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About impedance mismatch and reflections
c64:
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?
TheUnnamedNewbie:
--- Quote from: c64 on June 05, 2020, 06:04:15 am ---Would it be any reflections?
--- End quote ---
Yes.
Any mismatch will give you reflections.
--- Quote ---Would it depend on the length of this section?
--- End quote ---
Yes.
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) = 1/3
For 200->100 ohm this gives: T=(100-200)/(100+200)=-1/3
So 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).
RoGeorge:
--- Quote from: c64 on June 05, 2020, 06:04:15 am ---Would it be any reflections? Would it depend on the length of this section?
--- End quote ---
Yes, 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.
c64:
--- Quote from: TheUnnamedNewbie on June 05, 2020, 06:15:14 am ---So 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.
--- End quote ---
This is good to know :-+
RoGeorge:
--- Quote from: c64 on June 06, 2020, 01:42:03 am ---
--- Quote from: TheUnnamedNewbie on June 05, 2020, 06:15:14 am ---So 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.
--- End quote ---
This is good to know :-+
--- End quote ---
That'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.
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