Colpitts Oscillator Analysis

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  • Опубліковано 8 жов 2024

КОМЕНТАРІ • 17

  • @XiaolongHuang-y1b
    @XiaolongHuang-y1b 8 місяців тому

    It helps a lot. Easy to remember.

  • @Spottswoode
    @Spottswoode Рік тому

    Hey Jordan. Great video. Really helped me with my homework! I can see how this took 5 hrs lol.

  • @prateeksingh5718
    @prateeksingh5718 4 роки тому

    Thanks a lot. Just saw one lecture and got confused because he had taken - R smaller than R's for oscillations to start.

  • @leomurillo9108
    @leomurillo9108 Рік тому

    Your explanation was excellent. What book do you use to know these applications?

  • @danialmoghaddam8698
    @danialmoghaddam8698 4 роки тому +1

    Spectacular

  • @cayroo
    @cayroo 2 роки тому

    bro, that was really nice, thx

    • @JordanEdmundsEECS
      @JordanEdmundsEECS  2 роки тому +1

      bro, thx to u for the compliment 😎 Anything specific you liked?

  • @Shiny_Mewtwo
    @Shiny_Mewtwo 3 роки тому

    Thank you.

  • @luisilichvladimirguerrerol2321
    @luisilichvladimirguerrerol2321 2 роки тому

    Great. Thanks

  • @taimuongtean
    @taimuongtean 2 роки тому

    Just subscribed teacher🙂

  • @raingerliu319
    @raingerliu319 3 роки тому

    but where you draw an output? this analysis overlooks the output.

    • @JordanEdmundsEECS
      @JordanEdmundsEECS  3 роки тому +1

      Yeah that's the trick. You basically just pick any node and then attach a buffer to it (a very high input impedance amplifier, like a common source or the noninverting input of an opamp) so as not to disturb the oscillator.

  • @koggism
    @koggism 6 років тому +1

    Your analysis does not make sense. You mention the modules of -R, but R was defined in the first circuit as a resistor it has a real positive value, you say Rs is not a real resistor but R is negative, implying R is imaginary with a negative resistance and Rs is real with and positive. Then you introduce these new terms and I don't have a clue how they relate to the first circuit diagram you drew. Then at some point you start using the terms R,L,C1 and C2 as in the original circuit. But more confusing is your conclusion, when you say in order for the circuit to oscillate gm must = [(C1+C2)^2/C1C2]1/R. Well you are saying gm is equal to [(C1+C2)^2/C1C2]1/R then it must be so for ANY value of C1, C2 and R because you defined it as such.

    • @JordanEdmundsEECS
      @JordanEdmundsEECS  6 років тому +2

      Hey Paul, thanks for your feedback! I realize now there's a lot of notation floating around. Rs and Rp are meant to demonstrate the transformation relation between a series resistor and inductor and a parallel resistor and inductor. In this circuit, we have a parallel resistor and inductor which we want to 'transform' to a series inductor and resistor.
      R is the actual (circuit) resistance, and C1 and C2 are the (actual) known capacitor values. Setting gm equal to that mess of terms is the condition for oscillation. I did not mean that gm equals that by definition, but *if* that happens to be the case, the circuit will oscillate. Typically, you will design the circuit by choosing C1, C2, and gm such that this condition holds. Hope this helps, my apologies for the confusion.

  • @muppetpaster
    @muppetpaster 3 роки тому

    I hope I spelled his name correct??? How hard is it to look it up?? Or .....you just don't give a shit.......

    • @JordanEdmundsEECS
      @JordanEdmundsEECS  2 роки тому +1

      Easy enough, but when recording live lectures (as this one) such things can slip by.