Trying to determine dc current - 12v lead acid battery charger

[HankMcSpank]

Putting that together, each bit from the result of adc1 – adc2 will represent .048828 Amps (48.828mA).

Which isn't a whole lot of resolution.

I'm not sure which PIC you're using, but I use the latest range of PIC16f182x ....certainly with that suite of PICs (& mybe others...dunno), you can tie the ADC internal 'positive reference' to an internal derived fixed reference voltage.....so you'd then get over 4x resolution if you made the ADC's positve ref the 1.024V internal reference (vs a PIC 5V supply voltage)

Also just another way of looking at the above (which are obviously all correct, but sometimes it's nice to have another way of looking at things).

1. First establish the voltage drop across your current sense resistor.....adc1-adc2 ... multiply this result by your ADC 'volts per bit' in play (therefore if 1.024V fixed ADC positive ref I mentioned & 10 bits, that's 1mV per ADC 'bit')
2. Now you have the voltage drop ....you already know the resistor value, it's then just ohms law (I= measured voltage drop/0.1R) ....you'll need to scale the number up as PICbasic isn't keen on decimals.