Differential pairs, such as for ethernet or a CAN bus, can be designed to be loosely coupled or tightly coupled. For a loosely coupled differential pair, the odd-mode and even-mode impedances are roughly equivalent to the single-ended impedance (impedance of one conductor relative to the common, with the other conductor tied to common).

If you designed the single-ended impedance to be 50 ohm for a loosely coupled differential pair, then the odd and even-mode impedances are both about 50 ohm, and thus the differential impedance is 100 ohm, and the common-mode impedance is about 25 ohm. For Ethernet and CAN bus I have not seen schematics that use a 3 resistor topology for terminating both the differential and common-mode impedances. Is this because the design is for a loosely coupled differential pair such that only 2 resistors are needed to provide a matched termination for both the differential and common-mode signals?

I have never seen the application note for the physical layer of Ethernet or CAN discuss the coupling of the differential pair. For example, for Ethernet, the 50 ohm terminating resistors on each line to the power rail, would provide a differential load impedance of 100 ohm and a common-mode impedance of 25 ohm, which would match a loosely coupled differential pair in which each trace of the pair has a singled-ended impedance of 50 ohm. Similarly, I have seen for CAN that each line is terminated with 60 ohm to the split reference (which has a capacitor to the common).

posted almost 3 years ago

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3y ago by Andy aka‭

DavidE‭

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