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The whole multi-core debate got me thinking.

It's much easier to produce two cores (in one package) then speeding up one core by a factor of two. Why exactly is this? I googled a bit, but found mostly very imprecise answers from over clocking boards which do not explain the underlying Physics.

The voltage seems to have the most impact (quadratic), but do I need to run a CPU at higher voltage if I want a faster clock rate? Also I like to know why exactly (and how much) heat a semiconductor circuit produces when it runs at a certain clock speed.

Robotnik
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Nils
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  • there's a multi-core debate? I thought murphy's law ran out and the comp arch guys couldn't find anything else to do. –  Jul 14 '10 at 13:44
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    Moore's law. Murphy's law will only run out at the worst possible moment. :-) – O. Jones Jul 14 '10 at 13:56
  • There is an element of truth to Gary's comment though. At this point transistor counts are high enough that for larger CPUs designers can put *everything* on die instead of having to pick and choose, and adding cache eventually runs into diminishing returns. The 10-20% gains with new architectures indicate that designers are still managing to tweak performance but there's probably not anything revolutionary left that could be implemented if there were more transistors available to spend. – Dan Is Fiddling By Firelight Jul 14 '10 at 15:41

7 Answers7

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Each time the clock ticks you're charging or discharging a bunch of capacitors. The energy for charging a capacitor is:

E = 1/2*C*V^2

Where C is the capacitance and V is the voltage to which it was charged.

If your frequency is f[Hz], then you have f cycles per second, and your power is:

P = f*E = 1/2*C*V^2*f

That is why the power goes up at linearly with frequency.

You can see that it goes up quadratically with voltage. Because of that, you always want to run at the lowest voltage possible. However, if you want to raise the frequency you also have to raise the voltage, because higher frequencies require higher operating voltages, so the voltage rises linearly with the frequency.

For this reason, the power rises like f^3 (or like V^3).

Now, when you increase the number of cores, you're basically increasing the capacitance C. This is independent of the voltage and of the frequency, so the power rises linearly with C. That is why it is more power efficient to increase the number of cores that it is to increase the frequency.

Why do you need to increase the voltage to increase the frequency? Well, the voltage of a capacitor changes according to:

dV/dt = I/C

where I is the current. So, the higher the current, the faster you can charge the transistor's gate capacitance to its "on" voltage (the "on" voltage doesn't depend on the operating voltage), and the faster you can switch the transistor on. The current rises linearly with the operating voltage. That's why you need to increase the voltage to increase the frequency.

Nathan Fellman
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  • cpu doesn't have capacitors on it? it uses SRAM which is just transistors in a latch configuration. –  Jul 14 '10 at 13:10
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    Transistors have dynamic capacitance. In order to "open" a transistor, you need to charge its gate-capacitance. – Nathan Fellman Jul 14 '10 at 13:12
  • ah, still, is this different from leak current? I didn't get too far into analog :-). And capacitance wouldn't show up as heat, just the current through the impedance of the gate right? –  Jul 14 '10 at 13:17
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    I didn't even go into leakage, but that works pretty much the same. It's more like some sort of resistor, and less like a capacitor, so it consumes power based on v^2/r at the same frequency f. – Nathan Fellman Jul 14 '10 at 13:19
  • I still think it's confusing to blame capacitance as you describe, because ideal capacitors are just energy stores and wouldn't put out heat. –  Jul 14 '10 at 13:21
  • And semiconductor transistors work in the same way as dielectric transistors? –  Jul 14 '10 at 13:25
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    You're wrong. It's 100% about capacitance. The capacitance is charged and discharged through the source-drain resistance, which is definitely *not* leakage, but which consumes the power that you spent charging and discharging the capacitors. Also, not that these aren't ideal capacitors in any way. 1st of all they're real-world capacitors, 2nd of all they're MOS capacitors whose behavior is very different from regular capacitors, if only because their capacitance depends on the current voltage. – Nathan Fellman Jul 14 '10 at 13:25
  • @Nils, thanks, I forgot that the term was "dielectric" capacitors. – Nathan Fellman Jul 14 '10 at 13:27
  • I know I'm wrong about some stuff, but I know for a fact that capacitance has nothing to do with heat conversion. Capacitors do, but capacitance doesn't :-). If it's a pure capacitance, then the energy for charging has nothing to do with what's lost as heat, except if the power source has an impedance, which you do not blame. You blame the capacitance. So, I can tell you know what you're talking about, and know analog better than me, but I just think you could be more clear, especially since the asker of the question doesn't know as much as you. –  Jul 14 '10 at 13:29
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    ok... capacitance doesn't have anything to do with heat conversion, but it does have to do with energy consumption, because you need to invest 1/2CV^2 to charge the capacitor. That energy comes from some power source, so each time you charge the capacitor, you're spending 1/2CV^2 from that source. There's nothing keeping that source from heating up. The *capacitance* per-se isn't to blame. The environment that capacitor is in *is* to blame. – Nathan Fellman Jul 14 '10 at 13:38
  • "However, if you want to raise the frequency you also have to raise the voltage, because higher frequencies require higher operating voltages, so the voltage rises linearly with the frequency." While true in the abstract the default voltages on most CPUs are set conservatively so that you can often get a decent OC before having to boost the voltage. Also, in my experience with aggressive OCing, most of the total overvolt comes for the last few hundred mhz gained and that the response curve is not linear as it approaches the limit of stability. – Dan Is Fiddling By Firelight Jul 14 '10 at 18:34
  • @Dan, that's not necessarily true. Speedstep technology allows the platform to set the voltage to the lowest possible voltage, based on the current frequency. This is especially true for ULV notebooks and blade servers. Besides, I'm talking about the physics of this. The fact that OEMs set the voltage conservatively doesn't change it. On such systems the default setting is somewhat wasteful. – Nathan Fellman Jul 14 '10 at 20:41
  • AFAIK the speedstep values are equally conservative; running 24 hour burn tests at each multiplier/voltage stop would be prohibitively expensive so they set universal values that are good for 99.9...% of chips that pass whatever quick test they perform while binning the chips. I was commenting on the fact that in the real world the default settings will let you get a decent overclock at stock voltage, a decent undervolt at stock speeds, or operate with bad airflow in a hot room without failure. – Dan Is Fiddling By Firelight Jul 14 '10 at 21:03
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    @Gary You blame the capacitance because the heat scales with the capacitance. Say you need to raise the voltage on a transistor from 0.2V to 0.75V in a billionth of a second from an 0.9V source. The power you use to do that depends linearly on the resistance the current needs to pass through and the capacitance of the gate. If the capacitance was zero, no current would need to flow through the resistance. – David Schwartz Aug 07 '14 at 07:38
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Very basically :

