6
\$\begingroup\$

I have very little background in electronics, but suddenly I need some transmission line theory for a project I am working on. I have heard of various rules of thumb for how long a wire can be before you have to start worrying about its properties as a transmission line. I know these are pretty arbitrary, but the most common I have heard is that you have to start worrying about transmission line properties once the length of your wires are greater than about wavelength / 10, so I'll just go with that for this post.

My question is: what quantity or quantities are minimized by using wires that are short relative to the wavelength that makes it so that you can ignore transmission line theory? I.e., when our wire length is less than wavelength / 10, presumably there's some set of physical quantities that makes our transmission line look very wire-like in models. What are these?

As an example of the type of thing I'm looking for, here's what my first guess was: At short lengths, the phase difference for a wave would be small enough that the wave's amplitude at the two ends of the wire would be roughly the same. But this doesn't hold: sin(2 pi / 10) - sin(0) = 0.59, so if we use the wavelength / 10 rule of thumb, the amplitude change in the wave between the two ends of the wire could be more than half of the total amplitude.

As of now, I'm guessing that the actual answer is that the characteristic impedance of a wire is very near 0 (or whatever tiny resistance the wire has) at short wire lengths - i.e., you can ignore its capacitive and inductive properties. Is this correct? If so, are there some formulas relating Z_transmission_line to wire length which show that Z ~ R at lengths of about wavelength / 10? What I'm looking for is something where I can literally plug in length = wavelength / 10 and get some number close to what I would get if I plugged in length = 0, that sort of acts as a heuristic for how good a certain length is.

Thanks for any help!

\$\endgroup\$
  • \$\begingroup\$ sin(2pi/10) = 0.01, where did you get the 0.59 value? So there is very little phase difference as expected \$\endgroup\$ – sarthak Nov 18 '17 at 9:29
  • \$\begingroup\$ sin(2pi/10 radians) = 0.59, sin(2pi/10 degrees) = 0.01. Radians are the right one to use there. If I was using degrees, I would have wanted to look at the quantity sin(360/10 degrees). \$\endgroup\$ – user3558855 Nov 20 '17 at 18:02
2
\$\begingroup\$

When propagation delay exceeds 10% of wave time , and impedances are not matched throughout, reflections occur which alter the effective input impedance and Ohm's Law. The amount % depends on system tolerances for return loss, so this is a ballpark figure.

Remember that if the prop delay is 25% of \$\lambda\$ the output impedance if mismatched is inverted at source, ( short becomes open and open becomes short). So Ohm's Law applied to impedance ratios or in other words H(s) transfer functions will be false and more complex equations must be used.

Consider that all conductors ( wires , tracks, leaded parts ) have inductance based on the length to width ratio. So a perfect conductor rises in impedance with f. and if the applied f starts approaches 10% of λ then its impedance changes rapidly due to reflections.

\$\endgroup\$
  • \$\begingroup\$ I would note that in phase reflections can, under certain conditions, damage the output driver. \$\endgroup\$ – Peter Smith Nov 18 '17 at 10:21
  • \$\begingroup\$ Yes impedance inversions can cause open stubs to become shorts at that wavelength. \$\endgroup\$ – Sunnyskyguy EE75 Nov 18 '17 at 16:57
  • \$\begingroup\$ Thank you for the response. If I understand you correctly, what is relevant is voltage and current between the lines in a transmission line, either at the source or at other locations on the line. With a wire, that voltage and current will depend exclusively on what you have at the load via Ohm's law, and this is indeed the case for very short transmission lines. However, for longer transmission lines such as your L=lambda /4 example, this ceases to be the case - e.g., you can have a load of infinite resistance and still have full current across the source. Is this understanding correct? \$\endgroup\$ – user3558855 Nov 20 '17 at 18:09
  • \$\begingroup\$ Yes the wave is reflected to cause a short at source with a 1/4 wave stub such that after another 1/4 wave return it is now 180 deg out of phase at source. \$\endgroup\$ – Sunnyskyguy EE75 Nov 20 '17 at 22:26
0
\$\begingroup\$

It's to do with the time that reflections at the load of the transmission line take to return to the sending end.

For example, consider a step of direct voltage of \$\small 100 V\$ applied to an infinite length of \$\small 50\Omega\$ coaxial cable. There is no reflection, due to the infinite line length, so the voltage source always sees \$\small 50 \Omega\$, and a continuous current of \$\small 2 A\$ flows into the line.

In contrast, for a short transmission line with, say, an open circuit at the load end, the voltage and current reflections arrive back at the sending end almost instantaneously and the line voltage and current immediately assume their steady state values of \$\small 100 V\$ and \$\small 0 A\$. For an arbitrary load, \$\small Z\$, the source sees this impedance immediately.

Extend this analysis to sinusoidal forcing functions, and it's apparent that the source sees the load impedance for an electrically short transmission line length, and sees the line characteristic impedance for an infinitely long line.

For transmission line lengths between these two extremes, the source sees an impedance that depends on the line length, with respect to wavelength, and the degree of mismatch (if any) between line characteristic impedance and load impedance. The input impedance, for any given load impedance and transmission line length, can be found graphically from the Smith Chart

\$\endgroup\$
  • 1
    \$\begingroup\$ Please have the courtesy to explain briefly why you down voted. \$\endgroup\$ – Chu Nov 18 '17 at 9:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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