together as
a battery pack or something.

> It may be that none of this applies to you, but your project hasn't really
> been described so don't blame me if I didn't read your mind.

I've been purposely avoiding describing the whole project, so I could
do some reading and research w/o bothering everyone on the list.  But
since everyone is more helpful than anticipated (thanks, guys!), I'll
go into a little more detail.

I'd like to use a PIC to control access to a small spring loaded
drawer of some sort.  I'm no mechanical engineer, so my vision of this
may be way off.  But the way I see it, there is a latch holding this
drawer closed.  Then, the PIC will fire a solenoid (or something) to
release the latch.  The springs push the drawer out.  When done, the
user pushes the drawer closed, and it re-latches until next time.
This is similar to the way my laptop CD-ROM drive works.  I hit a
button on it, and the CD drawer pops out.  Once the CD is in, I push
it back, and it'll latch.  Note: the button is not physically or
mechanically connected to the latch on the CD-ROM drive.  There.. I
guess that wasn't so hard to describe. ;)

So, as you can see from this description, the solenoid is only fired
for a split second to release the latch.  Perhaps a motor is a better
fit for this scenaro.  Not a stepper motor, though, no?  I've no idea,
to be honest.  This is my first time interfacing the PIC to an
electromechanical device of any sort, and, like I said, I'm no ME.

-Dave

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads


2002\10\24@190741 by Bob Blick

face picon face
> I'd like to use a PIC to control access to a small spring loaded
> drawer of some sort.  I'm no mechanical engineer, so my vision of this
> may be way off.  But the way I see it, there is a latch holding this
> drawer closed.  Then, the PIC will fire a solenoid (or something) to
> release the latch.  The springs push the drawer out.  When done, the
> user pushes the drawer closed, and it re-latches until next time.

If it's a really small drawer, you might be able to use a solenoid from
an old answering machine. Solenoids are not very strong unless you put an
enormous amount of power into them. You might get one to experiment with
and see if it will work.

Another easy option is model airplane servos. They are cheap, about $10
mail order, and have motor, geartrain and all. Even if you rip the
electronics out of them, they're a great deal. And they run fine on 4.5
volts.

Lithium batteries, although they have greater capacity than alkaline, will
not produce the peak current you might need.

Cheerful regards,

Bob

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads


2002\10\24@192440 by William Chops Westfield

face picon face
   Maybe I'm incorrect here, but I was under the impression that lithium
   batteries had far more capacity (mAh) than conventional alkaline.

Well, yes.  But only for batteries of the same size.

   And rather than target the coin lithiums, there are some AA sized
   lithiums that I was looking at.  I think you can get a couple of these
   wrapped together as a battery pack or something.

A couple of 123A or CR2 sized lithium batteries (normally used in cameras,
and capable of providing impressive currents on demand) should work for
you. I'm not sure about the AA sized lithiums.  Some of these are 1.5V
for drop-in replacements of normal AA, and others are "long life but low
current" batteries that you wouldn't want.


   I'd like to use a PIC to control access to a small spring loaded
   drawer of some sort.  I'm no mechanical engineer, so my vision of this
   may be way off.  But the way I see it, there is a latch holding this
   drawer closed.  Then, the PIC will fire a solenoid (or something) to
   release the latch.  The springs push the drawer out.  When done, the
   user pushes the drawer closed, and it re-latches until next time.

This sounds doable.  In fact, it's been done.  In widely available toys,
such as http://shop.store.yahoo.com/iqkids1/ellocnsaf.html (Personally, I
hate it when my great ideas show up as mass-market toys before I get a
chance to implement them.  Grr.  I got a safe like this a couple of years
ago, full of candy.  Makes neat sounds as you push correct or incorrect
buttons.  Tiny magnetic latch...)

BillW

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads


2002\10\24@193514 by Russell McMahon

face
flavicon
face
> I'm trying to build a battery powered application.  I'd like to use
> long-lasting lithium batteries.  The problem... lithiums seem to come
> only in 3v flavors.  So I was gonna take 2 of those in series to get
> 6v.  Throw in a diode or two in there to feed 5v to the the PIC and
> other digital circuitry.  The next problem would then lie in getting
> 12v for a solenoid.  This is where I thought a step-up switching
> regulator would work.  I've read the data sheet on the LT1930:

Answers which would be useful:

What is the MINIMUM period between subsequent solenoid operations?
What is the longest on time?
What is the pull in current?
What is the hold in current?
Can you use a 5 volt one instead :-) ?

       RM

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads


2002\10\24@194806 by lexandre_Guimar=E3es?=

flavicon
face
   Seems like a perfect oportunity to use something like a photo flash
power supply with much less voltage. Charge a capacitor with a higher
voltage and dump it's charge on the solenoid. After that just put everything
back to sleep.

Best regards,
Alexandre Guimaraes

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads


2002\10\24@200129 by hard Prosser

flavicon
face
If you only need occasional & pulse operation - how about charging a
capacitor using a charge pump circuit & then discharging it  into the
solenoid
for the release.
This should be a minimal current drain solution while providing a healthy
current pulse when required.

Richard P

> I'd like to use a PIC to control access to a small spring loaded
> drawer of some sort.  I'm no mechanical engineer, so my vision of this
> may be way off.  But the way I see it, there is a latch holding this
> drawer closed.  Then, the PIC will fire a solenoid (or something) to
> release the latch.  The springs push the drawer out.  When done, the
> user pushes the drawer closed, and it re-latches until next time.

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads


2002\10\24@225347 by Wagner Lipnharski

flavicon
face
Dave Dribin wrote:
> I'd like to use a small, low power solenoid controlled by a PIC output
> in one of my applications.  I'm looking for information on how to
> correctly do this, including how to step up 5v or 6v up to 12v for the
> solenoid using a switching regulator.  Pointers to books are welcome
> (preferred?)  as I'd like to educate myself on this subject.  I'd like
> one that has some focus on the "real world" use of solenoids and
> switching regulators, not just theory, though.  I'd like to know
> specific switching regulators that work well in a low power
> environment.
>
> Also, what is a good low-volume source for solenoids?  Digi-key
> doesn't carry any (that I could find) and Jameco has a very limited
> set.
>
> Thanks!
>
> -Dave

Dave, contact Point Of Sale used equipment companies, you will find those
cash-registers with spring loaded drawers, lock by a small magnet that is
actuated by the small current supplied by the DTR signal from a computer
serial port...

The mechanical latch can hold an enormous spring loaded force, but a simple
mechanical trigger device can release the latch with minimum force, in this
case, by the small current from the COM port.

You can buy those latching mechanism with the solenoid probably by cents on
used parts suppliers.

The voltage converter shall be a second step on this issue.

Have fun.

Wagner Lipnharski - email:  .....wagnerKILLspamspam@spam@ustr.net
UST Research Inc. - Development Director
http://www.ustr.net - Orlando Florida 32837
Licensed Consultant Atmel AVR _/_/_/_/_/_/

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads


2002\10\25@034902 by Alan B. Pearce

face picon face
>And rather than target the coin lithiums, there are some AA sized
>lithiums that I was looking at.  I think you can get a couple of
>these wrapped together as a battery pack or something.

Look at the batteries used in cameras. Most of these are lithium for two
reasons, long shelf life, and high current capability to run motor drives.

>So, as you can see from this description, the solenoid is only
>fired for a split second to release the latch.  Perhaps a motor
>is a better fit for this scenaro.

Use a motor with a microswitch turn a cam one full revolution. The cam
pushes the latch spring to release the draw, and half a turn later pushes
the microswitch to turn the motor off. Use a small DC motor with a gearbox
to get the cam speed down to something suitable (check robotics stores for
suitable motor gearbox units). The PIC then needs only to turn the motor on
for long enough for the microswitch to short the drive transistor. Then turn
off the transistor, and motor continues until the cam pushes the microswitch
again.

--
http://www.piclist.com hint: The list server can filter out subtopics
(like ads or off topics) for you. See http://www.piclist.com/#topics


2002\10\25@052725 by Peter L. Peres

picon face
On Thu, 24 Oct 2002, William Chops Westfield wrote:

*>> I'm trying to build a battery powered application.  I'd like to use
*>> long-lasting lithium batteries.  The problem... lithiums seem to come
*>> only in 3v flavors.  So I was gonna take 2 of those in series to get
*>> 6v.  Throw in a diode or two in there to feed 5v to the the PIC and
*>> other digital circuitry.  The next problem would then lie in getting
*>> 12v for a solenoid.
*>
*>Solenoids, in general, are NOT "low power" devices, and I think you will
*>have a lot of trouble trying to drive any of the common types of solenoids

Not really. There are tons of solenoids in cameras and other small low
power devices. The secret is low on time and bistable solenoids where
power matters. A bistable solenoid has mechnical or magnetical bistable
(or multistable) positions and only short current pulses are required to
actuate it (example of multistable s.: solenoid + pawl + ratchet gear).

Peter

--
http://www.piclist.com hint: The list server can filter out subtopics
(like ads or off topics) for you. See http://www.piclist.com/#topics


2002\10\25@073508 by Olin Lathrop

face picon face
> Don't need a 12V.  My usual part sources only had 24V and 12V parts
> available.  I don't know the pros and cons of using different voltage
> solenoids, even if I could find out were to get 'em.  Plus, I was
> thinking that even if I go with a 6V or 3V solenoid, I'd still need a
> regulator due to the battery voltage variations.

12V solenoids will be easier to find, but it's worth looking around for
something that will do what you want with 6V.  The part that you need to
specify is the mechanical action.  At the least, you have to know:

*  Push or pull when power is applied.

*  Travel distance.

*  Minimum force required at the start of travel.

*  Minimum holding force required at the end of travel to keep it there.

These parameters depend on the mechanics of your device.  Once you have
these specs written down, look for a solenoid in the 6V range that does what
you want.  The solenoid force at any one point in travel is proportional to
the magnetic field, which is proportional to the current.  There is no magic
here.

