Can an AM Transmitter Work Without Negative Voltage?

[Firefox123]

Hi all...

I was looking over a simple AM transmitter from this website http://www.sci-toys.com/scitoys/scitoys/radio/am_transmitter.html

I have attached the diagram as well. (The battery is a 9 volt battery, so I am fairly sure the "-9 volts" is a typo)

I have three questions..


1. Am I correct in assuming that even though the bottom of the secondary coil is held at 9 volts by the battery the transformer still attempts to put an induced voltage over the secondary, which is why the voltages add?

The website has this to say about the transformer...

The transformer is connected to the power supply of the oscillator. The sound source causes the transformer to add and subtract power from the oscillator, just as it would have pushed and pulled on the speaker.

As the power to the oscillator goes up and down, the power of the electricity in the antenna goes up and down also. The voltage is no longer simply 9 volts. It is now varying between 0 volts and 10 volts, because the power from the transformer adds and subtracts from the power of the battery

2. If we are applying a oscillating voltage to the antenna that is between 0 and 10 volts, can we really say that the electrons move "up and down the wire"? It would seem to me that the electrons would accelerate and decelerate, but would not actually reverse direction.

Here is what the website has to say about the antenna...

The oscillator is connected to one end of a long wire antenna. It alternately applies 9 volts of electricity to the end of the wire, and then 0 volts, over and over again, a million times each second.

The electric charge travels up and down the wire antenna, causing radio waves to be emitted from the wire. These radio waves are picked up by the AM radio, amplified, and used to make the speaker cone move back and forth, creating sound.

As the power to the oscillator goes up and down, the power of the electricity in the antenna goes up and down also. The voltage is no longer simply 9 volts. It is now varying between 0 volts and 10 volts, because the power from the transformer adds and subtracts from the power of the battery.

The varying power in the antenna causes radio waves to be emitted. The radio waves follow the same curves as the waves in the antenna.

3. What does the electric field on the antenna or the radiated field look like if you apply an oscillating voltage to the antenna that never goes negative?




So for question #1 I want to make sure I am looking at this correctly with the battery applying 9 volts to the end of the secondary coil.

And for question #2 I am not seeing why the electrons would reverse direction in the wire, although they would accelerate and decelerate. If the voltage polarity never reverses, why would the directin change.

For question #3 I am not convinced that the picture of the radiated wave on the website is correct.





Thanks



Russ

 

[berkeman]

Zowie, that thing's going to spray something aweful. At least 1-10MHz is away from most emergency bands. Well, just don't be surprised if you get a visit from some HAMs or the FCC...

Anyway, to your questions. This design just uses a canned oscillator that is running off of the DC supply. The DC supply is getting modulated a bit by the audio coming in through the transformer, and that modulates the output of the 1MHz digital canned oscillator. The wording at the website is confusing, but I'd guess that you would aim for about 1Vpp modulation on top of the 9V battery. That sure looks like a 5V canned oscillator in their photo, so I'm surprised that it works up to 9-10V. But maybe it's a version that takes higher voltages or something.

Don't worry about DC and the antenna. The antenna just looks like an LC resonanct circuit with loss, and there is no DC current into the antenna even if you add a DC voltage to the RF AC voltage.

Keep in mind that the output of that oscillator is a square wave, so you will be spewing harmonic energy, mainly at the odd harmonics 3MHz, 5MHz, 7MHz, etc. In a normal transmitter, you will add filtering (and not use a square wave oscillator) to band-limit your emissions to fit into the band that the FCC is allowing you to transmit in.

And yes, the electrons do go up and down the antenna (not very far of course) in an oscillatory fashion. The best antenna you could use for your experiment would be a vertical 1/4 wave monopole, above some kind of ground plane or several horizontal elements to serve as the reference. Of course, the better your antenna, the better chance that you will get a knock at the door...

