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.

  • 5
    \$\begingroup\$ 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. \$\endgroup\$ Commented Jun 27, 2012 at 12:23
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    \$\begingroup\$ 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. \$\endgroup\$
    – miceuz
    Commented Jun 27, 2012 at 12:28
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    \$\begingroup\$ 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. \$\endgroup\$ Commented Jun 27, 2012 at 12:44
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    \$\begingroup\$ @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). \$\endgroup\$
    – stevenvh
    Commented Jun 27, 2012 at 15:22
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    \$\begingroup\$ 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. \$\endgroup\$
    – compumike
    Commented Jun 28, 2012 at 1:24

3 Answers 3


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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    \$\begingroup\$ It looks like he's trying to measure the DC gain, rather than calculate it, but his method presumes zero DC offset. \$\endgroup\$ Commented Jun 27, 2012 at 12:40
  • \$\begingroup\$ @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. \$\endgroup\$ Commented Jun 27, 2012 at 12:47
  • \$\begingroup\$ 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 \$\endgroup\$
    – MikeJ-UK
    Commented Jun 27, 2012 at 14:02
  • \$\begingroup\$ @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. \$\endgroup\$ Commented Jun 27, 2012 at 16:20

First, you are putting 18V on an op-amp with an absolute maximum supply voltage of 16V, which is ummmm.. not very good practice.

12nA would be an excessive leakage current unless you are doing something really silly, but I can't judge whether you've done something silly like cleaning the board with something that leaves a residue. The excessive supply voltage could be a factor too. Try +/-5V.

I would suggest paralleling the 10M resistor with a low leakage capacitor such as an MLCC NP0 ceramic or film type to prevent noise from affecting the reading. 10nF should be sufficient.

There are two unknowns in your circuit with the switch closed- the op-amp Vos and the gain, assuming your meter and the 1.00V source are both perfectly accurate. You will need more information to separate the two. Set the 1.00V source to zero with the switch closed to get the output offset voltage.

Your circuit is not capable of measuring the expected input current of 10fA (\$10^{-14}A\$). I suggest a much higher gain, and using a much higher value resistor. A gain of 100 and a resistor of 1G will give you 1mV output, which should be measurable to within ~1%. Get rid of the 1.00V source and measure the output with the resistor shorted and not shorted. Keep the time constant of the capacitor/resistor reasonable so you don't have to wait all day for it to settle. (maybe 500pF-1nF for 1G).


In this case, Input Offset voltage should be taken into account as some portion of amplified Input Offset Voltage will also be present in the output in both the cases(when SW1 is connected directly to supply and when connected through resistor). Here is my solution(including the offset voltage).![Offset Voltage is present at the non-inverting input at all times.][1]

  • \$\begingroup\$ Could you rotate the image and zoom it a bit? It's a bit hard to read at the moment. \$\endgroup\$
    – PeterJ
    Commented Mar 5, 2016 at 12:51

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