3 ms·

It is impossible for super-USB-C to carry that much power. The new 240V is at the limit of what USB-C can do and you are talking about order of magnitude differ

It is impossible for super-USB-C to carry that much power. The new 240V is at the limit of what USB-C can do and you are talking about order of magnitude difference. First, the wire thickness needs to be like regular wires to carry enough current. Changing the wiring in the walls is the hardest thing. Second, the voltage needs to be like regular service to carry enough power over regular sized wires. Higher voltages, like 500V, are better since DC loses more energy over distance than AC. Third, the plug needs to be similar size to power plugs to not arc, and DC arcs worse than AC. Is replacing everything worth the effort to increase efficiency by a little bit? You are optimizing for low-power DC devices at the expense of high-power AC appliances. The only way I can see DC power happening is in isolated community like the Moon or Mars.

londons_explore 3y ago
> It is impossible for super-USB-C to carry that much power. 5 amps, 2000 volts. 10 kilowatts. Made safe by milliamp precision leakage detection (you won't manage to kill anything with a milliamp of current, and more than a milliamp of current will trigger the protection). 2000 volt insulation made of PVC needs to only be 0.1mm thick (say 0.5mm for a 5x safety factor). That can easily fit in plugs and sockets of the existing size. The leakage detection would also detect current leaked between the power conductors, so dielectric breakdown can be protected against. The cable would be designed to deliberately cause dielectric breakdown when heated before catching fire, so that in cases of a fatigued conductor or wire draped across a stovetop, a fire is impossible. This design is only safe if implemented correctly, so each end of the connection will frequently test the other end of the connection to ensure it is adhering to the spec. It would do this by sending an encrypted message saying "In 37 seconds I am about to cause what looks to you like a baby chewing through the insulation". Then, in 37 seconds, that happens, the power gets turned off, and a few microseconds later the encryption key is send over proving that it was a test and the power is turned back on again. Both sides would have capacitors so the user wouldn't notice the power going off for a few microseconds per minute.

londons_explore 3y ago

> It is impossible for super-USB-C to carry that much power. 5 amps, 2000 volts. 10 kilowatts. Made safe by milliamp precision leakage detection (you won't manage to kill anything with a milliamp of current, and more than a milliamp of current will trigger the protection). 2000 volt insulation made of PVC needs to only be 0.1mm thick (say 0.5mm for a 5x safety factor). That can easily fit in plugs and sockets of the existing size. The leakage detection would also detect current leaked between the power conductors, so dielectric breakdown can be protected against. The cable would be designed to deliberately cause dielectric breakdown when heated before catching fire, so that in cases of a fatigued conductor or wire draped across a stovetop, a fire is impossible. This design is only safe if implemented correctly, so each end of the connection will frequently test the other end of the connection to ensure it is adhering to the spec. It would do this by sending an encrypted message saying "In 37 seconds I am about to cause what looks to you like a baby chewing through the insulation". Then, in 37 seconds, that happens, the power gets turned off, and a few microseconds later the encryption key is send over proving that it was a test and the power is turned back on again. Both sides would have capacitors so the user wouldn't notice the power going off for a few microseconds per minute.