Note that the voltage rating of a solenoid is only the maximum continuous
voltage you should apply.  There is no law saying you can't apply less.  For
example, you could find a 12V solenoid that has twice the force you need.
If you give it 6V it will draw half the current, which will result in
exactly the force you want.

Since power drain is a concern in your application, you can save some
battery life if the required holding force is less than the initial force
(often the case).  Also, solenoids tend to have higher force at the end of
travel than the start because the magnetic parts are closer together.  All
this means you can drive the solenoid with a high current initially, then
fall back to a lower current later.  You have to work this out carefully,
but this can be as easy as a resistor and large capacitor in the right
places.


*****************************************************************
Embed Inc, embedded system specialists in Littleton Massachusetts
(978) 742-9014, http://www.embedinc.com

--
http://www.piclist.com hint: The list server can filter out subtopics
(like ads or off topics) for you. See http://www.piclist.com/#topics


2002\10\25@115019 by Dave Dribin

flavicon
face
On Fri, Oct 25, 2002 at 07:32:53AM -0400, Olin Lathrop wrote:
> 12V solenoids will be easier to find, but it's worth looking around for
> something that will do what you want with 6V.

Sounds good.  I found this place:

 http://www.electromechanicsonline.com/

Anyone had experience with them or know of other solenoid sources?

{Quote hidden}

Ok, this may sound a little weird, but I don't know or care about
these requirements at this stage.  I wish I had a good explain of why,
but suffice to say, for this 'demo' if the solenoid pushes out into
thin air, it'll be ok.  Yes, I know it's weird. :) We're just looking
to investigate solenoids and get a feel for how the work.

> These parameters depend on the mechanics of your device.  Once you have
> these specs written down, look for a solenoid in the 6V range that does what
> you want.  The solenoid force at any one point in travel is proportional to
> the magnetic field, which is proportional to the current.  There is no magic
> here.

Given my previous "non" requirements... :) Let's say I chose this guy:

 http://www.electromechanicsonline.com/product.asp?pid=17

and get the 6V continuous DC version.  It's rated at 2W of power.
Does that mean this thing will always use 2W of power, under all
circumstances?  Is the current draw just 2W/6V or 333mA?  What if I
apply 3V, is the current draw then 667mA?

> Note that the voltage rating of a solenoid is only the maximum continuous
> voltage you should apply.  There is no law saying you can't apply less.  For
> example, you could find a 12V solenoid that has twice the force you need.
> If you give it 6V it will draw half the current, which will result in
> exactly the force you want.

Ok, I was under the impression that the voltage was the minimum
voltage to "enable" the solenoid.  After thinking about what you said,
though, it makes sense that this isn't the case.  I never really
thought about how solenoids worked before and I mistaken thought they
were binary: energized and not.  Maybe this comes from doing to much
digital logic.  I almost never have to deal with analog stuff. :)

> Since power drain is a concern in your application, you can save some
> battery life if the required holding force is less than the initial force
> (often the case).  Also, solenoids tend to have higher force at the end of
> travel than the start because the magnetic parts are closer together.  All
> this means you can drive the solenoid with a high current initially, then
> fall back to a lower current later.  You have to work this out carefully,
> but this can be as easy as a resistor and large capacitor in the right
> places.

Interesting idea.  I will definitely keep that in mind.

BTW, what applications would want a non-continuous duty cycle?

-Dave

--
http://www.piclist.com hint: The list server can filter out subtopics
(like ads or off topics) for you. See http://www.piclist.com/#topics


2002\10\25@143312 by Roman Black

flavicon
face
Dave Dribin wrote:

> Anyone had experience with them or know of other solenoid sources?
> Ok, this may sound a little weird, but I don't know or care about
> these requirements at this stage.  I wish I had a good explain of why,
> but suffice to say, for this 'demo' if the solenoid pushes out into
> thin air, it'll be ok.  Yes, I know it's weird. :) We're just looking
> to investigate solenoids and get a feel for how the work.

Hi Dave, your suggestion of using (say) 2x AA
cells and a low power solenoid for micro-power
use is perfectly valid.

I suggest activating the solenoid *only* from a
charged capacitor, especially as you only need it to
activate on rare occasions. I've used a similar system
with mechanical numerical counters, charge a cap
at fractions of a mA, then dump it into the solenoid
500mA coil to activate it and clock the counter.

I have a number of solenoids here with the manufacturers
charts, you will find coil "pull-in" current depends
on the distance it travels due to the magnetic path
etc of the solenoid. Proper mechanical design of your
latching mechanism will give very small travel, ie under
1mm total, with much reduced capacitor size to get it
to latch reliably. With a spring loaded drawer it WILL
spring open when the solenoid is dumped.

I suggest a solenoid rated for 5v or 6v, a buck regulator
(constant current) driven by the PIC to charge the cap
to your battery voltage and a low sat transistor or FET
to do the dump and pull the latch. Best if you only
charge the cap just before you use the latch so it is
normally in the discharged state and not leaking current.
:o)


> and get the 6V continuous DC version.  It's rated at 2W of power.
> Does that mean this thing will always use 2W of power, under all
> circumstances?  Is the current draw just 2W/6V or 333mA?  What if I
> apply 3V, is the current draw then 667mA?

No, what you do is charge a cap to your battery voltage.
This is done by a buck system that uses around 1/3 the
battery current as charging the cap from a resistor
based on my results. Takes about 1/2 a second.
Then the capacitor is dumped into the solenoid, and
the latch activates. The solenoid needs to work at
your expected minimum battery voltage, and the cap
size only needs to be big enough to give reliable
pull-in under all conditions.
-Roman

--
http://www.piclist.com hint: The list server can filter out subtopics
(like ads or off topics) for you. See http://www.piclist.com/#topics


2002\10\25@172224 by Jinx

face picon face
Have you considered magnetically latching relays ? These
need only pulses (eg charged cap) to change position, not
constant holding current. Google for "magnetically latching
relay"

eg http://www.metering.com/archive/972/08_1.htm

--
http://www.piclist.com hint: The list server can filter out subtopics
(like ads or off topics) for you. See http://www.piclist.com/#topics


2002\10\25@174143 by Jinx

face picon face
> Have you considered magnetically latching relays ? These
> need only pulses (eg charged cap) to change position, not
> constant holding current. Google for "magnetically latching
> relay"

Substituting "solenoid" for "relay" may get you closer to what
you were actually asking about !!!!!

Sorry, had relays on the brain this morning

--
http://www.piclist.com hint: The list server can filter out subtopics
(like ads or off topics) for you. See http://www.piclist.com/#topics


2002\10\26@155758 by Olin Lathrop

face picon face
> Given my previous "non" requirements... :) Let's say I chose this guy:
>
>   http://www.electromechanicsonline.com/product.asp?pid=17
>
> and get the 6V continuous DC version.  It's rated at 2W of power.
> Does that mean this thing will always use 2W of power, under all
> circumstances?

No.  2W is the amount of heat it can safely dissipate continuously.
Solenoids don't have an inherent current limit, but are limited by how hot
they can get before something fails.  Each solenoid will have a continuous
power rating at which the it can dissipate the heat continuously without
getting too hot internally.  That's why you can use higher currents for
shorter times.

You wanted a solenoid that would open a latch infrequently.  In that case
you can get a cheaper smaller solenoid and give it a lot of current for a
short amount of time.  The easiest way to do this is probably discharge a
large capacitor accross it.


*****************************************************************
Embed Inc, embedded system specialists in Littleton Massachusetts
(978) 742-9014, http://www.embedinc.com

--
http://www.piclist.com hint: The PICList is archived three different
ways.  See http://www.piclist.com/#archives for details.



'[EE]: Solenoid information'
2002\12\20@164252 by Dave Dribin
flavicon
face
Hi Roman,

I'm dusting off this old message you sent out a while back as I'm
gettting to play with this project again.  I bought a 6V solenoid with
a resistance of 20.8 ohms, the 36 AWG version of this:

 http://www.solenoidcity.com/solenoid/tubular/s-63-38hp1.htm

I also bought a 2-cell lithium battery pack, so that's my +6V power
supply, two BR-A cells:

 http://www.panasonic.com/industrial/battery/oem/chem/lith/cylin.htm

You mentioned using a cap to trigger the solenoid, using a buck
regulator to charge it.  What exactly is a buck regulator?  Some part
searches on Digi-Key turned up ICs that required a bunch of other
components.  Is there a particular setup you recommend?

BTW, could I do something simpler, like this, using a Darlington
transistor for Q1:

                   +6V (Battery)
                    |
                Solenoid (with reverse diode)
                    |
            C-------+
PIC-----R1---B Q1
            E
            |
           GND

The only issue I'm worried about is that the battery needs to source
300mA.  The battery data sheet mentions it can supply 1,000mA as the
maximum pulse current.  This would seem within spec, but perhaps it
will still pull the battery voltage down far enough where it will
affect the PIC.  I plan on sticking 1 diode between the battery and
the PIC to drop the voltage closer to +5V.

As usual, thanks for the excellent feedback.

-Dave

On Sat, Oct 26, 2002 at 04:26:45AM +1000, Roman Black wrote:
{Quote hidden}

--
http://www.piclist.com hint: The PICList is archived three different
ways.  See http://www.piclist.com/#archives for details.

2002\12\20@171027 by Wagner Lipnharski

flavicon
face
That's the point.

A simple larger capacitor connected to the lithium cells via a current
limiting resistor.
The resistor should be calculate to keep the cap charging current below the
lithium max current.

All the solenoid energizing current will be coming from the capacitor
rather than the cells.
So, the capacitor should also be calculated to have enough stored energy to
supply the necessary power to the solenoid.  Of course, during the solenoid
energize time, the lithium cell will also supplying some current via the
resistor, but most current will be supplied by the cap.

Remember that there is a cap charging time, gross calculated as time = RC,
what will limit solenoid consecutive activations to be longer than the cap
recharging time.

Wagner.


{Original Message removed}

2002\12\20@184555 by Bob Ammerman

picon face
In many electromechanical systems like this the holding current for the
solenoid can be quite a bit less than the pull in current. By correctly
sizing the R and C in this circuit you can have enough charge in the cap to
pull the solenoid in reliably, and then the solenoid is held in with the
current that flows thru R.

