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How transmitters work.

Discussion in 'Electronic Basics' started by Jason, Apr 20, 2005.

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  1. Jason

    Jason Guest

    Please forgive the pure ignorance of this question. I understand that
    this is very basic stuff that I should probably know, but I have never
    been to class to learn about this stuff, nor have I found any tutorials
    that explain this stuff in simple everyday terms. I passed by
    technician and general class amateur license exam by memorizing the
    right buzzwords for this stuff, but not actually understanding the
    concepts. I admit, I am ashamed of this fact. Here's what I have
    gathered so far.

    1. DC current flows through a crystal.
    2. The resonance of the crystal causes the current to form waves around
    the frequency of crystal
    3. This new "pulsating" current is then amplified and then sent through
    the crystal again. (This is called positive feedback)
    4. The pulsating current is sent through a low-pass filter to "filter
    out" unwanted frequencies. (That's why I said "around the frequency"
    of the crystal)
    5. From the filter, the current flows to an antenna which converts the
    current to RF radiation.

    Comments and help would be greatly appreciated.
  2. Andrew Holme

    Andrew Holme Guest

    Crystals, like capacitors, don't pass DC.
    The oscillator is an amplifier with a freqyuency-selective positive
    feedback network around it. Oscillation starts by amplifying noise.
    The crystal is a band-pass filter not a low pass filter. A low-pass
    filter is also required on the output to remove harmonics.
  3. Rich Grise

    Rich Grise Guest

    More or Less.
    Not Exactly.
    In a way, yes.

    Have Fun!
  4. No. A crystal is made of quartz, which isn't a conductor. It will
    never pass DC. Electrically, a crystal is the equivalent of an inductor
    and capacitor in series, hence it's ability to control frequency.

    Really, you need to get a book about radio and start reading. There
    is a big difference between asking questions based on points you
    don't understand, and asking us to write that material down here.

    You don't have the basics to understand how it works, and you
    have made too many leaps in your questions.

    Here's one place to start. Take an amplifier of some sort (even
    a computer with speakers connected to a soundcard), and plug
    in a microphone. Hold the microphone near the speakers, and
    start turning up the volume. You'll start getting a squeal,
    that's the positive feedback, ie oscillation.

    The signal goes through the amplifier and then is fed back to
    the input, not just once but an infinite amount. It's no longer
    amplifying an external signal, it is oscillating.

    The frequency of the howl is determined by the inherent characteristics
    of the setup, things like the response of the microphone and the speaker,
    and the response of the amplifier, and even the microphone's distance
    from the speaker. You don't have much control over the frequency, but
    it is a good demonstration of oscillation.

    Usually, one wants something more definite. So instead of the
    speaker/microphone combination, the output of the amplifier feeds
    the input directly. And between the output and the input is
    some selective element so the feedback is at that frequency. Put
    a bandpass filter in there that has a center frequency of1KHz,
    and the thing will oscillate at 1KHz.

    The crystal acts as the selective element, to ensure that the
    oscillator runs at that frequency.

    A crystal is one of the better filtering devices available, it
    doesn't generate unwanted signals.

    But the active elements are not so perfect, and they will often
    add unwanted signals. Indeed, the design of some transmitters
    will count on generating multiple signals, with the wanted
    one filtered out.

  5. Jamie

    Jamie Guest

    lets break it down in a different form.
    forget the crystal idea for now, that is only a source to generate
    a stable frequency reference!. there are other ways.
    any ways.
    lets first start with the signal generator..
    This generator will produce AC (alternating current/Voltage) at a rate
    per second that is equal to the frequency of what you want to transmit on.
    for example:
    1.6 Mhz, = 1600000 times per second will an alteration of a signal
    going from 0 Volts, to Full Volts+, back to 0 volts, then Full Volts-,
    and then back to 0 volts completes a full cycle.
    this signal is most likely then past to an amplifier to increase the
    output power abilities.
    it then makes it to the antenna as a sine wave (AC), at that point
    the AC will generate an electro magnetic field via the antenna..
    this field is powered by the AC energy you are creating.
    and since its now magnetic, the principles of Unlike and like
    polarities apply..
    minus and minus for example repel the magnetic energy that
    was there before gets pushed away.. this is how the energy actually
    travels in space.

    Known as a simple carrier, it really does nothing for you other than
    allows another
    receiver some where detect your transmitter as being there.
    CW (Morse code)is a good example
    systems like AM (amplitude modulation) will modify the level of this
    sine wave power output to reflect the patterns of the audio frequencies.
    because your transmitter frequency is so much higher than your audio
    freq, many cycles of your carrier freq will be sent out in space
    changing power levels in very gradual steps! but the receiver at the
    other end is converting this carrier into a DC (direct Current signal)
    where is the level of voltage that is changing depending on the Level of
    the signal which is influenced by the audio//

    if you need a deeper detailed than that i think i can also supply that
    also. this looked as simple as i could explain it.
  6. Jason

    Jason Guest

    Thanks for all of the help. I know that I do need to find a good book
    and just start from there. The problem is finding one that doesn't
    require huge leaps in understanding. For example, my ARRL handbook
    begins slowly enough, but then almost immediately jumps into formulas
    that describe what it going on. I recently found and I hope that helps out.
    I tried but didn't find it too useful. It
    wasn't straightforward like a textbook.

    Next time, I will try to make sure that my questions are a little more
  7. BobG

    BobG Guest

    While we're talking about transmitters, let me ask this question: Lets
    say I have a video opamp that will put out +-10V at 1 MHz into 377
    ohms... can I hook this to an antenna? P=E^2/R=100/377= 260mw In
    other words, the 'traditional' approach of making everything 50 ohms
    and using tuned class c amps is just how it was done with tubes in the
    40s right? If the impedance of the antenna is 377 ohms, and I can drive
    it direct, thats all it takes, right?
  8. John Smith

    John Smith Guest

    Impedance mismatch will mean that the antenna will not radiate 260mW, a
    significant amount will be reflected back to the op-amp. That is why the
    traditional approach exists.
  9. Andrew Holme

    Andrew Holme Guest

    If there is such a thing as a single-ended (unbalanced) antenna with an
    impedance of 377 ohms, you could drive it direct. If not, you can
    still drive it - through a matching network. Typically, though, your
    transmitter is not physically located at the antenna feedpoint - it is
    located at the end of a transmission line.

    The standard coax feedline impedance is 50 ohms, but this is not
    necessarily the impedance "seen" by tube or transistor output stage.
    The transmitter output contains an impedance matching network which
    transforms the 50 ohms up or down to the right impedance to suck the
    right amount of power from the transmitter.

    For class C, P = V^2/2R

    Where P = power "sucked" from PA stage
    V = DC power supply voltage
    R = Transformed load impedance as seen by PA transistor/tube

    This formula assumes the peak-to-peak AC voltage developed at the
    collector / drain is not far shy of 2*V.
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