Electric bill vs overclocking/How long does it take your OS to boot up?

icedbullet · 2 Jul 2009 at 19:12

Hey everyone. I recently got into overclocking nad have learned a lot from reading through old threads on these forums. Man, theres some ppl who sure know their computers out there!

I was wondering if my recently overclocked q6600 G0 cpu is going to pull more electricity then the stock speed. Now I ask this because i have NOT touched the stock cpu voltage, the overclock was simply altering the fsb up to 333MHz, I havent touched the RAM voltage which is steady at 2.1v (Corsair Dominator CM2X2048-8500C5D running now at 1000 MHz)

As I have kept the stock voltages I generate no extra heat which is fantastic given the damnable heat wave thats upon us this week. Although I know I could push the cpu further I wont in the interest of saving power consumption. A 3GHz q6600 is fine for what i need at the moment.

Another question I have is about bootup time of peoples pc's. After stripping my startup program list bare using tuneup utilities my vista ultimate x64 is booting in 59 seconds to 1min 30 secs (it varies for some reason).

So if you want to state your OS type and bootup time I would be very interested to see how long other people have to wait around (windows 7RC boots up so fast, its one of the main reasons I want the OS, also has anyone seen that asrock motherboard that boots to windows instantly? Looks amazing)

JonJ678 · 2 Jul 2009 at 19:18

Even at the same voltage, higher fsb will use more power. Think of it as X watts per calculation, do the calculations faster and you need more power. not much more though, as the vast majority of the chips power consumption goes into heat. Clock it as far as it can on stock volts is normally quite good advice. You can also undervolt it, where it will be cooler, use less power, but be slower. My old q9550 ran at 2ghz, 1V, passively at 42 degrees load.

Boot up was around 35 seconds. Most of that posting and initialising the extra sata ports and so forth. I'm not even going to time the amd Im on at the moment, it'll make me too sad.

That was ubuntu, which is hardly lean. The xpress gate os managed around 20 seconds but you couldn't do much with it once it had loaded. Debian minimal was around 10 seconds after post. All numbers from memory and inevitably approximate, I don't turn computers off often.

Mattus · 2 Jul 2009 at 19:57

The chip's maximum power consumption is directly related to the clock frequency. A chip at 4Ghz will use about twice as much power as at 2Ghz.

Voltage has a bigger effect, as power consumption increases with the square of the voltage.

And my Q6600 system boots to Vista in 25-30 seconds (excluding the BIOS).

setter · 2 Jul 2009 at 20:00

Mine takes about a minute or so to boot into windows, would be quicker if i switched of the cpu fan warning, full screen logo and then my login password.

DragonQ · 2 Jul 2009 at 20:09

1) I don't pay the electricity bill so dunno.
2) ****ing ages.

.:SAmSuNG:. · 2 Jul 2009 at 20:16

Mattus said:
The chip's maximum power consumption is directly related to the clock frequency. A chip at 4Ghz will use about twice as much power as at 2Ghz.

Voltage has a bigger effect, as power consumption increases with the square of the voltage.

And my Q6600 system boots to Vista in 25-30 seconds (excluding the BIOS).

So even at the same Vcore it will use more?

JonJ678 · 2 Jul 2009 at 20:41

Mattus said:
The chip's maximum power consumption is directly related to the clock frequency. A chip at 4Ghz will use about twice as much power as at 2Ghz.

Could you hazard a guess at why / link to a source for this?

The second half you're taking from P=VI & V=IR, assuming constant R. No arguement.

I believe, based on reasoning rather than research, that processors do not work in this way. For power consumption P, power used for calculations at 1ghz A, power wasted in electrical heating etc B, efficiency of calculation C.

P=AC+B where C is probably a nasty function of A, but is not 1 nor 0 and B is certainly not zero.
At 2ghz, we have P=2AC+B
At 4ghz, P=4AC+B
Now, I don't doubt that the power used increases for the calculations increases with clock speed, and probably linearly. What I doubt is that power used for joule heating and other misc waste processes all do the same.

