<< pm - please enable pm'ing >>
pm is now pm-able. Sorry about that - I forgot to re-enable PM's when I returned to the forums. In any case, you (or anyone else) are welcome to email me too.
<< - huh? can you explain that in simpler terms? while you're at it, although I think I understand the rest of that paragraph, a paraphrasing might help... So between clock cycles there is a huge change in power usage, and the cpu is actually drawing alterantely huge amounts of power followed by very little, hundreds of millions of times per second? >>
Your explanation is spot on. That's exactly what happens. On the rising edge of the core clock signal, a huge number of transistors all fire roughly simultaneously as they charge and discharges wires and transistors throughout the chip. Just switching the clock itself can account for 30% or more of the total power an entire CPU expends.
So, what you have is that millions (or billions nowadays) of time per second, the current load of the CPU changes dramatically. So without the caps you would see a localized (ie. on the chip, or even just on parts of the chip) voltage variations in time. To explain in simpler terms, capacitors are charge collectors that act to oppose changes in voltage by sourcing or discharging current. So if you increase voltage, the capacitor will act to oppose that change by requiring current to charge the capacitor, and visa-versa for discharging. The amount of "opposing capability" that a capacitor can have is known as it's capacitance (C). In equation form, this would be "i = dV/dt * C" or current is equal to the change in voltage in time multiplied by the capacitance. In effect, by storing charge capacitors can act like (very small, very poor) DC power supplies.