# Why P scales as C*V^2*f is so obvious (pt 2)

Por Taylor IoT Kidd, publicado em 25 de agosto de 2009, atualizado em 1 de janeiro de 2015

THE GORY DETAILS

Let’s continue from where we left off last time. Let’s figure out the why of the equation,

*P = C * V^2 * (a * f)*

To do this, we’re going to have to look at what is going on in one of the fundamental building blocks (a CMOS inverter) of an integrated circuit (IC).

So when and how does this circuit dissipate power?

Before getting into the math, let’s get our variables right.

*V _{dd}* is the voltage across the gate

*I _{peak}* is the peak short circuit current going through the gate when it switches state (0 to 1 or 1 to 0)

*I _{leakage}* is the current through the gate even when it is reverse biased (i.e. in a 0 or a 1 state)

*C _{L}* is the capacitance of one transistor

*t _{s}* is the switching time needed to change the state of the switch

*f _{g}*

*=*

*1/T*is the maximum rate that the gate can cycle at in our processor. In other words, it is the gate’s clock frequency.

_{g}Let’s start out with the basic CMOS gate, see above. There are three current paths, one going through the gate, one charging the capacitance of the gate, and one resulting from leakage through a reverse biased gate.

The one going through the gate results from the brief time that both semi-conductor transistors are closed causing a short circuit. In an ideal world, the switch would be instantaneous and there would be no current flow, and hence no power loss. But this isn’t a perfect world. There is a brief period of time, the switching time *t _{s}*, when we’ve a short circuit. The power is going to be the voltage across the open circuit,

*V*, multiplied by the current,

_{dd}*I*. (We’re using the peak current since this is going to give us an upper bound on the power dissipated.) Say the open circuit exists for

_{peak}*t*. Then the total energy lost is bounded by

_{s}*Energy loss due to open circuit <= V _{dd}* I_{peak} * t_{s}*

Let’s now look at the energy lost resulting from leakage through a reversed bias gate. Since *t _{s}* is small compared to

*T*we can approximate the energy loss as,

_{g}*Energy loss due to reverse biased gate ~ V _{dd}*I_{leakage}*T_{g}*

What about the total energy loss? If the energy loss due to the reverse bias leakage and short circuit current are so small, where is all that energy coming from that is heating our processor?

To get this, we need to look more closely at what a gate looks like in an analog sense. A reverse biased transistor is basically a capacitor, that is, two plates separated by an insulator / dielectric. From the figure above, it’s *C _{L}*.A forward biased transistor is a short. These plates charge and discharge like a capacitor because of the design of a gate. In the one state, one transistor is “open” and the other is the acting capacitor. In the other state, the roles reverse and other transistor is the acting capacitor. What I’m trying to say is that even though the circuit is essentially open, current still flows from one transistor / capacitor to the other. This current flow is going to cause resistive heating, and so consume power.

The equation for the energy stored in a capacitor (*C*) is

*Energy in a capacitor = ½ * C * V ^{2}*

At each transition, the capacitor dumps the energy stored in it to either to ground or to the other complement transistor, giving us the following.

*Energy flow due to a state transition = ½ * C _{L} * V_{dd}^{2}*

Remember that one cycle has two state transitions. So the complete equation for the energy loss caused by one cycle, which we’ll call *E _{tr}*, is,

*E _{tr}*

*=C*

_{L}*V^{2}_{dd}+ 2*V_{dd}*I_{peak}*t_{s}+V_{dd}*I_{leakage}*T_{g}I’m now going to do something that is blatantly wrong. It’s also a huge topic on its own that I’m completely unqualified to talk about. I’m going to ignore the last two terms. I’m pretty sure that they were 2^{nd} order in effect maybe 10 years ago. And the first, the one with *I _{peak}*, may still be. But that last term, oh that last term. Volumes have been written about it, generally in a language incomprehensible to us mere mortals. (Request: can anyone out there tell us more?)

After dropping those last two terms, we’re left with,

*E _{tr}*

_{}~ C_{L}*V^{2}_{dd}This is almost what we want. We’re missing that annoying little “*f*”. The little equation we wrote above is for one cycle of a gate. True, modern processors can do one heck of a lot in one cycle, but a one cycle application is still pretty uninteresting. Our gates above are switching all the time at a rate related to the frequency of the processor, which we’ll call “*a * f”*, where *f* is the frequency of the processor and *a* is some constant.

*Energy output of a gate/sec ~ C _{L} * V_{dd}^{2} * (a*f)*

And how many gates are in a high-end Intel processor today? Close to a billion for 45 nm. (And the next generation is 32 nm.) So we’ve 1.0E9 (1 billion) transistors per processor, running at frequencies of 3E9 Hz (3 billion). Let’s see, 1E9*3E9 is – scientific notation always confuses me – 3E18 transitions per second. Is there even a name for 1E18?

*Energy output of a processor/sec ~ C _{L} * V_{dd}^{2} * (a*f) * <number of transistors>*

Now before those of you who actually know this stuff start crabbing, let me make it clear that I am only attempting to help people understand where the equation comes from. Yes, the effect of the short circuit current contributes noticeably to the processor’s heating when we’re talking about, say, a billion transistors. And there’s that aforementioned leakage current. And there are the leakage and voltage issues related to smaller and smaller junctions. And there are a lot of circuit elements that aren’t strictly logic gates that contribute to the power. And there have been a lot of developments related to reducing the short circuit and leakage currents of a logic gate. And yes, I’m an ignorant software hack.

But for those of you who are neither processor architects nor researchers into modern IC materials, mayhap this gives you a little better understanding of where this (I hope) formerly mysterious relationship comes from.

Of course, I could be just blowing smoke, but then, one of you less than gentle readers out there will let me know.