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Politics : Formerly About Advanced Micro Devices -- Ignore unavailable to you. Want to Upgrade?

To: Elmer who wrote (82270)	12/7/1999 9:35:00 PM
From: Mani1	Read Replies (2) \| Respond to of 1575856

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

To: Elmer who wrote (82270)	12/7/1999 9:43:00 PM
From: Bill Jackson	Read Replies (1) \| Respond to of 1575856

Elmer, I am not sure that is an error, they often use the words, Max, min and typical for these voltages. It may well be that at 1.6 volts the current will drop a little and it will indeed use 43 watts. After all if you apply ohms law with E=IR 1.7=28.2 R gives an R=0.06028 Ohms and then using 1.6=I0.06028 then I=26.54 Amps and then 26.54 1.6 = 42.6 watts ~43 watts. What railroad are you engineer for? Bill

To: Elmer who wrote (82270)	12/7/1999 11:21:00 PM
From: Goutam	Read Replies (1) \| Respond to of 1575856

Elmer, So using Ohms law, 28.2A1.7V= 47.94 or 48 Watts yet AMD claims only 43 Watts @1.6v in their power dissipation section.* First off - this is not Ohms Law. Ohm's law is about the relationship among the three fundamental electrical entities - Voltage, Current and resistance. Power = VI is just an equation for power which in turn comes from the basic definitions of units Volt and Ampere . Applying Ohms law (I=V/R) and the power equation(P=VxI), you can deduce other equations for power = V^2/R or I^2R in terms of "R", Anyways, this is a very valid equation, and it doesn't make any difference for the main point you are trying to make here - the Athlons max power spec is shown to be different from the Max power based on Vcc and Iccmax of the chip. To make this discussion simple, lets assume the Vcc and Icc max values you are looking at are DC values. In this case, Vcc * Icc for sure gives you the power supplied by the power supply. Now does it equal to the power dissipated on the chip? The answer would be - it depends. Let's also assume we have a device with one output pin. The conditions are Vcc = 5V, Icc=10Amps while the output is at high - say 2.4V - sourcing 5Amps into a load. From this we can say that the power supply is supplying 50Watts to the chip, but the chip is dissipating only 38 Watts while the load is consuming 12 Watts. In other words, if a chip has I/O lines, then you can't say conclusively that the power dissipated by that chip equals Vcc*Icc. Regards, Goutama

To: Elmer who wrote (82270)	12/7/1999 11:23:00 PM
From: Petz	Read Replies (1) \| Respond to of 1575856

Elmer, re:<why Max Icc times 1.7v does not equal maximum power dissipation> The nominal core voltage is 1.6v, not the 1.7v at which Max Icc is measured. AMD does not want you to burn out the power supply if you supply the CPU with 1.7 volts instead of the specified 1.6. (overclockers like to do this). If you use Ohm's law and assume that current rises linearly with voltage, you will find that any discrepancy is eliminated. If you look at an Intel spec sheet, they don't even tell you how much current the device draws or what maximum power dissipation is at the maximum Vcc, only at the nominal Vcc. AMD's approach to the spec is more conservative. Petz