I have seen the answer for when to consider PCB trace as a transmission line in many places.

Typically, if the signal pulse rise time is ‘small’ compared to the time it takes for the signal pulse to propagate (e.g. ½ of the total time the signal takes to travel along the trace) then you need to consider your PCB as a high-speed circuit

(Source: EMA Design Automation)

No one explains how the signal integrity problems arise when the rise time is smaller than two times the propagation delay.

Please explain:

  1. What happens to a signal when its rise time is less than the propagation delay?
  2. How will this condition cause signal integrity issues?
  • 2
    \$\begingroup\$ Note that SI problems only arise when there are impedance changes or mismatches along the transmission line. If everything is matched (source, transmission line, destination), then there are no reflections and so no SI effects. \$\endgroup\$
    – SteveSh
    Jul 23, 2022 at 10:39
  • \$\begingroup\$ Thank you,but may i know what happens to a signal when it's RISE time is less than the propagation delay \$\endgroup\$
    – Hari
    Jul 23, 2022 at 10:41
  • \$\begingroup\$ Many things happen to distort the signal from complex impedance mismatched. The sensitivity to edge/phase error is affected as well as amplitude. It can affect eye pattern, jitter Return Loss, Transmission Loss , noise interference. Can you be more specific what you do not know about SI? \$\endgroup\$ Jul 23, 2022 at 11:24
  • \$\begingroup\$ Thank you.Do I need to terminate a transmission line if my rise time is larger than propagation delay \$\endgroup\$
    – Hari
    Jul 23, 2022 at 12:48
  • 2
    \$\begingroup\$ There is not one simple answer to this question. But it is a good idea to terminate long transmission lines at both ends if possible. Sometimes single-ended termination may be OK also (terminated at source but not at load). And there is also AC termination (R and C at load creating termination for high frequencies but open circuit for low frequencies). I think a full discussion would require a chapter in a book. \$\endgroup\$
    – user57037
    Jul 23, 2022 at 21:23

1 Answer 1


If the rise time is less than the propagation delay, and the line is not terminated ...

  1. The destination will overshoot, then undershoot, etc. This is often called ringing. Therefore, the signal will take longer to settle at a legal value. If your timing is critical, this could be important.

  2. If the destination is edge-sensitive, the positive edge after the undershoot can cause a double clock.

  3. The overshoot pulse can be higher than the recommended voltage input of the receiver. This can stress the receiving part or cause it to malfunction.

Here is a plot of the simple case described above. If you have multidrop nets, it is more complicated.

enter image description here

  • \$\begingroup\$ Do we need to terminate a transmission line if my rise time is larger than the propagation delay. \$\endgroup\$
    – Hari
    Jul 23, 2022 at 17:16
  • 2
    \$\begingroup\$ @HARITO that is exactly the question which leads to the rule of thumb with the ~quarter rise length.. The reflections always happen but their amplitude becomes negligible for slow rising edges. So for slow rises you can ignore termination for signal integrity (but not for EMC) \$\endgroup\$
    – tobalt
    Jul 23, 2022 at 18:50
  • \$\begingroup\$ There is no sharp dividing line where you need terminations. For borderline cases, you either add a terminator to be safe, or analyze/measure to convince yourself that it is OK without a terminator. We used Hyperlynx where I worked, a powerful but expensive tool. There are other tools, you could ask about them in a separate question (but it would be off-topic). \$\endgroup\$
    – Mattman944
    Jul 23, 2022 at 20:22
  • \$\begingroup\$ @tobalt " So for slow rises you can ignore termination for signal integrity (but not for EMC)" May I know how EMC issues come here \$\endgroup\$
    – Hari
    Jul 25, 2022 at 4:56
  • 1
    \$\begingroup\$ @HARITO an unterminated trace is basically RF-floating and can easily oscillate at its length-resonance frequency when excited by stuff around it \$\endgroup\$
    – tobalt
    Jul 25, 2022 at 5:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.