Thermal resistance of DO41 diodes

[cupoftea]

Hi,
Page 4 of the following tells that the shorter the leads of the DO41, the lower is the thermal resistance to ambient....which sounds odd, because it was thought that the leads acted as heatsink, so the longer the better?

https://www.onsemi.com/pdf/datasheet/1n4001-d.pdf

 

[D.A.(Tony)Stewart]

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.

 

[KlausST]

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

 

[cupoftea]

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")

 

[FvM]

If its the copper on the PCB that is the heatsink, then a DO41 would be no lower RthJA than a SMA diode?

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.

 

[D.A.(Tony)Stewart]

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.