Hello, nice to greet you all. I am starting to get into the topic of PLCs, and I am working on some projects. I hope you can help me, as I have been collecting information for days, but I still have some doubts. Let me explain:
It occurred to me to replace the connector of a tablet that has a micro USB port with a USB-C port. So far, everything is correct. I bought a module like the one in the image, but because the micro USB connector is smaller, I did not have enough space, and the insertion holes did not match to solder the new connector to the board. I decided to change its location, but since I have to make a new hole in the case to install the new USB-C port, it occurred to me to perhaps keep the previous micro USB port in its place and also install the new USB-C port in another part of the PCB board.
It is not an OTG port; it is a charging-only port. It does not transmit data and has a direct current input of 2.5A at 5V, which suggests that the charging signature would be a Dedicated Charging Port (DCP). Although I have not yet had the opportunity to read the SMD resistor that shorts the data paths, I have been reading about the mux and diode OR-ing setups. I realized it wouldn't be possible due to the forward voltage drop (Vf) of the diodes. Now, I think I could just make a parallel connection from the micro USB pins to the USB Type-C.
As far as I understand, and according to this post (How to replace mini USB with USB C? - #2 by kgray9), which describes a module similar to the one I plan to use, the 5th pin is ID. If I make the connection in parallel (I plan to use enameled copper wire and cover it with solder mask) and connect the voltage pins, data pins, ground, and ID pin like the schematic in the image, shouldn't it work fine as long as you don't use both ports simultaneously? Could there be any problems with negotiating power? If I am not mistaken, if the data pins on the micro USB or the CC pins on the USB-C are not connected, the device will not charge. Honestly, I wouldn't mind if it charged at the default speed, slower than it already does, as long as it charges.
Thanks in advance for all the help you can give me.
PS: I apologize for my possible grammatical errors in English; I am learning the language.