As long as the processor is not entirely efficient, which it isn't or it wouldn't get so damn hot, I do not think a 4ghz processor uses twice as much power as a 2ghz one, I think it uses negligibly more. I'm sure Ive seen this in my own computer, but can't replicate the results now

I would like to be told where I'm going wrong with this

fornowagain · 2 Jul 2009 at 20:45

Semiconductor power P, P=kLV²F

Where:
L = Load (from software)
V = Core voltage
F = Chip frequency
k = Empirical constant for the chip.
Pd = Total cpu power
TDP = Thermal Design Power (stock power)

Therefore delta P, i.e. change in power. The constants cancel, rearranged for total power.

Pd = TDP x (F2/F1) x (V2/V1)²

Or a little easier to understand

OCed Watts = TDP x (OCed Mhz / Default Mhz) x (OCed Vcore / Default Vcore)²

online version

Its actually a little more complicated in real life as this approximation takes no account of the leakage, i.e. power used with no load, this also scales. But its delta is tiny compared to the dynamic values above, so good enough.

JonJ678 · 2 Jul 2009 at 21:13

Legendary. Thank you

What sort of values does k take for todays chips? I take it the varience in k accounts for different operating temperatures at the same voltage, and reflects the slightly imprecise manufacuting approach of carefully dope then heat for a while.

L from 0 to 1, as leakage neglected? The first equation would have done, my maths isn't quite as shocking as my last post suggests.

Where did you learn this? I'm yet to find a text that covers solid state and electronics

Mattus · 2 Jul 2009 at 22:30

I didn't really know why - I just knew the second formula for CPU power consumption that fornowagain posted. Thanks for going into a bit more detail!

muon · 2 Jul 2009 at 23:28

The formula gives all the theoretical relationships but makes a big mistake.

TDP =! stock power. This will result in very large errors when calculating overclocked power consumption in a lot of cases.

user1453 · 2 Jul 2009 at 23:36

I was under the impression that TDP was maximum power at stock settings, far as I can see the formula involting TDP looks correct and the theoretical power consumtion formula looks loosely bassed on energy storage of a capacitor which would be quite logical as microprocessors are basied I belive on cmos

fornowagain · 3 Jul 2009 at 00:24

ghost101 said:
The formula gives all the theoretical relationships but makes a big mistake.

TDP =! stock power. This will result in very large errors when calculating overclocked power consumption in a lot of cases.

Well call it what you will, often TDP is the only value to go by, but yeah in reality TDP is just a generic value. Often means different things from different manufactures. The correct value is the stock wattage (if you know it) as I placed in brackets. Assuming reasonable control conditions I suppose you could, if you new the combined thermal resistance of a cooling system, work out the actual stock power from the delta temp with more accuracy.

user1453 said:
I was under the impression that TDP was maximum power at stock settings

It's more a loose value to guide Engineers designs for the thermal dissipation required. Also note all the power/temp specs they reference are external case e.g Tc. To be accurate, each chips actual power is different, as is the default Vcc tbh. Some TDP's are applied across a range of chips, use the same value and can't therefore be close to accurate. Or for instance the TDP envelope is set lower than the actual maximum power. But if you don't have anything its a place to start.

JonJ678 said:
Legendary. Thank you

What sort of values does k take for todays chips? I take it the varience in k accounts for different operating temperatures at the same voltage, and reflects the slightly imprecise manufacuting approach of carefully dope then heat for a while.

L from 0 to 1, as leakage neglecte The first equation would have done, my maths isn't quite as shocking as my last post suggests.

Where did you learn this? I'm yet to find a text that covers solid state and electronics

It's as old as the hills, bit like me. You don't see the derivation often though, you may see different symbols used of course. The constant k is determined to represent an average device, so combines constants from things like dope and node process et al. L is an activity factor, so depends on the percentage of Transistors getting busy. But yeah the leakage isn't constant, it also increases mostly with junction temperature iirc. Been awhile since I studied it. Try a paper search for Microarchitectural+Power+CPU+Static+Dynamic