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Understanding a Colpitts oscillator

  1. Feb 4, 2015 #1

    DrDu

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    As I was playing around with a small AM oscillator, I thought it would be time to understand how they actually work.
    I am referring to the 3rd picture on this page (base mode):
    http://elektroniktutor.oszkim.de/signalkunde/colpitts.html
    Given the working point of the transistor, I describe its small signal behaviour by two relations: ##U_{BE}=rI_B## and ##I_{CE}=B\; I_B## which should be the easiest model one can think off.
    Now using the schematic, I find
    ## I_{CE}/I_B=r[(1/r+1/R_E+1/(i\omega L))(1+C_1/C_2) +i\omega C_1)]##.
    This ratio should equal B, which represents two conditions:
    a) From the vanishing of the imaginary part, I find Thompsons formula for the frequency.
    b) The real part doesn't depend on ##\omega##. Hence I suppose we fulfill it varying the ratio of ##C_1## to ##C_2##.

    Now my questions:
    c) What happens, if the real part condition isn't fulfilled? Physical oscillators seem to work over a wide range of ratios of the capacities. I suppose one has to consider a more realistic model of a transistor. What would be the most important generalization?
    d) As the small signal analysis is linear in all the currents, one can't see why an arbitrary fluctuation gets amplified. I suppose this is mainly an effect of the nonlinear behaviour of ##I_B## on ##U_{BE}##?
    What would be the easiest way to take this into account?

    Thank you!
     
    Last edited: Feb 5, 2015
  2. jcsd
  3. Feb 4, 2015 #2

    LvW

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    I am not quite sure what your problem really is.
    Do you want to know how the oscillator works in principle?
     
  4. Feb 4, 2015 #3

    DrDu

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    Yes. I also wanted to get and idea how the components have to be dimensioned. I thought, a small signal analysis would be a good starting point.
     
  5. Feb 4, 2015 #4

    LvW

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    Well - it`s relatively easy to describe the oscillation principle: Between collector and the pos. supply (signal ground) we have a LC resonant circuit (L||C) withC=C1C2/(C1+C2).
    A certain percentage (feedback factor k) is fed back to the emitter. This resembles positive feedback (as required for an oscillator). For self-sustained oscillation we require that the openloop gain Ao>1/k. For analyzing the circuit (verification of the oscillation condition) it would be best to simulate the LOOP GAIN to show that the oscillation condition Aloop=Ao*k>1 is fulfilled. However, for the present case this task is somewhat difficult because there is no point in the loop which could be opened without disturbing the load conditions. (For opamp circuits this procedure is simple because the opamp output presents a low-resistive node).

    For a rough simulation you could try the following: Open the loop at the emitter node and inject a test signal (Vac=1V) into the emitter. At the same time, load the feedback node (between C1 and C2) with a resistor which mirrors the (disconnected) small-signal emitter input resistor with a value 1k||(1/g). The transconductance g is determined by the quiescent dc currrent Ic (g=Ic/Vt). Displaying the ac output voltage at the node between C1 and C2 gives you the loop gain (because 1V input).
    The oscillation condition requires that there must be one frequency wo with zero phase shift and a gain magnitude slightly above unity.
     
  6. Feb 5, 2015 #5

    DrDu

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    Thank you for your interesting reply!
    Having slept over my problem, I now have found the answer to my original problems: I was assuming ##\omega## to be real. Hence there will really be only a solution for a special feedback factor, which is the minimal feedback to support an oscillation. If I allow for complex frequencies, I can find solutions for all values of the feedback ratio C1/C2. if the feedback is too small, then the exponential will be dying, if it is larger than the minimal value, an exponential increasing solution is found.

    So setting ##\omega=\Omega-i\lambda##, i find
    ##\Omega^2+\lambda^2=1/(LC)##, which generalizes the Thomson equation (##C=(1/C_1+1/C_2)^{-1}##), and
    ##\lambda=1/(2rC_1) (B-(1+r/R_E)(1+C_1/C_2))##.
     
    Last edited: Feb 5, 2015
  7. Feb 5, 2015 #6

    LvW

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    I must admit that I do not understand the background of this "Ansatz". Is it equivalent to using the classical Laplace variable (s=sigma+jw) ?
    More than that, what is the meaning of the solution for "lamda" ?
     
  8. Feb 5, 2015 #7

    DrDu

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    Just plug it into the time dependence: ##\exp(i\omega t)=\exp(i\Omega t) \exp(\lambda t)##. I.e., depending on the sign of ##\lambda##, the solutions found are exponentially increasing or decreasing, the inverse of ##\lambda## defining the rising time.
    If ##\lambda<0## the oscillator won't oscillate, if ##\lambda>0## a small signal will increase with time, until the realm of small signal theory is left and the signal is limited by other factors. This is also an example of a Lyapunov stability analysis.
     
  9. Feb 5, 2015 #8

    LvW

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    Yes - that makes sense. However, in your post#5 it was λ that was made imaginary!. Therefore my question.
    Now you are in accordance with s=sigma+jw.
     
