2020-06-29 06:50 PM
Hello,
I'm using STMCubeIDE, with firmware version: STM32Cube FW_F4 V1.25.0
I am trying to implement a USB virtual COM port. My code is working on the STM32F4 discovery board. I then created a new project for the STM32F411CEU6 "Black Pill", but the USB device will not enumerate. Error message in windows is "Device Descriptor Request Failed"
For the project, I select "Device Only" for USB_OTG_FS, and in the USB_DEVICE middleware, I select "Communication Device Class (Virtual Port Com). The only other change I made to the generated code was to usb_cdc_if.c file to add the lineCoding.
I've attached my clock configuration screenshot.
Has anyone tried to implement a USB device on one of these black pills? If so, would you know if I'm making any obvious error with what I'm trying to do?
Thank you very much,
Garrett.
Solved! Go to Solution.
2020-06-29 08:30 PM
Hello Everyone,
I hope it's not bad etiquette to respond to my own question, but I believe I may have found out my issue. I am using a hub to connect to, and this hub works for the STM32discovery board but not the black pill. It's strange but if I do not go through the hub (i.e. directly to a USB port on the computer) then there is no issue. I wonder if It is a power supply issue? Maybe from the hub the black pill does not have stable power before it tries to enumerate. I'm just guessing.
I am OK with plugging into the computer, so I think I'm all good.
Thank you very much,
Garrett.
2020-06-29 08:30 PM
Hello Everyone,
I hope it's not bad etiquette to respond to my own question, but I believe I may have found out my issue. I am using a hub to connect to, and this hub works for the STM32discovery board but not the black pill. It's strange but if I do not go through the hub (i.e. directly to a USB port on the computer) then there is no issue. I wonder if It is a power supply issue? Maybe from the hub the black pill does not have stable power before it tries to enumerate. I'm just guessing.
I am OK with plugging into the computer, so I think I'm all good.
Thank you very much,
Garrett.