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Author Topic: Heatsink and fan design  (Read 1803 times)
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kiltro
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« on: February 25, 2023, 08:47:02 08:47 »

Hello everyone,

for a future project I'll be using two power mosfet, IRFP250
https://html.alldatasheet.com/html-pdf/22407/STMICROELECTRONICS/IRFP250/3247/2/IRFP250.html

They should disspiate 60W each... So if I'm not wrong heatsinking a mosfet would take

(Tj-Ta)/P - Rjc - Rcs

(100-50)/60 - 0.66 - 0.1 = 0.073C/W .... and I need to mount both on the same heatsink    Cry

Now, how do people more expert then me usually solve this?

I was thinking of mounting a big 12V fan on the heatsink but how can I estimate the new thermal resistance of the heatsink? Is there some formula for fans or similar?
« Last Edit: February 25, 2023, 09:33:50 09:33 by kiltro » Logged
Sideshow Bob
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« Reply #1 on: February 25, 2023, 10:12:05 10:12 »

This is a good write up.
https://www.re-innovation.co.uk/docs/heatsink-calculations/
I also used this online calculator assuming Case-sink resistance 0.8 C/W (common ordinary heat-paste)
http://mustcalculate.com/electronics/heatsink.php?tj=140&tamb=50&p=60&rjc=0.66&rcs=0.8
What is missing is something important. Will your setup be constant current or will it be switching?
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kiltro
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« Reply #2 on: February 25, 2023, 10:18:03 10:18 »

It's for an active current sink, 120W.

So even at max amp it will not have to stay on for hours, just to do some tests
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« Reply #3 on: February 25, 2023, 10:54:06 10:54 »

Is the voltage more or less constant? incandescent lamps may work well for dumping power. Not in all cases though
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kiltro
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« Reply #4 on: February 25, 2023, 11:00:27 11:00 »

You can choose if the sink works with constant current (set by a pot), any voltage you connect to the sink.
Or constant ohms, the current varies with the voltage you connect.

http://maxhobby.altervista.org/dummyload/SCHEMA_MAX_DUMMY_LOAD.pdf

Basically this is the schematic but I was going for 2 mosfet instead of 4
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h0nk
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« Reply #5 on: February 25, 2023, 12:26:40 12:26 »

Hello kiltro,

> for 2 mosfet instead of 4
Doing so, will effectively double the overall thermal resistance.
> 0.073C/W
Even good heatsinks with an attached fan are around 0.4 K/W:

https://de.elv.com/luefter-kuehlkoerper-lk-75-2-haelften-erforderlich-001762?fs=4001289229

I would recommend to use 4 FET's.
And You should check, that the current uses all paths of the circuit.


Best Regards
« Last Edit: February 25, 2023, 12:31:17 12:31 by h0nk » Logged
kiltro
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« Reply #6 on: February 25, 2023, 12:32:35 12:32 »

Hello kiltro,

> for 2 mosfet instead of 4
Doing so, will effectively double the overall thermal resistance.
> 0.073C/W
Even good heatsinks with an attached fan are around 0.4 K/W:

https://de.elv.com/luefter-kuehlkoerper-lk-75-2-haelften-erforderlich-001762?fs=4001289229

I would recommend to use 4 FET's.
And You should check, that the current uses all paths of the circuit.


Best Regards
Hello h0nk

I was thinking of using a single big heatsink, and make it one side of the enclosure, so that it disspitate to air directly.

Am I getting it right that you are suggesting of using one for each mosfet instead?
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h0nk
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« Reply #7 on: February 25, 2023, 02:08:23 14:08 »

Hello kiltro,

the thermal resistance of 0.4 K/W is achieved through the use of a fan.
The air flows through the heat sink. 0.4 K/W is an exceptional low thermal resistance.

> one side of the enclosure, so that it disspitate to air directly
"Directly" only means to use natural convection?

Even big heatsinks do not provide that thermal resistance to air without a fan.
The datasheet of You favorite heat sink should state a value for the natural convection.
I would not believe that they are much better than 2-3 K/W.

For the heat sink with a fan(!) per FET:
With 60 W per FET You get 0.66+0.4 = 1.1 K/W and You would reach approx. 90 C. (60 * 1.1 + 25)

The allowed power dissipation derates with 1.3 W/K so at 90 C the remaining allowed
power is: 180 - (66 * 1.3) = 95 W which is more than the 60 W.

When You use a heat sink with 2 - 3 K/W its obvious, that two FET's only will not keep it up.


> Am I getting it right that you are suggesting of using one for each mosfet instead?

I have not suggested this, but:
When You must mount the FET's isolated, it would be an option to mount the heat sinks isolated instead. This will keep the overall thermal resistance at the lowest level.


Best Regards
« Last Edit: February 25, 2023, 02:40:54 14:40 by h0nk » Logged
kiltro
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« Reply #8 on: February 25, 2023, 02:34:49 14:34 »

So for example, I was looking at this one

https://www.abl-heatsinks.co.uk/product/122ab

If I choose a length of 200mm it shows, 0,464 C/W with "forced" airflow.

Can I get the air flow value in m/s from a fan's datasheet? I usually see m3/min
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h0nk
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« Reply #9 on: February 25, 2023, 03:10:19 15:10 »


This is not exact science:

1 m3/min := 1000 l/min

1 l = 10 cm x 10 cm x 10 cm

So through a diameter of 10 cm x 10 cm it must move 1000 * 10 cm / min.
This giives 100 m/min or approx. 1.6 m/s.

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« Reply #10 on: March 04, 2023, 06:44:45 18:44 »

for sure you cannot dissipate 60w in a mosfet in to-247 package , the highest number i could achieve without fancy interface material is 30w per Fet for like 10 to 30 minutes continuously. if you assume ambient temperature is 40 degree and maximum junction temperature that you want to reach is 140 ( I know that maximum junction is 175 but you should leave some margin for transients). then the allowed temp rise is 100 degree . 100degree @ 60w means a thermal resistance of 1.6 K /w ( junction to ambient ) . we know from datasheet that junction to case is 0.66. Suppose interface material is silicon pad like this one (https://www.tme.eu/en/details/wk_247/heatsinks-equipment/fischer-elektronik/wk-247/) . Assuming best case senerio this leaves you with 1.66-0.66-0.45=0.55k/w for the heatsink, this is not cheap neither small. it is more cost effective to use more mosfets in parallel to share the load and reduce hotspots on the sink
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