[lclark2074]

There is no problem running at 133 instead of 166 but remember that you are 600mhz slower than before you started! This will help you save processor life but thy last so long this is a mute issue

[Kemon]

Very nice thread!! I was totally clueless as to what overclocking was, well aside from the fact it makes your computer faster, and now it makes total sense!! thanks alot!

[Swatman]

as soon as i get a good cooling system i will over clock w00t

[asusradeon]

awesome thanks m8

[T1Cybernetic]

Wow my brain hurts after reading through this lot great post though +1 wisdom, There is alot here i did not know.

[Verse914]

Great post! I always thought the opposite was true

Overvolting does not increase overclocking potential by giving our hardware "more juice" or "more fuel".

[Peepaw]

Very nice write up.

[burningcpu]

Nice, thanks for writing it, Feliz.

[catgoesmoo]

wow. cheers felinusz, excellent overview.

[RSDXzec]

nice guide, thanx

[ocping]

Originally Posted by felinusz

When we increase our voltage high value (overvolting), we force the signal/voltage to reach a higher voltage high, but in the same amount of time as before. We stretch out the ‘range of motion’ (the difference between VSS and VCORE), but we leave the transition time alone. The result is that it takes considerably less time for the signal to switch from VSS to a VCORE that is within transistor tolerance – this accommodates our faster switching frequency, and keeps our overclocked signal switching frequency strong (stable) and within transistor tolerance.

Sorry for the thread necromancy, but isn't there something wrong with this explanation?

The above passage is claiming that with a higher VCORE ("When we increase our voltage high value", voltage high has been definied as VCORE), the signal takes less time to switch from VSS to VCORE.

From what I have gathered by reading through, VSS to VCORE has been analogized to be a distance, so if the distance is increased between VSS to VCORE how would a higher VCORE result in a lower time allowed for the runner to reach VCORE from VSS within tolerance since the time allowed for the transition to occur is said to remain the same ("we leave the transition time alone")?

The only way I see this working out is if he meant that increasing VCORE allows the T (period for VSS to VCORE) to be lowered, forcefully increasing the speed of the runner (frequency of processor).

Any clarification appreciated.

[pokeatthedevil]

Two quick question about this:
Is there any way to increase the tolerance(e.g. 5% to 10%)?
If so, what stability issues would that cause?

[felix45]

thanks! I'll definitely be using this method!

[felinusz]

Originally Posted by ocping

Sorry for the thread necromancy, but isn't there something wrong with this explanation?

The above passage is claiming that with a higher VCORE ("When we increase our voltage high value", voltage high has been definied as VCORE), the signal takes less time to switch from VSS to VCORE.

Despite changes to the voltage high value, the time it takes the signal to switch from low to high does not change.

From what I have gathered by reading through, VSS to VCORE has been analogized to be a distance, so if the distance is increased between VSS to VCORE how would a higher VCORE result in a lower time allowed for the runner to reach VCORE from VSS within tolerance since the time allowed for the transition to occur is said to remain the same ("we leave the transition time alone")?

While the time that it takes for the signal to switch low to high remains the same, the signal is 'moving faster' than it was before the voltage high was increased.

Using the distance analogy, the runner is moving faster, at a greater meters per second rate than he was at a stock voltage high (or a stock 100 meter track). Running 110 metres in 10 seconds represents a faster transit than running 100 metres in 10 seconds, although the transit time is identical in both cases.

The only way I see this working out is if he meant that increasing VCORE allows the T (period for VSS to VCORE) to be lowered, forcefully increasing the speed of the runner (frequency of processor).

Any clarification appreciated.

The 'speed' of the signal is increased by raising our voltage high, and this allows the signal to potentially reach transistor tolerance faster than it did at a stock voltage high. Remember though, that transistor tolerance and VCORE/voltage high are not one and the same; while the signal will switch low to high in the same amount of time as before, with an increased voltage high it will reach transistor tolerance faster.

[rnldrc22]

really great guide, i'm more confident now on attempting the overclocking of my i7 thru overvolting, i'm going to be extra careful.

[devilDogbert]

Second thread necro, apologies.

This thread uses figures determined to be optimal for 2005 hardware, has technology and manufacturing changed enough in the past 6 years to suggest an update?

[Theocnoob]

No need. ITs so simplified now. Do not exceed safe volts or heat, you're good

[janas19]

AWESOMMMME. Thanks felinusz!

[Haider]

Impressive guide. Small question comes to mind I back in days of yore, I overclocked my Pentium 233 MHZ to 266 Mhz. In order to achieve that I ran vcore at a lower value than stock. I think it one increment lower on the motherboard. It was stable and produced less heat than when I was overclocked at stock vcore. What could account for that?

[Bobnova]

Voltage makes a much larger difference in power draw (and hence temperatures) than clocks do.