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Impedance of a Resonator

  1. Dec 23, 2008 #1
    I am trying to find the impedance of a resonator, but I am not quite sure where to begin. The resonator contains three LCL circuits which are then connected to a large capacitor.

    Should I calculate the impedance of each element in the resonator or should I use Kirchhoff's Laws to analyze the circuits piece by piece?

    The latter strategy raises another issue because the three circuits are attached in such a way that a wire (really a strip of copper foil) is connected right in the middle of an inductor, so I am not sure how to take that into account.

    Any help would be much appreciated and I can provide more information as well.

  2. jcsd
  3. Dec 24, 2008 #2


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    Both: when you apply Kirchoff's laws, you'll need to insert the impedance of each element. I can't follow your comment about a copper strip "in the middle". Is the inductor center-tapped (in which case it's really a transformer)?
  4. Dec 24, 2008 #3
    Sorry for the confusion. The inductor is center-tapped. So how does this effect Kirchhoff's Laws? Do I still take into account the whole inductor even though only three out of the five windings are contained within the loop when I apply Kirchhoff's loop rule? Thanks for the help.
  5. Dec 24, 2008 #4
    Post a schematic.
  6. Dec 26, 2008 #5
    Here is a schematic that I drew up on SPICE. I did not know how to put center taps on the inductors L2a and L2b, so I just had the wire continue into them. Also the RF inputs are not included. They would be in front of L1a, L2a, and L3a.

    Attached Files:

  7. Dec 27, 2008 #6
    Must you derive an analytical expression for impedance?

    You haven't shown which node is the one where you want the impedance. Also, you have shown no load. With no resistances or other losses anywhere, the impedance will not be well defined.
  8. Dec 29, 2008 #7
    First, I have added the RF input to the schematic attached below. L1, L2 and L3 are inductors with just a single loop and we have a function generator hooked up to this, so we can vary the waveform of the input.

    As you can tell I am new at this and I don't quite understand what you are asking.

    How can I choose which node I want the impedance in? This is a working resonator that I diagrammed, doesn't it have the impedance already built into it?
    Also, there are no resistors or anything else in the circuit, why will the impedance not be well defined?
    Just to let you know we are using this as a beam buncher on a particle accelerator, where the beam will pass through the main Capacitor, C (in the center of the schematic). Maybe that is useful for you to know.

    Thanks for all your help.

    Attached Files:

  9. Dec 29, 2008 #8
    If you will upload your image to www.freeimagehosting.net (or a similar image hosting site) and post the link here, I won't have to wait for your attachment to be approved to view it.

    When thinking about electrical networks, it is usual to let one node of the network be the reference node (often called a ground node). Impedance, like voltage, is a quantity that only exists between two nodes of a network; it does not exist at an isolated point in a network. It must be measured (or calculated) between some node and some other node (usually the reference node). This is the typical way of viewing "impedance" in a network.

    It is also possible to talk about the "differential impedance" between two nodes of a network, neither of which is the reference node; in this case there will typically be a third node designated the reference node.

    So, when I can see your latest schematic with the RF input shown, then we can proceed to determine the impedance.

    It will also be necessary to know the frequency of the sinusoidal input to determine the impedance, and the values of the various components. If you don't know the values of the components, it will be possible to derive an analytical expression for the impedance at a single sinusoidal frequency, but that expression will be very complicated.

    If your applied voltage is not a sine wave, then, traditionally, there wouldn't be a single "impedance". You would have to calculate the impedance for each Fourier component of the input waveform. Or, you could just calculate the voltage across your main capacitor, C, for whatever your input waveform is.

    Is this actually a homework problem?
    Last edited: Dec 29, 2008
  10. Dec 29, 2008 #9
    No this is not a homework problem. I originally posted it in the Electrical Engineering section, but it was moved.
  11. Dec 30, 2008 #10
    The attachment in post #7 is still pending approval, so I can't look at it. If you'll follow my advice about uploading to freeimagehosting, I will be able to see it and show you how to calculate the impedance.
  12. Dec 30, 2008 #11
    It has been approved now.
  13. Dec 30, 2008 #12


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    The circuit you provided has no load and no obvious function. Please provide more information. What is your circuit's function, and what are the application specs (frequency, bandwidth, power levels, Q, whatever)? Does the excitation truly float as you have drawn? As The Electrician noted, you have not included resistances so your model will not match real performance. What are the component losses?

    Suggest you talk to the person/people who designed the circuit for more information, then report back.
  14. Dec 30, 2008 #13
    The function of the circuit is to have an oscillating electric field at C with a frequency such that it bunches passing particles in the most efficient way.
    There are three different frequencies involved in this circuit. If all three pieces of the circuit are tuned to resonance then the L1a/b,C1 section will be at 12.125MHz, the L2a/b,C2 will be at 24.25MHz, and the L3a/b,C3 will be at 48.5MHz. I am not sure about the rest of the specs that you have listed.

    Can you explain what this means?

    I think this was designed at UC Berkeley back in the 70s and was re-created at the lab I am working at now about 20 years ago.
  15. Dec 30, 2008 #14
    In order to calculate the impedance seen at the driving terminals, you must know the values of each component. Do you know the inductance of each inductor, and the capacitance of each capacitor? You must also know the mutual inductance between L1 and L1a and L1b, between L2 and L2a and L2b, between L3 and L3a and L3b. You must also know the mutual inductance between the separate halves of L2a and the halves of L2b.

    It looks like some of the circuit is intended to provide a path for the harmonics of the driving frequency. Is the driving frequency 12.125 MHz? Is it a sine wave, or something else?

    To actually calculate the driving point impedance will be a complicated process, and if you don't know the loss of each component, your result will not likely be close to the real world value.

    I think you would be much better off to measure the driving point impedance (assuming you have actual hardware you can put your hands on). To do this you will need an appropriate LCR meter for a driving frequency up to perhaps 100 MHz.
  16. Dec 30, 2008 #15

    I think you circuit is way too complicated to calculate by hand even assumed the components are ideal. What are you trying to do with the circuit? If your job is to reverse engineer the circuit, put it in simulation and plot the gain and phase. Then design with more practical circuit according to the simulation. There is always an easier way to do the job. Try write out the equation of a 5 elements low pass filter using LC and you will have fun doing it already. This one is going to be much much harder. I notice you have L1a and L1b, that looked like a transformer, that complicate things even more.

    In real life, you have so many inductor going back and fore. The cross talk between inductors will affect the output to the point you can get very different result from the original intend.

    If this is not homework and not for work, let it go!!! If it is work that you have to reverse engineering. Do it on spice. I design frontend of a 64 elements phase array ultra sound imagining system with color dopler before. We have LC filter at the frontend of each channel, nothing like what you have, much simplier. The cross talk is so dominant that you have to be so careful the put inductors perpendicular to each other.

    Simplify is the key.
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