9
\$\begingroup\$

enter image description here

I understand that when I have a line transmitting a signal, it may be important depending on the length of the line and the wavelength of the signal to have matching impedances at the ends of the line to avoid reflections and standing waves. However, what I do not understand is why I wouldn't just have reflections before Zs and after ZL in the figure above, shouldn't "everything" in the circuits on both ends always be impedance matched?

\$\endgroup\$
2
  • 3
    \$\begingroup\$ Using ground symbols is misleading. Transmission lines have two conductors, and the characteristic impedance comes from the geometry of both conductors combined. There's nothing "after" ZL. \$\endgroup\$
    – Adam Haun
    Aug 7, 2023 at 21:23
  • 5
    \$\begingroup\$ Yes, in a real circuit (say a ~10 Ω CMOS output with an additional ~40 Ω resistor in series to match a 50 Ω microstrip), there will be reflections between the trace that connects the resistor and the CMOS chip. Thus, the design rule is to put the resistor as close to the output pin as possible, so that the distance is too short to matter and they can be seen as a lumped circuit as a whole. \$\endgroup\$ Aug 7, 2023 at 23:51

4 Answers 4

4
\$\begingroup\$

Let's first understand how a transmission line behaves:

Signals travelling forward

If we look at one point in the transmission line, and one moment in time, then the local, momentary voltage Uf and current If behave like a Zc resistor, meaning Uf = Zc * If. And this travels forward along the line with some speed v, specific to the cable type.

Signals travelling backward

If we look at one point in the transmission line, and one moment in time, then the local, momentary voltage Ub and current Ib also behave like a Zc resistor, meaning Ub = Zc * Ib (only now the current is counted in the opposite orientation!). And this travels backward along the line with v.

Signal combinations

If you have both, forward and backward travelling signals, they simply superimpose by addition. BUT: voltages add normally, currents have to be subtracted because of their different orientation.

Reflection

A forward-travelling signal now reaches the end of the line, where there's a Zl load. This travelling signal has both its voltage and its current. And they both must match the load, otherwise we'd break Kirchhoff's laws. They match if the transmission lines impedance Zc and the load's impedance Zl are equal.

If they don't match, an element must be missing so we comply with Kirchhoff, and this missing element is a backward-travelling signal. If the load impedance isn't zero or infinite, the backward-travelling signal will have a lower amplitude than the original one. Some time later, the backward one will arrive at the beginning of the line, and again, depending on the impedance found there, it can either be completely consumed there or cause another reflection, and so on and on and on. But, as the amplitude decays with every reflection, after some of these bounces, the reflected signal can be neglected.

Short and long lines

In principle, there's no difference between shorter and longer transmission lines.

So, even a millimeter of wire is a transmission line. But if we assume e.g. 10 reflections to be enough for the signal to drop below the noise level, then it takes 10 times the 1-millimeter travelling time, being maybe 0.05 ns. If your signal doesn't have components above 1 GHz, so short a distortion won't matter. The behaviour is close enough to that of an ideal connection that we ignore the "connection line" properties.

But if your line is one meter, then ten reflections will mean 50 ns, and already signals with a few MHz will see a noticeable distortion.

Your circuit

You can model the Zl side as having a few short transmission lines, e.g. from the cable to the Zl resistor, or from the Zl resistor's ground connector to the cable's ground connector. But we assume them to be short enough so they don't matter.

Having a matched impedance at the source side is a safeguard against a non-perfect match of the load side. the source-side impedance matters for signals travelling backwards through the cable, and decides whether they are again reflected at this source end of the cable. If the load-side is perfectly matched, there will be no backward-travelling signals, and then it doesn't matter what would happen to a backward-travelling signal at the source end.

As a safeguard, it might help to improve signal quality, but it also costs half the power (one half lost in the Zs impedance, only the other half getting into the cable).

Answering your questions:

why I wouldn't just have reflections before Zs and after ZL

If the connections are long enough to matter, you'll have reflections, and you should be careful to design the connections with a geometry that provides a constant impedance over the whole connection length.

And that's the reason why designing a 1 MHz circuit, where you can treat everything shorter than a meter as irrelevant, is much easier than reaching multiple GHz.

\$\endgroup\$
5
\$\begingroup\$

what I do not understand is why I wouldn't just have reflections before Zs and after ZL

You have to keep the length of those wires short compared to the wavelength of the signals used. By short, we usually mean that the load and source connection wires to the transmission line have a length less than one-tenth of the shortest wavelength.

I understand that when I have a line transmitting a signal, it may be important depending on the length of the line and the wavelength of the signal to have matching impedances at the ends of the line to avoid reflections and standing waves.

