# I can't understand parallel LC circuits

#### HarrisonG

Joined Aug 1, 2016
73
Hi. I've been trying for a whole day now to understand what's the deal with a tank circuit where both the Inductive and the Capacitive reactance are equal. I've learned only that the two cancel each other, the total inductance is infinite and the tank circuit acts like an open circuit. And I've also learned that the two currents overlap, cancel eachother and total current through somewhere in the whole circuit(I don't even know where) is zero. What I want to learn is how is that possible? What happens between the inductor and the capacitor? If no current from the power supply can enter the parallel lc circuit, then how will it be supplied with additional electricity to keep the oscillator going?

#### WBahn

Joined Mar 31, 2012
26,305
After the tank circuit is charged by the external supply, it shuttles energy back and forth between the capacitor and the inductor. If there is now resistance, then in theory you could remove the supply and the LC circuit would oscillate forever. But, of course, there is some resistance (not to mention that there will be other components attached to the circuit in any practical application) and so the energy in the tank will drop which will allow the supply to inject more energy into it. Just redo your analysis with a resistance in series with the inductor and you will see that the supply current is no longer zero.

• HarrisonG

Joined Jul 18, 2013
21,636
Hi. I've been trying for a whole day now to understand what's the deal with a tank circuit where both the Inductive and the Capacitive reactance are equal.
What you have is a parallel tuned circuit, when hit with energy it will ring at the resonant frequency, many uses are I.F. transformer in radio reciever, also usfull for detecting a shorted turn in a HF transformer, hit the winding with a high energy pulse and monitor the decay on a 'scope. if the resonant decay is very sudden instead of a linear progression, it points to a shorted turn.
Max.

• HarrisonG

#### BR-549

Joined Sep 22, 2013
4,938
You learned wrong. The reactance is only mathematically cancelled, not physically. If you physically cancel the reactance....it won't oscillate. In reality the reactance is balanced. NOT CANCELLED. You have a left hand and a right hand......they do not cancel one another. poppycock again.

• HarrisonG and #12

#### crutschow

Joined Mar 14, 2008
25,677
I've learned only that the two cancel each other, the total inductance is infinite and the tank circuit acts like an open circuit.
No.
At resonance the inductive reactance cancels the capacitive reactance, but the inductance is fixed and does not change.

• HarrisonG

#### WBahn

Joined Mar 31, 2012
26,305
No.
At resonance the inductive reactance cancels the capacitive reactance, but the inductance is fixed and does not change.
I don't think it's right to say that they cancel (in parallel resonance). That description clearly applies when they are in series and the total reactance goes to zero. But in parallel the reactance goes singular (to infinity). Although this is because the terms cancel in the denominator, I just don't know that describing the overall result as them canceling out is good because "canceling out" strongly implies that the only two reactances in the circuit make each other go away when in fact the reactance goes through the roof.

• HarrisonG

#### MrAl

Joined Jun 17, 2014
7,849
Hi. I've been trying for a whole day now to understand what's the deal with a tank circuit where both the Inductive and the Capacitive reactance are equal. I've learned only that the two cancel each other, the total inductance is infinite and the tank circuit acts like an open circuit. And I've also learned that the two currents overlap, cancel eachother and total current through somewhere in the whole circuit(I don't even know where) is zero. What I want to learn is how is that possible? What happens between the inductor and the capacitor? If no current from the power supply can enter the parallel lc circuit, then how will it be supplied with additional electricity to keep the oscillator going?
Hi,

I think you meant to say that the total impedance is infinite, or goes toward infinity.

If we drive the LC parallel circuit with a current source of 1 amp at the resonant frequency and analyze the circuit in time we get:
v(t)=(t/(2*C))*sin(w*t)

and with C=1/2 Farad we end up with:
v(t)=t*sin(w*t)

So with constant sinusoidal source at the resonant frequency we see an ever increasing sine voltage across the tank which means that the impedance must be approaching infinity.

The usual explanation for what happens physically is that if the inductor and capacitor are both idea components then they exchange energy over and over again. The inductor passes it to the cap, the cap passes it back, over and over again, and there are intermediate times when each has some of the energy but not all. To see this behavior in a simulator simply graph the two energies:
EL=(1/2)*L*i^2
EC=(1/2)*C*v^2

where 'i' is the current through the inductor and 'v' is the voltage across the capacitor.

• HarrisonG

#### HarrisonG

Joined Aug 1, 2016
73
Thank you all for the answers. I'm sorry I didn't reply earlier but I had no time.

#### MrAl

Joined Jun 17, 2014
7,849
Thank you all for the answers. I'm sorry I didn't reply earlier but I had no time.
Hi,

Oh ok I meant to also mention that one system mechanical analogy is a spring and mass where the spring supports the mass vertically in zero gravity or horizontally with zero friction.
The initial stretch of the spring contains energy and that is transferred to the mass on release, then the spring gets energy back from the mass when it compresses, then transfers it back to the mass when it expands again, then the mass transfers it back to the spring, etc., etc., and on and on forever with no energy dissipator in the picture. If there is even a tiny amount of friction it will stop eventually, but theoretically without any friction it will never stop.

