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

[david90]

From my understand, electrical current is required to generate RF. In the picture below, only one end of the antenna is connected L1. If only one end of the antenna is connected, how can electrical current travel up the antenna to generate RF?

 

[phyzguy]

The antenna is long enough that it needs to be treated as a transmission line, and not a simple open circuit. Think of shaking a rope with the far end free (analogous to an open circuit at the far end of the antenna). Waves will travel down the rope. This is basically what happens with an antenna. EM waves travel down the antenna and the energy is radiated into space.

 

[vk6kro]

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).

If you put a 100 MHz sinewave at one end of a piece of wire that is 75 cm long, the voltage takes about 2.5 nS to reach the other end. While it is doing that, the input keeps changing, so the voltage at the end never catches up with the input voltage.

There is always a voltage difference between the ends of the antenna. The end is always 90 degrees out of phase with the input voltage

So, there is a current flowing between the two ends of the wire even though one end is open circuit. This current causes some of the input power to be radiated.

This seems alien to everything you have learned about DC circuits, but it is what happens at these frequencies.

If you are thinking of building this circuit, the part of the circuit marked with a ground symbol should be connected to some large metallic object (not necessarily actual dirt) for the antenna to work properly.

 

[yungman]

I am currently studying antennas. This is half of the electric dipole. They assume it is on top of the ground plane or something close to it, the other half is the image. The way to calculate is to treat as if it is a dipole. The top half of the radiation pattern is exactly the same as if it is a real dipole but the bottom half is missing. Radiation power is half of the real dipole.

Regarding to where the current goes at the end, they use approximations by assuming current at the end is zero and charge is deposited at the end. This can work even if the dipole is very short( Hertzian dipole). For short dipole, it assume uniform current along the wire and actually model as electric dipole with opposite charges on either end(of cause the bottom is the image in your case). For longer dipoles, current along the wire is represented by different form depend on the driving input. Charge distribution along the antenna is very complicated and it is very difficult to get the exact expression even if it is only a wire. You have to give the exact dimension and the input current waveform before you can find the suitable approximation.

Yes, the explanation is a little out there, you can get more info in the EM book under image current and some antenna book. I am still studying this subject, so someone might have better explanation.

 

[jim hardy]

i'm real simple

What happens when you charge a parallel plate capacitor?
On one plate, electrons move closer together as more of them crowd onto that plate.
On the other plate, electrons move farther apart as some of them leave that plate.

So it appears that electrons inside a conductor can be somewhat compressed or rarified.

That opens the possibility of very brief direct current flow into one end of a conductor without any current flow out the other end.

Which i think was your question.

In a halfwave antenna electrons are alternately compressed and rarified at the ends while in the middle they just move back and forth a small distance. That's why current is highest in middle and zero at ends.

But...
there's people here who can continue the explanation better than i
it's just that i remember struggling with the same concept years ago.
and there's times when oversimplification is a necessary step to get our thinking straight.

old jim

 

[david90]

So if I take an antenna and I connect it to the positive lead (leaving the negative lead unconnected) of a function generator, there would be current in the antenna and RF radiation?

 

[Evil Bunny]

What happens when you charge a parallel plate capacitor?
On one plate, electrons move closer together as more of them crowd onto that plate.
On the other plate, electrons move farther apart as some of them leave that plate.

So it appears that electrons inside a conductor can be somewhat compressed or rarified.

Yes, but with the capacitor and a battery, for example, some electrons left one terminal of the battery and came back on the other terminal of the battery.

In other words, they had a place to go...

In this case they don't. So I'm having a hard time digesting that explanation.

And you must also be saying that if you connected a single wire to just one terminal of a battery that the electrons would rush to one end of the wire. Is this true?

 

[Studiot]

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).

If you put a 100 MHz sinewave at one end of a piece of wire that is 75 cm long, the voltage takes about 2.5 nS to reach the other end. While it is doing that, the input keeps changing, so the voltage at the end never catches up with the input voltage.

