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I am trying to determine the boltzmann constant by using a bipolar junction transistor.

In my circuit (apparently I don't have enough point to join a image sorry), the Ebers and Moll model gives the relation $ i_c = I_s\exp\left(\frac{V_{BE}}{V_T}\right)$


$ V_T= \frac{k_b\cdot T}{q}$ and $V_{BE}$ is the difference of potential between the emiter and the base.

I'm plotting the $\log(i_c)$ as a function of $V_{BE}$ and use the slope of the line ($\alpha$) to deduce boltzmann constant. So

$ log(i_c)= \alpha\cdot V_{BE} + b $

from which I deduce

$k_b = \frac{q}{\alpha\cdot T\cdot \ln(10)}$

I have done that for 4 different temperatures. I average the different values of $k_b$ to find $k_b = 1.57\cdot 10^{-23}$

How can I calculate the error margin for each of the $\alpha$ coefficient and how do I calculate the error for the mean coefficient $\bar\alpha$?

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Maybe this could help, or at least get you on the right track – Schlomo May 1 '13 at 17:54

If your data has error bars for the $y$-axis, you could use a least-squares/$\chi^2$ technique to fit your model:

You have $\{x_i,y_i\pm\epsilon_i\}$ for $i=1,\ldots,N$ data. You want to find $K_B$ and $I_c$ such that $y=f(k_B,I_C,x)$. Construct the function, $$ \chi^2 = \sum_{i=1}^N\frac{(f(k_B,I_c,x_i)-y_i)^2}{\epsilon_i^2}, $$ and minimize this function wrt to $k_B$ and $I_s$ to find your best estimates for these parameters, say $k_B^*$ and $I_s^*$. To find the confidence intervals for your estimates (at 68%), fix $I_s=I_s^*$, and vary $k_B$ up and down until $\delta \chi^2 = \chi^2(k_B,I_c^*) - \chi^2(k_B^*,I_c^*) = 1$. The interval for which $\delta \chi^2 \le 1$ is the 68% confidence interval for $k_B$.

This works because Wilkes' theorem shows that $\delta \chi^2$ is asymptotically $\chi^2$-distributed.

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You will want to first convert your current uncertainties $\delta i_c$ into uncertainties in $\log{i_c}$, $\delta(\log{i_c})$. You will need the uncertainties in the $x$ and $y$ values that you use in the fit in order to get the uncertainties in the fit parameters. Note that the uncertainty of the $\log$is not simply the $\log$ of the uncertainties.

Taylor derives the formulas for the uncertainties in the fit parameters in the case that all of the underlying measurements have the same uncertainties. He leaves the case where the uncertainties of the underlying measurements are not the same to the problems for the reader to derive. The derivation is not difficult for either case, but it is tedious for both of them. I don't think it's problematic to simply quote the results for you; they are well-known, and I'm assuming that you're just supposed to look them up. Frankly, I think they should be given to you by your fitting software, but everything I've used makes it difficult to find them.

Using your notation for the slope $\alpha$ and the y-intercept $b$, the uncertainties are: $$ \sigma_\alpha = \sqrt{\frac{\sum{wx^2}}{\Delta}}, \sigma_b = \sqrt{\frac{\sum{w}}{\Delta}} $$ where $x$ is the set of your $V_{BE}$'s, $w = 1/(\sigma_y)^2$, $\sigma_y = \delta(\log{i_c})$, $\Delta = N\sum{x^2} - (\sum{x})^2$, $N$ is the number of points you are using for the fit, and the summations are over all the points you are using for the fit.

When you have the uncertainty for each of your four values of $k_b$, you can get the uncertainty for their mean by adding the four uncertainties in quadrature, like you would for the uncertainty of a sum of measurements: $\sigma_\mathrm{total} = \frac{1}{4}\sqrt{\sigma_1^2 + \sigma_2^2 + \sigma_3^2 + \sigma_4^2}$.

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Very last equation: $\sigma_4 \to \sigma_3$? And more importantly, divide by $4$ to get the uncertainty in the mean rather than in the sum? – Chris White May 2 '13 at 22:38
Thank you for your answer I'll save it for future work. Unfortunately we were running out of time so we just approximated the error on the slope by taking the first and last point of the linear regression corresponding to a measured $V_{be}$ and draw two rectangle centered on these point with width = $\delta_{V_{BE}}$ and height = $\delta_{log(I_c)}$. Then by joining the 2 corners giving we found $\alpha_{min}$ and $\alpha_{max}$ and conclude that the error was about $\frac{\alpha_{min}-\alpha_{min}}{2}$. It's probably not totally right but it should give a idea off the error margin – Sylvain Blanco May 4 '13 at 11:57

For a linear regression fit of the form: $y=ax+b$, with $S_{yi}=S_i\neq const.$, then: $$ a=\frac{\sum{\frac{1}{S_i^2}} \sum{\frac{x_iy_i}{S_i^2}} - \sum{\frac{x_i}{S_i^2}} \sum{\frac{y_i}{S_i^2}}} {\sum{\frac{1}{S_i^2}} \sum{\frac{x_i^2}{S_i^2}} - \left(\sum{\frac{x_i}{S_i^2}}\right)^2} $$ with: $$ S_a=\sqrt{\frac{\sum{\frac{1}{S_i^2}}} {\sum{\frac{1}{S_i^2}}\sum{\frac{x_i^2}{S_i^2}}-\left(\sum{\frac{x_i}{S_i^2}}\right)^2}} $$ And from your expression for $\alpha$ you can determine the $S_\alpha$.

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