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Does matching impedance connectors matter in non-RF signals? Why?

Electrical Engineering Asked by Mister Mystère on December 17, 2021

I am designing a PCB which communicates with an encoder over RS422 at 2Mbps (15ns rise time typical in the MAX490’s datasheet), and I am currently selecting the connectors. I have found 120Ohm twisted cables online, but I have not found any connectors fit for purpose.

Does that mean the connectors don’t matter over such small distances? If so, how come there are no reflections from the impedance change?

Bonus question: does that mean I can pretty much use any connector (for example screw terminals) as long as I use a 120Ohm twisted cable, and solder a 120Ohm termination resistor across the sensor’s receiver pins and across the transmitter pins past the connectors on the PCB?

3 Answers

15 ns rise time and assumming a prop time of 2 ns/foot gives us a distance of 7.5 feet for the rise time, sometimes referred to as the spacial extent of the edge. Rule of thumb that many designers use is that you can ignore impedance discontinuities if they are less than 10% of the spacial extent of the edge. So 10% x 7.5 feet = 0.75 feet = 9 inches. This means that if the electrical length of the connector is less than 9 inches, you can ignore it for purposes of your 15 ns rise time interface.

Answered by SteveSh on December 17, 2021

Connectors do matter, as not all connectors provide impedance match. So there will be reflections at the connector. The connector just needs to be good enough so it does not distort the signal too much. Which means that the length of the connector parts that have mismatched impedance is small enough.

I believe screw terminals could work in your case. 2 MHz is pretty high, but then again, AES/EBU interface sends audio with RS-422 signaling over 3-pin XLR connector and 110 ohm cable, and that goes beyond 2 MBps easily. Gigabit Ethernet is also sent over 8P8C or M12 connectors and that uses 250MBps per pair.

RS-422 pair termination is important. As RS-422 has one transmitter at one end of the wire, the far end of the wire must be terminated with the termination resistor.

Answered by Justme on December 17, 2021

Impedance matching starts to matters when the following three criteria are satisifed, all relative to each other:

  1. the propagation time is slow enough
  2. the wavelength is short enough
  3. the connection length is long enough

In other words, when you can no longer assume the wave propagates down the line instantaneously and that the potential at both ends of the line are always equal. Imagine a single frequency component of a signal, a sinusoid, being sent down a line. It's not a constant voltage. Its amplitude varies as you go.

In more qualitative terms, it means that when these criteria are satisfied, the driver is placing "new bits of waveform" onto the line faster than it takes for conditions on far end of the line to propagate back to the driver. This means you get reflections bouncing back and forth as the two far ends "negotiate" with each other to form an equilibrium and the driver adjusts its output to accommodate the conditions at the far end of the line.

Actually, I lied. This doesn't only happen when the criteria is satisfied. It always happens since it always takes some amount of time for the conditions at the far end to propagate back to the driver.

The reason it doesn't matter so much when the criteria are not satisfied is that the reflections are able to settle down fast enough that it doesn't corrupt your signal or cause ringing that is high enough to be damaging. But as the line gets longer relative to the wavelength, the mismatch between the ends gets larger the magnitude of the reflections become more problematic. Imagine a sinusoid of any frequency and put it down a wire and freeze it in time. If the wire is 1/4 the wavelength, one end could be at Vpk and the opposite end could be as high as -Vpk for a total difference of 2Vpk. But if the wire is 1/100th the wavelength the difference is a lot smaller.

Same thing happens when you suddenly slam off the faucet in your house. You might hear the pipes rumble from the water hammer of the closed valve stopping the inertia of all the water flowing in the pipes rippling down the pipes because the far end of the pipe (and indeed all the water in the pipes) didn't know to stop flowing the instant you slammed the faucet shut.

A mechanical analog of this is if you pump water into a long pipe that narrows in diameter (or simply becomes a dead-end) far away from the inlet. At the inlet, the water appears to flows smoothly before the water reaches the blockage (as if the pipe was infinite, like an infinitely long transmission line). But when it reaches the blockage, the all the force behind the inertia of the water water splashes against the blockage and sends a ripple all the way back to the inlet, effectively giving you information about what is at the far end of the pipe and telling you that that you need to decrease your input flow rate.

So the reason it matters for RF is because RF is very high frequency and that usually means that pretty much practical all connection lengths long enough relative to the wavelength for impedance matching to matter.

But from the criteria, you can see that it is not just RF that is subject to this criteria. So 2MHz sine wave over a 30cm line is okay without impedance matching, but over 1000km, not so much.

Note that the 2MHz RS-422 you are talking about is a 2MHz square wave (more or less) but 2MHz is the fundamental; the lowest frequency present. However, from the first paragraph, it should be obvious that the highest frequency you care about is the one that will cause troubles first. The highest frequency sinusoidal component in your 2MHz RS-422 is found hidden within the rise time of the square wave, and therefore the rise-time (and by extension the highest frequency component) is the determinant in whether impedance matching is needed.

Answered by DKNguyen on December 17, 2021

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