Help me with instrumentation amplification

[Unicorn_tech]

Below is the proposed instrumentation amplifier to be used.

 

[Unicorn_tech]

But the problem I'm having with the instrumentation amplifier is that I need just 2 wires from the instrumentation amplifier to M and N. But I'm having 3, i.e two for floating voltages (V1 and V2) and the third one as the ground (GND). I need just two (the ground and another outlet for input) that will go to the M and N on the block diagram

 

[rjenkinsgb]

That's a Wenner earth resistance array:
https://en.wikipedia.org/wiki/Electrical_resistance_survey

You are effectively using the area of ground under the electrodes as a double-tapped potentiometer.

Any general instrumentation amp should work fine, just make sure the supply voltages are rather more than the outer electrode voltage - or divide the probe voltages first (with high value resistors) to keep them well within the amp common-mode range.

An INA128 is readily available and a good low-noise device.
https://www.ti.com/lit/ds/symlink/ina129.pdf

I'd add some form of protection diodes on the inputs, eg. diodes from each to the positive and negative supplies, so the input voltage cannot exceed a safe range.

 

[Unicorn_tech]

This is the voltage divider that measures negative and positive voltages. It works well in the lab but when connected to the ground, it is not that sensitive enough

 

[rjenkinsgb]

Re. your last post. (#22)

The system 0V or common "ground" should be the "centre" potential - eg. if the outer electrodes are at + and - 15V, system ground is the central 0V point.

That allows the greatest input range either side of zero.
The ground is also the zero reverence for the PIC ADC if you are using an analog input to capture the reading.

 

[rjenkinsgb]

Re. again (#24)

That will work when connected to the instrumentation amp output, as long as the resistor values are appropriate. R2 and R3 in parallel should equal R1, with the ratio between R2 and R3 setting the input sensitivity.

 

[ronsimpson]

A simple diode-capacitor is not a good way to see AC voltages that are small. It takes 0.6 volts to turn on a diode.

Here is a circuit from the data sheet of the LT1078. It will work with very small AC voltages.
If you build this circuit use a "R-R input & R-R output, low voltage op-amp". Or at least use a op-amp that will work when the input is at ground. See "input common mode voltage range".

Here is my modification. I added a switch for a gain of 1 or 10.

 

[Unicorn_tech]

A simple diode-capacitor is not a good way to see AC voltages that are small. It takes 0.6 volts to turn on a diode.

Here is a circuit from the data sheet of the LT1078. It will work with very small AC voltages.
If you build this circuit use a "R-R input & R-R output, low voltage op-amp". Or at least use a op-amp that will work when the input is at ground. See "input common mode voltage range".

Here is my modification. I added a switch for a gain of 1 or 10. View attachment 114392

View attachment 114393

Thanks so much for this... I appreciate your response

 

[rjenkinsgb]

For the "drive" to the outer electrodes, you can use two half-bridge circuits - or a couple of power opamps.

That allows you to switch each between the positive and negative supplies.
eg. If you use a pair of TLE2301 opamps, you can drive 1A out & control the switching polarity by PIC pins.

If you take each ADC reading a short time after the polarity change, to allow for settling, then change polarity after each sample, you can have "AC" out to avoid electrolysis.

You do not need a rectifier as you can can read both polarities directly using the instrumentation amp plus the ADC and the resistor network you posted previously.

That power opamp also has a "standby" function that allows the output stage to be turned on & off by a logic signal, so you also control that from the PIC to save power between readings.

Data: https://www.ti.com/lit/ds/symlink/tle2301.pdf