Olin,

- one: when you'll speak and write romanian language as I speak and write in
english and german, my hat off and I'll accept corrections. Nobody's perfect
except you. So it seems.

- two: I was not aware that this is a list for native english only. If so,
tell me to unsubscribe (as I can understand I can't)

 I was trying to do my best. If that's not enough for you, I'm really sorry.
And by the way, your spelling is not always perfect, or should I say
everyone makes typing mistakes ;).

As for my advice to Kieren I mentioned the zenner diode as an example on how
you could solve the problem using a zenner. I didn't said that it would be
the best solution to the problem. I also used the series diode hack for a
battery powered project but then even if the batteries voltage will drop
below 5.1V the voltage will be 4.4V not 5.1.

So the best solution in my opinion will be to split the supply rail in two:
one straight from the batteries to the series resistor for the LED (so the
brightness will last more at higher values) and the other with the series
diode to power the PIC. But this will also need the zenner to protect the
PIC because if accus are used instead of batteries then when a bit
overcharged the voltage can reach 6.5V (5*1.3V)-0.7(across the diode)=5.8V
across the PIC which is a bit too much.

I'm not sure on how low dropout regulator behave when the input voltage
drops below the intended output voltage and what current they use to try to
keep it that way.



No offense intended and none taken,

Mircea Chiriciuc

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