I want to size a water heat sink to dissipate the power of 16 MOSFET transistors of 300W each.
However, I don’t know how to go about it.
Can you help me?
I want to size a water heat sink to dissipate the power of 16 MOSFET transistors of 300W each.
However, I don’t know how to go about it.
Can you help me?
Hello quentin_c and welcome to the forum,
In practice you should consider what your target running temperature is. As electronics get hot the ability for you to cool them gets easier. Cui devices has a good write up of the concepts involved here: How to Select a Heat Sink | CUI Devices
Thanks for the info it will help me.
Are you familiar with Peltier modulus and heat pipe plates?
I wonder if these are alternative solutions to my application. Or are they made for very specific applications?
I found a post from one of the Engineers on using the Peltier devices. Here is the link:
This should help with some of the questions on using this type of device.
I would question whether the 300W figure mentioned is correct; this is a significant amount of power to burn off in a single transistor, and in practice it’s not very practical for devices in packages less beefy than an SOT-227.
A correct understanding of the actual amount of thermal power to be dealt with is the first step in identifying a suitable thermal solution. There’s no point in proceeding further until this information is understood clearly. A transistor that’s transmitting 300W of power electrically is a very different thing than one that’s dissipating 300W thermally.
I’d state with confidence however that peltier devices are not appropriate for anything even remotely approaching the amount of power mentioned. Heat pipes probably wouldn’t be a great solution either. If one’s actually needing to deal with 300W*16=4.8kW of thermal power, that’d be in the territory of cold plates and fan-equipped radiators of the sort found here.
A discussion on making thermal analysis calculations that might be informative can be found here.
Thanks Verna for the link.
I’m not a big electronics guy. I’m more of a beginner.
However, the transistor in question can support a maximum Pd=700 W with the SOT-227 package.
How can we mathematically distinguish a transistor that electrically transmits 300 W of power to the one that thermally dissipates 300 W.
I was told that there is a thermal balance when the dissipated power Pd is equal to the power consumed Pc. Is that true?
The sort of system which would call for 16 transistors of the scale in question is perhaps not a good context to develop familiarity with the considerations involved, much in the same way and for similar reasons that a Koenigsegg on the Nurburgring is not a good context for a student driver’s first behind-the-wheel experience.
Be wary of such figures; they often do not mean what one might assume them to mean, as illustrated in this resource.
Power dissipated by a transistor is the product of the voltage appearing across it with the current flow through it, integrated over time. Estimations of this quantity can be made by considering conduction losses (those occurring when the transistor is in a stable “on” or “off” state, and switching losses which occur as a transistor is transitioning between the two.
It’s unclear how you understand the meanings of the Pd and Pc terms, but the underlying principle is correct; if the rate at which thermal energy is deposited into an object equals the rate at which thermal energy is drawn out of it, the object is in thermal equilibrium and does not (in principle) experience a temperature change. Practical reality is somewhat more complex. The resource mentioned above and its prequel (linked therein) attempt to illustrate some relevant considerations and processes for calculation.
In fact I am rather specialized in mechanics than in electronics. But I have here a concrete case with electronic data of which I am unfamiliar. These are power MOSFET transistors used in linear. These transistors are used to simulate a load in an ATE. So the power consumed corresponds to the power dissipated, if I understood correctly. At the same time I wonder how we determine the thermal resistance of a heatsink when we have several transistors of different power?
Electronic folks are interested in energy in electrical form, and thus tend to use the term “power dissipation” in reference to the conversion of electrical energy to thermal form within a device. Others may understand “power dissipation” as the transfer of thermal energy from a device into the surrounding environment–this may be a point of confusion here.
Converting electrical energy into thermal form is akin to flushing a toilet; within a narrow buffer zone, if the flow into the bowl (electrical->thermal conversion) doesn’t equal the flow out (thermal energy from device-> environment) one ends up with an unpleasant result.
One sizes a heat sink to present a sufficiently low thermal resistance that the rate of thermal outflow it affords is sufficient to accommodate whatever inflow is expected without exceeding some critical limit; the maximum permissible temperature of the silicon within a transistor is similar to the rim of the toilet bowl in the analogy. Multiple transistors on the same heat sink burning different amounts of electrical power will experience different die temperatures, but the precise temperature of each ends up being both a function of its individual contribution as well as the combined effort of the whole.
On a different note, mention of using FETs in a linear mode raises an alarm bell, as contemporary devices don’t always do so well in that manner of use; do check the safe operating area, and make sure they’re being operated within prescribed limits.