There is always a voltage difference between the ends of the antenna. The end is always 90 degrees out of phase with the input voltage

So, there is a current flowing between the two ends of the wire even though one end is open circuit. This current causes some of the input power to be radiated.

This seems alien to everything you have learned about DC circuits, but it is what happens at these frequencies.

If you are thinking of building this circuit, the part of the circuit marked with a ground symbol should be connected to some large metallic object (not necessarily actual dirt) for the antenna to work properly.

What a good simple explanation!

 

[Evil Bunny]

So... When you touch a wire to only the negative terminal of a battery, electrons rush from the battery terminal and on to the wire?

If that's the case, then 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... All the available electrons would rush out of the battery and onto the wire.

This doesn't sound right...

 

[vk6kro]

So... When you touch a wire to only the negative terminal of a battery, electrons rush from the battery terminal and on to the wire?

If that's the case, then 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... All the available electrons would rush out of the battery and onto the wire.

This doesn't sound right...

If you had the other terminal of the battery grounded, yes.

The wire would have capacitance to ground and charging this capacitance could partially discharge a battery. The capacitance and the battery would eventually have the same voltage on them. This charge is not lost, though, as you could recover the charge from the capacitor and do useful work with it.

There would also be some power radiated as you connect the battery to the wire. This rising voltage would have harmonics in it and these would be radiated and heard as a "splatt" on nearby radios.

 

[jim hardy]

"Yes, but with the capacitor and a battery, for example, some electrons left one terminal of the battery and came back on the other terminal of the battery.

In other words, they had a place to go...

In this case they don't. So I'm having a hard time digesting that explanation.

And you must also be saying that if you connected a single wire to just one terminal of a battery that the electrons would rush to one end of the wire. Is this true?
"

Well let's think on that.
vk6kro nailed it.

Would the electrons rush to one end?
Or would they all move a little closer to the far end to make room for the ones being squeezed in by your battery?
I'd say the latter. Think of people squeezing into a subway car.
And the current would be so minute as to be almost immeasurable.
But if you ever grabbed the end of a co-axial cable that was recently carrying high DC voltage you know that a long piece of wire can hold quite a few electrons.


Now - recall what is capacitance

epsilon X A/D
where epsilon is dielectric constant, if no dielectric it's capacitance of free space,,
A is area of plate,,
D is distance to other plate.

So a piece of wire with surface area 'a' has a small amount of capacitance to the rest of the earth.
If your wire were long enough to have huge surface area, like long enough to wrap Earth like a ball of yarn, well yes it could probably hold all the amp hours in a small battery. The battery being small would need its electrons replaced via connection to Earth as our friend noted.

But it is hard to conceive of such an arrangement.
sorry for the confusion.

But what is significant is that electrons can squeeze into the wire.
They can also be induced to shift back and forth in the wire, alternately moving away from one end toward the other.
Like all physical phenomena that one has a natural frequency.
It's approximately twice speed of light divided by length of wire.
Speed of light, eh? Must be an electromagnetic phenomeon.


vk6kro sounds like ham radio call - if so he is well versed in antennas.

 

[rbj]

consider that we are standing some distance apart and facing one another. think of yourself holding a large electric charge and i am holding another one of opposite polarity (and about the same amount). we are restricting the motion of our charges so that they cannot move toward (or away from) one another on the line that connects them. but they are free to slide about on the plane that is perpendicular to that connecting line.

so i move my charge to my right (your left) and since your charge is attracted to mine it also moves to your left. then i move mine to your right and your charge follows. then i move mine up and your charge follows it up. that is an EM wave.

if i move it left to right and back a million times per second, your charge follows. my charge is essentially a transmitting antenna and your charge is a receiving antenna. and we are both tuned to the middle of the AM band.

anyway, that is all that an antenna is; at the simplest level, it is a conductor where charge is allowed to slosh back and forth in it. for a transmitting antenna, that charge is forced to slosh back and forth by the transmitter electronics. for a receiving antenna, the charge is influenced by the moving charge in the transmitter antenna which causes that charge to slosh around (which sensitive electronics in the receiver detects).