 

[Firefox123]

Zowie, that thing's going to spray something aweful. At least 1-10MHz is away from most emergency bands. Well, just don't be surprised if you get a visit from some HAMs or the FCC...

Im not sure if I am really going to try it out...I was looking more for information...

Anyway, to your questions. This design just uses a canned oscillator that is running off of the DC supply. The DC supply is getting modulated a bit by the audio coming in through the transformer, and that modulates the output of the 1MHz digital canned oscillator. The wording at the website is confusing, but I'd guess that you would aim for about 1Vpp modulation on top of the 9V battery. That sure looks like a 5V canned oscillator in their photo, so I'm surprised that it works up to 9-10V. But maybe it's a version that takes higher voltages or something.

Right...I guess I've just never really seen a transformer with a battery wired up to it like that. It looked a bit strange at first, but I guess you just sum up the voltage induced across the coil with the 9 volts from the battery.

Lets say we just have a normal transformer...the primary coil induces a voltage across the secondary coil, say 1 volt. So the 'top' of the coil is at 1 volt while the 'bottom' is a 0 volts.

Now hook up that same transformer as in the figure...
The battery holds the 'bottom' of the coil at 9 volts, so if the primary coil is going to induce 1 volt across the secondary, the 'top' of the coil must be a 10 volts. Is this correct?

Don't worry about DC and the antenna. The antenna just looks like an LC resonanct circuit with loss, and there is no DC current into the antenna even if you add a DC voltage to the RF AC voltage.

Its not DC that concerns me...its the fact that the voltage isn't really reversing direction. Its just going from 0 - 10 and back.

Keep in mind that the output of that oscillator is a square wave, so you will be spewing harmonic energy, mainly at the odd harmonics 3MHz, 5MHz, 7MHz, etc. In a normal transmitter, you will add filtering (and not use a square wave oscillator) to band-limit your emissions to fit into the band that the FCC is allowing you to transmit in.

Right...a square wave oscillator doesn't seem too common for this application.

And yes, the electrons do go up and down the antenna (not very far of course) in an oscillatory fashion.

Okay...this is what I am not getting.

Why would they reverse direction when the square wave input never really reverses polarity? The best we get is 0 volts...and the rest are positive voltages. So why would the electrons reverse direction? Doesnt that imply an electric field in the opposite direction?

All I see is an electric field of varying strengths, but with the same polarity.

The best antenna you could use for your experiment would be a vertical 1/4 wave monopole, above some kind of ground plane or several horizontal elements to serve as the reference. Of course, the better your antenna, the better chance that you will get a knock at the door...

Not that I would ever use do such a thing...

Thanks for the advise.




Russ

 

[berkeman]

First off, I realized in the shower after my noon workout that I said something pretty stupid in my previous post. If you made a 1/4 wave monopole element for 1MHz, that would be about 75 meters high. I don't think that's what you had in mind. If you do the experiment, start with just a couple feet of radiating element over a ground plane. That will put out enough of a signal to be picked up by AM receivers in the immediate area.

As for your DC/AC question, just think of the antenna as an inductor in series with a capacitor (the far side of the capacitor is grounded). When you charge that up to 9VDC, you are moving some electrons from one plate of the capacitor to the other, and they just stay there if nothing else happens. Then if you add a 1Vpp signal onto the 9VDC bias, the DC bias is basically out of the picture, and electrons are pumped back and forth between the plates of the capacitor by the 1Vpp AC voltage. The DC distribution of electrons and the AC response of the electrons are independent.

 

[pallidin]

The emmission of a radio wave by an electron does not necessarily depend upon complete reversal in travel. Rather, if it is stopped, pushed forward, stopped, pushed forward, etc... an EM field will be generated even on the first instance.

 

[Firefox123]

First off, I realized in the shower after my noon workout that I said something pretty stupid in my previous post. If you made a 1/4 wave monopole element for 1MHz, that would be about 75 meters high. I don't think that's what you had in mind. If you do the experiment, start with just a couple feet of radiating element over a ground plane. That will put out enough of a signal to be picked up by AM receivers in the immediate area.

