Post by thread:Power Off PIC Happenings

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

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

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/

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}> From: Antti Lukats <spam_OUTanttiTakeThisOuTSISTUDIO.COM> > To: .....PICLISTKILLspam@spam@MITVMA.MIT.EDU > Subject: Re: Power Off PIC Happenings > Date: Thursday, February 20, 1997 4:19 AM > > 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: > >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 {Quote hidden}> >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 > > > > > > -- Silicon Studio Ltd. > -- http://www.sistudio.com

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

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

Because, It doesn't run reliably on the crappy sensor input amps! And I want the uproccesor to reset! ---------- > From: Mike <erazmusKILLspamWANTREE.COM.AU> > To: .....PICLISTKILLspam.....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}> >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

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!" ==================================================================

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

Andy Kunz <EraseMEPICLISTspam_OUTTakeThisOuTMITVMA.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 - fastfwdspam_OUTix.netcom.com === Fast Forward Engineering - Vista, California === === Custodian of the PICLIST Fund -- For more info, see: === www.geocities.com/SiliconValley/2499/fund.html

> 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.

At 14:53 20/02/97 -0800, you wrote: >Andy Kunz <KILLspamPICLISTKILLspamMITVMA.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.

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

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 <spamBeGonePICLISTspamBeGoneMITVMA.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 - TakeThisOuTfastfwdEraseMEspam_OUTix.netcom.com === Fast Forward Engineering, Vista, California === http://www.geocities.com/SiliconValley/2499

At 16:55 20.02.97 -0500, you wrote: {Quote hidden}>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 > \ > | > | > ----- > --- > - > 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 | | RemoveMEliebchenTakeThisOuTffo.fgan.de | +-----------------------------------------------------+

At 12:29 21/02/97 +0100, you wrote: {Quote hidden}>At 16:55 20.02.97 -0500, you wrote: >>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 >> \ >> | >> | >> ----- >> --- >> - >> What is the idea, here ? I don't think I get it. Pedro.