Visual walkthrough — Energy efficiency (performance per watt)

We are going to earn, in order:

1. What a "capacitor" is and why a logic gate is one.
2. Why charging it costs energy $\frac{1}{2}C{V}^2$.
3. Why doing that $f$ times a second gives power $\alpha C{V}^{2}f$.
4. Why going faster secretly forces the voltage up.
5. Why that turns power into a cube and efficiency into $1/f^2$.

Nothing below assumes you've seen any of these before. Let's start from a bucket of water.

Step 1 — A logic gate is a tiny bucket (a capacitor)

WHY this picture? A capacitor is invisible and abstract, but a water bucket makes every later step physical: "voltage" becomes water height, "charge" becomes water amount, and "energy" will become the effort to lift the water up. Look at the red bucket — that redness will follow the "thing we're paying for" through every figure on this page.

Step 2 — Filling the bucket costs energy $\frac{1}{2}C{V}^2$

WHY the $\frac{1}{2}$? Look at the red triangle in the figure. Energy is the area under the push-vs-charge line. Because push rises linearly from $0$ to $V$ as charge rises from $0$ to $Q$, that area is a triangle: $\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}QV=\frac{1}{2}C{V}^2$. A triangle is half a rectangle — that is literally where the $\frac{1}{2}$ comes from.

Step 3 — Doing it $f$ times a second gives power $\alpha C{V}^{2}f$

WHY is $f$ only linear here? Look at the row of pulses in the figure — doubling $f$ just packs twice as many identical pulses into one second. Each pulse costs the same $C{V}^2$. So energy-per-second doubles: a plain, linear relationship. Contrast that with $V^2$, where each pulse itself gets more expensive. Hold this asymmetry — Step 5 detonates it.

Step 4 — The hidden coupling: faster secretly demands more voltage

WHY does higher $V$ mean faster switching? In the figure, the red curve (high $V$) reaches the switching threshold line sooner than the black curve (low $V$). A taller starting push charges the gate faster, so the bit is ready earlier — letting you clock the chip higher. Cutting $V$ slows the fill, so you must clock lower. This is why you cannot cheat: speed and voltage rise and fall together.

Step 5 — Substitute, and power explodes into $f^3$

Now the efficiency. Performance (throughput) rises only linearly with $f$ — twice the clock, twice the work: $\text{Performance}\propto f$

Divide performance by power to get performance per watt (which the parent proved is secretly operations per joule): $\frac{\text{Performance}}{\text{Power}}\propto \frac{f}{f^3}=\boxed{\frac{1}{f^2}}$

WHY this is the whole story. In the figure, the black performance line climbs gently and straight, while the red power curve rockets upward as $f^3$. The gap between them is wasted efficiency. Push $f$ up and you buy a little more work for a lot more heat — efficiency drops like $1/f^2$. This single divergence is why the industry stopped chasing clock speed and turned to Multicore and Parallelism, and why DVFS - Dynamic Voltage and Frequency Scaling saves so much by moving down this curve.

Step 6 — The degenerate cases (don't get ambushed)

Real chips don't live at the extremes, so let's check what happens there.

WHY show all three? The clean $1/f^2$ law is only the middle of the curve. At the low-$f$ end leakage floors you; at the high-$f$ end the cube ceilings you. The figure's red U-shaped total curve shows the real efficiency landscape — high efficiency lives in a valley, and Thermal Design Power (TDP) fixes the right wall you're not allowed to cross.

The one-picture summary

This compresses all six steps: a bucket ($C$) filled to height $V$ costs $\frac{1}{2}C{V}^2$ (the triangle); pay it $f$ times a second for $\alpha C{V}^{2}f$ watts; but $f$ drags $V$ up, so power becomes $f^3$ (red rocket) while performance stays linear (black line) — leaving efficiency falling as $1/f^2$, valid until leakage or the voltage floor ends the ride.