Bob Ammerman
RAm Systems

{Original Message removed}

2002\12\20@222645 by Wagner Lipnharski

flavicon
face
... or, you can have a double RC system, one to kick the solenoid, other
for the holding current, considering the holding time not to be to much
longer, sometimes the lithium cell can't supply the holding current.

           .---R1---o----R1A----.
           |        |           |
           |      +_|_          |
Lithium(+)--o       --- C1       o----> Drive
(-)         |        |           |
|          |       _|_          |
_|_         |       GND          |
GND         |                    |
           '---R2---o---->|-----'
                    |   Diode
                  +_|_
                   --- C2
                    |
                   _|_
                   GND


R2C2 = Recharge Time - Period between Kicks
R1C1 = Recharge Time - Period between Kicks
R1A  = Holding Current Limiter

C2   = Enough to supply kicking current
C1   = Enough to supply holding time current

R1, R2 = Current Limiter, under Lithium max current.

Diode = To avoid C1 current be diverted to recharge C2.



{Original Message removed}

2002\12\21@042150 by Roman Black

flavicon
face
Dave Dribin wrote:
>
> Hi Roman,
>
> I'm dusting off this old message you sent out a while back as I'm
> gettting to play with this project again.  I bought a 6V solenoid with
> a resistance of 20.8 ohms, the 36 AWG version of this:
>
>   http://www.solenoidcity.com/solenoid/tubular/s-63-38hp1.htm

> You mentioned using a cap to trigger the solenoid, using a buck
> regulator to charge it.  What exactly is a buck regulator?
>  Is there a particular setup you recommend?


Yes! :o) I drew this in a real hurry so please
forgive errors or sloppy parts values.

+6v (battery)
---------------------*-------------------------------
                     |
              BC327  |        L1=470uH
----,                E        ( <1ohm )
   C|----R---------B    Q1
    |   2k2          C
    |                |
   S|----R-----------|-----------------,
    |   270 ohm      |                 |
    |                |            A    |
    |                *----L1------*----|---*-----,
PIC  |                |            |    |   |     |
    |     ,----------|------------*    |   -    SOL COIL
    |     |          |            |    |   ^     |
    |     R          |            |    |   |     |
    |     |          |            CS   |   '-----*
   V|-----*          -       470uF|    |         |
    |     |          ^            |    |         C
----'     R   1N5819 |            |    '-------B    Q2
          |    scho  |            |              E  BC337
          |          |            |              |
----------*----------*------------*--------------*---
GND


PIC drives Q1 by LOW pulses on pin C (charge) these
pulses should be about 10uS long. Each pulse works
through L1 to put a "packet" of energy into CS, the
main charge cap. Try an off period (C=HIGH) of about
20uS but the on and off periods need to be tuned to
your inductor for max efficiency.

PIC pin V senses the voltage in the cap CS, when it is
charged (ie 6v or so) the charging process can stop.
For a cheap circuit you can just use a PIC input pin,
it will be within a couple of percent and close enough.
Obviously a comparator input pin is better.

When CS is charged, take PIC pin S HIGH to operate the
solenoid. :o)

This assumes you want to pulse the solenoid ONCE for a
drawer latch as you originally mentioned, not to KEEP
the solenoid engaged.

First find what size cap you need for CS, to work at
(say) 6v. Just play with a few caps and a 6v PSU, charge
the cap and touch to the solenoid connections. 470uF
or 1000uF will probably work ok.

To maximise for battery efficiency, put a decent (220uF
or so) cap before Q1, and decouple from your battery
via a small resistor, aim for about a 3% drop on this
cap voltage during the Q1 ON pulse.

Also test with different values of L1 and the ON and
OFF periods in charging. Effectively you are charging
the cap from zero to 6v, and the efficiency of the
buck system will vary throughout this cap voltage range.
Extra efficiency can be gained by varying the ON and
OFF times depending on the CS voltage at that point,
obviously at the expense of more complex software.

I've used a similar system with good results just by
using the right length ON pulse, and enough pulses
until the cap is charged. Because BC327, 1N5819 and
L1 all have low "on" resistance there is very little
places the pulse power can go apart from into the cap.
:o)
-Roman

--
http://www.piclist.com#nomail Going offline? Don't AutoReply us!
email listservspamKILLspammitvma.mit.edu with SET PICList DIGEST in the body

2002\12\21@095846 by Olin Lathrop

face picon face
> What exactly is a buck regulator?

A buck regulator is a type of DC to DC converter that creates the output
DC voltage from a higher unregulated input DC voltage, both with a common
ground.  This allows a certain topology that does not work when starting
with a lower voltage.  A simple type of DC to DC converter that makes a
higher output voltage is called a "boost" regulator.

I've put a schematic of a buck regulator from one of my projects at
http://www.embedinc.com/pic/temp.gif.  When the output voltage on C6 drops
below a threshold, Q2 is switched on.  This builds up current thru L2 and
charges up C6.  When the C2 voltage gets to another threshold, Q2 is
switched off again.  However, the current thru L2 must continue, so now it
comes from ground via D4.  This continues charging C6 a bit more.  The
voltage accross L2 is now reversed, so its current drops.  When it reaches
zero, D4 stops conducting, no more current charges C6, which is now being
discharged by load on the regulated output.  This whole process repeats
when the voltage on C2 drops below the lower threshold again.  IC1B and
the circuit surrounding it ensure that Q2 is switched on and off at the
appropriate times.  R2 provides some positive feedback (hysterisis), which
causes the difference between the ON and OFF threshold voltages on C6.  Q1
and Q3 are a voltage follower that provides much more current for
switching the gate of Q2 than IC1B could provide on its own.

The voltage on C2 is therefore has a sawtooth shape as it bounces between
the upper and lower threshold voltages.  L3 and C7 provide filtering to
reduce the high frequencies of this sawtooth as seen by the load.

The salient parts of this circuit that make it a buck regulator are Q2,
L2, D4, and C6, and the topology of how they are connected.

If all the parts were perfect (Q2 had no on resistance, L2 no DC
resistance, D4 no forward drop), then none of these parts could ever
dissipate power.  Therefore a buck regulator keeps the input and output
power the same, whereas a linear regulator keeps the input and output
current the same.  The former is obviously more efficient, which is the
big advantage of a buck regulator.  Note that this implies the output
current is greater than the input current.  You can see this must be true,
because the output current is the sum of the current coming thru Q2 and
D4, whereas the input current is only what comes thru Q2.

Do a net search if you want more details.  I'm sure there are many volumes
out there on buck converters, boost converters, and DC to DC converters in
general.


*****************************************************************
Embed Inc, embedded system specialists in Littleton Massachusetts
(978) 742-9014, http://www.embedinc.com

--
http://www.piclist.com#nomail Going offline? Don't AutoReply us!
email .....listservKILLspamspam.....mitvma.mit.edu with SET PICList DIGEST in the body

2002\12\21@135010 by Dave Dribin

flavicon
face
On Saturday, December 21, 2002, at 03:11  AM, Roman Black wrote:
> Dave Dribin wrote:
>> You mentioned using a cap to trigger the solenoid, using a buck
>> regulator to charge it.  What exactly is a buck regulator?
>>  Is there a particular setup you recommend?
>
> Yes! :o) I drew this in a real hurry so please
> forgive errors or sloppy parts values.

Wow, awesome.  I'm surprised (in myself) that I think I actually
understand the schematic.  My background is software w/ digital logic,
so all this analog stuff makes my brain hurt.  I've been pulling out
college textbooks and re-reading some basic principals, like how
transistors work. :)  That's the main reason I've been leaning towards
simpler.... I actually understand it.

Some questions about the circuit:

1) Why is Q1 PNP?  If it were NPN, then the PIC pin would be low for
off times.  Since this transistor is off unless charging, it will be
off most of the time.  It would seem that keeping the pin low would be
more power efficient.  Especially, when I put the PIC in sleep mode,
the pin can be set low, rather than high while sleeping.

2) Why is the sense pin necessary?  Once the time it takes to charge
the cap is found, it should not change, no?  Why not just charge it for
that amount of time (+ some error)?  This would also remove those two
resistors that drain the CS cap of some charge, too.

3) Why is Q2 not a Darlington transistor?  It still must handle the
full 300mA of the solenoid as if the solenoid was connected straight to
a +6V PSU.  Does the pulsing nature make the Darlington unnecessary?

4) What does the 1N5819 diode do?  It seems to be blocking a path that
is never taken.

And some questions in general:

1) How is this a buck regulator?  The output voltage (+6V) is the same
as the input, no?  In my newbie understanding of buck regulators, they
generally drop the voltage.

2) Why use an inductor at all?  Why not just have the transistor feed
right into the cap?  Just turn on Q1 until the CS cap is charged, then
dump it.  I assume the inductor is what makes this a buck regulator, so
there must be a good reason for it. :)

BTW, thanks to everyone who has helped out on the PIC list!  If you're
ever in Chicago, drop me a line and I'll buy ya a beer.

-Dave

{Quote hidden}

--
http://www.piclist.com#nomail Going offline? Don't AutoReply us!
email EraseMElistservspam_OUTspamTakeThisOuTmitvma.mit.edu with SET PICList DIGEST in the body

2002\12\21@193034 by Wagner Lipnharski

flavicon
face
Roman Black wrote:
{Quote hidden}

What is the PIC VCC in the above circuit?
If it is +5V, then Q1 will be always active, no matter what. It would only
cut if VBE would be lower than 0.6V, with +5V at VCC, the most the PIC can
deliver via the 2k2R is 4.95V, so, at least 1.05V will be Q1 VBE.  Even so,
that PIC VCC is at 6V, the 2k2R suggests 2mA as Base current, at best
Beta=80, Q1 will be conducting around 180mA maximum. If its load is lower
than 33 ohms (L1, CS, COIL), then Q1 will not saturate and its VCE will
heat it up.

The 270 ohms resistor suggests that 16mA will be injected at Q2 base, is
that right?

I am sorry, but except if you made a mistake about the +6V at battery, what
is the reason for a buck charge to the cap to +6V?  A simple resistor
woudn't do the same job?

