2 Replies Latest reply on Feb 24, 2015 8:01 AM by DiegoV_Intel

    USB FAULT and PSW pins in host mode

    Michael_Uniquest

      I want to attach external camera module through USB port of Edison module.

      Edison works as USB host and camera module works as USB device.

      In this case, how should I treat USB FAULT and PSW pins from Edison module? Just leave these pins floating will be OK?

        • 1. Re: USB FAULT and PSW pins in host mode
          rosenrot

          Here you can find an explanation what does pins do Treating Edison USB Fault pin (on the system that doesn't include USB)

           

          Short answer: From my opinion, you could leave them floating.

          • 2. Re: USB FAULT and PSW pins in host mode
            DiegoV_Intel

            Hello Michael_Uniquest,

             

            The FAULT and PSW pins are used to control the external power switch and monitor the overcurrent condition on VBUS line which will provide power to the USB device. So, in summary, these two pins are used to control how you power the USB device. The use of these pins is recommended but not mandatory. That will depend on your design.

             

            If you want to use them you have to use an external IC which actually will monitor the VBUS line. The Breakout Board and the Arduino Expansion Board use the MIC2039 IC for that purpose. You could use this one or another IC of your preference if you plan to use these pins.

             

            The other option is to not use protection for the VBUS line so it would not be necessary to use the FAULT nor the PSW pin. The FAULT pin is an input pin and it is active low so, if you won't use it you should connect it to a pull-up resistor to avoid activating it. Notice that these pins are 1.8V tolerant. If you connect the pin to 3.3V or 5V you will damage the module. In the other hand, the PSW pin is an output pin, so it could be leave floating but personally I never like to leave pins floating so you could connect it to a pull-down resistor.

             

            I hope your question has been answered with this explanation.

             

            Regards,

            Diego.