# Meaning of temperature coefficient for resistors

Is temperature coefficient of a resistor about the temperature of the resistor or ambient temperature or are they the same thing? What I mean to ask is that: Lets say I have a resistor and I have no data sheet about it and I want to find the temperature coefficient. Can I use the ambient temperature in calculations? Is the resistor temperature same as the ambient temperature after a while or is that always much higher?

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The temperature coefficient specification ostensibly provides a limit for the change in the resistance of a resistor from its nominal value at temperature $T_0$ to another temperature T. In fact it's usually defined using a box method at temperature extremes and does not really guarantee the slope of the temperature-resistance curve as you might assume.

Ideally, a maximum temperature coefficient of, say, 10ppm/°C would mean that if our 1.00K resistor measures 1.0015K at 25°C and the temperature changes to 35°C then the value should somewhere between:

$1.0015K + 1.0015K (35°C - 25°C) 10^6 \cdot 10ppm/°C$

and

$1.0015K - 1.0015K (35°C - 25°C) 10^6 \cdot 10ppm/°C$

Or 1001.5$\Omega$ +/- 0.1005$\Omega$

It doesn't matter why the temperature changes- ambient, self heating, nearby components, or some combination.

If you are trying to measure the actual temperature coefficient of a resistor, you can measure the resistance at two widely separated temperatures, using low enough current that self-heating is minimal (note that it cancels out to a first order if you allow it to settle out- also pulsed current can be used and the measurement made before the temperature changes much) and calculate the tempco as:

Temperature coefficient = $\frac{R_X - R_0}{R_0(T_X-T_0)}\cdot 10^6 ppm/°C$

If the tempco is large you might want to use the average resistance rather than $R_0$, but it shouldn't matter much in most cases.

Edit: Regarding the situation you mention - 0.2% change for a change in power dissipation of about 100mW.. you need a better resistor and probably a larger one that won't heat as much for a given dissipation.

Consider a 1206 249 ohm Susumu resistor. P/N: RG3216P-2490-B-T1. Tempco is +/-25ppm/°C, it will increase by 15-30°C at 20mA depending on layout (see the link). That should represent a change in resistance of about 375-750ppm or maybe 3-5x better than you are measuring. If you need even higher accuracy, you could use a bigger resistor, several smaller ones distributed with copper around, or use a smaller value resistor and amplify the signal so it doesn't get as hot.

You could also use a Z-foil style resistor such as Y1630250R000T9R that has only 0.2ppm/K tempco, but they are pretty expensive (>\$10 each).

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the reason i ask the question: i apply constant 4ma current to a resistor and i measure a voltage across. so the resistor R = V/I. but when i apply 20ma to the same resistor and i obtain the voltage then the resistor R= V2/I2 i expect the same. but in first case for 4ma i obtain 248.2 ohm and in 20ma case i obtain 248.7 ohm. if i have a device with a current loop output how can i then translate the voltages to currents? since i find different resistances. – user16307 Jul 29 '14 at 18:04
@user16307 (see edit) – Spehro Pefhany Jul 29 '14 at 20:10

It's not about the temperature of the resistor, its about the change in the resistance as the temperature changes,

so it's the ratio of the fractional change in resistance to the change in temperature at a certain temperature. In order to measure it, you need to measure the resistance at some particular temperature and then increase the temperature(may be by passing a current through it) and measure the resistance again, then put values in the formula and you'll get your coefficient of resistance.

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ok I can measure R1 and R2 by knowing current and voltages. but look at your formula. to know alpha one needs deltaT. how can I measure deltaT, in other words T1 and T2?. obviously they are not ambient temperatures. – user16307 Jul 29 '14 at 17:31
yeah, of course it's not ambient temperature, you need a thermometer for that, [Here's an experiment for measuring it](personal.tcu.edu/ystrzhemechn/Classes/2009/Summer/Labs/lab17.pdf) – Salman Azmat Jul 29 '14 at 17:35
your link doesn't tell how to measure the temperatures. if the T is not ambient temperature how to measure the resistor temperature? – user16307 Jul 29 '14 at 17:48
It does, you immerse resistors in the water along with a thermometer, heat up the water. In equilibrium, the temperature of the water will become equal to the resistors and you can measure it with the thermometer. In this case, temperature of the resistors is the ambient temperature(temp. of the water) because of the high specific heat of the water and the presence of a heating source so this keeps the temp of water to a constant level. – Salman Azmat Jul 29 '14 at 17:55

The temperature coefficient is GENERALLY a number reflecting how much you can expect the resistance to change when the temperature of the resistor changes. The resistor temperature can and will differ from ambient temperature, especially if the resistor is dissipating power. You might need fairly sensitive gear and wide temperature differences to measure the temperature coefficient

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but how can I measure the resistor temperature when a current flow through it? – user16307 Jul 29 '14 at 17:27
Generally, you don't. Why do you need this number? More specifically, what problem are you having that you need this info. Would it be enough just to know the resistance at any given time, regardless of temperature?? – Scott Seidman Jul 29 '14 at 18:21
i opened a new question here about why im asking: electronics.stackexchange.com/questions/123282/… – user16307 Jul 29 '14 at 18:24