I guess the battery voltage is not +6V, isn't it?


{Quote hidden}

--
http://www.piclist.com#nomail Going offline? Don't AutoReply us!
email @spam@listservKILLspamspammitvma.mit.edu with SET PICList DIGEST in the body

2002\12\21@200159 by Spehro Pefhany

picon face
At 07:18 PM 12/21/02 -0500, you wrote:
{Quote hidden}

<snip>

Presumably Roman would use a pin such as RA4 on the PIC16F628, which can
be pulled up to 14V (abs max). I'd like to see a base-emitter resistor
anyhow.

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
KILLspamspeffKILLspamspaminterlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

--
http://www.piclist.com#nomail Going offline? Don't AutoReply us!
email RemoveMElistservTakeThisOuTspammitvma.mit.edu with SET PICList DIGEST in the body

2002\12\22@081736 by Roman Black

flavicon
face
Spehro Pefhany wrote:
>

> Presumably Roman would use a pin such as RA4 on the PIC16F628, which can
> be pulled up to 14V (abs max). I'd like to see a base-emitter resistor
> anyhow.


Well actually he specified a micro power circuit
running from 6v ie two 3v cells. I always assumed
that he had a method of giving a pic-able voltage
like 5.5v with a dropping diode etc as has been
discussed on the list for micropower use, ie
everything runs from the same voltage.
:o)
-Roman

--
http://www.piclist.com hint: To leave the PICList
spamBeGonepiclist-unsubscribe-requestspamBeGonespammitvma.mit.edu>

2002\12\22@082635 by Roman Black

flavicon
face
Wagner Lipnharski wrote:

{Quote hidden}

Yes I should have been more specific, the "6v battery"
is the PIC Vdd voltage, probably two 3v cells and a
series diode to make it about 5.5v.


>the 2k2R suggests 2mA as Base current, at best
> Beta=80, Q1 will be conducting around 180mA maximum.

Not so, i've recently done a lot of testing of BC327
in buck regulators, and even at Ib=1mA Vce=0.2v@100mA
and even better at Ib=2mA of course.
This buck cap charger is *designed* to use a small
current from the battery, if the pulse on period is
correctly short enough the current drawn from the
battery (Q1 Ic) should be 30mA or even less. So 1mA
to 2mA Ib is about right for hard saturation.


> If its load is lower
> than 33 ohms (L1, CS, COIL), then Q1 will not saturate and its VCE will
> heat it up.

No, because it is a buck switch and the Ic current
is a function of L1 inductance and pulse length.

> The 270 ohms resistor suggests that 16mA will be injected at Q2 base, is
> that right?

10mA+ is about right, BC337 Ib=10mA Vce=0.1v@800mA,
he said the solenoid coil is about 500mA so I allowed
a decent margin.


> what is the reason for a buck charge to the cap to +6V?
> A simple resistor woudn't do the same job?

His original requirement was for a way to PULSE a solenoid
(drawer latch) and the whole circuit to run from a very
small battery that had limited current ability and a
small total capacity. The buck circuit should be close
to 90% efficient to charge the cap, a resistor is
simpler but much less efficient. This may equate to
3x more drawer openings before the battery is flat and
allows more flexibility to reduce battery current drain
and reduce total cap charging time. Since the PIC is
already available the cost is only one inductor, a
diode and two resistors more than using a resistor to
charge the cap.

> I guess the battery voltage is not +6V, isn't it?

Correct, the voltage is two 3v cells and whatever needed
to run the PIC, I assumed a drop diode as this is a
micropower circuit.
-Roman

--
http://www.piclist.com hint: To leave the PICList
TakeThisOuTpiclist-unsubscribe-requestEraseMEspamspam_OUTmitvma.mit.edu>

2002\12\22@084635 by Roman Black

flavicon
face
Dave Dribin wrote:

> > Yes! :o) I drew this in a real hurry so please
> > forgive errors or sloppy parts values.
>
> Wow, awesome.  I'm surprised (in myself) that I think I actually
> understand the schematic.

Thanks I think. ;o)

> Some questions about the circuit:
>
> 1) Why is Q1 PNP?  If it were NPN, then the PIC pin would be low for
> off times.  Since this transistor is off unless charging, it will be
> off most of the time.  It would seem that keeping the pin low would be
> more power efficient.  Especially, when I put the PIC in sleep mode,
> the pin can be set low, rather than high while sleeping.

Q1 is off all the time, except for when charging pulses
occur in the ON period. When the PIC is in sleep mode
Q1 will still be off. In this circuit PNP has advantages,
as does using an NPN for the solenoid.

> 2) Why is the sense pin necessary?  Once the time it takes to charge
> the cap is found, it should not change, no?  Why not just charge it for
> that amount of time (+ some error)?  This would also remove those two
> resistors that drain the CS cap of some charge, too.

The cost of two resistors and one PIC pin gives you
much simpler software and probable energy savings as
you don't need to "over" charge the cap to ensure it
is full. Remember battery voltage will change a LOT
over battery life, so the cap voltage test saves a
lot of potential problems. The resistor network can
be as high as 100k and only loses power during the
short charge cycle, total losses are much lower than
1% and less than power lost operating Q1.

> 3) Why is Q2 not a Darlington transistor?  It still must handle the
> full 300mA of the solenoid as if the solenoid was connected straight to
> a +6V PSU.  Does the pulsing nature make the Darlington unnecessary?

A darlington will saturate at about 0.85 volts, losing
significantly more than the 0.1v I expect you will get
with the BC337. This power saved with the BC337 is more
than you will lose driving its base at 10mA or so. The
diode across the solenoid coil causes slow-decay and keeps
the coil engaged longer than you expect from a short
current pulse, ie Q2 is only on for a very short time
and may only be a few mS.

> 4) What does the 1N5819 diode do?  It seems to be blocking a path that
> is never taken.

Umm, it s a buck circuit?? This diode is rather important.
<grin>

> And some questions in general:
>
> 1) How is this a buck regulator?  The output voltage (+6V) is the same
> as the input, no?  In my newbie understanding of buck regulators, they
> generally drop the voltage.

Buck is a topology, ie type of circuit. Ideally this
circuit reduces the total current drawn from the
battery to a small current, without power loss that
you would get from a series resistor. So it's really a
"buck current-reducer" more than a "buck regulator".


> 2) Why use an inductor at all?  Why not just have the transistor feed
> right into the cap?  Just turn on Q1 until the CS cap is charged, then
> dump it.  I assume the inductor is what makes this a buck regulator, so
> there must be a good reason for it. :)

<grin>
I think Olin suggested you do a net search??
-Roman

--
http://www.piclist.com hint: To leave the PICList
RemoveMEpiclist-unsubscribe-requestspamTakeThisOuTmitvma.mit.edu>

2002\12\22@133851 by Wagner Lipnharski

flavicon
face
Roman Black wrote:
{Quote hidden}

Even a Lithium Battery?
I thought they have a flat life voltage, dropping very fast at end of life
period.


{Quote hidden}

even that "buck" charge circuit is more related to capacitor transfer than
inductor pump. The "buck" expression means charge a buck with water, move
it to another location to fill up a reservoir, in this case, using
capacitors to move voltage from one battery to a capacitor, in most cases,
the circuit attaches the capacitor in parallel to the battery, then moves
the charged capacitor in series with the battery, so the resulting voltage
is almost double.  I rarely see the expression "buck" to be used with
inductors, even that the idea is almost the same, an inductor can't hold
charge longer. In a fast time window it doesn't make any difference, but I
thought to post this text just to avoid confusion in the mind of who is
learning.



{Quote hidden}

I would add that energy taken from the battery is the same (taken out the
losses in any conversion) to be consumed at the load.  One could convert
the energy to a different level of Volts or Current, but the Power would
always be the same (less the losses).  There is no magic, whatever you do,
is always thinking to get the necessary voltage or current, with the
smaller loss as possible.

When using a series resistor, some energy will be dissipated at that
resistor, when it is limiting the current, that energy is lost.  An
intelligent buck-charger or inductor based step-up or down, try to provide
a better conversion with less losses, even so, losses exist all around.
The ohmic resistance of the inductor wire will steal energy, the capacitor
leak or ESR (capacitive ohmic resistance) will also dissipate energy, what
makes all step-up/down converters productivity drop to less than 90%.

It needs to be remembered, that every component added WILL steal energy and
dissipate it in thermal way, a diode, transistor, whatever is used,
requires some voltage drop to work, multiplied by the electric current
gives you the Power in Watts, that is pure loss.

For one point, I would recommend to rethink the bases of any project that
requires a buck or step-up/down inductor based system.  In most of the
times, the requirement for those is to cover up a design mistake, not all
of them, but if you take 100 cases, probably will find out that 80 of those
could be done with a better base design.

For example, this cash drawer solenoid.  Why use lithium batteries?  Why
not simply connect 4 x C size 1.5V regular alkaline batteries? it will work
forever without replacing.  Even AA's size could do a good job for many
months.  Of course we need to keep thinking for better solutions and to
learn, and for that I am the first to say "go ahead", but we always need to
remember that the humans tend to complicate things.

:)


If considering the battery will be losing voltage and needing to keep the
capacitor voltage at 6V steady, this is another way to do the step up
converter:


   Inductor   1N5817
                _
(+B)--MMMM--o---->|----o----o--->
_   _      |          |    |
| |_|       C         _|_ + |
  ---R----B  NPN     ===   |
           E          |    |
           |          |    R
          _|_        _|_   |
                           |
<------feedback-------------o
                           |
                           R
                          _|_


This approach does not require any connection between the uC VCC and the
(+B) battery voltage, due to the open collector design.  The feedback of
course will tell the uC about the PWM values, or pulse shape, or whatever
you will be using to control the coil charging current.  Larger pulses will
generate higher voltage over the capacitor, but there is a limit, a very
large pulse will saturate the inductor (all magnetic field already
expanded), from that point on is pure energy loss, since your circuit will
be draining battery power without storing it at the inductor magnetic
field.

