Volts and Watts

[Danger30Q]

I thought this post would be appropriate here in WCG since most of us are concerned with wattage consumption of our crunchers.

I know the equation is (amps x volts = watts). I tried during some research to figure out how much wattage changes when you change the cpu volts but I realized it doesn't scale linearly.

Basically I'm curious to know how many more watts are consumed when increasing cpu volts from 1.35 to say 1.45 volts.

My Windows Home Server crunches 24/7 with a 65W Brisbane 3600+ X2. The stock volts I believe are 1.35V. Is there anyway to estimate the increase in wattage consumption from 1.35V to 1.45V?

How would this change when calculating it for a Core 2 Duo?

Thanks a ton!

[[XC] riptide]

I thought this post would be appropriate here in WCG since most of us are concerned with wattage consumption of our crunchers.

I know the equation is (amps x volts = watts). I tried during some research to figure out how much wattage changes when you change the cpu volts but I realized it doesn't scale linearly.

Basically I'm curious to know how many more watts are consumed when increasing cpu volts from 1.35 to say 1.45 volts.

My Windows Home Server crunches 24/7 with a 65W Brisbane 3600+ X2. The stock volts I believe are 1.35V. Is there anyway to estimate the increase in wattage consumption from 1.35V to 1.45V?

How would this change when calculating it for a Core 2 Duo?

Thanks a ton!

Right. I run a QX6700 on this rig. I'm gonna set its clocks back and change from 1.35V to 1.45V. Its always in a Kill=a=watt. BRB. Give me 2 minutes.

[[XC] riptide]

From 1.35-1.45 = 30Watts change. 100% loaded.
From 1.45-1.56 = 40Watts change. 100% loaded.


Now this is set via Windows app... not actual after droop.

[alexio]

Power draw increases +- exponentially with vcore. So: 1.45 / 1.35 = 1.074074....
1.074 ^2 = 1.154
The increase in power draw is 15.4%
Power draw increases a little worse than linearly with clockspeed.

Example: 65W TDP, 10% increase in voltage + 15% increase in clockspeed =
1.1^2 X 1.15 X 65W = +-90W.

It's a very rough calculation though....

[fornowagain]

Exponential? No, power comes from the square of the voltage.

I've simplified this and ignored static power leakage. But its pretty close in practical terms. A common expression for semiconductor power P, P=kLV²F

Where:
L = Load (from software)
V = Core voltage, default vcore V1 is VID.
F = Chip frequency
k = Empirical constant for the chip.
Pd = Total cpu power
TDP = Thermal Design Power

Therefore delta P, i.e. change in power Pd-TDP. Rearrange for Pd, the constants cancel.

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

[alexio]

Exponential? No, power comes from the square of the voltage.

Yeah, I used the wrong therm, but meant the same thing. It's almost 4.00 AM here now

[fornowagain]

Yeah, I used the wrong therm, but meant the same thing. It's almost 4.00 AM here now

Yeah, I saw your squared term after looking, 3.00AM here

[fornowagain]

From 1.35-1.45 = 30Watts change. 100% loaded.
From 1.45-1.56 = 40Watts change. 100% loaded.


Now this is set via Windows app... not actual after droop.

Also note that, that includes the PSU efficiency losses.

[alexio]

Also note that, that includes the PSU efficiency losses.

And don't forget CPU power circuity losses. Mofsets get warmer for a reason when the power draw of a CPU increases. It may just be a few watts normally, but it may be a bit higher when they are being pushed to the max.

[[XC] riptide]

Also note that, that includes the PSU efficiency losses.

I haven't. And I haven't forgot the extra heat load drop in efficiency, nor the PSU effeiciency... or the variable vdroop effect.

However, my simple measurements are real world for the purposes of the OT.

[Danger30Q]

Thanks a lot for the info guys. I'm thinking some others here would be interested in this as well. Some of you really know your physics calcs. I have a few more that you can hopefully answer.

Is there an easy way to factor in a very rough estimate of the total power losses from all circuitry and PSU efficiency?

Would the same formula work for an overclocked video card with increased voltage?

How do you calculate increased power consumption from adding volts to your overclocked RAM?

[[XC] riptide]

Thanks a lot for the info guys. I'm thinking some others here would be interested in this as well. Some of you really know your physics calcs. I have a few more that you can hopefully answer.

Is there an easy way to factor in a very rough estimate of the total power losses from all circuitry and PSU efficiency?

Would the same formula work for an overclocked video card with increased voltage?

How do you calculate increased power consumption from adding volts to your overclocked RAM?

Well lets see then..... BRB


2 x 512 Sticks Corsair XMS @ 395Mhz.

Set at 2.2 in Wondows. Change to 2.3 = 3W increase.


Change from 2.3 - 2.1 = 8W decrease. Hmm interesting. Full load of course running a VERY RAM intensive APP 900megs used.


I switched my X1900XT from OC'd 3d mode (vcore at 1.575 mem at 2.1 clocks at 756/820 core/mem) to 2d mode using atitool (vcore at 1.175, mem 2.1v , clocks at 500/600 core/mem). Kill-o-watt dropped by 25 Watts. Now GFX was not loaded at all.

[alexio]

Thanks a lot for the info guys. I'm thinking some others here would be interested in this as well. Some of you really know your physics calcs. I have a few more that you can hopefully answer.

Np

Is there an easy way to factor in a very rough estimate of the total power losses from all circuitry and PSU efficiency?

