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

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

 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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