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Is heat the only thing that keeps CPUs from having a faster clock cycle or are there other limiting physical factors which keeps CPUs under 3~5GHz?

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    \$\begingroup\$ Slew rate springs to mind. \$\endgroup\$ – Andy aka Mar 11 '17 at 16:20
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    \$\begingroup\$ One factor is scale... In order to go faster you need to make things smaller so the amount and distances electrons (and holes) that have to be moved is reduced. Unfortunately there is a limit to how small you can make things. The ability to do so has improved over the last 20years, but now it is getting to the point where the distances are measured in atomic levels and can not be made smaller. Hence all the research into quantum state computers. \$\endgroup\$ – Trevor_G Mar 11 '17 at 16:33
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    \$\begingroup\$ timing margin reduces until it faults. But they are cannot shrink much more, so speed increases demand a lower d substrate which costs more but for lower capacitance junctions. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 11 '17 at 16:40
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    \$\begingroup\$ Another issue of course is support. No point developing a 50GHz CPU, if the fastest memory chips you can buy is only 10GHz... Things like this tend to leap-frog over one another, as each tech company makes an advance the other follows suit. \$\endgroup\$ – Trevor_G Mar 11 '17 at 16:41
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    \$\begingroup\$ it 's also the cost limit of a smaller Lithography than the present limit. A cost which is measured in $B , but then the return is small since so many other factors make GHz speed meaningless. theguardian.com/technology/2002/feb/28/onlinesupplement3 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 11 '17 at 16:52
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Slew rate is actually one of the more significant problems to faster processors, the tiny gate capacitence of a FET means that its output cannot switch instantly. Therefore, its output edges are rounded, try to go too much faster, and those round edges will start to give the wrong value, because they weren't able to turn on the next transistor in time. This is related to heat; (almost) all of the heat dissipation in a cpu takes place during that rise/fall, since charging or discharging a cap requires moving current, and while being turned on or off, the resistrance of a FET is high, though current is flowing. This means the power it dissipates is significant only during rise/fall. If you had infinite slew rate, then clock rates could increase, AND heating would decrease.

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  • \$\begingroup\$ The amount of energy lost charging and discharging a gate once is going to be proportional to the capacitance times the square of voltage. Increasing slew rate by reducing capacitance will reduce energy consumption per cycle, but increasing slew rate by reducing resistance (a useful side-effect of raising supply voltages) will not. The latter will allow the device to be run faster while yielding correct results, but at the cost of increased power consumption. If a 10% increase in voltage causes a reduction in resistance that allows the clock to be run 25% faster, power consumption... \$\endgroup\$ – supercat May 9 '18 at 22:20
  • \$\begingroup\$ ...will be increased by a factor of about 1.1 x 1.1 x 1.25, i.e. a little over 50%. Actually, for a variety of reasons, a 10% increase in supply voltage may case more than a 21% increase in power, even if speed is held constant, but that's still a useful approximation for most cases. \$\endgroup\$ – supercat May 9 '18 at 22:21
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Distance, as mentioned previously, is a limiting factor. This can also be explained as the speed of electrons and/or the speed of light (with current computers).

Type of material also comes into play, but is another discussion entirely.

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  • \$\begingroup\$ Can you be more elaborate? \$\endgroup\$ – Umar Mar 12 '17 at 1:53

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