I am looking for an estimate on the relationship between the rate of increase of power usage as the frequency of the processor is increased.
Any references to findings on this would be helpful.
$\begingroup$
$\endgroup$
3
-
5$\begingroup$ This question appears to be off-topic because it is about power consumption of electronic components and not physics $\endgroup$– JimJul 21, 2014 at 14:27
-
5$\begingroup$ I'm voting to close this question as off-topic because it has nothing to do with physics; rather, it deals with power consumption of electronic devices which is off-topic for this site. $\endgroup$– YashasApr 22, 2017 at 8:20
-
7$\begingroup$ This is applied physics, isn't it? The site isn't called theoretical-physics.stackexchange.com, is it? $\endgroup$– Gerold BroserApr 23, 2017 at 19:11
Add a comment
|
5 Answers
$\begingroup$
$\endgroup$
1
Power consumption is about linear with frequency.
The processor contains millions of complementary FETs as shown. When the input goes low the small capacitance gets charged and it will hold a small amount of energy. A same amount is lost during the charging. When the input goes high again the charge will be drained to ground and be lost. So with each level change $n$ Joules are lost. If the frequency is 1 MHz then this switching occurs 10$^{6}$ per second, and $n$ 10$^{6}$ Joules will be lost per second. If the frequency is 1 GHz that loss will be $n$ 10$^{9}$ Joules.
Notice that the energy in a capacitor is $\frac{C \cdot V^2}{2}$, so the dissipation varies quadratic with voltage; running the processor at half the voltage will reduce power with 75 %.
This leads to the equation tuğrul also mentions:
$ P = c \cdot V^2 \cdot f + P_S $
where $c$ is a scaling constant, with the dimension of capacitance (F). $P_S$ is the static power dissipation Martin refers to, which is the power at a zero clock frequency.
-
1$\begingroup$ Note that modern CPUs dynamically scale their voltage along with frequency, running just over the minimum voltage necessary for correct operation at a given clock speed. That's approximately linear, so approximately $P ~ f^3$. (realworldtech.com/near-threshold-voltage discusses pushing voltages down so low that errors become more likely. See also realworldtech.com/power-delivery) Those aren't aren't explaining the f^3 scaling over normal voltage rages, but see Wandering Logic's answer on this question. $\endgroup$ Sep 24, 2021 at 1:44
$\begingroup$
$\endgroup$
2
For a given circuit in a given technology, power increases at a rate proportional to $f^3$ or worse. You can see by looking at the graph in @Martin Thompson's answer that power is superlinear in frequency.
$P=c V^2 f + P_S$ is correct, but only superficially so because $f$ and $P_S$ are functions of $V$ and $V_{th}$ (the threshold voltage.) In practice voltage, threshold voltage, and frequency are always changed together. Given a chosen voltage there is a maximum frequency at which you can run your circuit. Running any faster will result in bad data. But you would never set the frequency much below the maximum frequency for a chosen voltage because then you are just wasting power.
Let's ignore the leakage (static) power and just focus on the dynamic power, $cV^2f$. By the alpha approximation $$ f \propto \frac{(V-V_{th})^\alpha}{V}. $$
Here $\alpha$ is a technology-dependent constant that accounts for velocity saturation. $\alpha$ would be 2 for no velocity saturation (for example in 1000nm technology and older), and approaches 1 with complete velocity saturation. In 250nm technology it was somewhere between 1.3 and 1.5. In 45nm it might be somewhere between 1.1 and 1.4.
Before 1995 one might assume that $\alpha$ was 2 and that $V \gg V_{th}$, in which case $f \propto V$ so $P \propto f^3$. But in 2013 technology (45nm and below) not only is $\alpha$ more like 1.3 than 2, but $V$ is now only slightly larger than $V_{th}$.
Further the static power $P_S \propto e^{(-V_{th}/V_o)}V$, which means that choosing voltage, threshold voltage and frequency is now a nonlinear constrained optimization problem. (Given a fixed maximum power optimize for the highest achievable frequency or given a fixed required frequency optimize for the minimum power.)
