My sports watch has a 4-pin connector on the back used for charging and data exchange. The male connector can be plugged in 2 ways. This means that internally the power and ground are swapped. Same for the data lines ( either RX, TX or USB+, USB-)
I want to implement similar functionality for a project that i am doing. I have been doing research but cant seem to find a solution for this. Besides dealing with the reverse order of the pins, the solution should also provide short protection (the watch can be submersed in water which shorts the pins)

Are there IC solutions for this?
Does someone know of a reference design for this?
Can someone share schematics of a solution?

This is the type of connector that i want to use for my project:

enter image description here

  • 1
    \$\begingroup\$ On USB-C connectors and Lightning connectors it's entirely mechanical, no circuitry required (other than the connections to cross-wire the pins). Of course that won't work with the connector you want to use. \$\endgroup\$ Jan 2, 2021 at 2:58
  • 1
    \$\begingroup\$ maybe that connector cannot be reversed without forcing it ... it is possible that the magnets repel each other when aligned incorrectly \$\endgroup\$
    – jsotola
    Jan 2, 2021 at 3:08

1 Answer 1


I don't think there's an ic solution to such an easy problem: put and/supply on two "mirrored" contacts, have a simple diode bridge provide the right polarity regardless of orientation.

With that power, use microcontroller or the SoC that your device has anyway to detect the polarity.

Most microcontrollers have rather flexible pin assignments, so maybe software suffices.

If it doesn't, a pair of analog switches or bus switch ICs controlled by your software will do.

  • \$\begingroup\$ you could control the analog switched from the power input terminals. \$\endgroup\$ Jan 2, 2021 at 4:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.