I've asked this question a while back and user/Transistor told me that the circuit will be floating and drift without a common ground, he suggested a virtual ground. but in this case we already have a ground connection between two voltage sources via 3 series resistor.

To measure current on the low side of circuit, do we still need to have a common ground between V1 and V2? Do the multimeters make a common ground with the source too?


simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ If V2 is a floating supply voltage (like a battery for example), the circuit you show will work for low V2 frequencies. However, if V2 has capacitance to earth and the signal to be measured is high frequency in nature then you will get less-than perfect results dependent on how the capacitance to earth is imbalanced with your circuit components. If you want a proper answer you need to reveal details about V2, RL and the supplies that feed the op-amp and how they are earthed. \$\endgroup\$ – Andy aka Dec 6 '20 at 11:34
  • \$\begingroup\$ @Andyaka V2 can be a power source of any kind (battery or switching/linear PSU) without a connection to earth, RL is a BJT based dummy load. V1 is a center tapped transformer (rectified of course) without a connection to earth. \$\endgroup\$ – ElectronSurf Dec 6 '20 at 11:43
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    \$\begingroup\$ The devil is in the absolute detail that I mentioned in my 1st comment. To successfully design a differential amplifier requires a high level understanding of earth connections (galvanic or capacitive) for both the source and the measuring equipment. \$\endgroup\$ – Andy aka Dec 6 '20 at 11:53
  • \$\begingroup\$ @Andyaka It seems like a complicated subject to study. to make things easier for my hobby project, should I have common ground or not, if yes a direct common ground or a virtual one? \$\endgroup\$ – ElectronSurf Dec 6 '20 at 12:10
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    \$\begingroup\$ I can't tell you that because I cannot know the details of what I alluded to above. You know me by now; if I could make an answer, I would. \$\endgroup\$ – Andy aka Dec 6 '20 at 12:30

To measure current on the low side of circuit, do we still need to have a common ground between V1 and V2?

If you are absolutely sure that there are no hidden paths between the power supplies and any part of the circuit other than what is shown, you will be OK.

Do the multimeters make a common ground with the source too?

Multimeters are designed to ensure that the condition just mentioned, of no hidden paths is met. So, they often have plastic cases and battery power.

In addition, because there are no hidden paths between power supplies, it is safe for the ground ("common" really, because it isn't connect to "ground") of the multimeter to be connected to the negative probe. This is probably done, though I certainly cannot guarantee it is done in every case. It is probably done, because it is safe (because of the "isolation"), and because it eliminates some unnecessary worry about accuracy and proper operation of the multimeter.

V2 can be a power source of any kind (battery or switching/linear PSU) without a connection to earth

A problem arises when the assumption that the power supplies are truly isolated is mistaken. One may naively think a power supply is isolated, though it may not be. Even a very small current can disturb your circuit, as illustrated by the circuit below, in which 120AC "leaks" through a pair of capacitors between one rail of V1 and ground.


simulate this circuit – Schematic created using CircuitLab

The output, and the node marked "cm" are shown here. The "cm" node is the common mode signal present at the inputs of the op amp (assuming the op amp is still operating in its linear region).

enter image description here

One can see that the output appears to be a solid 2V. However, the peaks of V_CM are at about 10V. If the capacitors were just a little bigger, the peaks of V_CM would reach the rail voltage of 12V. A little higher and the op amp will work incorrectly, giving unknown results at the output -- probably one of the rail voltages. (The simulation does not show this, happily treating voltages at the inputs which are above the rails as if nothing untoward was happening.)

How much current was required by the noise source to bring this circuit to its limit? Only 1mA peak. That may seem very insignificant compared to the 10A that is flowing through \$R_{sense}\$, but it clearly is not insignificant.

The details of when a common mode noise source will drive the op amp inputs beyond the rail voltages depend greatly on the voltage of the common mode noise source, it's impedance relative to the circuit under test, and the resistors in the diagram. Change any one of those, and one will get different results. Different parameters, and any conclusions about immunity to noise need to be re-evaluated.

The impedance of 2 30nF capacitors in series at 60Hz is about 177k\$\Omega\$. Again, the current created by the noise source is only 1 mA peak, but the common mode signal at the op amp is 10V peak.

That is what could happen, if your voltage source is not truly isolated. However, if, like a portable ammeter, the voltage source is a battery, and the case is plastic, you can be quite confident that the impedance seen by your circuit by any common mode noise source will be much higher, (or the voltage of the noise source much lower) and your circuit will measure the voltage across \$R_{sense}\$ just fine.

