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Suppose we have a parallel termination like this on a high-speed trace:

enter image description here

If the line impedance \$ Z_o \$ and \$R \$ are matched, then there are no reflections from load to cause any ringing at the driver output or at the receiver input.

But if the source impedance \$R_s\$ of the driver doesn't match with the line impedance \$ Z_o \$, shouldn't there be reflections still at the driver output from the line? So isn't that impedance matching necessary as well?

I am asking this because, in serial termination scheme, we match the source impedance to the line impedance. But in parallel termination scheme, I just can't see this concept. So, does that mean we have to mix both the schemes to ensure signal integrity on a high-speed trace?

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  • \$\begingroup\$ What is the impedance looking into the line at the point where R is connected? \$\endgroup\$
    – Chu
    Apr 29, 2021 at 8:00
  • \$\begingroup\$ In series termination also, there is termination at the load and reflections will occur at load how they are hanled? \$\endgroup\$
    – user19579
    Apr 29, 2021 at 8:51
  • \$\begingroup\$ Yea, that's the drawback of series termination. The signal needs time to settle after looping back once after reflection and gets terminated by driver source impedance. \$\endgroup\$
    – Mitu Raj
    Apr 29, 2021 at 8:56
  • \$\begingroup\$ @Chu \$ R_s\$ plays a role in selecting the value of parallel termination \$ R \$ you mean? \$\endgroup\$
    – Mitu Raj
    Apr 29, 2021 at 9:01
  • \$\begingroup\$ Do you think that adding R at some point along the line affects the impedance seen at that point? \$\endgroup\$
    – Chu
    Apr 30, 2021 at 11:55

1 Answer 1

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But if the source impedance \$R_s\$ of the driver doesn't match with the line impedance \$Z_o\$, shouldn't there be reflections still at the driver output from the line?

Yes, there will be miniscule reflections between the driver output and the start of the transmission line. If that distance is greater than "very short" then the reflections will increase and could cause a problem. However, we try and ensure that the track distance between driver output and transmission line is short. This means that reflections are small and insignificant.

You can also get mismatches caused by the receiver input capacitance/impedance shunting the load resistor \$R\$.

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    \$\begingroup\$ "Very short" refers to the lambda/10 thumb rule? \$\endgroup\$
    – Mitu Raj
    Apr 29, 2021 at 14:35
  • 2
    \$\begingroup\$ @MituRaj yes it does. \$\endgroup\$
    – Andy aka
    Apr 29, 2021 at 15:58

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