I hope someone can clarify this. How can the USB common mode range be less than 3 volts if D and D- pulled down with 15k, differential resistance is 90 ohms, and pullup is 1.5k to 3.3 volts? Using simple network theory, the voltage must be
3.6 x (15k || (15k + 90)) / (1.5k+(15k || (15k+90))) = 3.0 volts
The STM32F103C6 data sheet specifies the characteristic impedance is 90 ohms, and states in table 44 on page 69 that the USB common-mode voltage range is 0.8 volts minimum to 2.5 volts maximum, measured from the local ground.
How can this be? What am I missing?