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I have a tablet for which I replaced the charging connector from micro USB to USB-C using a PCB module with a USB-C connector. Since there was no space on the PCB to mount the new connector, I had to change its location on the main board. This involved cutting a new hole for the USB-C connector and covering the old micro USB hole in the case. This raised a question: would it be possible to keep the old micro USB connector and add the new USB-C connector so that both can charge the same battery?

After doing some research and receiving a lot of help here, I successfully connected and wired the USB-C connector with pull-down resistors. Currently, it charges at 500mAh, but with the micro USB connector, I could charge at 2.5A. Is it possible to have two different connectors connected to the same battery?

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  • \$\begingroup\$ Possible, yes. Good idea, probably not. To make it work safely you'd have to include some electronics to prevent one connector backfeeding into the other, or some mechanical means preventing both from being plugged in at the same time. Your best bet is to figure out why it can't charge at 2.5 A with the USB-C connector (I would guess you just don't have the electronics to do the negotiation) and fix that, rather than trying ot have two connectors. Or just keep the micro-USB one and use that. \$\endgroup\$
    – Hearth
    Commented May 24 at 21:25
  • \$\begingroup\$ We can't know how you wired them. It is possible that the supply and tablet cannot negotiate now. Do you know what standard the tablet supports for more current over USB, and does your new supply support the same protocol too? Please note that different supplies may speak different protocols. 500mAh is a capacity, not rate. The connectors also do not connect to a battery, but to a charging circuit. \$\endgroup\$
    – Justme
    Commented May 25 at 7:43
  • \$\begingroup\$ @Justme I haven't put it into practice yet, as I want to fully inform myself before proceeding. I initially planned to create a mux configuration using diodes, but I realized that the diodes I intended to use would cause a significant voltage drop. Someone suggested connecting the two power pins in parallel instead. I'm still analyzing this option while waiting for the connectors to arrive in the mail. \$\endgroup\$
    – condor12
    Commented May 27 at 1:23

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The amount of consumed current is determined by DEVICE (provided that the source can supply it).

If the DEVICE takes 2.5A over micro-USB connector, it means that the DEVICE was able to understand/recognize capability of the source, and engage the corresponding charging mode. Most likely it was the Quick Charge protocol over D+/D- wires.

With Type-C connector, the 5k1 pull downs on CC pins only indicate to the source that a connection has been made, nothing else. The source then enables the VBUS. The source advertises its actual (primary) source capability by the strength of its pull-ups, see this topic. If the DEVICE does not have means to understand the meaning of CC voltage levels, it will default to basic 500mA charging.

Therefore, in order for your old tablet to take higher charge over Type-C port, you will need to instrument a CC controller/receiver (to measure the resulting voltage level on one of the CC line), and convert this information into QC serialized protocol (or BC protocol, whichever your old tablet is designed for) over D+/D- wires. This would be a pretty ugly (and challenging) solution.

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  • \$\begingroup\$ @ale.chenski I have a question. I was suggested the idea of connecting the pins in parallel. This should work, provided both power supplies are not connected simultaneously. What is your opinion on this? In that scenario, should I connect the voltage pins in parallel and also add pull-down resistors to the DC pins of the USB-C connector? It appears that the micro USB data pins are already shorted by the resistor in the BC protocol. Thanks again. \$\endgroup\$
    – condor12
    Commented May 27 at 19:05

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