I’ve been exploring the concept of impedance matching in transmission lines and the role it plays in reducing reflections and power loss. It’s clear that using termination resistors to match the impedance of the transmission line is important for signal integrity, but I’ve noticed that digital interfaces like Ethernet often use relatively low impedance transmission lines (e.g., 100 ohms).

My question is whether we can potentially reduce power consumption in digital interfaces by using high impedance transmission lines with correspondingly high impedance terminations. However, I haven’t come across any examples of this approach being used, which leads me to believe it may not be a practical solution. Could someone explain why this might not be a viable strategy for digital interfaces?

You may very likely reduce power consumption as you’ll no longer be sending data to the receiver. The reason you haven’t seen anything about it is because you can’t do that. Supposing you had a 50 ohm network. You replace the transmission path (coax, PCB trace, etc.) with a higher impedance, lets say one megohm. Then you change your termination resistor to one megohm. Now you connect that transmission to your transmitter, which has a 50 ohm output impedance. That’s a mismatch of 20,000 to 1. At a VSWR of 2:1 you lose 11.1% of your transmitted power. At 3:1 it’s a 25% loss and at 10:1 it’s 67%. At a 100:1 it’s 96.2%, so when you get to 20,000:1 I don’t see the receiving end getting any signal whatsoever, so you might save some power. Your system will not work however.

There is nothing to stop you using 100 Ohms transmission lines and terminating them with 100 Ohms resistors. But we generally don’t do this because of reasons of it not being very practical.

Consider a standard PCB with 4 layers. Let’s suppose the two inner layers are planes (assume they’re both 0V layers). Also suppose the outer layer to nearest plane is a prepreg of 0.1mm thickness. Also suppose you want some tracks to have an impedance of 50 Ohms because they’re RF tracks. What width track will give you that impedance? This works out to be about 0.16mm which is easily manufacturable.

The PCB company says they can do 0.1mm tracks as standard. What impedance are they? Answer 61.3 Ohms. What width do you need for 100 Ohms? Answer, approximately 0.015mm - which is certainly not manufacturable.

However, you’re not going to be put off by that. Is there another solution? Yes, increase the separation between the track and the plane. If you go with a track width of 0.1mm (thinnest that most board shops will process without counting it a “special processing board”) then you need 0.3mm between plane and track. This is possible, but what has it done to your 50 Ohms track? This has now got to increase in width, and 0.5mm will do. This works but is not helping your routing density.

The obvious question is how does Ethernet with its 100 Ohms impedance manage? The answer is simple - Ethernet is a differential interface, so this impedance is what is seen from signal to signal. If you neglect the contribution from the fact that the signals are usually adjacent to each other, your first-order approximation is to say that the impedance between them is the sum of the impedance of one signal to the 0V plane plus the 0V plane to the second signal. So if we go back to the original plane/track separation of 0.1mm, and we remember that a single 50 Ohms track is 0.16mm wide, then the pair will “see” an impedance of 100 Ohms. Well, not exactly.

We said that this is a first order estimate and it is close. But it does depend on the separation of those signals. The first order approximation is only true at an infinite distance apart. When you route your two 0.16mm wide tracks with a 0.2mm separation, the actual differential impedance is 78.9 Ohms. You really want 100 Ohms, so we make the track width smaller. It works out that if you have two 0.1mm tracks, separated by 0.2mm gap, and are 0.1mm above the plane, the impedance works out to 99.2 Ohms (which is near enough).

In the past, some buses like VME-bus used Thevenin termination with a 220 Ohms and 330 Ohms pair of resistors for every signal. This gives an equivalent impedance of 132 Ohms. And there was one resistor pair at each end of the bus, so it looked like driving a 66 Ohms load.

The real issue is what do we want termination for? The answer is that matched impedance terminations do not reflect anything back into the interconnect. This is important if there are several receivers on the signal.

Many tracks carrying digital signals have no termination. It follows that these must reflect. So why doesn’t this cause problems in all cases? It all comes down to rise/fall times and distance (note: it is nothing at all to do with frequency). The general rule is that if the signal has only managed to rise or fall 10% of the total swing by the time it reaches the point from which it reflects (i.e. the other end), then that is how big the reflection will be, and at this level we can ignore it. If the percentage rise or fall is higher than this, termination is usually somewhere between being a good plan to essential.

Going back to Ethernet again. Just consider a 100Mbps link. This has a signalling rate of 125Mbps thanks to 4b/5b coding. Its rise/fall time has to be less than about 4ns (if you are to have any reasonable eye opening). If we approximate the signal propagation velocity to be 2/3c (2x10^8 m/s) and we shall say that termination is needed if the signal has risen more than 10% of 4ns (4x10^-9 x 0.1 = 4x10^-10), then the distance beyond which we should terminate the signal is 2x10^8 x 4x10^-10 = 8x10^-2m = 8cm! As Ethernet cables are always much longer than 8cm, and can be up to 150m, it is obvious that this interface needs to be terminated.

Back to the initial concern of saving power by using high value termination (or none at all), the issue is not really one of power but whether the link can actually transfer data reliably at the data rate of interest, over the distance necessary; because in the end, it may be that the only way possible means trading that power for signal integrity when the transmission line impedance and termination impedance are matched.

