The LED says power, not success

What it is wired to

Follow the LED back and it does not come from the output at all. It comes from VDD — the small 3.3 V rail the chip regulates for its own use — through a 5.1 kΩ resistor to ground.

VDD exists because the CH224K needs a low-voltage supply for its own logic while VBUS is anywhere from 5 V to 20 V. The chip gets it with a shunt regulator: the 1 kΩ resistor from VBUS limits the current, and the chip holds its supply pin at about 3.3 V by sinking whatever it does not need. That is why the same single resistor works across the whole range.

So the LED comes on as soon as there is any VBUS at all — which, on USB-C, is the 5 V every source provides before anything is asked of it.

Which makes it almost useless as an indicator

It tells you three things:

• the cable is carrying VBUS;
• the socket's joints are good;
• the chip's supply is alive.

It tells you nothing about the voltage at the screw terminal. The board behaves identically, and lights identically, whether it asked for 20 V and got it or asked for 20 V and was refused. There is no second LED, no colour change and no flash.

So measure the terminal

Every question about whether this board worked is answered by a meter across VCC and GND at the screw terminal, and by nothing else on the board. That is the habit worth forming now, because it is the same measurement in when nothing happens at the end of the book.

About a quarter of a milliamp runs through the LED. At 20 V in, the 1 kΩ resistor feeding VDD is passing something over sixteen milliamps into the chip's shunt, so the LED is a rounding error in the board's own consumption — but it is also the reason not to add a brighter one.