High speed design noob here.

Resistance increases as wire gets longer, but I found in saturn pcb calculator the impedance depends only on track geometry and distance from plane.

Say for DDR3 single ended signals with controlled impedance, does a point to point long track (say 150mm) vs. a short track (say 10mm) have any effect on the signal quality?

  • 1
    \$\begingroup\$ Track impedance involves a perfect, zero-ohm conductor. In other words, a 50-ohm coax cable would still be 50ohm, even if it was made from superconductor. Also this: if your track was infinitely long, then you could directly measure its impedance with an ohmmeter. \$\endgroup\$
    – wbeaty
    Sep 29 '17 at 5:00

The characteristic impedance of a transmission line is not the same thing as a lumped resistance, it just happens to have same units. Similarly, certain amplifiers are designed to have a current as input and a voltage as output, so their gain is a ratio of volts to amps, with units of ohms. But that doesn't mean those amplifiers have much at all in common with resistors.

The characteristic impedance of a transmission line is the ratio of the voltage and current of a wave that can travel along the line without distortion. If you tried to inject a signal with a different ratio of voltage to current, you'd find that part of the injected signal travels one way on the line and the other part travels the other way --- you'd create a reflection.

Since this property of the line --- the type of wave that can travel along it without distortion --- is specified by a ratio of voltage to current, we can give it a value in ohms, and call it an "impedance". But just like the gain of a current-to-voltage amplifier, that doesn't mean it has any other behavior in common with a resistor, and you shouldn't expect it to.

  • 1
    \$\begingroup\$ Very well said. \$\endgroup\$
    – bitsmack
    Sep 29 '17 at 4:13
  • 5
    \$\begingroup\$ Your answer connected a lot of dots in my head. \$\endgroup\$ Sep 29 '17 at 6:35
  • 3
    \$\begingroup\$ It doesn’t behave like a resistor , rather it behaves like a resistor, inductor and shunt cap with slight leakage conductance per unit length AND you should expect it to, too. Since PCB dielectric constant decreases near & above 1 GHz , Zo also changes somewhat. \$\endgroup\$ Sep 29 '17 at 10:58
  • 2
    \$\begingroup\$ @TonyStewart.EEsince'75, The characteristic impedance does not act like a lumped impedance. For example, if a lumped impedance is real, it means real power is consumed by that element; if it's imaginary it means only reactive power is used. For a transmission line with characteristic impedance Z0, the behavior is the opposite: real Z0 indicates a lossless line, an imaginary component of Z0 indicates real power consumed. \$\endgroup\$
    – The Photon
    Sep 29 '17 at 15:35
  • 1
    \$\begingroup\$ @TonyStewart.EEsince'75, you wrote your answer, I wrote mine. If you think there's more to add, put it in your answer. \$\endgroup\$
    – The Photon
    Sep 29 '17 at 22:03

Distributed RLC is a per-unit-length variable which depends on track width and thickness-to-gap ratio to ground and dielectric e, which determines Zo.

  • minor changes occur due to conductivity and skin effects, but

Since RLC values are distributed, and impedance depends on ratios, length has no effect on Zo, but it does affect attenuation.

When there is a mismatched load and ω, the propagation delay is less than the rise time. The result is overshoot and when mismatched at the source, another reflection occurs which results in the classic dampened ring waveform at half cycle corresponding to this propagation delay time.

Inductance increases with lower track width to gap times length, while capacitance increases with conductor area to gap ratio times length.

  • Thus the impedance input and output of this transmission line Enter image description here becomes the ratio of \$Z_o=\sqrt{\frac{R+\omega L}{G+\omega C}}\$ which at high ω or small x, you can neglect R and G, and at DC you can neglect L and C.
  • If Zo is much lower than the load then C dominates the response with a fixed source resistance.

  • When Zo is higher than the load, then L dominates the response.

The math proof for the above exists, but it is not shown for the sake of brevity.