 

[Evil Bunny]

I always thought the sloshing around was in a complete circuit... Up into the antenna from the source and back to the source on the shield of the coax.

In fact, this is how we assemble transmission lines. If your shield isn't connected properly to the connector at the end of the cable, your antenna does not radiate.

 

[fleem]

I've always felt it was a mistake to imply to newcomers to electronics that a complete DC circuit is necessary for there to be current (before they've been taught about AC). Rather, I'd prefer they be given at least a hand-waving understanding of capacitance by use of plumbing analogies. The OP shows a common question that arises because it is not taught this way.

Electrons move away from other negative charge and toward positive charge. If there are extra electrons on an isolated chunk of metal, then the extra electrons flow to the extremities of that chunk of metal because they are repelling each other. If another chunk of metal that does not have an excess of electrons comes in contact with that first chunk, then some of the extra electrons on that first chunk will flow to the second to get away from the extra electrons on the first chunk. So here we certainly have electron flow without a complete circuit. The electrons will also slosh a bit (move to and fro) for a short time immediately after those two chunks come in contact, depending on how big those chunks of metal are and what their electrical resistance is. The sloshing can be acounted for by the inertia of the electrons (but with electrons, that kinetic energy is in the form of a magnetic field). An antenna is just a relatively big chunk of metal designed such that electrons can slosh to and fro along its various elements such that it has a real (not imaginary) impedance to the transmit circuit--which means the energy given to it never returns to the circuit, so it must go somewhere (the receive antenna, and everything else in the universe). (On a side note, if there were nothing else in the universe to couple to the transmitter antenna, then there would be no radiation from it.) We've analyzed the process of radio waves leaving an antenna pretty well. We have a classical understanding (the waves leave one antenna and eventually encounter another receiver antenna) and a quantum physics understanding (there is a sort of non-local coupling between the electrons in the transmit and the receive antennas). Also classically, we see that one can describe the magnetic field as simply a relativistic electric field and vice versa. However, we still really don't know what the electromagnetic field is or whether our current picture of it is best for gaining a deeper understanding (not that its possible to ever know what anything really "is").

 

[Evil Bunny]

So... are there two different types of antennas out there? ( I mean, I know there are all kinds of different styles and sizes and shapes, etc. but that's not what I'm talking about here).

One type is just an open circuit with electrons "sloshing" (I like that description, by the way) around back and forth on a wire...

And the other type is a completed circuit with electrons traveling in loops, leaving the source, traveling up the antenna and returning to the source on the sheild of the transmission line?

It's not that I don't understand these explanations, they have been helpful, but it seems that we usually talk about "static electricity" as being something different than "current electricity". We've all seen how electrons attract and repel each other in physics lab with the glass and the fur, etc... so it's easy to envision your examples of these charges moving back and forth on a wire with these external influences.

But it starts to get a little blurry when you start to marry these two concepts together... When you study circuits, it is wise to always remember that whatever leaves the source MUST return to the source.

Capacitive coupling always rears its head in these discussions, because the answer always turns out to be that the source must be connected to ground and that is how it's returning. But in most cases it seems that we're not talking about having the source connected to ground, so it all seems a little... unsatisying, I guess.

EDIT: http://amasci.com/tesla/tmistk.html"about Tesla and single line transmission.

 

[rbj]

I always thought the sloshing around was in a complete circuit... Up into the antenna from the source and back to the source on the shield of the coax.

In fact, this is how we assemble transmission lines. If your shield isn't connected properly to the connector at the end of the cable, your antenna does not radiate.

so, for a simple center-fed 1/2 wavelength horizontal dipole antenna, what is happening at the center where the inner conductor is soldered to the left half and the outer conductor (shield) is soldered to the right half? nothing else is connected to either halves anywhere along the element. does that count as a "complete circuit"?

