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Question about oscillator

Discussion in 'Electronic Basics' started by phil, Apr 20, 2004.

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  1. phil

    phil Guest

    Can anyone offer his/her explanation as to how the following
    oscillator works:

    http://members.tripod.com/~transmitters/begin.htm



    I think the tank circuit will oscillate at it resonant frquency. As for
    the positive feedback A x beta = -1 requirement I am not sure
    how much is the coil so I can't calculate. C1p8 has 500ohms at
    100mhz. Intuitively I can see that the feedback is positive because
    we have two inversions, one at the collector, and one at the emitter.
    How is the modulation performed?
     
  2. Its a function of L, Collector to positive supply, C collector-emitter,
    and the internal capacitance of the transistor Cbe.

    FM is achieved because of the *diffusion* capacitance of the emitter
    base. The Cbe is given by

    Cbe = Cbeo + gm/2.pi.ft

    gm = 40.IC.

    Varying IC by varying Vbe is what causes the capacitance to change.

    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.
     
  3. phil

    phil Guest

    It is funny you should say "No". why not then ?
    I can't see the relevance of capacitances outside of the tuned circuit, the
    tuned circuit will always want to settle on it's resonant frequency because
    then it has the highest impedance and the amplifier as a whole then has
    the highest amplification. (common emitter Av= Rc/Re)
     
  4. Because the circuit is effectively a Colpits oscillator. A colpits
    oscillator is an oscillator that has an inductance from base to
    collector and capacitances from collector to emmiter and emmiter to
    basee. Redra the circuit with the large capacitors s/c.

    We have an LC tank from collector to base. A cap from collector to
    emitter, and the internal Cbe, where it is the Cbe that tunes the
    oscillator. Its impossible for this circuit to oscillate unless the LC
    tank from base to collector is net *inductive*.
    That's because you don't understand how the circuits operates.
    This is nonsense. To achieve oscillation the loop gain must have the
    correct phase. Maximising the gain is irrelevant. Secondly, the tuned
    circuit consists of Cbe and Cce in series, across the inductor.

    The condition for oscilation is:

    gm*(Zce || (Zcb + Zbe)) = Zbe/(Zbe + Zcb)

    Calculated by knowing that the gain is gm.ZL and that Vi=Vo.Zbe/(Zbe +
    Zcb)

    Either Zcb is inductive with Zce and Zbe capacitive - Colpits
    or Zcb is capacitive with Zce and Zbe inductive - Hartley

    There are no other solutions for oscillation to the above equation.

    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.
     
  5. Rich Grise

    Rich Grise Guest

    Why? The tank circuit, you know, includes the transistor's base-
    collector capacitance.

    It's just a "tuned-circuit oscillator" - the transistor's running
    in common-base mode, and the 1p8 is the feedback cap.
    Yes, he already knows it's a function of L. He just said it's a function
    of L. He also said that he doesn't know the inductance of L. That's OK,
    it's not called out, and figuring it out from the physical construction
    of the coil, I'd trust my figures to be within maybe 50% of the actual
    value. If this were a construction article, I'd wind the coil the way
    the guy said to, and take his word for it.
    It explains that right there in the next couple of paragraphs on the
    very page you pointed out. Please finish reading the article, and you
    might not have to ask.
    This is in direct contradiction to what's printed right in the article.

    Have we been dipping into the warden's ale a bit much?

    Cheers!
    Rich
     
  6. I am referring to the *apparent* tank, i.e. the one L and one C.
    Secondly, the *most* important capacitance is missing as it is hidden,
    i.e. Cbe.
    This is misleading, sure its a "tuned circuit", but the impedance that
    is topologically connected from collector to base must be inductive. It
    is a Colpits oscillator. End of story. I already posted the oscillation
    condition:

    gm*(Zce || (Zcb + Zbe)) = Zbe/(Zbe + Zcb)

    in my other post.
    Ho humm... Your point would be?

    This is pointless as the guy writing the article does not know what he
    is talking about.
    Indeed it is. And I care a toss?

    The article is wrong. Period. Indeed I emailed the author to point out
    this fact.

    What part of:

    Did you fail to understand? At 1ma, ft=500mhz, Cbe ~12pf. Without Cbe it
    is simply impossible for the circuit to oscillate. Go and do the math on
    the above equation befor putting you foot in it.

    What part of "the small ac signals on the base emmiter has negligable
    effect on the large rail to rail Vcb swing, hence Cbe", do you not
    understand?

    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.
     
  7. phil

    phil Guest

    I think we need someone to draw the ac equivalent circuit
    to clear things up since I am not sure why it would be
    a colppitz nor why would transistor capacitances effect
    the LC tank circuit (the visible one). Are you all ignoring
    the emitter resistor to achieve this outcome or perhaps
    the base resistor, (47K is rather large I would say).
     
  8. phil

    phil Guest

    ....meanwhile I have understood. This is a common base configuration,
    as someone did point out, therefore the two colppitz capacitors are
    1p8 and Cbe, problem solved.

    I suppose we can then treat the C=6p8 and the L=6t as X and use
    X in the colpittz formula instead of L, is that the right way?
     
  9. phil

    phil Guest

    finally, there is a similar circuit where the 1n capacitor is connected to VCC
    instead of ground. Is there any difference ? See circuit with C to Vcc here:

    http://www.uoguelph.ca/~antoon/circ/fmt1.htm

    Can anyone offer his/her explanation as to how the following
    oscillator works:

    http://members.tripod.com/~transmitters/begin.htm



    I think the tank circuit will oscillate at it resonant frquency. As for
    the positive feedback A x beta = -1 requirement I am not sure
    how much is the coil so I can't calculate. C1p8 has 500ohms at
    100mhz. Intuitively I can see that the feedback is positive because
    we have two inversions, one at the collector, and one at the emitter.
    How is the modulation performed?
     
  10. phil

    phil Guest

    which transistor model is gm used in, the re model ? the h-parameters model
    ?
    I have gm=ICQ/25
     
  11. phil

    phil Guest

    which transistor model is gm used in, the re model ? the h-parameters model
    ?
    I have gm=ICQ/25
     
  12. Topologically, no.

    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.
     
  13. Both.

    hie = rbb' + (1+hfe)re.
    gm = 1/re = I/Vt = 40.Ic

    Vt = KT/q = 25mv at nominal T.

    The "real" model. If you slightly change the voltage, dv, across a
    diode, its current will change, di, by gm.I. where I is the bias. The
    base terminal simply allows one to impress this voltage at the diode
    junction, but without taking the full current, i.e. most of it flows
    into the collector..


    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.
     
  14. phil

    phil Guest

    Something doesn't add up. You say that Vt=25mV at nominal T.
    You also say that gm = I/Vt.
    There is should follow that gm= I/25 mV at nominal T for I in mA.
    Hence for IQ (operating point) gm should be IQ/25 which is
    approximately ICQ/25. Now you are saying 40ICQ, this is not the
    same
    When you say "real" model, I have a problem with that because in
    order to get Vbe you need re, and for re you need beta, and beta
    originates in the approximate model which describes a current
    controlled current source. So it's a catch 22 situation.
     
  15. Ahmmm. Its 25 *millivolts*, 1/.025 = 40.
    Nope. re = 1/gm = 1/40Ic. Beta doesn't come into re.
    Nope. In order to get Vbe, you need Vbe, i.e. Apply a voltage to the
    base emitter. Sure, there is a small drop of rbb'.ib, but it is not that
    significant, usually. Its error is ~ rbb'/hfe.re.

    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.
     
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