Finding the Surface Area of a Heat Sink: A Homework Guide

[ramox3]

Hello I've been stuck with this for ever, can't find the relevant formulas

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..

 

[berkeman]

Hello I've been stuck with this for ever, can't find the relevant formulas

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..

Welcome to the PF.

It looks like you have most of the numbers that you need, except you need some absolute limit on a temperature somewhere to finish the calculation of the heat sink total area.

Generally that absolute limit will be the highest pn junction temperature allowed. Are you given that?

If so, then just work out the numbers to figure out how big the area of the heat sink has to be to pull away the 25W...

 

[ramox3]

I wasn't given any other figures, I have managed to solve it this way, please correct me if I am wrong:


Q is the power dissipated by the device = 25W
TJ is the junction temperature in the device = 85˚C
TC is the temperature at its case=?
TH is the temperature where the heat sink is attached=?
TAMB is the ambient air temperature = 20˚C
RθJC is the device's thermal resistance from junction to case= 0.6˚C/W
RθCH1 is the thermal resistance from the case to the heat sink = 1.2˚C/W
RθHA1 is the thermal resistance of the 1st heat sink = ?
RθHA2 is the thermal resistance of the 2nd heat sink = ?

(b) If the surface are of the first heat sink is 500 cm2 = .05 m2

1
The thermal resistance between first heat sink and ambient, RθHA1 = ———
α × S2
1
= ———
8 × 0.05
= 2.5˚C/W



since the 2 heat sinks are in parallel :

RθHA2 = ( 1/ (TJ - TAMB) / Q) –( RθJC + RθCH1+ 1/(RθCH + RθHA1))
= (85-20/ 25) – (0.6+1.2+ (1/(2.5+1.2))
=1.23˚C/W


1
The thermal resistance between second heat sink and ambient, RθHA2 = ———
α × S2

1
1.23 = ————
8 × S2


S2=0.083 m2 = 830 cm2

How does this seem?