So... are there two different types of antennas out there? ( I mean, I know there are all kinds of different styles and sizes and shapes, etc. but that's not what I'm talking about here).

One type is just an open circuit with electrons "sloshing" (I like that description, by the way) around back and forth on a wire...

And the other type is a completed circuit with electrons traveling in loops, leaving the source, traveling up the antenna and returning to the source on the sheild of the transmission line?

It's not that I don't understand these explanations, they have been helpful, but it seems that we usually talk about "static electricity" as being something different than "current electricity". We've all seen how electrons attract and repel each other in physics lab with the glass and the fur, etc... so it's easy to envision your examples of these charges moving back and forth on a wire with these external influences.

But it starts to get a little blurry when you start to marry these two concepts together... When you study circuits, it is wise to always remember that whatever leaves the source MUST return to the source.

it's the same set of physics. Kirchoff's current law is not the most fundamental physics. it is the result of an assumption that no charge can build up at any node.

 

[sophiecentaur]

There are, indeed two ways of designing an antenna. Some are based on dipoles (or monopoles with an earth) and the others are based on loops or even slots cut in large sheets (many mobile phone antennae are the slot variety).

In all cases there are currents flowing and voltages appearing at various points. So, for instance, the current along a dipole gradually tapers to zero at the ends ('cos it can't go anywhere beyond that). The essential thing is that they will radiate Energy (as em waves). To radiate any significant amount of energy, the antenna has to be a 'reasonable' size compared with the wavelength. Round about 1/2 wavelength is good but you can get a reasonable amount of RF power from an antenna only 0.1λ if you 'match it' right.
What has been written above is fair enough but it doesn't really describe how Power is actually radiated from an antenna. A transmission line and a Capacitor don't radiate any appreciable amount of power because the currents and volts are in quadrature everywhere (90 degrees of phase) and so can dissipate no power. In a larger structure, the phases are, actually, not exactly in quadrature - there is a Real component of VI and that is due to actual radiated power. 'Something' happens to the fields around the antenna because it takes time for the changes in the fields to propagate over the space it occupies. The E and B fields add together so that wave is generated that carries energy away. When you look into the feedpoint of an antenna, you actually measure a resistive component, called the Radiation Resistance so space actually acts like a resistive load- taking power away.

 

[jim hardy]

""So... are there two different types of antennas out there? ( I mean, I know there are all kinds of different styles and sizes and shapes, etc. but that's not what I'm talking about here).

One type is just an open circuit with electrons "sloshing" (I like that description, by the way) around back and forth on a wire...

And the other type is a completed circuit with electrons traveling in loops, leaving the source, traveling up the antenna and returning to the source on the sheild of the transmission line?...""

EB nice summary!

Indeed that's so.
If you look in the back of an old high quality floor model radio you'll see a big (often square) loop of wire arranged around the cabinet. Usually it has several turns. That's your second type, 'completed circuit', antenna. It's called a "loop antenna".
It is the AM antenna, and is used because at frequencies of AM the length of wire required to achieve 'sloshing' at its natural frequency (that's "resonance") would be unmanageably long. So instead they use a tank circuit inside the radio to achieve resonance.

If you look at a classic TV antenna you'll see multiple horizontal metal tubes. They are your first type 'sloshing' antenna. Each is horizontal tube resonant somewhere in the TV broadcast band where frequencies are much higher, hence length is manageable.


Resonance in an antenna greatly improves its effectiveness.

That a simple piece of wire has an electrical resonant frequency is surprising when first encountered. My high school electronics teacher explained it to us boys using an analogy of kids in a hallway running back and forth, if you like i could try to recall it. It painted a simple mental picture that let me swallow resonance.

A more scientific explanation might start along the lines: a straight wire has self inductance, and its ability to comprress or rarify charge at its ends is not unlike capacitance. It is therefore capable of electrical resonance.

BUt there are much better theoreticians here than me so i'll not pretend to be an expert on antennas. Honestly i never quite grasped Maxwell's equations.

old jim

 

[jim hardy]

""When you study circuits, it is wise to always remember that whatever leaves the source MUST return to the source."

