After some reading, I have just learned how a voltage divider works
and I would appreciate feedback on this if I have it wrong:

Vout = Vin * (R1 / R1 + R2)

Where
        Vout is max 5V (connected to AN0)
        Vin  is the battery voltage

 bat+ ---|
         |
         <
      R1 >
         <
         |
         |---- AN0
         |
         <
      R2 >
         <
         |
 bat- ---|

If Vin is 7.2V and R1=10K then R2=4.4K

I must be dense - but I do not quite understand your "other idea". Vdd
is regulated through an LM2936 at appx. 5V - I need to get a rough
idea of the voltage "outside" the regulator.

On Wed, Mar 27, 2002 at 04:04:23PM -0600, Dale Botkin wrote:
> On Wed, 27 Mar 2002, Douglas Butler wrote:
>
> > Use a 2 resistor divider, no op amp needed.  Try a 22K from the battery
> > to the A/D, and a 47K from the A/D to GND.
>
> I'd recommend 10K max to Vss and Vdd, or a 20K pot.  MC recommends 10K max
> input impedance to the ADC.  With a pot you can set the input at full
> battery charge to your Vref, as pointed out by anothr poster.
>
> Another idea:  pot from Vdd to an otherwise unused I/O pin, wiper to the
> ADC input.  Now you drive the pin low to read the supply voltage, and set
> it as an input to avoid the current drain thru the resistors the rest of
> the time.  Yeah, your ADC reading is relative, but you said that's OK.
>
> Dale
>
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--
Glen Wiley
Application Technology, Core Processes IT
Capital One Financial Services

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