# Gas station without pumps

## 2014 September 11

### Thermal models for power resistors

Filed under: freshman design seminar — gasstationwithoutpumps @ 06:46
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I recently bought some power resistors, to use as dummy loads for testing PWM circuits and to use as heating elements in an “incubator” design for the freshman design seminar.  I bought 3 resistors: 10Ω 100W HSC10010RJ,  8.2Ω 50W THS508R2J, and 1.8Ω 50W THS501R8J.

I want to make simple models for the thermal behavior of these resistors when they are not mounted on a heatsink, but are just sitting on a low-thermal-conductance surface.  The simple model will have two parameters: a thermal mass M (in joules/°C) and a thermal resistance D (in °C/W).  If we just had the thermal mass, we would have $\frac{dT}{dt} = \frac{dE}{dt}/M = P/M$, where E is the thermal energy, and P is the power delivered to the resistor, and the temperature would increase linearly: $T(t) = T_{0} + Pt/M$. But as the temperature increases above the ambient temperature, the resistor loses energy at a rate proportional to the temperature difference from ambient: $\frac{dT}{dt} = (P - (T(t)-A)/D)/M$.  We can rewrite this as a standard first-order differential equation: $\frac{dT}{dt} + \frac{T(t)}{DM} = \frac{PD+A}{DM}$, which has the solution $T(t) = PD+A+(T_{0}-PD-A)e^{-t/(DM)}$.  Note that $\lim_{t\rightarrow\infty}T(t)= PD+A$, independent of the thermal mass, and the cool down with $P=0$ is dependent only on the initial temperature, the ambient temperature, and the product $DM$, not on D and M separately.

To find the parameters for each resistor, I connected each to my 9V 6A power supply, and measured the temperature at regular intervals with an IR thermometer.  For the 50W resistors, I blackened the bodies of the resistors with a felt-tip pen to make the IR thermometer more accurate—I had not done that with the 100W resistor, but it took so long to make the measurements on that resistor that I did not want to go back and remeasure it.  It had a colored finish and may have been closer to being a blackbody radiator than the 50W resistors, so the errors may not be too large.The errors due to not holding the gun in a perfectly fixed position probably contribute more error.

The fits are not too bad—this simple model seems to represent the thermal behavior of the resistors fairly well.

The 100W resistor, as expected, has a very high thermal mass and fairly low thermal resistance. It heats up and cools down slowly. With a low power input (8% of rated power), the equilibrium surface temperature is still quite low, only about 76°C—well below the 240°C melting temperature of styrofoam. Even with a 12V supply the temperature would only get up to (12V*12V/10Ω)*6.38°C/W + 25°C=117°C.

The 8.2Ω 50W resistor has a lower thermal mass but a higher thermal resistance than the 100W resistor. It heats up much faster, and cools down somewhat faster than the 100W resistor. It is being run at about 20% of the rated power, and it is supposed to be able to be run at up to 40% of rated power (20W) without a heat sink.

The 1.8Ω 50W resistor has similar thermal characteristics to the 8.2Ω 50W resistor (it is the same package in the same series), but because the power is much higher 86% of rated power, it heats up very fast and would exceed the temperature specs for the resistor if left on for more than a couple of minutes.

Adding a large heatsink would increase the thermal mass and decrease the thermal resistance of any of the resistors. If I want to use the 1.8Ω resistor, I will definitely need a heatsink! I can run the 8.2Ω resistor without a heatsink at 9V, but at 12V it would get up to 230°C, too close to the melting point of styrofoam. The 10Ω 100W resistor could be used safely even at 12V. I’ll try adding a 6″×12″ sheet of 0.063″ thick aluminum.  According to the Wikipedia article on heat capacity, the specific heat capacity of aluminum is about 2.422 J/°K/cm3, so the sheet should add a thermal mass of about 180 J/°C, but computing the thermal resistance is complicated, so I’ll just measure the temperature rise and fit the model.  Even if the heat dissipation were not increased (very unlikely), the greater thermal mass and resulting 7× slower response will make measurements easier and less likely to result in overheating the 1.8Ω resistor.

I’ve now tested that my power supply is capable of delivering 8.84V/1.81Ω = 4.88A. I still need to put the 1.8Ω and 8.2Ω resistors in parallel and see if I can get 6A from the power supply. The output impedance of the power supply seems to be about 78mΩ, given how much voltage drop there is with increasing current. Most of that may be the wiring from the power supplies to the resistor, as the power supply senses the voltage as it leaves the power supply, before the IR drop of the wiring.

1. […] night I fit a simple thermal model to temperature measurements of some power resistors: , where P is the power in watts, D is thermal resistance in °K/W, M is thermal mass in J/°K, A […]

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