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Why are characteristic impedances of traces not considered when the traces are shorter than half a wavelength? I've had the same issue with light diffraction, which happens when pinholes are smaller than half a wavelength - it sort of makes sense somehow, but I can't "see" it, I don't understand how wavelengths are related to reflections (which I assume are the only reasons why we care about impedance matching). I'm trying to make the ocean wave analogy work but... Well, the fact that I'm asking this says it all.

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    \$\begingroup\$ Nice question. Would love to see easily understandable answers \$\endgroup\$ – User323693 Jun 30 '15 at 16:49
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    \$\begingroup\$ Half a wavelength is wrong, it's 1/10 or less. \$\endgroup\$ – Leon Heller Jun 30 '15 at 17:17
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Some unscrupulous self promotion: Online Transmission Line Simulation

Adjusting the transmission line length vs. the signal frequency is equivalent to adjusting time delay (tDelay) vs. rise time (tRise).

Some interesting parameters: set tDelay=tRise/10. This is the case where the wavelength is much longer than the transmission line. Notice that the red trace will reflect from the far end multiple times before reaching the peak "on" level of 1V. However, each reflection is relatively small because the the voltage at the left of the red trace isn't significantly different form the drive level (blue trace). The signal was able to propagate to the target fast enough that the separation distance never became too significant.

Now repeat with a case of say tDelay=tRise/2. Notice that the driving source voltage separation from the red mismatched termination voltage significantly more. When the signal finally reaches the end of the transmission line, the reflection is quite severe. This mismatch between what the receiver thinks the drive voltage is and the true drive voltage dictates the magnitude of any reflections. Repeated reflections come because the reflection causes the line level to over-shoot the source level, but is smaller than the first reflection. The signal reflects repeatedly until the level settles to near the source voltage.

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    \$\begingroup\$ Amazing piece of software. Exactly what I was looking for when I was looking into signal reflections, and that fits perfectly the matter at hand here. \$\endgroup\$ – user42875 Jun 30 '15 at 22:08
  • \$\begingroup\$ The link helped me to visualize the reflection. Thanks! \$\endgroup\$ – abhiarora Apr 10 '18 at 18:40
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A 1/4 wavelength trace or shorter can also have a substantial effect. The usual rule of thumb I've heard and used is that you can probably neglect transmission line effects when the length is less than 1/10 or 1/20 wavelength.

For a simple example, say you terminate a 1/4 wavelength line with an open circuit and drive it with a single-frequency source. After the signal reflects back to the source (1/4 wavelength away), it will look to the source like it's driving a short circuit instead of an open. That's a pretty substantial effect.

For more usual situation in digital design, you design the line as 50 ohms, and terminate the line with 50 ohms, but the actual characteristic impedance of the line might vary in production between 45 and 55 ohms. You want to know how big an effect that will have on signal integrity.

If the line is long, the signal propagates to the end, and reflects back. Then it propagates back to the source (which might not be well-matched at all) and reflects again. And so on. This produces a voltage at the load with a substantial ring on each rising and falling edge. The time it takes for this ring to die out is longer if the trace is longer because it takes time for those reflections to propagate back and forth.

On the other hand, if the line is very short (less than 1/10 wavelength at the "critical frequency" related to the rise- and fall-time of the digital signals), these reflections will all happen within the time the rising or falling edge is still in progress, and won't produce very much ring (overshoot or undershoot) at the load.

This is why you will often hear a rule of thumb that impedance control isn't needed when the trace length is a small fraction of the wavelength.

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Long wavelength compared to the traces actually means that there is little voltage along the traces- one end is always almost the same voltage as the other end (compared to the magnitude of the signal) so the effect of reflections is minimal.

As @ThePhoton says you should be thinking 1/10 or 1/20 wavelength not 1/4.

If you think of waves of water in a narrow deep tank, and one side cannot be much higher than the other (10 times the wavelength, say), it becomes more like raising and lowering the water in the tank.

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    \$\begingroup\$ I like the water tank analogy :) \$\endgroup\$ – Dzarda Jun 30 '15 at 18:14
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A quarter wave unterminated cable will look like a short ciruit and this is to be avoided for obvious reasons. As the cable reduces in length things get better for the high frequency parts of your signal spectrum and, generally by about one-tenth of a wavelength terminations are forgotten about.

Here's what an open ended line looks like when its length is matched to a quarter wavelength of the applied voltage: -

enter image description here
(source: ibiblio.org)

And, if you really want to understand more about it, this site can help

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  • \$\begingroup\$ Outstanding article you linked in, it's cristal clear reading it. You deserve more votes. \$\endgroup\$ – user42875 Jun 30 '15 at 22:09
  • \$\begingroup\$ I'd like to accept yours and helloworld922's but for the work he's done on the simulation software I'll take his. Your link has all the answers though. \$\endgroup\$ – user42875 Jun 30 '15 at 22:11
  • \$\begingroup\$ I am still reading the linked article. I have to read it multiple times. Thank you for sharing \$\endgroup\$ – User323693 Jul 1 '15 at 2:26

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