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analysis of hartley oscillator

Discussion in 'Electronic Design' started by Ira Rubinson, Jun 12, 2005.

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  1. Ira Rubinson

    Ira Rubinson Guest

  2. Robert Baer

    Robert Baer Guest

    "Analyze"? Doesn't the simulation help??
    In any event, i think it would be better if you put the tank in the
    emitter-base instead of the collector-base configuration (emitter goes to tap).
    Wire the circuit in the grounded collector configuration.
    Then tie the collector to a low-Z tap of a second tank circuit and lightly
    couple an output follower to the top of that tank.
    In this configuration, any change of loading at the output (evenwithout the
    follower) will have no effect on the oscillator.
     
  3. If you are worried about the waveform distortion, the secret is to tweak the
    bias circuit to reduce collector current until the output comes clean. I
    would bypass the emitter resistor with a capacitor, then increase the
    emitter resistor until the output comes clean. You can do this in your
    simularion or in the real world.

    The amount of *gain* around the loop has little effect on distortion, as
    long as you have enough to make it oscillate. Rather, the DC collector
    current sets how much swing you get across the impedance of the tank
    circuit. What distorts the waveform is collector to emitter saturation,
    where the transistor becomes a low impedance, heavily shunting the tuned
    circuit, thus preventing the tuned circuit from producing a clean waveform
    and badly degrading frequency stability. A clean oscillator "runs out of
    current" before it hits saturation - in rough terms, the transistor puts
    current pulses into the tank circuit - something like V = I x Z .

    Wes Haywards "Introduction to RF Design" covers this with as a nice mix of
    theory and test bench tweaking.

    Interestingly,
     
  4. Ira Rubinson

    Ira Rubinson Guest

    Thanks Robert, Roger,
    I'll try implementing some of your suggestions.
    What I meant by analyze is to come up with an equation that will show that
    it indeed oscillates.
    For example, what circuit analysis technique does the simulation program use
    to derive the equation for the waveform.
    I assume that what I'm looking for will involve kirchoff loops, locus root,
    small signal parameters etc....
    My problem is that when I read about locus root, it is all math and I don't
    know how to apply it to this simple circuit.
    When I read about hartley oscillators I usually see "freq =
    1/(2*pi*sqrt(LC)) which is a little simplistic.

    Thanks -Ira
     
  5. The simulation program does non-linear analysis, so it does not work with
    equations you would use for root locus with small signal parameters.
    Instead, it tracks the movement of charge at small steps in time, at each
    step iterating to get a solution which satisfies the equations describing
    each circuit component, including the equations for the non-linear
    transistor. If it was linear analysis, you would only see sinewaves.

    Again, I recommend Wes Haywards "Introduction to RF Design. He will present
    you with equations, but also discusses the result you want to achieve. In
    the end, oscillator design means trading off many requirements, noise,
    amplitude, power, drift. It turns out that establishing that you have
    correct loop gain and phase is the easy part, at least below UHF. Haywood
    uses surprisingly simple transistor models which you will find give you a
    feel for the issues. He also has a Smith Chart design method for the higher
    frequencies. The book has a bit of everything, and will get you going in
    the RF world.

    If you want to, draw the hybrid pi model of the transistor, add the other
    components, then break the loop and calculate gain and phase around the
    loop. You will find this analysis in plenty of Uni library textbooks. It
    is useless for designing good oscillators at the 3 or 4 MHz where your
    oscillator functions, because those equations only tell you that the
    oscillator will start, and nothing about its performance.

    Feedback phase does not have to be perfectly 180 degrees at resonance,
    because oscillation can still occur at a frequency somewhat away from
    resonance, where the tuned circuit gives enough phase shift to give make up
    the exact 180 degrees. This is why oscillators like yours mostly work, and
    why the linear ?will it start? analysis is not much help.

    Your oscillator is not a linear device, once it gets started. The
    transistor puts spikes of current into the tank circuit and nothing the rest
    of the time.

    Reduce DC current until you are free of clipping and you will have a decent
    oscillator. Apart from avoiding clipping, the other issue is the impedance
    presented by the tuned circuit to the transistor base and collector. A
    lower impedance presented to the collector lets you run at a higher
    collector current, storing more energy in the tank circuit and swamping
    thermal and transistor noise. A lower impedance presented to the base is
    associated with less feedback, so it can't be too low, but lower means the
    base loads the tuned circuit less and allows higher Q, giving a cleaner
    sinewave.

    L1 and L2 are mutually coupled in a Harley circuit, and since L1=L2 in your
    circuit, there is very heavy feedback, meaning your oscillator can't fail to
    work. So reduce L1 and increase L2 until it stops oscillating, then back
    off. Then bypass R5 with a capacitor and increase R5 until clipping stops.


    Roger
     
  6. Roger Lascelles wrote:


    You appear to be stating here that mutual coupling is a required feature
    of a Hartley oscillator. If you are, this isnt correct. The basic
    Hartley doesn't have or require mutual coupling between its inductors.

    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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