Final Comments

  • Zo is distributed by the geometry impedance ratio of the path. Since skin effect affects effective thickness, R rises with f rapidly near and above skin depth while dielectrics Zc(f) lowers with rising f. Thus moist food with water’s high (80) dielectric constant absorb more current than dry foods and salt raises the G value to shunt more current.

  • In conductors and dielectrics, length does affect the signal level and frequency dependent the time delay and resonant frequency of the path in each dimension dependent on the geometry.

  • But don’t let anyone misdirect your thinking that length does not matter Just because Zo may not change; length and Zo certainly affects path current to a load, but at low frequencies it does not represent this as a load resistance, rather it becomes more dominated by length reactance.

This length has a strong effect on wave rise times, delay and current, which affects power transmission line currents, wireless equalizations, modem equalization, and logic level rise times.

The length and Zo affects path loss and special delays with frequency/length ratios like 1/4 wave impedance reflections (inversion) and all odd harmonics of same.

There is also a frequency-dependent loss pattern called transfer impedance, which is affected by impedance effects on coaxial weave patterns, foil secondary shields and quality of earth ground in distributed video.

In order to separate the contribution of loss to all these frequency dependent variables, it is necessary to use scattering parameters to define Zo with some source/load reference, which are also available for passive parts in microwave applications to better define the impedance and current flow in the part.

Since the values of L and C in Zo imply a group delay and the parameter values tend to change with the wave/length ratio, there are tolerances which cause unequal delays and dispersion or closure of eye patterns on digital signals that choices must be made to reduce these effects. If conduit cannot be improved with a rigid precision waveguide, then the signal can be split into any small audio bands and processed with their own equalization to achieve a much higher quality signal than the aggregated base signal.

Response to added input in the question

does a point to point long track (say 150 mm) vs. a short track (say 10 mm) have any effect on the signal quality?

As I indicated above, Zo depends strictly on geometry of the conductors length/gap-thickness ratio for a chosen dielectric constant, and tolerance on both is very important, so a TDR test ought to be paid for at the board shop to ensure D codes are tweaked to match the batch deviation on the D constant. Dwgs must define Zo for each D code trace needed.

After all it is the length/track width ratio that determines Zo and mismatch that will result. So a longer trace doesn't matter as long as the track width is also made wider or the dielectric gap made thinner.

  • The second critical difference is that the propagation delay affects RAS/CAS delays.
  • The third thing is the longer the line is, the greater the chance for bends, mismatch and etchback tolerance errors on the track so the slightly wider tolerance. Which is why a 50 m HDMI cable at 24 Gbit/s is harder to make than a 5 Gbit/s 1 m cable for 4K and high resolution 3D screens with high refresh rates. Not impossible, but maybe it costs twice as much /ft with stiffer cables, common-mode choke, and isolated secondary shields.
    • In PCB tracks, the skin effect has an effect on conductor inductance and resistance at UHF frequencies as well, so copper coatings make a difference too, as well as the loss tangent related to G and C affect signal quality and people move up to polyamide or Teflon or ceramic for higher GHz signals.

DDR3 has a nominal driver impedance of 34 Ω (30.5–38.1) according to the standards I read, but there are various Zo options for signals for nominal and dynamic writes.

  • 1
    \$\begingroup\$ Thank you Peter for your grammatical improvements. You remind me of the Quality Engineer I depended on during my early Aerospace R&D days in the 70's when I had the creative talents but weakness for syntax perfection. Probably why I later had challenges with Forth compilers. You deserve any points added to my answer with your improvements, if I could donate them and I will try to improve my punctuation and reverse my tendency to eliminate non-valued added prepositions , that make it smoother to read. \$\endgroup\$ Sep 30 '17 at 13:11
  • \$\begingroup\$ D codes? Is that a software variable to correct for physical imperfection? \$\endgroup\$
    – Pugz
    May 31 '19 at 20:50
  • \$\begingroup\$ @Pugz yes in Gerber files, the D codes are in a file with a list of track width values for each D number, so when controlled impedances need certain tracks say using D12=10 mil to be adjusted, only that value needs to change rather than every layer of gerber code using those tracks. \$\endgroup\$ May 31 '19 at 21:21

To a first approximation, the only effects of length are attenuation and delay. In the case where the line is terminated in its characteristic impedance (load matches transmission line impedance perfectly), the signal at the load will be smaller and smaller as the line gets longer and longer. Normally this is not an issue with digital signals on a PCB. The attenuation is normally not significant. It can be important in signals that go off board into long cables (DSL, LVDS, ethernet, video, etc).

