Charging a tuned L/C circuit

hitsfmdj

Jul 13, 2016
7
Hello, Just a silly question as I am learning here. Could any one provide me with an easy to understand explanation of how an inductor connected to a capacitor in parallel. Can charge the capacitor if essentially your shorting out the capacitor? Because all the experiments I done when I want to discharge and keep it discharge, I simply short the capacitor? Does this have to do with some fancy filtering going on because of the tuned L/C resonating frequency that if charged at the right input frequency the capacitor will not be seen in the resonating circuit as shorted? Thanks all!

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duke37

Jan 9, 2011
5,364
A pendulum swings from side to side. At one extreme, the bob is a little higher and has potential energy. When it is swinging in the centre, it has kinetic energy. This interchange of energy continues until friction damps the oscillation.

In the case of a tuned circuit, energy is interchanged between the inductor (E=I*I*L/2) and capacitor (E=V*V*C/2) until resistance damps the oscillation. Shorting the circuit dissipates the energy quickly so there is no voltage on the capacitor and no current in the inductor. Charging at the right frequency with an escapement in the case of a clock or an active device in the case of an electronic circuit makes an oscillator where the energy lost per cycle is replaced by the energy from the power source.

hitsfmdj

Jul 13, 2016
7
A pendulum swings from side to side. At one extreme, the bob is a little higher and has potential energy. When it is swinging in the centre, it has kinetic energy. This interchange of energy continues until friction damps the oscillation.

In the case of a tuned circuit, energy is interchanged between the inductor (E=I*I*L/2) and capacitor (E=V*V*C/2) until resistance damps the oscillation. Shorting the circuit dissipates the energy quickly so there is no voltage on the capacitor and no current in the inductor. Charging at the right frequency with an escapement in the case of a clock or an active device in the case of an electronic circuit makes an oscillator where the energy lost per cycle is replaced by the energy from the power source.

So even just a simple L/C circuit will eventually stop oscillating back and forth if a nearby field stops feeding it at the tuned frequency. Even if there is no physical resistor added into this circuit as the connecting wires and impedance value of the inducer etc.. the circuit will have a small resistance value of its own correct?

hevans1944

Hop - AC8NS
Jun 21, 2012
4,891
Could any one provide me with an easy to understand explanation of how an inducer connected to a capacitor in parallel.
I do not know what an inducer might be, nor is the quoted sentence an actual question, although it does begin with the interrogative phrase, "Could anyone provide me..."
Electronics is an exact science and the spelling and meaning of words are critical to understanding. Is it possible that you meant inductor instead of inducer?

Can charge the capacitor if essentially your shorting out the capacitor?
You cannot charge a capacitor that is shorted. An inductor is not a short. An inductor has electrical properties of inductance and resistance as well as distributed capacitance among turns of the winding, but it is never a short. Even a short length of straight wire is never really a short because it has inductance, resistance, and distributed capacitance from end to end. You might need microwave or greater frequencies to appreciate this. A short is a connection with insignificant resistance, capacitance, and inductance. A short with zero resistance, inductance or capacitance does not exist in the real world, but some things can come close.

Because all the experiments I done when I want to discharge and keep it discharge, I simply short the capacitor?
Well, whatever floats your boat. Or keeps your capacitor safe. I have seen large capacitors rated for high voltages (kilovolts) stored with wire connecting their two terminals, presumably to prevent accidental charging of the capacitor.

Does this have to do with some fancy filtering going on because of the tuned l/c resonating frequency that if charged at the right input frequency the capacitor will not be seen in the resonating circuit as shorted?
Yeah, it's pretty fancy alright. James Clerk Maxwell discovered some fancy equations that pretty much describes what's going on. They turned out to be so useful that the set of four equations were named after him. How's that for receiving rock-star status in all things electrical? In your parallel-connected L/C circuit, at the right frequency the two terminals appear to connected to an open circuit. But they are not. At the right frequency, electrical energy stored in an electrostatic field of the capacitor, and electrical energy stored in a magnetostatic field of the inductor, transfers back and forth between these two components. No components disappear or act shorted as a result. The capacitor does not appear as shorted in the resonating circuit, nor does the inductor. Some energy is lost in resistance and some may be radiated, so at resonance the lost energy must somehow be replaced or the oscillations will decay and vanish.

