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28th May 2006, 09:34 AM  #1 
diyAudio Member
Join Date: May 2005
Location: Mumbai

Loaded Tank circuit lossess  PSPICE Simulation
Hello friends,
I obtain the PSPICE simulation of a simple LCR parallel circuit, and formed some doubts regarding it. The voltage response across the LCR circuit at resonant frequency turned out to be dissimilar when i changed the L/C ratio. In such a tuned circuit it shouldnt have happenend if PSPICE considers L and C elements as ideal , with no internal resistances. I have posted the simulation results and the schematic of my tiny expperiment. Kindly post the answers that you guys have. Regards, Prasanna Rao. 
28th May 2006, 10:36 AM  #2 
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In cct #2 the capacitor is 10nF so it has 10x the impedance of cct #1; same with the inductor. If you make R 10x bigger too then it should work out right.
If you post a screenshot as a gif or png image they are *lots* sharper and clearer than a jpg for the same filesize.
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28th May 2006, 11:20 AM  #3 
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Yes, thats right.
The Q factor of the two circuits is different. The Q of a common tank circuit loaded by a resistor = R/X_{L}. In each case, X_{L} is different. It is true, that with ideal components and without the loading resistor, the Q's would be the same. 
29th May 2006, 05:31 AM  #4 
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Join Date: May 2005
Location: Mumbai

LCR At resonance
Ok, thanks for that info.
Forgive me with ideas, if i am wrong over here.. As far as i know, the Q factor of a tank circuit would affect its response at frequencies other than the resonant. At resonance, jXc = jXl, so the impedance of the circuit would be purely resistive (1k in our case). In short, if i maintain the same LC ratio, the response at RESONANT frequency should be the same. Regards Prasanna.. 
29th May 2006, 05:46 AM  #5  
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Join Date: Mar 2006
Location: nsw

Quote:
The response at the resonant frequency will increase with Q, the curve will also be more sharp. BTW, both of your curves are fairly high Q. Experiment with 1mH/1nF and 10mH/100pF. 

29th May 2006, 10:04 AM  #6  
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Join Date: May 2005
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I do agree with your 2nd statement regarding the curves. Everything is ok with the Q factor of the ckt, as it depends on the L/C ratio and R.
But you say : Quote:
I am afraid I cant understand that. To make it more confusing, the results in my experiment were exactly opposite. I have posted its frequency response in the first thread of this topic. At resonance here's what must happen : The impedance seen across the ideal LC parallel section is infinite. Now if I increase C and decrease L; or do it viceversa such that the L*C product remains the same; then it should still offer me with infinite impedance at that very same resonant frequency. So, at resonance the net current in LCR ckt is only due to R (since Z = R). Then Itot*R = Vresponse, must be the same even if i change the L/C ratio. Huh.. God help..! Regards Prasanna. __________________ 

29th May 2006, 10:52 AM  #7 
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One test would be to do it *without* the R and see what the difference is in the two cases.
I suspect that the sim has a hidden R // C to satisfy DC conditions, most Spice simulators have. Possibly you can even chnage it, see in the help file. Jan Didden
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29th May 2006, 10:53 AM  #8 
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Join Date: Mar 2006
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Humbly, you're right. It is the current that increases in a parallel resonant circuit at resonance. Voltage rises in a series resonant circuit. I'll get back shortly with details

29th May 2006, 11:42 AM  #9 
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Join Date: Mar 2006
Location: nsw

OK. I've checked this so it should be correct.
A tank circuit is a parallel capacitance and inductance. The Q of the tank circuit on its own should be infinite. In practice it is not. The inductors winding resistance damps the Q. The formula for the Q of the tank alone is: Q_{tank}=X_{L} / R_{winding}. In your circuit the tank is loaded by resistance. The resistance is the parallel combination of the 1k resistor and the source impedance. This is because the source is a short circuit to AC save for its own impedance. Since you are using a current source (infinite impedance), we need only consider the 1k. To calculate the circuit Q, first calculate Z_{tank}. This is X_{L}^2 / R_{winding} and since we have no winding resistance this is infinite. Then we need to put this in parallel with the load resistance. So in total we have 1k. We then divide this by X_{L}. If the resonant frequency is 159,155Hz , then X_{L}=10 ohms for the first circuit. The Q is then 100. When the Q of this type of circuit is reduced, the bandwidth is increased, and viceversa. Hopefully, this clears the air a little and answers your question BTW Jan, sorry for the overpost. 
29th May 2006, 05:34 PM  #10 
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Hello Sir ,
Yes everything is right in its place. But the mystery about different voltages at the same resonant frequency remains..! Q must not alter the o/p voltage at resonance. Infact that's what we all have shown in some way or the other, in our respective posts. I have asked this doubt to some of my coll. buddies and they too accept the same question mark. [QUOTE] One test would be to do it *without* the R and see what the difference is in the two cases. I suspect that the sim has a hidden R // C to satisfy DC conditions, most Spice simulators have. Without R, the Q remains around 55 in both the case. Even with R = 1K, the observed Q in case A is 54, whereas theorotically it comes out to be 10000..!! Surely PSPICE might be considering coil resistance and/or other parasites of L and C. But then no such attributes are defined in their models..? If someone knows the answer please reply. Regards, Prasanna Rao.. ____________________ 
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