Hey guys i cant find any usefull guide on how USB c charging works in depth. In particular i have bought a pair of Sony headphones which i would like to make wireless change so I also bought a crappy wireless coil meant to convert a phone into wireless charging. i opened the headphones, located the ground and 5v pin coming from the USB connected the circuit and surprise the charging led doesn’t light … The charging board is separated from the main board so I checked the flat cable that connects them, found the 5v and gnd ,spliced into it, and the led light lit as if it was charging. the next morning the led was of signaling the headphones are full, unfortunately after powering them on the battery status indicated was still 20% as the evening before … Have I done anything wrong ? What about that phase when they negotiate the power output with a magic resistor ? What should I try next? Thanks in advance 👍🏻

  • DNOSOP
    link
    fedilink
    English
    arrow-up
    1
    ·
    9 months ago

    Are you sure ? How does a standard phone fast charge then , its not the case anyway cause the headphones regularly charge at 5v so it’s enough… Thanks anyway for your comment 👍🏻

    • jeinzi@discuss.tchncs.de
      link
      fedilink
      English
      arrow-up
      1
      ·
      edit-2
      9 months ago

      There are other fast charging standards than Power Delivery. USB Battery Charging defines that when the data lines are shorted, a device can draw more current (up to 1.5A), but still at 5V. QuickCharge on the other hand uses the data lines to negotiate higher voltages, so an A-to-C cable can’t protect you from that.

      • DNOSOP
        link
        fedilink
        English
        arrow-up
        1
        ·
        9 months ago

        Thanks so assuming the headphones use the USB battery charging standard ( just because I have measured only 5v output and it seems logical) how should I comnect up the second charger ? Do I still need to find the cc line (guessing a 5 v pin the closest to the USB port itself) and add a resistor ? The diode must be there in each case right ?