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    How oscillations start in the oscillator?

    Let's take Colpitts oscillator for example. How does the oscillations start?

    Imagine that we have simple LC circuit. We charge the capacitor with a battery and then disconnect the battery. When we disconnect the battery the oscillations will start. In colpitts oscillator, the tank circuit is always connected to +V. That's the thing that confuses me. In LC circuit, the oscillations will only start if you remove power supply that you used to charge the capacitor.

    This is the LC circuit that I am talking about http://www.falstad.com/circuit/e-lrc.html.

    Can somebody explain what happens right then when we turn on Colpitts oscillator? Also, why is the tank circuit capacitor in Colpitts oscillator "split" into a 2 separate capacitors?
    Last edited by I14R10; 11th April 2016 at 01:16.

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    Re: How oscillations start in the oscillator?

    The LRC circuit (which opens automatically in Falstad's) is a mere ringing circuit. You admit a burst of current, which causes it to swing back and forth a few times. However it has no amplification or feedback, therefore it does not sustain oscillations.

    The Colpitts is an oscillator because it has feedback and an amplifying stage. The action is adjusted so that it gives a kick during each cycle in the LC (or LCC) tank circuit.

    I have collected a few oscillator circuits built around a single LC tank. They are often difficult to get going and to keep going. I believe the reason is because the LC tank does not provide a voltage swing which is easy to use at the precise moment, when it could turn on a transistor and admit the necessary small burst of current at the moment it is needed.

    The Colpitts has the split capacitor, which provides a node where the voltage swing rises at the precise moment, when it turns on a transistor at the precise moment, etc.


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    Re: How oscillations start in the oscillator?

    I understand how the oscillator works. Voltage is taken from the collector and fed to the base with a phase shift of 180 degrees. But, something has to start those oscillations?

    Does that mean that LC tank circuit is never "standing still"? Is it always oscillating, no matter how little energy it has? That would explain how the transistor could amplify those oscillations. They were always there and it just amplifies them, it doesn't start them.
    Last edited by I14R10; 11th April 2016 at 06:59.



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    Re: How oscillations start in the oscillator?

    The usual text book explanation says that electronic noise provides the energy to start the oscillations, see https://en.wikipedia.org/wiki/Electr...ack_oscillator

    In most real oscillator circuits, the power-up transients achieve a faster oscillation start than nano- or microvolt noise level ever can. But if you design an oscillator circuit with perfect DC balance and no initial transients, noise would still start the oscillation. An electronic simulator that doesn't model circuit noise often needs an explicit oscillator kick-start.


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    Re: How oscillations start in the oscillator?

    Each working oscillator must fulfill Barkhausens oscillation condition (loop gain>0 dB). In this case, existing oscillations are self-sustained.
    How and why can an oscillator start?
    In some documents we can read that the always present noise contains a frequency component identical to the oscillation frequency.
    Because the loop gain for this frequency is larger than unity (0 dB) the amplitude would be continuously rising (build-up rocess).
    This would be true, if there wouldn`t be a switch-on transient which also contains the oscillation frequency component.
    This phenomenon always dominates the noise.
    Therefore, because every oscillator needs a supply voltage which is switched-on at a certain time it is this switch-on process which allows a safe start of oscillations.
    Last edited by LvW; 11th April 2016 at 07:33.


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    Re: How oscillations start in the oscillator?

    Quote Originally Posted by I14R10 View Post
    Can somebody explain what happens right then when we turn on Colpitts oscillator? Also, why is the tank circuit capacitor in Colpitts oscillator "split" into a 2 separate capacitors?
    It is best to explain with a mechanical analog: weight hanging on a spring. You pull the weight a bit, the spring is stretched and energy is stored in the spring. Next you release the weight and oscillations continue.

    Because of frictions, the amplitude keeps on decreasing and you will get what is called a damped oscillations. The energy is lost and the motion ceases.

    To keep the oscillations going, you need to replace the lost energy. You *need* to supply the energy at the right time - like a pulse - with the right sense (phase). If the external pulses are of the right frequency, the system will vibrate with its natural frequency. If the external pulse frequency is not correct frequency, then we will get a forced vibration. If you are familiar with the dispersion and absorption of energy, we can now see the electrical analog.

