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Equal speed = equal heat output?

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kevral

Registered
Joined
Sep 13, 2003
Been wondering about this:
If you have two cpus of the same kind of core; let's say an XP2500+ and an XP2800+ (both Bartons), and you run both on identical setups at identical speeds -- should they generate equal amounts of heat?
In other words, does the 2800 get as hot as a 2500 overclocked to 2800-speed?
 
If they are from the same revision, then theoretically they do. Sometimes a lower speed model might require more voltage to reach the speed though, and then it would produce more heat.

If you go a bit further, and overclock them both to xp3200 speeds, the xp2500 might theoretically even produce less heat!

I'd say that as long as you don't add voltage, the diferance will be pretty small which ever way it goes.
 
People have mentioned that getting a barton with a newer stepping (for example) has given them lower temps then with the older stepping. This is the reason I mentioned the possibility of the lower speced processor actually running cooler, as they could be newer while the xp2800 could be older.

The diferance people have been mentioning has been 1-2 degreese, so well within the possibility of error due to applying thermal paste diferrantly / having diferant ambient temps due to autoum coming.
 
cool. my mistake then.

I used to think that newer steppings only improved overclockability. Guess that's not always the case.
 
dropadrop said:
People have mentioned that getting a barton with a newer stepping (for example) has given them lower temps then with the older stepping. This is the reason I mentioned the possibility of the lower speced processor actually running cooler, as they could be newer while the xp2800 could be older.

The diferance people have been mentioning has been 1-2 degreese, so well within the possibility of error due to applying thermal paste diferrantly / having diferant ambient temps due to autoum coming.

At identical speeds and voltages, the heat outputs should be identical. AMD didn't document any changes between Barton steppings, and I'm leaning towards human error in cooling mounting as causing the minimal temp difference.
 
Got the following number from benchtest.com.

2500+
stock speed 1830
stock Vcore 1.65
stock wattage 68.30
OC speed 2080
OC Vcore 1.65
OC wattage 77.63

2800+
stock speed 2080
stock Vcore 1.65
stock wattage 68.3

I dunno, I find it wierd that a 2500+ and a 2800+ have the exact same numbers, so this could be wrong. Time to hit the white papers.
 
Checked the white papers. 2500's and 2800's do indeed have the same maximun and typical wattage, so the above numbers are correct.

Here's the link to the white papers. It's table 1, in the list of tables.
 
this is from my tests!


i have a 2600+ t-bread and a 2100+ t-bread

these test were on air.ambiant temps 70F

the 2600+ hit [email protected] with temps 45c idle and 51c under load!

the 2100+ hit [email protected] temps same as the 2600+ at 2.4gig, even though the 2100+ used less voltage, at the same speeds, the chips temps run around the same!


now in conjunction with a barton. i realy don't know. don't have one. i would say, same core desighn 2500+ and 2800+, than at equal speeds the temps should be close to even! voltage may vary, but temps should be relativley close!
 
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