I think i understand it now. If 5A is the maximum and the output increases by 185mV for each amp then the output range should be 0-925mV. If vcc is 5 then you add 2.5 giving 2.5-3.425V for positive so the entire range must be 1.575-3.425V.

Ive read a datasheet for a PIC that has ADC but i didnt see anything about the lowest input voltage (unless i missed it or didnt understand it). Would i be right in saying that you take the output from the other chip and lower it so the range becomes 0-1.85V and set the PICs Vref to 1.85V? I think this is what the amplifier thing was about that i read somewhere. I think the idea is to stretch the 1.85V so the range goes from 0V all the way up to 5V so the PIC can read it easier

I dont think i should have this problem using the resistor method because i can set the PICs Vref to the vdd pin (5V) then use the potential divider circuit to make sure its output is 5V when a little over the maximum amount of amps are being drawn (im making it a little over so i can add overload warning and safety cutoff)

Where does 30 come into it? On the main page it said +/- 5A to 30A