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I've inherited a design that has a pair of PCBs mated with two-row mezzanine connectors (0.5mm pitch, 2.5mm mated height)

There are two differential pairs that travel through the connectors (RX and TX), and each pair passes through two conductors of the connector.

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It appears to have been done this way due to the mechanical nature of the PCBs, where the connectors are extremely close to the edge and the PCBs travel in opposite directions from one another. Granted it's a neat way to route with a flow-through type of design, and the signals are 100 Mbps, so not particularly fast, but this sort of loop-come-stub makes me feel uneasy. To my knowledge there are no problems, and I haven't had the opportunity to probe.

Does this behave like a stub?

Should I expect to see reflections?

Anything else to be aware of?

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    \$\begingroup\$ I think you're reading too much into it. IMO there's no significant high frequency issues here at all. I think they did this for mechanical reliability. If one connection fails, the other keeps the signal flowing. Sometimes pins are duplicated for power reasons, but seems not the case here. \$\endgroup\$
    – Kyle B
    Mar 31, 2022 at 17:25
  • \$\begingroup\$ No, it doesn't behave as a stub. It might be an impedance discontinuity, but at your speeds the effects of that discontinuity are invisible. \$\endgroup\$
    – SteveSh
    Mar 31, 2022 at 23:15

2 Answers 2

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I have no idea why they did that.

If this has been proven to work then I'd leave it be. At those speeds there's not much going on here.

At higher speeds I might start to worry about skew differences between the paths, and the reflections in the interconnect relative to what any equalization can handle. But at 100Mbps you shouldn't need more than about 75MHz of bandwidth and these dimensions are pretty small.

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  • \$\begingroup\$ Shouldn't that be ~300 MHz bandwidth, not 75 MHz? \$\endgroup\$
    – SteveSh
    Mar 31, 2022 at 23:06
  • \$\begingroup\$ As a rule of thumb, we design for ~1.5x Nyquist. For a 100Mbps data rate, alternating 1's and 0's yield a 50MHz square wave. Preserving 1.5x that bandwidth usually yields a sufficient eye. This depends on reflections, spacing of reflections, channel loss, equalization, etc. \$\endgroup\$
    – 65Roadster
    May 15, 2022 at 18:53
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Impedance match, is my guess.

Paralleling the 2 paths halves the impedance through the connector. That lower impedance may be a closer match to the impedance of the traces to the connector.

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    \$\begingroup\$ My first thought as well. Probe it with a scope and see if you're getting reflections. If not I would leave it as is. \$\endgroup\$ Apr 1, 2022 at 0:58

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