coupled LC resonance circuit

Discussion in 'General Electronics Discussion' started by toyonline, Jan 17, 2014.

1. toyonline

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Jan 17, 2014
Hi, I am working on fabrication of a coupled LC resonance circuit. It is like in attachment. From simulation, I know there are two resonant frequencies f1 and f2. Change capacitor C3 will give different ratio of f1:f2. From the simulation, I know as C3 approaches to 78 pFarad, the ratio is approaching 1:2, which is what I want.

But still I have some problems. How could I figure out a physical explanation of the existance of double frequencies in this coupled LC resonance circuit? Is it any qualitative way to 'predict' those two frequencies? Or at least a explanation of tendency of the ratio change under different capacitance and inductance.

The only thing I could understand is, if C3 approaches infinity, then C3 will behave like a open part, which will change the overall circuit as if C3 is not exist. Then the circuit could be understand as a simple series LC circuit, which possessing a single frequency. Am I right?

2. toyonline

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Jan 17, 2014
Here is the circuit attachment.

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3. Laplace

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Apr 4, 2010
Are you asking how to construct a Bode plot for this circuit. See attachment.

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4. toyonline

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Jan 17, 2014
Thank you so much.

The rational deduction is amazing. Could you tell me how do you do that?

Frankly, I understand little on that. For a LC circuit, is there any more intuitive way to illustrate? I mean, by a qualitative way, if one could understand why there are two frequencies rather than one in simple LC circuit.

Last edited: Jan 17, 2014
5. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
One simple way of looking at it is that there are two LC circuits connected together.

You would expect that if each one was resonant at different frequencies that the result would contain some artifact of that.

6. toyonline

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Jan 17, 2014
That's true. And more interestingly, at those two frequencies, the oscillation of the coupled circuit are either the same direction or opposed to each other. These two interesting oscillation modes could be analog to 'normal oscillation mode' of a coupled pendulum. If that holds more generally, one could expect to predict numbers of frequencies of LC circuit by numbers of 'normal mode'. I dont know if my explanation is reasonable and correct. Or is there any other qualitative interpretation instead?

7. CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
Personally, I think demonstrating the effects of two resonant circuits that are separated by close frequencies is best demonstrated and spiced using two parallel tanks that are loosely coupled to each other. It will produce the classic "bandpass" response curve where both resonant frequencies can easily be seen. Circuit "Q" can also be seen this way by varying the load resistance.

Chris

8. toyonline

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Jan 17, 2014

A coupled LC will behave as a bandpass filter in the situation of a unsplitted resonant frequencies. On the other hand, a frequency band of a circuit would also be changed to a number of splitted frequencies, which will be useful in magnification of several frequencies.

I will try to see if I could, by trial and error, modify the coupled LC circuit to produce three resonant frequencies.

9. CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
Where did you find that circuit?

Chris

10. toyonline

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Jan 17, 2014
From a journal publication. I am interested in fabrication a voltage magnification device by the circuit. The problem is I have no experience of doing that.

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11. CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
If this is an online journal please post the link. I want to see which node the author is using for an output. Are you picking the signal off across R1?

Chris

12. toyonline

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Jan 17, 2014
Exactly, the author use signal across R1 for voltage magnification. Here is the linkage for that circuit. I dont know if you can retrieve it.

http://www.sciencedirect.com/science/article/pii/S1044030510001133

13. CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
Thanks for posting the link. The purpose of C4 (0.1uF) makes much more sense now. It's a DC blocking cap when DC bias is injected into the system.

Have you spiced Fourier (spectrum analysis) or swept AC Analysis yet?

Chris

Edit: I can't find any reference to the value of R1 that says "See Note 1".

Last edited: Jan 20, 2014
14. toyonline

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Jan 17, 2014
For the DC part, I think both F1 and C4 (0.1uF, the original one in the publication is C7) are involved to filter high frequency signal. I guess this was due to extremely low reactance of both capacitor under high frequency signal, which means the DC part will mainly go through parallel resistor R3 and inductor L3 and L4. Am I right?