Having no knowledge of these technologies (peltier and heat pipe)
When you say, "I’d state with confidence however that peltier devices are not appropriate for anything even remotely approaching the amount of power mentioned. Heat pipes probably wouldn’t be a great solution either. "
How to justify this?
Peltier devices used for cooling purposes are hideously efficient as a general rule, and are best reserved for a few niche applications where burning 5 joules of electrical energy to move 1 of thermal energy is tolerable. The article Verna linked expands on this idea significantly if you’re interested in a longer-form answer.
Heat pipes do well at moving moderate amounts of thermal energy from point A to some point B that’s a few inches away, but become less appealing as distances increase. A serious look at some of the available products and their specifications should help illustrate the point.
Heat transport over longer distances at KW+ levels is a routine matter in automotive and domestic heating applications using liquid media. Sources from those industries would likely have useful parts of a solution to offer as well.
Thank you rick.
I understand better.
And so is it true to say that cooling by the Peltier effect is efficient and economical for applications requiring low cooling power (up to a few tens of watts per element) and operating at a temperature close to ambient (up to 'about 20°C apart)?
Those are conditions under which peltier devices are more reasonable, but I wouldn’t describe them as “efficient” even then. Their advantages lie in the areas of being compact with no moving parts, ability to provide a reversible heat pumping function that’s highly controllable, and ability to generate below-ambient temperatures with a minimum of equipment. If one wants to study a sample of something at very low temperatures under a microscope, or repeatedly cycle a small object between two very specific temperatures (common in DNA analysis) then peltier devices can be very useful tools.
For most “cooling” applications where one has a hot (above-ambient) object that one simply wants to be closer to ambient temperature, peltier devices are going to be a poor choice. For most refrigeration applications (generating below-ambient temperatures) a vapor-compression system as found in a typical refrigerator or air conditioning unit will move tens or hundreds of times more heat per unit of electrical input power than a peltier device will.
thank you for the explanation
Is it possible to correct my thermal resistance calculation of a radiator with fan? It’s just an exercise and checking if my calculations are correct.
I have a closed drawer of dimension 403x600x130.55 mm. Inside there are 16 modules (assembled printed circuits) each with 2 Mosfet transistors. The power dissipated is at the level of the transistors. (For a module the distribution of power in the 2 transistors is 48% and 52%).
There is a single heatsink capable of supporting a total power of 4800W. The thermal resistance of the heatsink is Rth=0.015°C/W and at the entrance to the drawer there are 3 fans with a flow rate of 310m³/h and an outlet to evacuate the hot air blown by the fans. The ambient temperature is 26°C in the room. Inside the drawer, the temperature is 45°C. The air leaving the drawer is around 60°C.
Characteristics of Mosfet transistors (datasheet):
The power distribution of the modules is done in rows of 4 modules:
→ 1st row: 4 modules of 375W each
→ 2nd row: 4 modules of 300W each
→ 3rd row: 4 modules of 275W each
→ 4th row: 4 modules of 250W each
For a total of 4800W
The goal is to check the value of the thermal resistance of the radiator by calculation.
I have a question concerning the ambient temperature to be taken into account in the calculations, is it the ambient temperature inside the drawer (45°C) or that outside (26°C)?
If it is the indoor ambient temperature that I must take into account in my calculations, is it the one with radiator and fan, ie 45°C or the one without radiator and fan (55°C-60°C)?
Results for ambient temperature Ta=45°C:
Knowing that for a module, the distribution of power in the 2 transistors are 48% and 52%. So we have for the 375W module, a distribution of 180W for transistor 1 and 195W for transistor 2.
Tj = Tambient + (Rthsa cooler * Total power in the cooler) + Rthcs contact paste + (Rthjc of the transistor from which the greatest power passes * Power dissipated by this transistor) <= 150
Rthsa=(150-44.85-45)/4800 = 0.0125°C/W
The value seems to be closer to that given at the start, which is 0.015°C/W.
Does my reasoning seem correct?
This calculation would appear reasonable as a first-pass process for estimating a maximum permissible thermal resistance of a heatsink required to dissipate some total thermal load distributed across multiple sources, based on limiting worst-case junction temperature to some specified value. 0.0125°C/W (the calculated result) is in the same ballpark as 0.015°C/W, but on the wrong side of the fence by 20% from where one would prefer to be.
