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 ππ»
That looks like a really simple USB C port in legacy A (with DP/DM charge signals) configuration. The single IC most likely does the USB negotiation and the CC/CV charge of the battery.
Often these devices are tightly coupled with the USB state machine. Just applying 5v without terminating the sense resistors won't do anything.
Also be very careful with Qi chargers. The resonant circuit produces voltage spikes in the tens of volts. It is usually regulated and smoothed to 5v out. But it's also very bursty. Make sure your output circuit after the coil is fully regulated 5v voltage. I've seen them be varying voltages and even current mode outputs.
Thanks is there a way to know if the voltage spikes are properly filtered out without a spectrum analyzer I only have a multimeter ( and It shows something around 5v )I also noticed the coils will randomly die, I have altrady bought 2 from china and now I'm wondering if it has something to do with me charging the headphones via USB c while the coil Is plugged in ... I Will add photos of the circuit as soon as possibile thanks again for your help
Have you checked what voltage arrives on the board if you use a regular USB-C charger? Maybe the headphones negotiate and need a higher voltage than 5V.
Does it charge with only 5V supplied? (could be forced by using an A-to-C cable in case of Power Delivery)
For Power Delivery, higher voltages are negotiated using the CC (configuration channel) pins. If you use an A-to-C cable, the A side does not have the CC pins and therefore you can't get more than 5V.
Two things come to mind with USB C charging that's usually different than micro USB or prior standards. First is that some USB C circuits are tightly coupled to battery charger circuitry. The idea is that if the pack voltage is higher than standard 5v, then it may be advantageous for USB C to request one of the higher voltages from the wall-power block. But I think this is unlikely, if the photo you've included is of the board in question; there would usually be the leads for the battery pack attached to the same board.
Second, USB C -- unlike every other USB spec preceding it, I think -- requires the sense resistors on the consume side before any power is supplied, even low-current 5v that we would otherwise expect. What's probably happening with the LED becoming lit is that it's probably indicating a data connection, as if you connected to a computer.
Thanks your comment is really helpful I will dig into the link you send tomorrow ... In the meantime what is the likely solution? find a resistor and permanently solder it on, the problem is that I would like to keep both USB and wireless charging options ( I don't think it has any data capabilities) again thank you so much for your answer
You might try the resistors first, although in such a way that it could be removed if that doesn't work. If you have correctly found the D+ and D- lines broken out from the USB C connector, and added resistors to those, I can't see how that would cause an issue with the wireless charging; they should be mostly independent, since if there's no USB power supply attached, the resistors won't be sensed anyway.