Intel 632xESB, 631xESB manual C ΨSA = ΨCA - ΨCS = 3.23 - 0.35 = W, T C - T La, Ψca =, = 4.84 W

Models: 632xESB 631xESB

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C ΨSA = ΨCA – ΨCS = 3.23 – 0.35 = 2.88°---

Thermal Solution Requirements—Intel®6321ESB ICH

C ΨSA = ΨCA ΨCS = 3.23 – 0.35 = 2.88°---

W

If the local ambient temperature is relaxed to 45° C, the same calculation can be carried out to determine the new case-to-ambient thermal resistance:

ΨCA =

TC – TLA

 

105 – 45

C

-----TDP------------

=

----12.4----------

= 4.84°W---

It is evident from the above calculations that a reduction in the local ambient temperature has a significant effect on the case-to-ambient thermal resistance requirement. This effect can contribute to a more reasonable thermal solution including reduced cost, heat sink size, heat sink weight, and a lower system airflow rate.

Table 4 summarizes the thermal budget required to adequately cool the Intel® 6321ESB I/O Controller Hub in one configuration using a TDP of 12.4 W. Further calculations would need to be performed for different TDPs. Since the results are based on air data at sea level, a correction factor would be required to estimate the thermal performance at other altitudes.

Table 4. Required Heat Sink Thermal Performance (ΨCA)

Device

ΨCA (º C/W) at TLA = 45º C

ΨCA (º C/W) at TLA = 65º C

 

 

 

Intel® 6321ESB I/O Controller

4.84

3.23

Hub @ 12.4 W

 

 

 

 

 

Intel® 631xESB/632xESB I/O Controller Hub for Embedded Applications

February 2007TMDG 15

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Image 15
Intel 632xESB C ΨSA = ΨCA - ΨCS = 3.23 - 0.35 = W, T C - T La, Required Heat Sink Thermal Performance ΨCA, Ψca =, = 4.84 W