Since the output Vpp is around 6kV, the resistance of R1 was estimated to be 440 kohm (10Watt). But in simulation, I found ~1mohm gives a sharper peak (better for resonance purpose?).

I have conducted AC analysis of the circuit. A Bode plot is attached along with the circuit used for simulation. Two frequencies were observed in the circuit. One thing that confusing me is whatever C3 I used (within 30 pF), the frequency ratio of two resonance frequencies is not 1:2, which is key parameter in for my purpose. One thing to improve is adding some parallel capacitor C4, e.g. 27pF, to obtain a 1:2 frequency ratio.

Also I found other components such as series capacitor and inductor will affect that frequency ratio.

I have tried to rationale my observation on those components influence on frequency ratio. What I figured out is changing either inductance or capacitance will produce different impendance and inherent resonant frequency of either series or parallel resonant section. And therefore will affect the frequency ratio.

One correction: the estimated resistance should be the boundary condition, i.e. R1 should be larger than 440 kohm.

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15. CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012

Your circuit does not reflect the values used by the author. All comments and component names below refer to the authors schematic.

I converted all series and paralleled components to their equivalent (effective) values. I varied C3 from 20pF.- 30pf. This produced only relatively small changes in my plots.

When two resonant circuits are coupled in this manner it forms a much more complex circuit, where each resonant block is highly dependent on the other.

My AC analysis results actually indicate 3 distinct points of resonance, (one negative going and two positive going) which doesn't surprise me at all. Change your sweep settings to logarithmic from 10Hz to 10.0MHz You will then see all three very sharp points of resonance.

I'd like to also mention that I don't understand the authors use of an RF Linear Amp as referenced in his design. This circuit works.. IE, generates very high output voltage because it's virtually unloaded on the output. The author is connecting the linear to the input of the circuit with a length of RG8 coax that's not terminated into 50Ohms on the input of the circuit. The output Z of the circuit is far to high to reflect any load back to the input that would look anything like 50Ohms.

This creates two extremely undesirable issues. The first being an unloaded linear is an unhappy linear and a good idea only if your doing stress analysis!

The second is an un-terminated transmission line is no longer a transmission line. It's a length of cable that looks like lumped capacitance, inductance and resistance. Without a 50Ohm termination on the input a 3 ft length of RG8 will produce very different results than a 6 ft. length. In other words we usually don't want our transmission line to become part of any circuit for any purpose other than being an RF pipe!

Chris

16. toyonline

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Jan 17, 2014
Thanks for you detailed explanation.

Yes, actually three resonant frequencies are observed, either in the original publication setup or the modified one. Two positive plus one negative. But for my purpose, the negative one does not magnify voltage and thus is useless. Varying C3 from 20pF to 30pF or even 80pF does not change the overall characteristics of those frequencies, but the only difference is the ratio of two positive frequencies shift slightly, which is a useful parameter for my purpose.

The purpose of linear amp is to amplify a RF signal from a special waveform generator. I am not sure whether a direct connection between that waveform generator with the circuit will work properly. This is a problem I need to take some thought.

I think that RG8 cable is not used for the connection between linear amp with the circuit, instead it was used in between the circuit and the downstream output, for signal transmission. But I do not understand your comments on impedance of RG8 coax. I have never consider that. Thanks for your pointing out that.

I am not sure whether my explanations to your comments are right?

1. An unloaded output with RG8 cable is actually loading combinations of capacitors and inductors of that cable. The total impendance of RG8 cable is around 50 Ohm. Since the impendance is so small, there will be large reflection effect from the RG8, which affects the circuit.

2. A loaded output with RG8 cable will be better, only if the loading impedance is far more than 50 Ohm. That larger impedance will produce negligible reflection effect. Therefore, the characteristic of the original LC circuit will be unchanged.

