Series-parallel tank circuit resonance condition and impedance

  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. vk6kro

    vk6kro 4,058
    Science Advisor

    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. 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. vk6kro

    vk6kro 4,058
    Science Advisor

    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. 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. vk6kro

    vk6kro 4,058
    Science Advisor

    Here is an LTSpice version:

  8. 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));


    Only one peak for me -_-
  9. vk6kro

    vk6kro 4,058
    Science Advisor

    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. 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. vk6kro

    vk6kro 4,058
    Science Advisor

    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. 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. vk6kro

    vk6kro 4,058
    Science Advisor

    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. Here is a diagram from Fukushima:
    I observe no series resonance which should be a low real impedance peak, not a dip.
  15. vk6kro

    vk6kro 4,058
    Science Advisor

    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. 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. vk6kro

    vk6kro 4,058
    Science Advisor

    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. 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. vk6kro

    vk6kro 4,058
    Science Advisor

    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. 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. vk6kro

    vk6kro 4,058
    Science Advisor

    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.
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