How can I estimate the thermal resistance of the thermal grease layer?

Microa

Enlightened
Joined
Mar 10, 2011
Messages
256
Location
Hong Kong
Hi CPF guys,

I am going to mount a 20mm star MCPCB to a 37x37x23mm heatsink. The thermal grease's heat conductivity is 6.0 W/m*K. Suppose I can apply the layer as thin as 0.4mm thick, can I estimate the thermal resistance of the layer?
 

Microa

Enlightened
Joined
Mar 10, 2011
Messages
256
Location
Hong Kong
Thanks for your valuable information. I estimated the thermal resistance this way. If I am wrong, please correct me.
Suppose the LED is an XML, the thermal pad area = 0.0028 x 0.0048 m = 0.00001344 sq. m.
R = t / K * A = 0.0004 m ( thermal grease thickness ) / 6 ( thermal grease's heat conductivity ) * 0.00001344 ( area )
= 4.96 K/W ( C/W )
 

IMSabbel

Enlightened
Joined
Dec 4, 2004
Messages
921
Hi CPF guys,

I am going to mount a 20mm star MCPCB to a 37x37x23mm heatsink. The thermal grease's heat conductivity is 6.0 W/m*K. Suppose I can apply the layer as thin as 0.4mm thick, can I estimate the thermal resistance of the layer?

You can not even just estimate it, you can calculate it exactly:

20mm star has about 3cm^2. So 6W/m*k multiplied by 3cm^2, divided by 0.4mm is the result you want.

BUT: 0.4mm grease thickness is beyond horrible. Ideally, you would only have a couple um left, with parts of the heatsink having direct contact (if its lapped). The uncertainty of the film thickness is by far the dominating source of errors in such estimations (people are very bad to guess small distances).
 

DIWdiver

Flashlight Enthusiast
Joined
Jan 27, 2010
Messages
2,725
Location
Connecticut, USA
You can not even just estimate it, you can calculate it exactly:

20mm star has about 3cm^2. So 6W/m*k multiplied by 3cm^2, divided by 0.4mm is the result you want.

BUT: 0.4mm grease thickness is beyond horrible. Ideally, you would only have a couple um left, with parts of the heatsink having direct contact (if its lapped). The uncertainty of the film thickness is by far the dominating source of errors in such estimations (people are very bad to guess small distances).

But you missed SemiMan's point that the heat does not spread uniformly throughout the star. Because the star is much thinner than it is wide, it will be hotter near the thermal pad than at the edges.

This assumes it is well attached to a heatsink. If it is sitting in air, the temperature will be much more uniform across the surface. If it is poorly attached to a heatsink, it will be somewhere in between.
 

Latest posts

Top