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I have extreme confusion in figuring out common mode voltage related topics. I would like to illustrate my understanding first: Below opamp has two inputs V+ and V- as usual. And lets say the differential voltage is:

Vd(t) = (V+) - (V-) = sin(w*t)

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And lets say there is 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 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??:

enter image description here

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 opamp is earth grounded here):

enter image description 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: 0V since - input is wired to GND/earth. So can the above system can be modelled as in below?

enter image description here

And in a book I read the following about this configuration:

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Here is a paragraph from the book for this configuration:

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

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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,

enter image description here

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.

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  • \$\begingroup\$ 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. \$\endgroup\$ – atmnt Mar 5 '17 at 2:41
  • \$\begingroup\$ "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. \$\endgroup\$ – atmnt Mar 5 '17 at 2:46
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    \$\begingroup\$ 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. \$\endgroup\$ – The Photon Mar 5 '17 at 3:21
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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.

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