Why P scales as C*V^2*f is so obvious

Published:06/29/2009   Last Updated:06/29/2009

I think that you've all seen this equation before:

P = a * C * V2 * f

Where P is power, C is capacitance, V is the voltage across the gate (typically, Vdd), f is the clock frequency and a is some constant.

Doing my reading, I see this equation a lot, usually prefaced with something like, “And everyone knows…” or “And of course…” When I read, “And it’s obvious,” I get really steamed because it’s never obvious to me. Maybe I’m dumb. Maybe they are way smarter than me. Or maybe they don’t know why either. Well, I’m here to say that it isn’t obvious, at least not to me.

So let’s see where this sucker comes from.

This is going to get into a little bit of math, but hey, we are all pretty comfortable with math. It comes with being a nerd. (OK, I didn’t mean to stereotype. I’m sure there are nerds that aren’t comfortable with math.-- at least theoretically. Not that I’m a nerd. I’m a perfectly well adjusted individual who just happens to be uncomfortable around non engineers, and who likes to hide in my cube and code all day.)

CHECKING THE UNITS

Before we go further, let’s check to see if the units match. This is always a good thing to do just to make sure we aren’t barking up the wrong tree.

Now, P is power (Watts). V is voltage (Volts). f is the frequency (cycles/sec), or equivalently, the number of seconds/cycle. And C is capacitance (Farads). A Farad is Coulombs/Volt, the charge divided by the EMF (Electromotive Force) potential.

Putting it all together and noting that P=V*I where I is the current in Amps, we’ve,

P = a * C * V2 * f
Volts*Amps = Coulombs /Volt * Volts2 * cyc/sec

Since amperage is charge flowing per unit time, i.e. Amps=Coulombs/sec, we’ve got

Volts* Coulombs/sec = Coulombs/Volts * Volts2 * cyc/sec
Volts*Coulombs/sec = Coulombs*Volts/sec

(note that both cycles and a are unit less)

Hey, it works! So at least we know that the units are correct. We’re not obviously barking up the wrong tree.

So that’s it folks. There you have it. Let’s now go onto something new.

What? You also want to know why? Wow, you are a hard crowd. OK, I’ll relent and explain why. Next time, that is.

Product and Performance Information

1

Intel's compilers may or may not optimize to the same degree for non-Intel microprocessors for optimizations that are not unique to Intel microprocessors. These optimizations include SSE2, SSE3, and SSSE3 instruction sets and other optimizations. Intel does not guarantee the availability, functionality, or effectiveness of any optimization on microprocessors not manufactured by Intel. Microprocessor-dependent optimizations in this product are intended for use with Intel microprocessors. Certain optimizations not specific to Intel microarchitecture are reserved for Intel microprocessors. Please refer to the applicable product User and Reference Guides for more information regarding the specific instruction sets covered by this notice.

Notice revision #20110804