Seriesparallel tank circuit resonance condition and impedanceby reakka Tags: circuit analysis, resonance circuit, tank circuit 

#1
Jul912, 10:58 AM

P: 13

I am simulating a seriesparallel 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
Jul912, 06:37 PM

Sci Advisor
P: 4,003

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. 



#3
Jul912, 06:41 PM

P: 13

Thank you for the reply.
I see see no peak for series resonance, I have swept 0100 Mhz for L = 8 nH, C1 = 600 pF, C2 = anything (doesn't matter). So I'm inclined to think you may be mistaken. 



#4
Jul912, 06:48 PM

Sci Advisor
P: 4,003

Seriesparallel tank circuit resonance condition and impedance
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. 



#5
Jul912, 06:57 PM

P: 13

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.




#6
Jul912, 07:08 PM

Sci Advisor
P: 4,003

Here is an LTSpice version:




#7
Jul912, 07:16 PM

P: 13

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 _ 



#8
Jul912, 07:37 PM

Sci Advisor
P: 4,003

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? 



#9
Jul912, 07:46 PM

P: 13

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. 



#10
Jul912, 07:54 PM

Sci Advisor
P: 4,003

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. 



#11
Jul1012, 07:14 PM

P: 13

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?




#12
Jul1012, 07:22 PM

Sci Advisor
P: 4,003

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.




#13
Jul1012, 07:54 PM

P: 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. 



#14
Jul1012, 08:04 PM

Sci Advisor
P: 4,003

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. 



#15
Jul1012, 08:14 PM

P: 13

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. 



#16
Jul1012, 08:34 PM

Sci Advisor
P: 4,003

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. 



#17
Jul1112, 01:54 PM

P: 13

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.




#18
Jul1112, 07:23 PM

Sci Advisor
P: 4,003

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 stepup 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. 


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