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So when exactly is termination of a digital signal required?

In my case, I'm dealing with a 1.8V signal that comes from an FPGA. This signal connects to a high impedance input of an OpAmp. The trace on the PCB is just about 5cm (about two inches).

The maximum frequency that I'll drive this signal will be 4Mhz. Right now I have no termination in place and everything works just fine, even with the huge impedance mismatch.

I want to learn the best practice and some rules of thumb when to care about termination and when it can be left out.

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It's not the frequency of your signal that matters - its rise time of your signal that matters. 1Hz and 100Mhz signals with the same rise time, will suffer the same problems because they have the same rise time.

A good rule of thumb is to aim for 20% the rise time.

If you have a signal with a rise time of 1ns, then the maximum time delay you can have without termination is 1ns * 0.2 = 0.2ns. In FR4, the speed of a signal is ~ 6inches/nsec, so you can have a 1.2 inch trace without much issue.

What that also means, is that if you have a trace that is longer than 1.2inches with a signal whose rise time is 1ns, then you need to terminate, otherwise you'll be running into signal integrity issues.

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  • \$\begingroup\$ Yes and no. Just because the rise shape of a signal can be detected as different by some instrument does not imply that anything actually connected to it is in a position to care. \$\endgroup\$
    – Chris Stratton
    Commented Jan 27, 2015 at 5:29
  • \$\begingroup\$ @ChrisStratton you referring to the last sentence ? \$\endgroup\$
    – efox29
    Commented Jan 27, 2015 at 5:35
  • \$\begingroup\$ You need to terminate if your signal integrity is marginal or the ringing is causing EMI problems. Data lines can be sloppy if setup and hold are satisfied. Clocks can generally not be sloppy (rise and fall should be monotonic to avoid double-clocking). Also, FR4 propagation speed depends on whether it is a surface trace or buried trace. Surface trace signals travel faster because they are partially in free-space. \$\endgroup\$
    – user57037
    Commented Jan 27, 2015 at 18:16

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