8
$\begingroup$

I will give the magmount antenna as an example:

The vertical element to the antenna is very critical, cut to a certain wavelength, often added with some loading coils, or a little spiral to make everything perfect for the specific frequency.

The other part of the antenna is a random car, just a chunk of metal.

On a dipole, both elements are equally important, both acting together to make the antenna.

Why on the magmount, and other antennas like AM towers is the ground or counterpoise not as important?

Can this be switched around, lets say I used the center coax as the RF ground, and the exterior coax as the vertical element, would this work equivalently?

| improve this question | |
$\endgroup$
  • 1
    $\begingroup$ One way to think about it is that a ground plane acts as a mirror for the active vertical element(s). As long as a mirror is big enough, you'll see your refection in it from a distance. Doesn't really matter if the bathroom mirror is round or square. $\endgroup$ – hotpaw2 May 25 '14 at 21:24
2
$\begingroup$

Why on the magmount, and other antennas like AM towers is the ground or counterpoise not as important?

It is important. For example, if you try to put a magmount on a fiberglass car, it won't work very well. It won't work well on a bicycle, either.

For the ground plane (roof of car, radials under antenna, etc) to work well, it must approximate an infinite plane of perfectly conductive material. Of course, we don't have infinitely conductive materials, but any reasonably conductive metal is a good enough approximation.

We also don't have infinite planes. However, as the distance away from the feedpoint increases, the electric and magnetic field strengths decrease. Rule of thumb: the ground plane should extend at least a quarter-wavelength in all directions. Beyond this distance there isn't much current or voltage, so the ground plane becomes less significant. You can extend it farther if you want, but you get diminishing returns in efficiency and only small changes to antenna parameters like feedpoint impedance and resonant frequency.

Can this be switched around, lets say I used the center coax as the RF ground, and the exterior coax as the vertical element, would this work equivalently?

Yes and no, depending on the details of the situation. Voltage is a relative measurement between two points, and electrically what we choose to call "ground" is arbitrary.

In the case of a magmount antenna on a car, were you to switch the center and shield connections, nothing much changes. The vertical element still oscillates at an AC voltage relative to the body of the car, which is the other half of the antenna. However, this mostly works because the car is floating. It runs on a battery and is on insulating tires, and is not connected to anything else except through some small, mostly negligible capacitance.

If you were to try this at a station where the ground plane is not a floating car body but rather radials firmly connected to Earth, then maybe bad things would happen, depending on how things are connected. For example, it's common to have the chassis connected to Earth. If you flip the shield and center connections where the radio meets the feedline you have a problem: the chassis of the radio is connected to Earth, and the center pin on the radio's antenna connector is also connected to Earth (because it's connected to the shield of the feedline, which goes to the antenna's radials). Thus, the thing between your radio's output isn't the antenna, but two points on the ground. Probably not what you wanted.

| improve this answer | |
$\endgroup$
  • $\begingroup$ "on a car, were you to switch the center and shield connections, nothing much changes" — Most cars bond the negative terminal of the battery to the frame. Isn't there a chance that the radio also ties its negative power supply terminal to this same ground? $\endgroup$ – natevw - AF7TB Mar 21 '16 at 18:00
  • $\begingroup$ Sorry last comment doesn't quite ask what I meant it to: might a radio tie its coax shield to its negative supply which would therefore also want to stay at the same potential as the car body? In that case connecting the center pin to the frame would be a serious "impedance mismatch", right? $\endgroup$ – natevw - AF7TB Mar 21 '16 at 18:09
  • 1
    $\begingroup$ @natevw-AF7TB what impedance would be mismatched? The radio doesn't care about the DC impedance of the antenna: it only cares about the RF impedance. You can even put an inductor directly across the feedpoint, which looks like a dead short at DC, but provided the inductance is significantly high it has negligible effect on the RF impedance as seen by the transmitter. $\endgroup$ – Phil Frost - W8II Mar 21 '16 at 18:18
-1
$\begingroup$

When you use a magnetic mount antenna the metal ground plane underneath (usually the vehicle) is extremely important to the signal. The better the ground plane the further the signal will travel.

As for your question about the coaxial cable, this is extremely important as well. The coaxial cable sheath is used to bounce/reflect the RF energy all the way down the line. Without it, the RF power would most entirely be lost. Therefore using the ground sheath for the vertical element would result in a pretty much a complete loss of transmit power.

The coaxial cable is also engineered to have a specific impedance at the frequency you are transmitting. This impedance match ensures the maximum transfer of your signal to the antenna with minimal losses. Again, using the ground sheath for the transmit versus ground would result in a very bad impedance mismatch to the antenna.

| improve this answer | |
$\endgroup$
  • $\begingroup$ While swapping the center and shield conductors of a coax feedline could have consequences, I don't think an impedance mismatch is one of them. The cable's capacitance and inductance per unit length, which define the characteristic impedance, would remain unchanged. $\endgroup$ – Phil Frost - W8II May 29 '14 at 12:51
  • $\begingroup$ So I was down voted on my response and there is a comment that impedance does not change. Yet no one has any scientific proof of it. Try it using a pulse generator and oscilloscope and see what actually happens. You will be surprised. $\endgroup$ – Warren Rox Jun 3 '14 at 15:52
  • $\begingroup$ yourlogicalfallacyis.com/burden-of-proof $\endgroup$ – Phil Frost - W8II Jun 3 '14 at 16:40

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.