Efficiency of a good PSU at load should be +-80%. Efficiency is rather consisted on most PSUs across different loads unless you put a very small load on it or you max it out. In those cases it can drop below 70%. The power circuity for the CPU should have 75-90% efficiency unless you overstress it. It's hard to find more accurate numbers than these.

Example: If the efficiency of the PSU is 80% and the efficiency of the power circuity is 85% then the real efficiency (from wall outlet to CPU) would be:
0.80 X 0.85 X 100 = 68%

Would the same formula work for an overclocked video card with increased voltage?

For the GPU alone it would. The extra stress on the vRAM from the faster running GPU is neglectable. What you do have to keep in mind is that a GFXs power circuity isn't build to run at higher voltage in most cases. You may be pushing the components which can add quite a few watts.

How do you calculate increased power consumption from adding volts to your overclocked RAM?

Exactly the same as with the CPU and GPU. The temperature factor may become very important if you don't actively cool your RAM while running high volts trough them. This can significantly increase the power the chips consume.

[Danger30Q]

Thank you riptide for testing this out - I really should buy myself the kill-a-watt.

Is my math correct using a 6700 with stock of 2.66ghz and 1.35V stock and OC'd to 3.4ghz and 1.45V:

65 TDP x (3.4/2.66) x (1.45/1.35)^2 = 95 watt increase due to OC

The default watt consumption at stock everything is 65 TDP x 1.35 = 88W

[alexio]

Thank you riptide for testing this out - I really should buy myself the kill-a-watt.

Is my math correct using a 6700 with stock of 2.66ghz and 1.35V stock and OC'd to 3.4ghz and 1.45V:

65 TDP x (3.4/2.66) x (1.45/1.35)^2 = 95 watt increase due to OC

The default watt consumption at stock everything is 65 TDP x 1.35 = 88W

Default wattage is 65W (1.35V X 48A), no need to multiply it by the vcore, because that is included in the wattage already. The new power draw is +- 125W, so the increase is 60W. 65 x (3400/2667) x (1.45/1.35)^2 = 125W
I think you forgot the "^2" part.

[Danger30Q]

Now I'm confused with my math...I get the same as my old calculation:

(1.45/1.35)=1.07407 and 1.07407^2=1.1536

65 x 1.275 x 1.1536 = 95.6 W?????

[STEvil]

you're leaving out the mhz increase.

[Okda]

isn't this calculator is accurate enough ?

http://www.extreme.outervision.com/tools.jsp#cpuoc

[Mad_Man]

Now I'm confused with my math...I get the same as my old calculation:

(1.45/1.35)=1.07407 and 1.07407^2=1.1536

65 x 1.275 x 1.1536 = 95.6 W?????

one thing to remember, 65W tdp cpu doesnt use 65W, 95W tdp cpu doesnt use 95W at load. and so on.
If cpu is flagged as 95W tdp, it simply means it is under that. But above next lower tdp rating. you've got the point.
Actually, without knowing the stock full load power consumption, you can calculate only theoretical increase in percentage.
in other words, 1.47*65W is the same 47% increase of 1.47*125W.
but if your cpu uses 64W this increase of 47% will translate to 30W increase
and if your cpu uses 48W (the same 65W category) this same increase of 47% will yield in ~22W increase.
and the diff is even larger with higher tdp ratings.

[littleowl]

isn't this calculator is accurate enough ?

http://www.extreme.outervision.com/tools.jsp#cpuoc

Hey SaFrOuT Thanks!!!

[Mad_Man]

isn't this calculator is accurate enough ?

http://www.extreme.outervision.com/tools.jsp#cpuoc

hell no... imho :P
it says my 4000 brisbane has stock Vcore 1.35V and 2000Mhz
sad its 2100mhz and mine was 1.3V
and it says its 65W stock.
well, if you read my previous post you got the point...

[littleowl]

hum ok so I am stupid please explain all that in english!

[STEvil]

one thing to remember, 65W tdp cpu doesnt use 65W, 95W tdp cpu doesnt use 95W at load. and so on.
If cpu is flagged as 95W tdp, it simply means it is under that. But above next lower tdp rating. you've got the point.
Actually, without knowing the stock full load power consumption, you can calculate only theoretical increase in percentage.
in other words, 1.47*65W is the same 47% increase of 1.47*125W.
but if your cpu uses 64W this increase of 47% will translate to 30W increase
and if your cpu uses 48W (the same 65W category) this same increase of 47% will yield in ~22W increase.
and the diff is even larger with higher tdp ratings.

AMD rates that way, yes. Intel rates on average processor useage... so "65w" processors can actually put out more if you push them to full potential, which is why Intel laptops run so hot a lot of the time.

[Mad_Man]

hum ok so I am stupid please explain all that in english!

thats what im trying to do all the time lol... english is sometimes very hard, especialy if i try to explain smth more... educated rofl

AMD rates that way, yes. Intel rates on average processor useage... so "65w" processors can actually put out more if you push them to full potential, which is why Intel laptops run so hot a lot of the time.

my bad, i didnt know that.
maybe its long time since i had intel setup.
if i had spare $425 ill buy new Q and mobo

[littleowl]

thats what im trying to do all the time lol... english is sometimes very hard, especialy if i try to explain smth more... educated rofl

my bad, i didnt know that.
maybe its long time since i had intel setup.
if i had spare $425 ill buy new Q and mobo

now see that is just bull! I asked you to come here and crunch! and then you say I am stupid...... OH wait I said that! Thanks that gave me a nice lol at 515 am!