  • \$\begingroup\$ V2 voltage can be anywhere from 1.5V to 50V, updated the question schematic with resistor values. current passing through the RS is not going to be more than 10A and the op amp is OP07. \$\endgroup\$ – ElectronSurf Dec 12 '20 at 19:10
  • \$\begingroup\$ As you can see from my simulations, the circuit you have is quite sensitive to capacitative coupling to the AC mains. I don't know what you are going to do with the output signal. Does it need to have a rapid response? If not, you can add a decoupling capacitor from the low side of the sense resistor to ground. 1\$\mu\$F is good. 3.3 is better etc. This will drastically reduce the sensitivity of your circuit to AC mains coupling. At the same time, it will decrease your AC response and your response time. It's a trade-off that requires knowledge of how the output signal is actually being used. \$\endgroup\$ – Math Keeps Me Busy Dec 12 '20 at 21:34
  • \$\begingroup\$ Thanks for simulating the circuit, it's 10m ohm not 100m ohm. I know that it's easier to have a common ground, the only reason I want to measure V2 voltage floating because if for any reason V2 is connected with reverse polarity it might damage the whole control circuitry. it can be prevented with reverse polarity protection but then it will require additional components and the current is high so I have to manage the heat etc... \$\endgroup\$ – ElectronSurf Dec 12 '20 at 21:41
  • \$\begingroup\$ If V1 and V2 voltage is provided with main voltage via a transformer (rectified) without a connection to earth ground, am I going to still have that noise current? \$\endgroup\$ – ElectronSurf Dec 12 '20 at 21:44
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    \$\begingroup\$ I made what I think is my last edit on this answer. I believe I have answered the question posed, and if you would like suggestions for how your circuit might be improved, given your requirements, I think they should be addressed in a separate question/answer. \$\endgroup\$ – Math Keeps Me Busy Dec 13 '20 at 1:23

Common ground?

All voltages are related one to another in some way, and in its simpliest form it depends on impedances and interconnections.

When we say that one voltage is "floating" with regards to the other, we simplify our world by forgetting that they still are related through capacitive coupling and electromagnetic fields - in most cases we can effectively ignore these other couplings.

In this particular case you are measuring the current in the load using "direct connection(s)", this forces a relationship between your circuit and your input.

You are are correct that there are multiple resistors contributing to a "common ground" - \$R_1\$ and \$R_2\$ pretty much force the negative side of \$R_L\$ to the "ground". However this is not a "hard" force as you can connect the positive end of V1 directly to ground and force about -50V on your opamp's input terminals likely destroying it.

IMHO that shows that your "common ground" can be considered virtual.

Do the multimeters make a common ground with the source too?

Before connecting the multimeter to your circuit it is isolated(!) from it and "floating".

When you connect the multimeter the electrostatic charges are balanced so that the terminals voltages are perfectly leveled with the measured circuit. This initial connection can induce some very small currents into the measured circuit. After that initial interconnection, the multimeter pretty much acts as a very small resistor when making current measurements, like the circuit that you propose yourself.

If one of the multimeters is connected to ground in the measured circuit, then it has a common ground, otherwise it does not have common ground. But the multimeter voltages and the circuit voltages will still have a strong relationship.

To measure current on the low side of circuit, do we still need to have a common ground between V1 and V2?

If you add another current path between \$V1\$ and \$V2\$, you're likely going to disturb your measurement.

\$R2\$ is important though because otherwise your opamp input voltages can easily exceed the limits of the acceptable voltage range for your opamp. The value of \$R2\$ can surely be higher, it only needs to allow the electrostatic balance to be installed and maintained.

I would avoid a path to ground through the feedback circuit \$R_1\$ and \$R_2\$. Any unwanted current from the \$R_{load}\$ input to ground through these resistors impacts the measurement. To avoid this path, a circuit change would be needed. It's only "needed" if you confirm that this hinders you in practice.

  • \$\begingroup\$ "The value of R2 can surely be higher" and "You could improve your circuit by connecting R2 to Rsense". Actually, R1 and R2 play an essential role in a differential amplifier of making the gain of the + and - inputs equal. If R1 is eliminated, or R2 increased, that will mean the gain of the two inputs are not equal, and common mode signals will be amplified. IMHO a better solution, if the OP is absolutely 100% sure that there are no hidden (such as capacitative) paths between the power supplies, then what reason is there not to connect their grounds? But he doesn't ask for circuit improvement. \$\endgroup\$ – Math Keeps Me Busy Dec 17 '20 at 14:37
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    \$\begingroup\$ You're right, I had my own circuit too much in mind. The idea is that stray ucrrent do not influence the measurement. I'll edit the answer. \$\endgroup\$ – le_top Dec 17 '20 at 19:25

Forget about V2 for a moment. The circuit looks like a current sensing amp. The output of the network is referenced to earth via R2. It could work.

V2 and RL make a current source going almost entirely into RS (assuming RL is small).
Looking at it that way, V2 and RL do not provide a low impedance to the rest of the network at all. They only provide a particular impedance (RL) between the top and bottom of RS. Is this a problem? I think not necessarily.

The other answers mention CMRR and capacitances. In addition, if RS is small enough, such as the 10mohm example, then its parasitic inductances might become significant compared to its resistance. If RL is really a transistor as mentioned above that too can be inductive. Anyway, "it depends".

