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PICList Thread
'Power Off PIC Happenings'
1997\02\20@105000 by myke predko

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Hi Folks,

Here's a question that somebody probably has a real quick answer to:

I have a circuit in which I independantly power a PIC (and it's reset line)
with respect to the components that it interfaces to.  The Power is
controlled to the PIC via an Opto-isolator and _MCLR is isolated by a
forward biased diode (the PIC is in an ISP Programmer attached to the
circuit).

One of the lines the PIC is attached to is an LED, connected to a Current
Limiting Resistor and Vcc.  The PIC sinks the LED current.

The problem comes in when I turn off the Power and _MCLR to the PIC (Ground
is still connected) - the LED lights!  Is there anything that I can do to
eliminate the current path?

Thanx,

myke

"I don't do anything that anybody else in good physical condition and
unlimited funds couldn't do" - Bruce Wayne

1997\02\20@120534 by Gvran Mvrk

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part 0 96 bytes
HTML>
Let the PIC source the LED !
gm

1997\02\20@121929 by Antti Lukats

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Myke,

You should always think any IC as a bunch of diodes, going to GND/VCC
terminals, most inputs have protection diodes.

It is possible to have PIC chips working with open VCC pin, if any input
is powered by external circuitry.

pin that do not have diodes to VCC rail are MCLR and RTCC, and maybe some
more, but you can always chack that out.

I have seen many circuits working with no power! :)

You cant rely PIC VCC be at low level unless _all_ io pins are
at low level (not driving active high)

antti


At 10:45 AM 20/2/97 EST, you wrote:
{Quote hidden}

-- Silicon Studio Ltd.
-- http://www.sistudio.com

1997\02\20@131211 by Bob Blick

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At 10:45 AM 2/20/97 EST, you wrote:
>The problem comes in when I turn off the Power and _MCLR to the PIC (Ground
>is still connected) - the LED lights!  Is there anything that I can do to
>eliminate the current path?


Hi Myke,

Two ways around this - either use the switched power that supplies the PIC
to also supply the LED(but if you are using an optoisolator alone to switch
the power, I doubt that you have much surplus current!), or add a
transistor on the output pin to sink the LED current(open-collector). In
that case the state of operation will be inverted, so you'll either have to
change your code or add another inversion in hardware.

Perhaps some of the real experts will be able to venture an opinion on the
use of the RA4 pin(which I believe is open drain with no diode to +), but I
have no  official opinion on the use of that pin.

Those were sort of the answers you didn't want to hear, eh? Sorry.

Bob

http://www.bobblick.com/

1997\02\20@142658 by Shawn Ellis
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Well, I've had a similar problem with a PIC16C55 running itself off
external sensor input lines!  I finnally had to use a relay that shut the
Vcc to GND after the power was off (the inputs didn't have enough current
to drive the relay without main power)  In the future I plan to use optical
isolation to eliminate the problem...  I wish there was a slicker way...
Anybody know how to ensure the PIC will shut off even with voltage applied
to its inputs/outputs?

----------
{Quote hidden}

line)
> >with respect to the components that it interfaces to.  The Power is
> >controlled to the PIC via an Opto-isolator and _MCLR is isolated by a
> >forward biased diode (the PIC is in an ISP Programmer attached to the
> >circuit).
> >
> >One of the lines the PIC is attached to is an LED, connected to a
Current
> >Limiting Resistor and Vcc.  The PIC sinks the LED current.
> >
> >The problem comes in when I turn off the Power and _MCLR to the PIC
(Ground
> >is still connected) - the LED lights!  Is there anything that I can do
to
{Quote hidden}

1997\02\20@143238 by Mike

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At 02:22 PM 20/02/97 -0500, you wrote:
>Well, I've had a similar problem with a PIC16C55 running itself off
>external sensor input lines!  I finnally had to use a relay that shut the
>Vcc to GND after the power was off (the inputs didn't have enough current
>to drive the relay without main power)  In the future I plan to use optical
>isolation to eliminate the problem...  I wish there was a slicker way...
>Anybody know how to ensure the PIC will shut off even with voltage applied
>to its inputs/outputs?