You don't need to have matching impedances at both ends. Only one is necessary to maintain signal integrity. For instance if ZL's impedance matches the cable's Z0 then there won't be a reflection. Alternatively, if ZL is open circuit then there will be a reflection but, that can be totally absorbed by ZS if its impedance matches Z0. This latter situation is used by high-speed digital drivers all the time.

\$\endgroup\$
2
  • \$\begingroup\$ On CSMA/CD systems like coaxial Ethernet, you cannot afford reflections on either side of the line since transmission will be aborted if an incoming reflection.is detected. And of course with anything but point-to-point connections, one-sided reflections will be problematic. Even with non-sensing point-to-point connections, reflections may cause driver voltages/currents to be exceeded. resulting in clipping. \$\endgroup\$
    – user107063
    Aug 9, 2023 at 9:53
  • \$\begingroup\$ Yes, what I said applies to point-to-point data connections. \$\endgroup\$
    – Andy aka
    Aug 9, 2023 at 9:56
5
\$\begingroup\$

You are perhaps missing a not-oft-told, subtle aspect of schematic notation.

When we draw a schematic (that is, a pictorial connection of component pins and wires or nets), it is implied that traditional circuit analysis techniques (such as nodal analysis) can be used on it.

In practical cases, it may be less about analysis at all (good luck doing nodal analysis on some random unspecified IC), and just used to generate a connectivity map for drawing the PCB, or the wiring inside a machine. And indeed in these cases, it may be that "nodes" are not in fact nodes, but transmission lines (PCB traces, cabling), and connections cannot be made randomly but must be in a specific order (which may simply be implied through experience, or can be described by various means).

In this case, it seems implied to me that there is some element in the middle, whose physical nature eludes traditional analysis and is therefore drawn as a graphical element (the transmission line) rather than an RLC component, meanwhile the circuits (or subcircuits if you like) at either end are expressed as schematic symbols and wiring, implying their suitability for analysis, at least locally if nothing else.

The critical thing is this: nodal analysis has no visibility of dimensionality, of physical size or the speed of light. Such circuits/analyses are pointlike, or alternately the speed of light is infinite.

This is a direct contradiction to the transmission line in the middle, which has a linear dimension (length), and a speed of light (c * velocity factor). At best, the two ends can be analyzed with nodal analysis, but the two sides together cannot, not at all frequencies anyway. (It does reduce to the usual case at low frequencies, where the line can be converted into equivalent capacitance and/or inductance, and resistance.)


Related: are the ground symbols the same on the left and right sides? Should they be? Does it matter? -- When would it matter? Presumably, we should join all left-side grounds together, and all right-side grounds together, but something interesting happens when joining left and right: specifically, at frequencies where the shield conductor's resistance matters, we no longer have current returning strictly through the transmission line (signal and shield currents equal and opposite), but there is a common mode current (the sum of currents) that prefers to follow the zero-ohm path; or equivalently, there is a small voltage drop between left and right sides, at low frequencies.

Going back to practical cases -- critical length depends on frequency. If we can make a length of transmission line shorter than the maximum wavelengths we're working with, it can be reduced to an equivalent L/C element, and we can use a pointlike analysis in the local area. This is why it is suggested to place termination resistors adjacent to transmitter/receiver pins on the board, for example: the small transmission line stub between pin and resistor can be of negligible length.

\$\endgroup\$
4
\$\begingroup\$

In the diagram above, the source is probably an ideal voltage source, so the voltage at Vs is equal to Vs at all times, also Vs can source infinite amounts of current. This also means no reflections affect Vs, they might affect the voltage other side of Zs, but not Vs. So that is why you don't get reflections to Vs.

The series resistance Zs gives Vs some source resistance and matches it to the transmission line. It also makes Vs look more like a real world voltage source with limited current sourcing.

However what I do not understand is why I wouldn't just have reflections before Zs and after ZL in the figure above, shouldn't "everything" in the circuits on both ends always be impedance matched ?

You will get reflections at VL if Zl is not matched to Zc, that voltage is dependent on the ratio of Zc and Zl. You will also have reflections between Zs and Zc if they are not matched.

You can think of a transmission line kind of like a trough of water, if the impedance is matched the water is flat and you can send a wave from one end to the other. If you don't match impedances, it's like putting a bar in the water, the wave will reflect and possibly go over it to the other end but it will not be a perfect wave.

Also keep in mind that most real world transmission lines are lossy, so you lose some of the wave to heating and magnetic losses as it moves along.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.