MOSFET power dissipation. Can a heatsink take +300W?

Thread Starter

ballsystemlord

Joined Nov 19, 2018
150
You cannot have an "infinite heatsink". I made them by adding liquid cooling and keeping the temperature at 25C when 1000 watts was added to it. An infinite heatsink does not need fins or airflow. It is stuck at room temperature no matter how much energy you add. You can't have one. It is math not reality.

After saying you can't have one, I build them. I build machines that test power transistors so we can make data sheets. We have a heatsink that can be set to any temperature from -60C to +200C and will hold that temperature even when the power transistor is trying to lift the temperature. Our heating/cooling machine is the size of a washing machine. It takes huge amounts of power. It could heat and cool your house. I am testing transistors that power the drive motors in cars and trucks. I know that before the test the heatsink is 25C and after the test it is at 25C. Most data sheets are made using math not real-world tests. But if you don't trust the math, I can prove the numbers.
I understood that much...
I was talking about the math behind real calculations for heatsinks. From what I have read, it's an arcane art.

https://thermtest.com/thermal-resources/thermal-conductance-calculator
Here's a thermal conductivity calculator. According to it, the conductivity of aluminum is 225.94 W/m K. According to it, a 1mm thick piece of aluminum has a conductivity of 225940.00 W/m2 K; which seems too high to me. Note: I don't actually use these values below.

Ohmite's examination of their F and R series heatsinks (pdf).
Here's a work by Ohmite. I'm using this as a basis to judge airflow.

Basics of thermal resistance and heat dissipation (pdf).
I'm working off of the above number provided by the conductance calculator, and the formula on page 4 of this guide.

I'd like to add, that the CFM and the M/S required by that pdf are incompatible. That is to say, air is three dimensional whereas plain old meters is a one dimensional value system. So I have assumed that the author got their value wrong, just like where "Representative length" is specified as meters, but other places I've read it's square meters in the other people's calculations. In fact, how would you choose which dimension to use when measuring a heatsink if you were to use only meters?

Trying to convert CFM to cubic meters a second, doesn't look very promising. Looking at this pdf from ohmite, my own 44CFM fans, would have such a low value on this chart (0.0207656878 CM/S), that even zooming in to get the value would be impossible.
So, I chose the low value from their chart of 300 m/s. Again, I'll have to find some way to test this supposition.

So the calculation ends up where I'd be able to dissipate ~ 61watts of heat with my best plain old aluminum heatsink and fans.

Equation: 6*(300/(((150*69+35*69*26)/1000)/1000)**0.25)**0.8

6 [turbulent flow in the presence of fins. A flat heatsink would not produce turbulent airflow] * (300 [already explained above] / (((150 [mm long] * 69 [mm wide] + (36-1 [mm for height. Subtract 1 mm for height as it was already used just before here.] * 69 [mm wide] * 26 [fins] )/1000 [mm2->meters]) /1000[mm2->meters2]) **0.25[part of the formula] )**0.8[part of the formula]

The result is 970.6Watts; which seems way too high.
OTOH: If I use the 44CFM value and convert it to cubic meters and use that instead of 300m/s, I get 0.22W; which seems way too low for a heatsink that size.





It's as big as a large chest freezer, and about as exciting to look at.
And the top of it gets used as a "temporary" repository for all sorts of junk.
Ok, I'll write every question I can think of about the setup off the top of my head down. Sorry it's so excessive, but what else can I do to understand how these high heat dissipation devices work?

What is the fin length? Are the fins all one piece, or are they segmented like long columns as some heatsinks are? Do the fins have parallel gouges in them to increase the surface area? Is there are special design to the fins that allows more surface contact for airflow, or better movement of air, like those special swimsuits in the Olympics that mimic a shark's skin?
Does it use heatpipes? What is the diameter of the heatpipes? What are their lengths? Are the heatpipes flat, or round? Are they sintered heatpipes? Do the heatpipes have wicks? Are the heatpipes grooved? Are they more than one of the above?
Does it use vapor chambers? What are the vapor chamber's dimensions? What is the thickness of the vapor chamber's copper connection to the heatsource (MOSFETs)? What is the spacing of the pillers in the vapor chambers? What do the vapor chambers connect to?



Not sure which datasheet you ere looking at, but the IXYS datasheets (attached is that for the device I use) have extended SOA down to DC and that's critical for linear operation. Here are the relevant charts for the IXTX110N20L2 at case temperatures of 25 and 75degC...
My mistake.