I didnt even do the math with regards to length, but you are correct.

As for your DC/AC question, just think of the antenna as an inductor in series with a capacitor (the far side of the capacitor is grounded). When you charge that up to 9VDC, you are moving some electrons from one plate of the capacitor to the other, and they just stay there if nothing else happens. Then if you add a 1Vpp signal onto the 9VDC bias, the DC bias is basically out of the picture, and electrons are pumped back and forth between the plates of the capacitor by the 1Vpp AC voltage. The DC distribution of electrons and the AC response of the electrons are independent.

I like this line of thinking although I am going to move away from the cap/inductor analogy for now.

So are you basically saying that when we apply a voltage (say ten volts) to an antenna (which is like an open circuit) we get a non-uniform charge distribution on the antenna. Then when we drop the voltage back down to zero those charges can move in the opposite direction?

Is this what you believe is happening when we apply a 0 to 10 volt square wave to te input of the antenna?






Russ




Russ

 

[berkeman]

The 0V-10V square wave drive to your antenna is like 5VDC plus a 10Vpp square wave. The important thing for the antenna is the AC component, since it takes accelerating charges to generate the EM wave. The static 5VDC signal only gives a static charge distribution and static E field to the surrounding earth/ground. The shape of the static E field will depend on the geometry of the antenna with respect to the ground potential surface.

I googled antenna design tuturial, and got lots of good hits. Here's the hit list for your reading pleasure:

http://www.google.com/search?hl=en&q=antenna+design+tutorial

 

[Firefox123]

The emmission of a radio wave by an electron does not necessarily depend upon complete reversal in travel. Rather, if it is stopped, pushed forward, stopped, pushed forward, etc... an EM field will be generated even on the first instance.

Correct.

Any acceleration will create a dynamic electric field (the radiation field and the induction field) and we will get an EM wave propagating.

But these changes in the electric field will then create currents in a receiver antenna and this is where my question arises.

The website implies that this field (the one created by the circuit in question) would create alternating current (as in positive and negative voltages) on a receiving antenna.

But I am not seeing where the negative component is ever created in the trnasmitting antenna since the minimum voltage is 0 and doesn't drop below ground.

If you look at charge distribution on an antenna driven with an AC voltage and compare the periodicity of the input volatge with the periodicity of the free space alternating electric field you will notice a direct relationship between them.

This alternating electric field then induces currents in a receiver antenna, and the received currents closely resemble the currents that created the original electromagnetic waves.


I am wondering how a voltage, that alternates between 0 and 10 volts, can create an electric field that would induce a voltage that alternates in polarity on a receiver antenna?

While a 0-10 volt square wave does change in strength the polarity never changes. So what would the output of an antenna look like when driven with such an input?

 

[berkeman]

Correct.

Any acceleration will create a dynamic electric field (the radiation field and the induction field) and we will get an EM wave propagating.

But these changes in the electric field will then create currents in a receiver antenna and this is where my question arises.

The website implies that this field (the one created by the circuit in question) would create alternating current (as in positive and negative voltages) on a receiving antenna.

But I am not seeing where the negative component is ever created in the trnasmitting antenna since the minimum voltage is 0 and doesn't drop below ground.

If you look at charge distribution on an antenna driven with an AC voltage and compare the periodicity of the input volatge with the periodicity of the free space alternating electric field you will notice a direct relationship between them.

This alternating electric field then induces currents in a receiver antenna, and the received currents closely resemble the currents that created the original electromagnetic waves.


I am wondering how a voltage, that alternates between 0 and 10 volts, can create an electric field that would induce a voltage that alternates in polarity on a receiver antenna?

While a 0-10 volt square wave does change in strength the polarity never changes. So what would the output of an antenna look like when driven with such an input?

