Thermal resistance of DO41 diodes

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I think a Board Ground Plane with 1.5" X 1.5" Copper Surface is better than 1/2"L x 0.031"D wire for radiating heat.
So the tombstone method appears to be best case into a ground plane, that nobody hardly creates. with 4 diodes side by side (worst example)

But the datasheet was not clear about case 1 & 2 so if you do not derate power by 50%, expect the PCB to look a little brown after a while and if no ground plane then I agree method 1 looks best.
 
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Hi,

no, not the leads atc as heatsink, it´s the copper on the PCB that acts as heatsink.

Datasheet mentiones: "P.C. Board with1−1/2 X 1−1/2 Copper Surface"

Klaus
 
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If its the copper on the PCB that is the heatsink, then a DO41 would be no lower RthJA than a SMA diode? (considering both are on "Minimal pads")
 

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cupoftea said:
If its the copper on the PCB that is the heatsink, then a DO41 would be no lower RthJA than a SMA diode?
Click to expand...
Recognize no logic behind this statement. RthJA is a sum of several terms, if one term is the same for both diodes, what does it tell about the other terms? In addition, DO41 case surfaces reduces total RthJA a bit.
 
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DO-41 RthJC = 42 'C/W
PCB RthCA = 8 'C/W for 15 cm^2 2oz, 2s
Total RthJA = 50 'C/W for method 3 Tombstone

This takes a very large board area and 4 diodes together would need 60 cm^2 area top and bottom of Cu is an expensive solution.
 
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