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I understand the basics behind the concept of common mode voltage and CMRR and friends, in the context of differential amplification.

What I don't understand is the determining if the whole concept of "common mode voltage" actually applies to a given circuit. That is, sometimes the signal being measured (i.e., the V+ and V-) doesn't have any particular relationship to the supply voltage. In that case, how can a common more voltage even be determined? The V+ and V- might be 1001V and 1000V or 1V and 0V - since voltage is relative I can't see how the amp would even know the difference.

For example, consider the case of a handheld multimeter - it's seems pretty clear that there the concept of common mode voltage wouldn't apply here. The voltmeter should perform identically in the 1001 V- and 1000 V+ case as the 1 V+ and 0 V- case (indeed, they aren't really distinguishable, are they)?

So if a dirty cheap voltmeter can simply avoid the whole question of staying inside the supply rails, and have a "perfect" CMRR, why is it such an issue for most components? I.e., components with very high CMRR are expensive, and require special designs, etc.

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You are right about the multimeter. The difference between that and an amplifier circuit is that the amplifier circuit is supplied from power rails which are not "floating". Common mode voltages refer to voltages at the input terminal with respect to the power supply rails.

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  • \$\begingroup\$ How does one determine whether a measured signal is floating wrt to the supply voltage? I suppose it's simple in the case of a battery powered device - it is clear there is no external connection there, but I'm not sure about more complicated cases. E.g., imagine cases of a USB powered device which may be powered from a laptop, which itself may be on mains power or running on battery. In that case does the "floating" state depend on whether the laptop is plugged in? Can a circuit be designed to accommodate both modes of operation? \$\endgroup\$ – BeeOnRope Jan 21 '16 at 23:15
  • \$\begingroup\$ It is simply a case of whether the source is supplied from the same source or the supplies are connected e.g. through the ground connection. It is possible to design for both cases \$\endgroup\$ – user1582568 Jan 21 '16 at 23:33
  • \$\begingroup\$ Let's say you have a circuit where the supply is from a battery, passed through a linear regulator (like the arduino), and the signal you are measuring is the voltage of the same battery. Is the signal floating with respect to the supply in this case? \$\endgroup\$ – BeeOnRope Jan 21 '16 at 23:38
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    \$\begingroup\$ Sorry for the delay, in the case you describe the signals are not floating. The question you have to ask is "if we connected one of the inputs, would there be a path for current to flow between the source and the measurement circuit?" If so then the signals are not floating. As @Andyaka pointed out, even the multimeter has some leakage at AC due to capacitance, ultimately back to the signal source. If you try to design a differential amplifier, you will see the problems. \$\endgroup\$ – user1582568 Jan 22 '16 at 10:06
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One difference is that your floating multimeter has no ground reference, but in many applications such as balanced microphones, for example, the signal feeds into an amplifier that has a ground reference. The amplifier must amplify the difference but reject the common-mode signal.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Differential transmitter into differential amplifier with ground reference.

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The voltmeter should perform identically in the 1001 V- and 1000 V+ case as the 1 V+ and 0 V- case (indeed, they aren't really distinguishable, are they)?

Theoretically and on the face of it they will perform identically and you wouldn't notice the difference. Theory and practise are a bit different.

If both probe leads of the multimeter are connected through to its "measurement silicon" with the same series resistance (always present in a multimeter so that you can choose different voltage ranges) then any leakage currents to ground (you are at ground holding the meter), would cause an equal volt drop in both these series resistances and you would indeed measure 1V.

I'm referrring to "earth impedance balance" i.e. how balanced are both inputs with respect to each other in terms of impedance? It's a bigger deal with AC measurements because the capacitance to earth from a measurement terminal (or any point in the signal path) comes into play - if the capacitance to earth is not identical in every way at every point in the circuitry feeding the "measurement silicon" on BOTH connections, then you'll get an error if there is a significant common mode AC signal.

Of course, if the "measurement silicon" is basically grounded then trying to measure a signal on top of 1000 volts is probably going to cause a fire.

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  • \$\begingroup\$ I see - so even the multimeter has some leakage to ground and so there may be a practical difference with high common differences. I guess a lot of what I'm getting at given that a cheap multimeter can at least in theory (and probably mostly in practice given some reasonable isolation to ground) measure small voltages on a large carrier, can a circuit be designed that uses the same principle (but the same power source as the signal) to avoid the various issues of CMRR and common mode range. Context is the battery monitor I am building. \$\endgroup\$ – BeeOnRope Jan 22 '16 at 0:11
  • \$\begingroup\$ I'm not sure I understand what you are asking - try a simpler approach but I'm going bed soon (England) and that may be part of the problem - tiredness - but still try a more direct and simpler question - I may just about be able to answer it tonight! \$\endgroup\$ – Andy aka Jan 22 '16 at 0:38

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