Calculating resistor self-heating. It’s a fairly fundamental concept that is often unfamiliar to, or overlooked by, many engineers.

I was reminded of its importance while explaining the theory behind a precision resistance temperature detector (RTD) acquisition system I recently designed. The design is featured in TI Designs – Precision reference design TIPD152. In the design, simplified in Figure 1, the errors due to the self-heating of the resistors in the signal path need to be considered to prevent them from causing undesired error levels.

**Figure 1: Simplified Ratiometric RTD System**

The design is configured for a ratiometric measurement, so the final analog-to-digital converter (ADC) conversion result is directly dependant on the absolute value of a reference resistor, R_{REF}. Because an excitation current flows through R_{REF}, it will dissipate power and therefore self-heat, causing a change in resistance affecting the system accuracy. The effects of resistor self-heating are also important in many other applications, such as current sensing or power-metering, that depend on the absolute value of a resistor that may change in value as it dissipates power.

The temperature coefficient, or TC, of a resistor specifies how much the resistance will change when the resistor temperature changes. Resistor TC is typically specified in parts per million per degree Celsius, ppm/°C. A 1% resistor typically has a TC around +/-100ppm/°C, while precision metal foil resistors offer TCs less than 0. 1ppm/°C.

Equations 1 and 2 show an example of how you can use a resistor’s TC specification to calculate the change in resistance, ΔR_{TC} of a 1kΩ, ±100ppm/°C resistor when the temperature changes from 25°C to 125°C.

In general, smaller surface mount components (0201, 0402, 0603, etc.) are less effective at dissipating power resulting in very high self-heating coefficients, θ_{SH}, up to, and sometimes over 1000°C/W! Granted these smaller resistors are usually rated for power levels <0.1W, but their temperature will change very quickly with dissipated power.

Equation 3 calculates the increase in resistor temperature, ΔT_{SH}, due to the power dissipation. Equation 4 plugs ΔT_{SH} into Equation 1 in place of ΔT to determine the resistance change due to self-heating with a modest self-heating of 100°C/W and 0.5W of power dissipation.

Although you won’t usually find the self-heating coefficient in a resistor datasheet, you can back-calculate it from the power derating curve, which is typically included.

The power derating curve dictates the maximum power dissipation versus ambient temperature that the resistor can dissipate without exceeding the maximum specified temperature. Figure 1 shows an example of a resistor power derating curve for a 0.5W resistor.

**Figure 2: Example Power Derating Curve for 0.5W Resistor**

You can easily determine the maximum operating temperature, T_{MAX}, from the curve in Figure 1 as the value on the x-axis when the rated dissipation = 0%. In the example shown, the maximum operating temperature is 150°C.

Further, the resistor can no longer operate at 100% of the rated dissipation, T_{MAX_PWR100%}, at 85°C. Using this temperature, the maximum operating temperature and the power rating of the resistor, you can calculate the value for θ_{SH} as shown in Equation 5.

With the coefficient of self-heating calculated, you can now determine the increase in heat, and therefore change in resistance, due to the power it is dissipating using Equations 3 and 4. From there, you can determine the effects on the final system accuracy based on the change in resistance.

So, next time you’re designing a system that requires precision resistor values, be sure to take resistor self-heating into account!

Thanks Collin. Very informative post which is not into account most of time by designer.

Thanks, I'm glad you enjoyed it. Check out the other blogs in the series, they're great as well!

Hi Collin,

Thanks for the calculation. Although the calculation is very informative, I found a little bit confusing.

First of all the values used in the example perhaps were not the best chosen because the used resistors is overloaded at 125°C ambient temperature.

The max. ambient temperature is 125°C and contribution due to self-heating is 50°C. This means that the resistor from example is working at 175°C, correct?

However the article has showed, that the max. temperature working point of device is not max. working temperature endpoint of used resistor due to self-heating coefficient that impact temperature rise.

Br,

Hello Josko,

Thanks for your comments. The calculations were actually not meant to be connected and were intended to explain two separate concepts. I didn't do a great job noting that these were separate concepts so I understand the question. EQNs 1/2 show how to calculate the change in resistance based on a 100C temperature change and EQNs 3/4 explain how to calculate the expected temperature change in a resistor due to self-heating and the resulting resistance change with the self-heating effects.

Hello Collin.

Thank you for the insights.

However, I do not understand how to calculate Resistor Case Temperature when Ambient Temperature= 25°C, Power Dissipated= 0.1W, Max. power dissipation = 0.5W (1/2W)

As per the derating graph and equation (5) : θ_SH= 130°C

As per equation (3): ΔT_SH= 130 x 0.1= 13°C

So does the calculations mean;

(a) At T_amb=30°C: Temperature of resistor(case) = ΔT_SH+ T_amb= 13 + 25= 38 °C

(b) At T_amb=50°C: Temperature of resistor(case) = ΔT_SH+ T_amb= 13 + 50= 63 °C

Best Regards,

Sarang

Hello Sarang,

Your calculations are correct. With 130C/W and 0.1W the expected temperature increase will be 13C. This will add to the ambient temperature yielding the total temperature you've calculated.