I'm still waiting for an EE to calculate the thermal properties of a power resistor and show me how to calculate the heat generated of various wattage resistors run at the same current and voltage. What specs do I need to use and what's the math? All I can think of doing is using a thermocouple temperature sensor and feed the data into a logging DMM and then upload the temp data into an excel spread sheet and graph the data.
Well, the answer to this is both simple and not so simple. Firstly, the heat generated is simple. The resistive power law says heat is current times voltage, which is the same as voltage squared over resistance, which is the same as current squared times resistance.
So if I put my 10 ohm resistor across a 5 V power supply, the heat generation will be (5 x 5) / 10 = 2.5 watts.
What is not so simple is what temperature the resistor must reach to dissipate 2.5 W into the surroundings. This will depend quite a lot on the size of the resistor body, the air flow around the resistor, and the temperature of the surroundings. Or in my case any heat sink the resistor is touching. Probably the most reliable way to proceed is to connect the resistor to a regulated power supply and feed the expected worst case current through it, then measure the surface temperature with an IR temperature probe. Experimental data always trumps calculations, especially when some parameters are uncertain.
The temperature of a hot resistor is given by the heat transfer law:
(heat transfer) = (heat transfer coefficient) x (surface area) x (temperature difference)
In this equation the heat transfer coefficient is the thing we generally don't know very well. There may be some estimates given on the resistor data sheet to assist, but these are only estimates. It will all depend on how and where the resistor is installed.
Here's an example calculation.
Let's suppose our big ceramic resistor is 50 mm long and 10 mm on a side. Then its surface area will be 50 x 10 x 4 = 2000 mm2.
Further, let's presume the heat transfer coefficient will be 10 W/m2/degC.
Lastly, let's assume the ambient air temperature is 25 C.
Then to dissipate 2.5 W we have:
2.5 W = (10 W/m2/degC) x (2000e-6 m2) x (T - 25 C)
Solving for T we get:
T = 2.5 / 10 / 2000e-6 + 25 = 150 C
Quite toasty
It's important to stress here that the value of 10 W/m2/degC for the heat transfer coefficient is an estimate. We really don't know it with any accuracy unless we measure it for our specific build. But it's clear that even a piddling 2.5 W of heat dissipation has the potential to make things quite hot.