LC series oscillator with DC supply

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SUMMARY

The discussion centers on the behavior of LC series oscillators when a DC supply is applied. The analysis reveals that the circuit oscillates with a voltage amplitude of 2*Vdc, confirmed through Pspice simulation. The transient response of the circuit, when subjected to a step input, generates a broadband frequency pulse that initiates oscillation despite the theoretical expectation of zero current in steady state due to infinite impedance at ω=0. The presence of a DC component on the capacitor leads to permanent oscillation, as the step input signal includes a frequency component at ω=1/sqrt(LC), where impedance is zero.

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  • Understanding of LC circuit theory
  • Familiarity with differential equations in electrical engineering
  • Knowledge of Pspice simulation software
  • Basic principles of impedance and transient analysis
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  • Explore the mathematical modeling of LC circuits using differential equations
  • Learn about the Fourier transform and its application in circuit analysis
  • Investigate the effects of resistance on oscillation in real circuits
  • Study Pspice simulation techniques for analyzing transient responses
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Electrical engineers, circuit designers, and students studying oscillatory circuits and transient analysis will benefit from this discussion.

Shinji83
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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.
 
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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.
 
f95toli said:
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.

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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