Uncertainty estimation of a quantity after its calibration

Question

I have to calibrate power output of a diode to its input voltage. The voltage source has negligible error but the power meter has significant uncertainty (a couple %). I take a number of power readings at successive voltage inputs, plot a power vs voltage graph, and do a curve fit using least squares. I have the variances in the fit parameters. Now I put away the power meter and want to use the curve fit for calculating the power output as a function of input voltage. How do I estimate the uncertainty in the power determined this way?

Answer 1

I am a little unclear from your comment, but if you have not done so already, you should linearize your data. In other words, if you have a theoretical equation from the literature that says $P = \alpha V^2 + \beta$, where $\alpha$ and $\beta$ are constants, then you should plot $P$ on your vertical axis, and $V^2$ on your horizontal axis so that the slope of the line is $\alpha$, and the y-intercept is $\beta$. If your theoretical equation is more complicated than that, you may have to work a little harder -- regardless, the first step is to put your equation in $y=mx +b $ form.

Perform your weighted linear least-squares fit (so that the fit takes into account the uncertainties in your measurements), and obtain an $R^2$ value for the fit.

The uncertainty in the slope can than be found using $$ \sigma_m = m\sqrt{\frac{(1/R^2)-1}{n-2}}, $$ where $m$ and $\sigma_m$ are the slope and uncertainty in the slope respectively, and $n$ is the number of data points used to make the fit. Careful, the $R^2$ is already squared! Don't square it again!

The uncertainty in y-intercept can be found using the above result: $$ \sigma_b = \sigma_m \sqrt{\frac{\sum x^2}{n}} $$

So, now you have an equation such as $P = \alpha V^2 + \beta$, and you have uncertainties in the constants; namely, $\sigma_\alpha$, and $\sigma_\beta$. Use standard propagation of uncertainty in quadrature to propagate uncertainty from $\alpha V^2 + \beta$ to $P$ for each value of $V$.