In principle any impedance is possible. The decision of using high or low impedance depends on the application. Why 50 ohms (100 differential) is used? If you want to go a little more in depth, take a look at www.microwaves101.com. In the early radio broadcasting days (also RADAR), when most of the microwave theory was developed, the goal was to minimize looses and transmit the maximum power. The maximum power handling occurs when characteristic impedance is 30 ohms and you can get minimum looses at 75 ohms. 50 ohms is a tradeoff, something in the middle. In any case I will give you two reasons why you should not use a very high impedance.

The first one is to minimize loop inductance. If you want to increase impedance there are two options: reduce trace width or increase the separation between trace and the return path. Trace width has a limit so sooner or later the separation has to be increased. If you do that the loop inductance increases and this is something you should avoid. Keep the signal and the return path as closer as possible.

The second reason is the cross talk. Cross talk decreases when the impedance characteristic decreases. This is quite important in a digital system especially if the rise and fall time of the signals is small.

While in theory, high impedance might seem like an attractive solution for reducing power dissipation with the same voltage swing. However, there are several key considerations you need to consider during practice.

Firstly, it’s essential to recognize that it’s the power, rather than the voltage alone, of a signal that influences the signal-to-noise ratio. While increasing impedance could potentially reduce power dissipation for full rail swings, if you’re dealing with specific power requirements, low impedance may not pose a significant issue; adjusting the swing accordingly could suffice.

Secondly, achieving impedances much higher than 100 ohms on a board isn’t practically feasible. This limitation stems from the physical constraints involved; attempting to increase impedance significantly would necessitate excessively thin signal conductors and unmanageably large spaces to the ground plane. As impedance varies logarithmically with the ratio of spacing to the center, the scope for improvement diminishes rapidly.

Moreover, there are additional factors favoring a substantial center conductor. Copper losses, for instance, are inversely proportional to the conductor’s surface area, with most RF flowing on the surface. Notably, 75 ohms represents the geometry with the lowest loss, commonly utilized in receive antenna feeds. Conversely, the highest power-handling geometry typically hovers around 35 ohms, contingent on heating and surface electric fields. The choice of 50 ohms as the ‘standard’ impedance for test gear reflects a compromise between these competing criteria.

Lastly, in high-speed detectors, input impedance emerges as a critical parameter. Opting for lower impedance lines proves more manageable, given similar geometric constraints. Much like the difficulty in creating high impedance lines on a board, designing a high impedance line receiver IC poses similar challenges.

You could reduce power consumption, at least in the interfaces, but this introduces a variety of other challenges. On a printed wiring board, higher impedances require narrower line widths and greater spacing between layers. Narrower line widths can negatively impact manufacturing yield, while larger spacings between layers increase the amount of material used and result in thicker boards. Both factors drive up the cost of the PWB.

Even with these adjustments, achieving a single-ended impedance much above 60 or 70 ohms is difficult. For example, with approximately 5.6 mils between layers, which is typical for our designs (0.134" total board thickness and 24 layers), it’s challenging to exceed these impedance values. Additionally, most high-speed test equipment is designed around a 50-ohm characteristic impedance, which would complicate measurements on interfaces with different impedance values.

The purpose of receiver-end termination is to dissipate all the power to prevent reflection. Imagine a “pulse” traveling down the line, carrying a certain amount of energy. Some energy is dissipated due to the non-ideal nature of the transmission line. At the receiver end, the pulse must maintain a sufficient amplitude to be distinguished from noise. By working backwards, we determine the amount of energy required for the pulse “launch” to ensure it arrives cleanly.

If the receiver is not impedance-matched, some of the signal will reflect, complicating signal detection. This is analogous to the requirements for radio signal-to-noise ratio (SNR) because a transmission line is similar to a radio wave, though mostly contained in a wire rather than a waveguide or free space.

To reduce energy requirements, consider the following solutions:

- Improve SNR: Enhance the shielding of the transmission line and minimize its exposure to noise sources.
- Enhance Signal Discrimination: Use better channel coding schemes and more sensitive receivers.
- Reduce Margin: Instead of selecting a power level guaranteed to work under adverse conditions, train the transmitter to the minimum reliable power level. While this training is common in wireless interfaces, it is also widely used in wired interfaces like DRAM.
- Reduce Path Loss: Improve the transmission line’s shielding and decrease its external losses.
- Lower Ohmic Resistance: Use materials with lower resistance for the transmission line.

While other answers have addressed your question, there’s an underlying assumption that needs clarification:

“I’ve been exploring the concept of impedance matching in transmission lines and the role it plays in reducing reflections and power loss. It’s clear that using termination resistors to match the impedance of the transmission line is important for signal integrity”

This method is known as parallel termination. However, there’s also something called series termination. In series termination, if the source is connected to the transmission line through 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. This reflection will be minimal and will be absorbed when it reaches the source, thanks to the matching impedance acting as a parallel terminator.

So why isn’t series termination more commonly used? The main issue is that logic devices typically have different output impedances for high and low outputs, making simple matching difficult. High-quality transmission setups often use both series and parallel termination to minimize reflections, despite the guaranteed 3 dB power loss this approach entails.