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It's common for a lot of power circuits to bolt to a piece of aluminum flat stock. How large does the stock need to be?

Say that I'm mounting a tip122. The worst case condition is for it to have a 24V drop at 3A and a 50% duty cycle. So it is dissipating 36W.

Looking at the datasheet, at 35W, the max case temp is ~80C. Assume a 25deg ambient temperature.

Temp drop = 35W * Tr = 55 delta or 1.57C/W for the plate.

So how much surface area do I need to achieve that?

Did I approach it correctly?

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  • \$\begingroup\$ Do you have forced air? Without it, this is only feasible with a finned extrusion, unless you want to have a hot plate with dimensions in feet. \$\endgroup\$ – Kevin Vermeer Jul 30 '10 at 17:29
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You can use this calculator to run some numbers for a flat plate (your design should have fins). Beware that you have a limited number of tries, and that it requires an air velocity parameter. Plug in your desired case temperature from the derating curves, which I'll go into further later.

Are you stuck with the TIP122? How about the TO-220 package? Both the TIP122 and TO-220 are only designed for medium power applications. This type of application would be better served by a high power transistor and a metal can package.

The difference between a high power, medium power, or small-signal transistor is not only in their packages-it's in the device's construction as well. The maximum ratings table for the [TIP122 datasheet] show that it has a maximum collector power dissipation Pc of 2W in 25°C air, or 65W with Tc=25°C. The second stat assumes that you could have an infinite heat sink, connected with the ultimate heat sink compound to the tab (technically the case, but the tab's all that matters) on the TO-220, such that the heat sink tab is at 25°C. Even in that case, the transistor junction, which is what you're concerned about, will go over 150°C. There is thermal resistance between the junction and the tab. (Sidenote: I'd agree with jluciani - I like my silicon 125°C or cooler). (Sidenote 2: Metal heatsinks on BJTs are usually connected to the collector, so you'll have a 3A source connected the case, at a voltage greater than the emitter/ground, and won't want it to be somewhere that it can short out.)

Take a look at the derating curves (Figure 5 in the TIP122 datasheet):
TIP122 Derating curve
If you need to dissipate 72W, you simply can't do it. If you need 36W, you would have to keep your heat sink less than 50°C above ambient (25°C. It's this 50 degree temperature gradient that gives you power dissipation). Compare that curve to a high-power transistor such as the MJ11022 [datasheet]:
MJ11022 derating curve

Your heatsink can now be a burn hazard long before the transistor is damaged. 72W corresponds to almost 100°C above ambient, and 36W to an absolute operating temperature of almost 150°C. Beware of thermal cycling if you want to run it really hot.

I'd strongly suggest that you use a high-power TO-3 or TO-204 transistor instead of your TIP122.

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    \$\begingroup\$ I'd like to point out that, while my answer is arguably the better solution to the problem, jluciani's answer, which calculates Rsa (Thermal resistance, sink-to-ambient) to be 1.05, and links to a source which supplies finned sinks with various thermal resistances, is arguably the better answer to the question. \$\endgroup\$ – Kevin Vermeer Aug 1 '10 at 16:46
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You will probably need a very large plate or a fair amount of moving air.

What is the on-time of the TIP122? If the on-time is greater than 100mS then you are dissipating 72W not 36W. You need to look at the transient thermal response curves to determine the derating.

You need to allow some thermal resistance for the interface between the case of the transistor and the sink (or plate).

Assuming that your on-time is less than 1mS you are dissipating 36W. Looking at the On-Semi datasheet --

Rjc = 1.92 degC/W max. Absolute maximum junction temperature = 150degC (I would not exceed 125degC)

T = (Rjc + Rcs + Rsa) * Pd

125 = (1.92 + 0.5 + Rsa) * 36

Rsa = 1.05degC/W (which agrees with your calculation when you subtract Rcs)

If you take a look at the heatsink vendors datasheets you can get an idea of sizes. Checkout http://www.aavidthermalloy.com/

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To calculate power dissipation or changes of Rθjc or Rθja in dynamic environment (i.e pulse current), it is not so straightforward process. You have to see at the so-called “Typical Thermal Response” curve provided by the manufacturer. From this curve you can get the “Transient Thermal Resistance” (normalized or actual Ζθ). Anyway I cannot do the detailed calculations right now. Roughly, in an environment of 35oC, if you want to dissipate 35W from a TO-3 case and to keep the heatsink temperature at around 55oC using natural cooling, you need a gray aluminium plate, 3mm thick, with an edge of 16cm (i.e 210 gr). This plate should be free to radiate from both sides in vertical arrangement, with the divice tight mounted on the center of the plate. Do not forget to include in your calculations the thermal loss caused in two metals contact. In practice the 35W it is close to the maximum power that you can dissipate using metal plates and natural cooling (i.e Al metal plate 400 cm2, 5mm thickness, 0,5Kg, in vertical arrangement one side free, or 50W both sides). Above this powers, you have to use finned heatsink (natural or forced), which is not difficult to calculate and to construct