I would use a small N Channel FET in place of the NPN

Wagner.

{Quote hidden}

--
http://www.piclist.com hint: To leave the PICList
EraseMEpiclist-unsubscribe-requestspammitvma.mit.edu>

2002\12\22@144438 by Olin Lathrop

face picon face
> For one point, I would recommend to rethink the bases of any project
that
> requires a buck or step-up/down inductor based system.  In most of the
> times, the requirement for those is to cover up a design mistake, not
all
> of them, but if you take 100 cases, probably will find out that 80 of
those
> could be done with a better base design.

This is pure nonsense.  (Where DO they dream this stuff up?!!)

Buck converters and other efficient DC to DC converters are used more and
more, mainly because of the efficiency.  The electronic parts to implement
them well are now smaller, cheaper, and better than they were just a few
years ago.  This has opened up many more uses that were previously living
with lower efficiency.  Efficiency is almost always an issue in battery
operated systems, and battery voltages vary and don't always come at
convenient values.  An efficient DC - DC converter is a good answer to get
the most out of a battery and to provide constant voltage to a circuit
regardless of the battery level.

Efficiency is also useful in high power circuits where getting rid of the
heat wasted by inefficient circuits is a challenge.  A well designed DC to
DC converter is usually smaller and cheaper than heat sinks, cooling, and
the added cost of the thermal stresses.


*****************************************************************
Embed Inc, embedded system specialists in Littleton Massachusetts
(978) 742-9014, http://www.embedinc.com

--
http://www.piclist.com hint: To leave the PICList
RemoveMEpiclist-unsubscribe-requestEraseMEspamEraseMEmitvma.mit.edu>

2002\12\22@163853 by Wagner Lipnharski

flavicon
face
Olin Lathrop wrote:
{Quote hidden}

Dear Olin, sorry, this was not what I tried to say.

I am not talking about projected systems where efficient conversion is a
must.  You are correct, right now almost anywhere you can find DC/DC
converters, and I agree with "most" of them.  In one of our products, we
replaced the bulky wall wart adapter for a small, not so heavy, switching
power supply, that not only consume less power, supply a fixed voltage to
the equipment, and cost the same of a heating pig transformer, and also
save $0.60 in shipping weight.

It is almost impossible to use the most possible energy from a power solar
cell panel without DC/DC converters, and much other designs.

What I said, is that doesn't make sense to use something just because it is
possible to use it.  I said that many designs using DC/DC converters are
just used because the designer (?!?) thought it was to be a nice solution.

I already saw DC/DC converters to supply 6Vdc @ 100mA to a device into a
car, just because somebody thought it will save energy equivalent in one
liter of gas in 10 years of use.  At this level, it really doesn't make
much sense, mostly when the tires pressure is lower than the optimum and
the waste fuel to bend the rubber and drag those heavy tires spend hundreds
of kW a day in fuel, or, just walking to the bakery at the corner, instead
of driving the car 2 miles to go to the market and buy the same kind of
bread, it would save hundreds of liters of gas in 10 years.

But I really don't see a real and strong reason to use a DC/DC converter to
charge 6V to a capacitor just to pick a solenoid, mostly when you have
space into the cash drawer to install different types of batteries.  For
sure 2 x 3V Lithium batteries will cost much more than several C size
Alkaline cells.

As you can think, the solenoid WILL pick with 6, with 5.8 or even with
5.5V, if not even with 5V, if enough current is available from a low
impedance supply (capacitor).  A solenoid is not a device that needs a
strict regulated voltage.

I already demonstrated in some papers that "some" projects are just waste
of time and money.  One example was a design that was sent to me to refine
for mass production. Power supply was based on a DC/DC that converts 2xAA
Alkaline cells (3V) to 5V @ 150mA.  The project was made to supply an AVR
device, that by the way, runs starting from 2.6Vdc up to 6Vdc.  First that
there was no reason to use exactly 5V to that board, no other component on
the board would require fixed 5V, no LCD, no other memory or glue logic
that didn't work with lower voltages.  The point is that the DC/DC
converter running at 85%, demanded 5*0.15 / 3*0.85 = 300mA from the
batteries.  When the batteries discharge a little, and present 2.8V, the
current demand goes to 315mA, it rapid drain the cells and voltage goes
down fast to 2.7, 2.6, what demanded 340mA and this really sucked battery
life.  Ok, that's the way it is, right?  wrong.  This project used an AVR
running at full 8MHz clock, some external logic sampling an input logic at
2MHz, at 5V this really needed the 150mA.  The battery life in this project
was not more than 4 hours.

Observing the hardware, software and results, it was clear that the same
circuit could be running at 2MHz, with a sampling rate of 10kHz, with the
same exactly results.  At that new clock and sampling rate, the same
circuit now consumed only 62mA.  The same DC/DC conversion now result in a
consume of only 121mA from the 3V battery.  Immediately we triple battery's
life span.

Than, we just used the available space in the box, including the one used
by the step-up converter, to add a third AA cell.  Then, now we have power
that would travel from 4.5Vdc down to 3.7Vdc when cells would be almost
discharged.  This lower VCC reduced the current consume from 62mA to less
than 55mA (4.5V) and 46mA (3.7V).  Then we redesigned the software and
reduced external glue logic, replaced by few CMOS components, the final
current consume at 4.5V was lower than 22mA.  Now we expanded battery life
to more than 40 hours. It was almost 9 times better than the original
promect.  We could keep going further, but the customer was very happy, so
we stopped there.

Then we asked the original designer, why he used fixed 5V and why the uC
clock of 8MHz and sampling rate of 2MHz... answers?  He didn't have them,
he just thought it was ok to use those values. He said it worked at the
first try and then they kept the design as it was.

This is why I say that certain solutions are made to fix original bad
designs.  I am not saying that we should not use DC/DC converters, what I
am saying is that if you need to climb to your house roof often, or you
first look for a high ladder (and probably hospital expenses for a broken
leg sooner or later) or rethink the idea of having a two-store house.  How
many times you already saw in movies old people using some kind of lifting
motorized chair to go upstairs? the question is: "what the hell an old
person suffering with arthritis is living in a duplex house, and worst of
all, sleeping upstairs?"

Happy Holidays.

Wagner.

--
http://www.piclist.com hint: To leave the PICList
RemoveMEpiclist-unsubscribe-requestspam_OUTspamKILLspammitvma.mit.edu>

2002\12\22@214834 by Russell McMahon

face
flavicon
face
> For one point, I would recommend to rethink the bases of any project that
> requires a buck or step-up/down inductor based system.  In most of the
> times, the requirement for those is to cover up a design mistake, not all
> of them, but if you take 100 cases, probably will find out that 80 of
those
> could be done with a better base design.

I'll try and put it a little less bluntly than Olin did :-).
Your point is well made and I agree with the general spirit it was made in -
thinking smarter usually helps simplify designs.
I agree that a proportion of switching regulator applications could be
avoided with more thought.

But there's a real place for them and I'd guess that more than 20% of the
applications are well enough justified.

Anywhere where -

- The input and output voltages are set by other parameters outside your
control (eg 12v system or auto supply is the specified input).

- Input voltage varies widely either across lifetime or during operation (eg
battery supply, alternator/wind/solar/wave energy supply).

Then a voltage conversion system allows best efficiency. If efficiency
counts then a switching regulator is worth looking at.

As a reasonably typical example, imagine a portable device which wishes to
use alkaline batteries (cost/energy density etc) and needs a 5v supply.
Cells are deemed to have a voltage range of 1.5v to 0.9v across lifetime.
Mean voltage for energy delivery purposes is delivered at an average of
(say) 1.2v (will vary with actual cells used). We need at least 6 cells for
this (6 x 0.9 = 5.4v).  Depending on lifetime needed we may HAVE to use C or
D cells but most applications will allow AA or AAA. 6 x AA is annoying for a
handheld so we may go to AAA. AAA is abomination incarnate in terms of cost
per energy and energy density relative to its bigger brothers. (IMHO of
course). 6 x 1.2 V average mean that's 7.2v for the pack or an energy
efficiency of 5/7.2 = 69%.  A switching regulator with an 80%  mean
efficiency across the battery range would give 80/69 = 15% better battery
life. At 90% efficiency it would give 30% better efficiency. With modern
buck designs (> 1 MHz, synchronous rectification etc) even more than this is
possible. The ability to use 4 rather than 6 cells (or 2 or 3 or 5) may
allow the use of AAs rather than the dread AAAs with substantially lower
battery cost and improvements in life / energy per battery volume. The
extension of battery life may allow a smaller design or longer battery life
or ... .
If you manufacture or sell batteries you may not appreciate these arguments
:-).

Some other possible factors:

Off the shelf parts rated at voltages other than those readily available.

Heat dissipation (rather than energy availability)

Physical size.

Ability to heatsink in available environment.


There's still certainly a large place for linear regulators though.
(Amongst other reasons because, amongst other things,  of cost, noise level,
availability and simplicity).




           Russell McMahon

--
http://www.piclist.com hint: To leave the PICList
RemoveMEpiclist-unsubscribe-requestTakeThisOuTspamspammitvma.mit.edu>

2002\12\23@142322 by Dave Dribin

flavicon
face
On Mon, Dec 23, 2002 at 12:03:33AM +1100, Roman Black wrote:
> Yes I should have been more specific, the "6v battery"
> is the PIC Vdd voltage, probably two 3v cells and a
> series diode to make it about 5.5v.

Yes, using 2x3V plus one diode drop to power the PIC was my plan.

{Quote hidden}

Ok, so the pulsing plus inductor limit the current to 30mA?  Is this
quantifiable based on duty cycle, or only through experimentation with
real parts?  Also, from the BC327 datasheet I found, Vce is 0.7V,
unless I'm mis-reading something.  That would put it over its 625mW
power limit, no?

