About impedance mismatch and reflections - Page 1
Forum section: electronics / projects

author: c64
author_rank: Frequent Contributor
timestamp: June 05, 2020, 06:04:15 am
content: Lets say I have a nice differential pair on my PCB with impedance perfectly matched to driver and receiver of the signal. No reflections.Now let's suppose there is a very short section on this differential pair which has lower or higher impedance. Would it be any reflections? Would it depend on the length of this section?

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: June 05, 2020, 06:15:14 am
content: Quote from: c64 on June 05, 2020, 06:04:15 amYes.Any mismatch will give you reflections.QuoteYes.You will actually get two reflections: One at the start of the new impedance (in your case the transition from 100 to 200 ohm) and one at the end (200 back to 100 ohm). The reason why people say you can 'ignore' the mismatch if it is very short is because the two reflections are opposite and cancel each other out.The math: Reflection coefficient for going from a line with Z1 to Z2:T=(Z2-Z1)/(Z2+Z1).For the 100->200 ohm this gives: T=(200-100)/(200+100) = 1/3For 200->100 ohm this gives: T=(100-200)/(100+200)=-1/3So if the electrical length between the two sections is very short, the two reflected waves just cancel each other out and you get no total reflections. As the length of the 200 ohm line gets longer, they don't fully cancel each other out any more and you gradually see more of the reflections (peaking at a line-length of 1/4 wavelength, where the two reflected waves are exactly inphase and add up again).

author: RoGeorge
author_rank: Super Contributor
timestamp: June 05, 2020, 06:25:18 am
content: Quote from: c64 on June 05, 2020, 06:04:15 amYes, there will be two reflection places, one at each step change of the impedance.No, the length doesn't play the main role. The amount of reflection is given by the ratio between Z1 and Z2, and a reflection happens at each boundary where the impedance changes. In that pic, there will be two reflection places, one where the impedance changes from 100 to 200 ohms, and the other reflection place is where the impedance changes again from 200 to 100 ohms.If the length of the 200 ohms line is very short, then there will be hard to distinguish between the two reflection places, but the reflections will still be there, just very close to each other.

author: c64
author_rank: Frequent Contributor
timestamp: June 06, 2020, 01:42:03 am
content: Quote from: TheUnnamedNewbie on June 05, 2020, 06:15:14 amThis is good to know

author: RoGeorge
author_rank: Super Contributor
timestamp: June 06, 2020, 07:29:08 am
content: Quote from: c64 on June 06, 2020, 01:42:03 amThat's not how it works.The wavefront can not guess the future, so it will reflect from the first impedance change (100 to 200 ohms step), no matter how long or short the 200 ohms trace is. Only a part of the energy will be able to continue forward past the 100 to 200 ohms change, and an even lower part of that energy that managed to travel through the 200 ohm trace will bump into the 200 to 100 ohms boundary and reflect back (the second reflection of our initial wavefront).From the second reflection, a fraction of the energy will continue to travel to the receiver, and a small part will reflect back, where will meet again the first impedance step, and it will split again, back and forth, and so on.Notice that the second reflection indeed will have the opposed sign relative to the first reflection, but it will also have a small time delay relative to the first reflection (because of the length of the 200 ohms trace), but what's most important, the second reflection will have a much lower amplitude than the first reflection, so the second reflection won't cancel the first one.

author: Marco
author_rank: Super Contributor
timestamp: June 06, 2020, 08:51:29 am
content: Quote from: RoGeorge on June 06, 2020, 07:29:08 amIt's been eons since college and my intuitive understanding of physics was never that good, so lets just use spice (mplab mindi to be specific). Maybe a good analogy is that the double reflecting signal in the high impedance section, sucks energy out of the preceding section and throws it forward again?

author: RoGeorge
author_rank: Super Contributor
timestamp: June 06, 2020, 10:13:51 am
content: The visible descending steps in the red signal are what I was trying to say by "not canceling".In fact, there are infinitely many steps there, just that the first two steps are clearly visible, with the third red step only barely visible in the chart just above the 500mV level. In theory next reflections are there, too, just above the 500mV level, but they are not visible in the chart because of the scale. If we zoom in, the following steps should be visible too.A funny consequence of that (by "that" I mean transmission lines and infinite number of reflections): a capacitor will always charge (or discharge) in stairs, and not as a continuous exponential we use to think about (by "continuous exponential" I mean as in "not stairs"). Not kidding with that, capacitors always charge/discharge in stairs voltage, because capacitor's plates also form a transmission line.

author: Marco
author_rank: Super Contributor
timestamp: June 06, 2020, 05:14:33 pm
content: But for all extents and purposes the reflections do cancel out, obviously not immediately but he never suggested that, but for a step the second reflection almost cancels out the first one completely already.If the high impedance section is short relative to the rise time of the signal you won't even see a difference.