Ahhhh here's the little mental trick.

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 current 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."

Well your one wire antenna isn't a node - it's just a volume of metal. Only one way in and one way out.
Nobody said we couldn't rearrange the charge in a volume of metal, or for that matter add to or subtract from it like static electricity does.

In your original post the "Node" where Kirchoff's Law applies is the junction of L1 and the antenna.
If current enters inductor L1 at bottom and flows up, when it enters the node at antenna it has two ways out. It can continue up the inductor or hang a left to enter the volume of antenna wire.
So at any instant Kirchoff is satisfied.Those three currents will add to zero.

For antennas it is really only practical to consider the AC , so practically speaking any electrons that enter antenna will come back out next half cycle. . Your "return to source" requirement will however eventually be satisfied - when the radio is turned off any excess charge squeezed into the antenna at power on time will come back out.
I guess that fact helped me swallow this business when first learning it.

In antennas we don't consider the DC at all. It's just impractical. But it is nice to keep one's thinking rigorously straight and resolve these apparent conflicts.

keep it simple
old jim

 

[mheslep]

It might also be helpful to note that from the perspective of the EM wave the the antenna does not terminate as might be suggested by the schematic but is 'connected' to the http://en.wikipedia.org/wiki/Impedance_of_free_space"

 

[gtacs]

Very interesting topic, and great comments.

I found these "Antennas for dummies" books off a google search, and thought someone might enjoy them.

http://www.hottconsultants.com/pdf_files/dipoles-1.pdf
http://w4trc.org/dipoles/dipoles-2.pdf

 

[sophiecentaur]

So... are there two different types of antennas out there? ( I mean, I know there are all kinds of different styles and sizes and shapes, etc. but that's not what I'm talking about here).

One type is just an open circuit with electrons "sloshing" (I like that description, by the way) around back and forth on a wire...

And the other type is a completed circuit with electrons traveling in loops, leaving the source, traveling up the antenna and returning to the source on the sheild of the transmission line?
.

The problem with talking about electrons actually "sloshing about" is that electrons move through a metal at a few mm per second. Bearing in mind that you would have to think of them moving back and forth at a frequency of several (or even thousands of) megaHerz, they will not be moving any significant distance at all - less than the diameter of a metal atom, in fact. So please don't try to paint a picture of anything "sloshing" anywhere. It's the very slightest oscillation that's involved.
Why not just use the good old word 'Current' and stop rushing in where angels fear to tread? Anyone would think that using the electron model could actually help with understanding this stuff!

 

[jim hardy]

"Anyone would think that using the electron model could actually help with understanding this stuff!"

actually it does help a lot.



"Bearing in mind that you would have to think of them moving back and forth at a frequency of several (or even thousands of) megaHerz, they will not be moving any significant distance at all - less than the diameter of a metal atom, in fact. So please don't try to paint a picture of anything "sloshing" anywhere. It's the very slightest oscillation that's involved. "

Flow of current down a wire is as you describe a quite slow drift of electrons.
Like pushing peas through a straw - it takes a long time for any given pea to traverse entire length of the straw.
But if the straw is full the delay between inserting a pea at one end and getting one out the other end is quite short.

Electrons are as alike as peas in a pod.
Push some in one end of a wire and almost immediately some identical ones will ty to exit the other end . The "push" between electrons moves at nearly speed pf light.
So we say that current flows at nearly the speed of light, and do not try to track individual electrons. That's easy to forget.

That "Antennas for Dummies" was a good find. It broached the topic of the external electric field between the ends of the dipole antenna. See fig 2 of first link:
http://www.hottconsultants.com/pdf_files/dipoles-1.pdf
the dotted lines might be thought of as indicating the E-field
Bravo - that electric field is there due to the rarefaction and compression of charge going on inside the antenna wire. Note it goes from end to end through space. And an electic field contains energy - voila there's half the answer to original question how does an antenna radiate.