-Dave

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

2002\12\23@161321 by Dave Dribin

flavicon
face
Thanks for all your suggestions on this topic.  I like the buck
circuit provided, but I am not quite confident enough in my knowledge
that I could actually build it and, more importantly, debug it.
Something *will* go wrong, and I don't want to just sit there
scratching my head.  I'm planning on buying "Switching Power Supply
Design" by Abraham Pressman to help me learn about more on this topic:

 http://www.amazon.com/exec/obidos/tg/detail/-/0070522367/

In the meantime, I'm going to go for a simpler, even if less
effecient, circuit, like the following:


                 +6V (Battery)
                  |
                  R1
                  |
                  C
PIC 1-----R2-----B    Q1
                  E
                  |
                  +---------+-----+
                  |         |     |
           470uF  C1        -    SOL
                  |         ^     |
                  |         |     |
                 GND        +-----+
                                  |
                                  C
PIC 2-----R3---------------------B    Q2
                                  E
                                  |
                                  |
                                 GND

Notes:

1) Is R1 needed at all?  I think it's needed to limit Ic of Q1 to
  300mA.  R1 would need to be 20 ohms to limit Ic to 300mA.

2) Assuming a gain of 100, Ib needs to be at least 6mA.  Assuming a
  PIC logic 1 goes as low as +4, R2 should be about 330 ohms.

3) R3 should also be about 330 ohms because Ic of Q2 is also 300mA.

4) I could probably do the voltage sense stuff to detect when the cap
  is charged, but for now, I'll just use a simple software delay.
  I'm not quite sure how the sense pin simplifies the software
  (what's simpler than a delay? :).

I think all bases are covered, no?

-Dave

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

2002\12\23@163012 by Dave Dribin

flavicon
face
{Quote hidden}

Duh, I can be so stupid sometimes. :) The whole point of this was to
limit the current draw from the battery because 300mA was too much!
So R1 should be chosen to limit Ic to, say 30mA, thus R1 is about 220.

> 2) Assuming a gain of 100, Ib needs to be at least 6mA.  Assuming a
>    PIC logic 1 goes as low as +4, R2 should be about 330 ohms.

Ok, Ib is now about 0.6mA, thus R2 of about 3K should do.

> 3) R3 should also be about 330 ohms because Ic of Q2 is also 300mA.

R3 is still 330.

Whew....

-Dave

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

2002\12\23@170133 by Dwayne Reid

flavicon
face
At 03:12 PM 12/23/02 -0600, Dave Dribin wrote:

{Quote hidden}

Eliminate Q1 and R2 - you should not need them.  Keep R1 - it helps ensure
that the solenoid does not load down the battery when you fire it.  Add a
pull-down resistor to the base of Q2 (10k or so).

Check to make sure that C1 is large enough to operate the solenoid when the
battery has dropped below its end-of-life voltage.  In other words, feed
the circuit with a power supply set to the lowest voltage that you want the
system to work and ensure that the solenoid still operates correctly.

dwayne

--
Dwayne Reid   <EraseMEdwaynerspamspamspamBeGoneplanet.eon.net>
Trinity Electronics Systems Ltd    Edmonton, AB, CANADA
(780) 489-3199 voice          (780) 487-6397 fax

Celebrating 18 years of Engineering Innovation (1984 - 2002)
 .-.   .-.   .-.   .-.   .-.   .-.   .-.   .-.   .-.   .-
    `-'   `-'   `-'   `-'   `-'   `-'   `-'   `-'   `-'
Do NOT send unsolicited commercial email to this email address.
This message neither grants consent to receive unsolicited
commercial email nor is intended to solicit commercial email.

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

2002\12\23@172630 by Olin Lathrop

face picon face
{Quote hidden}

ensure
> that the solenoid does not load down the battery when you fire it.

I agree.

> Add a
> pull-down resistor to the base of Q2 (10k or so).

I disagree here.  The PIC pin will be driven actively low when set to a 0.
That will be good enough to ensure that Q2 is always off when the PIC pin
is low.  A B-E resistor would speed up the turn off a bit, but that is not
an issue in this case.


*****************************************************************
Embed Inc, embedded system specialists in Littleton Massachusetts
(978) 742-9014, http://www.embedinc.com

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

2002\12\23@175235 by Dwayne Reid

flavicon
face
At 05:26 PM 12/23/02 -0500, Olin Lathrop wrote:

> > Add a
> > pull-down resistor to the base of Q2 (10k or so).
>
>I disagree here.  The PIC pin will be driven actively low when set to a 0.
>That will be good enough to ensure that Q2 is always off when the PIC pin
>is low.  A B-E resistor would speed up the turn off a bit, but that is not
>an issue in this case.

The added resistor simply ensures that the transistor is held OFF if the
PIC pin should be tri-stated for any reason.

dwayne

--
Dwayne Reid   <RemoveMEdwaynerKILLspamspamplanet.eon.net>
Trinity Electronics Systems Ltd    Edmonton, AB, CANADA
(780) 489-3199 voice          (780) 487-6397 fax

Celebrating 18 years of Engineering Innovation (1984 - 2002)
 .-.   .-.   .-.   .-.   .-.   .-.   .-.   .-.   .-.   .-
    `-'   `-'   `-'   `-'   `-'   `-'   `-'   `-'   `-'
Do NOT send unsolicited commercial email to this email address.
This message neither grants consent to receive unsolicited
commercial email nor is intended to solicit commercial email.

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

2002\12\23@180320 by Olin Lathrop

face picon face
> The added resistor simply ensures that the transistor is held OFF if the
> PIC pin should be tri-stated for any reason.

Yes, it would, but proper firmware should insure that won't happen.  Bad
firmware could leave the solenoid always on, never turn it on, or
whatever.  Also, this transistor is driving a solenoid that requires
significant current.  Even if a little noise is picked up on the base,
there is no danger that the solenoid will be energized.


*****************************************************************
Embed Inc, embedded system specialists in Littleton Massachusetts
(978) 742-9014, http://www.embedinc.com

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

2002\12\23@182609 by Schultz, Peter

flavicon
face
Try to tri-state the PIC and put your cell phone antenna close to the transistor.........
Pull-downs are often neglected, using nice reasons why we do not need them.
Few months later nobody can explain the field complaints and the unexpected behavior.

Peter Schultz

{Original Message removed}

2002\12\23@194834 by Matt Pobursky

flavicon
face
On Mon, 23 Dec 2002 18:02:13 -0500, Olin Lathrop wrote:
> > The added resistor simply ensures that the transistor is held OFF
> > if the PIC pin should be tri-stated for any reason.

> Yes, it would, but proper firmware should insure that won't happen.
> Bad firmware could leave the solenoid always on, never turn it on, or
> whatever.  Also, this transistor is driving a solenoid that requires
> significant current.  Even if a little noise is picked up on the
> base, there is no danger that the solenoid will be energized.

How does firmware help if the MCLR pin is pulled low continuously? ;-)
Also, while the solenoid may not be turned fully on, it certainly would
kill the battery quickly.

Matt Pobursky
Maximum Performance Systems

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

2002\12\23@195920 by Dwayne Reid

flavicon
face
At 06:02 PM 12/23/02 -0500, Olin Lathrop wrote:
> > The added resistor simply ensures that the transistor is held OFF if the
> > PIC pin should be tri-stated for any reason.
>
>Yes, it would, but proper firmware should insure that won't happen.

Yeah - it shouldn't.  But I use them anyways.  It doesn't add much to the
cost and helps me and my clients to sleep better at night.

I usually size the pulldown resistors in my stuff such that the output
voltage from the PIC has to be around Vdd / 2 before the transistor turns
ON.  This gives good noise immunity as well as dealing with tri-state issues.

One reason I tend to be so anal about such things is that the majority of
my higher end systems use an external BOD / watchdog / reset controller
(Xicor X5043).  This can cause the PIC to be held in reset which in turn
causes all i/o pins to be tri-stated.  But I've been designing industrial
controls for a really long time now and little things like pull-down and
pull-up resistors are just part of how I design all my stuff.

To each their own . . .

dwayne

--
Dwayne Reid   <dwaynerSTOPspamspamspam_OUTplanet.eon.net>
Trinity Electronics Systems Ltd    Edmonton, AB, CANADA
(780) 489-3199 voice          (780) 487-6397 fax

Celebrating 18 years of Engineering Innovation (1984 - 2002)
 .-.   .-.   .-.   .-.   .-.   .-.   .-.   .-.   .-.   .-
    `-'   `-'   `-'   `-'   `-'   `-'   `-'   `-'   `-'
Do NOT send unsolicited commercial email to this email address.
This message neither grants consent to receive unsolicited
commercial email nor is intended to solicit commercial email.

--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

2002\12\23@220459 by Mike Singer

picon face
part 1 3705 bytes content-type:text/plain; (decoded quoted-printable)

Wagner Lipnharski wrote:

> For one point, I would recommend to rethink the bases of any project
> that requires a buck or step-up/down inductor based system.  In most
> of the times, the requirement for those is to cover up a design
> mistake, not all of them, but if you take 100 cases, probably will
> find out that 80 of those could be done with a better base design.
.
Olin Lathrop wrote:

{Quote hidden}

Wagner Lipnharski wrote:
.
> Than, we just used the available space in the box, including > the one used by the step-up converter, to add a third AA cell.  > Then, now we have power that would travel from 4.5Vdc > down to 3.7Vdc when cells would be almost discharged.  > This lower VCC reduced the current consume from 62mA to > less than 55mA (4.5V) and 46mA (3.7V).  Then we redesigned > the software and reduced external glue logic, replaced by few > CMOS components, the final current consume at 4.5V was > lower than 22mA.  Now we expanded battery life to more than > 40 hours. It was almost 9 times better than the original promect.  > We could keep going further, but the customer was very happy, > so we stopped there.
>
  If you "could keep going further", do it!
Keep on going further back to DC/DC converter!
Look at 21372b.pdf TC125-126 PFM Step-up DC/DC regulator. (US$1.67 retail here in the Ukraine). (Attached .gif)
  You could use only one thick 1.5v battery.
With TC125301 you get 3.0V _CONSTANT_ output voltage.
Power consumption is less then at your 4.5V-3.7V.
TC125301 eats less then 40 microA at 3V@30ma.
I was told Seiko's are even better.
You needn't carry out rather expensive tests on reliability under floating voltages (oscillator start-up or overdriving conditions for example).

