Re: [SI-LIST] : Heat sink and radiated emissions

From: Doug McKean ([email protected])
Date: Tue Jul 25 2000 - 14:06:56 PDT

Wayne Gibson wrote:
>
> Hello,
>
> I have an EMC oriented question. Currently, investigating the use of
> aluminum, ceramic and enhanced plastic heat sinks for possible use on a
> TBPA package. Does anyone have information regarding different heat
> sink material with respect to radiated emission levels? As well as heat
> dissipation for different material.

Here's something I wrote for another newsgroup ...

-------------------------------------
Heat Transfer from an Exposed Surface
to Ambient by Convection:

Q = h * A * DT

(originated by Sir Issac Newton)

h = Heat Transfer Coefficient [ W/m^2-C ]

Still Air, h = 23 to 28
Turbulant Air, h = 85 - 113

A = Area [ m^2 ]

DT = Temperature Differential [ C ]

-------------------------------------
From: International Thermoelectric, Inc.
131 Stedman Street
Chelmsford, MA 01824

-------------------------------------
Still Air: h = 23 to 28

-------------------------------------
h A DT Q
W/m^2-C m^2 C Watts
------- --- ---- -----
23 1 10 230
...
25 1 10 250
...
28 1 10 280

-------------------------------------
Turbulant Air: h = 85 to 113

-------------------------------------
h A DT Q
W/m^2-C m^2 C Watts
------- --- ---- -----

85 1 10 850
...
100 1 10 1,000
...
113 1 10 1,130

-------------------------------------
Example #1: (if needed)

A microprocessor with a surface area
1 square inch (6.45161 E-4 m^2) of
exposed material is speced out to
have a max surface temp of 80 C
absolute temp. The 1 sq. inch surface
area is what would otherwise be in
actual contact with a heatsink.
Heatsink is not used.

What is the maximum wattage it can
dissipate in still air?

Given: Q = h * A * DT (watts)

Substitute: h = 28 (best case still air)
A = 6.45161 E-4 m^2
DT = 80C - 20C = 60C

Compute: Q = 28 * 6.45161 E-4 * 60

-------------------------------------
Example #2: (if needed)

The microprocessor in example #1 above
will generate 35 Watts. Given the max
surface temp differential of 60C, what
is the minimum surface area required
of a heatsink?

Given: Q = h * A * DT (watts)

Solve: A = Q/(h * DT) (sq. meters)

Substitute: Q = 35 watts
h = 113 (best case turb air)
DT = 60C

Compute: A = 35/(113 * 60) (sq. meters)

A = 5.162 E-3 (sq. meters)

= 51.62 sq. centimeters

Answer: Or approx 8 sq. inches

-------------------------------------

Comment: Which appears to be just a
little on the low side for the surface
area (total fin surface area) of the
heatsink for my Pentium. But not much.

And which assumes all things considered
such as matched thermal resistances,
junction temp and surface temp diff
remains the same, and so on ...

Doug McKean
-------------------------------------

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