JedL0
Mechanical
- Mar 25, 2015
- 1
Hello,
I'm prototyping a circuit that involves several (6) lossy components mounted on a large heat sink (18"x7").
Most heat sink data sheets list R[sub]θ[/sub] in C/W/3 inch (0.8 C/W/3 inch). My understanding is that this calculation assumes a point heat source located in the center of a 3" length of heat sink at full width (so in this case, a 21 in[sup]2[/sup] area). I understand it also assumes a 75C temp gradient. I've found sufficient literature to correct this R[sub]θ[/sub] for heat sink length, temp gradient, and forced air convection (
My problem lies in generating thermal circuits for each of these components. I've calculated power dissipation for each component; I know the junction-to-case thermal resistance, the heat wink grease thermal resistance, and the footprint area. Heat sink grease R[sub]θ[/sub] is listed as C*in[sup]2[/sup]/W, which is great, as I can calculated the absolute R[sub]θ[/sub] [C/W] by dividing by footprint area. However, it doesn't make any sense to me to use the length-corrected, temp-correct R[sub]θ[/sub] value for the whole 18" heat sink (0.397 C/W). And it doesn't make sense to calculate a local R[sub]θ[/sub] [C*in[sup]2[/sup]/W], multiplying by area, and then dividing by footprint area as I did with the heat sink grease -- the local R[sub]θ[/sub] calculated this way is ludicrously large, and it also seems to contradict the length correction method of computing full-heatsink absolute thermal resistance.
So my question is: how do people calculate a local R[sub]θ[/sub] for heat sinks, when computing power loss calculations for several dissipating components on a large heat sink?
My hunch is that one must use some kind of reverse length correction chart, but then how could one reduce the width of the heat transfer area as well?
A separate, but related question, is how would one calculate a local R[sub]θ[/sub] on a liquid cooled cold plate? Lytron is the only cold plate/heat sink company I've seen present a local thermal resistance in C*in[sup]2[/sup]/W. They seem to think it can in fact be done by simply multiplying the absolute thermal resistance [C/W] by area: . But as I mentioned before, applying that method to heat sinks doesn't make any sense (ludicrously high local R[sub]θ[/sub], plus the inherent contradiction to the length correction method.)
Thanks to anyone who can provide any advice! I feel like I generally have a pretty good handle on thermal analysis, but this problem, seemingly so simple, has proved very frustrating.
Jed
I'm prototyping a circuit that involves several (6) lossy components mounted on a large heat sink (18"x7").
Most heat sink data sheets list R[sub]θ[/sub] in C/W/3 inch (0.8 C/W/3 inch). My understanding is that this calculation assumes a point heat source located in the center of a 3" length of heat sink at full width (so in this case, a 21 in[sup]2[/sup] area). I understand it also assumes a 75C temp gradient. I've found sufficient literature to correct this R[sub]θ[/sub] for heat sink length, temp gradient, and forced air convection (
My problem lies in generating thermal circuits for each of these components. I've calculated power dissipation for each component; I know the junction-to-case thermal resistance, the heat wink grease thermal resistance, and the footprint area. Heat sink grease R[sub]θ[/sub] is listed as C*in[sup]2[/sup]/W, which is great, as I can calculated the absolute R[sub]θ[/sub] [C/W] by dividing by footprint area. However, it doesn't make any sense to me to use the length-corrected, temp-correct R[sub]θ[/sub] value for the whole 18" heat sink (0.397 C/W). And it doesn't make sense to calculate a local R[sub]θ[/sub] [C*in[sup]2[/sup]/W], multiplying by area, and then dividing by footprint area as I did with the heat sink grease -- the local R[sub]θ[/sub] calculated this way is ludicrously large, and it also seems to contradict the length correction method of computing full-heatsink absolute thermal resistance.
So my question is: how do people calculate a local R[sub]θ[/sub] for heat sinks, when computing power loss calculations for several dissipating components on a large heat sink?
My hunch is that one must use some kind of reverse length correction chart, but then how could one reduce the width of the heat transfer area as well?
A separate, but related question, is how would one calculate a local R[sub]θ[/sub] on a liquid cooled cold plate? Lytron is the only cold plate/heat sink company I've seen present a local thermal resistance in C*in[sup]2[/sup]/W. They seem to think it can in fact be done by simply multiplying the absolute thermal resistance [C/W] by area: . But as I mentioned before, applying that method to heat sinks doesn't make any sense (ludicrously high local R[sub]θ[/sub], plus the inherent contradiction to the length correction method.)
Thanks to anyone who can provide any advice! I feel like I generally have a pretty good handle on thermal analysis, but this problem, seemingly so simple, has proved very frustrating.
Jed
