Hmm, this seems to be just another question on line impedances.

I understand that when we say "transmission line" effects we talk about things like cross talk, reflections and ringing (I guess that is just about it). These effects are not present at low frequencies where the PCB trace behaves like an "ideal" transmission medium, more like we expect a wire to behave in our early school days.

I also understand that the 50 ohm value comes not from the line resistance which is going to be very small and less than 1 ohm. This value comes from the ratio of L and C on the line. Changing C by changing the trace height above ground plane or changing L by changing the trace width shall change the impedance of the line.

We all know that the reactance of L and C is dependant on the signal frequency as well. Now my questions:

  1. Why should we not call this as line reactance only rather than line impedance?

  2. How can it be just 50 ohm? It has to be signal frequency dependent right? E.g 50 ohm at 1 MHz

  3. Will the world end if I chose to do a 100 ohm or 25 ohm trace instead? I know that while we like to say 50 ohm as a magic number, it will be within some range around 50 ohm and not 50.0000 ohm exactly.

  4. Is there any time when the actual resistance of a PCB trace may matter?

  • 1
    \$\begingroup\$ If you know Zo comes from the ratio of L and C surely it doesn't take much more thinking to realize that it isn't frequency dependent (above 1MHz or thereabouts). Ditto for length. -1 \$\endgroup\$
    – Andy aka
    Jul 9, 2013 at 11:40

5 Answers 5


Let's look at the formula and equivalent circuit for a transmission line.

Enter image description here

(1) Impedance rather than reactance.

Reactance refers to the opposition to the change in current (of an inductor) or voltage (for a capacitor) - single components. The transmission line has \$R,L\$ and \$C\$ components - impedance is the ratio of voltage phasor to current phasor.

(2) It is \$50\Omega\$ because the ratio of inductance to capacitance per unit length produces that value. As \$R << j\omega L\$ and \$G \to 0\$, these values can be ignored and so the expression reduces to \$\sqrt{L/C}\$ (frequency independent).

(3) Nope, but it's generally a good idea to keep things as standard as possible. You may find it difficult to find a suitable connector for your \$167\Omega\$ transmission line. There's also a lot of information available for designing standard transmission lines on PCBs, etc. The magic number in my book is 376.73031... the impedance of free space. Now without that one we'd live in a different universe.

(4) Going back to the formula. At low frequencies \$R\$ may be significant as the reactance of the inductor will be small). At very high frequencies the dielectric losses may become significant.

  • \$\begingroup\$ Other points are clear, but what do you mean by dielectric losses? \$\endgroup\$
    – quantum231
    Jul 9, 2013 at 14:57
  • \$\begingroup\$ @quantum231 The dielectric is just a fancy name for the insulation between the two conductors of the transmission line. In other words the middle bit of the capacitor, C. Like all capacitors its not 'ideal' check out en.wikipedia.org/wiki/Loss_tangent \$\endgroup\$ Jul 9, 2013 at 18:00

A transmission line has distributed inductance and capacitance along its entire length. We can think of it as infinitely many little inductors and capacitors along the line:


simulate this circuit – Schematic created using CircuitLab

Each inductor serves to limit the rate at which the capacitor can charge. But, as we divide the line into increasingly many parts, the inductors and capacitors each become smaller. So, does the number of them matter? We can choose to divide the transmission line into however many segments we want, from one to infinity. Thus, we can make the capacitors and inductors arbitrarily small.

Thus, the value of these inductors and capacitors must not matter. Indeed, it is only the ratio of inductance to capacitance that matters, because this does not change as the transmission line is divided. And if the characteristic impedance doesn't change as the line is divided, it follows that it also doesn't change as we make it longer.

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    \$\begingroup\$ My dear Phil, your answer is SPOT ON I must say. You made my day :D \$\endgroup\$
    – quantum231
    Jul 9, 2013 at 14:58
  • \$\begingroup\$ I need to check out how we derive the value of Zo for a transmission line. \$\endgroup\$
    – quantum231
    Jul 9, 2013 at 15:03

Adding to what Phil said:

Now imagine everything starts out at 0 Volts and Amps in this long chain of inductors and capacitors, then you put a voltage step in one end. The way the inductors slow down how the capacitors are charged, a steady current will flow, which will be proportional to the voltage you put in. Since you have a voltage and a current proportional to that voltage, you can divide the two to find the resistance this infinite transmission line mimicks. In fact, for a ideal infinite transmission line, you can't tell the difference between the transmission line and a resistor from the outside.

