I can't quite get to the bottom of a USB peripheral auto-switching circuit diagram.
Here's the scenario: I'd like to share a USB hard disk enclosure between two devices - a media player and a games console. Since only one device will be in use at a time, I need a USB switch of the sort people sometimes use to share printers - here's one for sale on Amazon UK
But, since I'm a cheapskate, and since I have drawers full of electronic components - it's a fairly recent hobby - I'd like to build my own automatic USB switch. The plan is to use the +5V from the USB input of device B to activate a relay, switching the disk to that device. When the device is unplugged, the relay switches the disk back to device A.
I found a circuit - here it is
- that appears to do what I want. After puzzling for a minute over the extra connections on the USB sockets I'm guessing that 5 and 6 are the spare ID pin found on mini-USB and the shielding on the cable, both to be connected to ground. Right?
What I don't understand, however, is why it's only the D+ and D- that are being switched. Shouldn't I be using a 3PDT relay and switching the +5V as well? What am I missing?