I have a board(BoardA) with a MCU which are powered from a battery. the board also has a buffer circuit for the ADC.

I would like to measure a signal from an other board(BoardB) which is powered from an plug. One possible solution is to connect the two grounds and measure the signal.

Let say because of some reason I am not able to connect the two grounds. How can I achieve accurate measurement? If first I measure the ground on the BoardB, so In this case I can have the potential difference between the grounds on the BoardB and BoardA, the I can measure the signal on the BoardB.

Can I achieve accurate measurement with this solution? What can be the disadvantages?

  • 1
    \$\begingroup\$ Maybe you need an opto-isolator/optocoupler? \$\endgroup\$ Commented Sep 23, 2013 at 9:44
  • \$\begingroup\$ Actually I don't need an optocoupler, I just cant have access to one of the ground, so It would be hard to connect together. \$\endgroup\$
    – Mokus
    Commented Sep 23, 2013 at 9:53
  • \$\begingroup\$ When measuring voltage you are measuring Potential Difference. I don't believe there is anyway you can measure that without 2 connections to the circuit. I'd love know if there is a way with 1 connection. \$\endgroup\$
    – Spoon
    Commented Sep 23, 2013 at 11:19
  • \$\begingroup\$ It's important to stress that voltage is not an absolute value. Think of it this way, trying to measure voltage at a single point is like trying to determine how far you are. The only way to give a meaningful answer is to ask "how far from what?". \$\endgroup\$
    – Samuel
    Commented Sep 23, 2013 at 16:06

2 Answers 2


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.


Here is how I see what you have: -

enter image description here

If you can measure the ground on board B from board A then that's fine but only if the measurement remains accurate when you are measuring the signal. And this won't happen - it's likely that noise between the two boards will seriously inhibit your chances of making an accurate measurement. The noise will come from: -

  1. AC mains pick-up
  2. Electromagnetic interference
  3. power currents changing on the two boards
  • \$\begingroup\$ So the best way to connect the two grounds. Thanks. \$\endgroup\$
    – Mokus
    Commented Sep 23, 2013 at 11:39
  • \$\begingroup\$ @run. The minimal way is to connect grounds. A better way is to use a differential measurement input on the ADC buffer circuit - one input goes (as shown) to the signal source and the other input goes directly to the other side of the signal source i.e. the first place it connects to local ground on board B. The best way is to locally measure on board B then transmit digitally to board A. \$\endgroup\$
    – Andy aka
    Commented Sep 23, 2013 at 11:43

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