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Wattage increase with OC?
- Thread starter Nearsite
- Start date
BillParrish
Supreme [H]ardness
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- Aug 25, 2006
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yes, however I cannot tell you by how much, I could not find a power vs cpu frequecny curve in the AMD documentaion.
To clarify, under load the cpu will both pull more current from the motherboard/powersupply and dissipate more heat. Same at idle but that is not a concern as it will pull less at 2.5 and idle than it would at full load at 2.3.
When OCing it is important to consider proper cooling of both the cpu and motherboard cpu voltage regulation components to ensure long life and reliability. It is not difficult, just needs to be done. Modest OCing usualy requires only a better heatsink and good case airflow to be very successful.
To clarify, under load the cpu will both pull more current from the motherboard/powersupply and dissipate more heat. Same at idle but that is not a concern as it will pull less at 2.5 and idle than it would at full load at 2.3.
When OCing it is important to consider proper cooling of both the cpu and motherboard cpu voltage regulation components to ensure long life and reliability. It is not difficult, just needs to be done. Modest OCing usualy requires only a better heatsink and good case airflow to be very successful.
Thanks for the info Bill! The reason why I ask is becuase I'm building an HTPC with a 2350BE at 2.3GHZ and thought I'd bump up the power a little bit, but I didnt' want to increase the power usage, so I guess I'll just leave it at stock.
Another question for you, how did you get your E6300 to 3.3ghz. I can only get mine up to 3.0 (7 x 429mhz). I have the same speed RAM as you PC6400, but it seems like my RAM craps out on anything above that. I'm using Patriot value RAM. Do you think that's what's messing me up? CPU V = 1.2 / RAM V = 2.1
Nenu
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When you change the voltage of any component, its easy to work out how much extra power it will use.
(note that increasing the frequency of your CPU / memory / memory controller etc.. reduces the effective resistance which also increases power consumption, this isnt considered in the calculation as we dont have the frequency power curve data)
Power is proportional to Voltage squared and for our means is expressed as
P = Vsquared / R
Where R is the resistance of the device.
The resistance is considered constant if the frequency remains the same so we can now find out how much extra power is used by a higher voltage.
Assuming R = 1 for ease (it doesnt matter what value this is, the final difference in ratios is what matters)
Therefore:
Power = V squared.
So if you take the original voltage of the device and square it you will get P1.
Take the new increased voltage you wish to use, square it and you will have P2
Divide P2 by P1 and you have the ratio of how much extra power is being used.
% Overall power = P2/P1 x 100%
(this will give a result like 130% overall)
% Extra Power used = (P2/P1 - 1) x 100
(this will give a result like 30% extra)
(note that increasing the frequency of your CPU / memory / memory controller etc.. reduces the effective resistance which also increases power consumption, this isnt considered in the calculation as we dont have the frequency power curve data)
Power is proportional to Voltage squared and for our means is expressed as
P = Vsquared / R
Where R is the resistance of the device.
The resistance is considered constant if the frequency remains the same so we can now find out how much extra power is used by a higher voltage.
Assuming R = 1 for ease (it doesnt matter what value this is, the final difference in ratios is what matters)
Therefore:
Power = V squared.
So if you take the original voltage of the device and square it you will get P1.
Take the new increased voltage you wish to use, square it and you will have P2
Divide P2 by P1 and you have the ratio of how much extra power is being used.
% Overall power = P2/P1 x 100%
(this will give a result like 130% overall)
% Extra Power used = (P2/P1 - 1) x 100
(this will give a result like 30% extra)
CallMeIshmael
Limp Gawd
- Joined
- Aug 20, 2003
- Messages
- 174
To simplify Nenu's post:
Power dissipation increases linearly with increased frequency.
Power dissipation increases exponentially (square) with increased voltage.
Increasing frequency by 10% (without increasing voltage) increases thermal output by ~10% (1 x 1.1 = 1.1);
Increasing voltage by 10% (without increasing frequency) increases thermal output by ~21% (1.1 x 1.1 = 1.21);
Increasing frequency and voltage both by 10% increases thermal output by ~33% (1.1 x 1.1 x 1.1 = 1.331).
IMPORTANT DISCLAIMER: the preceding formulas are rough approximations, not an exact calculation. The actual calculations contain many more factors than I use here.
Power dissipation increases linearly with increased frequency.