    You are right when you say that a tank circuit will not oscillate if the two ends are tied (V and ground). So we need to relax one- the V is allowed to change a bit and then we can have oscillation. We also need a split capacitor or inductor so that the we know when to put the voltage *kick* so that the oscillations can be sustained.

    It is interesting to study how the phase changes and what happens if you give a big *kick*.


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    Re: How oscillations start in the oscillator?

    Thanks for the explanation, as I said, I understand the principle of oscillation, I have done a differential equations for LRC circuit and solution for that differential equation.

    Another question - how much does the supply voltage has to change in order to allow the oscillations? Take that example from falstad simulator. Could we have oscillations if we reduced the voltage by 1V ?



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    Re: How oscillations start in the oscillator?

    Quote Originally Posted by I14R10 View Post
    how much does the supply voltage has to change in order to allow the oscillations? Take that example from falstad simulator. Could we have oscillations if we reduced the voltage by 1V ?
    Exactly like the case of the mechanical oscillation: you pull the weight by a amount and then let it go. If you do not let it go, it will refuse to vibrate. Afterwards, you need to apply force in a periodic manner and the external periodic force should match with the natural force of vibration of the spring system.

    Any change in the supply voltage should start the oscillation. The condition is that the change must be sufficiently rapid compared to the natural vibration of the system. In electrical terms, it means that the rapid change in the supply voltage will produce a potential in the LC system and the oscillations will appear on the top of the new voltage.

    - - - Updated - - -

    I love the mechanical analogy: if you tightly hold the weight and move it to a new position, that does not give rise to oscillations. In electrical terms, it is tightly coupled.

    If you allow the weight to slip a bit (mow much is much??) and loosen your grip, you will see that it is trying to vibrate. If you study the transfer of energy you will find more similarity between the two.



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    Re: How oscillations start in the oscillator?

    Another question - how much does the supply voltage has to change in order to allow the oscillations? Take that example from falstad simulator. Could we have oscillations if we reduced the voltage by 1V ?
    I don't see an oscillator in the falstad circuit, only an LC resonant circuit. It needs a feedback amplifier to generate sustained oscillations. Minimal supply voltage is a matter of the amplifier properties. You can e.g. operate a JFET transistor based oscillator with 100 mV or less supply voltage.



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    Re: How oscillations start in the oscillator?

    FvM: I accidentally swapped the terms, but I know the difference between those two (oscillator and resonant circuit).

    c_mitra: so, in electrical oscillator, will there be oscillations if the supply voltage suddenly drops very fast to, for example, half of what it was?



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    Re: How oscillations start in the oscillator?

    I accidentally swapped the terms.
    O.k., but what are you asking? The linear RLC circuit works with any voltage, volts or microvolt.



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    Re: How oscillations start in the oscillator?

    Quote Originally Posted by FvM View Post
    O.k., but what are you asking? The linear RLC circuit works with any voltage, volts or microvolt.
    You already answered me in post #4.
    If you are referring to my question to c_mitra, I asked if you have to remove the supply voltage in order to have oscillations in RLC circuit?



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    Re: How oscillations start in the oscillator?

    Quote Originally Posted by I14R10 View Post
    c_mitra: so, in electrical oscillator, will there be oscillations if the supply voltage suddenly drops very fast to, for example, half of what it was?
    Sure.

    But if the supply voltage is stiff, it will show only a damped movement to the new voltage.

    I hope you understand that the oscillations will be damped because of loss of energy and no feedback, and the appearances can be interesting.

    Depending on the coupling, it will move to the new voltage with or without oscillations.

    - - - Updated - - -

    Now we can take up the falstad circuit. Will it work without the 100R resistor with the power supply? Convince yourself that this resistor is responsible for the loose coupling of the power supply.

    Also remember that the equations have a trivial solution with all the voltages being zero.

    The "Q" of the RLC circuit is determined mostly by the R (10R in the pic). What will happen if you swap 10 and 100 R resistors?



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