5
$\begingroup$

There have been a number of questions regarding the issue of feed line loss and ladder line eventually gets discussed. So I am curious, how does one begin to estimate the loss in ladder line?

So, for the case of 2 bare wires "somehow" spaced an appropriate distance apart acting as a feedline with SWR=1.000, what loss could be expected? (Loss due to SWR was very well covered here: What is the actual loss in a feed line with high SWR? )

I would think that spacing the wires farther apart would increase the impedance, lowering the current and thus the loss/100m (at a cost, perhaps, of more complicated and lossy impedance transformers at each end which I am ignoring). I would think that adding an insulator would (1) lower the velocity factor --somewhat increasing the loss by somewhat increasing the number of wavelengths at a given frequency-- and (2) add some sort of additional loss mechanism in addition to that which could be attributed directly to the velocity factor.

Is there a way to quantify any of this?

$\endgroup$
  • 2
    $\begingroup$ You mean, other than by looking it up or determining it empirically? $\endgroup$ – Phil Frost - W8II May 3 at 23:43
  • $\begingroup$ @PhilFrost-W8II - yes. If a spec sheet is available, I would prefer that of course. But if one were, say, considering the potential merits of a home brew line with 10 AWG wire spaced 55mm apart vs. 14 AWG with 644mm spacing (!!) (450 ohms vs 800 ohms), how would one begin to estimate which has better loss characteristics? There are calculators out there for many things, but I have not seen one like this. $\endgroup$ – Chris K8NVH May 4 at 0:15
5
$\begingroup$

For most feedlines, dielectric loss is very low, and at HF where ladder line is practical, negligible. So significant losses are due to resistance and associated Joule heating of the copper conductors.

The characteristic impedance $Z_0$ gives the ratio of voltage to current in a matched line, so we can always find voltage if the current is known:

$$ E = Z_0 I \tag 1 $$

and the power $P$ transmitted by the line is the product of current and voltage between the conductors:

$$ P_\text{tx} = IE $$

substituting equation 1 for $E$, we get

$$ P_\text{tx} = I^2 Z_0 $$

and solving for $I$ we can determine the current in a matched line as a function of power and characteristic impedance:

$$ I = \sqrt{P_\text{tx} \over Z_0} \tag 2$$

Now, Joule heating which is most of the loss is the product of resistance $R$ and current squared:

$$ P_J = I^2 R $$

substituting equation 2 for $I$ yields

$$ P_J = {P_\text{tx} \over Z_0} R \tag 3 $$

therefore as a first order approximation, we can say losses are inversely proportional to characteristic impedance:

$$ P_J \propto {1 \over Z_0} $$

In other words, doubling the impedance halves the power lost.

Keep in mind "loss" in transmission lines is actually the ratio of power in to power out: it's power not lost. So don't make the mistake of thinking a feedline with 1dB of loss, if the impedance is doubled, becomes 3dB better with a result of 2dB of gain!

Rather, a loss of 1dB means $ 1 - 10^{-1/10} = 20.6\% $ of the power is lost. Doubling the impedance would change that loss to $ -10 \log_{10} (1- .103) = 0.47\:\mathrm{dB} $.

Equation 3 provides a start for calculating the loss, but the devil is in the details, especially in calculating $R$. Both the skin effect and proximity effect must be taken into account, and the math is not nearly so simple.

For practical purposes I suspect the approximation is good enough to appreciate the potential benefits of a feedline change. If more precise loss numbers are required then perhaps an empirical method is easiest.

$\endgroup$
  • $\begingroup$ Interesting. So R could be measured at the open end, with the far end shorted? $\endgroup$ – Mike Waters May 4 at 19:03
  • 1
    $\begingroup$ @MikeWaters If you had some way to compensate for the transmission line effects, I suppose it could. Though at that point you're basically measuring the loss empirically. More likely R could be estimated from copper's conductivity and the conductor geometry, multiplied by some factor to account for skin depth and proximity effect. $\endgroup$ – Phil Frost - W8II May 4 at 19:53
  • $\begingroup$ The OQ asked about "estimating", so my question isn't really valid. "Measuring" is easier. $\endgroup$ – Mike Waters May 4 at 21:26
  • $\begingroup$ Thank you for adding that part better explaining what 3dB loss really means. This is the level of detail and complexity I was looking for. $\endgroup$ – Chris K8NVH May 4 at 22:41
4
$\begingroup$

There is a program called TLDetails, available from https://ac6la.com/tldetails1.html, that will calculate the loss for many standard transmission lines. One enters the frequency, the type and length of transmission line, and the load impedance and TLDetails will calculate the line loss, SWR, and impedance seen at the source plus other parameters. More information on k0, k1, and k2 is covered under https://ac6la.com/zplots1.html.

$\endgroup$
  • 1
    $\begingroup$ Wow!! Thanks for pointing out that excellent program, Cecil. I use a lot of other tools from AC6LA, but this one flew under my radar. $\endgroup$ – Brian K1LI May 4 at 0:19
  • 1
    $\begingroup$ This is very close. For an arbitrary line, I would think there is some way to estimate K0, K1, and K2. Presumably that could be fed into that software. K0 is just DC resistance. Maybe K1 can be derived from the wire gage? Maybe K2 is too uncertain to estimate? (That is okay, K2 is mostly from the insulation IIRC) $\endgroup$ – Chris K8NVH May 4 at 0:20

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.