I am controlling a device using a programmable current $I_1$ that behaves as $A = 1+\frac{I_1}{I_2}$, where $I_2$ is a non-adjustable (sort-of unknown) current that's a characteristic of my setup.

I have used a calibration device that is known to give measurements that I know is is "correct" (i.e. in agreement with the physical definition of the unit of current) to within 0.1% over its entire range.

Using this device, I have measured $I_2 = (200 \pm 0.1)\mu A $ and for all values of $I_1$ that I program, the current source is correct to within 0.05% (or 5 $\mu A$ maximum).

I wish to calculate the uncertainty on the value $A(I_1)$, $\delta A$. Naively I could calculate $\sqrt{\left(\frac{\delta I_1}{I_1}\right)^2 + \left(\frac{\delta I_2}{I_2}\right)^2}$, where $\frac{\delta I_1}{I_1} = \frac{\delta I_2}{I_2} = 0.1\%$. The 0.1% I use from the calibration device, because this is the precision to which I know the "true" value.

It seems to me that this will be too conservative, as the errors are not independent, but will tend to cancel out.

Does anyone have a suggestion for how better to approach calculating $\delta A$?

  • 1
    $\begingroup$ Your expression seems to suggest that $A=1 + I_2$ - is that actually the expression you intend to use? It doesn't seem to be quite dimensionally correct $\endgroup$
    – Floris
    Commented Sep 14, 2016 at 13:11
  • 1
    $\begingroup$ Assuming you mean 1+$I_1$, I think that this is just how SE has typeset my post. I have changed it a bit to make it easier. $\endgroup$
    – Gremlin
    Commented Sep 14, 2016 at 13:28
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    $\begingroup$ There's also something unclear about the errors. You say $\delta I_1 = \delta I_2 = 0.1%$, but for one thing $\delta I_1$ and $\delta I_2$ should be absolute uncertainties (with units of current), not relative ones, and also they are in conflict with the $(200\pm 0.1)\mu A$ (or 0.05%) uncertainty you give elsewhere in the question. $\endgroup$
    – David Z
    Commented Sep 14, 2016 at 13:35
  • $\begingroup$ @DavidZ, sorry it should have been the relative errors. I have used the 0.1% uncertainty of the calibration device because this is the uncertainty in the "true value" of these variables. $\endgroup$
    – Gremlin
    Commented Sep 14, 2016 at 13:38
  • $\begingroup$ If the errors are not correlated, the expression you use (independent sum of squares) should be correct; if they are correlated, you would have to sum the errors themselves (rather than take the square root of sum of squares). Why do you think your estimate is too conservative? Why do they cancel out? Is it because they are using the same calibration standard, and that is the one that has an error? $\endgroup$
    – Floris
    Commented Sep 14, 2016 at 13:42

1 Answer 1


If you say the error is in the scaling of the calibration, so that


And $$\hat{I_2}=I_2(1+\epsilon)$$

Then it should be obvious that that aspect of the calibration factor cancels out exactly in the expression $\frac{I_2}{I_1}$

In reality your uncertainty is a combination of gain, offset and noise. Take out the gain calibration error and you are just left with the other uncorrelated errors. Which you then combine in the usual way.


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