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Relation between heat output and wattage - Page 2

post #11 of 15
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Theres quite a lot of misunderstandings. I'm not talking about stock voltages and speeds - or TDP, I'm asking: "does one chip run hotter than another if HWMonitor shows the same wattage for the two chips?" be it overclocked or stock.

Duckieho understood what I meant. And gave a very good explanation, too. Thanks.
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post #12 of 15
Quote:
Originally Posted by ScurK View Post
Theres quite a lot of misunderstandings. I'm not talking about stock voltages and speeds - or TDP, I'm asking: "does one chip run hotter than another if HWMonitor shows the same wattage for the two chips?" be it overclocked or stock.
Let me take a stab at this. Assuming both cpu's have identical cooling and both consume 150W, then the cpu temp will depend on overall cpu efficiency.

If the cpu is 100% efficient, then the cpu temp should be same as ambient temp. Temperature above ambient means wasted energy which is being converted into heat.

In general, 32nm cpu will have the lowest temperatures (because they are more efficient and waste less energy). Next is 45nm, then 65nm and so on.

If both cpu's are 45nm, then it mainly depends on the cpu design. I would guess the i7 would be slightly more efficient because it's a newer architecture but that's only a guess. Overall, I guess temperatures would be close.
Edited by Partol - 3/2/11 at 2:18pm
     
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post #13 of 15
More thoughts on this ....

Basically, the cpu which is more efficient will have the lower cpu temperature at 150W input and identical cooling systems.

For example:
i7 920 (45nm) at 3GHz 1.4Vcore
Q9550 (45nm) at 3GHz 1.3Vcore

Even though the i7 920 should have better efficiency (because it's a new, higher performance architecture), since it runs on 0.1Vcore more than the Q9550, its possible that its overall efficiency is worse than the Q9550.

The i7 920 (3-channel memory) has an "unfair" advantage over the Q9550 (2-channel memory). The extra ram sticks on the i7 920 board consume power (and create heat) but that power is outside the cpu. It is not included in the 150W. There are other chips on the motherboard which produce heat, but they are not in the cpu, therefore, not included in the 150W.

In my opinion, a better way to calculate efficiency is to use total motherboard (including cpu) power consumption. Or maybe even use the total system power at the wall.

Another thing .... for applications which only use 1-4 threads, the i7 920 may become less efficient due to hyperthreading. In this case, turning off hyperthreading may increase the efficiency of the i7 920 (and thus reduce cpu temperature). disable some cores if less than 4 cores are used and this should improve efficiency (and reduce cpu temperature too).

As you see, the temperature will heavily depend on the bios settings, the architecture, and especially, the voltage settings. Did I forget anything?
Edited by Partol - 3/3/11 at 6:39am
     
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post #14 of 15
Quote:
Originally Posted by Partol View Post
As you see, the temperature will heavily depend on the bios settings, the architecture, and especially, the voltage settings. Did I forget anything?
Physical size of the chip
Transistors type used
Layout of the chip
Location of the sensors
Once again...
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Once again...
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post #15 of 15
I'll just post this, and leave it at that... it was in answer to sort of the reverse question...

Quote:
Originally Posted by purpleannex View Post
It's due to power leakage. Different cpu designs and manufacturing processes cause current to leak in tiny amounts from the circuits. This varies from chip to chip. The less leakage a chip has, the less power it requires to operate at a given frequency. However since it's leaking less power that power gets converted to heat, it has to go somewhere.

A cpu that leaks more power requires more current (because it's losing more, like pumping up a tyre with a puncture) to operate at the same frequency as the other cpu, and stays cooler because it's leaking the current and not converting it to heat.


...or something like that! LOL
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