# Calculating Boltzmann constant using semiconductor

My task from the latest laboratory exercise in physics is to calculate Boltzmann constant from known temperature, current and voltage. We were given this circuit: By changing the resistor value, I measured different voltages and currents. It's obvious that the current is exponential of the voltage - see this result plot: Calculated by pasting my results into this exponential regression form. We obtained a very rigorous instructions paper - it's really a nice reading with lots of additional interesting information about semiconductors - but I have some problems gathering the needed equation out of it. In my opinion, the important part is this one:

To meassure Boltzmann constant we'll use theoretical equation for volt-amper characteristics of the PN transition. Because in our experiment $I \gg I_0$ we can reduce the equation like this:

$$I = I_0 exp(\frac{eU}{kT})$$

The measures twins of values $U_i$, $I_i$ where $i = 0, ..., N$ will be used to create exponential function using the least squares fittings method:

$$I = Ae^{\alpha U}$$ The text above has been translated from Czech and may be incomplete or imprecise due to my lack of English terminology.

The exponential regression calculator gave me the following:

$$I = 2.064880249*10^{-8} e^{36.57803404U}$$ $$A = 2.064880249*10^{-8} \\ \alpha = 36.57803404$$

Now what? What are these values? Of course, the paper I have obtained is full of equations, but finding something in them is like finding needle in the haystack.

Besides, even though reading the paper instructions was enlightening and I learned a lot about semiconductors, I'm still not so sure why did I get the results I got. Since this was the point of the whole homework, I think I need help.

Please, focus on explanation rather than actual calculations.

## 1 Answer

The inherent idea is, from that equation $$I = \exp (\frac{eU}{k_B T})$$ if you plot $(\ln I)$ versus $U$, that would be a straight line (of the form $y=mx$), with a slope $$\alpha = \frac{e}{k_B T}$$ That's all you have to do in the experiment, use least square fitting to find an accurate value of $\alpha$ and then find the Boltzmann constant using the relation $$k_B = \frac{e}{\alpha \ T}$$ You know $e$, and all you need is to measure $T$. Not being too fussy, you could just use the value obtained from a lab thermometer.

You have done the job well, $\alpha = 36.578$, and the relation gives $k_B = 1.4581 \times 10^{-23} {\rm JK^{-1}}$ at a temperature of $T=300K$, which is roughly the same order as (though a tad higher than) the documented value of the Boltzmann constant $1.38 \times 10^{-23} {\rm JK^{-1}}$. Just find out the temperature and go ahead.

• Thank you a lot. Could you additionally give me some leads to the explanation what do these things mean? I mean, how is the conductivity exactly affected by temperature and what does boltzmann constant do with it. – Tomáš Zato - Reinstate Monica Oct 18 '14 at 15:06
• $k_B$ enters here because $k_B T$ is the typical order of thermal energy, with this Boltzmann constant serving as nothing more than a proportionality constant. At larger temperatures, there are more charge carriers liberated out of a semiconductor material, which explains the larger current. That equation you have, describes how exactly it grows, so it is exploited in a fairly simple experimental set up to estimate the value of $k_B$. If you want a more details of how exactly this originates (apart from the essence that I've tried to convey here), I point you to Millman, Halkias :) – 299792458 Oct 18 '14 at 15:15