I was reading a text about instrumentation amplifiers. I couldn't find any easy explanation what really common mode voltage means and its importance.
Electronic – What does common mode voltage stand for in an instrumentation amplifier
instrumentation-amplifiertheory
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I haven't seen an InAmp chip with a dedicated common mode output.
I've used the circuit below to gleam the common mode voltage. Notice that gain-setting resistor RG is split into 2 equal parts R5 and R29.
I haven't thoroughly tested the accuracy of this common mode output. I only needed an approximate common mode voltage for a driven guard.
You didn't show the all-important value of C1, but it together with R1 looks more like a low pass rolloff than a deliberate phase shift. Without knowing C1, we can't tell what the rolloff frequency is, and therefore whether it produces significant phase shift over the valid frequency range or is just there to cut down high frequency drive to the right leg.
The basic principle is to drive the right leg so as to null out the common mode variations in the signal accross the heart. The body is going to pick up whatever the ambient electric fields are, particular the power line "hum".
The signal to noise ratio of skin voltages is horrendously bad. The common mode power line hum can easily be a few orders of magnitude higher than the signal you are trying to measure. A ideal inamp amp will eliminate that, but nobody has made one of those yet. They all have some real upper limit on common mode rejection and common mode range. By trying to null out the common mode part of the signal, it helps the inamp do its job better. If nothing else, it cuts down on the common mode range that the inamp must be able to handle, even if common mode rejection ratio isn't the main issue.
Best Answer
The common mode voltage is a voltage offset that is "common" to both the inverting and noninverting (i.e. "+" and "-") inputs of the instrumentation amp. An instrumentation amplifier is set up as a difference amplifier, so it measures the difference between these two inputs and so rejects any voltage that is common to the two. In other words, if you have two signals v1(t) and v2(t) on the two inputs:
v1(t) = f1(t) + Vcm(t)
v2(t) = f2(t) + Vcm(t)
what the instrumentation amp will measure is:
vo(t) = v1(t) - v2(t) = (f1(t) + Vcm(t)) - (f2(t) + Vcm(t)) = f1(t) - f2(t)
Note that Vcm(t) (the common mode voltage that appears in both input signals) is cancelled out. Also note that this doesn't have to be a DC signal, but can vary with time.
Now why do we care about common mode voltage when selecting a difference amplifier? As other folks have said, there are two key characteristics of the amplifier to consider, the common-mode rejection ratio (CMRR) and the common mode range.
The CMRR is important because the instrumentation amplifier is not an ideal difference amplifier. An ideal difference amplifier would reject 100% of the common mode voltage in the input signals, and would only measure the difference between the two signals. In a real-world instrument amp, this is not the case, and there is a measurable (although typically very very small) amount of the common-mode voltage on the input that gets into the output.
The common-mode range is important, because it limits how far away from ground the measured input signals can be. This is a limit because typically you can't measure signals outside the supply voltages (often referred to as "rails) of the amplifier. There are exceptions to this, but in general the voltage of each input signal must remain within the supply rails of the amplifier. So if you are supplying your amplifier with rails of +/-12V, you may be unable to measure the difference between two signals with a common-mode offset of 15V, even if the difference between the two signals is only 20mV. For example, if your two signals are completely DC and are:
V1 = 15 + 0.010
V2 = 15 - 0.010
Vo = V1 - V2 = 0.020
You would not be able to measure these if your instrumentation amplifier had a common-mode range of +/-12V.