# Confusion about common-mode voltage in a single-ended system

I have trouble figuring out common-mode voltage related topics.

I would like to illustrate my understanding first:
The below op-amp has two inputs: V+ and V-, as usual. And lets say the differential voltage is:

Vd(t) = (V+) - (V-) = sin(ωt)

Let's say there is a DC common-mode voltage. When we measure the voltage between V+ input and the system GND (could be earth) the oscilloscope shows this difference signal as: (V+) + Vcm. When we measure the voltage between the V- input and the system GND (could be earth) the oscilloscope shows this difference as: (V-) + Vcm. So the above circuit can be modelled as below:

Is the above model only valid when the - input is not system or earth grounded?

Now imagine we have a single-ended system with a floating signal source like a battery powered transducer. At the differential amplifier end the GND of the transducer is wired to the system ground as below (you see - input of the op-amp is earth grounded here):

In this case when we measure the voltage between V+ input and the system GND (could be earth) the oscilloscope shows this difference signal as: (V+) + Vcm. When we measure the voltage between V- input and the system GND(could be earth) the oscilloscope shows this difference as 0 V since the - input is wired to GND/earth. So can the above system be modelled as below?

Here is a paragraph from the book for this configuration:

What does that mean here? Is common-mode offset error higher in this configuration than in differential signalling? I'm completely lost.

One way to calculate the common mode voltage is very simple:

$$V_{cm} = \frac{V_+ + V_-}{2}$$

From this formula you can find the common-mode voltage in any configuration where you can work out the voltage at the inverting and non-inverting inputs.

For example,

Here, the common-mode voltage is $\frac{V_\rm{cm} + V_\rm{d}}{2}$ ($V_\rm{cm}$ being the value of the source in the diagram, not the actual common-mode voltage) because the inverting input is fixed at 0 V. If the amplifier "OA" is actually an op-amp, the result is likely that the output is driven to one or the other power supply rails (or as close as that op-amp is capable of driving it), because there is no negative feedback to reduce the circuit gain.

So the above circuit [your first diagram] can be modelled as below [your second diagram]??

The two diagrams are not equivalent. The first one shows no means of establishing the common-mode voltage, so the internal bias networks of the amplifier will drive it to some value. But they may do so only very weakly, and that the common mode voltage might even drift around as a few electrons are blown onto or off of the circuit by passing breezes (depending on the exact design of the op-amp).

Your second diagram shows a voltage source establishing a common mode voltage with a low impedance, which is probably a much better way of driving an op-amp circuit.

• I dont understand. "Your second diagram ... probably a much better way of driving an op-amp " That diagram is just an equivalent of the first. The input in my question is always floating(like 9V battery powered) single ended signal. Vcm is the DC voltage wrt system/earth ground. Vcm is theoretical I just wrote. In real its wired like the first diagram. Mar 5, 2017 at 2:41
• "Vcm being the value of the source in the diagram, not the actual common-mode voltage" Vcm is not source, its just I added it to model single ended input. Mar 5, 2017 at 2:46
• You drew it as a source, therefore in your model it's a source. If you want a model that doesn't include a low-impedance common-mode source, draw that model. Mar 5, 2017 at 3:21
• If the OPA is working, the input terminals will be nearly identical. So in practice, when used typical feedback, the Vcm is that voltage. For the OPA. Now what's it relative to? Good question, depends on context. Dec 22, 2020 at 17:08
• Hi @PeteW, 1. Nowhere does the question mention optical parametric amplifiers (OPAs). 2. The question is 3 years old. 3. The input pins of the actual op-amp are not shown in this diagram, they are hidden within the generic amplifier symbol. If we design a differential amplifier circuit with an op-amp, the input pins to the whole circuit won't be driven equal to each other. Dec 22, 2020 at 18:28

That is a long question.

Variation in common mode voltage will induce error or instability on the system.

It all depends on how accurate and stable you wanna be. Some op-amp are better than others at handling common mode voltage.

One important thing, is to avoid having the input of the opamp floating, as this will introduce noise and instability in your circuit. This is why most of the schematics you refer always have an input tied to the ground.

If you want to keep high impedance, you can use a high value resistor divider to either one of the input to keep it from swinging around. Then you can do a calibration to compensate the common mode voltage error.

The latest schematic refer to the error introduced by the resistance of the cable, which will create a voltage drop on the line and thus the opamp will see the Ves + I*RLead. I ILead is small, this error will be insignificant. I think the term common mode voltage is mistaken on that definition explaining your confusion.

To avoid errors due to line resistance, you can use a 3 or 4 point connection (called kelvin connection) where you can sense the voltage with a second high impedance line directly at the voltage source and have 1 cable for current path and 1 for voltage sense.