Post by thread:[PIC] Power on with momentary pushbutton

Rochester, 4 november 2005. Not sure if this would work, but here is what I came up with: http://horta.urmc.rochester.edu/~mhofman/PIC/poweronoff.gif The button will supply power to the PIC and circuit when pressed (note that the power supply of the circuit and PIC will need some stabilization as the switch will bounce). It will also provide a "high" signal to PIN1, which is normally pulled "low". When the PIC starts running, it should immediate output a "high" on POUT1, which will cause the MOSFET to switch and power the PIC and circuit instead of the push button. It should then wait for the signal on PIN1 to become low, indicating that the user has finished pressing the button. Later, whenever PIN1 suddenly becomes high again, the PIC can shut down gracefully. I would appreciate it if anyone found any flaws in my design (my electronics skills are slightly stale compared to my software skills). Greetings, Maarten Hofman.

Way back in 2002, I also needed to be able to use a PIC to turn off it's own power. I received an outstanding response to my question from Roman Black. He uses an inexpensive SCR. (Thank you, Roman). Check out his page at: http://www.romanblack.com/self_swi.htm Andrew ________________________________________________ Get your own "800" number Voicemail, fax, email, and a lot more http://www.ureach.com/reg/tag ---- On Fri, 4 Nov 2005, fred jones (spam_OUTboattowTakeThisOuThotmail.com) wrote: > Hi all, > Hope I don't get flamed for this because I recall it being covered here > before, however I have searched and can't find it. Does anyone have > info/schematic on using a momentary pushbutton to turn on the PIC and > circuitry and then another press of the button to turn it off? Thanks for > the help. > FJ > > > --

Maarten Hofman <.....cashimorKILLspam@spam@gmail.com> > Not sure if this would work, but here is what I came up with: > > horta.urmc.rochester.edu/~mhofman/PIC/poweronoff.gif > > The button will supply power to the PIC and circuit when pressed (note > that the power supply of the circuit and PIC will need some > stabilization as the switch will bounce). It will also provide a > "high" signal to PIN1, which is normally pulled "low". When the PIC > starts running, it should immediate output a "high" on POUT1, which > will cause the MOSFET to switch and power the PIC and circuit instead > of the push button. It should then wait for the signal on PIN1 to > become low, indicating that the user has finished pressing the button. > > Later, whenever PIN1 suddenly becomes high again, the PIC can shut > down gracefully. > > I would appreciate it if anyone found any flaws in my design (my > electronics skills are slightly stale compared to my software skills). Pretty close. But the MOSFET needs to be a P-channel device, and its gate is driven low to turn it on. An N-channel device would need to have its gate driven *above* 5V to work in this configuration. Since the PIC outputs are effectively shorted to ground by its protection diodes when it is unpowered, you'll also need a separate driver transistor for the MOSFET -- a low-power NPN device with the emitter grounded, the collector connected to the gate and also pulled up with a high-value resistor to the unswitched +5V rail, and its base connected through a resistor to the POUT1 pin. Then the circuit will work as you describe. -- Dave Tweed

From: "fred jones" > Hi all, > Hope I don't get flamed for this because I recall it being covered here > before, however I have searched and can't find it. Does anyone have > info/schematic on using a momentary pushbutton to turn on the PIC and > circuitry and then another press of the button to turn it off? Thanks for > the help. I'm sure there must be many ways to accomplish this. The PIC can wake up from sleep mode by a change on a PORTB pin or an RB0 interrupt. Depending upon the power requirements of the external circuitry, you might be able to simply power it directly from a PIC pin. I have successfully done this with a temperature datalogger that I built. When it was awake, the PIC powered an LM35 temp sensor and a couple of serial I2C EEPROMs with no problem. Otherwise, a MOSFET should do the trick. Just remember to not let input pins float when your external circuitry is turned off. As far as shutting down on a button press, your software could easily handle that by waiting for a second or so (to allow the user time to release the button so you don't immediately wake back up) and then going to sleep after powering down the external circuitry.

