Arduino Nano USB connection with external power

Hi I know this has been touched on a few times but i'm really trying to understand how this is actually going to work.

I have built a small 7805 external power supply with a distribution bus for all the items I need to power outside on the nano.

I would also like to use the usb port for serial data...

How does the zener autoswitch? Is it really highest voltage wins? I can see from the schematic that is its across the VCC and USBV. How best to ensure that the external regulator always wins, and will i then have problems powering the ftdi chip?

When I power the nano from the usb I see I get +5v on the vin pin wrt ground on the nano.

Thoughts with pics below (ive drawn on the power lines as I planned)

https://www.dropbox.com/s/0uugy1infu6su6s/2014-12-30%2017.27.59.jpg?dl=0 https://www.dropbox.com/s/2lrmo9r8rym1iya/2014-12-30%2017.32.42.jpg?dl=0

thanks in advance...neil

So i buzzed out what I think is the zener and one side is on the +5v usb and the other is on the +vcc of the Arduino pin.. so I think that answers my own question as to the identification of the zener...

Still need help figuring out best practice on power the Arduino

Thanks in advance

Neil

If the 7805 output is > Vusb + 0.45V (the drop across a MBR0520), then diode will be reverse biased and regulator current will be supplied to all the parts on the board, including the FTDI chip. Otherwise, Vusb will supply up to 500mA of current to all the parts on the board. The diode is mainly there to keep the regulator from back-powering the Vusb source in a computer.

If your external source connects to Vin instead of 5V, the regulator won't regulate (it needs ~ 6V to 6.5V), it will have some loss from input to output, and if USB is connected then Vusb will power the board.

If you want to ensure regulator power is always used, then remove D1.

Great thanks crossroads its looking good so far :) Like the tip of removing D1 in the future.