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It seems odd to me that overclocking a CPU causes it to consume more energy.

The amount of energy flowing through the CPU should be related to the average number of 'gates' (or transistors) that are open, not how fast you open and close them.

Does changing a transistor's state cause heat to be generated or is there another factor?

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    $\begingroup$ This is perhaps a better question for computer science or electrical engineering StackExchange sites. $\endgroup$ – Jim Apr 1 '14 at 20:32
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Modern processors are built from CMOS technology. These digital circuits consume relatively little current when sitting in one state or the other. However, there is some inevitable capacitance on every node. When the output of a digital gate changes state, that capacitance is charged or discharged, which means current has to flow. The total average current used by a digital gate is therefore largely proportional to how often it switches, which is to say it is proportional to frequency.

The supply current times the supply voltage is the power used by the digital circuit, which it dissipates as heat. So yes, clocking a modern processor faster causes it to dissipate more power, which can cause it to overheat if this power isn't removed properly. Note the big heat sink with dedicated fan clamped to the processor chip. Getting rid of the heat of such chips is a major design consideration.

Newer processors also draw significant power just because they are on. One way to reduce the power required for switching is to lower the voltage. That reduces the charge that is moved onto or off of the little capacitors each digital state change, which reduces the overall average current and therefore the heat dissipated. However, when the transistors are made for too low a voltage, they don't really turn off that well. This results in a little leakage current thru each gate. A few million gates here, another few million gates there, and pretty soon you have some real currents to worry about. This tradeoff between less current due to switching versus more current when not switching is a carefully balanced design decision.

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  • $\begingroup$ I would only add that the charging of the capacitors through the inevitable parasitic source or sink resistances is the ultimate cause of dissipation (heating), but as Olin Lathrop has said the amount of current is determined by the capacitors and their rate of charge/discharge. $\endgroup$ – hyportnex Apr 1 '14 at 21:48
  • $\begingroup$ Thanks! I'm still curious, does that transistor itself generate a significant about of heat when it changes state? $\endgroup$ – Hoytman Apr 1 '14 at 22:37
  • $\begingroup$ @Hoytman CMOS has 4 main sources of power dissipation: switching, short-circuit power, diode leakage power and gate leakage power. The transistor itself IS what is generating all the heat. $\endgroup$ – user6972 Apr 1 '14 at 22:58
  • $\begingroup$ Sure it does because during the switching transient neither the drain current nor the drain-source voltage is zero. The product of the two is the power that the transistor dissipates. $\endgroup$ – hyportnex Apr 1 '14 at 22:59
  • $\begingroup$ @Hoyt: Transistors don't inherently dissipate power when switching, but that is the time when they aren't fully off (no current, so no power) or fully on (no voltage so no power) in the ideal case. Switching losses happen because for a little while both transistors of a pair are partially on, and the current from charging the node capacitance goes thru that transistor that is being switched to on while there is still some voltage accross it. This voltage accross the transistor times the current thru it is the power it converts to heat. $\endgroup$ – Olin Lathrop Apr 2 '14 at 14:42

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