- #1
ramox3
- 12
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Hello I've been stuck with this for ever, can't find the relevant formulas
Given that the surface area of the first heat sink, S1= 500 cm2 = 0.05 m2
2nd heat sink = ?
The thermal resistance between p-n junction and case, RTjb = 0.6˚C/W
The thermal resistance between the case and any heat sink, RTbh = 1.2˚C/W
Power loss, P = 25W
The ambient air temperature, Ta= 20˚C
Heat transfer coefficient, α = 8 W/m2˚C
so how do I find out the area of the second heat sink?
Don't know how to start..
Homework Statement
Given that the surface area of the first heat sink, S1= 500 cm2 = 0.05 m2
2nd heat sink = ?
The thermal resistance between p-n junction and case, RTjb = 0.6˚C/W
The thermal resistance between the case and any heat sink, RTbh = 1.2˚C/W
Power loss, P = 25W
The ambient air temperature, Ta= 20˚C
Heat transfer coefficient, α = 8 W/m2˚C
Homework Equations
so how do I find out the area of the second heat sink?
The Attempt at a Solution
Don't know how to start..