2024-03-19 07:58 AM
I am designing a self-powered device with an STM32G473 and a USB-C connector, and want to be certain about the connections.
I have made PCBs with USB-C connectors before, but this is the first time that I have neither a VDDUSB, nor a VBUS Sense input. As I understand it, this microcontroller uses the USB-C CC signals instead of the non-existing VBUS Sense input, and two 5k1 pull-down resistors (for CC1 and CC2 in the connector). My understanding is that I should connect CC1 and CC2 directly to the microcontroller's UCPD1_CC1 (PB6) and UCPD1_CC2 (PB4) pins, and that the microcontroller has built-in pull-up and pull-down resistors for the CC ports.
Can anyone confirm this?
I don't think that I need to use the UCPD1_FRSTX (PA2) signal in this case, as there is no power delivery. However, I find this part quite confusing when reading the STM32G473 datasheet and reference manual together with application note 5225.
Can anyone shed some light on this?
2024-04-26 07:16 AM - edited 2024-04-26 07:23 AM
Hi @EThom.3
In datasheet, we have a note describing the procedure,
UCPD1_FRSTX signal is used for Fast Role Swap transmissions in USB Power Delivery. If you are not using USB Power Delivery dual role, this signal is not required.
If dead battery is not required, UCPDx_DBCC1 and UCPDx_DBCC2 pins must both be tied to ground.
In AN5225, we took an example of STM32G071 since it has 2 UCPD controllers.
Internal Ticket 180168 is submitted to clarify Figure 15. Device pinout example applicable to STM32G071 LQFP64 package
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2024-05-14 06:13 AM
@FBL wrote:
(...)If dead battery is not required, UCPDx_DBCC1 and UCPDx_DBCC2 pins must both be tied to ground.
I assume that I can just disable the signals...?