English is currently available.
USB DISCUSSION
I replaced the micro-USB socket on my board with a 16-pin USB-C receptacle and wired only VBUS, GND, D+ and D-. The board powers up from a USB-A to USB-C cable, but with a USB-C to USB-C cable from a laptop or a phone charger it stays dead.
I have read that the CC pins need resistors. What exactly is required for a simple device that only wants 5 V and USB 2.0 data, and how does the device know how much current it may draw?
A USB-C source does not put 5 V on VBUS until it detects that a sink is attached, and it detects that through the CC pins. The device must have a 5.1 kΩ pull-down (Rd) from CC1 to ground and a second 5.1 kΩ from CC2 to ground. The source has pull-ups (Rp) on its CC pins; when one of them is pulled down into the right voltage range through the cable's single CC wire, the source recognises a sink, works out the plug orientation from which pin it was, and turns VBUS on.
With the CC pins left open, a C-to-C source sees nothing attached and never enables power. An A-to-C cable works anyway because a USB-A port has VBUS on permanently, and the cable itself contains the pull-up.
Use two separate resistors, one per CC pin, and do not tie CC1 and CC2 together to share one. A cable has only one CC conductor. In electronically marked cables the plug's other CC pin connects to the cable's marker chip, which presents its own pull-down (Ra, about 1 kΩ). If your CC pins are joined, the source sees Rd and Ra in parallel on one line, interprets the combination as a different kind of accessory and refuses to supply power. The symptom is exactly that some chargers and cables work and others do not.
On tolerance: the specification asks for 5.1 kΩ within ±10 % if the sink wants to read the current advertisement, so ordinary 1 % parts are fine.
The source advertises its current capability through the value of its pull-up, and the sink can read it as a voltage on the active CC pin. With a pull-up to 5 V the values are 56 kΩ for default USB power, 22 kΩ for 1.5 A and 10 kΩ for 3 A. Against your 5.1 kΩ that gives 5 × 5.1 / 61.1 = 0.42 V, 5 × 5.1 / 27.1 = 0.94 V and 5 × 5.1 / 15.1 = 1.69 V. The detection bands are roughly 0.25 to 0.61 V, 0.70 to 1.16 V and 1.31 to 2.04 V.
So a device that wants more than the default current feeds both CC pins into ADC inputs or comparators, takes whichever is higher, and limits its load accordingly. Default means 500 mA for USB 2.0 (900 mA for USB 3.x) after enumeration; a device that never reads CC should stay within that.
One caution on A-to-C cables: a compliant one has a 56 kΩ pull-up from CC to VBUS inside the C plug, which advertises default current so that the device does not overload an old USB-A port. Some early or cheap cables used 10 kΩ instead, falsely advertising 3 A. If your device draws according to the CC reading, such a cable can make it overload a weak port, so conservative designs also watch VBUS for sag.
For anything above 5 V, resistors are not enough. Higher voltages are negotiated with USB Power Delivery messages on the CC line, which needs a PD controller IC; small standalone sink controllers exist that request a fixed voltage without firmware.
While you are revising the board, check the data pins. A USB-C receptacle has two D+ pads (A6 and B6) and two D- pads (A7 and B7) so that the plug works either way up, because a USB 2.0 plug has the pair on one side only. Join A6 to B6 and A7 to B7 on the PCB with short traces right at the connector; if only one set is wired, data works in one orientation and not the other.
The 16-pin receptacles carry exactly this subset: VBUS, GND, CC1, CC2, the two D+/D- pairs and SBU1/SBU2, which you can leave unconnected. The 6-pin power-only versions have just VBUS, GND and the two CC pins, and they still need both 5.1 kΩ resistors.
Back to USB community
