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Politics : Formerly About Advanced Micro Devices
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To: Elmer who wrote (82270)12/7/1999 9:35:00 PM
From: Mani1  Read Replies (2) of 1575738
 
Elmer Re <<So using Ohms law, 28.2A*1.7V= 47.94 or 48 Watts yet AMD claims only 43 Watts @1.6v in their power dissipation section.

Maybe it was the same guy who came up with the Intel Performance benchmarks posted on AMD's website.>>

Elmer as I have stated the last time this issue was brought up power dissipation of a processor is not determined the way you just did. Neither AMD nor Intel does it this way.

This time I give you a more elaborate response, perhaps you can refrain from ridiculing me this time around.

Here is how its done:

When a manufacture says their chip dissipates XX.X Watts under "normal operating conditions", it means they determine that in test. They use an industry standard, ASTM approved "cut bar" method. They "operate" the chip under "normal conditions" and connect it to a sink. After the system reaches "steady state", they back track the heat flux using the delta T across a known material of known geometry placed right after (in series to) the sink. I used to do this test for electronic packages all the time.

As far as using the ohms law to determine the power dissipation, it just does not make sense. The current (electrical) peaks, are not nearly long enough to require a thermal design around them. Remember that the time constant for a electrical charge is several orders of magnitude shorter than time constant for thermal charge. A high current peak (even for a short time) requires a power supply capable of generating it. But a high thermal peak for a short period of time will simply get "absorbed" by the thermal capacitance of the heat spreader or even the die itself.

Agian, AMD's method are 100% industry standards.

Mani
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