Hello, Just a silly question as I am learning here.
There are no silly questions, but there may be plenty of silly answers (depending on who you ask). Welcome to Electronics Point @hitsfmdj. Ask all the questions you want.

hitsfmdj

Jul 13, 2016
7
Yes I am so sorry that was a typo, I did mean to say induction coil, English is not my main language, Sometimes I get it wrong with speech to text even!

So could I rectify this energy within the tuned L/C circuit by the help of a simple diode and capacitor tapping into one side of the L/C circuit and I am taking another guess here but using a real earth ground as ground - and get some DC? assuming of course some field nearby is feeding the L/C circuit at the proper frequency? more or less an AC to DC converter?

Thank you!

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davenn

Moderator
Sep 5, 2009
14,300
So could I rectify this energy within the tuned L/C circuit by the help of a simple diode and capacitor tapping into one side of the L/C circuit and I am taking another guess here but using a real earth ground as ground - and get some DC?

yes
this is what is done with a radio receiver. Take the most basic one, a crystal set radio

cheers
Dave

hitsfmdj

Jul 13, 2016
7
I was just going to get to that, Let's say I want to rectify the 50kw 105.1 FM signal from less then 1/4 mile away, Where would I place the antenna? I guess trail and error using multi taps on the coil? And is one really needed, An Antenna if the circuit already is resonant? I read a story of a guy who used an over sized AM transmitter coil and without an antenna was able to power a small speaker to audible volume. Crystal radio style.
Thanks!

hevans1944

Hop - AC8NS
Jun 21, 2012
4,891
It all depends on the area of your antenna that intercepts the transmitted radiation. Bigger and closer to the source of radiation is better.

hitsfmdj

Jul 13, 2016
7
I was thinking more about where the antenna connects to the induction coil, I'm sorry , Probably my bad terminology again.

hevans1944

Hop - AC8NS
Jun 21, 2012
4,891
That's a function of matching impedances for maximum power transfer. Not usually something you can easily do at 105.1 Megahertz. An L/C combination resonant at that frequency is very tiny... a few picofarads of capacitance and a few turns of wire on the coil. Good luck trying to couple significant energy into that from an antenna.

hitsfmdj

Jul 13, 2016
7
I already did at 170 mhz VHF commercial band, Works not bad with just a tiny coil of a few turns and was getting voltage spikes as two way radio nearby where transmitting. But that don't really impress me. I am more interested in the low frequencies, I built something already but I am not 100% sure what I am doing but I do observe an effect, My L/C meter shows a frequency of around 180 hertz , Don't ask me what is around here in this small town at that frequency but there is enough of it with an antenna, I used a wire out the window horizontal about 100 feet length and fed it to one of the sides of what used to be a tesla coil secondary as part of my L/C circuit and used a bunch of capacitors to try and bring the frequency down as much as I could with the parts I got here. So once I rectify it, I get a solid 25 volts DC taped into the antenna side of the high inductance coil with the help of my capacitors tuned at around 180 hertz. Even Burt out a few LEDs with it! I was just hoping to come on here and understand better what is going on here and you have answered many question very well, Thank you so much.

I would like to take this idea one more step and somehow short this tuned l/c circuit in sync with the input frequency, The sorting of a tuned l/c and have maybe a secondary low winding low impedance pick up coil and rectify those discharge spikes to get even more potential out of it? like current wise?

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duke37

Jan 9, 2011
5,364
If you look at the crystal set circuit, the aerial and diode are tapped into the circuit at different places. The idea is to match the aerial to the tuned circuit so that maximum energy is transferred and the diode load tap is chosen to get a compromise between selectivity and sensitivity..

Your FM transmitter will not give any useful output since it is at a constant level. You need a complicated circuit to demodulate it.

You are in Canada with a mains frequency of 60Hz. If this wave is distorted you get harmonics. The third harmonic is at 180Hz. Possibly due to switch mode power supplies

hitsfmdj

Jul 13, 2016
7
If you look at the crystal set circuit, the aerial and diode are tapped into the circuit at different places. The idea is to match the aerial to the tuned circuit so that maximum energy is transferred and the diode load tap is chosen to get a compromise between selectivity and sensitivity..

Your FM transmitter will not give any useful output since it is at a constant level. You need a complicated circuit to demodulate it.

You are in Canada with a mains frequency of 60Hz. If this wave is distorted you get harmonics. The third harmonic is at 180Hz. Possibly due to switch mode power supplies

You probably hit it dead on,

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