So I'm trying to measure the signal from a potentiometric sensor, that has a working and reference electrode. Electrically it can be summarized with the picture below. Note that my circuit is intended to be mounted on a PCB and powered by a battery.

enter image description here

Now my problem is that the sensor is left floating. I initially thought that's not a problem with a differential input, but I realized that if it "floats away" it will saturate my internal nodes and kill operation. I know that its not going to drift millions of volts away due to leakage currents, but I am still concerned that it will move too close to one of the rails and distort. I not sure how to estimate how far it will float away other than by physically building it and seeing what happens, hence why I'm looking for a solution beforehand.

I was thinking of the following solutions:

  1. I can't use a high pass filter because I want to read what is essentially a constant voltage.
  2. I thought of grounding the reference electrode to my PCB ground, but I'm fairly sure that this defeats the purpose of a differential input. Am I right to assume that the common mode noise would then be transmitted to my signal via ground?
  3. Do the same thing as above, but via a pull-up/-down resistor. Does adding a resistor change anything? I don't understand why.
  4. Add a 3rd electrode and connect it to my circuit via a buffer.

I'm getting really confused with this problem and would appreciate any suggestions or pointing me to potential solutions.

Thank you.

  • \$\begingroup\$ Signal voltage needs a source and load Z defined and Stray voltage , current needs to be defined to estimate stray capacitance or induction noise attenuation requirements . So one may consider LPF or active guarding or receiver ground shield pair , where balanced shielded pair is grounded only at one end. Also What is BW above DC? \$\endgroup\$ Jun 2, 2018 at 12:35
  • \$\begingroup\$ What is the nature of the common mode voltage Vcm? Can you tell us what the principle of the sensor is? Can you post a link to something similar that has already been published? Without knowing these things, we are unlikely to give you the right solutions. \$\endgroup\$ Jun 9, 2018 at 0:27
  • \$\begingroup\$ Maybe put your sensor in a bridge circuit? Or maybe instrumentation amplifier isn't the right front-end? Why did you choose it? \$\endgroup\$
    – HKOB
    Jun 12, 2018 at 20:02

1 Answer 1


The normal method is to tie one or both inputs to the mid rail (0 volts) of the instrumentation amp via typically 100 kohm resistor. If your sensor has reasonably low output impedance this won’t create an error due to 100 kohm mismatch in value. To supplement this (should common mode input noise be a problem) then a capacitor in parallel with each resistor is typically used. Maybe 1 nF to 100 nF won’t kill your bandwidth too much?

  • \$\begingroup\$ When you say "reasonably low" you mean that the impedance is hopefully lower or higher than 100kohm? The thing we are working with has output impedance around 1 megaohm, so I'm wondering if how that affects it? Also, what is the difference between tying one vs both? \$\endgroup\$ Jun 2, 2018 at 14:46
  • \$\begingroup\$ 1 Mohm is high in my book. You should use a single resistor from the reference electrode to 0 volt on your design. Leave a link for the sensor and state what your InAmp power rails are. Also explain where the common mode noise comes from. \$\endgroup\$
    – Andy aka
    Jun 2, 2018 at 15:32
  • \$\begingroup\$ I can't leave a link to the sensor, because this is a uni project and the sensor itself is underdevelopment. I wasn't aware of the significance of common voltage until we run a simple test in the lab and I realised that this could be a problem, hence started wondering about possible solutions. I guess the problem with the resistor connection is the limited amount of charge available, in our case it comes from a chemical reaction catalysed by enzyme coating on the electrode. \$\endgroup\$ Jun 2, 2018 at 16:36
  • \$\begingroup\$ A single resistor connected to any node of a floating (galvanically isolated) source will not affect the signal unless the signal is of significant frequency and has an unbalanced impedance to ground. \$\endgroup\$
    – Andy aka
    Jun 2, 2018 at 16:41

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