  Mike.
------------------------

Wagner: I want you to tell me the names of the fellows        on the team.
Olin: I'm telling you. Who's on first, What's on second,        I Don't Know is on third
Wagner: You know the fellow's names?
Olin: Yes
Wagner: Well, then who's playing first?
Olin: Who.
Wagner: The fellow's name on first base?
Olin: Who.
Wagner: The guy on first base?
Olin: Who is on first base.
Wagner: What are you asking me for?
Olin: I'm not asking you, I'm telling you. Who is on first.
Wagner: Then I'm asking you. Who's on first?
Olin: That's the man's name.
Wagner: That's whose name?
Olin: Yes.
Wagner: Well, all I'm trying to do is find out what's the        guys name on first base?
Olin: No, no. What is on second base.
Wagner: I'm not asking who's on second.
Olin: Who's on first.
Wagner: What's the guy's name on first base?
Olin: What's the guys name on second base.
Wagner: I don't know.
Olin: He's on third. We're not talking about him.
....


--
http://www.piclist.com hint: PICList Posts must start with ONE topic:
[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads




part 2 3335 bytes content-type:application/octet-stream; (decode)

part 3 2 bytes
-

2002\12\24@075153 by Olin Lathrop

face picon face
> How does firmware help if the MCLR pin is pulled
> low continuously? ;-)

OK, if the PIC is going to be held in reset for long periods of time, then
you should have a resistor there.  However, I thought that wasn't the
case.

> Also, while the solenoid may not be turned fully on, it certainly would
> kill the battery quickly.

This can only happen if the PIC pin is floating.


*****************************************************************
Embed Inc, embedded system specialists in Littleton Massachusetts
(978) 742-9014, http://www.embedinc.com

--
http://www.piclist.com hint: The list server can filter out subtopics
(like ads or off topics) for you. See http://www.piclist.com/#topics

2002\12\26@162917 by Wagner Lipnharski

flavicon
face
Dave Dribin wrote:
{Quote hidden}

You are missing an important point here.
Name is transistor polarization.

Suppost the PIC high logic goes down to 4.6V, with R2 as 3k, and C1 fully
discharged, you will have a R2 current of ( 4.6 - VBE - VC1 ) / 3000 =
1.3mA, suppose a gain of 100, Q1C possible current will be 130mA, but it
will be limited by R1 (220 ohms) to (VCC - VCE - VC1)/220 = (6 - 2 - 0) /
220 = 18mA... ooops, confused? yeah!  why VCE = 2V ?  That's the thing:
Because the Q1 Emitter voltage will always be (in a Silicon NPN unit) 0.6V
below  base voltage.

So, if your PIC supply 4.6V to R2, the base currente will be dictated by
what rest from this 4.6V via the resistor (less R2 Vdrop), less VBE (0.6V)
divided by the emitter resistance seen by the base point of view (emitter
resistance multiplied by the gain).

What is the emitter resistance ?  Just the C1 discharged resistance - in
this case almost zero, what from this resistance appears at the base?
multiply by the gain, almost zero times 100 = almost zero.

So, the base current will be 4.6V less 0.6 (VBE), divided by the sum of R2
plus the emitter resistance multiplied by the gain, it will be then

4.6V - 0.6V  / 3000 + 0

= 1.3mA

So, what voltage rest at the Q1 base?

Easy,

4.6V - R2Vdrop

4.6V - (3000 x 0.0013) = 0.6V

What is the Q1 Emitter voltage?

0.6V - VBE = Zero.

ooops... where is the +6V to charge C1?

There isn't

WHat happens, is there is a current via Q1, 1.3mA x gain = 130mA, ooops,
limited by R1 220 ohms.

Let see;

220 x .13 = 28.6V, impossible, so, it will limit Q1 current up to
saturation, that will be around 0.5V, it means, VR1 = 5.5V, 220 Ohms,
current = 25mA.

Then, Q1 will supply 25mA to C1.  But as soon C1 starts to hold up voltage,
everything changes.
Now, the formula that specifies Base current needs to consider C1 voltage.

Q1 Base current =

PIC Voltage (4.6V) less VBE (0.6V) less VC1 (increasing).
---------------------------------------------------------
                    3000 ohms


Suppose VC1 is already 1.5V

Q1 Base current = (4.6 - 0.6 - 1.5) / 3000 = 833uA

Ok, now multiply by gain (100), Q1 max current = 83mA, but you have a
current limit by R1 (220 ohms).

Let see.  Now, V1 = 1.5V, VCE = 0.5V, VR1 = 6 - 1.5 - 0.5 = 4V / 200 ohms =
18mA.

Oooops, when C1 was discharged, its charge current was 25mA, now that it
has 1.5V charge, the charging current is 18mA, and it will be decreasing
until...

until it reaches 4.01V, when the Q1 cuts out... why?

Because Base-Emitter polarization goes below 0.6V... PIC 4.6V - 4.01V (C1)
= 0.599 V as VBE, transistor cuts out.

But what was the C1 charging current before that?  when C1 was 3.9V?

Q1 Base Current = (4.6 - 0.6 - 3.9) / 3000 = 33uA multiplied by gain = Q1
collector current of 3.3mA... ooops, pretty low, isn't it?

If the top voltage over C1 is 4V, you can think as an equivalent circuit of
4VCC and a resistor charging C1

4V o-------RX----o---->
                |
               _|_ CX
               ===
                |
               _|_
               GND

To have only 3mA when CX is 3.9V, RX = (4 - 3.9) / 0.003 = 33 ohms.
To have 25mA when CX is 0V, RX = 4 / 0.025 = 160 ohms.

So, as you can see, RX is a variable resistor that changes resistance
according to CX charge.
Lets consider a middle value of 130 ohms (max current of 30mA).  If the
capacitor is a 1000uF, the RC will be 130 x 0.001 = 130ms, even so, the
time is not bad for charging such capacitor.

So, at your original circuit, Q1 being a NPN is just messing with your
life.
It not only charges C1 with max of 4V (even with 6V as VCC) and a simple
resistor could do better.


+6V o----R1----o--------o------.
       100    |        |      |
             _|_      _|_     3
             === C1    A      3 SOL
              |        |      |
              |        '------o
             _|_              |
                              C
PIC o-----------R2---o-------B   Q1 NPN
              390   |         E
                    R3        |
                    |2k2      |
                   _|_       _|_

R1 will charge C1 to almost 6V, always, will keep C1 top charged.
R1 value should be calculated to avoid suck current from battery more than
it can supply, so it top 60mA is available, then a 100 ohms resistor is ok.
At this moment (C1 discharged) R1 will dissipate 6V x 60mA = 360mW, but it
is so fast that a half watt resistor is ok, even a 1/4 W will do the job.
When the solenoid will be active, this resistor will be supplying 60mA to
all the circuit too, so, a 1/2W is recommended.

PIC sends 4.6V to R2+R3 Vdivider, forget R3 now.  4.6V - 0.6V (VBE) = 4V
available over R2 to limit base current.  What is the NPN current? Lets
suppose 500mA will be ok to activate the solenoid.  Then, 0.5 divided by
transistor gain (100) = 5mA, but lets push a little bit to 8 or 10mA at the
base, to make sure the transistor will saturate and deliver as much as
possible current from C1 to SOL.

Then, R2 = 4V / 0.01 = 400 ohms, commercial 5 to 10% resistors will find
390 ohms.

R3 can or can not be used. If you have false Q1 trigger, than use R3.
Value should be a minimum of 5 times R2.

Calculate R1 to not suck more than 50% of maximum battery current.

Wagner Lipnharski - email:  spamBeGonewagnerSTOPspamspamEraseMEustr.net
UST Research Inc. - Development Director
http://www.ustr.net - Orlando Florida 32837
Licensed Consultant Atmel AVR _/_/_/_/_/_/

--
http://www.piclist.com#nomail Going offline? Don't AutoReply us!
email KILLspamlistservspamBeGonespammitvma.mit.edu with SET PICList DIGEST in the body

2002\12\26@171525 by Wagner Lipnharski

flavicon
face
Mike Singer wrote:
{Quote hidden}

I use to think that a battery is just a pack that supply Watts/Hour, the
same technology, lets say Alkaline, will deliver the same Watts/Hour no
matter what cell voltage or current, the same volume means the same
Watts/hour.

Lets see:

Suppose the circuit drains 15mA at 3V (full charge) and 11mA at 2.2V (end
life).  The average consume will be around 13.5mA.  Using two AA cells
(1A/h) you can have around 74 hours of operational battery life.

Now, suppose you use both AA cells in parallel, and use the TC125301 unit
to generate 3V fixed.
The conversion current will be 2x when batteries are fully charged (1.5V)
and 2.7x when at life end (1.1V). This gives you an average of 2.4x.
Consuming 15mA, the circuit will drain 36mA in average, from the cells. As
they are in parallel, now you have 2A/h, result in 55 hours of operational
battery life.  Not even counting the fact that the 125301 efficiency is not
100%. Suppose it is 85%, battery life will be down to 46 hours.

To have the same battery life (74 h), the battery drain should be around
13.5mA constantly (per cell), it means the 125301 could not drain more than
27mA (parallel batteries).  Considering a worst case of 2.2V VCC, 11mA
consume, the 125301 will drain 1.46x current from batteries (fully charged
1.5V), and 2x at the end of life (1.1V).  Applying 85% efficiency, it goes
to 1.64x and 2.35x rate.  Then, the 11mA circuit consume, will be draining
(11mA x 1.64) 18mA (fully charge) and (11mA x 2.35) 25.8mA (end life) from
the cells. The average here can go to around 17mA, what gives 2A/0.017 =
117 hours.

This extra gain, even with the 15% loss of the efficiency of the dc/dc
conversion, is based on the higher current drain when battery is above 2.2V
(up to 3V), when the circuit is directly connected to 2xAA (no dc/dc).

One needs to consider if this extra operational hours (from 74 to 117) is
accepted by the customer, considering the final product price increase due
the dc/dc conversion (can not cost less than $2), final product can
increase $12 or more.  For some products that customer uses no more than
one or two hours per month, he really doesn't care about the extra battery
time, but $12 more or less, can really makes a difference.

