
Home  Forums  Rules  Articles  diyAudio Store  Gallery  Wiki  Blogs  Register  Donations  FAQ  Calendar  Search  Today's Posts  Mark Forums Read  Search 
Analog Line Level Preamplifiers , Passive Preamps, Crossovers, etc. 

Please consider donating to help us continue to serve you.
Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving 

Thread Tools  Search this Thread 
28th May 2006, 08:34 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, 09:36 AM  #2 
diyAudio Member
Join Date: Jun 2002
Location: Melbourne, Australia

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.
__________________
Bestever T/S parameter spreadsheet. http://www.diyaudio.com/forums/multi...tml#post353269 
28th May 2006, 10:20 AM  #3 
diyAudio Member
Join Date: Mar 2006
Location: nsw

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, 04:31 AM  #4 
diyAudio Member
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, 04:46 AM  #5  
diyAudio Member
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, 09:04 AM  #6  
diyAudio Member
Join Date: May 2005
Location: Mumbai

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, 09:52 AM  #7 
diyAudio Member

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
__________________
Whether we like to think of it this way or not, an audio engineer shares the professional goal of a magician  Richard Heyser Linear Audio Vol 12 is out! Check out my Autoranger and SilentSwitcher on Kickstarter! 
29th May 2006, 09:53 AM  #8 
diyAudio Member
Join Date: Mar 2006
Location: nsw

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, 10:42 AM  #9 
diyAudio Member
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, 04:34 PM  #10 
diyAudio Member
Join Date: May 2005
Location: Mumbai

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.. ____________________ 
Thread Tools  Search this Thread 


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
How big difference of noise between the realistic circuit and pspice simulation?  goodnight521  Software Tools  1  20th June 2008 11:58 AM 
How big difference of noise between the realistic circuit and pspice simulation?  goodnight521  Analog Line Level  4  19th June 2008 05:22 PM 
PSPICE simulation  sunknu  Everything Else  0  21st September 2006 09:30 AM 
PSpice Simulation  annex666  Everything Else  1  25th November 2005 03:56 PM 
New To Site?  Need Help? 