Hi, I´m reading the following document : "an4879-introduction-to-usb-hardware-and-pcb-guidelines-using-stm32-mcus-stmicroelectronics" to implement USB in my application.
In section 2.6.1 Simple Resistor Divider it says "This solution works well as long as GPIO maximum operating conditions are >= VDD+ 3.6 V" this must be meet to use the following voltage dividider
I´m going to use PA9 pin, in the datasheet of my MCU STM32L4P5RGT6 it states this pin is 5V tolerant
As I understood this pin doesnt meet the condiciton mentioned before because my VDD=3.3V and so GPIO maximum operating condition should be >= VDD+3.6V so we have 3.3+3.6V=6.9V so 5V < 6.9V so I can't use this configuration do I understand this well ?
Should I use this other configuration?
I´m a bit confused about it hope some one could help me to clarify this.
It'll work.
PA9 is a 5V tolerant pin.
So incoming voltage limits are VDD + 4.0 V:
Voltage there will never be more than 3.75 V, as long as VBUS respects USB limits (4.75 V to 5.25 V).
The issue here is ensuring the voltage on the pin is acceptable when the device is off. This is satisfied.
PA9 is a five volt-tolerant pin, including an additional function natively dedicated to VBUS sensing.
The absolute maximum ratings table of the datasheet indicates that the five volt-tolerant pin voltage cannot exceed VDD+ 4V.
You can check the schematics of STM32L4P5 Disco board provided as reference:
Thanks for your replies @FBL @TDK
Okay, I think I got it, so just one question why in one parte of the" introduction to USB hardware" it shows the following
"five volt-tolerant pin voltage cannot exceed
VDD + 4 V."
And then in the MCU datasheet the the following appears with respect to five volt tolerant pins
So it´s VDD+4V or min(VDD) +4V ??
Hope you can clarify this, thanks for your time.
