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This is a thought experiment. I would like know and what will be the bit-rate in differential channel when signal is driven differentially and in single ended mode? Lets there is differential trace in a PCB, first signal is driven through the channel as differential signal, then one of the trance is excited with one of the differential signal and the other trace is left as it is or grounded. Is there going to be any change in the bit-rate of the signal?

I have been listing this webinar and that triggered this question.

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    \$\begingroup\$ Signal speed as in signal propagation speed or how much data you can send per unit time? \$\endgroup\$ – crossroad Mar 5 at 5:28
  • \$\begingroup\$ @syntax, signal speed, not the bit or baud rate \$\endgroup\$ – student7 Mar 5 at 14:13
  • \$\begingroup\$ Like DKNguyen mentioned, signal speed is not a thing. You're talking about propagation speed. \$\endgroup\$ – crossroad Mar 5 at 17:14
  • \$\begingroup\$ My bad, I am talking about signal propagation, though I am not sure, how the term signal speed is wrong. \$\endgroup\$ – student7 Mar 5 at 19:39
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    \$\begingroup\$ Propagation speed is how more like the time it takes for something to travel down the line and register at the other end. Data rate is how much data you can cram down the line once the ball gets rolling. For example, when I speak, the speed of sound is the propagation speed, but the data rate is how many words per minute I can speak. At very long distances, it take the sounds of my voice a long time to initially reach you (latency), but when it finally does, you will hear the words at the same rate which I originally spoke them at (throughput). \$\endgroup\$ – DKNguyen Mar 5 at 19:50
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I assume you are asking about data rate (not propagation speed, and not signal speed which is a term that doesn't mean anything). But clarify if you are asking about propagation speed and whether the return current causes the single-ended signal to take twice as long to register as the differential signal. I don't think it does but I can't explain it well. The answer below is about data rate only, not the speed of propagation of the electrical signal down the wire.

For the same slew rate and voltage transition levels, a differential pair has double the noise margin of a single-ended signal.

That means that you can reduce the voltage transitions (and associated logic thresholds) on the differential pair and reduce the noise margin until it is then again equal to the single-ended. The result is that the the slew rate has remained the same, the noise margin has remained the same, but the voltage transition on each wire has been reduced so when you flip the signal on each wire, it is able to finish its transition faster which lets you send data at a higher rate.

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  • \$\begingroup\$ "but now you only have one signal that must travel the entire distance on its own rather than two that only need to travel half way", I not sure what this line means, can you just add a quick explanation? \$\endgroup\$ – student7 Mar 5 at 15:00
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    \$\begingroup\$ @student7 See edit. \$\endgroup\$ – DKNguyen Mar 5 at 17:34
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    \$\begingroup\$ @student7 Actually, something got confused in my head and I made a mistake now that I am looking at the graph. Each signal in the differential pair still traverse the same distance as the single ended. So the distances traveled is the same. What it is, is that the differential signal requires two signals to travel in opposite directions so you have twice the noise margin. That means you can reduce the slew rate and reduce the voltage thresholds while maintaining noise immunity so you can transition faster for faster data rate. \$\endgroup\$ – DKNguyen Mar 5 at 19:26
  • \$\begingroup\$ +1 this comment make it easier to understand. \$\endgroup\$ – student7 Mar 5 at 19:36
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    \$\begingroup\$ Whenever I say "distance" I mean voltage transition, not something moving along the length of the wire. \$\endgroup\$ – DKNguyen Mar 5 at 19:38

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