From the data sheet, the thermal resistance, junction to case Rthjc for the plus247 package is 0.13degC/W, so for a case temperature of 75degC and a junction temperature of 150degC the maximum dissipation is (150 - 75)/0.13 = 576W, ie the same as the SOA chart and the SOA example in the datasheet (200v, 2.88A, the RH limiting line on the SOA chart). But is that case temperature realistic? Well, for an ambient of 25degC the thermal resistance of of the thermal paste and the heatsink together must be less than (75 - 25)/575 = 0.087degC/W. The thermal resistance of a good thermal paste, a few microns of Arctic Silver 5, is around 0.05degC/W so we are looking at a heatsink of around 0.037degC/W! Such heatsinks exist, for a price, but as always there's a catch; heatsinks are rated assuming the heat load is across the whole surface not at one point. The upshot of all this is you actually need 3 devices to hit around 550W.

Here is a CHT simulation of 8 devices on a pair of identical heatsinks, each device handling 150W for a total of 1200W with a junction temperature of around 80degC with an airflow of 7.5m/s.
That's clear. And thanks for the picture.
One question, why the 75C graph? I mean, 75C is a fine assumption for case temperature, but why not 65C or 85C?


Now I need to read Janis59's post; several times and the rest of the pdf...
 
Last edited:

Irving

Joined Jan 30, 2016
3,845
I'd like to add, that the CFM and the M/S required by that pdf are incompatible. That is to say, air is three dimensional whereas plain old meters is a one dimensional value system. So I have assumed that the author got their value wrong, j
On the face of it, yes. But a 120mm fan has an active 'airflow' of about 120 mm diameter, so 7.5m/s translates to (roughly) 7.5 x (0.120/2)^2 x π = 0.085 cu m/s =5.08cu m/m = 180cfm..

BTW 44CFM = 0.02cum/s = 1.26m./s for a typical 60mm fan

One question, why the 75C graph? I mean, 75C is a fine assumption for case temperature, but why not 65C or 85C?
Because its a JEDEC standard that allows direct comparison between parts across manufacturers.
 

Thread Starter

ballsystemlord

Joined Nov 19, 2018
150
Theoretically you can, practically it is hard to do.
Theory says that R(th)=400/A(cm2). Thus, if you alter T to say +30C above ambient then Rth must be about 0.1 and so the area would be around 4000 cm2 or 65x65 cm. Isn't that bit large for practical applications?
The other problem, is that if the heat impact is coming in one single point, the thermal flow may not travel so much to the sides. Thus, one will have a heat spot on the center of the radiator and a cold perimeter. Thus, effective cooling is expected to be very weak. The cure is to take a radiator that is abnormally thick, like 8...10...12 mm. You would think it is over zealous. Thus, one cure is to use fans. A plethora of computer fans have 10...20... 30 CFM. Thus, to have 0.1 C/W, one needs to use a heatsink that is 200 in^2 at 200CFM. It would sound like a farming tractor! At more normal values, 10 in^2 and 40 CFM gives 0.6 C/W. Generally, the fan may be calculated by CFM=1.76*N(W)/dT (C). Another example is a fan of 20 CFM on a radiator of 4.5x1.5x3 in (80 in^3) gives you 0.58 C/W but at 10 CFM you get only 0.77 C/W. In air without using a fan, a 170 cm2/W radiator gives you a dT of 10C, 65 cm2/W gives you 20C, 26 cm2/W give 40C, 10 cm2/W gives you dT=70C. So, Pentium (an Intel CPU brand that is currently their low end and are typically outfitted with aluminum radiators,) radiators may give you slightly less than 100W, but more modern round types with a copper core may give you 200W or, with extra strong fans from both sides, near 300W. Generally, from the 100W and to one kilowatt range the custom is to use water cooling. Just drill the holes in 1 cm thick aluminum block and then screw the channels closed where needed to from a passage for the water in what is to be your cold-plate. If it is not good, you may use heatpipes. They might cost near 100 USD, but are highly effective until half kilowatt, and they demand no water. Moreover, they are the ultimate solution in cases when high power to area load is necessary (W/cm2) for semiconductors, as an example Peltier cooler plates.
Sorry for the delay, @Janis59 . I took the liberty of trying to improve the English of your answer. Please look it over to make certain I did not mess it up at all.

What is N?
Is 1.76 a constant?
CFM=1.76*N(W)/dT (C)

The original text reads, "Other example 20 cfm at 4.5x1.5x3 in (80 in^3)..."
4.5*1.5*3 == 20.25 in^2. Did I misunderstand, or is there an incorrect value in there?

