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AM Transmitter question

  1. Jul 18, 2006 #1
    Hi all...

    I was looking over a simple AM transmitter from this web site 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 Im 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...

    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...

    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 Im 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 Im not convinced that the picture of the radiated wave on the website is correct.



    Attached Files:

  2. jcsd
  3. Jul 18, 2006 #2


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    Zowie, that thing's gonna 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....
  4. Jul 18, 2006 #3
    Im not sure if Im really going to try it out...I was looking more for information...

    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?

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

    Right....a square wave oscillator doesnt seem too common for this application.

    Okay...this is what Im 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.

    Not that I would ever use do such a thing...o:)

    Thanks for the advise.

  5. Jul 18, 2006 #4


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    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.
  6. Jul 18, 2006 #5
    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.
  7. Jul 18, 2006 #6

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

    I like this line of thinking although Im 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?


  8. Jul 18, 2006 #7


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    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:

  9. Jul 18, 2006 #8


    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 Im not seeing where the negative component is ever created in the trnasmitting antenna since the minimum voltage is 0 and doesnt 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?
  10. Jul 18, 2006 #9


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

    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-
  11. Jul 18, 2006 #10
    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 realise 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).

    But doesnt 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?

    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 dont remember ever exciting an antenna element with a voltage that never dropped below zero volts.

  12. Jul 18, 2006 #11
    Please dont treat me like Im an idiot. Maybe you need to dig deeper into your answers and explain what is going on in more detail. Im not looking for a pat answer here.


    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.

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

    Maybe Im 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.

  13. Jul 18, 2006 #12


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    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.
  14. Jul 18, 2006 #13
    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. Im not seeing how the two are equivalent in terms of electron motion.

  15. Jul 18, 2006 #14


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

    [tex]I(t) = I_0 + I_1 sin(\omega t)[/tex]

    [tex]I(t) = I_1 sin(\omega t)[/tex]

    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.
  16. Jul 18, 2006 #15

    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,
    Later you said this,
    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.

  17. Jul 18, 2006 #16


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    The electrons are not moving due to the DC applied to the antenna. Since it is an open circuit, they cannot be moving.
  18. Jul 19, 2006 #17
    Well, they do move...just not in the same way that they do in a "closed circuit". It isnt 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" isnt sustained like it is in a circuit.

  19. Jul 19, 2006 #18


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    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.
  20. Jul 19, 2006 #19
    No Im 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 doesnt 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.
  21. Jul 19, 2006 #20


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