How do I calculate the heat sink value for this MOSFET design?

AI Thread Summary
To calculate the heat sink value for the specified MOSFET, the formula (Tmax - Tamb)/Pd is a starting point, with Tmax at 150°C and ambient temperature at 40°C, leading to a calculated value of 0.4°C/W. However, this calculation does not account for the thermal resistances from junction to case (Rθ_JC), case to sink (Rθ_CS), and sink to ambient (Rθ_CA), which are crucial for accurate thermal management. It's important to determine the actual power losses during operation, as the device will not handle the maximum rated power continuously. For effective heat sinking, consider using commercially available options designed for the TO-247 package, which can provide better thermal performance than custom designs. Understanding thermal dynamics is essential, and consulting with local vendors or using thermal simulation software can aid in achieving optimal results.
sayeen
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Hi, i want to design a heat sink for following mosfet

http://au.element14.com/vishay-formerly-i-r/irfp360pbf/mosfet-n-400v-23a-to-247ac/dp/8649359

its technical data sheet is provided here

http://www.irf.com/product-info/datasheets/data/irfp360.pdf

i am not sure how to design it...by design i mean just the mathematical value for C/W not schematic.

i read somewhere that the formula for finding the value for heat sink is (Tmax - Tamb)/Pd

the ambient temperature is 40C

from my calculation i found the heat sink value to be (150-40)/(280) = 0.4 C/W...is this correct...i located the value for Tmax and Pd (power dessipated) from the datasheet. Is this correct? please any help is appreciated.
 
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sayeen said:
Hi, i want to design a heat sink for following mosfet

http://au.element14.com/vishay-formerly-i-r/irfp360pbf/mosfet-n-400v-23a-to-247ac/dp/8649359

its technical data sheet is provided here

http://www.irf.com/product-info/datasheets/data/irfp360.pdf

i am not sure how to design it...by design i mean just the mathematical value for C/W not schematic.

i read somewhere that the formula for finding the value for heat sink is (Tmax - Tamb)/Pd

the ambient temperature is 40C

from my calculation i found the heat sink value to be (150-40)/(280) = 0.4 C/W...is this correct...i located the value for Tmax and Pd (power dessipated) from the datasheet. Is this correct? please any help is appreciated.

I did a Google Images search on the package and added the word heatsink, and got lots of good images. Maybe follow some of the picture links to get to websites with more information on heatsinks for that device package:

http://www.google.com/search?hl=en&...source=og&sa=N&tab=wi&ei=i5Z0UKyfK8mUiAKKg4FA

.
 
Thanks, but i am looking for the ratings of the heat sink first...i want to confirm if 0.4C/W rating for the heat sink is correct.
 
sayeen said:
Thanks, but i am looking for the ratings of the heat sink first...i want to confirm if 0.4C/W rating for the heat sink is correct.

No, you are not approaching this correctly. Look at the datasheet where it lists the Rθ_JC and Rθ_CS. What do they add up to? And that doesn't include the heat sink's Rθ_CA (case to air). And you are not going to run that part at 280W, if that's what you are implying in your previous post.

What real power level are you planning on running this FET at? What is the application? Do you have forced air available to blow over the heatsink?

You should google Heatsink Tutorial, and you will get some very good hits, including ones at heatsink manufacturers and also at FEA software vendors like COMSOL. To "design" this heatsink, you can either look at what is available for that package already (and see what their values of Rθ_CA are, or you can use COMSOL or ANSYS or some other FEA software package to do the thermal simulations...
 
Hello Sayeen:

The Basic formula you are looking for is Tj = T Ambient + (Rth(total) ( Rth(j-c) + Rth(c-s) + Rth(s-a) ) * Watts Loss) Sorry hard to show in text. th-theta ... Rth thermal resistance, j=junction, c=case, s=sink, a=ambient.

So you can rearrange to get an Rth needed for the Heatsink. Your basic formula is the right idea not not complete & you come up with Rth of 0.4 Deg C/W needed. You are not including the Heatsink - so the device's rating of 0.45 C/W ( Rth(j-c)+Rth(c-s)) is really the result of your calculation.

Berkman is correct - you will NEVER get 240W loss "out of" this device - that data point is at Tcase of 25C ( a standard measurement point for comparing devices - but essentially never a real applicable rating) . However if your converter is 240W and you are 95% efficient, then you have 12W Loss Possible - so you NEED to know what you LOSSES will be in Watts.

TO make an effective heatsink will not be as easy as it seems. I would first go to Newark's website This is a TO-247 Case - and search for TO-247 and HeatSink - they have a bunch for Wakefield ( Wakefiled also has a lot of good info on their site) - for Natural Convection ( no fan) these run 3-15 C/W -- and if you notice these ratings are often expressed as Temp C @ 6W ( I take that to mean that is a typical level of heat these will see w/out a fan is in the 6W range !) With a Fan N you get a 3 to 10X improvement - depending on the design. (I know you may want to do this yourself - but a real HS at $5 or less will be very hard to beat and less than the cost of the MOSFET)

So try to estimate your Losses first - but note an accurate number will not be available until you use a REAL Heatsink Rth. Once a HS concept or ballpark is determined you will need to go back over the calculations - all of the values are interrelated ( The higher the temp the higher the losses - I always start with Tjmaz 125C ( unless a newer 175C rated device - then 150C). - I refer to this as my design limit. Estimate losses, Tambient, NAtual, Fan or Liquid cooled - pick a "typical" heatsink and look at the result. You can "add silicon" as we say reducing losses with bigger, better or more devices, or add cooling with bigger better heatsinking.
 
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Thanks for the info Berkman and Windact...Thermodynamics is not really the field of my study so i got lost there a bit...but you cleared it up..i will try to follow your advice...I will try to calculate Rth for sink by following the formula you have provided...I understand i can ask any local vendor regarding the info for the heat sink for a particular MOSFET...but I have made it a point to do it all by myself! Thanks again
 
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