Emitter Coupled Oscillator

  1. Hello folks,

    I've been trying to understand how crystals work in crystal oscillator circuits. I understand the piezoelectric effect to the following extent: If we apply an electric field to the crystal it will deform and when the field is removed, the crystal will generate an electric field in the opposite direction. I get stuck at the point where the electric field is removed. I can't seem to understand how the electric field is removed in crystal oscillator circuits. I've attached a diagram of a crystal oscillator circuit. I'm looking for a qualitative analysis of the circuit without using the RLC model of a crystal. I'm trying to understand what exactly happens to the crystal during oscillation. What I'm looking for in terms of an analysis, is an explanation similar to the following one: base voltage of Q1 increases, which increases voltage at the emitter which increases current through R1, etc, etc, etc, around the loop.

    Thanks.
     

    Attached Files:

  2. jcsd
  3. Baluncore

    Baluncore 2,514
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    A crystal is an insulator so it can be operated with a permanent bias voltage.
    The electric field is not really removed, it is cyclically reversed about the average bias point.
     
  4. The key to the oscillation is the capacitor. When you apply power to the circuit, current flows into the base of Q1 which causes V(Q1c) to drop. The rate of change of that voltage is what passes through the capacitor, not the voltage itself. Eventually the voltage at Q1c stops dropping and at that point the base of Q2 stops seeing the negative voltage from the capacitor. V(Q2c) now starts to fall, lowering V(Q1b), lowering the voltage across the crystal and raising V(Q1c) which causes V(Q2c) to increase even more.
     
  5. sophiecentaur

    sophiecentaur 13,384
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    Gold Member

    With the crystal in the emitter, the gain of the stage goes to a maximum at the series resonance frequency so the (positive) feedback C supports the oscillation.
     
  6. For an even more generic analysis... think of the crystal as a freq-variable emitter degeneration resistor. At the crystal freq, it permits (ac) current flow, at all other frequencies it presents a high impedance and through negative feedback (emitter degeneration) it blocks Q1 from amplifying.
     
  7. sophiecentaur

    sophiecentaur 13,384
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    The point about a crystal is that it has an extremely narrow frequency band where the impedance goes very low. At other frequencies (+/- 0.0001%) it looks like either a very small Capacitor or a very large Inductor. (This is just another way of saying it has a very narrow resonance bandwidth.)
     
  8. sophiecentaur

    sophiecentaur 13,384
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    Reading that through again, I think it's the wrong way round. The gain of the first stage is -0.8 off resonance as the crystal is high impedance and the emitter resistor is R1. It then goes to whatever β lets it be, at resonance.
     
  9. Baluncore

    Baluncore 2,514
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    Correct. The frequency of oscillation will not be at the crystal's resonant frequency because at resonance the crystal impedance will be a maximum, so circuit gain will be a minimum. The oscillation frequency will therefore be offset slightly onto the shoulder of the resonance where the crystal behaves more like a phase shifting trap. For that reason this design of oscillator will not be frequency stable.
     
  10. sophiecentaur

    sophiecentaur 13,384
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    Crystals have two resonances - series and parallel. I know that oscillator circuits can be confusing (and it's been a long time. . . . .) but why do you assume the crystal is operating in its parallel mode? The maximum gain will be when the resonance is series and the emitter load is nearly zero. I am not arguing - just asking.
     
  11. I went through this exact process but had difficulty seeing how it would BEGIN oscillating. The moment when power is applied, what exactly happens to the crystal? (i.e. Does it begin deforming? Does it fight back with an opposite E field?)
     
  12. Baluncore

    Baluncore 2,514
    Science Advisor

    Sophiecentaur; Voltage gain is not the only criteria. Phase is also important. A crystal in that circuit is most unlikely to ever operate at either it's parallel or it's series resonant frequency. It will slide to some operating point part way between the two, hence the poor frequency stability.

    I have a theory that the vast majority of oscillators are named after unsuccessful amplifier inventors who, in their search for the Holy Grail of a perfect amplifier, experimented with feedback. They find that there is always some frequency where they cannot control the regeneration.
    There are a great many eponymous oscillators, yet very few named amplifier configurations.
     
  13. sophiecentaur

    sophiecentaur 13,384
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    Yep And when you try to design an oscillator, it can often turn out to be no more than an amplifier. (Or go off at some other, unexpected frequency)

    I take your point about the crystal. I will stop worrying about it. Suffice to say that the open loop gain is small in magnitude and there will be a frequency for which the phase around the loop is right to make it oscillate. Unlike in a 'proper' crystal oscillator, the feedback capacitor value becomes more relevant than the presence of the crystal might suggest.
     
  14. Yes, looking at it again, and assuming series resonance... which apparently we can't?
     
  15. Baluncore

    Baluncore 2,514
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    The circuit in the OP is from; CRYSTAL OSCILLATOR CIRCUITS. Revised Edition, 1992, by Robert J. Matthys.

    The title "MODIFIED MEACHAM" on the diagram refers to discussion of a different design beginning at the bottom of page 53.

    The text on page 53 that accompanies the diagram reads as follows:
     
  16. sophiecentaur

    sophiecentaur 13,384
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    Hiyo Silver away. :smile:
     
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