Why does it HAVE to shut off if there's enough power to run it from
the sensor lines ? Unless its a power waste - just use another I/O pin
to read the power etc So if it keeps running then effectively make it
do nothing by a continuous loop - or think up some clever way for it
to do something (useful) when its not running - like self tests }:o)

Rgds

Mike


Some say there is no magic but, all things begin with thought then it becomes
academic, then some poor slob works out a practical way to implement all that
theory, this is called Engineering - for most people another form of magic.
                                                                      Massen

1997\02\20@150900 by Antti Lukats

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The parasitic power from PIC I/O pins may be a real problem
you remove mains power, however PIC will not be in Reset state
and if there is no Brown-Out Circuit it may never start running
when power is applied again.

One should not hope that PIC gets enough power to run from
input pins, and if there is heavy laod on +5 Power rails,
then this situation (parsitic power) may even damage the PIC.

OC Drivers help out, and series current limiting resistors
on PIC pins. Sometimes when there is not enough load on VCC
a ballast resistor from +5 to GND will help too.

and what goes RA4 then I think it is safe to use it as OC
ouput (not for higher than +5/VCC)

{Quote hidden}

-- Silicon Studio Ltd.
-- http://www.sistudio.com

1997\02\20@154227 by Shawn Ellis

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Because, It doesn't run reliably on the crappy sensor input amps!  And I
want the uproccesor to reset!

----------
> From: Mike <erazmusspamKILLspamWANTREE.COM.AU>
> To: .....PICLISTKILLspamspam.....MITVMA.MIT.EDU
> Subject: Re: Power Off PIC Happenings
> Date: Thursday, February 20, 1997 2:32 PM
>
> At 02:22 PM 20/02/97 -0500, you wrote:
> >Well, I've had a similar problem with a PIC16C55 running itself off
> >external sensor input lines!  I finnally had to use a relay that shut
the
> >Vcc to GND after the power was off (the inputs didn't have enough
current
> >to drive the relay without main power)  In the future I plan to use
optical
> >isolation to eliminate the problem...  I wish there was a slicker way...
> >Anybody know how to ensure the PIC will shut off even with voltage
applied
{Quote hidden}

becomes
> academic, then some poor slob works out a practical way to implement all
that
> theory, this is called Engineering - for most people another form of
magic.
>
Massen

1997\02\20@161609 by Andy Kunz

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At 10:45 AM 2/20/97 EST, you wrote:
>The problem comes in when I turn off the Power and _MCLR to the PIC (Ground
>is still connected) - the LED lights!  Is there anything that I can do to
>eliminate the current path?

Myke,

I don't have an answer, but a neat that _might_ be related.  Since the PIC
uses so little power, did you know you can actually operate it using the I/O
pins rather than the Vdd and Vss lines?  Try it sometime - it really works.

Andy
==================================================================
Andy Kunz - Montana Design - 409 S 6th St - Phillipsburg, NJ 08865
         Hardware & Software for Industry & R/C Hobbies
       "Go fast, turn right, and keep the wet side down!"
==================================================================

1997\02\20@170040 by Bill Durocher

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Please forgive the ASCII art, but could something like this work?  ( haven't
tried it, just throwing it out....)

                        o  V+
                        |
                        |
                        o |
                          |)  Power switch
                        o |
                        |
   +--------------------+
   |                    |
   |                 +--------+
   |                 |    PIC |
   |                 |        |
   |                 +--------+
   |                    |
   |                ___ /
   |              /    / \
   +--/\/\/\-----+----    |    <--- NPN transistor
        R1       |    \   |
                  \ ___\ /         Select R1 for current limiting
                        \
                        |
                        |
                      -----
                       ---
                        -

1997\02\20@170044 by myke predko

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Thanx to everybody who replied (and so quickly too).  I was pretty sure that
the answer would be that there's a path that will allow the current to flow,
but I just thought I'd ask as well.