I'll try one last time, then it's off to the google hit list with you

The drive voltage into an antenna creates a current in the antenna. It is the changes in the current in the antenna (the acceleration of the electrons) that creates the EM wave that propagates away from the antenna. Static fields or static currents fall out of the differentiation, right? You'll enjoy the hit list. Read some more about antennas, then come on back and post follow-up questions in a new thread. -Mike-

 

[Firefox123]

The 0V-10V square wave drive to your antenna is like 5VDC plus a 10Vpp square wave.

Right...but a ten Vpp square wave with a 5 volts DC offset is a different input into an antenna then a 10Vpp square wave without any DC offset.

With respect to ground, the one with the 5 volts offset never drops below ground potential, while the one without the offset send -5volts and +5volts to the antenna.

Wouldnt the radiated waves, in terms of voltage levels, be different?

I understand what you are saying about looking at the "ac component" and not worrying about the DC component, but a DC offset can cause a current that was once reversing direction to now flow in only one direction.

I do realize that "current" in an antenna is not quite the same as current flowing in a digital circuit, for example, but I am trying to keep this simple and avoid discussions about mroe complicated topics (like transmission line theory, standing waves, waveguide theory etc).

The important thing for the antenna is the AC component, since it takes accelerating charges to generate the EM wave. The static 5VDC signal only gives a static charge distribution and static E field to the surrounding earth/ground. The shape of the static E field will depend on the geometry of the antenna with respect to the ground potential surface.

But doesn't that offset affect what polarity the antenna is seeing and which "direction" the current is flowing?

Doesnt the current in an antenna change direction because the applied voltage reverses polarity?

I googled antenna design tuturial, and got lots of good hits. Here's the hit list for your reading pleasure:

http://www.google.com/search?hl=en&q=antenna+design+tutorial

Thanks. There are some good links in there.

I am not completely clueless though..I have had both antenna and microwave engineering courses, but I don't remember ever exciting an antenna element with a voltage that never dropped below zero volts.




Russ

 

[Firefox123]

I'll try one last time, then it's off to the google hit list with you

Please don't treat me like I am an idiot. Maybe you need to dig deeper into your answers and explain what is going on in more detail. I am not looking for a pat answer here.

The drive voltage into an antenna creates a current in the antenna.

Yes

It is the changes in the current in the antenna (the acceleration of the electrons) that creates the EM wave that propagates away from the antenna.

Yes, I know.

But there is a difference between an electron accelerating in one direction, slowing down, stopping, and then accelerating in the opposite direction and an electron accelerating in one direction, slowing down, stopping, and then accelerating in the same direction.

Static fields or static currents fall out of the differentiation, right? You'll enjoy the hit list. Read some more about antennas, then come on back and post follow-up questions in a new thread. -Mike-

I will. But in the mean time why not answering the question if you know the answer?

Maybe I am not being clear enough in what questions I am asking (this happened often as an undergrad when I asked questions that were "beyond the scope of the class"). If you need further clarification just let me know.



Russ

 

[Averagesupernova]

Right...but a ten Vpp square wave with a 5 volts DC offset is a different input into an antenna then a 10Vpp square wave without any DC offset.

Russ

No it isn't. The radiated signal will look NO different than with a 'true' AC signal driving the antenna. While technically it is different, it is irrelevant to the antenna.

 

[Firefox123]

No it isn't. The radiated signal will look NO different than with a 'true' AC signal driving the antenna. While technically it is different, it is irrelevant to the antenna.

Could you please explain why this is?

If I take a 10Vpp signal and then 5V DC offset that signal we now have a signal that goes from 0 volts to 10 volts.

If I hook this up to a simple circuit (say a few resistors) I have a voltage that alternates from 0 to + 10 volts and the current in the circuit flows in one direction.

But if I hook up a 10 Vpp signal to the same circuit, now I have a voltage that alternates between -5 volts and + 5 volts and the current in the circuit flows in two different directions.

