Intel Enabled Thermal Solutions; Thermal Solution Requirements - Intel Celeron D Thermal Design Manual

775-land lga package for embedded applications
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Intel Enabled Thermal Solutions

5.0
Intel Enabled Thermal Solutions
5.1

Thermal Solution Requirements

The thermal performance required for the heatsink is determined by calculating the
case-to-ambient thermal characterization parameter, Ψ
Metrology" on page
and compare different thermal solutions in similar boundary conditions. For the Celeron D, an
example of how Ψ
Equation 6. Case-to-Ambient Thermal Characterization Parameter
Ψ
In this calculation, T
datasheet.
Note: In this calculation, the T
ambient temperature (T
Table 4
shows an example of required thermal characterization parameters for the thermal solution
at various T
are subject to change, and in case of conflict, the specifications in the processor datasheet
supersede the T
Table 4.
Thermal Characterization Parameter at Various T
Intel
Frequency
2.93 GHz
Figure 8
further illustrates the required thermal characterization parameter for the Celeron D
Processor in the 775-land LGA package at various operating ambient temperatures. The thermal
solution design must have a Ψ
temperature.
Intel
30
18. This is a basic thermal engineering parameter that may be used to evaluate
is calculated is shown in
CA
o
T
(
C
)
T
(
=
C
max
LA
CA
TDP
(
W
)
and TDP are taken from the thermal profile specification in the processor
C max
and TDP are constant, while Ψ
C max
).
LA
s. This table uses the T
LA
and TDP specifications in this document.
C max
®
®
Celeron
D Processor 341
TDP
T
C MAX
84 W
67.7 °C
less than the values shown for the given local ambient
CA
®
®
Celeron
D Processor in the 775-Land LGA Package Thermal Design Guide
, as explained in
CA
Equation
6.
o
o
o
C
)
67
7 .
C
38
C
=
84
W
and TDP from the processor datasheet. These numbers
C max
LA
Required Ψ
(°C/W) of Thermal Solution at T
CA
44.2
40
0.280
0.330
Section 3.0, "Thermal
o
C
=
. 0
354
W
will vary according to the local
CA
Levels
LA
35
30
0.389
0.449
Order #303730
= (°C)

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