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


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