In the case that the line is terminated in an impedance which does not match, then there will be reflection from the load back to the driving source. This reflection will disrupt the waveform back at the source. Because of the delay, any reflection from the far end will come at a different time depending on transmission line length, so it can definitely affect signal integrity. It is possible that in some cases lengthening a line can improve signal integrity by moving the reflection to a place where it is harmless. This could be an issue any time you have bidirectional signalling on one line.

There is one other issue. Signal integrity also includes timing. Long traces, because of the delay they add, can cause timing failure just by taking too long. For example, the memory chip must receive a read command, then assert valid data on the lines, then that valid data signal needs to propagate back to the host, and finally be read by the host. If the "flight time" of the signals is too long, the memory chip will not be able to assert valid data fast enough to satisfy host setup timing requirement. So, long transmission lines can effect signal integrity in this fashion also.


Just to reinforce what @ThePhoton said in his excellent answer, even vacuum has its own value of characteristic impedance, which is called usually \$Z_0\$ (a.k.a. characteristic impedance of free space):

$$ Z_0 = \frac {E} {H} = \sqrt{ \frac {\mu_0} {\epsilon_0}} \approx 377 \Omega $$

It has nothing to do with currents, ohm's law and resistance, but it is the ratio of E to H fields amplitudes for a plane wave traveling in free space, and just happens to have ohms as measurement unit!

  • \$\begingroup\$ Actual there are are currents flowing in waves in free space but being an insulator, the currents are reactive vs conductive, which due to gap to far end electrode or antenna reduces to extremely low currents. Think about wireless power transfer (WPT) Zo being matched, to maximize the current flow and thus cannot exceed 50% effic at MPT but can improve efficiency of transfer only slightly below max due to slight mismatch. So False conclusion \$\endgroup\$ Sep 29 '17 at 11:14
  • \$\begingroup\$ @TonyStewart.EEsince'75 I'm not completely sure about what you mean with "false conclusion". Anyway, when I said "nothing to do with currents, ohm's law and resistance" I thought it was obvious I was talking (rather informally) about conduction currents inside lumped circuits. If you object that whenever you have a time-varying D field (which is equivalent to E field in vacuum) you have a displacement current density, well, that's obvious from Maxwell's equation, but I wasn't trying to condense the whole EM theory in a short answer... \$\endgroup\$ Sep 29 '17 at 12:32
  • 1
    \$\begingroup\$ @TonyStewart.EEsince'75 ... I just wanted to show the OP that even in vacuum you can have "something" measured in ohms (and my whole answer is a sort of appendix of that of @ThePhoton). So I don't think I reached a false conclusion, I just reported an example taken from a difficult theory trying to get the thing simple. \$\endgroup\$ Sep 29 '17 at 12:34
  • \$\begingroup\$ Yes, the simple L/C ratio is a constant AC impedance for any length, but the real RLC values and Ohm's law will have some effects with DC losses and distributed RLC, frequency dependent losses per unit length will exist. \$\endgroup\$ May 31 '19 at 21:40

While the other answers are not wrong, I don't think they adequately address the fundamental conceptual hurdle to understanding characteristic impedance.

Imagine you are a wave. You propagate by taking a step - these steps are always the same size. This is your wavelength.

The characteristic impedance is the impedance, or resistance, you will feel taking each step. Low impedance might feel like walking normally, while high impedance might feel like walking through mud - there is a lot more viscosity resisting the movement of your leg every time you take a step.

Now, the total energy or loss or however you'd like to look at it is going to very much depend on length, and it does. But it doesn't matter how far you have to go, it is going to be a certain difficulty taking one step through air, and one step through mud. The characteristic impedance is the impedance felt taking one step. The number of steps there are to take does not change this value.

