USB-C doesn't provide power by default which is better design than USB-A always providing some power. The device needs to signal that it is a sink with 5.1k ohm resistors so the charger can turn on the power. There is also resistor to signal legacy USB charging to work with USB-A ports.

How is the standard supposed to work with defective devices that don't implement the standard?

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seba_dos1 19 hours ago | parent [-]

> There is also resistor to signal legacy USB charging to work with USB-A ports.

There's no such thing on the device. A power sink with a USB-C receptacle has two 5.1kΩ resistors, one on each CC line. A power sink with a USB-C plug has a single 5.1kΩ resistor. Always. That's it - anything else in on the power source (or cable) side.

You only need to bother with anything more on the sink side once you want to either sense what's the max current provided by the source (which you can do simply by measuring the voltage on CC pins), or implement PD.

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ianburrell 19 hours ago | parent [-]

I think you corrected me before. Maybe this time it will stick.

I don't understand why the USB-C module that they are using doesn't include the resistors. Then they can replace the microUSB module with USB-C module and have it work.

	▲	crote 18 hours ago \| parent [-]
		Because putting a somewhat-precise resistor inside a connector is not trivial, and would likely double the price of the connector due to the increased complexity of manufacturing. On the other hand: the resistor itself is basically free, and you are already putting a few dozen of them on your final PCB anyways. Adding one more resistor value barely changes the assembly process. The people who are willing to skimp on USB-C resistors are definitely not going to splurge on a custom drop-in connector module, so it would not solve any real problem. The resistor-less devices are broken because their manufacturers do not care that they are broken. You can't engineer yourself out of that kind of greed and apathy.