The Schottky generally turns OFF (Recovers) a lot faster than a generic diode.... but the low voltage drop is the key.
QUESTION: What is the voltage you are feeding into the PIC? +/- 10V +/-2V ???
Your ground will need to be common to what you are measuring... otherwise you could very well have an offset voltage and potential ground loop issues.
You could use an OPAMP circuit that gives you a balanced input, and does the clipping (Low side and high side) but that is more complex that the simple Resistor/diode/Resistor combo I showed prior.
For Diode look at the BAT85
http://www.nxp.com/#/pip/pip=[pip=BAT85_4]|pp=[t=pip,i=BAT85_4][pip=BAT85_4]|pp=[t=pip,i=BAT85_4]
Or surface mount BAT54A (Single) or BAT54S (Dual in series if you want to clamp min and max voltage)
http://www.nxp.com/#/pip/pip=[pip=BAT54_SERIES_4]|pp=[t=pip,i=BAT54_SERIES_4][pip=BAT54_SERIES_4]|pp=[t=pip,i=BAT54_SERIES_4]
Both of the above are about 0.5uA leakage current at 5V @ 25C
If you used a 10K input resistor, this current would drop 5mV across it (10K/0.5uA=5mV)
Lower resistor values have less effect but more current will flow when diode conducting.
Also at low current it will conduct/clamp at a lower voltage, maybe 0.2V or less
Give it a try, but note it is a very simple "Crowbar" type solution.
Mike
post edited by mimemike - 2009/02/24 16:41:56