=20
Hello,

in a new project I need to operate a PIC16F684 at a 9V battery.=20
I saw, that e.g. a MCP-1702 voltage regulator gives me a current consumptio=
n
of around 2=B5A while the PIC is in SLEEP mode, OK, some nA more for the PI=
C
but this is more than OK for my application.

My problem ist, that I need to check the battery voltage, because there are
some more externals which need at least 6V for proper operation.

How is this done? Sure, using an ADC input of my 684 but for the battery
voltage is well above 5V I cannot simply connect my battery to the PIC port=
,
but I need a voltage devider.
When looking at the data sheet I see a leakage current for an analog input
in the region of 500nA, so I assume, that I need a current through my
voltage divider which is about 10 or 20 times the leakage current, which is
5 to 10=B5A!

But this much more than my 1702 needs!

Does anybody have a tip how I can reduce the current through my voltage
divider?

Thanks and have a great day.

Peter


--=20
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