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'[PIC]: Membrane latching power switch for PIC circ'
2002\08\10@080349 by Les Otter

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Hello
I am trying to move from a toggle power switch to a NO momentary membrane
switch.

My circuit thoughts are :-
http://www.saurel.demon.co.uk/files/Membrane.bmp

(Caps not shown)
where the membrane switch S1 supplies power to the circuit at startup, which
boots the pic.
The first thing the PIC does is sets Pin B0 high, so S1 can be released.
When the PIC detects button press of another membrane switch ("OFF" - not
shown) pin B0 can be set low, thus powering off the circuit.

Is this a sound design concept, and anyone suggest how to do it better?
Can't find anything on Google / PIClist Archive or FAQ.
Thanks in advance for any help.

Les

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2002\08\10@092649 by Rick C.

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part 0 44 bytes
his is a multi-part message in MIME format.
part 1 1424 bytes content-type:text/plain; charset=us-ascii (decoded 7bit)

You have the right idea. If you only want one button but can tolerate a few more
discretes, use the following circuit. Pushing the button will power up the pic
just like you said, and at bootup set portb.1 high. Pressing the button again
momentarily, sensed by portb.2, (polled or interrupt) and with a small timing
routine will shut the circuit off. Depending on your routine, if you hold the
button down too long, it may restart the power.
Rick

Les Otter wrote:

{Quote hidden}


part 2 2469 bytes content-type:image/gif; (decode)


part 3 154 bytes
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2002\08\10@094345 by Matt Pobursky

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Yes, the concept is good -- however there is at least one basic
problem with your circuit.

You'll never get more than about 4.4V at the emitter of the transistor,
since the PIC can only supply ~5V of base drive to the transistor. Use
a PNP transistor (or P channel FET) with NPN driver stage (or N channel
FET) to supply the regulator circuit.

Matt Pobursky
Maximum Performance Systems

On Sat, 10 Aug 2002 13:04:31 +0100, Les Otter wrote:
{Quote hidden}

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2002\08\10@104022 by Dale Botkin

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I've done this (in a non-PIC circuit, but the same sort of thing) using a
P-MOSFET with the gate tied high with a high-value resistor.  PIC sets the
pin LOW to keep power on, sets the pin as input (hi-Z) to turn off.

Dale
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Fusistance is retile.
Your ass will be laminated.

On Sat, 10 Aug 2002, Les Otter wrote:

> Is this a sound design concept, and anyone suggest how to do it better?
> Can't find anything on Google / PIClist Archive or FAQ.
> Thanks in advance for any help.

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2002\08\10@111727 by Dale Botkin

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Here's what I'm talking about:

http://madmax.botkin.org/dale/FETpower.png

On Sat, 10 Aug 2002, Les Otter wrote:

> I am trying to move from a toggle power switch to a NO momentary membrane
> switch.

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2002\08\10@113513 by mike

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Don't forget that in some cases you don't actually need to turn the
PIC off at all - with the right voltage reg, sleep mode draw is on a
par with battery shelflife. The PIC can usually pretty easily power up
other parts of the system as required. e.g. with a 9V PP3/MN1604 battery (500maH), and a micropower reg like
a 78LC50 (ON/Mot), the 30uA or so draw in sleep with WDT enabled
equates to about 15,000 hours, or about a year and a half.


On Sat, 10 Aug 2002 10:16:13 -0500, you wrote:

>Here's what I'm talking about:
>
>http://madmax.botkin.org/dale/FETpower.png
>
>On Sat, 10 Aug 2002, Les Otter wrote:
>
>> I am trying to move from a toggle power switch to a NO momentary membrane
>> switch.

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2002\08\10@125601 by Les Otter

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Matt Pobursky said..
> You'll never get more than about 4.4V at the emitter of the transistor,
> since the PIC can only supply ~5V of base drive to the transistor. Use
> a PNP transistor (or P channel FET) with NPN driver stage (or N channel
> FET) to supply the regulator circuit.
>
>
Thanks for the help..but I don't understand why I will only get 4.4v.
I thought that if I supply 5v to the NPN base from a logic high PIC output
pin, the transistor would act as a simple switch and pass the +9v across the
C-E junction.  Why only 4.4v?
Thanks
Les

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2002\08\10@131559 by Les Otter

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Dale
Thanks, but that link is currently broken.  Will try later.
Rgds
Les

{Quote hidden}

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2002\08\10@145115 by Philip Pemberton

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"Dale Botkin" <dalespamKILLspamBOTKIN.ORG> said:
> Here's what I'm talking about:
> http://madmax.botkin.org/dale/FETpower.png
Er... Proxomitron says the connection to your server timed out. i.e. the URL
is bad. This one works though:
http://www.botkin.org/dale/FETpower.png

Later.
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Phil.
.....philpemKILLspamspam.....dsl.pipex.com
http://www.philpem.dsl.pipex.com/

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2002\08\10@161215 by Dale Botkin

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Whoops.  Sorry about that, mere mortals can't reach madmax on port 80
thanks to my stupid damned ISP.  Try this:

http://www.botkin.org/dale/FETpower.png

Sorry for the confusion.

Dale
---
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Fusistance is retile.
Your ass will be laminated.

On Sat, 10 Aug 2002, Les Otter wrote:

> Thanks, but that link is currently broken.  Will try later.

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2002\08\11@115426 by T.C. Phelps

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Hi Les,

In response to your question (below), you're right in
that applying +5V to the base of your NPN transistor
will make it act like a switch; but the transistor is
basically a controlled _current_ source. So you design
the circuit to pass the desired current into the
collector and you get almost the same current out the
emitter (depending on your beta -- for your purposes,
you can probably assume it's the same) when the
transistor is on. BUT, the voltage difference between
the base and the emitter is 0.6-0.7 volts, like a
diode (there is a pn junction between the base and
emitter pins after all). So, if you apply +5 volts
from a PIC pin to the base then your emitter will be
at 4.4 volts.

Regards,
Todd.

{Quote hidden}

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2002\08\11@141524 by Les Otter

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Todd and everyone who replied

Thanks for all your help. I really need to swot up on my analog stuff.  I
DID read Horowitz & Hill Art of Electronics (transistor section)  before I
posted the question, but I find it heavy going.  I need to find a book which
explains things in more basic terms - rather like the responses I have
received from my posts to the PICLIST on this subject.

So, if I understand your response Todd et al, that means I can drive the
base of a PNP BJT with the collector of a NPN , the emitter of which goes to
ground, and the base of which will connect to my PIC pin - rather like the
circuit that Rick C posted on this thread, and Matt Pobursky pointed out?

Appreciate your time
Les

PS So my circuit has +9v at the collector, +4.4V at the emitter...have I
invented a logic controlled 4.6v heater? :o)

> {Original Message removed}

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