Some of you might get a kick out of this. i5 2500k @ 5ghz stable

[Zulthus]

Originally Posted by Rbstr

Sure, but you're going to need calipers, yeah?

You and me ain't done, Zeke...

[NePaS]

Originally Posted by NumbaOneStunna

I am using Speedstep, the voltage never goes above 1.45v unless running an artificial load (Prime, IBT, Toast, Etc) Another words, under real world use I haven't seen it go above 1.45v.

Most of the time its between 1.4v to 1.45v while gaming depending on the game. Also I don't fold or anything so like 95% of the time its idleing at 0.975v

Get it validated then!

[NumbaOneStunna]

Originally Posted by NePaS

Get it validated then!

like this?

http://valid.canardpc.com/show_oc.php?id=2452296

[NePaS]

Originally Posted by NumbaOneStunna

like this?

http://valid.canardpc.com/show_oc.php?id=2452296

nice one,am surprised it let you seeing as the vcore reading is off and its an old version of cpuz

[NumbaOneStunna]

Originally Posted by NePaS

nice one,am surprised it let you seeing as the vcore reading is off and its an old version of cpuz

The Vcore is correct, as I said before speedstep is enabled.

VCore is scaled according to the demand of the CPU, in this case I was running Heroes Of Newerth so there was not much demand.

Also im not updating CPUZ, I like the ROG one that came with my motherboard, it has fancy colors.

[NePaS]

Originally Posted by NumbaOneStunna

The Vcore is correct, as I said before speedstep is enabled.

VCore is scaled according to the demand of the CPU, in this case I was running Heroes Of Newerth so there was not much demand.

Also im not updating CPUZ, I like the ROG one that came with my motherboard, it has fancy colors.

have you checked the ROG forums to see if it has been updated?I know that CPUID have said they will stop allowing validations from old versions soon(which means the big red X on the pic if you try)

Can I ask why you validated whilst not running prime or IBT?most people do that,as if there is a slight instability CPUZ will not validate.
O and the vcore is WRONG on that cpuz btw,that is the vcore for a stock [email protected] under load,not the vcore for the loaded turbo@5ghz.

[Ailos]

I'm just kind of curious what kind of PSU you've got and what is your power draw at this point?

I've been starting to mess my with Phenom II X4 955 BE and have stepped it up to 3.8 GHz (from the original 3.2) with no sweat, but I'm holding back on pushing further because my PSU is only a 560 W, and when I load the system up with benchmarks I hear that PSU's fan starting to earn its paycheck. I know I can push the system a bit further, but I don't want to brick it in the process, so atm, I'm thinking of just holding onto that $150 and just buying a whole new rig on Haswell's launch (in Feb as opposed this time next year).

[NumbaOneStunna]

Originally Posted by NePaS

have you checked the ROG forums to see if it has been updated?I know that CPUID have said they will stop allowing validations from old versions soon(which means the big red X on the pic if you try)

Can I ask why you validated whilst not running prime or IBT?most people do that,as if there is a slight instability CPUZ will not validate.
O and the vcore is WRONG on that cpuz btw,that is the vcore for a stock [email protected] under load,not the vcore for the loaded turbo@5ghz.

I dont know why I didnt validate while running prime or IBT, I have never validated anything before. (dont really see the point, except to prove something to haters.)

The vcore is correct, obviously you have never heard of speedstep.

Here is another, this time with IBT running. Notice the Vcore jumped because of the increased demand? http://valid.canardpc.com/show_oc.php?id=2452361

Originally Posted by Ailos

I'm just kind of curious what kind of PSU you've got and what is your power draw at this point?

I've been starting to mess my with Phenom II X4 955 BE and have stepped it up to 3.8 GHz (from the original 3.2) with no sweat, but I'm holding back on pushing further because my PSU is only a 560 W, and when I load the system up with benchmarks I hear that PSU's fan starting to earn its paycheck. I know I can push the system a bit further, but I don't want to brick it in the process, so atm, I'm thinking of just holding onto that $150 and just buying a whole new rig on Haswell's launch (in Feb as opposed this time next year).

My PSU is 750watt.

[Rbstr]

Originally Posted by Ailos

I'm just kind of curious what kind of PSU you've got and what is your power draw at this point?

The power dissipation scales with voltage squared all other things equal (P = (V^2) * 1/R in watts, volts and ohms respectively)...so if you double your voltage you're going to use four times as much power.

But it may actually go up more than that, because the higher temps mean less resistance in a semi conductor. I'm going to handwave that effect away as small (likely not that small...but whatevs) and also say that all other processor features behave the same .

For a ballpark, take your current voltage, divide by the "normal" voltage, square that number and then multiply your TDP by the result.
Tada, an "overclock adjusted" TDP...which should probably be close the the maximum consumption under normal loads.

This doesn't account for increased frequency. That, I think, would be a simple ratio multiplier. Divide new frequency by the old. Multiply TDP by that number.

In real life...it can't be anywhere near that simple...but I think it's a decent "worst case" ballpark.

(Wiki says P = C V^2 f , c is capacitance [farads] and f is frequency [hz]. God only knows how the capacitance scales. Maybe it's constant. In that case, my guesstimate is correct. Science, it works bitches.)

