0
\$\begingroup\$

I am quite new in this field and I started recently by reading the first books in electrical engineering where I learned the concept of current, voltage, impedance, reactance (inductive and capacitive).

I was reading yesterday about coaxial cables where the conductive elements are the core and the shield. I though that the the core and the shield are the "extension" of plus and minus poles of the power source (that in the ac source alternate).

The reason I am asking is that the shield is typically connected to the ground ("The cable is protected by an outer insulating jacket. Normally, the shield is kept at ground potential and a signal carrying voltage is applied to the center conductor."). So somehow I am a bit confused on if I understand precisely how one can think a coaxial cable being a part of a simple RLC circuit schematic.

\$\endgroup\$
7
  • 4
    \$\begingroup\$ I am a bit confused on what you are actually asking... \$\endgroup\$
    – PlasmaHH
    Sep 12, 2017 at 11:21
  • \$\begingroup\$ Can you help me perhaps understand if I have an AC power source a coaxial cable and a lamp how I will connect the AC power to the core and the shield. Perhaps this might help me apprehend better the concept. Alex \$\endgroup\$
    – Alex P
    Sep 12, 2017 at 11:29
  • \$\begingroup\$ I think he's asking how coax can be used for antennas, which transmit from AC. And the ground bit seems to be confusion about whether it's okay to let the sleeve become '+'. \$\endgroup\$
    – user2497
    Sep 12, 2017 at 11:29
  • \$\begingroup\$ @AlexP Don't use coax for lamps. The shield is there to filter out interference, coax isn't for anything except radio/audio signals, in- or outgoing. Nice coax can handle over 100W, but that's not the point. \$\endgroup\$
    – user2497
    Sep 12, 2017 at 11:30
  • \$\begingroup\$ I do not use coax for lamps. Lets assume that I indeed have an antenna connected to some AC source. How one can think of connecting core and shield to the two "pins" of the AC source? Regards Alex \$\endgroup\$
    – Alex P
    Sep 12, 2017 at 13:59

3 Answers 3

1
\$\begingroup\$

An AC signal can be applied across any two conductors, it has no intrinsic polarity (because that keeps alternating!).

However it is often convenient to hold one wire at a constant voltage, while the other one does all the oscillating. For example in domestic power distribution the constant one is called Neutral and the oscillating one Live. In lower-voltage applications, they are often called the Signal and Return lines, just as with a DC circuit.

In electric circuits, it is common practice to use the Ground or Earth connection as the Return.

The coaxial setup should now be obvious enough; the outer Earth braid provides a dual function as both a shield and the Return.

\$\endgroup\$
0
\$\begingroup\$

first you have to understand that R, C and L are electric parameters that can be built up from special structures and materials. For example, a capacitor is a couple of paralel plates isolated with a dielectric, and its capacitance is determined from plate's area (A) and separation (d), kind of isolation (E), so C=EA/d, in general from material and geometry. In this idea one can divide any transmission medium like a coax into several segments, a very short segment whose length tends towards zero, if you look carefully to one segment you will find out that its structure in esence is quite similar to that of a capacitor, plates no planed but curved, and dielectric, so that short segment should have capacitance too. Similarly, one can say the same for the other electric parameters: resistance and inductance, and deduce that what has in a cable is really a AC circuit with RLC components.

Well, the RLC parameters in a cable are significant only if the transmitted signal through it has a wavelength comparable to the cable length, otherwise one can just look down on them, and this occurs generally at low frequencies in communication applications. So when you are working a low frequencies and need RLC parameter you have to work with resistor, inductor, and capacitor components (lumped elements) but at high frequencies you just need to consider only the RLC component in the line itself (distributed elements), and some times you can use a hybrid approach: lumped and distributed elements in order to built a AC circuit with RLC.

Hope you can understand the concept.

\$\endgroup\$
2
  • \$\begingroup\$ thanks for the reply. Unfortunately my upvotes do not count still. Thanks for the reply. How then the core and the shield relate to the two polarities of an power source? Is it too wrong to thing of the shield and the core as two simple cables transfering energy.? \$\endgroup\$
    – Alex P
    Sep 13, 2017 at 6:16
  • \$\begingroup\$ Not at all, you can think of shield and core in that way, nevertheless coax cable are designed to transfer low power signals not power energy, though there is a cathegory of coax, outdoor coax, that can transfer both low and high power signal. In such a way you can relate core to line and shield to ground, but trying to make the experiment can damage the coax cable due to its power limits \$\endgroup\$ Sep 14, 2017 at 13:23
0
\$\begingroup\$

The two output terminals of an RF transmitter, considered as an AC source, would be its RF tap-off point and ground.

Here's the schematic of a very simple low power transmitter, using coaxial cable to feed a dipole antenna. Coaxial connectors have not been shown. The antenna and feeder would not be any different even for a more complicated transmitter, with an output of a couple of hundred watts.

enter image description here

RF output of the transmitter is fed to the core of the coaxial cable. The braid (shield) is connected to ground.

In general the ground symbol need not necessarily denote actual ground but only a common reference.

In this case it would be actual ground, which would not only provide safety earthing for the mains power supply, which is not shown, but also facilitate draining of static charges from the antenna to earth.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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