Want to do a comparison in the real world?

Cell phones.

Using top of the line dc/dc and battery technology, standby time of 80
hours, talking time of 6 hours.
Ok, what about if it was 90 hours and 7 hours?  what difference it really
does make?
I talk less than 1 hour/day, and ALWAYS will keep the cell in the charger
during the night.
Why I will pay more $$$ for a phone or a battery that can gives me 8
talking hours?
Obviously some people requires that, but not me, and not a bunch of other
people too.

Other than that, who talks 8 hours a day in a cell phone, needs to do a
brain damage catscan, for two reasons, first, the RF damages the brain,
second, what a hell is he doing 8 hours/day in the phone?

Wagner

--
http://www.piclist.com#nomail Going offline? Don't AutoReply us!
email EraseMElistservspamEraseMEmitvma.mit.edu with SET PICList DIGEST in the body

2002\12\30@190238 by Mike Singer

picon face
  Wagner, sorry for the late response.
  To shape our discussion: What is better for battery-operated systems?

1. To feed them from say two or three 1.5V cells directly.
2. To feed them from step-up DC/DC regulator powered
    by _one_ 1.5V cell.

You advocate first approach, I do second. Factors to be considered:

  "Reliability" section:
- loss of  reliability due to extra contacts; - loss of  reliability due to using two cells instead of one; - loss of  reliability due to using extra components (DC/DC    regulator); - some vagueness with reliability under floating voltage -  How do you like this from AN601:
"...The number of devices that can fail before a particular
endurance criteria is not met is also somewhat flexible.
Even the most quality conscious manufacturer will
occasionally have a failure, so a failure level is defined.
The industry standard conditions for many types of
reliability tests are set by JEDEC (the Joint Electronic
Device Engineering Council). JEDEC defines that if 5%
or less of a given sample fails at a given endurance
goal, then that goal has been met... "

  "Cost" section:
-extra cost due to extra contacts for extra cells; -for a given "Watts*Hour" extra cost when using more then one cell;
(one 1.5V 1A*H costs less then two 1.5V  0.5A*H and takes less room)
- extra cost due to using extra components (DC/DC regulator, room cost);
- extra cost of tests on reliability under floating cells voltage.

"The Roads we Take" are on our own. And I am not sure your road is cheaper: extra reliable contacts are not cheap; for a given "Watts*Hour" extra cells cost more;
tests on reliability under floating cells voltage are tricky business and cost a lot;

I agree with Dale:
"All operated well within spec, and all still fail.  Stuff happens.  The difference between an engineer and a top-notch engineer is often the ability to design systems that take these things into account, and balances reliability against cost."

  Mike.

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

Wagner Lipnharski wrote:
{Quote hidden}

the
> same technology, lets say Alkaline, will deliver the same Watts/Hour
no
> matter what cell voltage or current, the same volume means the same
> Watts/hour.
>
> Lets see:
>
> Suppose the circuit drains 15mA at 3V (full charge) and 11mA at 2.2V
(end
> life).  The average consume will be around 13.5mA.  Using two AA cells
> (1A/h) you can have around 74 hours of operational battery life.
>
> Now, suppose you use both AA cells in parallel, and use the TC125301
unit
> to generate 3V fixed.
> The conversion current will be 2x when batteries are fully charged
(1.5V)
> and 2.7x when at life end (1.1V). This gives you an average of 2.4x.
> Consuming 15mA, the circuit will drain 36mA in average, from the
cells. As
> they are in parallel, now you have 2A/h, result in 55 hours of
operational
> battery life.  Not even counting the fact that the 125301 efficiency
is not
> 100%. Suppose it is 85%, battery life will be down to 46 hours.
>
> To have the same battery life (74 h), the battery drain should be
around
> 13.5mA constantly (per cell), it means the 125301 could not drain more
than
> 27mA (parallel batteries).  Considering a worst case of 2.2V VCC, 11mA
> consume, the 125301 will drain 1.46x current from batteries (fully
charged
> 1.5V), and 2x at the end of life (1.1V).  Applying 85% efficiency, it
goes
> to 1.64x and 2.35x rate.  Then, the 11mA circuit consume, will be
draining
> (11mA x 1.64) 18mA (fully charge) and (11mA x 2.35) 25.8mA (end life)
from
> the cells. The average here can go to around 17mA, what gives 2A/0.017
=
> 117 hours.
>
> This extra gain, even with the 15% loss of the efficiency of the dc/dc
> conversion, is based on the higher current drain when battery is above
2.2V
> (up to 3V), when the circuit is directly connected to 2xAA (no dc/dc).
>
> One needs to consider if this extra operational hours (from 74 to 117)
is
> accepted by the customer, considering the final product price increase
due
> the dc/dc conversion (can not cost less than $2), final product can
> increase $12 or more.  For some products that customer uses no more
than
> one or two hours per month, he really doesn't care about the extra
battery
> time, but $12 more or less, can really makes a difference.
>
> Want to do a comparison in the real world?
>
> Cell phones.
>
> Using top of the line dc/dc and battery technology, standby time of 80
> hours, talking time of 6 hours.
> Ok, what about if it was 90 hours and 7 hours?  what difference it
really
> does make?
> I talk less than 1 hour/day, and ALWAYS will keep the cell in the
charger
> during the night.
> Why I will pay more $$$ for a phone or a battery that can gives me 8
> talking hours?
> Obviously some people requires that, but not me, and not a bunch of
other
> people too.
>
> Other than that, who talks 8 hours a day in a cell phone, needs to do
a
> brain damage catscan, for two reasons, first, the RF damages the
brain,
> second, what a hell is he doing 8 hours/day in the phone?
>
> Wagner

--
http://www.piclist.com hint: The PICList is archived three different
ways.  See http://www.piclist.com/#archives for details.


'[EE]: Solenoid information'
2003\01\03@164018 by Dave Dribin
flavicon
face
On Thu, Dec 26, 2002 at 04:24:09PM -0500, Wagner Lipnharski wrote:
{Quote hidden}

Hmmm... the reason I was using Q1 was to only charge C1 when
necessary, since the solenoid fires only occasionally.  Looks like I
got too caught up in the "transistor is a switch" line of thought and
did not pay attention to the details of what was going on.  Thanks for
pointing out what I overlooked.

I don't think a PNP would work well, either, since the collector
voltage (Vc) would still be dependent on the base voltage (Vb), which
is still on the order of 4.6V, right?  It seems the only way to fully
charge C1 to the full +6V battery voltage is to use a resistor like
you have shown.

Which sorta leads me back to the buck regulator Roman showed.  I don't
understand how that circuit can charge the cap to +6V when it's going
through a transistor, too.  But I'm not gonna worry about that for
now. :) At this stage of the game, I'll take a simpler, but not quite
as efficient, circuit and worry about the better version later.

-Dave

--
http://www.piclist.com hint: The list server can filter out subtopics
(like ads or off topics) for you. See http://www.piclist.com/#topics

2003\01\03@183446 by Wagner Lipnharski

flavicon
face
Dave Dribin wrote:
{Quote hidden}

so, what is the problem in have the capacitor charged all times?
Just choose one with low leak and it will keep quiet, no drain the
batteries even in 10 years... ooops, batteries will not hold charge so
long... :)

Wagner

--
http://www.piclist.com hint: The list server can filter out subtopics
(like ads or off topics) for you. See http://www.piclist.com/#topics

2003\01\03@184100 by Dave Dribin

flavicon
face
On Fri, Jan 03, 2003 at 06:29:41PM -0500, Wagner Lipnharski wrote:
> Dave Dribin wrote:
> >> +6V o----R1----o--------o------.
> >>         100    |        |      |
> >>               _|_      _|_     3
> >>               === C1    A      3 SOL
> >>                |        |      |
> >>                |        '------o
> >>               _|_              |
> >>                                C
> >> PIC o-----------R2---o-------B   Q1 NPN
> >>                390   |         E
> >>                      R3        |
> >>                      |2k2      |
> >>                     _|_       _|_
>
> so, what is the problem in have the capacitor charged all times?
> Just choose one with low leak and it will keep quiet, no drain the
> batteries even in 10 years... ooops, batteries will not hold charge so
> long... :)

I was under the impression that the cap would leak enough to cause
significant drain.  If that's not the case, then all the better. :)

Thanks again for your help,

-Dave

--
http://www.piclist.com hint: The list server can filter out subtopics
(like ads or off topics) for you. See http://www.piclist.com/#topics

2003\01\03@184858 by Olin Lathrop

face picon face
> I was under the impression that the cap would leak enough to cause
> significant drain.  If that's not the case, then all the better. :)

That depends on the cap and can vary a lot with temperature.  Check the
spec sheet intead of guessing.


*****************************************************************
Embed Inc, embedded system specialists in Littleton Massachusetts
(978) 742-9014, http://www.embedinc.com

--
http://www.piclist.com hint: The list server can filter out subtopics
(like ads or off topics) for you. See http://www.piclist.com/#topics

2003\01\04@025148 by Roman Black

flavicon
face
Dave Dribin wrote:

> Which sorta leads me back to the buck regulator Roman showed.  I don't
> understand how that circuit can charge the cap to +6V when it's going
> through a transistor, too.  But I'm not gonna worry about that for
> now. :) At this stage of the game, I'll take a simpler, but not quite
> as efficient, circuit and worry about the better version later.


For this stage why not just charge the cap through
a 100k resistor and leave it charged all the time?
Then just one transistor to pulse the solenoid.

A good quality 25v electro will have about 1uA or
so leakage at 6v, and if you don't have one it's
worth buying a uA meter and testing it. I don't
trust cap or diode spec sheets after some of the
crud we've bought lately.

Or just have a go at the scary buck regulator, heck
its only 2 parts more than your circuit! ;o)
-Roman

--
http://www.piclist.com hint: The PICList is archived three different
ways.  See http://www.piclist.com/#archives for details.

More... (looser matching)
- Last day of these posts
- In 2003 , 2004 only
- Today
- New search...