As for running a PIC from the Input Lines...  Hmmmm....  That could be an
interesting thing to remember for an application some time.

myke
{Quote hidden}

"I don't do anything that anybody else in good physical condition and
unlimited funds couldn't do" - Bruce Wayne

1997\02\20@174859 by Andrew Warren

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Andy Kunz <EraseMEPICLISTspam_OUTspamTakeThisOuTMITVMA.MIT.EDU> wrote:

> Since the PIC uses so little power, did you know you can actually
> operate it using the I/O pins rather than the Vdd and Vss lines?
> Try it sometime - it really works.

Andy:

The reason it works has little to do with the PIC's power consumtion
and a LOT to do with the fact that most of its I/O pins have
protection "diodes" to Vdd and Vss.  When you apply a voltage to an
I/O pin while leaving the Vdd pin open, that voltage powers the PIC
through those diodes.

-Andy

=== Andrew Warren - fastfwdspamspam_OUTix.netcom.com
=== Fast Forward Engineering - Vista, California
===
=== Custodian of the PICLIST Fund -- For more info, see:
=== www.geocities.com/SiliconValley/2499/fund.html

1997\02\20@183243 by peter

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I use just a NPN transistor (open collector) to sink the led
current

what could be easier !

--
Peter Cousens
email: @spam@peterKILLspamspamcousens.her.forthnet.gr
snailmail: Peter Cousens, karteros, Heraklion, Crete, 75100, Greece,
phone: + 3081 380534,    +3081 324450   voice/fax

After Bill Gates announced to the world that he was Microsoft,
his wife was asked to comment. She said that as his wife, she
had been the first to notice this problem

1997\02\20@203249 by John Payson

picon face
> The reason it works has little to do with the PIC's power consumtion
> and a LOT to do with the fact that most of its I/O pins have
> protection "diodes" to Vdd and Vss.  When you apply a voltage to an
> I/O pin while leaving the Vdd pin open, that voltage powers the PIC
> through those diodes.

This is true.  From my experience, the PIC will 'never' operate usefully in
this way when you actually want it to, but it will 'always' operate this
way when you don't want it to.

For example, some types of I/O devices will manage to keep a PIC powered to
about a volt even if it has no other source of power; once the PIC's voltage
drops to that point it starts drawing extremely little power so even just a
cap on an I/O pin can do this.  Unfortunately, the power-on reset will often
fail if the voltage doesn't go really close to zero, and so the ability of
the PIC to power itself through the I/O lines can be a pain.

1997\02\20@212042 by engmessi

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At 14:53 20/02/97 -0800, you wrote:
>Andy Kunz <KILLspamPICLISTKILLspamspamMITVMA.MIT.EDU> wrote:
>
>> Since the PIC uses so little power, did you know you can actually
>> operate it using the I/O pins rather than the Vdd and Vss lines?
>> Try it sometime - it really works.
>
>Andy:
>
>The reason it works has little to do with the PIC's power consumtion
>and a LOT to do with the fact that most of its I/O pins have
>protection "diodes" to Vdd and Vss.  When you apply a voltage to an
>I/O pin while leaving the Vdd pin open, that voltage powers the PIC
>through those diodes.
>
>-Andy


Andies:

Isn't the power consumption important, here ? What I mean is, if the PIC can
be powered through low current diodes, as those for port protection, isn't
it thanks to its low consumption ? (Well, I am assuming the protection
diodes are low current and fast. Is it right ?)


Pedro.

1997\02\20@213327 by Dwayne Reid

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>I have a circuit in which I independantly power a PIC (and it's reset line)
>with respect to the components that it interfaces to.  The Power is
>controlled to the PIC via an Opto-isolator and _MCLR is isolated by a
>forward biased diode (the PIC is in an ISP Programmer attached to the
>circuit).
>
>One of the lines the PIC is attached to is an LED, connected to a Current
>Limiting Resistor and Vcc.  The PIC sinks the LED current.
>
>The problem comes in when I turn off the Power and _MCLR to the PIC (Ground
>is still connected) - the LED lights!  Is there anything that I can do to
>eliminate the current path?

Greetings from chinook country, Myke

The obvious solution is to arrange the LED so that it is sourced by the PIC
pin.  I'm guessing that you may not be able to do this because of limited
current from your opto.  If this is the case, you may be forced to add
another transistor, either as a switch (2n4401, E to port pin, C to LED, B
to PIC VDD via a 10 K resistor) or as a standard driver (2n4401, E to gnd, C
to LED, B to port pin via 10 K resistor).  The first method does not require
inverting the port in software.

