At 08:04 AM 11/2/2005 -0800, you wrote: >I have...what I am told....a 4-20mA current loop, two terminals at least >so the assumption is that its a 'constant current source' in the reguards >that any load placed on it won't affect the current being driven. If >thats the case, then the current loop itself should be isolated from the >driving circuitry, but since I do not have a schematic thats kind of an >unknown. Lets assume it is. > >So, I use a high precision 248ohm resistor across the two terminals that >will generate a 0.992 to 4.96V over the range. Next...getting it to the >A/D of the PIC. Typically overrange up to 22 or even 30mA is allowed, so be careful about this. > I planned on using a voltage follower, simple circuit using the MCP601 > but as I started to sketch out the schematic, I wonder if this is really > single ended. > >If I use the (-) side of the resistor as the ground reference, and it is >truly isolated on the transmit side then just hooking the (+) side to the >positive input of the opamp (configured as a vf) should work. If its NOT >isolated....then what? Then you have a problem, or at least a more demanding specification. The general solution for process control signals is to have a transformer-isolated supply and an opto-isolated, transformer isolated or differential capacitor isolated digital or analog signal path across the isolation barrier. That will allow a large common mode voltage on either input pin without problems. Any other kind of solution is going to be situational, but may be cheaper (*if* it works reliably in all cases). >Best regards, Spehro Pefhany --"it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com ->> Inexpensive test equipment & parts http://search.ebay.com/_W0QQsassZspeff -- http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist