# Tag Info

13

In the absence of common-mode currents, then the optimum feedline length is 0, because a longer feedline only increases your feedline losses. These losses are due to the resistance of the wire, dielectric losses, etc. and are specified in dB per unit length in the coax datasheet. At VHF and up, these losses can be significant even at car lengths, especially ...

10

A balun matches a balanced load to an unbalanced line, but it can also do other useful things. A current balun can present a high impedance to common-mode signals, which will help reject noise. Common mode signals are the same on both conductors, so are not "balanced" or differential. An unun is an impedance transformer, usually 4:1 or 9:1, which matches an ...

9

Your question seems as much about psychology as much as technical concerns. We mainly favor the technical questions, but I'll take a stab at the psychological aspects also. All of you, please feel free to disagree with my conclusions! Coaxial cable is fairly inexpensive for many, compared to our time, even for LMR400. (Your mileage may vary.) For many, ...

8

For many modulations, the modulation is very slow compared to the propagation delay of the feedline. For example, SSB is typically limited to no more than 4 kHz. That corresponds to a wavelength of of 75 km. As long as the feedline is significantly shorter than this, then the delay due to the feedline is negligible. It may be easier to understand ...

8

Balanced lines (of which twisted pair is a special type) really have an upper frequency limit; you can't use them to transport 1 GHz (well, you can, but the smallest variation in direction or distance would have catastrophic effects, and the conductor distance would get pretty small). This can be seen in technical practice: 100 Mbit/s Ethernet (Fast ...

7

It looks like you're looking for an intuitive, practical understanding rather than precise definitions, so I'll see what I can do with that, with my own recent learning. The reason you care about impedance matching is that impedance mismatches cause the signal to be partially reflected — some of the energy is going the opposite direction than you want it to ...

7

The most important thing about the cable is how much loss are you willing to accept in the feed line. Figure out how much cable you will need, and then determine what the loss on said cable will be. As I mentioned on my website, here's a few good rules of thumb: Keep the power loss to no more than 3 db in the cable. If you can, use the same impedance as ...

7

I am sure you will get many good suggestions for your situation. The higher quality coax cables do have a specification for their minimum bend radius during install. For LMR-400 it is 1 inch (25.4 mm) [ref. https://www.timesmicrowave.com/documents/resources/LMR-400.pdf]. I talked with a Times Microwave technical sales specialist and here is his answer ...

6

1% here is just an overly precise way of saying "really small compared to the wavelength". The problem is that when the conductors are close together, their electrical and magnetic fields cancel; as they get farther apart, there's a larger volume around the line where the fields don't cancel, and the ladder line starts looking more like an antenna than a ...

6

When the SWR is 1:1, the matched line loss of ordinary ladder line is lower than the matched line loss of ordinary coax because at HF, most of the loss is $I^2R$ loss, and the current magnitude is inversely proportional to the characteristic impedance of the feedline. Of course, a high SWR results in higher standing wave currents and thus increases the $I^2R$...

6

The short answer is that it can't. A shielded transmitter, connected to an ideal piece of coax, does not generate common mode currents. The inner and the outer of the cable look connected (and for Direct Current they are), but high frequency currents really cannot pass through the thick metal, they are confined to the inside or the outside. Real coax can ...

5

First of all, acid-core solder is never used for electronics any more, that just isn't what it's for. I think it's still used for plumbing or other soldering where there aren't sensitive components involved. Second, aluminum solder isn't a thing. The best I can find is that there's some solder specifically for soldering aluminum to other stuff, but you ...

5

Twin-lead transmission lines don't radiate because the opposite fields from each conductor cancel, but when the spacing is far apart this does not happen. First, let's consider the magnetic field around an infinite, straight conductor with uniform current throughout. In this image, the conductor is just right of center, and the current is coming straight ...

5

OK, let me try to answer this, but this answer may also be qualified as unqualified. If you have a 50 Ohm receiver, and connect a perfectly (Z=R) 75 ohm antenna system, then your VSWR would be 1.5, and the "load mismatch attenuation" will be about 0.177dB. (with antenna system I include feedline) I doubt that you would actually notice this. However, you ...

5

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 = ... 5 That is a great idea. In fact antennas of this design are quite common, and you probably have a much smaller antenna of similar construction on your Wi-Fi router: Effectively, the lower dipole element, which is a tube, forms a "bazooka" or "sleeve" balun. See also W8JI's description of sleeve baluns. I'll let you follow the links for ... 4 The primary advantages of coax with respect to ladder line are most transceivers are equipped with coax connectors, whereas using ladder line requires a balun or balanced tuner coax is not affected by nearby metal objects, unlike ladder line the impedance of coax doesn't change when it rains or snows, unlike ladder line The primary advantages of ladder ... 4 No. The cause of power loss in a feed line is not due to the characteristic impedance, it actually varies among transmission lines of the same impedance. Otherwise all 50 ohm coax would be the same - but there are many different types and qualities of cable. The actual loss is measured not in ohms but in decibels per foot, and varies with both the type of ... 4 At most amateur radio frequencies, there are really only two feedline types that are readily available and effective: twin-lead coaxial The goal of the feedline is to move electromagnetic energy from one end to the other without much loss. This means not absorbing it, and not radiating it. Not absorbing EM energy is pretty easy. Use good conductors, like ... 4 There is, but it depends on the characteristics of the cable and the frequency of interest. There are online calculators that can do the work for you. For instance the two lengths of your cable have the following attenuation (almost none for RG-58) and phase delay. The phase delay is noticeably different, but as I haven't evaluated the antenna design I ... 4 Avoid common-mode sheath currents Common-mode sheath currents on the coaxial feed line should be avoided because: At the antenna feed point or the transition point of balanced line to coax, common-mode currents interfere with your radiation pattern. In the shack, common-mode sheath currents will interfere with your electronic equipment (RFI). Origin of ... 4 I'm no expert on balun designs, but I think they mean to construct a folded balun, as follows. The coax's outer conductor is soldered to the same part of the antenna, but it is left on the coax, not peeled away, so the final balun structure includes a length of the coax. (The interpretation you proposed in your question is unlikely because if the outer ... 4 TV coax typically has foam dielectric and very low loss. On receive it does not matter if you use 50, 60 72 or 75 ohm cable. With short cables, losses are also not important. The cheap TV balun could be problematic, particularly if you want to transmit. It is trivial to make a 1:2 transformer to get the 4:1 impedance transformation you need, but it is hard ... 4 The concept of "electrical length" begins with the concept of wavelength, \lambda - the distance over which the value of a periodic phenomenon repeats, expressed as:$$\lambda=\frac{c}{f} When describing electromagnetic wave propagation in free space, c is the speed of light in a vacuum - 300 million meters per second - and $f$ is the frequency of the ...

4

It is well understood in the amateur radio community that ladder line, window line, etc. have lower losses per foot than say RG-213 or LMR400, particularly on HF frequencies. Is it though? I'd say only sometimes. From the LMR-400 datasheet: From DX Engineering's 300 ohm ladder line datasheet: Compare the loss at say, 30 MHz: 0.7 versus 0.668 dB/100 ft. ...

4

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 ...

4

It is not normal to have and use tuners on both ends of a feed line. The purpose of an antenna tuner is to provide an adjustable impedance transformation. The reason we want that transformation is so that the transmitter can work into the (usually) 50 Ω load it was designed for. If you have an antenna tuner located (only) at the antenna, then this is the ...

4

Let's say the antenna impedance on a given frequency is 100 Ohm, the feedline is lossless 50 Ohm, the transceiver input impedance is 50 Ohm. Between the antenna feed point and the feedline SWR = 2, 11% reflected power. This is a bit of dangerous thinking, because the distance between the antenna feedpoint and the feedline is zero. As such there can be no ...

3

Here's the problem: information can only travel at the speed of light. So, say you have a resistor, connected to your ohmmeter by a pair of very long wires. When you connect the ohmmeter to the wires, it applies some voltage, but how much current flows? When this voltage is first connected, it sets off a wave in the electric field, but that wave hasn't yet ...

3

Transmission line theory is complicated by many factors. Here's one simplified explanation that might help you understand a little better what you're asking about. When you send a wave down a line, it doesn't appear instantly on the other end, the electrons push each other, and in a measurable, finite amount of time the wave is present on the other end of ...

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