**From:** Doug McKean (*dmckean@corp.auspex.com*)

**Date:** Tue Jul 25 2000 - 14:06:56 PDT

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Wayne Gibson wrote:

*>
*

*> Hello,
*

*>
*

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

Answer: Q approx 1 Watt

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

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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**Next message:**IL SEONG: "[SI-LIST] : Pls make Recommandation for IBIS model service"**Previous message:**Varshney, Poonam: "RE: [SI-LIST] : Assistance with HSPICE"**Next in thread:**Doug McKean: "Re: [SI-LIST] : Heat sink and radiated emissions"**Maybe reply:**Doug McKean: "Re: [SI-LIST] : Heat sink and radiated emissions"

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