why are usb 2.0 differential traces terminated with 45ohms each to ground, instead of being terminated with a 90ohm resistor across the pair? and if it is equivalent, why would we choose one way over the other?

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    \$\begingroup\$ It is necessary to use differential to null the effects of common mode bias and CM noise and use impedance matched to transmission line source and termination. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 16 '19 at 19:34
  • \$\begingroup\$ USB 2.0 uses single-ended signalling for end of packet, reset, and some other signals. My guess would be that would be why you would terminate using 45 ohms to ground \$\endgroup\$ – Oscillonoscope May 16 '19 at 19:43

USB 2.0 signaling is actually a single-ended signaling but "driven differentially" on major data transfer functions. The original idea of USB physical drivers was a 18-mA current source into a transmission line terminated with 45Ω on both ends. As result, the impedance as the driver sees is is 22.5Ω, times 18 mA = 400 mV. During the data packet transmission the driver drives 18 mA into one wire, and drives nothing into another, and then vice versa, making the voltage differential. Each "shoulder" either drives up, or is OFF, making each data line to be alternatively into positive swing. As it was mentioned in this site already, the signal is not completely "balanced", leaving 200 mV of common mode. Because of this way of "one-side driving" and resulting common mode, the termination must be split and "center grounded".

In recent time Intel was driving an implementation of USB 2.0 drivers in a form of "voltage-driven" driver, which initially resulted in quite a few hiccups in interoperability with older classic devices.


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