[Ailos]

Originally Posted by Rbstr

The power dissipation scales with voltage squared all other things equal (P = (V^2) * 1/R in watts, volts and ohms respectively)...so if you double your voltage you're going to use four times as much power.

But it may actually go up more than that, because the higher temps mean less resistance in a semi conductor. I'm going to handwave that effect away as small (likely not that small...but whatevs) and also say that all other processor features behave the same .

For a ballpark, take your current voltage, divide by the "normal" voltage, square that number and then multiply your TDP by the result.
Tada, an "overclock adjusted" TDP...which should probably be close the the maximum consumption under normal loads.

This doesn't account for increased frequency. That, I think, would be a simple ratio multiplier. Divide new frequency by the old. Multiply TDP by that number.

In real life...it can't be anywhere near that simple...but I think it's a decent "worst case" ballpark.

(Wiki says P = C V^2 f , c is capacitance [farads] and f is frequency [hz]. God only knows how the capacitance scales. Maybe it's constant. In that case, my guesstimate is correct. Science, it works bitches.)

I am not even going to attempt to follow that because that's a simplification that can be applied to one transistor, not billions. The background principles are sound, but decreased resistance effects are much more pronounced. I am not an electrical engineer (I'm an ME), but I do know enough about electronics engineering to know that my understanding of circuitry is not enough to make a reasonable calculation of what the new power consumption will be. And even then, most engineers in this situation will prefer to solve this problem empirically. That's our grown-up term for "I don't know what the fuck'll happen."

Also, that frequency equation from Wikipedia doesn't apply here: the frequency we're talking about is not of AC current, but how fast we're switching transistor gates, and that's not a sine wave.

[Rbstr]

Originally Posted by Ailos

Also, that frequency equation from Wikipedia doesn't apply here: the frequency we're talking about is not of AC current, but how fast we're switching transistor gates, and that's not a sine wave.

Well, I swiped it directly from the CPU Power consumption article on wiki. It's sourced from an Intel white paper on the Pentium M.
ftp://download.intel.com/design/netw...s/30117401.pdf

Frequency obviously has something to do with it. There is a finite amount of work involved when you cycle a transistor on and off...and that should be linear in frequency, like that power equation. And it has to be quadratic in voltage.

Here's what wiki says on CMOS switching (unfortunately I'm at home and lazy so I don't have better sources for you)

Dynamic Dissipation

Charging and discharging of load capacitances.

CMOS circuits dissipate power by charging the various load capacitances (mostly gate and wire capacitance, but also drain and some source capacitances) whenever they are switched. In one complete cycle of CMOS logic, current flows from VDD to the load capacitance to charge it and then flows from the charged load capacitance to ground during discharge. Therefore in one complete charge/discharge cycle, a total of Q=CLVDD is thus transferred from VDD to ground. Multiply by the switching frequency on the load capacitances to get the current used, and multiply by voltage again to get the characteristic switching power dissipated by a CMOS device: P = C V^2 f .

Since most gates do not operate/switch at every clock cycle, they are often accompanied by a factor \alpha, called the activity factor. Now, the dynamic power dissipation may be re-written as P = \alpha C V^2 f .

A clock in a system has an activity factor α=1, since it rises and falls every cycle. Most data has an activity factor of 0.5. If correct load capacitance is estimated on a node together with its activity factor, the dynamic power dissipation at that node can be calculated effectively.

Anyway, it seems that equation actually over estimates the increased power output of the Pentium M as it goes though its speed stepping in the paper (C isn't constant it changes between ~.0109 at low power to ~.0069 at high). So...it's in an order of magnitude. Fermi would be OK with that as a guess.

(I'm a ChemE, I do materials science. I'm pretty familiar with semi-conductors and IC fab technology, though I'm not an EE...I taught Engineering Phys 2 for a while, which was Electromagnetism and optics. So while I'm being handwavy and it is much more complicated I'm not just herp derping around here)

[Eyeklops]

Originally Posted by Rbstr

Well, I swiped it directly from the CPU Power consumption article on wiki. It's sourced from an Intel white paper on the Pentium M.
ftp://download.intel.com/design/netw...s/30117401.pdf

Frequency obviously has something to do with it. There is a finite amount of work involved when you cycle a transistor on and off...and that should be linear in frequency, like that power equation. And it has to be quadratic in voltage.

Here's what wiki says on CMOS switching (unfortunately I'm at home and lazy so I don't have better sources for you)


Anyway, it seems that equation actually over estimates the increased power output of the Pentium M as it goes though its speed stepping in the paper (C isn't constant it changes between ~.0109 at low power to ~.0069 at high). So...it's in an order of magnitude. Fermi would be OK with that as a guess.

(I'm a ChemE, I do materials science. I'm pretty familiar with semi-conductors and IC fab technology, though I'm not an EE...I taught Engineering Phys 2 for a while, which was Electromagnetism and optics. So while I'm being handwavy and it is much more complicated I'm not just herp derping around here)

LOL...thats why I just bought a Zalman ZM-MCF2 5-1/4 fan controller. It has a watt meter built in.