In one circuit the current goes from maximum to zero while in the other the current goes from maximum in one direction to maximum in the other direction. I am not seeing how the two are equivalent in terms of electron motion.





Russ

 

[berkeman]

What do you get when you differentiate the following two equations?

I(t) = I_0 + I_1 sin(\omega t)

I(t) = I_1 sin(\omega t)

No different, right? An antenna, like a transformer, is an AC device. I didn't intend to imply that you're an idiot, BTW. Your questions are okay, you just need to do a little reading to help your understanding.

 

[Firefox123]

What do you get when you differentiate the following two equations?

I(t) = I_0 + I_1 sin(\omega t)

I(t) = I_1 sin(\omega t)

No different, right? An antenna, like a transformer, is an AC device.

Right.

I didn't intend to imply that you're an idiot, BTW. Your questions are okay, you just need to do a little reading to help your understanding.

Before asking the question here I read through all my antenna and electromagnetics books and also did an internet search. If those fail to provide an answer then I come here.

I understand that both acceleration and deceleration create a radiation field.

In my first post I said that it didnt seem like the electrons would "reverse" direction in the antenna when driven by the 0-10 volt square wave. You replied and said,

And yes, the electrons do go up and down the antenna (not very far of course) in an oscillatory fashion.

Later you said this,

As for your DC/AC question, just think of the antenna as an inductor in series with a capacitor (the far side of the capacitor is grounded). When you charge that up to 9VDC, you are moving some electrons from one plate of the capacitor to the other, and they just stay there if nothing else happens. Then if you add a 1Vpp signal onto the 9VDC bias, the DC bias is basically out of the picture, and electrons are pumped back and forth between the plates of the capacitor by the 1Vpp AC voltage. The DC distribution of electrons and the AC response of the electrons are independent.

I can convince myself that mathematically the signal you describe works fine, but it is the physical movement of the electrons that I am struggling with.

What do you think of this explanation?

When you first apply the 10 volts it will create an acceleration of charges and will create a charge distribution slightly different from the conditions before any voltage was applied. The electrons will tend to accelerate in the same direction.

When the 10 volts is relaxed to 0 volts those same electrons (which are "clumped" together due to them slightly moving in one direction during the acceleration) now move in the opposite direction due to their mutual repulsion.



Russ

 

[Averagesupernova]

The electrons are not moving due to the DC applied to the antenna. Since it is an open circuit, they cannot be moving.

 

[Firefox123]

The electrons are not moving due to the DC applied to the antenna. Since it is an open circuit, they cannot be moving.

Well, they do move...just not in the same way that they do in a "closed circuit". It isn't a continuous movement. Electrons can still "flow" in an open circuit, but you end up with standing waves and such instead of normal "current".

Electrons will react to an applied voltage regardless of whether we have a complete circuit or not. Depending on the situation this could either be a standing wave situation or we could just get very short lived charge flow (like a change in the distribution of charge).

But without a "circuit" the "flow" isn't sustained like it is in a circuit.





Russ

 

[Averagesupernova]

But without a "circuit" the "flow" isn't sustained like it is in a circuit.

Russ

This is my point exactly. Are you arguing the point just to be arguing? The EM field is generated ONLY when a current flows. If it is an open circuit for DC, then no magnetic field is generated from the DC. The DC bias on the antenna has no effect on the radiated wave.

 

[Firefox123]

This is my point exactly. Are you arguing the point just to be arguing? The EM field is generated ONLY when a current flows. If it is an open circuit for DC, then no magnetic field is generated from the DC. The DC bias on the antenna has no effect on the radiated wave.

No I am not arguing to argue.

No one in this thread gave me the answer I sought...perhaps I wasnt doing a good job of explaining my question.

It doesn't matter now anyway, because after thinking about it for 10 minutes I figured it out myself.

The answer to my question of why the electrons will reverse direction when we apply 0 volts is "self repulsion". It has nothing to do with DC versus AC.

Thanks everyone for the replies anyway.