Here are three very good papers that discuss the optimization procedures and their consequences:
This paper clearly shows the nonlinear increase of power consumption with an increase of frequency:
-
$\begingroup$ This answer could be more persuasive if the research you cited wasn't conducted (now) 13-20 years ago. $\endgroup$– ColumboMay 19, 2018 at 19:05
-
1$\begingroup$ @Columbo: Lots of research that we still trust is thousands of years old, though. The papers being old, does not mean that the phenomena they describe are not relevant to modern processes. That said, I haven't read the papers, but I think it seems like Wandering Logic knows what he's talking about. I could be wrong though. $\endgroup$– nijoakimJan 10, 2020 at 16:11
$\begingroup$
$\endgroup$
8
To add to the "linear with frequency" point, there is also an additional factor. As that "dynamic power" increases, the temperature of the die will increase and this will also increase the leakage current through the millions of transistors, which will cause more dissipation (termed "static power")
There's a long Anandtech thread taking lots of values and pulling them apart into their static and dynamic contributions which results in the following graph:
The slight kick up in static power at higher clock speeds is (as I understand it) as a result of the higher die-temperature.
-
2$\begingroup$ That's why I said linear, and not proportional :-). $\endgroup$– stevenvhAug 23, 2012 at 12:28
-
1$\begingroup$ @stevenvh: All things being equal, it's not quite linear (although I note you did say "about linear" ) as the static power changes with temperature and hence total power (or am I missing your point)? $\endgroup$ Aug 23, 2012 at 13:10
-
$\begingroup$ No, you're quite right, but the power due to the leakage current is only a small fraction of dynamic power. Even though it rises exponentially with temperature total power at 85 °C will be a few orders of magnitude larger, so the static power will bend the curve only slightly. $\endgroup$– stevenvhAug 23, 2012 at 13:20
-
$\begingroup$ Is it linear to the left? Normal frequencies are likely right on the edge of the linear regime (if they weren't, they would just increase the frequency to get performance increase at linear power costs) $\endgroup$– YrogirgAug 23, 2012 at 13:45
-
$\begingroup$ I'm not sure about the accuracy of the graph, (and the reference bottom right doesn't seem to give me much reason to :-)). The graph says dynamic power would be around 10 W at 1.6 GHz, and 100 W at 4.8 GHz? That's a factor 10, where we would expect a factor 3. Also, if it's linear below 1.6 GHz it would consume 10 mW at 1.6 MHz, which is extremely low for a processor with around a billion transistors, like the i7. $\endgroup$– stevenvhAug 23, 2012 at 13:58
$\begingroup$
$\endgroup$
3
For example, the Pentium 4 2.8 GHz has 68.4 W typical thermal power and 85 W maximum thermal power. When the CPU is idle, it will draw far less than the typical thermal power. The power consumed by a CPU, is approximately proportional to CPU frequency, and to the square of the CPU voltage: $P=CV^2f$
Taken from here: http://en.wikipedia.org/wiki/CPU_power_dissipation
answered Aug 23, 2012 at 9:50
-
1$\begingroup$ This site supports TeX code, e.g. $P=CV^2f$ by using dollar signs left and right. You don't have to inport a picture of a formula. :) $\endgroup$ Aug 23, 2012 at 10:14
-
$\begingroup$ Testing: $E=mc^2$ and $-((h^2)/2m)$ $\endgroup$ Aug 23, 2012 at 10:19
-
$\begingroup$ @tuğrulbüyükışık : you can also use \frac{h^{2}}{2m} in dollar signs to get nice fractions. You can do a lot more. Google for MathJax or LaTeX, and you'll see a whole array of math you can do. $\endgroup$ Aug 23, 2012 at 17:23
$\begingroup$
$\endgroup$
0
See also a video that covers this topic presented by one who knows:
Tim Mattson (Intel): Introduction to OpenMP: 02 part 1 Module 1, YouTube, December 6, 2013
Starting at 3:34 he derives the equation:
$$ P = C V^2 f .$$
Then he says at 5:14 referring to halving the frequency:
[...] frequency scales with voltage, but you know [leakage...], so let's say the voltage goes not half, lets say it goes to .6.
The equation on the slide shown there is as follows:
$$ P_{reduced} = 2.2 C ⋅ (0.6 V)^2 ⋅ f/2 $$
$$ = 2.2C ⋅ 0.36 V^2 ⋅ f/2 $$
$$ = (2.2 ⋅ 0.36 ⋅ 0.5) CV^2f $$
$$ = 0.396CV^2f. $$
That means a power reduction to 40 % at half of the frequency in his case.
