It's only an idea, but could use the regulators with a shutdown control pin.
Just put a cap across the power supply pins of the PIC, shut off the
regulator and put the PIC to sleep. Wake the PIC up to control the regulator
when the voltage drops to a pre-defined level and charge the internal caps
again.  Most of the regulators with the shutdown draw 1ua in shutdown.
Haven't tried it, so I don't know what problems you might get with doing
this.

Brian.

-----Original Message-----
From: pic microcontroller discussion list
[mailto:PICLIST@MITVMA.MIT.EDU]On Behalf Of Dave Dilatush
Sent: 22 November 2003 15:51
To: PICLIST@MITVMA.MIT.EDU
Subject: Re: [EE:] Quiescent current ?


James Nick Sears wrote...

>I am looking at the feasability of running my circuit from 2 high capacity
>coin cells in series for a total output at full charge of 6V.  I need to
get
>this down to <= 5.5V for the well being of the PIC and other ICs in my
>circuit.  The circuit sleeps most of the time consuming uA current.  If I
>use a regulator with (for instance) 50uA quiescent current and my circuit
is
>drawing 30uA, does the battery have to supply 50uA or (50 + 30) = 80uA?  In
>other words is quiescent current a minimum draw or a constant amount that
is
>wasted?

It's what's wasted- that is, the current the regulator itself
consumes to do its job.  So 80 uA would be the correct number.

Seems to me there was a thread here on this same topic a year or
so ago, with a lot of good ideas in it.

Dave D.

--
http://www.piclist.com hint: To leave the PICList
mailto:piclist-unsubscribe-request@mitvma.mit.edu

--
http://www.piclist.com hint: To leave the PICList
mailto:piclist-unsubscribe-request@mitvma.mit.edu