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Comment by KaeseEs | Hacker News Reader

As the wires get smaller and smaller and smaller, resistance becomes a huge problem, and you wind up needing a crapload of repeaters and wasting a ton of power.

Consider that r= (rho * L / A), and that you need a ~4nm tantalum sheath around copper wire to keep it from poisoning silicon. What happens to r when the width starts getting really small? 40nm - (4 + 4) = 32nm, 10nm - (4+4) = 2nm, so for a 4x process shrink we've seen a 16x reduction in area assuming height remains constant and that the cross section of the wire is roughly rectangular (neither of those assumptions is perfect).

So yes, superconductors would be nice.

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The trend I've seen in Intel processors is that every time they jump down a manufacturing size (eg: 32 vs. 22nm), the power usage (per flop, at least) has gone down. Why's that?

Smaller transistors take less energy to switch. The problem with heat dissipation is not that the amount of energy wasted goes up, it is that the [i]ratio[/i] of energy wasted goes up. There is a point where the increase in resistance will nullify the decrease in switching costs, it is at this point where we would need superconductors.

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Smaller gates mean lower capacitance, so you need to drive less current into them to switch.

The nice thing about CMOS is that you only need to flow current when switching them (at other times current is blocked by an "OFF" transistor somewhere in the logic), so resistance doesn't matter as much as it would for a motor or a light.

This is not entirely true, in an ideal world a CMOS device just sitting there not switching would use basically no power, but in actual application there is some leakage (both form not being fully off and some current jumping from the gate). This leakage gets worse as the transistors get smaller and when you have billions of transistors even the tiniest amount of leakage adds up, the result is leakage current can be a significant amount of the power used by a modern processor.

IIRC the voltage can drop with the size, and power loss drops in proportion to the V^2

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