  • A transistor switches faster when you apply more voltage to it.
  • modern IC consume most power when swithing from one state to the next (on the clock tick), but consume no power to stay in the same state (well, there is leakage, so not exactly no power) so the faster you switch, the more switch per seconds you have, the more power you consume.

A very good book on all the details of processor architecture :Computer organization and design by David A. Patterson, John L. Hennessy.

Guillaume
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Every time a transistor switches state, current is spent. Higher frequency means faster switching, more current wasted. And the impedance of everything converts it to heat. P=I^2*R and all that. And P is V^2/R. In this case though, you'd really want the average V and I over time to be able to calculate, and it'd be quadratic to voltage and current both.

gtrak
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1) two cores vs. speeding up one core
To speed up one core you need new technology to speed up the transistors switching from one state to another. To add another core you just need more of the same transistors.

2) Heat
The power dissipation is in the form of heat. Power = Voltage * Current. Voltage = Resistace * Current. Power = Voltage^2 / Resistance. So the heat disipated is proportional to the voltage squared.

Robert Deml
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  • Yes i know that the heat increases quadratic with the voltage, what I don't get is what influence the voltage has on the clock speed? Do I need higher voltage for higher clock speed? –  Jul 14 '10 at 13:10
  • I know the Microchip's PICs have a graph for Voltage vs Frequency. There's a minimum voltage that the chip will run at a low frequency. It's a linear scale to the maximum voltage and maximum frequency. – Robert Deml Jul 14 '10 at 13:25
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    @Nils, higher voltage gives sharper and less ambiguous transitions and states, and makes it more likely a 1 will be interpreted as a 1 and not a 0. And higher frequency makes transitions less square-ish. Remember square waves don't exist. –  Jul 14 '10 at 13:38
  • you mean more square-ish – Nathan Fellman Jul 14 '10 at 14:19
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Well, in electrical power, there are two kinds of power, reactive power and real power. Some people call reactive power dynamic power. Reactive power is never consumed or lost. For instance, if an ideal capacitor is connected to an AC voltage source by ideal lossless wires, the capacitor will charge and discharge, taking energy from the generator in one cycle, and returning energy to the generator in the next cycle. The net loss is zero.

However, if the wires are nonideal and resistive, then energy is dissipated in the wires during the charging and discharging of the capacitor. This dissipated power is real power loss, and cannot be recovered. As the clock rate goes up, the rate of charging and discharging goes up, increasing the power loss in the wires.

The gates of transistors behave like capacitors. As the clock rate goes up, more reactive power is delivered to the capacitors. The fraction of which is lost in the resistive wires also goes up.

Weng Cho
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One thing not mentioned so far - chips get faster and the lithography process to make them gets components smaller. They've gotten so small that they're a few atoms wide in some cases. There's significant current leakage now, which gets dissipated as heat generally.

Rich Homolka
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To switch the state of a circuit rapidly requires more current than to switch it slowly. To achieve that current you need a higher voltage and/or larger, more power-thirsty components. And, of course, larger components need more drive current, causing a snowball effect.

(Interestingly, there was an article in the latest Scientific American (July 2011) that covers this topic for the human brain. Same principles, and one way the human brain packs in more power is to partition the brain in to separate sub-processors, so to speak.)

Daniel R Hicks
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