Power dissipation increases exponentially (square) with increased voltage.
Increasing frequency by 10% (without increasing voltage) increases thermal output by ~10% (1 x 1.1 = 1.1);
Increasing voltage by 10% (without increasing frequency) increases thermal output by ~21% (1.1 x 1.1 = 1.21);
Increasing frequency and voltage both by 10% increases thermal output by ~33% (1.1 x 1.1 x 1.1 = 1.331).
IMPORTANT DISCLAIMER: the preceding formulas are rough approximations, not an exact calculation. The actual calculations contain many more factors than I use here.
Alright guys, are you all effin with me?!?!?! I have a bachelors degree in Social Science and math is not one of it's disciplines. Can someone tell me in laymen's terms what the hell all those calculations mean? I know if I spent like half an hour really thinking about them, I could eventually figure it out but can someone simply tell me how much more wattage a 2.3 ghz 45 watt CPU will use going to 2.5 ghz
BillParrish
Supreme [H]ardness
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- Aug 25, 2006
- Messages
- 7,519
Nearsight - about the ram, I think you may be right. My ram was rated 4 4 4 12 @800 MHz and as you can see from my sig I had to loosen the timings to get it to run that fast. I actually think the ram has a little more but my board becomes unstable around 485 MHz so I just backed off some. I am guessing your value ram starts off at 5 5 5 15 or 5 5 5 18. You could try CL6 settings but I dont know if it would be worth it (increase in cpu speed offset the memory penalty). I also bought my cpu the day they became available and back then there where not nearly as many versions and I beleive they where not speed sorting the raw dies as hard as they do now. I think I got a "good one" before Intel implemented process and manufacturing changes to "refine" the models. (pure speculation on my part but I did a stint in telecom manufacturing and usually thats how it goes, overbuild and get the new ones out of the door on time, then go back and squeeze every dime out of them you can on the factory floor).
If true (Power dissipation increases linearly with increased frequency.) for your cpu, again I cannot find sufficient detail in the AMD docs/specs to go on but Intel chips exhibit linear power vs frequency in their operating range and that AMD would be very strange if it acted much differently.
If it is putting out 45W at 2.3 GHz that is 45 / 2.3 = 19.56 watts per GHz
so if the target is 2.5GHz the power dissipation might be 19.56 x 2.5 = 48.9 call it 50w as engineers like to round up. As you can see this is not a huge increase and if we take the 50W number it represents a (50 - 45) / 45 x 100 = 11% increase.
Of course this would only happen if the cpu was under 100% load all the time. Sleep states etc. would have a big impact on power usage and I have not kept up with what AMD has done recently with that. eh, you can always OC it later if you need to.
A lot of assumptions I am not sure about in the above but I think its close enough of an estimate for the OP intentions.
If true (Power dissipation increases linearly with increased frequency.) for your cpu, again I cannot find sufficient detail in the AMD docs/specs to go on but Intel chips exhibit linear power vs frequency in their operating range and that AMD would be very strange if it acted much differently.
If it is putting out 45W at 2.3 GHz that is 45 / 2.3 = 19.56 watts per GHz
so if the target is 2.5GHz the power dissipation might be 19.56 x 2.5 = 48.9 call it 50w as engineers like to round up. As you can see this is not a huge increase and if we take the 50W number it represents a (50 - 45) / 45 x 100 = 11% increase.
Of course this would only happen if the cpu was under 100% load all the time. Sleep states etc. would have a big impact on power usage and I have not kept up with what AMD has done recently with that. eh, you can always OC it later if you need to.
A lot of assumptions I am not sure about in the above but I think its close enough of an estimate for the OP intentions.
CallMeIshmael
Limp Gawd
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- Aug 20, 2003
- Messages
- 174
Nearsight: Apologies if I confused you; I'm a technician by trade and sometimes forget not everyone works daily with numbers and formulas.
BillParrish's calculations are good. Again, these are only approximations, but they're close enough for what you asked. Keep in mind that that ~50 watt figure only applies if you left your CPU core voltage stock. Voltage increases ramp heat up faster than clock increases.
BillParrish's calculations are good. Again, these are only approximations, but they're close enough for what you asked. Keep in mind that that ~50 watt figure only applies if you left your CPU core voltage stock. Voltage increases ramp heat up faster than clock increases.