 

[berkeman]

It's not self-repulsion. Google displacement current. What you are describing is not an exercise in static electricity.

 

[Firefox123]

It's not self-repulsion. Google displacement current. What you are describing is not an exercise in static electricity.

Displacement current is not "real current" it is a changing electric flux through a surface.

If you want to try and make an argument for an induced current, perhaps due to some inductance in the antenna I would be interested in hearing it...


If I apply 10 volts to an wire antenna and then remove the voltage source the electrons will return to their original configuration within the wire due to the mutual repulsion between the electrons.

If we remove the voltage what other forces are present to cause the motion?

Unless we want to try and invoke some kind of back EMF due to inductance (which might be a difficult argument to make for a straight wire antenna) I am not sure what other forces we are left with.




Russ

 

[berkeman]

Displacement current is real, and the current in the wires leading to the capacitance is just as real and of the same value (it's a series circuit after all). If you apply 10V to a capacitor (like the wire of the antenna with respect to ground) and remove the voltage source, the capacitance stays charged to 10V and slowly leaks off via whatever the leakage current mechanism is. For DC and low frequencies (before the antenna reaches its first 1/4 wave resonance), the metal antenna is an equipotential surface. Putting a 10VDC voltage on the antenna does not "bunch up" the electrons at the far end or something, where they would want to repel each other after the voltage source is remove.

You're asking okay questions, but you should do a little reading in an electrostatics and E&M book to get you farther along. I can tell that we're getting nowhere trying to help you in this thread. Check out "Applied Electromagnetics" by Plonus or "Engineering Electromagnetics" by Hayt. Your first course in electromagnetics will help you to make a lot more sense out of what is going on.

 

[Firefox123]

Displacement current is real, and the current in the wires leading to the capacitance is just as real and of the same value (it's a series circuit after all).

No it isnt..at least not in the same sense as "normal current." Displacement current has to do with electric flux and induced magnetic fields...not with actual moving charge like normal current.

It has untis that are mathematically similar, but physically it is something different from real current.

If you apply 10V to a capacitor (like the wire of the antenna with respect to ground) and remove the voltage source, the capacitance stays charged to 10V and slowly leaks off via whatever the leakage current mechanism is.

The electric field between the plates produces attraction from one plate to the other. The electrons on the "negative" plate are, in part, held there by the positive charge on the "positive" plate.

For DC and low frequencies (before the antenna reaches its first 1/4 wave resonance), the metal antenna is an equipotential surface. Putting a 10VDC voltage on the antenna does not "bunch up" the electrons at the far end or something, where they would want to repel each other after the voltage source is remove.

Im not applying a DC voltage (although I would argue that even if you did apply a DC voltage of 10 volts there would be some kind of short lived transient charge movement at the moment of turn on).

Im applying 10volts then 0volts then10volts etc.

Antennas do have a charge distribution (or "bunching up") along the antenna as a result of the applied voltage waveform.

You and I agree that the electrons move up and down (vibrate) in the antenna.

This is easy for me to explain physically if I am applying +5volts and then -5volts, because I can imagine the "current" flowing in one direction and then the other due to the polarity of the voltage.

But with +10 and 0 I have a difficult picturing what "force" causes the electrons to "reverse direction".

You're asking okay questions, but you should do a little reading in an electrostatics and E&M book to get you farther along. I can tell that we're getting nowhere trying to help you in this thread. Check out "Applied Electromagnetics" by Plonus or "Engineering Electromagnetics" by Hayt. Your first course in electromagnetics will help you to make a lot more sense out of what is going on.

Stop assuming I haven't had these courses. I have had antenna engineering, microwave engineering, and engineering electromagnetics.

I worked as an EMI/EMC engineer working on antennas and radar systems for 3 years and I am currently involved in electronic warfare work.

Im looking for someone to explain to me at a fundamental level what is going on, i.e. what is causing the reversal of direction, in this antenna.

