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.
Electronic – Why do characteristic impedances matter only when traces are longer than half a wavelength
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Best Answer
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.