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I am using micro electrodes in a conductive solution (PBS: Phosphate Buffered Saline) and I would like to be able to measure the voltage with a National Instrument measuring device (PXIe6345).

I apply a voltage in the bath (a sinusoid 10 Hz and 50 mV amplitude). I have a very high impedance due to the conductive solution and the small size of the electrode (about 10-20M ohm). My measurement system switches very quickly between several inputs, so I need a low impedance so that the internal capacitor has time to reach the right voltage. To do this, I want to make a voltage follower with an LM324N and a +/-5 V voltage source (RT-50A Mean Well USA).
The circuit is very simple, but I have a result that I don't understand, and I hope that people here can help me.

I have a voltage offset because my generated signal is centered on zero volts, plus I have a drift effect towards the positive which seems to be abruptly compensated after a certain value, see the picture.

voltage measurement

opamp circuit

Edit 1:

Following the advice I received here, I bought an LMC660, but I'm still having the same problems.

I did some measurements with an oscilloscope to try to find the problem. This time in my setup I added resistors to put a gain of 5 to the output signal (16K ohm and 4K ohm resistance). I also added a 100 ohm resistor on the output. Here are some photos I took of the circuit and the oscilloscope (sorry for the reflections from the sun).

breadboard circuit

input and output

The signal seems to be correct apart from the offset still present on the op amp output signal.

in+

I think the IN+ signal looks like this because of my oscilloscope's ability to read high impedance values rather than this being the true IN+ signal because the op amp output is correct.

in-

(If it's not a question of current bias as suggested, could it be due to breadboard ?)

Edit 2:

I've just tried to see if there could be a galvanic battery effect, but I haven't measured any voltage between two electrodes once they've been disconnected from everything.

Edit 3:

Finally, there seems to be 10mV between certain electrodes on the electrode matrix that I use when the system is not connected to anything. However, even with a gain of 5 this alone cannot explain the almost 1V offset in the previous photos.

Edit 4:

I've also just realised something: when I want to measure the voltage between the reference electrode and the electrode I'm measuring with a voltmeter in parallel, the offset gradually decreases until it almost cancels out - see the images below.

generator

circuit

signal centered in zero

I imagine this is due to the high impedance of my signal after passing through the bath, which means that the current passing through the voltmeter is able to influence it.

Thanks again to those who will take the time to reply.

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  • \$\begingroup\$ You need several things. If you are putting a voltage on the conductive solution, we need to know what it is. Also the graph has no axis labels, please put them on. \$\endgroup\$
    – Voltage Spike
    Commented Jul 23 at 18:09
  • \$\begingroup\$ What is PBS? I assume not lead (ii) sulfide. \$\endgroup\$
    – Hearth
    Commented Jul 23 at 18:52
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    \$\begingroup\$ @Hearth PBS stands for phosphate-buffered saline. \$\endgroup\$
    – Jad Rabhi
    Commented Jul 23 at 19:00
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    \$\begingroup\$ Is there some galvanic battery effect between the liquid and the electrode material? \$\endgroup\$
    – Jens
    Commented Jul 25 at 20:22
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    \$\begingroup\$ You should check the galvanic DC potential between these electrodes. Gold has a relative high potential. \$\endgroup\$
    – Jens
    Commented Jul 25 at 20:52

2 Answers 2

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LM324 is a poor choice for such a high source impedance due to it's heavy bias current (typically 20nA, but could be as high as 500nA). The bias current is also temperature sensitive. A 20nA bias current will cause an offset of +300mV with a 15MΩ source impedance, so it's well within your signal range. The bias current is a (relatively constant) current that comes out of the input pins on an LM324.

It would be better to use a CMOS-input op-amp such as LMC662 which can operate from your relatively high voltage +/-5V supply, is available in the antediluvian through-hole DIP-8 package for ease of construction, and has a maximum bias current in the pA range (typical is in the fA range).

There may be something else going on due to the multiplexing of the NI board. I would suggest a small series resistor (perhaps 100Ω) on the op-amp output (between output and the NI input), and making sure the signal and power grounds are properly connected. It would be easy to pick up mains hum with such a high impedance, so good construction is important.

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  • \$\begingroup\$ Hello and thank you for the detailed answer, I am quite interested in the calculation to know that the result will be +300mV with 15MΩ \$\endgroup\$
    – Jad Rabhi
    Commented Jul 23 at 18:32
  • \$\begingroup\$ 20nA * 15MΩ = 300mV. The current is out of the inputs so the non-inverting input will be positive wrt ground. There's no series impedance on the inverting input (connected directly to the output) so the bias current has no effect on the inverting input and therefore the bias current causes a +300mV offset in the output. \$\endgroup\$ Commented Jul 23 at 18:37
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    \$\begingroup\$ Thanks you ! I will try with the LMC66X \$\endgroup\$
    – Jad Rabhi
    Commented Jul 23 at 19:49
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    \$\begingroup\$ Hello, I've just tried again with LMC660 and a 100ohm resistor before the PXIe6345, and I'm having the same problems. When I bypass the conductive solution the op amp works correctly, so I still think the impedance of my signal is the problem. I'm going to take measurements with an oscilloscope of my input, my output and the voltages at the IN- and IN+ terminals to try and debug and post the graphs here if I still have no other leads. (also maybe the breadboard or the conductor load problem) \$\endgroup\$
    – Jad Rabhi
    Commented Jul 25 at 13:38
  • \$\begingroup\$ I've just edited my post with the new photos \$\endgroup\$
    – Jad Rabhi
    Commented Jul 25 at 18:32
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Microelectrodes in liquid have an extremely high capacitance. This is because in situ there are two conductors separated by a thin non-conductive layer.

The way they are commonly dealt with is with a negative capacitance amplifier to null out the electrode capacitance.

Further, there is a junction potential every time a surface of one type meets a surface of another type. In glass microelectrodes in physiological systems, this is minimized by using saturated KCl solutions in the electrode, between the electrode tip and the silver-silver chloride wire in the electrode.

The issues can get complicated. I recommend finding a good discussion in a medical physiology book like Webster's Medical Instrumentation.

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  • \$\begingroup\$ Thank you for your reply. I'm going to look into this as well, but is this video about what you're talking ? youtube.com/watch?v=33c6Str_JQg \$\endgroup\$
    – Jad Rabhi
    Commented Jul 26 at 15:19
  • \$\begingroup\$ And with my multimeter i measure 50nF between my ref electrode and another one. \$\endgroup\$
    – Jad Rabhi
    Commented Jul 26 at 15:26
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    \$\begingroup\$ @JadRabhi Yes, but I believe there's a better treatment in wikipedia. \$\endgroup\$ Commented Jul 26 at 15:44
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    \$\begingroup\$ Try personal.utdallas.edu/~tres/microelectrode/… \$\endgroup\$ Commented Jul 26 at 15:47

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