"Mike Monett" <[EMAIL PROTECTED]> wrote: [...]
> Gentlemen, > The problem is very simple. If the sensor lead that contacts the > mercury has a diameter of 0.010 inch, with a flat end, and is 100 > microinches away from the mercury, the capacitance is 0.35 pf. Sorry, I used a bad formula for the area of a circle. The actual capacitance is 0.1767 pF, still well within the capability of the AD7747. Regards, Mike Monett _______________________________________________ time-nuts mailing list -- [email protected] To unsubscribe, go to https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts and follow the instructions there.