It depends on which “ambient” the heatsink in question is actually exposed to. If it’s sitting in front of the cool air intake that’s one thing, if it’s sitting at the exhaust and only getting air that’s been pre-heated by other co-located equipment, that’s quite another.
That math standing as it is, there are some other observations on the larger scenario that seem to warrant mention:
If one’s taking that number off the nameplate of the fans in question, it’s very likely to be inaccurate in practice. This post may help illuminate the reasons why. Insofar as heatsink thermal resistances tend to be based on some prescribed airflow condition, it’s a rather important detail.
As written, this would appear to imply that the the exhaust airflow is increasing in temperature by 15°C as a result of passing through the exhaust port. In practice, such results would likely suggest that the internal temperature probe is not well placed to capture the maximum internal air temperature.
Radiator…? Additional 10-15°C temperature? This causes me to question the degree of match between the idea in my mind and that which lies in yours. Pictures might be worth a thousand words, but one should take inflation into account…
thank you for the explanations.
“Radiator…? Additional 10-15°C temperature? This causes me to question the degree of match between the idea in my mind and that which lies in yours. Pictures might be worth a thousand words, but one should take inflation into account…”
The ambient air in the room is 26°C (the room is ventilated), the air passes through fans and then passes inside the cabinet. When calculating the thermal resistance of the radiator, should I take into account the room temperature (26°C) or a higher temperature (45°C for example)?
Other questions, by what mathematical relationship can I determine a fan knowing that according to the thermal resistance of the radiator found, the speed of the air at a speed of 5m/s?
“Ambient” refers to whatever environment a device is exposed to; it could be understood differently as the temperature of the airstream just before it begins passing over the device. Note that thermal resistance figures for heatsinks are based on some particular test condition, and that actual results in a particular application can vary significantly depending on the fluid flow properties within an enclosure, which are not the easiest thing in the world to predict. Leaving oneself some wiggle room is usually a good idea.
The question as I understand it might be rephrased as “how does one select a fan to produce a linear flow speed across a heat sink of 5m/s?”
It’s not an easy question to answer precisely. Assuming one’s heatsink is contained within a tight -fitting duct, the cross-sectional area of that duct multiplied by the desired flow speed yields some minimum volumetric flow rate (m2*m/s=m3/s). That’d represent an absolute minimum CFM rating required.
But because fan CFM ratings are given at zero static pressure, and forcing air through an enclosure at some useful rate will present a non-zero static pressure to the fan, a fan with a higher CFM rating than the estimate above will be necessary. (See the post on fan & system curves)
Estimating static pressure drops across a straight, round, unobstructed duct using hand calculations is doable to a reasonable degree of accuracy; but for any system of significantly greater geometric complexity one’s going to be relying on some rather expensive fluid mechanics modeling software, and doing physical tests afterward to validate the results.
It’s pretty common for folks to skip the fancy software part, and move on to testing a system designed using generous assumptions, estimated guesstimation, and decent self-protection mechanisms.
Thank you for the explanations. They are very complete and interesting.
I do have another question though:
If in an existing assembly, the thermal resistance of a finned radiator is 0.015°C/W intended for use with fans and that no component exceeds its maximum temperature.
And if I verify it by calculation but obtain a thermal resistance of 0.011°C/W, does that mean that the choice of the heatsink has been badly evaluated (undersized)?
Knowing that in my calculations I did not take into account a safety factor (which would therefore mean that I will obtain instead of 0.011°C/W, a value of 0.009°C/W after safety).
By using a 0.015°C/W heatsink, are the components likely to have their lifespans reduced despite not exceeding their temperature limits?
The internal temperature of the devices being cooled is the endpoint, the goal; the ratings of any devices used to achieve that goal are incidental, so long as the goal is achieved. If you’ve got some heatsink that’s rated for 0.015°C/W and careful measurements of an existing design indicate an observed performance of 0.011°C/W, that could well be a result of airflow conditions in practice being different than those under which the heatsink was characterized. Error in measurement is always a possibility as well.
The relationship between maximum device temperature and longevity is complex, since multiple failure modes exist. Temperature alone is not the only factor; temperature changes also play a role. Mechanical stresses are created within a device as parts of it warm and expand relative to others, which can cause things like bond wire or die attach failures. Increases in magnitude and abruptness of temperature changes increases the likelihood of such failures occurring, to the extent that a device rated for 150°C max would likely last longer if run at 155°C continuously than it would if cycling between 30 and 145°C thirty times an hour.