I want to measure the current that goes into a noninverting opamp input. I've setup my experiment as follows:
I can't measure the voltage drop across the 10M (R1) resistor directly since my DMM has an input impedance in the megaohm range. So, i have to measure the voltage on the ouput of the opamp. R2 is a potentiometer, so the gain is not exactly 4, and V1 is 1.19V.
When I connect the input to voltage source of 1.19V, I get 4.04V on the opamp's output. When I connect the input via the 10M resistor (R1), I get 3.64V on the output.
Now, my gain is:
$$
4.04/1.19 = 3.39
$$
so, voltage that the opamp sees when connected via the 10M resistor is
$$
3.64/3.39 = 1.07V
$$
this means, I have a voltage drop across the resistor R1:
$$
1.19 – 1.07 = 0.12V
$$
so, the current into opamp's input is
$$
0.12V/10M = 12nA
$$
Am i doing it right?
Another question is: Could 12nA leakage current go thru a home made PCB that was vigorously cleaned from flux residue?
LMC6001 has a stated input current in femtoampere range, it might be that it was overheated while soldering and went kaput. But first, I want to be sure that I'm doing the measurements the right way.
Best Answer
No, you are not doing this right. Your concept is good but your calculations are flawed. The gain of the opamp circuit is not R2/R3, but (R2+R3)/R3. Your gain is therefore (400 kΩ)/(100 kΩ) = 4. I'm using the values in your schematic because I shouldn't have to go looking elsewhere. If you don't like that, put the real values right on your schematic next time.
You see a change of 4.04V - 3.64V = 400 mV on the output by switching in the 10 MΩ resistor. Divided by the gain of 4, this means a 100 mV change at the opamp positive input. By Ohm's law, (100 mV)/(10 MΩ) = 10 nA current thru the resistor, which is the opamp input current in this case.