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Series-parallel tank circuit resonance condition and impedance

  1. Jul 9, 2012 #1
    I am simulating a series-parallel tank circuit in MATLAB. This means parallel resonance (inductor and capacitor in parallel) with an added capacitor in series. I would assume the resonance condition is w^2LC = 1 and the impedance is 1/(1/(z_L+z_R)+1/z_C1) + z_C2 with C1 being the parallel capacitor and C2 being the series capacitor.

    The part where I am getting confused is why Fukushima Experimental Pulsed NMR (pg. 413) is telling me the resonance condition is close to w^2L(C1+C2) = 1. The capacitors are not in parallel so this makes no sense to me.

    It is apparent that C2 has no contribution to resonance, it simply cancels reactance. What am I missing here?

    Thank you!
     
  2. jcsd
  3. Jul 9, 2012 #2

    vk6kro

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    It is apparent that C2 has no contribution to resonance, it simply cancels reactance. What am I missing here?


    That is what causes series resonance.

    As you increase frequency, you get a series resonance first then a parallel resonance. So, a dip first then a peak.
     
  4. Jul 9, 2012 #3
    Thank you for the reply.

    I see see no peak for series resonance, I have swept 0-100 Mhz for L = 8 nH, C1 = 600 pF, C2 = anything (doesn't matter). So I'm inclined to think you may be mistaken.
     
  5. Jul 9, 2012 #4

    vk6kro

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    Try it with both capacitors 600 pF and the inductor 8 uH, and a series 5000 ohms from the signal generator.

    I get a dip at 1.62 MHz and a peak at 2.3 MHz.
     
  6. Jul 9, 2012 #5
    Hmm, I must be doing something wrong. Is my impedance expression (first post) correct? I can email you my code if you don't mind.
     
  7. Jul 9, 2012 #6

    vk6kro

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    Here is an LTSpice version:

    http://dl.dropbox.com/u/4222062/series%20parallel%202.PNG [Broken]
     
    Last edited by a moderator: May 6, 2017
  8. Jul 9, 2012 #7
    syms w;

    L = 8000; %nH
    C1 = 600; %pF Cs
    C2 = 600; %pF Cp
    R = 1; %Ohms

    z_L= 1i*w*L*10^6*10^(-9);
    z_R = R;
    z_C1 = 1/(1i*w*10^6*C1*10^(-12));
    z_C2 = 1/(1i*w*10^6*C2*10^(-12));

    z = 1/(1/(z_L+z_R)+1/z_C1) + z_C2 + 5000;

    f = (0:20);

    X = real(subs(z,w,2*pi*f));
    Y = imag(subs(z,w,2*pi*f));

    figure;
    plot(f,X);
    figure;
    plot(f,Y);

    Only one peak for me -_-
     
  9. Jul 9, 2012 #8

    vk6kro

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    I get the reactance of 600 pF at 1.62 MHz to be 163 ohms and the reactance of 8 μH to be 81.43 ohms.

    Are you taking the output at the junction of the 5000 ohms and the first capacitor?
     
  10. Jul 9, 2012 #9
    We are measuring reactance? In that case, yes I get 2 peaks. But the resonance is determined at the peak of the real impedance. What is the significance of peak reactance?

    I am not taking any output, I am just measuring the real/imag impedance at discrete 1MHz intervals.
     
  11. Jul 9, 2012 #10

    vk6kro

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    The resonances are quite sharp so you would miss them with only 1 MHz resolution.

    Could you increase it to 1 Hz resolution and take an output from after the resistor?

    If you have to use the impedance of the whole circuit, leave out the resistor and you should be able to see a dip folowed by a peak as you go up in frequency.
     
  12. Jul 10, 2012 #11
    I ran again with 1 hertz resolution. Only peak I see is at 2.3 Mhz, and this is real impedance. Why would there be a dip in real impedance?
     
  13. Jul 10, 2012 #12

    vk6kro

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    The top capacitor is in series with the coil and this forms a series tuned circuit where the two reactances cancel each other out, just leaving any resistive component.
     
  14. Jul 10, 2012 #13
    Here is a diagram from Fukushima: http://i.imgur.com/mTkIH.png
    I observe no series resonance which should be a low real impedance peak, not a dip.
     
  15. Jul 10, 2012 #14

    vk6kro

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    What would you get if you removed "C" ?

    This is a real effect which you can measure with suitable equipment and you can simulate it with simulators.

    Incidentally, there is no "real impedance". Impedance is the resultant of reactance and resistance.
    If you want to refer to resistance, then just call it resistance.
     
  16. Jul 10, 2012 #15
    I seriously believe you, but whats the balls is going wrong with my simulation. Is LTspice assuming that the inductor has some resistance? If so, what is this resistance (I assumed it to be 1 ohm). Why am I still getting one peak?

