It occured to me to look into the USB-C spec. A USB-C device can negotiate up to 20V @ 5A which is 100 watts of power. Compare this to USB 2’s 5V @ 500mA = 2.5W and even USB 3’s paltry 5V @ 900mA = 4.5W.
This means USB-C can potentially output 4 times the voltage at 5 times the current = 20 times the power of USB 3. Has anyone stopped to think of the failure modes here? It didn’t take me long.
Failure 1 – Intermittent Cable
What’s going to happen when you plug in your USB-C power guzzler which negotiates the port to 20V/5A. Then the 3 year old cable that’s become mangled, pinched and chewed partially shorts out? 100W will generate a lot of heat real quick (try holding a 100W incandescent light bulb).
Failure 2 – Loss of Voltage Regulation
USB 1 and 2 are a fixed 5 volts. This is most likely derived from a power rail supplying many other components, a fixed 5 volts carefully regulated by the system power supply.
USB-C offers negotiated variable voltage which will be generated by a local power converter (each port would need its own). The negotiation occurs when a device is connected to set the voltage/current limit. I sure hope they put a good checksum on that negotiation.
So, what is your USB drive or battery charger (which happily negotiated 5V upon connection) going to do when the inevitable g-g-glitch, power spike, electrostatic discharge or silicon failure ramps the port to its potential 20V limit (which is still ‘valid’ for the port)? What about unplugging a 20V device and the port not realising it (or latching up) before you plug in a 5V device?
With all those available amperes of current, the result will be spectacular. Voltage-overloaded devices love more amperes.
Maybe the “Safely Remove Hardware” option will take on a new importance.
Well publicised recalls of dodgy USB-C cables have already occured, and these are problems out-of-the-box. As hardware ages in the real world, I predict stories of fireworks and a market for bogus USB-C circuit breaker dongles.