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I'm trying to understand the application of the water electrical conductivity measurement circuits based on transimpedance amplifier.

Here the picture of general TIA configuration here the picture of general TIA configuration

And here what I'm talking about And here what I'm talking about

In the second picture, the op-amp is feeded with dual supply. Two metal and equal pins are dipped in the water and connected to the battery as shown.

Will the current be flow in correct way like figure 1 to accomplish TIA?

p.s. Just to be simple I used a battery as current source, this question is not about the perfect water electrical conductivity activation signal.

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  • \$\begingroup\$ In the second circuit you must connect the bottom wire to the ground too. Furthermore it would be better not using a DC voltage to measure water conductivity. The electrodes will likely corrode over time (at least much faster than using AC). \$\endgroup\$ – next-hack Sep 24 '17 at 14:08
  • \$\begingroup\$ You mean the ground of the op-amp's supply? (V- for opamp) \$\endgroup\$ – user30878 Sep 24 '17 at 14:11
  • \$\begingroup\$ Connect the "-" of the battery and the non inverting input of the operational to ground. Those circuit assume a dual power supply. \$\endgroup\$ – next-hack Sep 24 '17 at 14:12
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    \$\begingroup\$ Metal electrodes in condducting water form an EDLC and yields a non trivial effective circuit. Using an AC source is a better solution if you want reasonably accurate results \$\endgroup\$ – Peter Smith Sep 24 '17 at 14:56
  • \$\begingroup\$ I know it and wrote in p.s. \$\endgroup\$ – user30878 Sep 24 '17 at 14:58
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Assuming that:

  • You connect the ground as shown below.
  • You use a dual supply.

enter image description here

Then, thanks to the the negative feedback, the inverting input will be at a virtual ground. Therefore the voltage applied to the "water" will be the same value as the battery.

What you will measure is:

$$V_{OUT}=-R_1 \cdot I$$

Where \$I\$ is the current flowing out of the battery, as indicated.

As suggested in the comments, do not use a DC voltage, use AC instead. You can either:

  • put an AC voltage in place of the battery
  • OR you can remove the battery (replace it with a direct connection to ground) and put an AC signal to the non inverting input (instead of ground). The negative feedback will have the inverting input to assume the same (AC) voltage as the non inverting one.
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  • \$\begingroup\$ Can you explain the second choice more \$\endgroup\$ – user30878 Sep 24 '17 at 18:46
  • \$\begingroup\$ As I said, the negative feedback will make the inverting input to have the same voltage of the non inverting one. This does not only apply to DC voltages, but also to AC/varying voltages (provided that the bandwidth of the OA is large enough, or that the AC signal frequency is not too high). Therefore if you apply an AC source to the non inverting input, and if the "water" is connected between ground and the inverting input, then, the "water" voltage will be the same as the AC voltage you're feeding to the non inverting input. \$\endgroup\$ – next-hack Sep 24 '17 at 18:53

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