Originally posted by: taltamir
there is no way to tell based on mhz and voltage... there is no linear growth or some mathematical function to plug into, neither are all parts identical in their behaviour.
Get a power meter and measure it.
The dynamic power consumption is linear with frequency and squared with the voltage.
http://www.duke.edu/~kaf3/lowpower/slide4.html
Static power dissipation depends linearly on the voltage. (specifically see page 1 slide 4)
http://www.tkt.cs.tut.fi/kurssit/3516/chap11.pdf
Since for the OP's purpose we are talking about the same chip for both voltage and GHz scenarios this robustly enables us to assume some coefficients are the same.
Originally posted by: Katana
I have a Phenom 8450 undervolted and underclocked to 1600mhz@1.015v but noticed that with CnQ enabled the cpu speed would go down to 1000mhz but the v would go up to 1.250v. This is for a file server so raw speed isn't really needed and I'd like to consume as little energy as possible, so which would consume less; without CnQ and running constantly at 1600mhz@1.015v or with CnQ at 1000mhz@1.250v when idle(which it is at most of the time)?
btw-1.015v is the lowest I could go, anything lower and I would get errors in prime95
So we have two equations:
(A and C can be assumed to be the same here as the OP has identified the loadings on the system to be the same in both cases, i.e. mostly idle with minimal processing activity)
Power1 = A*C*(1.015^2)*1600 + 1.015*I = 1648.36*A*C + 1.015*I
Power2 = A*C*(1.250^2)*1000 + 1.250*I = 1562.50*A*C + 1.250*I
(And Power1 - Power2 = 85.86*A*C - 0.235*I)
Now we don't know what the leakage current is specifically, but what we can say is that the difference in the dynamic power consumption is 1648.36/1562.50 ~= 5.5% (that is the 1600MHz system has 5.5% more dynamic power consumption than the 1000MHz system when accounting for the voltage differences) while the static power consumption of 1.015/1.250 ~= 27% lower.
And with this we can make an if/then argument - in order for the total power consumption (static and dynamic) of the 1600mhz setup to be less than the total power consumption of the 1000Mhz setup then the static power losses on the OP's Phenom is must be >365.36*A*C...see footnote...or to put it in terms of dynamic power consumption we are saying the static power consumption must be (365.36*A*C)/(1562.50*A*C) >= 23.4% of the dynamic power losses of the 1000mhz system for the 1600mhz system to be the power conservative one.
The more the overall percentage of power consumption for Phenom is coming from static power losses from leakage in excess of 23% the more favorable it is to run the higher MHz lower Vcore setup of the 1600MHz variety shown here, but if static power losses of Phenom are not >23% then the OP will see lower power consumption (for the loosely defined usage patterns alluded to in the OP) by going with the lower MHz higher Vcore setup noted as 1000Mhz setup.
So the question is...for a 65nm tri-core Phenom 8450 is the static power consumption likely to be >23% or <23%?
If it is <23% then the OP is better off using the 1GHz/1.250V setup.
My bet is that it is <23%, in fact I'd be surprised if it was above 10%, meaning the power consumption really ought to favor the 1Ghz setup quite favorably.
footnote: This is arrived at by setting both power equations equal to one another and solving for I, that is to say solve the value of Ileakage that would give you identical power consumptions for these two setups, then rationalize what a lower or higher value of Ileakage would mean to the difference between Power1 and Power2. Standard solution process for two linear equations with three unknowns.
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