  10. Feb 5, 2015 #9

    DrDu

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    I don't see what you mean: I use ##\omega=\Omega-i\lambda## as in #5. So ##\exp(i(\Omega-i\lambda)t)=\exp(i\Omega t)\exp(\lambda t)##,
     
  11. Feb 5, 2015 #10

    LvW

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    OK - for my opinion, it was a bit confusing to have an imaginary quantity jw and to replace w by another complex variable.
    But this gives the same result as the classical way: Extend the real frequency variable w by a complex variable s=sigma+jw.
    Of course, this leads to an exponenetially increasing amplitude in case if sigma>0 (in your case lamda>0).
    No problem - just a confusion (on my side) caused by an uncommon way of deriving the result.

    Comment: By the way, you can proof you result for lamda with the help of the loop gain simulation (ac analysis). For lamda>0 the loop gain magnitude must be > 0db (for zero phase) and the gain must be<0 dB for lamda<0.
     
  12. Feb 5, 2015 #11

    DrDu

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    Now that I understood the high frequency part: What are the criteria to set the working point of the transistor, i.e. the resistors R1, R2 and RE? What is unusual for me, in comparison with other amplifier circuits, is that the collector is at U source without a collector resistor.
     
  13. Feb 5, 2015 #12

    LvW

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    In short:
    1.) Select a suitable Ic (mA range);
    2.) Select RE=1k (following the example in the ref. document);
    3.) Calculate VE=Ic*RE and design the base voltage divider to produce a voltahge of app. (VE+0.65)V.
    For calculation of both base resistors take the base current IB=IC/beta into consideration.
    Hint: If allowed, use resistors for the base divider in the lower kohm range (not as large as in the document).
    _____________
    A collector resistor RC is not necessary because the LC tank realizes a dynamic resistance.
     
  14. Feb 5, 2015 #13

    DrDu

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    But that's not possible! If the oscillator is not oscillating VE=VC=U source. Hence it is not possible to chose VB above VE.
     
  15. Feb 5, 2015 #14

    LvW

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    I don`t understand your reply. In your post#11 you spoke about the DC operating popint - and I gave you some hints regarding DC values.
    What is "not possible"???
     
  16. Feb 5, 2015 #15

    DrDu

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    Sorry, I see. While at the DC operating point, VC=V source, this does not mean that VE has to be VC.
     
  17. Feb 6, 2015 #16

    DrDu

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    From what I understand now, we would choose the working point so as to maximize the amplitude of oscillation. While the DC potential of the collector equals the source potential, the DC potentials of the base and emitter can be chosen.
    I think a reasonable condition is that the transistor works always in the normal region, hence ##U_C^\mathrm{min}>U_B>U_E^\mathrm{max}##. Where the appearing potentials are the DC+AC minima and maxima, respectively. As a lowest order approximation, one could replace the > by =.
    We know the relative AC amplitudes of the emitter and collector. So the DC ##U_B## and ##U_E## can be determined. It would be interesting to learn how to work out the precise conditions using the characteristics diagrams of the transistor.
     
  18. Feb 6, 2015 #17

    LvW

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    No we cannot.
    Always we require Vc>Vb (to keep the CB diode always reverse biased) and Vb>Ve (for "linear" operation of the BJT).
    I am afraid, you are mixing DC and AC values (potentials). DC values are chosen (and realized by the various resistors) independent on AC values, unless you have certain requirements regarding output amplitudes. In my post#12 I gave you some help for biasing the BJT properly.
     
  19. Feb 6, 2015 #18

    DrDu

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    So based on what consideration do I chose a "suitable" IC? And why should I accept 1k as a reasonable value for RE?
     
  20. Feb 6, 2015 #19

    LvW

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    In electronics, everything is a trade-off. A larger Ic (let`s say 20mA) allows large gain values (higher transconductance), but causes corresponding DC voltage drops across the resistors Rc resp. Re. This - in turn - requires larger supply voltages and/or power transistors.. Therefore, for most of the small-signal applications some milliamps are suitable values for Ic.
    Moreover, the DC stabilizing resistor Re should be (a) as large as possible and (b) should cause a DC voltage that is not more than (10..20)% of the supply voltage. Both requirements together result in Re values between several hundreds of Ohms and some (low) kOhms.
    As I have mentioned - it is a trade-off, which results in one possible solution - out of 99 other design alternatives.

    The same applies to the voltage divider providing the base bias:
    (a) It should be as low-resistive as possible. Only then it creates a "stiff" base voltage which depends only a little on the DC base current Ib , which is temperature-sensitive and has relatively large tolerances. For some rough calculations, we even can neglect the base current and it´s influence on these two resistors.
    (b) On the other hand, a chain of small resistors requires a larger power consumption; moreover, it degrades the input resistance at the base node (an undesired property).
    Hence, again a trade-off is necessary between conflicting requirements - resulting in mid-range kOhm values.

    Does this help?
     
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