> -----Original Message----- > From: piclist-bouncesKILLspammit.edu On Behalf Of fred jones > Sent: Friday, November 04, 2005 8:35 AM > <snip> > Does anyone have info/schematic on using a momentary pushbutton to turn > on the PIC and circuitry and then another press of the button to turn it > off? Thanks for the help. > FJ It appears to me that you want the first button press to turn on the circuit and then a second press of the same button to turn off the circuit. If I've interpreted this correctly, then are you really locked into a momentary push button? A standard push on - push off push button switch placed in the power line will do this without any other circuitry. Paul

Thanks, you are correct. Sorry if I wasn't clear on this. I want to be able to press a membrane power switch to turn on the PIC/device and then press the same membrane momentary switch to turn it back off. I think I came up with a circuit to do this when I re-phrased my google search today and a solution from another forum came up. It really may not be the best solution though so any others would be appreciated. Thanks, FJ ================================================== It appears to me that you want the first button press to turn on the circuit and then a second press of the same button to turn off the circuit. If I've interpreted this correctly, then are you really locked into a momentary push button? A standard push on - push off push button switch placed in the power line will do this without any other circuitry. Paul

er... I am stumped by Roman Black's SCR scheme. The 7805 is NOT turned off by the SCR, the 7805 simply no longer regulates at 5V when the SCR drops out due to low current. In addition, SCR specs are VERY unreliable, changing drmatically with heat, voltage, and load current. --Bob Andrew Kieran wrote: {Quote hidden}>Way back in 2002, I also needed to be able to use a PIC to turn >off it's own power. I received an outstanding response to my >question from Roman Black. He uses an inexpensive SCR. (Thank >you, Roman). Check out his page at: > >http://www.romanblack.com/self_swi.htm > >Andrew > > > >________________________________________________ >Get your own "800" number >Voicemail, fax, email, and a lot more >http://www.ureach.com/reg/tag > > >---- On Fri, 4 Nov 2005, fred jones (@spam@boattowKILLspamhotmail.com) wrote: > > > >>Hi all, >>Hope I don't get flamed for this because I recall it being >> >> >covered here > > >>before, however I have searched and can't find it. Does >> >> >anyone have > > >>info/schematic on using a momentary pushbutton to turn on the >> >> >PIC and > > >>circuitry and then another press of the button to turn it off? >> >> > Thanks for > > >>the help. >>FJ >> >> >>--

Gerhard Fiedler wrote: >Bob Axtell wrote: > > > >>er... I am stumped by Roman Black's SCR scheme. The 7805 is NOT turned >>off by the SCR, the 7805 simply no longer regulates at 5V when the SCR >>drops out due to low current. >> >> > >The input of the 7805 may not be switched off, but the return path seems to >be. No return current == no input current, no? Where would the current >flow, if the 7805 is not turned off, as you say? > >Gerhard > > > No, Gerhard, that's not how a 7805 works. If the GND pin of a 7805 is opened, the pass element inside is 7805 is switched on fully, and the OUT pin equals the IN pin except for a small PN drop. Try it! --Bob -- Note: To protect our network, attachments must be sent to KILLspamattachKILLspamengineer.cotse.net . 1-520-777-7606 USA/Canada http://beam.to/azengineer

At 10:30 AM 11/5/2005, Bob Axtell wrote: >Gerhard Fiedler wrote: >>Bob Axtell wrote: >>>er... I am stumped by Roman Black's SCR scheme. The 7805 is NOT turned >>>off by the SCR, the 7805 simply no longer regulates at 5V when the SCR >>>drops out due to low current. >> >>The input of the 7805 may not be switched off, but the return path seems to >>be. No return current == no input current, no? Where would the current >>flow, if the 7805 is not turned off, as you say? >>Gerhard >No, Gerhard, that's not how a 7805 works. If the GND pin of a 7805 >is opened, the >pass element inside is 7805 is switched on fully, and the OUT pin >equals the IN >pin except for a small PN drop. What I think that you've missed, Bob, is that the SCR completes the ground leg of the circuit for both the regulator AND the load. IOW, the Vss connection to the PIC and all the rest of the load current is returned to the anode of the SCR. That's also where the reference (Gnd) leg of the regulator is connected. You are right: when the SCR is off, the Vdd line floats up to the unregulated input with respect to the incoming power ground. But the voltage across the regulator output is zero because there is no ground return and the device is OFF. Also note that the voltage drop across the SCR is outside the regulation loop - any voltage drop across the SCR simply appears as a voltage drop at the input of the regulator. So long as there is sufficient input voltage that the regulator remains above dropout, the PIC and the rest of the circuit are operating at 5V. dwayne -- Dwayne Reid <RemoveMEdwaynerTakeThisOuTplanet.eon.net> Trinity Electronics Systems Ltd Edmonton, AB, CANADA (780) 489-3199 voice (780) 487-6397 fax Celebrating 21 years of Engineering Innovation (1984 - 2005) .-. .-. .-. .-. .-. .-. .-. .-. .-. .- `-' `-' `-' `-' `-' `-' `-' `-' `-' 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.