Because electrons can't get past the ends of the wire there's zero current at ends.
In the middle the current is highest .
So there's a magnetic field that's strongest about middle of antenna. Apply your right hand rule around center of the antenna -
magnetic field will be in a plane perpendicular to current flow hence perpendicular to the E field.
It might be called the B field.
Now we're getting into fields - E and B fields always coexist and are perpendicular, if i remember right.
If you create those two fields in space you have radiated electromagnetic energy.
There's the other half the answer of 'how does an antenna radiate'.

"Sloshing" then involves not complete end-to-end traverses of the wire by individual electrons ,

but as i said many posts ago, enough motion to cause alternate compression and rarefaction of charge near ends of the wire.
Like pushing peas back and forth in a soda straw.


Does this approach help people understand?
When I took my antennas and fields course i was overwhelmed by horrific multi-term vector calculus equations. I would not have passed that course without taking the terms in those equations and relating them one by one to physical effects like above.
I was blessed to have had an extremely practical high school electronics instructor who taught us boys how antennas and transmission lines work from an electron-by-electron approach then tookus into the formulas. Of course in high school we just used algebra (he taught us complex algebra) and Smith charts (and slide rules - it was early 60's). He enabled me to figure out why those fearsome college equations worked. I could relate each term to something physical. Else i couldn't have believed them.

Since as you point out electrons do actually move, it is VERY useful to use an analogy when helping beginners. Just got to remember it's the push between electrons that moves fast and the electron you put in over here isn't the same one you get out over there.
I actually prefer the analogy of water in pipes.
Speed of sound in a medium like water is sqrt (elasticity/density)
and speed of light is 1/ sqrt (permittivity X permeability)
so i think modern physics has parallels to classic.
But I'm a picture thinker.
There may exist people who can think in equations and if so, i envy them.

I enjoyed all the posts here and would welcome any enlightenment anybody has for me.

old jim

 

[Evil Bunny]

With much respect to you Sophie, thinking of electrons as "sloshing about" is actually quite helpful in understanding the concept of what is happening here (to me, and apparently many others)... You're bringing up http://en.wikipedia.org/wiki/Drift_velocity" , which is a valid point, but thinking about how far any individual electron may travel in the process is a little irrelevant in this discussion, in my humble opinion. We're talking about the movement of charge, not the movement of a single electron in particular.

I don't think that it is causing confusion... To the contrary, I think it is a helpful description of what is taking place to better understand the concept.

Anyway... your point about terminology is a good one... "current" is technically the correct term to use here, but "sloshing" helps some of us understand it a little better.

Just saying...

 

[sophiecentaur]

If you use the word "sloshing" and seriously expect people to bear in mind that the actual movement involved is sub-atomic then I think that is naive. You are hijacking the electron into being something it isn't. They don't "slosh" anywhere.
Otoh, Current ( why is it out in the wilderness these days?) can quite justifiably be used in any of these explanations because it doesn't involve a basic misconception.
Pictures can be very handy but, like many analogies, are incomplete. Without this thing called maths, Science wouldn't't have got off the ground. Without it, one is just a passenger.

 

[mheslep]

The movement of charge is not sub-atomic.

 

[sophiecentaur]

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.

 

[yungman]

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.

 

[sophiecentaur]

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

 

[gtacs]

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 a lot from your posts. I've worked on a lot 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.


http://www.antenna-theory.com/basics/whyantennasradiate.php
http://www.microwaves101.com/encyclopedia/absorbingradar1.cfm

 

[yungman]

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!).

 

[yungman]

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 a lot from your posts. I've worked on a lot 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.


http://www.antenna-theory.com/basics/whyantennasradiate.php
http://www.microwaves101.com/encyclopedia/absorbingradar1.cfm

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.

 

[Evil Bunny]

You are hijacking the electron into being something it isn't. They don't "slosh" anywhere.

Mine slosh. I know because I asked them :)

 

[gtacs]

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 current 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 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.