> Is it as simple as it seems....4-20mA current loops....put across a > precision resistor and then feed to the A/D ? Or better....buffered > thru a unity gain amp...? > > Or is it a little more tricky than that? Yes, it is (or, can be) that simple. For example, a 250 Ohm resistor converts 4-20mA to 1-5V, ideal for input into your PIC's ADC. The current loop topology is important, but if your sensor sources the 4-20mA signal and if it's OK to put your resistor to 0V then this is indeed the simplest way to go. Why buffer it with a unity gain Op-Amp? If it's for "protection" then it's better to do it properly with an appropriate network of voltage clamping and series protection devices. You could however use the Op-Amp with some gain to allow you to reduce the current shunt resistor and thereby the voltage drop in the loop. Or go to the next level and use a differential Amp to allow your resistor to "float" between the supply/common mode limits making it's placement in the current loop more flexible. But that just complicates things...just use a resistor and be amazed how simply it converts current to voltage! -- Brent Brown, Electronic Design Solutions 16 English Street, Hamilton 2001, New Zealand Ph: +64 7 849 0069 Fax: +64 7 849 0071 Cell/txt: 027 433 4069 eMail: brent.brown@clear.net.nz -- http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist