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FM transmitter

Discussion in 'Electronic Basics' started by fred, Feb 10, 2004.

  1. fred

    fred Guest

    I have found an fm radio transmitter circuit diagram,
    can't remember where it was so I put it here:

    http://www.geocities.com/x_file_space/fmt-1.gif

    This circuit looks too simple to be any good, but just
    out of interest I was wondering how the fm modulation
    is occuring. Obviously the tuned circuit will determine
    the main frequency, but how varying the base of Q2
    varies the frequecy this I don't see, any ideas?
     
  2. Not really. That is, it does effect the frequency of oscillation, but it
    is not part of the LC tank.
    The circuit is a colpits oscillator. There is a capacitor from collector
    to emitter. There is a capacitor from base to emitter (the transistor
    diffusion capacitance). At the oscillation frequency the tank is
    inductive (it is not at resonance) and is connected from the collector
    to the base via C3.

    So...changing the bias, changes the bias...very informative:)

    Ice changes thus effecting not only the Z on the C5
    The base emitter diffusion capacitance, Cbe is a direct function of the
    emitter current, i.e. Cbe=40Ic/2.pi.ft, hence the oscillation frequency
    will change with input signal.

    Kevin Aylward

    http://www.anasoft.co.uk
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.

    "That which is mostly observed, is that which replicates the most"
    http://www.anasoft.co.uk/replicators/index.html

    "quotes with no meaning, are meaningless" - Kevin Aylward.
     
  3. Jamie

    Jamie Guest

    C5 is the Cap that is making the majority of the freq shift.
    as C5 is part of the tuned circuit. with out it you can
    no oscallation.
    since the circuit biasing the emitter with the Tuned Tank
    giving you the phase shift needed you get your oscallation state.
    if you start varying the bias on the transistor it will cause this
    emiter bias to change thus causing the rate of cycle time to change
    due to current Ice changes thus effecting not only the Z on the C5 but
    changing its return path Z..
     
  4. Shouldn't you capitalize "i"?
    What is a "lamen"? Some sort of mammal from South America? Or maybe
    you mean "layman".
    Still short of capital letters, I see. I will send you a shift lock
    code.
    "Ediot"?? Man, you should have finished third grade.
    I looked all over my jar of peanut butter but could not see your origin.
    I'm not certain if you meant "Jif" or "gif". I suppose I should give you
    the benefit of the doubt and assume that English is not your first language.

    Putting on the asbestos tee-shirt...

    Chip Shults
     
  5. Jamie wrote:

    Don't top post.

    Obviously, not. What you said was incorrect.
    I did.
    It isn't part of the oscillator frequency setting components. As I
    already stated in my first post, it connects the LC tank circuit from
    the collector to the base. The tank operates off resonance so that the
    collector is connected to the base via an effective inductive reactance.
    I never said it was. What part of "..is connected from the collector
    to the base via C3..." did you not understand?
    You manage that quite well on you own, with no help from me.
    I'm sure they will.

    Kevin Aylward

    http://www.anasoft.co.uk
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.

    "That which is mostly observed, is that which replicates the most"
    http://www.anasoft.co.uk/replicators/index.html

    "quotes with no meaning, are meaningless" - Kevin Aylward.
     
  6. Jamie

    Jamie Guest

    I may not be an expert on the subject at hand but
    i do know enough that about what i said while i was
    trying to put it lamens terms.
    Please look at the URL Link that was talked about.
    http://www.geocities.com/x_file_space/fmt-1.gif
    tell me that C3 is part of the osc and not a
    simple RF by pass.
    as you notice the value is 0.01 ! kind of a long
    shot for being used as part of the resonant circuit.
    so since you like chopping others messages and trying to make
    them look like an ediot, i just chopped your reply.
    i hope those that know better after looking at the
    Gif can see where i am coming from.
     
  7. Jamie

    Jamie Guest

    i rest my case.
     
  8. fred

    fred Guest

    Good discussion, one last point:

    since the LC circuit is not a tuned tank circuit,
    why do we need C4 anyway ?
     
  9. Dbowey

    Dbowey Guest

    Fred.... posted:
    << Good discussion, one last point:

    since the LC circuit is not a tuned tank circuit,
    why do we need C4 anyway ? >>

    Wow! The discussion must not have been that good. C4 IS used to tune the
    circuit.

    Don
     
  10. fred

    fred Guest

    Wow! The discussion must not have been that good. C4 IS used to tune the
    It was established that the LC tank circuit does not
    operate in resonance, read other replies please
    before you comment.
     
  11. Although the tank is not in resonance, the capacitor is still used to
    "tune" the oscillator frequency. The tank needs to be +jX as far as its
    reactance goes, and this reactance is a function of both L and C.

    Kevin Aylward

    http://www.anasoft.co.uk
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.

    "That which is mostly observed, is that which replicates the most"
    http://www.anasoft.co.uk/replicators/index.html

    "quotes with no meaning, are meaningless" - Kevin Aylward.
     
  12. Dbowey

    Dbowey Guest

    Fred posted:
    <<
    It was established that the LC tank circuit does not
    operate in resonance, read other replies please
    before you comment.--------
    Y'all may "establish" anything you like, but it does not make it correct. The
    inductor and C4, being in a state of Xc = Xl, is what permits the circuit to
    oscillate at a particular frequency.

    Don
     
  13. Here is my long, boring guess about how this oscillator works:

    http://www.meridianelectronics.ca/circ/fmt1.htm

    I don't think its a colpitts oscillator, at least as I understand them. A
    colpitts oscillator will use two series caps in the tank as a phase shifting
    device, to allow the phase of the excitation waveform to vary by 180 degrees
    from the feedback waveform. This is done so it can be fed back into an
    inverting transistor amplifier, such as a common emitter amplifier.

    For this one, however, I think that the way to analyze this is to see that
    Q2 is configured as a common base amplifier, which is NOT an inverting
    amplifier. Also, the capacitors that influence the frequency are all in
    parallel, not in series.

    For a common base amplifier, the signal is applied to the emitter, and the
    output is at the collector, in phase (mostly) with the input. (There is a
    slight phase shift through the transistor, which I believe will limit the
    max output frequency.)

    The gain of the amplifier increases with the impedance of the tank, and the
    tank impedance increases as the frequency aproaches the resonant frequency.

    The capacitor C5, in parallel with the capacitance between the collector and
    the emitter in the transistor, form the feedback of the system. They modify
    the voltage at the emitter, which is then amplified back to the collector,
    forming a feedback loop.

    The gain of the amplifier is maximum when the impedance of the tank is
    maximum. That will happen when the tank is in resonance, ie, when the
    frequency is equal to

    1
    f0 = ----------------
    2.pi.sqrt(L1.Ct)

    where Ct is the total capacitance in the tank... Thus, the system will
    oscillate at frequencies near the point where there is maximum impedance,
    ie, the tank resonance frequency.

    Ok, that allows it to oscillate, and sets the carrier frequency. Now lets
    talk about modulation.

    Since the inductance of the tank is known, varying the capacitance of the
    tank will vary the carrier frequency, thus modulating the output.

    The capacitance of the tank is due to the parallel combination (ie, the sum)
    of C4, C5, and the capacitance between the collector and emitter of the
    transistor. Thats C4+C5+Cce. The last term, Cce, is a series term of Cbe,
    which is proportional to Ic, as Kevin Aylward mentioned in his message. Ic
    is modified by the base voltage. Consequently, the total capacitance of the
    tank, Ct, is influenced by the bias voltage of the transistor, which is how
    the output signal is modulated.

    Unfortunately, the capacitance of the transistor is hard to predict.
    Consequently, the variable cap is needed to tune the system to a particular
    frequency. I believe Cce will be influenced by temperature as well, so the
    system will require tuning at different temperatures, and probably as it
    warms up as well.

    Additionally, I believe the antenna will also contribute slightly to the
    capacitance of the tank in this transmitter. However, I don't have any way
    to predict how much of an effect that will have.

    Comments requested... :)

    Regards,
    Bob Monsen
     
  14. fred

    fred Guest

    I don't actually see, but not to worry i think this
    circuit is too complicated for a on-the-hoof analysis
    in a newsgroup, probably the complete solution would
    be spread over a whole chapter in text books. I doubt
    if this is even tought in text books because this
    circuit is a "trick circuit" because it has hidden
    capacitors, and so is a bad example.
     
  15. You misunderstand colpitts oscillators. It is indeed a colpits
    oscillator. It is not debatable.

    Draw a floating transistor. Put 3 impedances across it, Zce, Zbe, Zbc.
    Now do a nodal analysis, e.g. equate real and imaginary terms for there
    to be a sustained current. You will find that in this completely general
    circuit there are only *two* possible ways for such a transistor circuit
    to oscillate. One where Zce and Zbe are capacitors with Zbc an inductor,
    and one where Zce and Zbe are inductors with Zbc a capacitor. The first
    is a Colpits, the second is the Hartley. It is trivial that Q2
    *topologically* satisfies the Colpits connection. Don't be confused as
    to where the ground is connected. It makes no difference.
    Ho hummm....

    Completely irrelevant.
    No. Go and do the actually sums as I suggest above.




    Kevin Aylward

    http://www.anasoft.co.uk
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.

    "That which is mostly observed, is that which replicates the most"
    http://www.anasoft.co.uk/replicators/index.html

    "quotes with no meaning, are meaningless" - Kevin Aylward.
     
  16. Not at all. Its a very simply circuit.
    I agree that it is a somewhat subtle circuit, but once one is told how
    it works, it is trivially obvious after the fact. The relation between
    Cbe and Ic is a very basic one. Colpits oscillators with tuned circuits
    are also very well covered.

    Kevin Aylward

    http://www.anasoft.co.uk
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.

    "That which is mostly observed, is that which replicates the most"
    http://www.anasoft.co.uk/replicators/index.html

    "quotes with no meaning, are meaningless" - Kevin Aylward.
     
  17. Thanks for attempting to educate me. I see that your view of colpitt/hartley
    is far more general than mine.
    Sorry, my misunderstanding of your definition of colpitts as one of the two
    modes of oscillation of a transistor. I'm unfortunately thinking from a few
    examples.
    I'd like to see your closed form analysis of the circuit that predicts the
    frequency, given in terms of the capacitance and inductance values, Vcc, and
    the base voltage. That might help me understand what you are trying to say.

    For me, given my simpleminded analysis above, it seems fairly easy to
    predict the oscillation frequency using the formula

    f0 = 1/(2.pi.sqrt(L1.(C4 + C5||Cbe + Ccb)))

    This matches pretty well with what spice predicts for a similarly configured
    circuit, when using Cbe=CJE and Ccb=CJC. I don't know what the relation
    betwen Cbe and Vb is, though, so I can't really predict the frequency given
    the base voltage. Thus, I'm not able to predict the modulation.

    Regards,
    Bob Monsen
     
  18. fred

    fred Guest

    http://www.geocities.com/x_file_space/fmt-1.gif
    I agree, also in Colpitts, the center tap of the two capacitors
    is connected to ground whereas here it's connected to the
    emitter (which is not bypassed to ground).
    I agree, this is a common base. I don't think all the capacitors
    are in parallel though, C4 and C5 are in parallel? unless you are
    ignoring R7, are you ?


    Again, as I mentioned above this is not clear to me unless R7 is ignored.




    series? series with what you forgot to mention ?
     
  19. See my other post on this.

    You need to see the *big* picture.
    It is also common emitter and emitter follower. The oscillator is all
    configurations at once!
    You need to actually do the full, general sums of the small signal
    equivalent circuit. I have given a bigger outline in the other post.



    Kevin Aylward

    http://www.anasoft.co.uk
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.

    "That which is mostly observed, is that which replicates the most"
    http://www.anasoft.co.uk/replicators/index.html

    "quotes with no meaning, are meaningless" - Kevin Aylward.
     
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