alt Hacker NewsGuides like this explain why there are so many broken USB-C devices. The guide mentions that you do not need a PD chip for 5Vs, but then tells you that USB C is a cold connector meaning 0V is on VBUS when nothing is connected and jumps straight into the complexities of the PD protocol running over the CC pins instead of explaining how to get the 5V without the PD chip first.
Then in the section where it tells you how to do that, it fails to properly explain how to connect a load switch (10 cent component at 100 units) to get around the 10uF limit. The vast majority of applications will require less than 15 W and a good chunk of them can't get away with 10uF between VBUS and GND so a schematic how to do it in the lowest cost way would have helped here.
Edit: After reading until the very end I got the impression that this is just an ad for Texas Instruments PD controllers.
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USB-C uses resistors to signal the power level for 5V. There is 5.1k for USB legacy, and I forget for 7.5W and 15W. Lots of cheap devices leave out the 5.1k resistors and depend on the same resistors in the USB-C to USB-A cable.
It is perfectly fine for USB-C device to signal USB Legacy since every charger needs to support it for microUSB to USB-C cables. The other way is for device to dynamically negotiate power level with the charger, but this is different than USB PD for higher voltage.
My phone's USB-C port is worn out, and so now it's a brick if I don't have a wireless charging dock around.
Is it physically impossible to get bandwidth and power out of something as durable as a Magsafe connector, even a larger-scale version?
Basic 5V devices just need a 5.1k resistor on each CC pin to GND. It is pretty obvious when reading the actual USB-C Connector Specification.