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It seems to me that there are three ways to terminate a transmission line:

  1. at both ends
  2. at the load only
  3. at the source only

schematic

simulate this circuit – Schematic created using CircuitLab

the coax shield is grounded, and the buffers have power supply connections as usual, and these are \$50\Omega\$ transmission lines.

What are the advantages and disadvantages of each, considering:

  • we might want to transfer power (as to an antenna) and not information (as in a digital circuit)
  • the signal may be analog
  • the transmission line might not be ideal (discontinuities in the middle, etc.)

What is common practice in typical situations:

  • high-speed digital circuits
  • low-power RF (between stages, receivers)
  • high-power RF (transmitters)
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  • \$\begingroup\$ The diagram you drew has some issues. The shields need to somehow end up being connected back to the GND of the buffer / driver circuit. They would not just be connected as you show. Also note that terminations would be used in more situations than the coax that you show. You could have parallel conductor cables, twisted pair cables, parallel traces in a PC board, or any of these inside a common shield jacket. \$\endgroup\$ – Michael Karas Jun 22 '13 at 15:43
  • \$\begingroup\$ Please consider that assumptions lead to all sorts of problems. A proper circuit diagram would also show the power/ground connections of the buffer circuit. You may have heard the old adage about ASSUME? \$\endgroup\$ – Michael Karas Jun 22 '13 at 16:15
  • \$\begingroup\$ My point is simply that as shown in the diagrams the shield is not connected to anything but the termination resistors. As such no current will flow and the termination resistors will not do much of anything. Until that is clarified there is not much point of talking about effectiveness of the terminations. \$\endgroup\$ – Michael Karas Jun 22 '13 at 16:38
  • \$\begingroup\$ The third example does not work for me unless you don't care about distortions of the signal due to no terminator at the far end. Is there any application where this might be valid? (1) and (2) are both OK but this I'm repeating myself from your previous question where I felt I may have missed some subtlety in your question? Did I? \$\endgroup\$ – Andy aka Jun 22 '13 at 18:17
  • \$\begingroup\$ @Andyaka I got the impression that if \$R_L >> Z_0\$, then the resulting reflection exactly cancels the 1/2 voltage divider set up by \$R_S\$ and \$Z_0\$, and the reflection, when it returns, is absorbed by \$R_S\$ and you end up at a stable state. This is how the simulator is set up, by default. I guess this is common in digital electronics, but really I was more interested in RF, and the goal might be to put power in the antenna, not transmit a bit. \$\endgroup\$ – Phil Frost Jun 22 '13 at 19:36
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If you are talking about high speed digital and not RF signals then you can choose one of the following schemes (all assume continuous ground planes). Keep the stub section as short as possible and you can choose a transmission line impedance that works well for your layout (Zo=50 ohms is not a requirement).

• Simple parallel termination: In a simple parallel termination scheme, the terminating resistor (Rl) is equal to the line impedance. Place the termination resistor as close to the load as possible to be efficient -- keep the stub section as short as possible.

schematic

simulate this circuit – Schematic created using CircuitLab

• Thevenin parallel termination: An alternative parallel termination scheme uses a Thevenin voltage divider. The terminating resistor is split between R1 and R2, which equals the line impedance when combined--(R1||R2)=Zo. Although this scheme reduces the current drawn from the source device, it adds current drawn from the power supply because the resistors are tied between VCC and GND.

schematic

simulate this circuit

• Active parallel termination: An active parallel termination scheme, the terminating resistor (Rl=Zo) is tied to a bias voltage (Vbias). In this scheme, the voltage is selected so that the output drivers can draw current from the high and low-level signals. However, this scheme requires a separate voltage source that can sink and source currents to match the output transfer rates.

schematic

simulate this circuit

• Series-RC parallel termination: A series-RC parallel termination scheme uses a resistor and capacitor (i.e., series-RC) network as the terminating impedance. The terminating resistor (Rl) is equal to Zo. The capacitor must be large enough to filter the constant flow of DC current. However, if the capacitor is too large, it will delay the signal beyond the design threshold. Capacitors smaller than 100 pF diminish the effectiveness of termination. The capacitor blocks low-frequency signals while passing high-frequency signals. Therefore, the DC loading effect of Rl does not have an impact on the driver, as there is no DC path to ground. Not all drivers can handle the dynamic current requirements for larger capacitor loads.

schematic

simulate this circuit

• Series termination: In a series termination scheme, the resistor matches the impedance at the signal source instead of matching the impedance at each load. The sum of Rl and the impedance of the output driver should be equal to the Zo. Because silicon IC output impedances are low, you should add a series resistor to match the signal source to the line impedance. The advantage of series termination is that it consumes little power. However, the disadvantage is that the rise time degrades due to the increased RC time constant.

schematic

simulate this circuit

RF and microwave terminations are another animal and depend greatly on the physical 3d parameters of your design, input and output impedance, operating frequency range. They rarely depend on resistive elements. They are designed by reactively moving the input and output impedances into the proper 50ohm match.

However these are passive networks so both of your cases you were curious about don't really matter -- just get the matching correct (of course your components have to be sized to handle the voltage/current demands):

low-power RF (between stages, receivers)

high-power RF (transmitters)

As for

What are the advantages and disadvantages of each, considering:

we might want to transfer power (as to an antenna) and not information (as in a digital circuit)
the signal may be analog
the transmission line might not be ideal (discontinuities in the middle, etc.)

Antennas are passive loads so design for max power transfer, but in their operating bandwidth they will need a matching network to do this. Analog signals are the same as RF. Just match the impedance. Non-ideal transmission lines is too vague to answer, but any discontinuity causes a reflection and a loss of power.

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  • \$\begingroup\$ @PhilFrost is this not what you're looking for? I spent too much time putting it together for you... \$\endgroup\$ – user6972 Jun 26 '13 at 22:57
  • \$\begingroup\$ Well, all but the last case are just variations on case 2 from the question, and case 1 from the question wasn't covered at all. The most interesting part, how these behave in real-world conditions, where there are connectors in the transmission line, maybe they are wet, maybe there are birds nesting on them, etc, wasn't covered at all. It's useful information, but it doesn't really answer the question. \$\endgroup\$ – Phil Frost Jun 28 '13 at 17:34
  • \$\begingroup\$ @PhilFrost I would like to help. Can you edit your question to ask specifically what you want? Are you interested in discontinuities only? It isn't clear if you are asking about Power, Digital or RF signals because those are all different animals. Are you talking about power transmission lines now with birds, rain? Where to put connectors? Etc=what exactly? \$\endgroup\$ – user6972 Jun 28 '13 at 18:41
  • \$\begingroup\$ Your answer has a lot of specific information about a very small subset of my question. I know to minimize the stub length. The question wasn't about that. I know there are many ways I could realize a \$50\Omega\$ impedance at the load. The question wasn't about that, either. The question is about the three cases clearly illustrated in the question, and their specific advantages and disadvantages, and how these properties benefit or not the common use cases enumerated in the question. \$\endgroup\$ – Phil Frost Jul 5 '13 at 17:49
  • \$\begingroup\$ @PhilFrost The 3 cases you posted are not valid ways to do a transmission line. All of them introduce unnecessary discontinuities because your transmission line ground plane has discontinuities at the source. Read my 1st sentence about the ground planes for the examples I show. I don't understand your drawings either because the transmission dotted line isn't grounded for case 1. \$\endgroup\$ – user6972 Jul 6 '13 at 5:23

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