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Questions about colpitts oscillator design?

Discussion in 'General Electronics Discussion' started by samy555, Aug 28, 2012.

  1. samy555

    samy555

    63
    0
    May 11, 2010
    [​IMG]

    (1) How much the ratio C2/C1 should be in order to start oscillations?

    (2) If the output signal is taken from emitter through a suitable capacitor, how will be the feedback path to the base if the transistor?

    (3) We all know that there is a stray capacitance came of external connections and the internal capacity of the transistor itself. If I use a 2n3904 transistor which has the following data:
    [​IMG]
    My question is how much capacitance I should add to the series C1 and C2?

    (4) Are there any effects from Cc cap on the oscillation frequency?

    Thanks
     
  2. duke37

    duke37

    5,211
    718
    Jan 9, 2011
    I canot answer you questions wilh certainty but here goes.

    A. The resistors R1 and R2 should be connected to the base. The wires show crossing without connecting.

    B. The loading on the emitter should be small so that the oscillator is not affected.

    C. C1 and C2 will be much higher than the transistor capacitance so you can forget this.

    D. L, Cc, C1 and C2 form the tuned circuit. Cc will be much smaller than C1 and C2 in series and so will be the primary frequency setting capacitance.
     
  3. Sage

    Sage

    7
    0
    Aug 24, 2012
    1)
    The ratio C2/C1 can be anywhere from around 0.1 to 1.0. About 10% (that is, C2/C1>=0.1) is a rough estimate needed to sustain oscillations; excess feedback results in distortion and selection affects the loaded Q.

    2)
    Loading can affect behavior and performance. Perhaps you can rephrase your question if you were asking something different.

    3)
    Add capacitance in series for what purpose? Are you trying to model this circuit or concerned that the internal capacitance will not allow you to operate at your selected frequency?

    4)
    Coupling capacitor Cc prevents DC shorting of R2 through the inductor (thereby nullifying your biasing) and determines the coupling of the resonator to the transistor and affects the loaded Q. In this circuit Cc is in the tank and will have an effect on frequency; you'll need to consider this when you compute Ct.
     
  4. samy555

    samy555

    63
    0
    May 11, 2010
    Ther are connected ti the base

    What do you mean exactly, give me example?

    What if I want 200MHz freq?

    No, Cc will be much more than C1 and C2 in series and so will not be the primary frequency setting capacitance
    Ctotal = [(C1 series C2)// Cstray] series Cc, so if Cc is big enough we can ignore its value.
    Give me your opinion and comment, thanks
     
  5. samy555

    samy555

    63
    0
    May 11, 2010
    Is there even approximate formula to calculate that percentage?
    How that ratio affects the loaded Q?
    I design this circuit
    [​IMG]
    tell me please what to measure to see the effects on the loaded Q?
    For my above design Cc has a very small effect on frequency, here is my calculations:
    [​IMG]
    thank you
     
  6. duke37

    duke37

    5,211
    718
    Jan 9, 2011
    After 50 years, the memory gets a bit dim!
    I looked at the circuit and forgot that this circuit is that of a Clapp oscillator with a small series capacitance. If you are running it as a Colpitts, you should take that capacitor out of the tuned circuit and place it between C1 and the base, then it will only be passing base drive current and will have less loss.

    With your very large C6, the loading will be across C2 which is part of the tuned circuit. Any variation in loading will affect the frequency, if the loading is capacitive, the frequency will be lower. At your frequency I would suggest that C6 should be less than 10pF.

    The Clapp circuit means that C1 and C2 can be made much larger so that variation of the transistor capacitances will have less effect.

    You could try different loading capacitance and see how much the frequency varies.

    I see that the resistors are connected correctly in your latest diagram.
     
  7. samy555

    samy555

    63
    0
    May 11, 2010
    I was trying to imitate this circuit from:
    http://hem.passagen.se/communication/bug_rf.html
    [​IMG]

    When I stand in order to explain it to my colleagues, I expect to ask me
    1 - Why did you choose this ratio for C2/C1 = 0.5?
    2 - How will feedback path from emitter to base?
    3 - How was stray capacitance value calculated using the 2n3904 transistor datasheet?
    4 - How the capacity divider works? i.e how it matching the impedances between low emitter and high base?
    5-What relationship of loaded Q to this ratio (C2/C1)?
    So I need your help so I do not feel overwhelmed
    I do not want to understand how oscillator works , because dozens of books and hundreds of articles talking about it and I'm sick of them
    I want to design one
    I want clear answers and specific, and not the words of not fattening nor of hunger.

    thank you
     
    Last edited: Aug 29, 2012
  8. Sage

    Sage

    7
    0
    Aug 24, 2012
    First, on my tiny screen yesterday your original circuit labels didn't come across; for this setup you generally want C2 larger than C1 (that is, C1/C2>=0.1). This proportionality is not always the case, but is typical; you increase C2 to reduce the amount of feedback.

    The ratio depends on what you're trying to do with the circuit, the requirements, the bandwidth, and so forth; if you decrease the ratio you increase the loaded Q. To compute this you need the source and load impedances for the section in question; transform this into an equivalent parallel circuit and compute the loaded Q via Ql=(w0)C(Rs||Req||Rl); this is susceptance/conductance; a higher loaded Q gives you a narrower bandwidth BW=(w0)/Ql, where w0=1/(sqrt(LC) in a parallel resonator. A large C increases the loaded Q and reduces the bandwidth.

    I'm writing this very quickly because I'm running to a meeting, but I know you have other questions. I'll try to see if I can give you more detailed answers later, but if I can't get back here I'm sure there other people on the forum here who can help (so try to be patient).
     
  9. samy555

    samy555

    63
    0
    May 11, 2010
    ========================
    I think the opposite is true, i.e C1 larger than C2 (that is, C2/C1>=0.1)!!!!
    Do you really mean it or you've made ​​a mistake inadvertently?

    Here I want to explain what I understand, you have to tell me if it is true or false?
    The feedback signal voltage coming from emitter reaching the divider point beyween C1&C2
    If this feedback signal voltage = 400mVac, and C1 = 2 C2 (C1= 20pF and C2=10pF)
    Voltage is distributed in a manner opposite of resistors, i.e the large cap has the lower voltage.
    now the 400mVac is applied across C2, so the voltage across C1 = 0.5*400mVac = 200mVac and the feedback signal

    voltage at the base = 600mVac.

    How did you know that if I decrease the ratio I increase the loaded Q. Where did you look in the circuit?
    How do you calculate it in your head and concluded that the loaded Q increases?

    Thank you for the kind feelings

    No choice but patience

    thanks
     
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