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I want to measure the current that goes into a noninverting opamp input. I've setup my experiment as follows:

circuitlab schematic https://www.circuitlab.com/circuit/za4q74/input-bias-current-test/

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.

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4  
Folks, we are seeing a lot of schematics drawn with circuit lab lately. That's fine since they are generally readable, but let's loose that annoying and distracting banner at the bottom. Also, put the real values in the schematic, not a note later saying they are different. Surely this program lets you type in whatever values you want. There is no excuse for not putting the right values directly in the schematic. – Olin Lathrop Jun 27 '12 at 12:23
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I'm not sure about ethics of that - they provide us with reasonable tool to draw our schematics quickly with a web friendly output, so we can just paste URL and save all the trouble of exporting and cropping it from whatever other source. So yeah, it's annoying for me too, but i want to give them the credit for doing this tool. – miceuz Jun 27 '12 at 12:28
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There are various free schematic tools out there. We don't want every one cluttering up the result with a banner. Nobody else cares how you made the schematic, only that it be clear and readable, which the banner detracts from somewhat. The free version of Eagle, for example, doesn't impose a banner on its output. – Olin Lathrop Jun 27 '12 at 12:44
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@miceuz - Your schematic links to the CircuitLab site, so we know that it's a CircuitLab schematic. For me the banner isn't necessary either, and a schematic can often be cropped on all sides too (too much white). – stevenvh Jun 27 '12 at 15:22
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FYI, CircuitLab generates the schematic images with banners for use on its own site (and allows other sites to hotlink those), but also has Export PNG/PDF/EPS/SVG buttons that don't include any banners. – compumike Jun 28 '12 at 1:24
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1 Answer

up vote 3 down vote accepted

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.

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It looks like he's trying to measure the DC gain, rather than calculate it, but his method presumes zero DC offset. – Alfred Centauri Jun 27 '12 at 12:40
@Alfred: My understanding is that he is trying to measure the opamp input current and is using a known circuit gain to do that. His circuit setup and general concept are good, he just messed up a bit in the calculations. – Olin Lathrop Jun 27 '12 at 12:47
I have to agree with Alfred. If the OP was miscalculating the gain he would get 300k/100k = 3. Instead, he measures 4.04/1.19 = 3.39 – MikeJ-UK Jun 27 '12 at 14:02
@MikeJ: Like I said, the OP's calculations are flawed. I think his overall intent is to measure the opamp input current. To get that he needs to know gain, not calculate it, at least if there are only two data points. With more data points it is possible to derive both gain and input current, but it's a lot easier to set up the circuit to know gain and find input currents from two measurements, as I did in my answer. – Olin Lathrop Jun 27 '12 at 16:20

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