Antenna Radiation: How Does it Work with Only 1 End Connected?

[vk6kro]

I didnt see where the wavelength was given. vk6kro, figured out, by looking at the circuit diagram in the first post, what frequency the circuit would operate at. I was just curious how he got that answer from looking at the diagram.

The transmitter would operate over a wide range of frequencies, but to receive the signal you would use a commercial FM receiver covering the FM band of about 88 MHz to 108 MHz.
Anywhere in this band would be OK as long as there was not a transmission there already.

Transmitting in this band without a license is illegal in any country, but low powered devices like this are unlikely to radiate very far.

100 MHz is convenient for calculation but you could calculate the wavelength from the usual formula:
Wavelength = Speed of light / Frequency
eg Wavelength = 300 000 000 m/s / 95 000 000 Hz = 3.157 meters
or more conveniently,
Wavelength = 300 / F in MHz
eg Wavelength = 300 / 95 MHz = 3.157 meters.

 

[jim hardy]

rbj ---

Well i looked at a couple new texts and found some authors agree with you.

So i have to concede to you that point.

I was taught ca 1961 that a "one way in one way out node" is just a trivial wire unworthy of Kirchoff's attention, and to consider it a node amounts to unnecessarily re-naming a current midwire.
I1>-------------o------------>I2

why not just stick with I1 ?

Kindly do not think ill of this old dog for being slow to pick up new tricks.

Maybe that's why we're allotted threescore and ten , in that time the world just changes too much for us.

Thanks for the update. And thanks for reading the thread !



old jim

 

[vk6kro]

I doubt if Kirchoffs Laws apply where any or most of the energy in a wire is being radiated.

 

[sophiecentaur]

Kirchoff is a subset of Maxwell which works for 'most' circuits. It assumes that the energy in a circuit is sourced and dissipated by the visible components.

 

[yungman]

Guys, before you get too much into current and KVL, remember one thing, this is all EM. electrons and current don't travel nearly at this speed. It is the EM wave that travel. The current and charge are only the consequence of the boundary conditions where free charge and free current density appeared. I've gone through discussion in the Classical Physics to verify this. It is the EM wave that travel, electrons travel in walking speed!

All the theory about the basic dipole antenna has been written in most of the engineering EM books at the last chapter. Take a look at Balanis Antenna design, it has one chapter concentrated on this.

The antenna rod really do not end at the end, wave launched into the space(air). It is specified that the current distribution is very difficult to get exact, so various approx is used. It is all covered in books.

Any dipole antenna start out with the Hertzian dipole model. For longer antenna, it is approx by many sections of the tiny Hertzian dipoles. The complication is the phasing of the current and also the phase of each Hertzian arrive at any given point. Formulas are developed for different length which I posted in the former post.

The pattern of a Hertzian dipole is very much to same as the electric dipole in EM books. The difference is they simplify it for far field so the E does not consist of both $\theta, \phi\;$ but instead only $\theta$.

 

[gtacs]

gotcha! tyvm

 

[vk6kro]

Kirchoff is a subset of Maxwell which works for 'most' circuits. It assumes that the energy in a circuit is sourced and dissipated by the visible components.

Yes.

The radiation can be factored in by using the concept of "Radiation resistance".
http://en.wikipedia.org/wiki/Radiation_resistance
This is measurable at the input terminals of an antenna but it is the accumulated effect of all radiation from the surfaces of the antenna.

Although it is represented by a resistor, it is frequency dependent and varies with the geometry of the antenna.

A few meters of wire will have a very low radiation resistance (much less than 1 ohm) at 100 KHz, but maybe several hundred ohms at 100 MHz.

Hence at 100 KHz, it becomes very difficult to produce much power across this radiation resistor. Very large currents are required and this has to be achieved while cancelling out any reactance of the antenna.

At 100 MHz, the current does not have to be very high, but a much bigger voltage is required. So, it is much easier to cancel out any reactance of the antenna because this involves putting an inductor in series with the antenna and it can have more resistance if the impedance of the antenna is higher.
Also, other resistors in the circuit, like the resistance of the wire or the imperfections of the grounding system become less important.

Once you do get the power into an antenna, though, it will be radiated.

 

[rbj]

no sweatsky, old jim.

i'm no young spring chicken, myself.

 

[cmb]

if you had a long enough coil of wire, you could drain the battery by only attaching it to the negative terminal of the battery...

This doesn't sound right...

Consider a single 1mm wire held 1m above ground, that'd have ~10pF/m wrt ground, that is connected to one side of a 1000mAh 9V battery, the other is grounded.

The energy of the wire when fully charged is 0.5 x 10x10-¹² x length x 81 = 4x10-¹º J/m.

Let's say the available energy in the battery is 1Wh per useful volt drop - call it 10kJ.

So the length of wire that would drain the battery due solely to its length would be 2.5 x 10¹³ m long.