Hi,
From thread ( https://community.nxp.com/t5/i-MX-Processors/i-MX8-USB-VBUS-voltage/m-p/1266085?lightbox-message-ima... ) It is clear that VBUS feeds an internal LDO that inturn feeds the USB PHY for IMX6. However, for IMX8 it is not clear from the picture what this pin is connected to on the inside of IMX8. Can you help to clarify this for me?
Is there anyway to configure the USB block to skip VBUS connection and still use USB?
Br
Anders
Good point, it is not at all clear how this Pin functions.
It is clear what voltages it works on, but the functionality of the pin is not clear.
What happens in situations of different voltages at the input? What happens if VBUS is not provided? What happens with unstable VBUS?
In the case of an embedded hub that is connected directly to the IMX8MP, do I need to connect the VBUS pin to the HUB? Or enough to connect to constant VBUS voltage?
Sorry we do not have that. If you using the type-C design please use the design as our reference design. If USB connector is MicroAB or MicroB, USBx_VBUS MUST not connect directly to the 5V VBUS voltage
of connector; Instead, this pin must be isloated with an external 30K 1% resistor.
For the i.MX8MP you can see our reference design:
Thanks for reply!
Yes, I understand that this is how you should connect it. My question is for deeper understanding what this pin actually does. If this pin is low, is the entire USB block disabled? is this true even for boot? A block schematic of how this pin is connected on the inside of IMX8M plus would be nice.
Br
Anders
My experience is NXP engineers don't know their chip than you do.
So let me give this NXP engineer some basic knowledge. then this nxp engineer can understand what you are talking about.