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Series resonant circuit help

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chadj2

New member
Jul 5, 2007
22
Hi I am new to the forum. I have been studying and recently doing experiments with (LC) series resonant circuits at home. I have been referencing numerous electrical engineering books for information on series resonance, but unfortunately I have been unable to find any books that break the phenomenon down to the most basic level as to what is really going on in a series resonant circuit to reduce impedance to to zero. I have seen the vectors and the math but that really doesn't explain to me how for example the capacitor is able to interefere with the inductor's ability to control current rise in itself. All I can do is theorize that the capacitor's voltage rises at the same rate as the inductor's reactive voltage which results in no voltage drop across the the LC network. If anyone has had it explained to them or understands it more deeply and is willing to share the information please respond.

Thanks
 
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If you think that the vectors do not explain it, then I do not know how to explain it to you. Complex algebra. Would that be better for you? Vectors, math, algebra there's not much left, really.



Gunnar Englund
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
 
If you think series LC resonance (Lo-Z) is a mind bender, then try parallel LC resonance (Hi-Z).

 
How about: Reactance (X) can be either positive (+X) or negative (-X). In series, they add (+). Thus: +X + -X = 0.

 
I can understand parallel resonance a lot better than series resonance. When I say I would like to have a deeper understanding of series resonance I mean to take for example one 12 volt pulse of power. If I had a inductor at a value of 1 henry and 3 ohms it would take aprox 333 milliseconds to build up a current of 4 amps through it. Now when I add a capacitor of 0.001 uF in series behind the inductor the current builds to 4 amps in approx 100 microseconds. what I am trying to understand is how the capacitor enables the current to build to its full amount in the inductor in a shorter time than if the inductor was by itself. I speculated that it had something to do with the capacitors voltage rise since I know that a inductor controls current flow through itself by inducing a voltage to counter the current that is rising in itself. If the capacitor is rising in its voltage at the same rate that the inductor is inducing the counter voltage (back emf) wouldnt that mean there is no voltage drop across the two components?

thanks
 
I am sorry about a miscalculation I did in the above post. In the first inductor example it would take current 1.665 seconds to rise to 4 amps.
 
Are you clear on the difference between transient response and steady-state ac response?
 
Consider using a sine wave rather than a pulse. The current through both the cap and inductor is the same. The voltage across the cap will lag the current by 90 degrees, the voltage across the inductor will lead the current by 90 degrees. Thus, the voltage across the cap and inductor will be 180 degrees out of phase. When at resonance, the magnitudes of the voltages will be the same, thus cancelling out and causing a net voltage of 0.
 
dpc,

I am not really clear on the difference between the two definitions. I have played around with parallel resonant circuits and was able to drive them with pulsating dc at the resonant frequency. My example about was based on the calculated time constant for current to rise to maximum through the inductor. I assumed that current through the inductor rises to maximum in series resonance in just a shorter period of time but I may be mistaken.
 
geekee,

Thanks for your response. I have read that explanation before the the cap and inductor work together to bring current and voltage into phase. I may be on a pointless quest. I was just hoping to learn more deeply what happens in that small instant of time when current flow though the inductor is rising much faster than it would without a capacitor in series, or how it is that the capacitor is appearently able to interfere with the inductor's back emf. I mess around with inductors and capacitors a lot as a hobby. thanks again
 
The problem is the difference between a steady state response to AC and a transient response to a switch-on (for example). You can be assured that AC transients are very hard to think about so electronics engineers tend to not think about them too much! The answer is to simulate the circuit with SPICE and you can then plot out the voltage and current waveforms. This may help to explain it without using maths.
 
logbook,

That maybe the direction I need to explore next. thanks
 
My suggestion is to first analyze the transient and steady-state response to a dc input. The steady-state portion of the dc case is trivial, but that won't be the case for ac.

The ac case will also have the same second-order differential equation to solve with both a transient and steady-state component. The standard descriptions of resonant LC circuits for ac systems generally only deal with the ac steady-state case.

Actually, a pretty good reference source is the ARRL's Radio Amateur's Handbook - just about any edition. There is a pretty good section on basic ac and rf theory explained without need for calculus. You should be able to find a recent edition at a local library.
 
dpc,

Thanks for the suggestion. I happen to have the 2006 version of the arrl handbook. I must not have looked close enough for a good explanation. Thanks again to taking the time to respond.
 
A spring with a mass connected to the spring is an equivalent to a series LC circuit (Spring = C, Mass = L). Put driving functions into the spring support and observe what happens.
 
Dear chadj2,

If I had understood that you did not even understand the difference between transient DC and steady state AC! Then I had redflagged this thread as inappropriate at once instead of letting so many people try to make you understand.

Your last post "That maybe the direction I need to explore" says it all.

It may not be somthing that you need to explore. These things are well explored and well understood by most engineers that has taken basic electricty at school. You are an engineer? Aren't you?

Gunnar Englund
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
 
Skogsgurra,

I am terribly sorry for wasting your time. I would not have posted if I had known that my question would upset some members. I am not an electrical engineer, I just really love electronics and talking about it. Once again sorry.

 
Basic electricity is included in all engineering schools - not only EE schools. So, you should know about this.

The problem is not that you waste my time. The problem is that threads like these make people wonder about the level of engineering as such and if the Eng-Tips is for hobbyists or professionals. This has recently been discussed in Pat's Pub and other places.

Gunnar Englund
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
 
Well, not so much "discussed" as "ranted about" it by me for a few precious minutes before it was yanked ;-)


Dan - Owner
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There are also other discussions going on. It is a policy thing. Besides, there are hobbyist sites.

Gunnar Englund
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
 
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