Why is temperature coefficient divided by reference resistance?

Question

In a book I am following, it is assumed that the temperature resistance relationship is linear.

I would expect, then, that the resistance at a given temperature would be given by the formula

$$ R_{t} = mt + R_{0} $$

Where m is increase in resistance per unit temperature. Therefore by taking another known resistance at temperature A we could calculate this m

$$ m = \frac{R_{A} - R_{0}}{A - 0} $$

However the book (and everywhere else) defines the temperature coefficient as the above gradient divided by the resistance at temperature 0.

$$ \alpha = \frac{R_{A} - R_{0}}{R_{0}(A - 0)} $$

and then the formula for resistance at temperature t as

$$ R_{t} = \alpha R_{0}t + R_{0} \equiv R_{0}(\alpha t + 1) $$

My question is: what is the purpose of dividing by resistance at temperature 0 just to multiply by it again later, and what does this alpha coefficient actually 'mean' (in the same way I understand gradient m 'means' resistance increase by increasing temperature by 1 unit)?

Answer 1

If we used your scheme, \$ R_{t} = mt + R_{0} \$ then we would need a unique m for every resistance value. This would be a pain.

A more useful parameter is the temperature coefficient expressed as a percentage or factor. So, given a temperature coefficient of resistance of 0.2%/K and, say, a ΔT of 12°C then I know that a 1k resistor will increase by \$ 1k \cdot 0.002 \cdot 12 \$ and 47 Ω resistor will increase by the same ratio or \$ 47 \cdot 0.002 \cdot 12 \$.

The big advantage is that the one parameter can be applied to all the resistors in a catalogue series.

I would write the formula as

$$ \alpha = \frac {R_{T1} - R_{T0}}{R_{T0} (T_1 - T_0)} $$

where \$ R_{T0} \$ is the reference resistance at temperature \$ T_0 \$, \$ R_{T1} \$ is the resistance at temperature \$ T_1 \$ and \$ \alpha \$ is the temperature coefficient.