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LC series oscillator with DC supply

  1. Aug 21, 2015 #1
    Hi, I have a question about LC series oscillators. Specifically when a DC supply is applied.
    Solving the differential equation for such circuit using as initial conditions at t=0 that Vc=0 and I=0 I get as a solution that the circuit oscillates and the voltage across the capacitor has an amplitude of 2*Vdc.
    Solution is:
    Vdc*(1-cos (t/sqrt (LC))
    This is also confirmed using Pspice simulation.

    Now my doubt is that if I consider the circuit impedance j (ωL -1/ωC) of course for ω=0 it goes to infinite which means that with a step signal applied to the curcuit (starting from zero state at t=0) after a transient I should get zero current in steady state. The circuit is not an oscillator but acts like an open circuit in steady conditions according to circuit analysis theory.

    So what's happening? Thanks.
  2. jcsd
  3. Aug 21, 2015 #2


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    Science Advisor
    Gold Member

    What is the resistance of your circuit?
    If you suddenly turn on a DC source you do not really have a static situation, the transient in fact a broadband frequency pulse and that is what is causing your oscillator to get going.
    In any real circuit this will be rapidly damped out by the resistance.
  4. Aug 21, 2015 #3
    Thank you for your reply.
    I was considering the ideal case with R=0.
    But you gave me the right hint and I think that I get it now.
    Impedance at ω=0 (ideal costant signal of infinite duration) is infinite and gives me a DC component on the capacitor.
    The transient in this case doesn't go to zero on steady state but it's a permanent oscillation because the step input signal has a fourier transform with a component at ω=1/sqrt (LC) and at that frequency impedence is zero and the circuit oscillates.
    So on the capacitor I get an oscillation on top of a DC component which is what I was expecting.
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