Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

First of all: I am a computer science student, so I don't have much knowledge of physics. So please keep your answers simple.

I recently learned something about circuit design and its problems (differend kinds of hazards). To model the problems, we introduced a "dead time model" ("Totzeitmodel" in German, I have if it is "dead time model" in English)
We added some dead time to each element of the circuit, but we didn't add dead time to the wires of the circuit. I asked the prof. why we didn't add dead time to the wires. He responded that the signal is moving much faster and you can neglect the time that signals need to pass the wires.

Now I would like to calculate the speed of the signal (is this the wave propagation speed?) for some very simple settings:

  • assume we have a copper wire
  • the wire is a perfect cylinder with diameter of 1mm
  • the current is 2A
  • the voltage is 12V

Can you help me with this? Do you need something else to calculate the speed?


Notes: I found the wikipedia article Wave propagation speed and some questions on physics.stackexchange.com, but the questions and answers were either too complicated or didn't directly give numbers (like that one)

A little side question: When I think about the electric signal, I imagine some elastic balls. When there is a signal at one end, you push the ball. It gets compressed and expands later, which compresses the next ball a bit and it expands, ... This way, the last ball gets moved and the signal arrives at the end. Do I have to get another thinking-model for simple circuits or will I be able to understand simple circuits with this model in mind?

share|improve this question

2 Answers 2

up vote 1 down vote accepted

To calculate the propagation speed, you need to specify the return current path in addition to the "forward" path. The reason is that the electromagnetic fields that determine the propagation characteristics fill the space between the two conductors. [If you try to calculate the inductance of a single wire, you get an infinite result.]

The filler material between the conductors matters too: its electric polarizability (quantified by the dielectric constant $\epsilon$, which is typically 2-5 times the value for free space $\epsilon_0$) slows down the signal speed. Typically the filler is magnetically neutral, so its susceptibility $\mu$ is the same as for free space.

For a coaxial conductor, the wave speed formula ends up being very simple:

$$ v = \frac{1}{\sqrt{\mu \epsilon}}$$

For a relative dielectric constant ($\epsilon/\epsilon_0$) of 3, one calculates a velocity of 58% of the speed of light.

Finally, your elastic ball analogy is good to zeroth order, but I don't think you can use it to think about propagation velocity. There are two independent (but coupled) fields (electric and magnetic) at play here.


UPDATE: It turns out that the geometry of the conductors doesn't matter much; the main determinant of the propagation velocity is the filler material properties. For parallel conductors of arbitrary (but constant) cross-section, the propagation velocity is: $$ v = \frac{c}{\sqrt{\mu_r \epsilon_r}} $$

Here the relative permeability of the filler $\epsilon_r = \epsilon/\epsilon_0$ (typically 2-5) and relative magnetic susceptibility $\mu_r = \mu/\mu_0$ (usually 1), while $c$ is light-speed. So the formula for the coaxial geometry turns out to be quite general (note $c=1/\sqrt{\epsilon_0 \mu_0}$).

As Jaime mentions in the comments below, there will be some additional "internal" inductance due to the magnetic fields within the conductors which will reduce the velocity; that bit is geometry-dependent.

share|improve this answer
    
I believe the infinite self inductance is an artifact due to infinitely thin wires, that disappears when you consider a finite radius. It is after all self inductance of a single wire that causes things like skin effect. –  Jaime Dec 26 '12 at 21:13
    
@Jaime: Even a finite radius wire has infinite inductance, at least for the infinite-length case. (The B-field goes like 1/r, so its integral goes to infinity logarithmically at both 0 and infinity). It's true that the internal inductance (inductance from B-fields within the conductor) is finite. –  Art Brown Dec 26 '12 at 21:33

I understand you are already aware that the signal, i.e. the electromagnetic wave, propagates much, much faster than the actual electrons move. You want to read about the telegrapher's equations, which in a first lossless approximation yield a speed of propagation $v = 1/\sqrt{LC}$, where $L$ and $C$ are the inductance and capacitance of your circuit.

In this article, referenced in the comments to one of the answers for the question in this site that you mentioned, there are, starting in page 9, formulas to calculate the exact $L$ and $C$ for infinite pairs or wires (one going, the other returning), either parallel or coaxial, and thus the speed from them.

You could also try to figure out the inductance and capacitance of a single infinite wire, and use those values to come up with a transmission speed for your setup, but I am not sure it will be very relevant in a any real setting. Or you can simply go with the "anywhere between 40% and 90% of the speed of light" ballpark approximation, which is still ridiculously fact, and probably proves your professor right in not needing to worry about the wires Totzeit...

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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