    Oh, and sorry I'm used to saying "real impedance" for the real part of the impedance, but I guess that is just resistance so I will call it that from now on. Thanks for sticking with me through this btw.
     
  17. Jul 10, 2012 #16

    vk6kro

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    The behaviour of a series tuned circuit depends on how you drive it.

    If the capacitor and the inductor were perfect, then the current would just depend on the resistance of the power supply or any series resistance.
    That is what the 5000 ohm resistor was for. It sets up one component of a voltage divider so that if the impedance of the circuit involving the capacitors and inductor vary then the voltage across the resistor will also vary.
    So, you get a different voltage out as the frequency varies.

    I can't help you much with MatLab as I have never used it or owned a copy of it.

    However, if you would consider downloading a free copy of LTSpice, I could show you how to get some useful results with that program. I'm not a real expert at that either, but I have found a few useful tricks.
     
  18. Jul 11, 2012 #17
    Okay. I can't use LTspice because I am on a Macbook. But what I did do was build the circuit, and lo and behold: the series capacitor does NOT affect the resonance peak for parallel resonance and there is NO peak for series resonance. Can somebody tell me whats going on now? -_- It seems my Matlab simulation was correct, but why is Fukushima telling me otherwise.
     
  19. Jul 11, 2012 #18

    vk6kro

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    Demonstrating series resonance depends on the output impedance of the signal generator. So, to avoid this problem, it is necessary to add some capacitance across the signal generator so that the internal parameters of the signal generator don't matter.

    Try it with just the series tuned circuit (no capacitor across the inductor), no resistor and a larger capacitor directly across the signal generator.

    For the capacitor, I would suggest about 5 times as high a value as the one in the tuned circuit. Maybe 1000 pF.

    At the junction of the inductor and the smaller capacitor, you should get a large step-up in voltage at the resonant frequency of the tuned circuit.

    Once you are getting this, add the other capacitor across the inductor. You may have to add some resistance between the signal generator and the added capacitance to see the parallel resonant peak.
     
  20. Jul 12, 2012 #19
    I am using a network analyzer, so I think the output impedance is 50 ohms. I don't think I need to do any of what you are suggesting. I am measuring S11.
     
  21. Jul 12, 2012 #20

    vk6kro

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    Absolutely. You don't have to do anything.

    But if you want to see how tuned circuits work, you should try a capacitor across the signal generator.
    If you don't, then the 50 ohms is in series with the series tuned circuit and that may be enough to damp out the resonance of the tuned circuit.
     
  22. Jul 12, 2012 #21
    I completely understand what you are saying. I should expect a dip in the reactance for the series resonance. I will try this out when our equipment is free tomorrow. Thank you
     
  23. Aug 15, 2012 #22
    i am bumping this. I am still confused on the matter, and no matter what I have tried in experiment and simulation I am still only getting one resistance peak. I am convinced that C2 does not produce a second peak (it does shift the reactance along the vertical axis). Can anyone else help me with this?
     
  24. Aug 16, 2012 #23

    The Electrician

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    I think you are doing a couple of things wrong.

    I soldered together a circuit like the one you are concerned with. I didn't have any 600 pF capacitors, so I used 511 pF. I also used a 9.4 μH inductor. These aren't exactly the values you are using, but they are close enough to show what's going on.

    In post #7 I see from your matlab code that you are plotting the real and imaginary parts of the impedance. If you do this, you won't see a notch and a peak; you'll only see a single peak. You need to plot the magnitude of the impedance, not its real and imaginary parts. Matlab should have a built-in function such as ABS (absolute), or you can do it yourself by taking the square root of the sum of the squares of the real and imaginary parts.

    Also, measuring S11 (the reflection coefficient) isn't going to give you the same plot as a plot of the impedance.

    I've attached 3 images. The first image shows a plot of the measured real and imaginary parts of the impedance. The real part is plotted on a log scale, and the imaginary part is plotted on a linear scale, with zero in the center of the image (negative values below the center line).

    The second image shows the magnitude of the impedance plotted on a log scale. These two measurements were made with an impedance analyzer.

    The third image shows the measured reflection coefficient, measured with a VNA. Some VNAs can show the impedance as calculated from the reflection coefficient. I'll attach an image to the next post showing a plot of impedance magnitude as measured by the VNA.
     

    Attached Files:

  25. Aug 16, 2012 #24

    The Electrician

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    Here's an image showing the impedance magnitude as calculated from S11 on the VNA.
     

    Attached Files:

    Last edited: Aug 16, 2012
  26. Aug 16, 2012 #25
    Nice work!
     
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