UPDATE a way to look at it

  • The current in the two 20k are equal.
  • The current has nowhere else to go, it just goes in one loop around clockwise, starting from the OPA output, ending at the ground node.
  • V2, RL, RS (if RS and RL are small) are roughly equivalent to a voltage source and a small series R. The equivalent voltage is roughly RS/(RS+RL).
  • The circuit output multiplies this equivalent voltage by 20:1
  • \$\begingroup\$ see update above \$\endgroup\$ – Pete W Dec 18 '20 at 19:33

The main issue is how do you measure REAL CMRR comes from how you model your REAL environment.

The other answers are correct but incomplete. To add further, you must consider that the CMRR is not just the Op-Amp but the ingress EMI from all common-mode noise on your target, the wiring imbalance and the impedance imbalance of each wire coming from the current sense resistor.

It is relatively easy to get > 100 dB CMRR in an Op-Amp which is 5 decades of the magnitude of balanced gain, but nearly impossible to get this even from any cable, twisted pair or coax. If you imagine two 120 Ohm wires twisted to the 240-ohm differential, imagine how hard it is to get each wire with a 5% tolerance on insulation down to 10 parts per million in both dielectric constant consistency and thickness as well as wire gauge for inductance and proximity to the noise source for the differential of mutual coupling to each wire. This is your real challenge and is why Shielded Twisted Pair (STP) rather than unshielded UTP.

The combination of both improves the result of EMI ingress attenuation.

CM Chokes

Yet even this is not enough so a high impedance CM choke is used for ethernet, HDMI and VGA RGB signals to raise the impedance of each line to reduce the effects of EMI for induced ingress of common-mode noise current and unintended radiation of EMI (egress) while keeping a controlled differential low impedance.

High impedance lines are susceptible to unbalanced common-mode E-Field EMI translating into a differential noise signal as well as unbalanced common-mode H-field EMI. A CM Choke is only useful if the differential impedance is low as the ferrite impedances involved are typically <<1 kOhm.

SMPS Noise interferes with CMRR

Often an isolated switched-mode supply has terrible common-mode noise due to the crosstalk in the switching transformer from interwinding capacitance. Sometimes simply an RF cap between the "isolated" grounds will reduce the EMI, if you are less worried about earth ground noise than the SMPS noise. This can happen also due to the reduced gain in Op-Amps are RF EMI range and thus lower CMRR or it can be coupled into the interconnecting wires from power to signal again from poorly balanced or lack of shielded STP signal wires.

There is no magic bullet until you know where your EMI comes from and it's characterization for impedance and spectrum. Is it a low frequency AC line or a HF noise or a modulated RF etc. Each may required slightly different solutions.

In cases of EEG low level signals the virtual ground must be an active ground often called RLD Right Leg Drive which is a standard common-mode signal from all sources and used to drive the RLD electrode thus reducing the differential CM voltage in each signal wire.

In sensing SMPS noise and motor current noise, the current loops might be large and become a loop antenna radiating on your signal wires. Then you must consider orthogonal orientation to minimize the loop antenna effective gain in that crosstalk in addition to other methods.

None of these methods are new and are all well-documented in Henry Ott's Book on handling EMC which you may find older copies on archive.org.

But in all cases, the CMMR of the Op-Amp is not the worst case. You must identify the root cause and path of EMI, then the solution is easy. Often making it worse temporarily helps one find the sensitive area and so it can be attenuated.


Previous answers don't address CMRR.

Yes you do. While you do not need a ground plane, you still need a fairly low ground path, I wouldn't go above a few hundred ohms.

The reason is that op-amps are not perfect, the output voltage is not only relative to the differential input, but also to the absolute input voltage. So you want your input voltage to be tied to the ground of the opamp.

It is called the common-mode rejection ratio or CMRR.

This is why you don't want your op-amp input to be floating around as this will make the circuit non-linear and not predictable.

Also, common-mode noise will come into play, and the input side can easily be shifted fairly high if it is not on the same ground.

Another problem is that your input voltage may float outside the range of the op-amp input, destroying it on the way.

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    \$\begingroup\$ "Math answers doesn't address the elephant in the room, CMRR." Perhaps you could expand on this more. The CMRR of an OP07 is 106 dB minimum. As long as the op amp is operating in linear mode, the op amp's CMRR should be fine, no? Do you mean a lower CMRR arising from some other source? \$\endgroup\$ – Math Keeps Me Busy Dec 17 '20 at 13:56
  • \$\begingroup\$ Definitely not. If you don't have common ground, the input will bounce around at several kHz due to common-mode noise, especially when the system is powered through hashing DC supplies. Opamp Protection diodes will kick in and the linearity will be very poor. Within multimeter, the opamp is connected to the V-, but is galvanically isolated from earth. The plastic casing has nothing to do with electronic implementation. only battery-powered multimeter has plastic cases..... \$\endgroup\$ – Damien Dec 21 '20 at 9:40

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