> Nope, if you amplify the input by two you'll get a 0 - 1023 value that
> maps to TWICE the full range of the (original) input voltage, and
> represents twice what it would normally.  In other words, with a 2V
> reference and a 1V unamplified input signal, one ADC step equals (roughly)
> 1.95mV, and you won't see an ADC reading over 512.  Amplify the input
> signal x2 and you will get ADC readings up to 1023, but now each ADC step
> represents 3.9mV.

No, this is backwards.  Each A/D step will still be 2 / 1023 = 1.95mV at the
PIC pin because nothing has changed there (still using 2V reference).
However, that A/D step size gets *divided* by the opamp gain when projected
back to the original signal.  Each step now corresponds to a 978uV change in
the input signal.  Another way to see this is that the 0 to 1V input signal
maps to the full range of the A/D, regardless of how we got there.  1V /
1023 = 978uV.

Unless I needed very high accuracy, I would multiply the input signal by a
little less than 5 and dispense with Vref altogether.  In otherwords, use
the 5V power regulator as the reference.


********************************************************************
Olin Lathrop, embedded systems consultant in Littleton Massachusetts
(978) 742-9014, olin@embedinc.com, http://www.embedinc.com

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