To bring things back into reality a bit and less analogy, the characteristic impedance is the impedance one wave length of a propagating electromagnetic wave will 'feel' through a given transmission line. This is why it is called the characteristic impedance - it's an impedance that characterizes the bulk nature of it. At any given step, the signal is going to see the same impedance between it and the next step.

This is why one can terminate a 50 Ω transmission line with a 50 Ω resistor on one end regardless of length - one can view the termination as the final 'step' the wave makes on its transmission journey, so a lumped 50 Ω resistance across the transmission line pair is perfectly acceptable - because the wave has already been experiencing 50 Ω of impedance at all times.

Now, let's take this conceptual understanding as context, and touch on The Phonon's excellent answer.

Knowing that the characteristic impedance is, in fact, the impedance felt at any given time when traveling down a transmission line, it becomes obvious that this is also the ratio of voltage to current that won't cause a reflection.

However, this might still be confusing. Wouldn't that mean that higher frequencies, having to take more steps, would experience that much more resistance for the same length of line? Well, attenuation down a transmission line does generally increase with frequency, but not because of this.

Let's assume you get the 'characteristic' part of characteristic impedance. But, you also need to get the impedance part. Impedance is a complex value, meaning it has both real and imaginary components.

Imaginary in the mathematical sense - don't fall into the trap of taking imaginary in a mathematical context literally. It's a name, that's all. Imaginary numbers are named as such as sort of a play on words compared to the name we gave the opposite basis number line - real numbers. All numbers are, technically, imaginary. Likewise, no numbers are real. But some are imaginary. And some are real.

Real numbers and imaginary numbers form the complex plane, which can be imagined as two axes at right angles, one being the real number line, stretching from -∞ to ∞, the other being the imaginary number line, stretching from -∞*i to ∞*i. And, we know they exist and we need them because there are equations whose solutions demand imaginary numbers. Without them, you simply ignore the ability to answer an entire category of equations. At the simplest, imaginary numbers allow us to give an answer to this equation: \$x^{2} + 1 = 0\$. x, of course, is equal to i.

OK, that was a bit of a tangent, but a valid understanding of complex numbers is absolutely required before one can understand impedance.

Impedance is made up of a real component, which is simply DC resistance, and an imaginary component called reactance. Reactance is apparent resistance, but it is not due to the dissipation of energy as heat (as with resistance), but rather the temporary storage of energy that is later released. If you see energy being siphoned off because it is getting stored in an electric field (aka a capacitor) or a magnetic field (inductor), at that moment, it appears just like energy that is simply lost as heat due to resistance.

It depends on the transmission line, but they of course do suffer increasing losses with length. You will usually find this somewhat indirectly given as 'attenuation per foot' or attenuation per 100 meters or similar, in dB/. This will include losses due to real resistance (which is not even as simple as measuring with an ohmmeter - frequency will change things like skin depth, making the same conductor appear more resistive, etc., etc.), dielectric loss, whatever other things cause a true dissipation of energy into entropy/heat.

Characteristic impedance is, generally, almost entirely due to reactance. So 50 Ω of reactance and 0 Ω of resistance wouldn't actually cause any loss - it would only be a temporary loss as energy is stored, but then later released, back into the line. If you have a voltage and current ratio that is not such that the voltage drop (energy stored) at a given current is equal to the voltage across the transmission line, then you do not perfectly balance the energy stored with the energy released, and you get the bane of signal integrity's existence, REFLECTIONS!! Oh noes!

This cycle of energy storage and release forms a standing wave in our transmission line. Any excess voltage forces more current to flow, which means we've exceeded the energy storage capacity of the cable, so the phase gets thrown off, and our standing wave gets destructively interfered with. Our signal is, to varying degrees, destroyed.

  • \$\begingroup\$ Your next-to-last paragraph is the key. Hilite that? And your final paragraph: "exceeded energy storage capacity" misleads. \$\endgroup\$ Sep 29 '17 at 14:09

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