Also, aren't Peltier plates very limited in the wattage they can move?

Thanks!
 
You CANNOT use switching MOSFETs for linear job. They fail at several watts in this regime. Period. I suggest using bipolar transistors instead, old fashion rugged bipolar, like 2N2055, BDxxxx, TIPxxx, etc. Several in parallel (using them at quarter of their catalogue power is a good approach) , and with equalizing resistors in the emitter
Regarding heatsink, an old, 2002-2006 era processor heatsink (Athlon and such) will be a good start, they are designed to let go a hundred watts (with fan attached) at reasonable temperature.
 

Ian0

Joined Aug 7, 2020
9,680
The older the MOSFET the more suitable it is! IRFP140/ IRFP240 have much more linear capabilites than more modern switching devices.
I achieved 240W using a moderately sized CPU cooler and 6 IRFP140 MOSFETs, but it was designed for use with 12V batteries (i.e. not lower than 10V) and I dissipated half the power in resistors in series with the MOSFET sources.
 

Janis59

Joined Aug 21, 2017
1,834
Theoretically, yes, it can. Practically, nope, it cannot.

So... if the target is about 40 C warmer than air (expecting the air is 40 C and max T for un-degraded work is 80 C) then free air circulation demands a 26 cm2 for each W. Thus 300 W demands a 7800 cm2 or meter to meter large cooler. Bitt to heavy....

Okay, catch the fun taking the fan. Then cfm=1.76*N/dT. Or cfm=1.76*300/40=13.2. Just about what was needed as the 12-20 cfm is PC fan range depending on size. But may stuck into problem that cooler not only must pass the wattage but also the wattage per square mm of contact place.

Thus, for so large power density as 300 W per cm2 (about) the very thick COPPER core is needed as the play-computers have custom to use.
Third choice - take the copper or aluminium brick (thickness at least 8 or better 10-12 mm), drill the long-holes every some 10-15 mm one, and in every of ends one perpendicular shorter caching all "longs". The unneeded holes must be covered with conical thread screws (sorry, my experience tells the tubular thread is always leaking except the special glues used). Two of the holes may be sed for incoming and outcoming water pipeworks.

Not like? Then take the heatpipe. Play computers have brilliant heatpipes for videocharts, just the typical size of transistor is 3x4 cm and most largest 5x5 cm. Up to 300 W most of it are operable. The little dark secret - the air side is rather massive and demands the fan - and it`s too small a fun.
 
Last edited by a moderator:

Thread Starter

ballsystemlord

Joined Nov 19, 2018
150
This guy's brilliant, but his English is poor. I'll translate for those who can't follow.
@Janis59 , feel free to correct me if I make a mistake. I'm trying my best here. I love your posts, but they are hard to read.

Theoretically, yes, it can. Practically, nope, it cannot.

So, if the MOSFET's internal target temperature is before it's performance degrades is 80C and the air temperature is 40C, then a free air cooled heatsink would require a 26cm2 heatsink surface for each watt of dissipated power. Thus, 300W demands a 7800cm2, which is almost 1 square meter in surface area. That's a bit large.

Now, you will probably suggest that we use a fan instead of free air. Then cfm=1.76*N/dT. Or cfm=1.76*300/40=13.2. (N is number of watts dissipated). PC fans operate at 12-20 cfm, depending on their size. But you also have to take into account that you still need a cool with a large enough surface area for the fan to blow on to dissipate 300W and to transfer the heat from the MOSFET to the surface which the fan blows on.
For so large of a power density as about 300 W per cm2, a very thick COPPER core is needed such as is used in custom PC cooling hardware. A normal PC CPU cooler will not work.

For your third choice, you can take a copper or aluminum brick which is at least 8mm, or better yet, 10-12mm, and drill long-holes every 10-15 mm apart parallel with the surface the the MOSFET will contact. Perpendicular to the holes you just drilled, you drill on the opposite face (again, parallel with the surface that the MOSFET will contact), holes at the same spacing to intercept the holes you drilled previously. Only two holes are left exposed on the surface. One for the inlet, one for the outlet. The unneeded holes must be covered with conical thread screws (sorry, my experience tells me that tubular threads are always leaking, except for when special glues are used).

