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by maho 1048 days ago
Note that the resistance of superconductors is 0 only for DC currents. The AC resistance is not 0 and typically gets larger for higher frequencies!
3 comments

I think this terminology is really confusing for people not familiar with digital electronics. The term AC to a layman usually means a single sine wave like mains. The point is though that when square waves get really fast, they have a lot of high frequency content.
That becomes relevant for frequencies that are high enough to break Cooper pairs. But this material is claimed to be in the superconducting phase up to 400K, which corresponds to a superconducting gap of 8.3THz.
Yes, but computer chips are powered by DC.
At the frequencies computer chips operate, it acts more like AC. Sure, you place some nice bypass capacitors capacitors very close to the chip so you can feed it nice clean DC power, but when you start switching those transistors in the GHz range, the signals inside the chip rapidly start to look a lot like AC.

You see the same thing with external signals, like a connection between CPU and memory. Operate the bus at 1MHz and it is effectively DC. The signal has a wavelength of 300 meters, so when the signal travels a distance literally two orders of magnitude smaller across your motherboard the AC behavior is negligible. Operate that same bus at 1GHz and your wavelength is down to 30cm. Got a 40cm-wide motherboard? Better treat it like a transmission line or it isn't going to work!

The latest versions of USB and PCIe are no longer binary digital signalling at all. It is a modulated radio signal [1] carried over a wire, similar to how DSL and cable modems work. Processor and memory busses will be next to switch over. Most digital systems will probably move in this direction as speeds continue to increase.

[1] https://en.wikipedia.org/wiki/Pulse-amplitude_modulation

Interesting.
High-end CPUs are actually not powered directly by DC. Basically all server (and growing portion of consumer) CPUs are powered by multiphase buck regulators[1] which split the power from from the DC PSU rails into a parallel set of modulated buck regulator power stages. The outputs of the parallel regulators are recombined to generate DC (as the combined waveform of the phased AC parts).

The reason for this multiphase design is because it offers better power efficiency and better transient response to the CPU as the CPU moves between it's different power states (high vs low load).

[1]: https://www.ti.com/lit/an/slva882b/slva882b.pdf?ts=169087951...

Even on the DC input which ostensibly isn't doing anything it will probably be wiggling around a lot. You can attack electronic safes by analyzing the power draw into their computer as you type into
Power dissipation in chips isn't an issue with the DC power supply but switching the current billions of times a second.
I thought that was the switching loss of the transistors themselves, not the interconnects?
No? Transistors have a voltage drop in operational mode, and P=IV.
As long as you don't keep switching the DC current on and off, which is what computers are all about