At 05:11 PM 11/4/2005 -0700, you wrote: >er... I am stumped by Roman Black's SCR scheme. The 7805 is NOT turned >off by the SCR, the 7805 simply no longer regulates at 5V when the SCR >drops out due to low current. > >In addition, SCR specs are VERY unreliable, changing drmatically with heat, >voltage, and load current. > >--Bob I'm with Bob on this one- I looked at one of the suggested SCRs, the C103B, and the *maximum* holding current is rated at 5mA. There is no minimum spec, so there is no guarantee it will drop out with 4 or 5 mA current from a 78x05 (in fact, it's practically guaranteed that it *won't*. However, the idea is feasible. I suggest an "artificial" SCR made from a couple of BJTs *with* base resistors to control the drop-out current, and using a low-Iq regulator (some LDOs draw more current as the dropout voltage approaches-- we don't want one of those. Best regards, Spehro Pefhany --"it's the network..." "The Journey is the reward" spamBeGonespeffspamBeGoneinterlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com ->> Inexpensive test equipment & parts http://search.ebay.com/_W0QQsassZspeff

Dwayne Reid wrote: {Quote hidden}> At 10:30 AM 11/5/2005, Bob Axtell wrote: > >> Gerhard Fiedler wrote: >> >>> Bob Axtell wrote: >>> >>>> er... I am stumped by Roman Black's SCR scheme. The 7805 is NOT turned >>>> off by the SCR, the 7805 simply no longer regulates at 5V when the SCR >>>> drops out due to low current. >>> >>> >>> The input of the 7805 may not be switched off, but the return path >>> seems to >>> be. No return current == no input current, no? Where would the current >>> flow, if the 7805 is not turned off, as you say? >>> Gerhard >> >> No, Gerhard, that's not how a 7805 works. If the GND pin of a 7805 is >> opened, the >> pass element inside is 7805 is switched on fully, and the OUT pin >> equals the IN >> pin except for a small PN drop. > > > What I think that you've missed, Bob, is that the SCR completes the > ground leg of the circuit for both the regulator AND the load. > > IOW, the Vss connection to the PIC and all the rest of the load > current is returned to the anode of the SCR. That's also where the > reference (Gnd) leg of the regulator is connected. > > You are right: when the SCR is off, the Vdd line floats up to the > unregulated input with respect to the incoming power ground. But the > voltage across the regulator output is zero because there is no ground > return and the device is OFF. > > Also note that the voltage drop across the SCR is outside the > regulation loop - any voltage drop across the SCR simply appears as a > voltage drop at the input of the regulator. So long as there is > sufficient input voltage that the regulator remains above dropout, the > PIC and the rest of the circuit are operating at 5V. > > dwayne > I guess I just have to breadboard it to understand it. If it works, it certainly is an unusual way to accomplish this. I always do power switching schemes by a flipflop, usually by crosscoupled gates, using a CD4001, so I switch it at 5-17V of the raw power. A PIC can turn it off by pulsing a voltage-doubling charge pump (a diode and two caps), driving an extra gate which turns it off. A momentary PB in the input of the turns it on. I have also found a slick way to do it using a latching relay, a PB switch, and two caps. The caps store the energy to pulse a change in the relay. At 5V, this even works when driven by a PIC signal. --Bob -- Note: To protect our network, attachments must be sent to TakeThisOuTattachEraseMEspam_OUTengineer.cotse.net . 1-520-777-7606 USA/Canada http://beam.to/azengineer

Spehro Pefhany wrote: >> In addition, SCR specs are VERY unreliable, changing drmatically with >> heat, voltage, and load current. > > I'm with Bob on this one- I looked at one of the suggested SCRs, the > C103B, and the *maximum* holding current is rated at 5mA. There is no > minimum spec, so there is no guarantee it will drop out with 4 or 5 mA > current from a 78x05 (in fact, it's practically guaranteed that it > *won't*. I don't have much experience with SCRs, but I think the idea here is not that the SCR will drop out closely below 5 mA, the idea is that it won't drop out at 5 mA or more, and that it will drop out at currents in the range of a sleeping PIC -- in the microampere range. Is that something that's doable with SCRs, even with varying specs? Gerhard