However, this all works only if the voltage step can keep propagating down the transmission line. But, and here is the aha moment, if you have a short line but put a resistor of the characteristic resistance accross its end, it will appear like a infinite transmission line at the other end. Doing this is called terminating the transmission line.

  • \$\begingroup\$ Thanks Olin, so on the infinite transmission line the signal is attenuated to 0 on the other hand. This is what we want to make happen and this is what I understood from your description. \$\endgroup\$
    – quantum231
    Jul 9, 2013 at 15:00
  • \$\begingroup\$ In a ideal tranmission line, the signal remains intact indefinitely. In a real line, the resistance of the conductors dominates after a while, and the signal is attenuated and low pass filtered with distance. \$\endgroup\$ Jul 9, 2013 at 15:57
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    \$\begingroup\$ on that point: Do long transmission lines degrade rise/fall times, and if so, by what mechanism? \$\endgroup\$
    – Phil Frost
    Jul 19, 2013 at 12:11
  • \$\begingroup\$ Best answer I've seen around. How does the generator "know" that the transmission line is open ended and that it needs to increase the voltage? Do electrons bounce back? I'm trying to answer that here (there is a diagram): electronics.stackexchange.com/questions/165099/… \$\endgroup\$
    – user42875
    Apr 16, 2015 at 12:39

Jim had a very good answer. To expand on a few, however:

2) 50 Ohms is 50 Ohms (kind of). The dielectric constant of a material IS slightly frequency dependent. Therefore, the trace height and width you choose for 1 GHz will be a slightly different impedance at 10 GHz (if you need to worry about the difference, you probably already know about the difference!)

4) For standard PCB FR4 material, the dielectric loss will become a concern around 0.5 to 1 GHz. The RESISTANCE, however, does become important when you have higher current lines. For example: If you have 1 Amp going on a 6 mil wide trace of 1 oz copper for 1 inch of length, that's .1 Ohms of resistance. You'll have a drop of about 0.1V and around a 60C temperature. If you can't handle that 0.1V drop, you need to obviously widen the trace or thicken the copper.

As a rule of thumb, if you have lengths under 1 inch, most DC resistances can be ignored.

  • 1
    \$\begingroup\$ Good point about the pcb substrate material (+1) from me picking up on the dielectric losses. \$\endgroup\$ Jul 9, 2013 at 12:38
  • \$\begingroup\$ It seems that I need to read about this dielectric loss at high frequencies. Is it somewhere in the High Speed Digital Design (Black Magic) book of howard johnson? \$\endgroup\$
    – quantum231
    Jul 9, 2013 at 15:02
  • \$\begingroup\$ If you cannot find what you need there, you'll probably get enough information just looking up information about the loss tangent of materials vs. frequency. Many RF simulators will also have these calculations built in. Furthermore, datasheets of your circuit board material will oftentimes show you the graph vs. frequency. If you need to start worrying about Dk at higher frequencies, you'll want something like a Rogers or Taconic board material that has a flatter loss profile over frequency. \$\endgroup\$
    – scld
    Jul 9, 2013 at 15:09

There is a simple hand-waving explanation why the effective impedance of a (ideal) transmission line is a constant. Other explanations leave some confusion on how do we "select" Li and Ci in the model of transmission line. What are these Li and Ci exactly?

First, once we say "transmission line", we are talking about long wires. How long? Longer than the length of an electromagnetic wave that is transmitted along the line. Therefore, we are talking either about very long lines (miles and miles), or a very high frequencies. But the concept of wavelength relative to trace length is fundamentally important.

Now, as people mentioned, a trace has certain inductance per unit of length, and, correspondingly, certain capacitance, again proportional to length. These L and C are inductance and capacitance per unit length. So, the actual inductance of a wire segment would be L = L * length; same for C.

Now consider a sine wave coming into the trace. Waves propagate at the speed of light (in particular dielectric/air media it is about 150ps/inch). At every and each moment the particular charge deviation (waveform) interacts with a section of wire equal to the corresponding length of this wave. Slower frequencies have longer wave lengths, while faster frequency components have proportionally shorter lengths. So, what do we have? Longer waves "see" a longer trace and therefore a bigger L, and bigger capacitance C. Shorter (higher frequency) waves "see" the shorter effective line length, and therefore smaller L and C. So, both effective L and C are proportional to wavelength. Since the impedance of the line is Z0=SQRT(L/C), the dependence of L and C on length cancels, and that's why waves with different frequencies "see" the same effective impedance Z0.


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