You continue to assume that I have little or no electromagnetics knowledge or experience, which is false. I have tried to correct this and explain that I am not looking for a "see google" answer. I am also looking for something deeper and more physical than just someone telling me to look up Maxwells equations or displacement current as if I don't know what they are.

Just because someone knows the equations or textbook descriptions of some topic like quantum physics, or relativity, or electrodynamics, or fluid mechanics, etc doesn't imply they really understand what is going on. Equations often describe what is going on, but don't always answer the why.

You have made a lot of assumptions about my knowledge and experience, yet I have not done the same to you. I have treated you as if you were an "expert" just because you took it upon yourself to try and answer my question.

So if you know what "force" is causing the electrons to reverse direction when we are applying 0 volts (just like they do when we apply +5 and -5) then please tell me what force this is because I am stumped.




Russ

 

[berkeman]

I guess all I can say is that the force that moves the electrons is the EMF, same as in the inductor-capacitor approximation of the antenna that I mentioned early on. Maybe it's the antenna part that is causing you to not be able to directly apply what you know about how currents in capacitors work. Did your antenna book talk about the capacitor plate antenna approximation? That model helped me conceptually when I was first learning about antennas. If you can find a copy of "Antenna Theory and Design" by Stutzman and Theil (great book, BTW) at the library, check out Figure 2-3 of the Capacitor Plate Antenna. It shows the moving charges and resulting currents. I'll see if I can find a similar figure with google...be right back.

 

[berkeman]

I googled "capacitor plate antenna" and got several hits. This one has a figure:

http://www.ece.msstate.edu/~donohoe/ece4990notes4.pdf

 

[Firefox123]

I googled "capacitor plate antenna" and got several hits. This one has a figure:

http://www.ece.msstate.edu/~donohoe/ece4990notes4.pdf

Thats a great link...thanks for the information.

I just spoke to another antenna engineer and he set me straight.

My explanation wasnt completely off apparently, but I was focusing too much on the level portions of the square wave.

During the transient from 0 - 10 volts we get current flow. If we kept the voltage at 10 volts after this we would have an equipotential surface (since the antenna is an open circuit) and the electrons would go back to electrostatic equilibrium and distribute themselves throughout the conductor due to the repulsive forces between them.

But if we continue to alternate the voltage from 0-10 volts and then back to 0 volts then this transient change in voltage will reverse their direction.


For some reason I was applying what happens during the transient portions of the square wave to the "level" portions of the square wave.


I think I am good now.

Thanks for the help berkeman.




Russ

 

[Averagesupernova]

Russ the last part of your last post is pretty much what I tried to explain to you. The transition part of the waveform is not considered DC.

 

[Firefox123]

Russ the last part of your last post is pretty much what I tried to explain to you. The transition part of the waveform is not considered DC.

Right...the problem was that I was trying to picture where the electric field was coming from that causes the charges to move.

For charge to move via an electric field there has to be a potential difference, otherwise everything will just sit there as an equipotential surface (like shortly after we hook up DC to an antenna).

I was forgetting to take into account the time it takes for a change in voltage (or electric field) to "reach" the length of the wire.

Before we connect the antenna to anything, it is an equipotential surface sitting at 0 volts (Im assuming no E-fields being received by the antenna) and no charge is moving.

Then we connect one end to a source, say 10 volts, and during this transient change we have one end of the wire at 10 volts, while the far end is still at 0 volts.

Since changes in the electric field travel at the speed of light it takes (approximately) t= L/c (where L is the length of the antenna) seconds for the voltage/electric field change to reach the far end of the wire.

So for t<L/c we have essentially put a voltage across the length of the antenna and the electrons will move a bit. Then when we apply 0 volts we have the opposite situation and the electrons will move again, but in the opposite direction.

My error was in applying the voltage changes instantaneously over the entire antenna with a square wave input. I forgot to take into account the fact that it takes time for the changes in the electric field to propagate down the wire.

Stupid mistake.

Russ