Something that you may not be aware of is that the "sneak" current lighting
your LED is also partialy powering your PIC.  This can cause all manner of
power-up problems.  One of the most serious flaws with the PICs that I have
used is that if VDD does not fall below 0.4 - 0.5 Vdc, the part may fail to
initialise properly and lock up such that even a hard reset will not start
it going.  PICs that have complex peripherals (timers, uarts, etc) may
appear to init except that some of those peripherals may not work.  I had a
real problem with a 16C71 locking up so solid in one project that the
external watchdog couldn't reset it (MCLR was being pulled solidly to gnd by
the watchdog but the PIC wouldn't respond).  Andrew Warren got me on the
right track with that one: it turned out that my supply was only collapsing
to about 0.75 Vdc when power was removed.  Adding circuitry to ensure that
VDD went all the way to gnd with power off fixed that one.

Bottom line: if you are supplying independent power to the PIC that might be
removed but leaving the remainder of your circuit powered, make sure that
signals on your port pins can't power the PIC.  Hope this helps.


Dwayne Reid   <RemoveMEdwaynerTakeThisOuTspamplanet.eon.net>
Trinity Electronics Systems Ltd    Edmonton, Alberta, CANADA
(403) 489-3199 voice     (403) 487-6397 fax

1997\02\21@011132 by Andrew Warren

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I wrote:

> > The reason [a PIC can be powered from its I/O pins] has little
> > to do with the PIC's power consumption and a LOT to do with the
> > fact that most of its I/O pins have protection "diodes" to Vdd
> > and Vss.  When you apply a voltage to an I/O pin while leaving
> > the Vdd pin open, that voltage powers the PIC through those
> > diodes.

and Engenharia Mestra de Sistemas Sociedade Ltda <spamBeGonePICLISTspamBeGonespamMITVMA.MIT.EDU>
replied:

> Isn't the power consumption important, here ? What I mean is, if
> the PIC can be powered through low current diodes, as those for
> port protection, isn't it thanks to its low consumption ? (Well, I
> am assuming the protection diodes are low current and fast. Is it
> right ?)

Pedro:

It's a factor, I suppose (which is why I said that power consumption
has "little", rather than "nothing", to do with the behavior).

However, the "diodes" (which are actually transistors) can handle
currents as high as 20mA, so the behavior would probably be observed
even if the PIC needed much more power than it actually does.

In any case, this is NOT a good way to power your PICs.

-Andy

=== Andrew Warren - TakeThisOuTfastfwdEraseMEspamspam_OUTix.netcom.com
=== Fast Forward Engineering, Vista, California
=== http://www.geocities.com/SiliconValley/2499

1997\02\21@045305 by Kieran Sullivan

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Sounds like being able to power a PIC through one or more of its IO pins might just be useful sometime..Maybe not very wise, but something to remember. Now to get around the problem of the PIC still working when Vcc has been removed...well what is the brown-out circuit used for? The brownout circuit is designed to pull MCLR to Gnd whenever Vcc falls below, say, 4V. When MCLR has been grounded, the PIC will stop running!

Kieran

1997\02\21@063050 by Wolfram Liebchen

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At 16:55 20.02.97 -0500, you wrote:
{Quote hidden}

Hm, I think this circuit ends in a catastrophe, because R1 is critical
if the U-BE of the transistor is not constant (change in temperature).

If by some reason the current through the collector increases (e.g. the
temperature rose), then the voltage across the PIC will increase. This
will increase the current into the basis of the transistor --> current
through collector increases....

Wolfram


+-----------------------------------------------------+
| Wolfram Liebchen                                    |
| Forschungsinstitut fŸr Optik, TŸbingen, Deutschland |
| RemoveMEliebchenspamTakeThisOuTffo.fgan.de                         |
+-----------------------------------------------------+

1997\02\21@142146 by engmessi

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At 12:29 21/02/97 +0100, you wrote:
{Quote hidden}

What is the idea, here ? I don't think I get it.


Pedro.

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