• HarrisonG

#### HarrisonG

Joined Aug 1, 2016
73
By the way, if you don't mind could you answer this too: So in an article in this website about Resonance it says that "Resonance occurs when when capacitive and inductive reactances are equal to each other" and this got me thinking what if Xc is not equal to Xl, would it still oscillate at one particular frequency based on what Xc and Xl are? And if so would that frequency still be a Resonant frequency?

#### crutschow

Joined Mar 14, 2008
25,677
what if Xc is not equal to Xl, would it still oscillate at one particular frequency based on what Xc and Xl are?
If the two are not equal, then you are not at the resonant frequency.
If will oscillate at the resonant frequency where Xc equals Xl.
There is always a frequency when the two are equal, and that is where it resonates.

• HarrisonG and Sinus23

#### BR-549

Joined Sep 22, 2013
4,938
Yes and yes. It's how analog radio circuits are tuned.

#### MrAl

Joined Jun 17, 2014
7,849
By the way, if you don't mind could you answer this too: So in an article in this website about Resonance it says that "Resonance occurs when when capacitive and inductive reactances are equal to each other" and this got me thinking what if Xc is not equal to Xl, would it still oscillate at one particular frequency based on what Xc and Xl are? And if so would that frequency still be a Resonant frequency?
Hello again,

As you can see by the previous two posts and sorry to have to say, there is a common misconception about resonance. That is the misconception that there is only one type of resonance. The reason this idea comes into view sometimes is that usually in a circuit only ONE type of resonance is used as the main operating principle and physical resonance is the one that is usually talked about in textbooks.

There are actually three types of resonance, and even resonant peaking in a radio circuit does not have to follow physical resonance. The three resonant types are related to Cassini Ovals but you'd have to look that up to find out that relationship and that was pointed out some years ago by a researcher.

As noted, one type is physical resonance, and that is where XL=XC and is more often used when efficiency is important.
Another type is where the voltage or current response reaches a maximum, also known as "peaking".

There is a pdf file out there on the web somewhere that talks about the three types and their relation to Cassini Ovals but i havent seen it for some time now so you'll have to do a search if you want to read it. The main idea though is that something about the circuit reaches a maximum or possibly a minimum depending on how you look at it (such as with impedance).

• HarrisonG

#### KL7AJ

Joined Nov 4, 2008
2,226
Hi. I've been trying for a whole day now to understand what's the deal with a tank circuit where both the Inductive and the Capacitive reactance are equal. I've learned only that the two cancel each other, the total inductance is infinite and the tank circuit acts like an open circuit. And I've also learned that the two currents overlap, cancel eachother and total current through somewhere in the whole circuit(I don't even know where) is zero. What I want to learn is how is that possible? What happens between the inductor and the capacitor? If no current from the power supply can enter the parallel lc circuit, then how will it be supplied with additional electricity to keep the oscillator going?
I recently wrote what I believe is the definitive answer to this. Enjoy!
Eric

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• HarrisonG

#### KL7AJ

Joined Nov 4, 2008
2,226
I recently wrote what I believe is the definitive answer to this. Enjoy!
Eric
Try this one...the other one was missing the graphics. #### Attachments

• 193.9 KB Views: 23
• HarrisonG and Sinus23

#### HarrisonG

Joined Aug 1, 2016
73
Thanks folks! KL7AJ that's a great work, thank you!

#### Janis59

Joined Aug 21, 2017
1,232
RE:""If no current from the power supply can enter the parallel lc circuit, then how will it be supplied with additional electricity to keep the oscillator going?""
1)Just energy is CIRCULATING from L to C, from C to L and again and again.
2)Full compensation is mathematical fiction. In the sorrow praxis the cancellation happens only partial - as the Q-factor of real schematics may stand between 20 and 100, and only very partial cases the 200 or slightly more. That means - (in case of Q=100) the losses will be 1/100=1%, what must be added in each cycle to sustain the same amplitude.
3)The small battery powered fed parallel tank may have to circulate in it kiloamperes, the same way that serial tank fed by mosfet gate voltage (10-20 V) may easily give out the several kilovolt voltage. This is named a voltage multiplication (of Q-fold) for serial tank and current multiplication for parallel tank.
4)The resistance causing the Q may be easily re-calculated from serial representaton to parallel representation. If serial parasythic resistance is drawn in series with L and has value of miliohms typically, then its equivalent parallel to coil analog has value typically megaohms or it parts.

• HarrisonG

#### HarrisonG

Joined Aug 1, 2016
73
Hi again it's me. I don't like to look on this like spamming, but rather Understanding better . So I took all your words in mind and I have pretty good idea what's happening. Thank you for that. Can we say that in an ideal tank circuit since the real resistance is 0, and the reactances balance each other, there is no opposition to the current flow and that's why tank circuits would oscillate forever. Whereas if the reactance was not 0, then that would impede the flow and eventually cause it to stop?

#### MrAl

Joined Jun 17, 2014
7,849
Hi,

In a word, "yes". Oh wait that's four words now, geeze • HarrisonG

#### KL7AJ

Joined Nov 4, 2008
2,226
Hi again it's me. I don't like to look on this like spamming, but rather Understanding better . So I took all your words in mind and I have pretty good idea what's happening. Thank you for that. Can we say that in an ideal tank circuit since the real resistance is 0, and the reactances balance each other, there is no opposition to the current flow and that's why tank circuits would oscillate forever. Whereas if the reactance was not 0, then that would impede the flow and eventually cause it to stop?
Yes, that would be true.

• HarrisonG