# Gas station without pumps

## 2014 September 11

### Thermal models for power resistor with heatsink

Last night I fit a simple thermal model to temperature measurements of some power resistors: $T(t) = PD+A+(T_{0}-PD-A)e^{-t/(DM)}$, where P is the power in watts, D is thermal resistance in °K/W, M is thermal mass in J/°K, A is the ambient temperature in °C, and T0 is the initial temperature.

I ran into problems with the 1.8Ω 50W THS501R8J resistor, because it heated up very fast and I could only get a few measurements when delivering power, before I had to turn it off.  I proposed adding a heatsink, a 6″×12″ sheet of aluminum 0.063″ thick, to increase the thermal mass M and decrease the thermal resistance D.  I estimated that the thermal mass should increase by the heat capacity of that much aluminum (74.33 cm3 at 2.422 J/°K/cm3, giving 180 J/°K), but I did not have a good way to estimate the change in thermal resistance.

The 6″×12″ plate is much larger than the power resistor, which is bolted in the center with M3 screws (American 6-32 screws are a little too big for the holes in the resistor). I used a thin layer of white thermal grease to get better thermal conduction between the resistor case and the aluminum plate.

I do not expect the simple thermal model to work well, because it assumes that you have an isothermal object—all the aluminum at the same temperature.  But a large flat plate is going to have significant thermal spreading resistance, so that the resistor in the center is hotter than the edges of the plate.

With a heatsink the time constant DM is about 260s, only a little faster than the 347s without the heatsink, but the thermal resistance is much lower, so the maximum temperature (PD+A) is much lower.

As expected, the fit is not great. When cooling off, the initial temperature of the resistor is higher than of the surrounding plate, so the initial cooling at the resistor is faster than the eventual cooling, when resistor and the plate are closer in temperature, because heat is being transferred to the plate as well as to the air. The increase in thermal mass  (about 100 J/°K) was less than my crude estimate based on the heat capacity of the added aluminum (180 J/°K)—this is probably also due to the thermal spreading resistance and the non-uniform temperature of the heatsink.

resistance rated power heatsink? test power M [J/°K] D [°K/W] DM [s] T [°C]
10.10Ω 100W No 8.288W 101.7 6.38 649 75.7
8.21Ω 50W No 10.169W 32 11.58 371 143.7
1.81Ω 50W No 43.174W 31.9 10.87 347 495.8
1.81Ω 50W Yes 43.174W 131.7 1.97 260 110.3

Note: the asymptotic temperature T in the table above is with the 9V power supply I have, which does not have quite constant voltage over the range of powers tested. With a 12v supply, temperatures would be much higher: $D V^2/R +A$.  The asymptotic temperature is also the maximum when the resistor is sitting in still air that is unconfined.  A fan would reduce thermal resistance and make the asymptotic temperature lower, but confining the resistor in a box (like in the incubator design) would make the “ambient” temperature not be constant—the relevant thermal resistance is how slowly the air in the box loses heat, which for the thick-walled styrofoam boxes we’ll use is a very high thermal resistance.  Without a feedback loop and PWM to keep the power down, even the 10Ω resistor would get very hot in a styrofoam box.

I should probably test the 10Ω 100W resistor on the heatsink also, to see if that reduces the time constant DM.  I expect the thermal mass to go up by something between 100 and 180 J/°K, but the thermal resistance to drop to around 1–1.5 °K/W, getting DM in the ballpark of 300s.  I don’t think I’ll do that today, though, as making measurements every 20 seconds for 2000 seconds is tedious and leads to cramping in the hand that aims the IR thermometer and keeps the trigger pulled.

Which raises a pedagogical question: Should I have students do the measurements?  Should I show them how to make a recording thermometer with a thermistor first? They’ll need to figure out how to use a thermistor for measuring air temperature anyway.

The thermistors I have at home (NTCLE100E3103JB0) only go up to 125°C, and I’d want them to have one that goes to at least 175°C for this lab, which means using something like NTCLG100E2103JB (10kΩ, ±5%, ±1.3% on B-value, -40°C to 200°C), which is only 35¢ in 10s, so still cheap. I should get myself some of these higher temperature thermistors and test out the recording . (Or the tighter tolerance NTCLE203E3103SB0, which only goes up to 150°C, or the wider temperature range 135-103LAF-J01, which goes to 300°C.)

How will I attach the thermistor to the resistor for temperature measurement? tape?  (I have to be sure not to short out the thermistor leads on the aluminum case of the resistor.)

Air temperature sensing poses less of a mounting challenge, but the thermal delays will be quite large—I have to look at how difficult it will be to tune a PID or PI controller with large delays—we really don’t want huge overshoot.  If the students have multiple temperature measurements (resistor temperature and air temperature, for example), they may need a more complicated control loop than a simple 1-variable PID controller.  How much can we simplify this?  (Perhaps a PI or PID controller based on the air temperature, with over-temperature shutdown on the resistor temperature?  Then tuning the PID controller with the constraint that the gain be kept low enough to keep the over-temperature shutdown from kicking in?)

### Thermal models for power resistors

Filed under: freshman design seminar — gasstationwithoutpumps @ 06:46
Tags: , , , ,

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.