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Maybe this question is obvious, but I'll try.

I am trying replicate this potentiostat without R5:

enter image description here

In the interests of calibrating it I've joined the reference and the counter electrode to resistor of 10k ohm in the other extreme of the resistor I've connected the working electrode.

Later I set Ard D9 such as the potential drop was -0.58V measured with the multimeter (with positive pin in CE-RE and negative pin in WE) so the question is:

The current in the resistor as Ohm's law states is the output of the voltage follower V divided by R_6, which must be equal, in magnitude, to 0.58/10000 = 58uA.

How can I measure the voltage in WE with the multimeter? It must stay constant.

I know a little bit more handling Arduino but instrumentation it is too hard for me.

JRE
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Elí Flores
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  • R5 provides a constant current into the summing node of the transimpedance amplifier (TIA), that is presumably balanced by a current from WE in the electrochemical cell. It doesn't appear you can ditch R5, without a significant redesign. What's the problem that you think removing R5 is the answer to? The voltage at WE will be nominally zero volts when the TIA is operating correctly, as it servoes it to ground. – Neil_UK Sep 26 '22 at 08:47
  • Yes the current from R5 act so that output of the voltage follower (VF) remains in the between 0-5V. My wish to momentarily eliminate R5 was read the output of the VF and check Ohm's law in my interests to calibrate, thanks for your comment it is the great aid. :D – Elí Flores Sep 26 '22 at 13:04

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