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Transmission lines must be terminated with the same impedance as the transmission line has itself to get an impedance matching and avoid reflections at a high impedance end point. This termination resistance closes the current loop between two differential signal lines and creates a decent amount of power loss so that low voltage amplitudes are required to reduce power consumption.

But still digital interfaces use quite low impedance transmission lines like 100 Ohm in Ethernet communication.

Can we reduce power consumption of digital interfaces by using high impedance transmission lines with high impedance terminations? Because I don't know of any examples I assume that it is not a good way to go. If that is true, why so?

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    \$\begingroup\$ lack of suitable conductive wire materials for one. and I suspect parasitic capacitance would not change too much resulting in very slow rise/fall times with such high impedance \$\endgroup\$ – DKNguyen Dec 3 at 14:34
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    \$\begingroup\$ As far as I know, when the impedance of a transmission line is increased, its bandwidth will be decreased. My guess is that the low characteristic impedance of the cable/transmission line is needed to achieve a certain bandwidth. If the bandwidth is limited (low data rates) then proper termination might not be needed.When termination is needed, it means the data rates are high so a high bandwidth is needed so we need low characteristic impedances. \$\endgroup\$ – Bimpelrekkie Dec 3 at 14:39
  • \$\begingroup\$ Have you considered series fed transmit-end terminations with an open circuit receive end? I ask this because this may be what you should look into. \$\endgroup\$ – Andy aka Dec 3 at 16:31
  • \$\begingroup\$ Have you calculated what the net savings would be? Negligible, especially set against the performance loss. \$\endgroup\$ – Mast Dec 4 at 14:25
  • \$\begingroup\$ @Andyaka Why should this change the power consumption? The power might not be dissipated at receiver end but still gets startet as a pulse to the line. It only gets dissipated again at the sender side, right? \$\endgroup\$ – jusaca Dec 5 at 7:32
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While, in theory, high impedances would reduce power dissipation for the same voltage swing, there are several important issues in practice.

1) It's the power, not the voltage, of a signal that determines signal to noise ratio. If you must swing the full rail, then you'd win by increasing the impedance. However if you launch a specific power, then low impedance is not so much of a problem, just reduce your swing.

2) It's not physically practical to get impedances of much more than 100 ohms on a board. The signal conductor needs to get unmanufacturably thin, the space to the ground plane space-consumingly large. The impedance goes as the log ratio of spacing to centre, so you rapidly run out of improvement.

There are other reasons we like a fairly meaty centre conductor, as well as the fab being able to make it. The copper losses vary inversely with the conductor surface area (all the RF flows in the surface), and in fact 75 ohms is the lowest loss geometry (which it why it's used for receive antenna feeds). The highest power handling geometry is around 35 ohms, dependant on heating and surface electric fields. These two figures are why 50 ohms was chosen as a compromise between the two competing criteria as the 'standard' impedance for test gear.

3) In a high speed detector, input impedance is a critical parameter. It's easier to handle with a lower impedance line, for much the same geometrical reasons that you can't make a high Z line on a board, you can't really make a high Z line receiver IC.

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  • \$\begingroup\$ That is a great answer, thank you! \$\endgroup\$ – jusaca Dec 3 at 21:43
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This termination resistance closes the current loop between two differential signal lines and creates a decent amount of power loss so that low voltage amplitudes are required to reduce power consumption.

I think there's a misconception here about how transmission lines work. The purpose of receiver-end termination is to dissipate all the power, in order to avoid it being reflected.

Think of it this way: there is a "pulse" travelling down the line. This pulse embodies a certain amount of energy. Along the way, some of that energy is dissipated by the non-idealness of the transmission line. At the receiver end, the pulse has to have a sufficient amplitude in order to be distinguished from noise. Working backwards from that gives you the amount of energy that has to be put into the pulse "launch" in order to ensure it arrives cleanly.

If the receiver is not impedance-matched, some of the signal will reflect, which worsens your signal-detection problems.

If this sounds like the requirements for radio SNR, there's a very good reason: a transmission line is very similar to a radio wave that's (mostly) contained in a wire rather than a waveguide or allowed to radiate into free space.

The solutions for reducing energy requirement are similar:

  • improve SNR: improve the shielding of the transmission line and reduce its coupling to noise sources
  • improve signal discrimination: better channel coding schemes, more sensitive receivers
  • reduce path loss: improve the shielding of the transmission line and reduce its coupling to external losses
  • reduce ohmic resistance of the transmission line
  • reduce margin: rather than choosing a power level that's guaranteed to work under adverse circumstances, "train" the transmitter to the minimum reliable power level. Currently the most widely used wired interface with any kind of training is for DRAM, but it's more common in wireless interfaces.
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  • \$\begingroup\$ I'm aware of what you describe, but it does not really explain why you would not increase the impedance of the transmission line. This would make it possible to "launch" a pulse of less energy and still get a distinguishable amplitude at receiver side. \$\endgroup\$ – jusaca Dec 3 at 15:40
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Well you could reduce power, at least in the interfaces, but this brings up a whole host of other issues.

On a printed wiring board, higher impedances mean narrower line widths and larger spacing between the layers. Narrower line widths negatively impact manufacturing yield, and larger spacings between layers mean more material and thicker boards. Both of these drive up PWB cost. Even with these heroics it's tough to get the single-ended impedance much above 60 or 70 ohms, assuming ~5.6 mils between layers, which is typical for our designs - 0.134" total board thickness and 24 layers.

Then there's the issue that most high speed test equipment is designed around 50 ohm characteristic impedances, which would complicate making measurements on interfaces with other than 50 ohm impedance.

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  • \$\begingroup\$ I see, this makes sense. So you think we COULD reduce power consumption with high impedance transmission lines, but it would be to challenging or to expensive for production. \$\endgroup\$ – jusaca Dec 3 at 15:42
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While other answers have addressed your question, there is an assumption which you have made which is not necessarily true:

Transmission lines must be terminated with the same impedance as the transmission line has itself to get an impedance matching and avoid reflections at a high impedance end point.

As it happens, this is called parallel termination. If you're familiar with the terminology of EEs, this should immediately lead to speculation about "why parallel? Is there such a thing as series termination?" And, yes, Virginia, there is such a thing.

If the source is coupled to a transmission line by a series impedance equal to the line impedance, and the receiving end has a high impedance relative to the line impedance, the receiving end will produce a reflection. However, the reflection will (a) be very small, and (b) will be absorbed when it reaches the source, due to the matching impedance acting as a parallel terminator.

So, if this is so great, why isn't it used more? Traditionally, the biggest problem is that logic devices have different output impedances at high and low outputs. This makes simple matching impossible.

Very high-quality transmission setups use both series and parallel transmission for maximum reflection suppression, while accepting the guaranteed 3 dB power loss which this implies.

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    \$\begingroup\$ "If the source is coupled to a transmission line by a series impedance equal to the line impedance, and the receiving end has a high impedance relative to the line impedance, the receiving end will produce a reflection. However, the reflection will (a) be very small, and (b) will be absorbed when it reaches the source, due to the matching impedance acting as a parallel terminator." This is not quite right. A high impedance or open circuit at the receiver will reflect close to 100% of the signal back to the source, where it is absorbed by the combined resistance of the resistor and the source \$\endgroup\$ – SteveSh Dec 3 at 19:18

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