Consider a transmission line which is terminated with characteristic impedance at one end but other end is kept open, the reflections are not observed. I don't understand the mechanism behind. Please explain how it is happening?

What I think is that, since other end is kept open, the signal will get reflected and reaches the other end where there is proper termination so no reflection occurs but the reflected signal from the open end is going to the load so it should see that reflection, but it is not seen. Why?

I don't understand the flaw in my explanation, please help me understand the mechanism.

  • 2
    \$\begingroup\$ draw a diagram. You have mentioned a line with open and terminated ends, and a load, but no source. Whether you see a reflection or not depends on the configuration, and where you're looking. \$\endgroup\$ – Neil_UK Feb 23 '20 at 6:26
  • \$\begingroup\$ It sounds like what you really mean to ask is "How does series termination work?", regardless of whether the opposite end is open or not (often it is high impedance so might as well be open). If so...the answer is: There is a reflection, but there is a resistor sitting between you and the reflection. Remember, a resistor has two terminals which means there are two different places which you can probe to search for the reflection. \$\endgroup\$ – DKNguyen Feb 23 '20 at 6:27

There are two common situations where you might have an open / terminated transmission line. The first is in a series terminated logic line running from one point to another. The second is using a 'high impedance' probe on microstrip.


simulate this circuit – Schematic created using CircuitLab

In the first example, let's have everything start at 0v. IC1 output now goes high to 5v. The PCB trace is a 100 ohm transmission line, and so looks like a 100 ohm resistor to ground. This forms a voltage divider with R1, so a 2.5v step gets launched into the line travelling rightwards, with IC1 seeing a 200 ohm load. The current flowing from IC1 is 5/200 = 25mA, which is also the current wave flowing down the line, 2.5v/100 ohms. The current wave supplies the current to charge the line up to 2.5v.

After a while, it takes about 1.5nS per foot of trace on FR4, the step will reach IC2 input, whose high impedance is high enough to consider it an open circuit. The current flowing in the line has nowhere to go, so generates a reflected wave of -25mA, +2.5v, now travelling leftwards. Both waves add at IC2 input, so you get no current, and 5v. The return 2.5v wave, sitting on top of the existing 2.5v level, charges the line to 5v, and eventually arrives back at IC1. Here it's absorbed by R1.

Note that this series terminated line requires the reflection at the open end to work properly. This configuration can only be used for a single receiver at the far end. If you put intermediate receivers along the line, they will see a period of sustained 2.5v from the outward wave, before going up to 5v with the return wave, gaurranteed to mess with any logic. If you want a line to serve several receivers along its length, then you must voltage drive the line to get the full logic swing on the outward wave, and terminate it to avoid reflections.

The second picture illustrates the 'poor man's' RF probe. 1k is high enough to avoid causing problems on many nodes in a 50 ohm microstrip circuits. The terminated coax looks like a 50 ohm load to ground, so R2 forms a roughly 20:1 (roughly -26dB) divider with 50 ohms. The signal travels along the coax, where it's absorbed without reflection in the spectrum analyser. Although we could use a shunt 50 ohm resistor at the probe end to doubly terminate the coax, it's not necessary, and it introduces 6dB more loss.

  • \$\begingroup\$ A bit late for church lol. \$\endgroup\$ – Andy aka Feb 23 '20 at 10:44

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