By the way, would you like to provide me more hints on stress analysis? I don't know anything of that.

17. CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
Electronics, like most scientific fields of study, is made up of subsets, that in their own right can be viewed as their own field of study. RF Theory & Design is definitely a valid subset. Within the RF subset, Antenna / Transmission Line Theory & Design could and does fill textbooks. Fortunately, you can access all the information you can absorb by doing a Google search.

That said here are some transmission line tidbits:

(1) The reason 50Ω was chosen as the characteristic impedance of coax used in radio communications is because a quarter wave vertical ground plane antenna presents a natural 50Ω load impedance. It is also the simplest of all antennas and requires no matching networks at all.

(2) The term "Transmission Line Characteristic Impedance" is the most misunderstood and misapplied term in the electronics lexicon. I would venture a guess that only 1 out of every 100 people that think they understand it actually do! I've actually seen posts where the OP was trying to measure it with an Ohmmeter!

(3) I've forgotten more about transmission lines than I remember.

(4) Other than loss per foot; Transmission lines should play a transparent role but they can only do this when properly terminated at both ends. When it isn't it becomes something other than a transmission line. IE it's no longer transparent to the circuit driving it or the circuit it's piping to.

(5) Transmission lines are analogous to water pipes. Ideally we don't want them to leak at all. Except for resistive and dielectric loss (which we can't eliminate) we want all the RF fed into them to come out the other end. A mismatched load or source will cause the transmission to (radiate) leak. If the mismatch is severe enough it will resemble an antenna more than a transmission line.

(6) In a classroom environment there are few test setups that demonstrate the idiosyncrasies of transmission lines better than (TDR) Time Domain Reflectometry.

Linear Amplifier:

Yes, I understand that the author is using a linear as a driver but that circuit and the device it drives does not present a proper load for it. Since the ionizer does not appear to dissipate any real power it stands to reason that a real load won't be reflected back to the linear. Linear's or for that matter, any RF power amp does not like this and can destroy either. A linear with Fold-Back may protect it from destruction but would also prevent full power output. The only thing I see dissipating power is R1 @ <10W.

Hey, I'm not suggesting that his circuit doesn't work. I'm just concerned that there's no mention of the above.

Stress Analysis:

These are tests that manufacturers perform much like the auto industry does crash tests. All components have voltage, current, frequency, temperature and humidity maximums. Some may even have atmospheric pressure, vibration, shock impact maximums spec'd out. These are not tests that the home shop engineer gets involved in because it invariably results in the destruction of the (DUT) Device Under Test.

Chris

18. davennModerator

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Sep 5, 2009
No if the load is anything other than the characteristic impedance of the cable, there will be reflection of power back to the transmitter. ie. the worse the missmatch, the moer power that is reflected.

not really correct, Chris. actually has nothing to do with the load on the end of the cable
The quick answer is that 50 ohms is a great compromise between power handling and low loss, for air-dielectric coax. This means between 30 ohms (best power handling) and 77 ohms (lowest loss) is 53.5, the geometric mean is 48 ohms. Thus the choice of 50 ohms is a compromise between power handling capability and signal loss per unit length, for air dielectric.

cheers
Dave

Last edited: Jan 22, 2014
19. CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
I'll take your word for it Dave but it seems like quite a big coincidence that a 1/4 wave ground plane has a natural 50 Ohm input Z.

By the way this stuff wasn't always standardized. I have a very old 1932 signal generator made by General Radio for the U.S. Navy. It originally had a 10 Ohm output Z, yet I've never seen a 10 ohm coax.

Chris

20. CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
Hey Dave, I investigated it a bit further. This link expands a bit on what you posted. http://en.wikipedia.org/wiki/Coaxial_cable#Choice_of_impedance

I want you to know that this revelation really sucks for me. I sure am glad that we didn't have internet forums for the entirety of the last 48 years!. After all, I've spent those years preaching the 1/4 wave ground plane theory!

Chris