Colpitt Oscillator Baffeling Question

Discussion in 'General Electronics Chat' started by KCHARROIS, Jul 4, 2012.

  1. KCHARROIS

    Thread Starter Member

    Jun 29, 2012
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    Hi I've been analyzing colpitt's oscillator for quite a bit and theres a question no one seems to be able to answer, hopefully one of you can.

    If you look at the circuit they say that capacitor C1 will charge and discharge in L. But how could C1 discharge if there is constant DC going to it from the supply?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Any stored DC charge on C1 &/or C2 would quickly discharge via the inductor L. Only the base coupling feedback capacitor would see any steady state DC charge.
     
  3. Sensacell

    Well-Known Member

    Jun 19, 2012
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    In the circuit shown, there is no DC path through L, it's isolated by the capacitors.
     
  4. KCHARROIS

    Thread Starter Member

    Jun 29, 2012
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    So what your saying is the capacitor near the base of the transistor blocks DC path from going through the inductor?
     
  5. KCHARROIS

    Thread Starter Member

    Jun 29, 2012
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    But the supply voltage is still going towards C1 even once its charged.
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Trace the DC pathway from the +Vcc supply point.....

    Low resistance path via the RFC -> Low resistance path via L -> base feedback capacitor blocks the path. C1 & C2 are effectively in parallel with the low resistance path through L. The non-earthed ends of C1 & C2 are effectively at the same DC potential [the collector DC voltage ≈ Vcc]. Hence C1 & C2 can't charge to a steady state value.

    Sorry - my mistake. Disregard most of the previous rubbish from me. C1 & C2 will have an equal DC potential difference because of the earth point at the capacitors and the bridging inductance L.

    It's possible for an AC component to be superimposed on the DC value - which is what would happen during oscillation. Presumably this AC component is what the explanation is referring to as the charge & discharge of C1 via L.

    I'll head off to bed now. I obviously need sleep.
     
    Last edited: Jul 4, 2012
  7. KCHARROIS

    Thread Starter Member

    Jun 29, 2012
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    What do you mean by superimposed?
     
  8. t_n_k

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    Suppose the inductor has several volts peak-to-peak across its terminals when oscillation is established. That same inductor AC voltage is across C1 & C2 in series. If C1 & C2 are equal they will have almost equivalent AC magnitude values offset by [or superimposed on] an average DC value of around +Vcc. VC1 & VC2 would have equal magnitude AC components, in anti-phase, but with equal DC offsets.
     
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  9. KCHARROIS

    Thread Starter Member

    Jun 29, 2012
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    Thanks for the replies the circuit makes a little more sense then what it did before.
     
  10. upand_at_them

    Active Member

    May 15, 2010
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  11. KCHARROIS

    Thread Starter Member

    Jun 29, 2012
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    Ok now that I understand how that works how do you DC bias the circuit such as which valued resistors to use and the gain an such.
     
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