Thanks yes, each of the two signals needing a ground for twisted pairing.Note says two grounds are required. Why? TX side, the grounds are
common (by symbol-tie).
Thanks, you mean an "LCLCLCLC" type thing for the distributed L and c of the txmission line?i don't see a cable (transmission line) in the simulation circuit. In so far a no-question.
You can believe what you want, but your "common sense comms" are not going to give the results you want in a real system.i believe what is needed now is "common sense comms"....not all this "balanced signalling", " differential signalling", etc
Thanks, in that case, the twisted diff pair has to get twisted with the ground.Taking a differential pair and then separating the pair and twisting the individual wires with grounds makes zero sense-you’re defeating the whole common-mode rejection ability of diff pairs.
...thanks, i believe that my method agrees with ethernet...if you see point 6 of the following...If you want to know how it's done in the real world, look up how HDMI or high-speed Ethernet signals are sent.
“As I know” sending just the diff pair is completely possible. It’s done ALL THE TIME. A ground does NOT ‘have to go as well’.As you know, sending just the twisted diff pair is not possible......a ground has to go aswell so that both PCBs are both grounded tot he same potential.....and then you have to mind the area of that diff signal__ground loop.
There is no choice in this, i believe many would agree?
...Page 6, Fig 11.1 of this“As I know” sending just the diff pair is completely possible. It’s done ALL THE TIME. A ground does NOT ‘have to go as well’.
...Thanks, just noticed this, will trace out the currents in the loops to see it. It seems strange that if the current sums to zero, why then is slla272d above going on about ground return current in page 6, fig 11.1?Voltage and current of differential signal lines adds to zero, therefore no return path is required as such.
No. That is not what that diagram is showing. It is showing that a difference in ground potential in the two different systems manifests itself as a common-mode offset. There is NOTHING in the diagram or the text that says you need a ground return for the differential signals. Nor is there anything in the document saying you should run ground “closely alongside each signal”. Or run ANY ground....Page 6, Fig 11.1 of this
..shows that the ground is used as a return by each differential signal. As such, ground should go closely alongside each differential signal. (to reduce the current loop area)
Thanks, page 11 , section 11 states...There is NOTHING in the diagram or the text that says you need a ground return for the differential signals
Thanks, the attached shows differential signalling, and the currents do not sum to zero, as the waveform shows...the shown currents are the currents in each differential signal wire.Voltage and current of differential signal lines adds to zero, therefore no return path is required as such
...Thanks, do you mean the signals in the LTspice sim of post #18 aren't inverted with respect to each other?Normal practice is to use TWO wires carrying inverted signals with respect to each other,
...Thanks, though do you think the dead time i inserted does harm?....the signals will be read in the centre of their "high time"......well away from the dead time bit.If you use a conventional differential transmitter it wont have the gap between the signals.
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