For your fourth option you could use heatpipes. PCs have brilliant heatpipes for videocards. Just the typical size of a GPU is 3x4 cm, and the largest are 5x5 cm. Most GPUs operate up to 300W. A little dark secret of GPU coolers is that their air side is rather massive, demands a fan, and they're equipped with too small of fans.
@Janis59 , I'm confused on 2 things.
What are tubular threads?
Also, per your calculation of airflow above, what's the calculation for the surface area of a forced air cooled heatsink?

Thanks!
 
Last edited:

Janis59

Joined Aug 21, 2017
1,834
Tubular thread = cylindrical standartd thread (nitting) You may find on any screw on this Planet.
Conical thread = nitting having about 1/2 mm smaller diameter in the beginning and then diameter is widdening up to 1/2 mm wider than nitting in hole.

Airflow: when speaking about small radiators in airflow where absolutely dominates the Q transported by airflow, in the first approximation radiator surface have far less impact than airflow value. In more exact form (of course) the area plays some role, but then this simple formula brokes up to much more complicated one. I found this formula in one old russian radioamateur booklet and until now it works rather well, however about it accuracy I am not 101% sure, indeed. Because of this lack of area. Just, may think in the way like when after crossing a small area air is already warmed up until metal temmperature, the further increase of area is giving no increase in transported Q, because both temperatures (air and metal) are equal.

In more exact figures I can give the table data and the rest may be interpolated. Figure in brackets - C/W, figure without brackets volume of radiator in^3. At free natural convection: (0.4)=300; (0.6)=150; (1)=70; (2)=20; (4)=6.5; (6)=3.4; (10)=1.6. At forced cooling by 250 cfm (0.2)=250; (0.4)=80; (0.6)=40; (1)=18; (2)=5.5; (4)=1.8; (6)=0.8. At 500 cfm it will be like (0.2)=140; (0.4)=40; (0.6)=20; (1)=9; (2)=3; (4)=4. And last with an enormous windblow 1000 cfm (0.1)=200; (0.2)=60; (0.4)=20; (0.6)=10; (1)=4.8; (2)=1.6. The figures was given for 10 C ambient. If ambient is 20 C use factor 0.8; at 30 factor 0.7 and at 40 factor 0.63.

When radiator is cooled by natural convection instead, the table is bit other: 1/8 inch thick Alu at air gives R(th)=400/A(cm2) or in form of table: 3 (in^2)=11 C/W; 5=9; 10=6.5; 15=4.8; 20=4.2; 30=3.5; 40=3.0; 50=2.7; 80=1.9; 100=1.8; 120=1.5; 140=1.4 for copper. If radiator stays like a pancake factor 1.0, if stays vertical, must use the factor 1.8; if stays horizontal then factor 2.2. If its made from blackened aluminium then factor 0.8, if white aluminium then factor 0.5.

Some more tables: 4.5x1.5x3 inches=80 in^3 in air by fan: 2cfm=1.6 C/W; 4=1.2; 6=0.95; 8=0.86; 10=0.77; 12=0.70; 16=0.61; 20=0.58; 24=0.57.

Another version by surface area: natural cooling in air. dT=10 C @ 170 cm2/W; 20=65; 30=40; 40=26; 50=19.5; 60=12; 70=9.8.

Or by formulas: Vol(liters)=0.8/(Rth)^1.47 and dT=10(0.8/Vol (liters)^0.68)*N(W)^0.5

So.... the datasets are rather different and contradicting in prognoses, yet for practice exact enough.

And about languages... sorry of course. I still learn up to express me in English. My native language is one of very few ancient minus minus languages having no more any language versions alive, and is famous with very complicated grammar and ultra-wide expression style possibilities full with a nuances. Sometimes it isnt easy to translate not loosing a true fragrance. All the time I thought a Finnish is one awfully difficult language with 18 declinations. Yet now I had occassion to speak with one Finn to realize how they count their declinations. By the same methodic We her use in our European continent oldest language the 7x3x3x2x6x2 declinations for nouns and plus verbs etc etc. I managed to take a part to teach the language for Ukrainian war refugees, just wondered how is difficult to explain even the basic laws.... Maybe that is the true reason why we have so difficult with an economics as well?? English is even impossible to tell another way in Minister Cabinet room as : Minister James, You have to have do this and that to next week six o clock evening and rapport the results here at one hour later. At us, the minister would feel insulted by such too straightforwarded command, because more polite form would be to say, Minister James, do You agree the problem we met demands a certain solution somewhen. Please, try to find a some alternatives to discuss here collectively how to solve the problem by different manners and rapport to someone somewhen.... So, nothing had been never done. :) :)
 
Last edited:
Top