Does the Bluetooth WiFi card M.2 connector have the I2S signals wired up to the chipset or audio codec?
The signals involved are described in section 3.1.9 of the PCI Express M.2 Specification:
PCM_CLK/I2S_SCK - pin 8
PCM_SYNC/I2S_WS - pin 10
PCM_IN/I2S_SD_IN - pin 12
PCM_OUT/I2S_SD_OUT - pin 14
My question is specifically for the AMD 7640U but it could be interesting for the other motherboards as well.
Welcome to the forum.
For questions such as this, it’s likely the best information you’re going to get can be found in Framework’s github. They have basic overview schematics for the mainboards. If exactly what you want isn’t there, then you might be down to hoping that an active forum user has happened to also explore what you’re interested in. Whatever you find, please post it for others who might also be interested & who come across this thread you started.
Here is the (public, reduced) schematic: Framework-Laptop-13/Mainboard/Mainboard_Interfaces_Schematic_7040_Series.pdf at main · FrameworkComputer/Framework-Laptop-13 · GitHub
Thanks!
The schematics show it connected to page 10 (if the resistors are installed). Pages 6~14 are not public but the index page shows these contain the CPU so it seems the I2S is connected to the CPU/chipset there.
The Intel series motherboards do not connect the I2S, instead the CNVio is likely used.
It notes above the resistors “DVT1 change BOM to install”. You might know the meaning, but for anyone else who’s interested in these pins, DVT is one of the stages to production manufacturing in full volume. It goes EVT (Engineering Validation Test), DVT (Design Validation Test), PVT (production validation testing), the first small production run. BOM is Bill Of Materials, the list of components that will be used in manufacture. So “DVT1 change BOM to install” says that a change was made to add the resistors for the next stage. A good indication that they should be there in production mainboards. And to be clear, since the resistors a place in series, if there are not there, then those pins would be left not connected.
