The equation for temperature dependence on resistivity:

$$ \rho(dT) = \rho(0)(1+\alpha dT) $$

where \$dT\$ is change in temperature from some reference and \$rho\$ is resistivity. This leads to temperature dependence on resistance:

$$ R(dT) = R(0)(1+\alpha dT) $$

This seems terrible to me. I would much prefer something like:

$$ \rho(dT) = \rho(0) + \alpha_1 dT\\ R(dT) = R(0) + \alpha_2 dT $$

This way seems much simpler to me. It looks like a constant offset plus a slope. This way, the slope doesn't depend on the initial value, which is mathematically more pleasing. Why isn't it this simpler, more natural way?

  • \$\begingroup\$ \$dT\$ or \$\Delta T\$ ?n The latter one is \Delta T \$\endgroup\$ – jippie Mar 2 '14 at 21:01

First, it's to emphasize that (in most cases) \$\alpha\$ << 1.

Secondly, \$\alpha\$ is a function of the material and temperature, it is not a function of the construction of the resistor. Any resistor of that material, at that temperature, would (ideally) have the same \$\alpha\$.

Your \$\alpha_2\$ is dependent on \$R_0\$ so it would be different for every resistor. That's very inconvenient.


While your [my] way is more mathematically appealing, my guess is that it simplifies common calculations that you need to do with resistivity and resistance.

For example, since \$R(dT) = \frac{\rho(dT) L}{A}\$, then the resistivity equation gets transformed into \$R(dT) = R(0) + \alpha_1 \frac{L}{A}\ ,\ dT = R(1) + \alpha_2 dT\$

Notice that you now need two different \$\alpha\$ coefficients! With the established way, they are the same.

It also simplifies the units: \$^\circ\text{C}^{-1}\$ instead of \$^\circ\text{C}^{-1} \Omega \text{m}\$ for resistivity and also \$^\circ\text{C}^{-1} \Omega\$ for resistance.

One might argue that having two coefficients and two different units makes more sense, since you're computing two different quantities. Maybe someone with more experience actually knows why the choice was made.


Spehro's answer is correct. Simple and intuitive, with a good use of Occam' s razor for rationale of the equation form.

Since there are good mathematical reasons for its form it seems worth while to show its origin though. First, the thermal coefficient of resistivity is defined by the differential equation

\$\frac{\text{d$\rho $}}{\text{dT}}\$ = \$\alpha \rho (T)\$

When solved with a boundary condition of \$\rho (T_o)\$ = \$\rho _o\$ an exponential form for resitivity as a function of temperature is obtained:

\$\rho(T) \$ = \$\rho _o e^{\alpha \left(T-T_o\right)}\$

This exponential form contains an approximation, namely that the thermal coeffcient \$\alpha\$ is a constant. For the simpler structured materials (mostly metals), and with a restricted temperature range (like between ~250K and 350K) this will be nearly true. \$\alpha\$ is not really a constant for more extreme temperatures, especially low (cryogenic) temperatures.

A second approximation is made by taking a power series expansion of the exponential form, only keeping the first two terms to get a first order (linear) form. No secret, this is the usual way to obtain a first order model of anything.

\$ \rho(T) \$ = \$\rho _o\left[1+\alpha \left(T-T_o\right)\right]\$

So, the linear form expressing resistivity as a function of temperature is really an approximation of an approximation. Usually the linear form is good enough for the region of interest in electronics. For example for Cu \$\alpha\$ is 0.004, and if \$\text{$\Delta $T}\$ is 50K, the difference of resistivity between the exponential and linear equations is less than 2%. Of course, the smaller the exponent, the more accurate the linear form will be.


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