mheslep
Gold Member
The movement of charge is not sub-atomic.

sophiecentaur
Gold Member
The movement of charge is not sub-atomic.
Tell me what actual distance you think an electron would move if the mean drift speed is 1mm/s and the frequency is 100MHz.

According to Balanis and Cheng books, they use different models for dipole antennas depend on the length. For total length much less than wave length, current is approx to be uniform along the line and the model is an electric dipole where opposite charges on each end connected by a very thin wire. The radiation pattern can be approx by the electric dipole.

For longer dipole antennas, open ended transmission line model is used and the current is approx by something like a standing wave pattern.

The EM radiation is from the current traveling along the wire
to create the Vector Magnetic Potential. Assuming dipole is along z axis. For short dipole where current is approx to be uniform:

$$\tilde A = \frac {\mu_0 I \tilde {d l'}}{4\pi}\left(\frac{e^{-j\beta R}}{R}\right) sin \theta \;\hbox { for very short dipole.}$$

From this we calculate $\tilde H = \frac 1 {\mu_0} \nabla \times \tilde A \;\hbox { and } \tilde E=\frac 1 {j\omega \epsilon_0} \nabla \times \tilde H$

$$\hbox { For longer dipole where length is somewhere close to }\; \frac {\lambda} 2,\; F(\theta)=\frac{\cos(\beta h \cos \theta)-\cos\beta h}{\sin \theta} \;\hbox { replace }\; sin\theta\;\hbox { of the short dipole equation.}$$

h is the length of one side of the dipole. $\theta\;$ is same as in spherical coordinates.

With these, we can simplify the radiation pattern by assuming $\beta d \;$>>1 to simplify to far field EM.

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sophiecentaur
Gold Member
Actually, the current on a short dipole tapers, uniformly, to zero at each end (triangular distribution), as opposed to roughly sinusoidal for longer elements.

This circuit would operate at about 100 MHz and, at this frequency, one wavelength is about 3 meters. So, a quarter wave length is about 75 cm (29 inches).

I am just curious how the frequency of this circuit was calculated?

Here are a couple of websites that provide a good visualization of E and H fields. Thanks for the answers Jim, I learn alot from your posts. I've worked on alot of systems that use waveguide as a transmission line and now I have a better understanding of how the RF travels down the transmission lines/ waveguides.

Actually, the current on a short dipole tapers, uniformly, to zero at each end (triangular distribution), as opposed to roughly sinusoidal for longer elements.
The book approx uniform current( same along each half of the line) for very short dipole, and the current flow to the end and charge up the end to form an electric dipole. This is called Hertzian dipole!!!

I just repeat from the book. I just start studying antenna design for about two months, still have ways to go!!! The book specified that the exact calculation of current at each point is very hard to calculate. All are just approx only. Antenna is all about EM, I spent 4 years studying from Calculus II to PDE, three different books of EM to prepare for this big day ( more like year!!!:rofl:).

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This circuit would operate at about 100 MHz and, at this frequency, one wavelength is about 3 meters. So, a quarter wave length is about 75 cm (29 inches).

I am just curious how the frequency of this circuit was calculated?

Here are a couple of websites that provide a good visualization of E and H fields. Thanks for the answers Jim, I learn alot from your posts. I've worked on alot of systems that use waveguide as a transmission line and now I have a better understanding of how the RF travels down the transmission lines/ waveguides.

Speed of light is 10EE8 m/s. If the line is quarter wave of certain length, then the wave length is 4 time the length. Divide the speed of light by the wave length give you the frequency. You can reverse calculate the length of the wire you need for quarter wave also.

You are hijacking the electron into being something it isn't. They don't "slosh" anywhere.
Mine slosh. I know because I asked them :)

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.

rbj
Indeed Kirchoff's current law requires that the sum of currents at a node be zero.
This antenna stuff appears to violate that. So let's take a coser look.

Remember , a NODE is where curent has a choice of paths - i explain it to people: "Gotta have three wires to make a node. A node has one way in and two out, or vice versa."
i wouldn't say that, jim. you can certainly have a circuit and identify nodes with one wire going in and one wire coming out and have KCL apply to it, just as much as if 3 or more wires were connected. you could even have a node with one wire going in and that's it. but KCL would say that the current in that wire is zero.

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 licence 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
Gold Member
2019 Award
Dearly Missed
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.

old jim

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vk6kro
I doubt if Kirchoffs Laws apply where any or most of the energy in a wire is being radiated.

sophiecentaur
Gold Member
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.

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 $\theta$.

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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".
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.