Author Topic: LC oscillator  (Read 602 times)

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Offline 00Topic starter

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LC oscillator
« on: July 28, 2020, 09:49:01 am »
I know that LC oscillators don't run forever so in circuit they are connected to a power supply and preamplifier etc.
Can someone explain why the oscillators still oscillate even after attaching power supply? Why doesn't all the current flow through inductor and away or why capacitor isn't getting charged only one way? How capacitor discharging in reverse can still oppose power supply?
 

Offline TimFox

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Re: LC oscillator
« Reply #1 on: July 28, 2020, 01:25:37 pm »
A "normal" LC oscillator starts by amplifying noise (which, like the poor, is always with us).  The resonant circuit selects the frequency and the signal grows (due to positive feedback) until it reaches the equilibrium level of oscillation (set by the amplitude stabilizing method, which varies from design to design).  Another starting mechanism is the shock of applying power to the circuit, which starts an oscillatory voltage  in the LC circuit. 
Unless it doesn't:  a corollary of Murphy's Law states that "self-starting oscillators won't".
One method of limiting the oscillation is merely when the voltage swing reaches the maximum allowed by the supply voltage.  Another is a reduction in loop gain as the average bias on the active element increases.  Sometimes the output is rectified and used as a control voltage in a DC feedback circuit to the bias voltage.  A useful oscillator requires both a frequency-determining network (LC in this case) and an amplitude-determining network, since the oscillation, left to its own devices (sic) would increase without bound in the simple mathematical analysis, and that could never happen.
« Last Edit: July 28, 2020, 01:28:16 pm by TimFox »
 
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Offline ledtester

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Re: LC oscillator
« Reply #2 on: July 28, 2020, 01:42:18 pm »
The physical equivalent to an LC tank circuit is a swinging pendulum or a mass vibrating on the end of a spring.

A real pendulum will eventually come to rest due to friction and that's why a pendulum based clock needs to kick it every so often to keep it moving.

In an LC-tank energy is lost due to resistances that are inherent in real inductors and capacitors. With ideal components the energy would slosh around between the inductor and capacitor forever like in this video:

https://youtu.be/5Fg8Nqsh3QE&t=25s

(Note that in this simulation the LC tank is not connected to the power supply except at the very start to give it an initial impulse.)

But if you add just a little energy to the tank circuit at the right time (timing is important -- exactly how a clock spring is used to give an impulse to a pendulum at the bottom of the swing) you can easily sustain the oscillation.
 
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Offline TimFox

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Re: LC oscillator
« Reply #3 on: July 28, 2020, 03:50:25 pm »
The positive feedback in an oscillator circuit gives the energy addition to the L-C "tank circuit" at the right timing with respect to the oscillating waveform.  The power required to add the energy comes from the power supply to the oscillator circuit.  Depending on the circuit, the energy losses in the circuit can include external loss from the active devices and output load, as well as the inherent loss from the inductor and capacitor.  An alternate way to understand an oscillator circuit is to treat the active devices as a negative resistance in parallel with the tank circuit:  "transitron" and tunnel-diode oscillators are examples.
The general solution to the oscillation in a tank circuit is
V(t) = A x eat x cos (wt)
The decaying oscillation found without external power applies is for a < 0.  If a > 0, the oscillation would increase without bound, but the physical situation does not permit infinite voltage and the practical solution forces a to 0 by clipping or feedback.
 
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