In this article, they are explaining the benefit of using a technique called remote ground sensing by using a single-ended sensor and a differential amplifier or an instrumentation amplifier. Below is the excerpt regarding that part:

enter image description here

But I don't understand what text means here and how does this work better than the following (my own edited drawing) connection below:

enter image description here

In both diagrams above the sources are grounded in the measurement side. The only difference the one in the article uses an extra wire to the ground of the differential amplifier(or instrumentation amplifier).

  1. Why is the ground difference is removed in the first case? Obviously I don't get the text's explanation.

  2. If first one is using 3-wires like a shielded twisted pair cable, and my drawing is using a 2-wire coaxial cable; would there be any difference in terms of common mode noise rejection? And why/why not? I'm asking this because I'm trying to understand whether the so called "remote ground sensing" only has advantage for ground offset errors but not common mode interference. In other words are both of ways of connections unbalanced by the same amount?

Regarding the below drawing representing the original diagram above:

enter image description here

Considering the source impedance is zero or negligible, and considering the line/wire impedances are equal \$Z_1 = Z_2\$, is the diagram above a balanced system?

"The impedance to ground" seen by common mode interference for the first hot wire is:

\$Z_1+Z_{\text{amp}}\$ (the path from the + terminal of the source through hot wire and \$Z_{\text{amp}}\$ to GND of the diff amp.)

But what is the path taken by the source's GROUND SENSE?

  • \$\begingroup\$ I notice in your edit drawing, you place a Z1 and Z2 on the sense lines. But you do not put a Z3 on the line from Vs to ground. Why not? \$\endgroup\$ Commented Jan 22, 2018 at 2:03
  • \$\begingroup\$ Yes I see I opened a question and asked about that: electronics.stackexchange.com/questions/351345/… \$\endgroup\$
    – user1245
    Commented Jan 22, 2018 at 2:22
  • \$\begingroup\$ Should the Z3=Z2? \$\endgroup\$
    – user1245
    Commented Jan 22, 2018 at 2:23
  • \$\begingroup\$ Normally I think Z3 is the shield so I dont know if a resistor should be added to limit the current caused by a common mode voltage . \$\endgroup\$
    – user1245
    Commented Jan 22, 2018 at 2:32
  • \$\begingroup\$ no no no, you're not adding a physical resistor. You're modeling as if there was a low-value resistor there to try to accurately model the fact that real conductors do have impedance, albeit small. Being savvy to this impedance is essentual to understanding sense lines and what the article says. \$\endgroup\$ Commented Jan 22, 2018 at 4:59

3 Answers 3


With ground at two remote points there will inevitably be ground currents (from other equipment also) circulating that cause interfering volt drops between sending end and receiving end. This produces an error and degrades your signal. That’s usually regarded as the primary issue.

A secondary issue is that the impedance seen on hot and common wires from the perspective of electric or magnetic field interference is vastly different. This means interference cannot be coped with to any significant degree. In the original circuit, the impedance to ground on both wires is generally equal.

  • \$\begingroup\$ "In the original circuit, the impedance to ground on both wires is generally equal." Do you mean the original circuit is balanced? But the source's GND is tied to diff-amp's ground through the wire called "GROUND SENSE". How come they are balanced(impedance to ground equal)? Thanks \$\endgroup\$
    – user1245
    Commented Jan 21, 2018 at 19:42
  • 1
    \$\begingroup\$ @user134429 a line is balanced by virtue of equal impedances on both wires. Transmitting a differential signal is nice and can ensure balance but, it is the equality of impedances that makes a system balanced. If the driven line is from a low source impedance then that type of circuit is fairly well balanced. However, if driven via (say) 50 ohms then a same value resistor is also needed in the ground wire to make it balanced. \$\endgroup\$
    – Andy aka
    Commented Jan 21, 2018 at 19:44
  • \$\begingroup\$ Hmm I will try to make a drawing to the question about this maybe I can figure out my confusion. \$\endgroup\$
    – user1245
    Commented Jan 21, 2018 at 19:47
  • 1
    \$\begingroup\$ Get a sim tool and push a current through the ground wire to simulate the sensor power current and you’ll see what I mean. \$\endgroup\$
    – Andy aka
    Commented Jan 21, 2018 at 19:49
  • \$\begingroup\$ Thanks I will try that but I made an edit if you read and make a comment to the question would be great basically "But what is the path taken by the source's GND(not GROUND SENSE)?" (Im confused with direction of the common mode interference) \$\endgroup\$
    – user1245
    Commented Jan 21, 2018 at 20:01

Replace the wires to the sensor with resistors of value R and consider the current drawn by the sensor and you will see that the voltage drop along the wire to ground will add a voltage Is*R to the sensor output (the sensor sees a total supply voltage V+ -2*Is*R but we assume the sensor regulates its internal supply voltage.

The suggested circuit subtracts the error signal from the real+error, leaving just the real signal (ideally).

The impedance is unbalanced on the wires so you cannot expect improvement to common mode noise. For that it's better to use loop powered current techniques.

  • \$\begingroup\$ please see my edit \$\endgroup\$
    – user1245
    Commented Jan 21, 2018 at 20:02

Copper foil of the standard thickness, 1 ounce/square foot, has 0.000500 ohms resistance at room temperature, with 0.4%/degree Centigrate temperature coefficient.

Thus 100 squares of foil (1mm by 100mm, or 4" long) has 50 milliOhms resistance. A 1mA current flowing in imbalanced resistances causes 50 microVolts voltage drop, which in 22-bit ADC (+-5v/2^22 = 10v/4Million = 2.5 microvolts per quanta) is 20 quanta error.

In which case you can drop the ADC performance to 18 bits and save money.

You need to design the grounding-system of your signal-chain. VDD return currents, signal currents that overlap between output and input, and external aggressors from 117VAC rectifiers, from low-voltage ripple-reduction capacitors, from switching regulators, etc all contribute to Ground noise.

If you have 10 amp stepper motor in a robot drive-wheel, with a couple squares of 1-ounce foil (1 milliohm), the voltage drop is 10 millivolts with ugly fast edges on the Ground error voltage. With 5 volt ADC, you have 0.01/5 = 9 bit ADC.

Life gets tough, when you punt and home. Make sketches of the Ground current flows.


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