Electronic – Why is USB D+ tied to VDD (for PCM2707 at least)


I'm trying to implement an audio DAC using PCM2707 as a receiver and, looking over a couple schematics of previous implementations – at least one being verified as correct, I noticed D+ being connected to VDD through a 1.5k resistor. Considering I trust the source of the schematics, I take this to be correct.

But why is implemented that way? I'd really want to understand how it works… and Google has not helped me so far.

Best Answer

It is part of how USB negotiates the speed at which the host and device plan to communicate, as well as allowing for presence detect of a device being plugged in. Basically, the host has a 15K pull-down on both D+/D-, and the device has a 1.5K pull-up on one of the 2 lines to indicate full or low-speed. You can read more at the following link: USB in a NutShell.