I am studying thermocouples. In a text book, the author said that, the electromotive force can be written as

$$ E= \alpha \theta + \beta \theta^2 \tag{1}$$

where $\alpha$ and $\beta$ are constants and the thermocouple is connected at $0 \;\mathrm{°C}$ on one end and at $100 \;\mathrm{°C}$ at the other end.

My problem is that I didn't find a good logic for the equation (1). Where does it arise from?

  • $\begingroup$ Could you maybe clarify what $\theta$ is referring to. I assume that $E$ is the electromotive force. You refer to thermocouple, so I assume you would be dealing with (the steady state case of) the seebeck effect. But since the temperature difference is given, you could only change which materials are used. $\endgroup$
    – fibonatic
    Commented Jan 2, 2014 at 6:41
  • $\begingroup$ The non-linearity of the voltage-temperature relationship is described with greater detail a the thermocouple Wikipedia page. It appears you text is only using the first two terms. $\endgroup$
    – DWin
    Commented Jan 2, 2014 at 6:48

1 Answer 1


The voltage generated by a thermocouple arises from the Seebeck effect. At the risk of over-simplifying, the electrons in a metal can be thought of as a gas of nearly free electrons, and when you heat a gas its pressure changes. So if you have a metal wire and there is a temperature difference between the ends electrons will tend to move from the hot end to the cold end. This sets up a potential difference, and this potential difference is what is measured in a thermocouple.

If the electrons behaved as an ideal gas the voltage would be simply proportional to the temperature difference, and indeed over small temperature ranges the voltage generated is approximately proportional to the temperature difference giving you:

$$ E \approx \alpha \theta $$

However it shouldn't surprise you to find that the electrons in a metal are only approximately free, and the potential difference is actually some complidated function of the temperature difference:

$$ E = f(\theta) $$

where $f(\theta)$ is the complicated function. Any function can be written as a Taylor series:

$$ f(\theta) = f(0) + \frac{f'(0)}{1!} \theta + \frac{f''(0)}{2!}\theta^2 + \frac{f'''(0)}{3!} \theta^3 + ...$$

This is an infinite series, but it's often the case that $\theta$ is small and so $\theta^2$ is even smaller and can be neglected. That gives us the linear approximation:

$$ f(\theta) \approx f(0) + \frac{f'(0)}{1!} \theta $$

Over a larger range of $\theta$ the value of $\theta^2$ becomes significant but $\theta^3$ is still small enough to be ignored and we get the quadratic approximation:

$$ f(\theta) \approx f(0) + \frac{f'(0)}{1!} \theta + \frac{f''(0)}{2!}\theta^2 $$

The quadratic approximation is being used in your formula for the voltage. The voltage is zero when the temperature is zero, so the first term $f(0)$ is zero and that leaves:

$$ E \approx \alpha \theta + \beta \theta^2 $$

where the constants $\alpha$ and $\beta$ are equal to $f'(0)$ and $f''(0)/2$ respectively. In practice the constants are empirical i.e. we would measure the voltage over a large range of temperatures and fit a quadratic equation to the experimental data.


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