I remember reading this somewhere but cannot find the ratio of the rise time vs. the propagation time (i.e. the trace length?) of a signal when transmission line effects come into effect?

For example - if I have a SPI bus running at 12.5MHz. If it runs a few inches through the PCB trace, it is not a transmission line - but at what length goes it become a transmission line (at least in theory). How to calculate that?

  • \$\begingroup\$ It is somewhat arbitrary and may depend on what you are trying to accomplish. \$\endgroup\$
    – user57037
    Sep 20, 2016 at 2:50
  • \$\begingroup\$ I am trying to determine what reflections I might get. I have a SPI master and I want to run it at 12.5MHz over about 5" of FR4 PCB trace and about 6 feet of 26awg wire. I dont know if that is even possible and how to do it. \$\endgroup\$ Sep 20, 2016 at 2:52
  • \$\begingroup\$ It will probably work. Use a dedicated GND wire for each signal wire and twist the signal with the GND. Add a series resistor on the source board (start with 0 Ohms, and change it if needed) and an AC termination on the receive board. AC termination is a cap and resistor in series. For MOSI, add the series resistor on the master, and AC termination on the slave. For MISO add series resistor on the slave board, and AC termination on the master. Don't install any AC termination to start. You might not need it. Clock is same as MOSI. \$\endgroup\$
    – user57037
    Sep 20, 2016 at 2:59
  • \$\begingroup\$ pericom.com/assets/App-Note-Files/AB023.pdf \$\endgroup\$
    – user57037
    Sep 20, 2016 at 3:00
  • \$\begingroup\$ In theory, it is a transmission line at any length. Termination always works, but sometimes you can leave it out and ignore the resulting overshoot/ringing. \$\endgroup\$
    – Whit3rd
    Sep 20, 2016 at 4:37

2 Answers 2


The boundary between lumped and distributed systems is not clear-cut but there are some commonly used values. For distributed systems transmission line theory is required.

The distinction is usually made based on the effective length of a signal or the feature of a signal like an edge. So it's important to consider the rise- and fall time of a signal and not the frequency. Nevertheless the frequency imposes an upper limit on the risetime.

In air a signal travels with about 85ps/in (~ 33ps/cm). The propagation delay depends on the dielectric constant, it is proportional to the square root of it. For a PCB with a dielectric constant of 4 (like FR4 which is in the range of 3 to 5) the propagation delay doubles.

A rising edge with a risetime of 1ns would occupy a trace length of 1ns/(2*85ps) ~ 6in (~ 15cm). At the driving side the signal is already high when at a 6in distance it just starts to rise.

So a 6in (15cm) track clearly is too long, since the potential varies from low to high along the track.

If the length of the track is between 1/6 or 1/4 of the effective length of a feature like an edge a system can be regarded as lumped.

So the upper limit for the example given above is between 6in / 6 (= 1 in, ~2.5cm) and 6in /4 (= 1.5in, ~4cm) for a trace on a PCB with a dielectric constant of 4.


For analog signals (sine waves, let's say), the rule of thumb I was taught is when the line length is 1/8 of the wavelength, you should treat it as a transmission line. Practically speaking, the propagation speed of an electromagnetic wave is determined by the dielectric the wave is traveling in. For PCB's it is typically around C/2, where C is the speed of light in space. Your wavelength will be shorter by the same scale factor (wavelength in free space / 2). So a 100 MHz signal would be 3 meters in free space. 1.5 meters in a PCB. And 1/8 of that is 18.75 cm.

For digital signals, there are several alternative ways to look at it. One way is rise time and bandwidth. The basic idea is, you use the rise time to estimate the bandwidth, then use the analog signal rule above, but with the bandwidth as the frequency. The justification for this is that the wave can be somewhat accurately reproduced provided that you have the majority of the BW. So the highest significant frequency in the signal is given by the BW.

The rule of thumb I have seen is that BW = 0.35/TR. BW is bandwidth in GHz, and TR is rise-time in nanoseconds. This formula uses the 10/90 rise time. So if your rise time is 10 ns, then your BW is 0.35/10 = 0.035 GHz = 35 MHz.

Another way to look at it is that you want the round trip flight time of your signal to be substantially less than your rise time. This means that the reflections will return to the source while the signal is still rising or falling. Reflections like this will not cause the rising and falling edges to be jagged or have "shelves" on them. They will still be smooth.

Hope this helps.

  • \$\begingroup\$ Thank you. This is the kind of 'rule of thumb' i was looking for. \$\endgroup\$ Sep 22, 2016 at 4:29

Your Answer

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

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