# A question about an instrumentation amplifier and common mode currents

Regarding my previous question, an unbalanced(single-ended) source Vs is connected to inputs of an instrumentation amplifier and ground as follows:

I have seen above connection in many texts.

For simplicity let's assume Vs is DC and common mode voltage Vcm is DC. And all impedances are real and they are resistances.

And regarding the above diagram, let's say the line resistances Rl1=Rl2=100 Ohm. The input impedance of the inAMp is Rx=10 MegOhm. And the source impedance is 200 Ohm.

What I understand is the common mode current Ic1 loops through the following resistances:

Rs-->Rl1-->Rx

and Ic2 loops through the following resistances:

Rl2-->Rx

If I'm not wrong the impedance seen from the hot terminal is: Rs+Rl1+Rx and the impedance seen from the negative terminal is Rl2+Rx.

My two questions are:

1-) It seems like Rx>>>Rs in an inAmp because of its buffers, does that mean the source impedance matching is less important than in a difference amplifier? Can Rs in practice can be neglected when using an inAmp this way? Let's say Rs is 50 Ohm and Rx=10Meg, would you still add a series 50 Ohm resistor to the negative terminal? Or would you neglect it because Rx dominates?

2-) Look at the third wire at the bottom from the negative terminal of the source to the ground of the inAmp. There is no resistor on that wire. In practice should there be a resistor there not to cause a short circuit?

Edit:

• do you understand this now? If not, please explain. – Andy aka Mar 22 '18 at 18:02
• I have to read it and think about it again. That time obviously i didnt get it. – user1245 Mar 22 '18 at 18:35

Here's the problem as I see it: -

• Irrespective of the InAmp being present, if electromagnetic interference hit both signal conductors equally, a higher interfering voltage will be present on the upper wire because it has a bigger impedance due to Rs. This scheme only works when there is the same (balanced) impedance in both signal wires.
• Linking remote grounds is (usually) bad because any earth fault currents can flow down the cable and pose a problem for the cable integrity AND induce voltages in the two signal cables that may add to the common mode problem and make the InAmp unable to deal with the common mode voltage all the time.
• Both the presense of Rs (unbalanced) and the ground link make earth currents flowing in the remote ground connection induce an asymmetrical interference onto the two signal wires feeding the InAmp.
• 1-) Rs will affect unbalance but my question about neglecting it when Rx is huge comparing to Rs. Isn't the amount of ubalance is linked to how big is Rx? Imagine Rx is infinite theoretically. Wouldn't the both lines be balanced even there Rs exists? – user1245 Jan 22 '18 at 13:44
• 2-) Please see my edit, or here: i.stack.imgur.com/HPbOr.png Now the source is floating I mean it is only grounded at receiver side. Is that still problem? Now it is exactly how they recommend: i.stack.imgur.com/FEVCr.png you hadnt say anything about this last time. – user1245 Jan 22 '18 at 13:46
• @user134429 you are missing the point about Rs imbalancing the system - it does so with respect to EMI hitting the cable - EMI will produce an interfering voltage on the wires that is proportional to impedance to ground and if both impedances are identical then the interfering voltage will be identical on both conductors and the diff amp will ignore those equal voltages. With different impedances it will produce a differential noise. Rx has nothing to do with this effect because it is, in effect an open circuit compared to Rs. – Andy aka Jan 22 '18 at 13:49
• But what is that total impedance wtr to ground? That is I wanna know. For example for the upper wire is that Rs+Rl1+Rx? – user1245 Jan 22 '18 at 13:50
• Oh okay the effect is then about Rs+Rl1 only? Rx shouldn't be in the balancing effect. – user1245 Jan 22 '18 at 13:52
1. In practice, the need for impedance matching depends on your signal. Differential amplifiers are available with very high gain-bandwidth product, sometime higher than 5 Ghz. At those speed, impedance control is definitely required. Instrumentation amplifier are much slower, and rarely go beyond few MHz. By definition, an instrumentation amplifier is a differential amplifier with input buffers. Those buffer Op-amps eliminate the need for impedance matching at low speeds, and give you very high input impedance.
2. I'm not sure what this schematic is trying to show with the $V_{CM}$ source. Maybe it's to explain that the ground potential is not exactly the same at different point in the circuit? Were did you got that circuit from? In any case, $V_{CM}=0$ V so I think that the wire between the positive terminal of $V_{CM}$ and the ground of the amplifier should not be here. Note that instrumentation amplifiers have very high common-mode rejection ratio, so they are very good at rejecting $V_{CM}$ while amplifying $V_S$.
• You wrote: "eliminate the need for impedance matching at low speeds". I was trying to demonstrate that with Rx>>>Rs in the circuit, is that correct approach for DC-like signals? And the circuit from here:i.stack.imgur.com/FEVCr.png Vcm in mine is common mode voltage between grounds. – user1245 Jan 22 '18 at 9:08