Gerhard Fiedler wrote: > I don't have much experience with SCRs, but I think the idea here is not > that the SCR will drop out closely below 5 mA, the idea is that it won't > drop out at 5 mA or more, and that it will drop out at currents in the > range of a sleeping PIC -- in the microampere range. Is that something > that's doable with SCRs, even with varying specs? I haven't seen the schematic so I'm only going on what people are saying about it. However if there is a 7805 regulator in there, it won't matter if the PIC is taking 0 current. The quiescent current of a 7805 is quite high, probably above the minimum sustaining current of most SCRs, when such a thing is even specified. ****************************************************************** Embed Inc, Littleton Massachusetts, (978) 742-9014. #1 PIC consultant in 2004 program year. http://www.embedinc.com/products

Bob Axtell wrote: >>> er... I am stumped by Roman Black's SCR scheme. The 7805 is NOT turned >>> off by the SCR, the 7805 simply no longer regulates at 5V when the SCR >>> drops out due to low current. >> >> The input of the 7805 may not be switched off, but the return path >> seems to be. No return current == no input current, no? Where would the >> current flow, if the 7805 is not turned off, as you say? >> > No, Gerhard, that's not how a 7805 works. If the GND pin of a 7805 is > opened, the pass element inside is 7805 is switched on fully, and the > OUT pin equals the IN pin except for a small PN drop. As Dwayne wrote, it seems you imagine a current path from the regulator output to input supply ground. Which I don't see in the circuit. > Try it! I just hooked up the input of a 7805 to 12 V and a load resistor between the other two pins (and nothing else). It did not create a current through it... just kidding :) I didn't do that, but I doubt you can have a current flowing there. If you have, you have just invented the perpetuum mobile! Gerhard

At 07:43 PM 11/5/2005 -0200, you wrote: >Spehro Pefhany wrote: > > >> In addition, SCR specs are VERY unreliable, changing drmatically with > >> heat, voltage, and load current. > > > > I'm with Bob on this one- I looked at one of the suggested SCRs, the > > C103B, and the *maximum* holding current is rated at 5mA. There is no > > minimum spec, so there is no guarantee it will drop out with 4 or 5 mA > > current from a 78x05 (in fact, it's practically guaranteed that it > > *won't*. > >I don't have much experience with SCRs, but I think the idea here is not >that the SCR will drop out closely below 5 mA, the idea is that it won't >drop out at 5 mA or more, and that it will drop out at currents in the >range of a sleeping PIC -- in the microampere range. Is that something >that's doable with SCRs, even with varying specs? > >Gerhard Yes, the problem is that the quiescent current of a 78x05 can be as high as 6mA at 25°C, so even if you unplug the PIC and throw it in the trash the thyristor current can't fall below that. Only large insensitive-gate SCRs have holding current that's reliably above that level, and of course you'd have to spend significantly more current than *that*, controlled by the PIC, to have it reliably stay *on*. There are lower quiescent current regulators, of course, though some bipolar LDOs have an Iq that shoots through the roof as you approach the dropout voltage. Best regards, Spehro Pefhany --"it's the network..." "The Journey is the reward" RemoveMEspeffTakeThisOuTinterlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com ->> Inexpensive test equipment & parts http://search.ebay.com/_W0QQsassZspeff

Spehro Pefhany wrote: >> I don't have much experience with SCRs, but I think the idea here is not >> that the SCR will drop out closely below 5 mA, the idea is that it >> won't drop out at 5 mA or more, and that it will drop out at currents >> in the range of a sleeping PIC -- in the microampere range. Is that >> something that's doable with SCRs, even with varying specs? > > Yes, the problem is that the quiescent current of a 78x05 can be as high > as 6mA at 25°C, so even if you unplug the PIC and throw it in the trash > the thyristor current can't fall below that. Oh, of course... <slap on forehead> :) I missed this one. I did like the other approach he found better, and if I had to do something like this, that would be probably close to what I would come up with. I just liked the parts count of the SCR solution... but of course if it doesn't work, that doesn't help much :) Thanks, Gerhard