Crossed coupled oscilator

Discussion in 'Homework Help' started by screen1988, Apr 4, 2013.

  1. screen1988

    Thread Starter Member

    Mar 7, 2013
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    Hello, I know this is a silly question but it strikes me again!
    The attached file is crossed coupled oscilator. The circuit oscilates at f= 1/2π√LC . At the frequency the equivalent resistance of R, L, C is equal to R, then how is it possible to create a sin signal at the output as in the picture?
     
  2. t_n_k

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    For the parallel R,L,C network to 'appear' as a purely resistive equivalent requires the entire network to be in resonant condition. The purely resistive condition and resonant condition must therefore be concurrent. They are mutually dependent or more accurately stated : the former depends on the latter.
     
    Last edited: Apr 4, 2013
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  3. #12

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    Another point of view is: Don't worry. It will be square waves in a few seconds if you don't provide a way to limit the amplitude.

    That's probably what Rp is for, but in reality, it's almost impossible to do that across time and temperature with no adjustments or self compensating components.
     
    Last edited: Apr 4, 2013
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  4. t_n_k

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    The circuit may not go into amplitude distortion if correctly designed - without resort to amplitude limitation measures.

    Attached is a 500kHz JFET type I tried in simulation which "works" without apparent distortion.

    Not sure if a simple mosfet type as shown by the OP is all that easy to coax into stable oscillation.
     
  5. screen1988

    Thread Starter Member

    Mar 7, 2013
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    I want to check of I understand what you said correctly.
    I think that of the oscillator work then it will make a periodic wave at the output. If two conditions purely resistive condition and resonant condition happen concurrently then the circuit will not oscillate? I am confused because if this is the case the equivalent of three components R, L, C equal to R and now it is impossible to make a sine wave which is created by L and C.
     
  6. screen1988

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    Mar 7, 2013
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    If there is no resistor Rp, I can see that the circuit will oscillate and its amplitude will be amplify to infinity.
    I think I have a problem here, in the case without Rp charge flows back and forth between two plates of the capacitor and a sine wave will be created between the capacitor. But the circuit oscillate at f=1/2π√LC then equivalent of the two components is zero therefore the voltage between them is zero???
    What I am wrong here?:confused:
     
  7. #12

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    Never mind. screen explained the question well enough.
    Back to T_N_K?
     
  8. t_n_k

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    The conditions are "concurrent" because the circuit is already oscillating at the resonant condition.

    The RLC network can't attain the purely resistive impedance state until the circuit is at resonance. So oscillations must have already built up to enable this condition to be reached. It's unlikely then that the circuit having reached steady state (or quasi-steady state) resonance would then immediately cease oscillation upon achieving a purely resistive state in the RLC network. One would then have some equally disconcerting bi-stable mode of oscillation, where the circuit forever toggles between oscillation and pre-oscillation or non-oscillation - which is a nonsense.

    Note also, that if the circuit shifts even slightly away from resonance then the impedance immediately becomes either capacitively or inductively reactive rather than purely resistive.

    EDIT: It is perhaps likely that with a very low 'Q' RLC network one may have some other oscillation mode in which the circuit cannot attain stable amplitude sinusoidal oscillations. As I implied in an earlier post one has to overcome the problem of getting the circuit to commence oscillation in the first instance - which has little to do with the RLC network appearing to be purely resistive.
     
    Last edited: Apr 4, 2013
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  9. screen1988

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    Mar 7, 2013
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    If the circuit alternatively swich between oscillation and pre-oscillation or non-oscillation then the output wave will be not stable and it is not a sin wave???
    I think it because in the pre-oscillation the amplitude will increase with time and not have sine wave form.
     
  10. t_n_k

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    There is always a process where an oscillator circuit, after initial powering up, progresses from pre-oscillation to steady-state oscillation conditions.

    At this stage I'm unsure whether you are still arguing for your case or not. It's all becoming a bit unclear to me as to where your understanding now lies.

    Another thing to keep in mind about the parallel RLC network at resonance is that there is varying energy storage in both the L & C branches at any instant of time. Therefore it is unwise to simply regard the network as being modeled as a pure resistance without having regard to the energy distribution.
     
    Last edited: Apr 5, 2013
  11. #12

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    Think: What if the oscillator starts to slow down? Some of the energy in the LC circuit will change, instantly. This is analog we're talking about. It won't wait for a clock cycle. It is constantly self correcting. As long as it has enough energy from the power supply, it will self correct to resonance.
     
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