This is my way for thermal design. Never understand the thermal resistance concept. It is full of assumptions!! Anyway if you want to proceed your calculations using thermal resistance, it is necessary to have measures of the actual case temperature as a function of time at full or half load.

enter image description here

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I know this is an old thread, but I discovered it researching this subject and wanted to correct/add a couple of things. The formula to find the required thermal resistance of the heat sink given by jluciani is basically correct but is missing a term for the ambient temperature (Ta). The equation should be:

Tj = (Rjc + Rcs + Rsa) * Pd +Ta

Where Tj is the maximum target temperature of the junction. I will use 125 degC as the maximum temperature of the junction to allow a safety margin in the event the ambient temperature goes over the standard 25 degC. This gives:

125 = (1.92 + 0.5 + Rsa) * 36 +25

Rsa = (125-25)/36 - 1.92 - 0.5 = 0.3577 degC/W

The next part to find the size of the aluminium plate required to achieve this low a thermal resistance is much more complicated, but this blog https://engineerdog.com/2014/09/09/free-resource-heat-sink-design-made-easy-with-one-equation/ gives a very simple rule of thumb aproximation given by:

Area = (50/Rsa)^2 cm2

Unfortunately this formula applies to passive heat sinks with fins and I believe the author made a typo and meant area = 50×(1/Rsa)^2. The fins make a big difference. After looking at the results of of this online calculator https://www.heatsinkcalculator.com/free-resources/flat-plate-heat-sink-calculator.html and the data sheets from a range of passive heat manufacturers I did a bit of curve fitting and came up with this more comprehensive ball park formula:

Area = (20*1/(1+flow)*1/(0.25+h)*1/Rsa)^2 cm2

Where flow is any flow from a cooling fan in cfm and h is the height of any fins.

For the situation in the OP there is no forced cooling so flow= 0 and there are no fins, so h = 0 and the formula simplifies to:

Area = (80/Rsa)^2

Given that we require a thermal resistance <= 0.3577 the size of plate required to cool the transistor in the OP is:

Area = (80/0.3577)^2

      = (223.6 cm)^2

This is probably too large to be practical.

As Kevin Vermeer pointed out, this particular transistor in this service is not really suitable for passive cooling. However, a dramatic decrease in the heat sink size can be obtained by adding fins and a fairly modest cooling fan as shown by the chart at the bottom of this link https://www.designworldonline.com/how-to-select-a-suitable-heat-sink/#_

Staying with a flat plate and adding a fairly good PC cooling fan of 100cfm air flow, the plate size could be reduced to:

Area =(80/(0.3577*(1+100/8)))^2

      =(16.56 cm)^2

Extruded aluminium can be bought in long strips with fins and using such a finned plate with 3cm fins and no cooling fan would require a heatsink size of:

Area = (20*1/(0.25+3)*1/0.3577)^2

      =(17.2 cm)^2

Finally, combining forced cooling of 100cfm and 3cm fins gives:

Area = (17.2/(1+100/8))^2

     =(1.27 cm)^2

Notes:

Pressure drops and proximity of other hot components in the cabinet can reduce the efficiency.

Dust ingress can insulate heat sinks and cause cooling fans to slow down and fail over time.

Heat sinks that are much much larger than the contact area of the component they are cooling loose efficiency due to the distance the heat had to travel to spread to the extremities of the heat sink

Follow the usual guidelines on ensuring good contact with component to be cooled using a thin layer of a suitable heat transfer compound between the contact surfaces.

Results from this formula for extremely small or large heat sinks should be treated with suspicion. For example in the last result the cooling fan radius is much larger than the heat sink and so most of the airflow would not be flowing in close proximity to the fins and so the result is suspect. Otherwise, it is a pretty good approximation.

Probable best to add 25 degrees to whatever you think the ambient air temperature is and deduct a 25 degrees margin of safety from the maximum target temperature of the component when carrying out the calculations, just to be on the safe side.

Dont use this formula to design the cooling for a nuclear power station.

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There is a great blog article located at http://www.heatsinkcalculator.com/blog/how-to-design-a-flat-plate-heat-sink/ that provides a detailed explanation of the calculations needed to size a flat plate to be used as a heat sink. They also provide the spreadsheet with the calculations, however you will have to provide your email address to get the download link.

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  • \$\begingroup\$ Welcome to the site! Link-only answers are discouraged here because the links tend to break. Can you summarize the information at the link? \$\endgroup\$ – Adam Haun Nov 16 '15 at 22:48

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