Electronic – Difference between analog signal and battery voltage

As long as you can tolerate tying each end of the resistor to ground via a large resistance (500K or so), without disturbing the circuit, you can use an arrangement like this:

where you use two channels of the ADC (or two ADC's), and compute the difference between them.

The voltage dividers as shown allow only 20% (100K/(400K+100K)) of the voltage to appear at the inputs to the microcontroller. Choose other values as needed.

As Olin as pointed out, this arrangement exceeds the maximum recommended impedance for PIC ADCs, due to the time constant of the ADC input capacitance and the source impedance. It might still be okay as long as the voltage across the resistor doesn't change rapidly and you can tolerate a long acquisition time. It might be worth trying and see if the values you obtain are suitable for your application.

First: Voltage at any point in a circuit is measured with reference to some point within a circuit, not some absolute voltage.

There are two possible scenarios here:

1. Isolated Boards:
   There is some form of isolation between the grounds of Board A and Board B. This would happen, for instance, if either Board A and Board B are not connected to each other at all, or are connected via either a transformer, or an isolated power supply.

   In this scenario, measuring the "difference between grounds" is meaningless: The ground of one board would vary arbitrarily with reference to the ground of the second board. More specifically, if one of the boards is a mains power adapter or wall-wart, its mains-side ground is approximately at building earth potential, while the "ground" of the user-side board may see a voltage of as much as the line AC voltage signal.

   Thus the signal to be measured can only be measured with reference to the ground on the board containing the signal. Attempting to measure signal on Board B relative to the isolated ground on Board A would have no meaning.

2. Interconnected boards:
   There is no isolation between the two boards, they are interconnected, but perhaps via some net other than ground. This could happen, for instance, if the only connection between the boards is some kind of differential pair signal wires, or say an audio cable.

   In this scenario, the theoretical difference between the "ground" potentials between the two boards, in the absence of any EMI or noise sources, would be some constant DC voltage. Thus, once this DC voltage difference between the grounds is known, it would seem that measuring the signal on Board B with reference to its own ground, would also provide the signal voltage with reference to the ground on Board A, plus whatever the measured difference in ground voltages is.

   However, this is not exactly what happens in practice: The ground potentials between the boards would vary constantly based on any electromagnetic noise picked up by the interconnecting wires, as well as a function of the current flowing through whatever the reference wire is between the two boards, i.e. varying difference in ground voltages, V = I (current through reference interconnection) x R (resistance of reference interconnection).

   Thus again, measurement of the signal must be relative to its local ground, whereas measurement of picked up noise could be done between the two boards.

Solution: Measure the signal on Board B relative to its own ground line for any meaningful result.