| ▲ | mort96 20 hours ago | ||||||||||||||||||||||
I know how cheap resistors are. I design PCBs as a hobby and occasionally semi-professionally. Leaving off parts should result in behavior which is identical whether you connect an A-C cable or a C-C cable. Detecting the difference between them should require adding parts. | |||||||||||||||||||||||
| ▲ | crote 19 hours ago | parent | next [-] | ||||||||||||||||||||||
There is nothing to detect on the sink side, for basic operation. All the smarts are in the power source. It's like saying USB cables should still carry 10Gbps data when only the VBUS and GND wires are present, because some AliExpress scammers leave out the data wires: you can't leave out trivial core parts and still expect it to work! | |||||||||||||||||||||||
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| ▲ | Ajedi32 19 hours ago | parent | prev [-] | ||||||||||||||||||||||
I don't think you understand the problem. Unlike a USB-A power supply, a USB-C power supply cannot just assume by default that the device you plugged into it is not another power supply. That would risk causing a short and potentially starting things on fire. I suppose that instead of a $0.01 resistor to request power they could have mandated some kind of (probably more expensive) circuit breaker that detects the short and shuts things off before it causes a fire, but I suspect then instead of broken devices having missing resistors you'd get ones with missing circuit breakers... It pretty hard (maybe impossible) to design a spec that still works when manufacturers randomly omit critical components for no reason. | |||||||||||||||||||||||