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  #1  
Old 10-03-2010, 05:18 PM
finley finley is offline
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Default problems with RLC bandpass filter simulation in LTspice

Hello there!
I have a project for school that is aimed towards engineering students (which I am not) to build an RLC bandpass filter which will pass 10 Hz--1.0kHz where those same points are the -3dB points. I have calculated that the natural frequency should be 495 Hz and, from a bandwidth of 990 Hz I have gotten these values: L=.161 H C=.642uF and R=1k. I started by deciding R=1k.
My problem comes when I simulate this in LTSpice. I get the correct natural frequency, but my corner frequencies are 200 and 1201 Hz respectively.
Can someone tell me why this might be occurring and how I could go about correcting for it?
finley
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Old 10-03-2010, 05:30 PM
finley finley is offline
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oh, and additionally, my setup is a series RLC circuit with a capacitor, inductor, and resistor in that order. I don't know how to post pictures of the schematic and frequency response graph but if someone tells me how I could definitely do that.
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Old 10-03-2010, 06:22 PM
Jony130 Jony130 is offline
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It seems that you forgot about Q factor.
Becaues in LCR filter Q factor determine bandwidth of a filter.
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Old 10-03-2010, 06:40 PM
finley finley is offline
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sorry if this is the most obvious thing in the world, but I've been working with this problem for about three grillion hours by now and my brain is fairly mushy. can you be a little bit more specific as to how I should use the Q-factor to solve this problem?
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Old 10-03-2010, 07:10 PM
Jony130 Jony130 is offline
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This equation show everything
Q = XL/R = Fo/ (F2 - F1)
Fo = √(F1*F2)
http://www.allaboutcircuits.com/vol_2/chpt_6/6.html

Last edited by Jony130; 10-03-2010 at 07:18 PM.
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Old 10-03-2010, 08:06 PM
finley finley is offline
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What I meant was could you please tell me in words why this is important. it's not hard to find equations, what's difficult for me is that I don't have enough background theory in this area to understand what precisely I should be looking for. My problem is that while I have the correct natural frequency, my corner frequencies seem to be shifted up by about 200 Hz. I'm assuming that this is due to some sort of internal resistance or something in the inductor, but I don't know how to apply adjustments to my circuit in an efficient enough way to correct for this. have you got any advice along those lines or clarification on how the Q-factor applies?
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Old 10-03-2010, 08:27 PM
finley finley is offline
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Okay, I was a little confused with one of the equations that your page gave me, but after looking it up elsewhere I realized that what you mean is that I calculated the wrong natural frequency for my circuit. Is this correct? and that it should be more around 100 Hz. In this case, my question is now a little different. Everywhere else I've looked it said that the natural frequency should be 1/sqrt(LC) so could you please explain to me why f=sqrt(fh*fl) is valid? As I said, I'm missing a whole lot of theory here and kind of trying to patch together stuff that I have learned a couple of years ago. I appreciate the fact that you even replied to me at all.
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Old 10-03-2010, 08:55 PM
Jony130 Jony130 is offline
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Quote:
Originally Posted by finley View Post
Is this correct? and that it should be more around 100Hz.
Yes, fo =100Hz

Quote:
Originally Posted by finley View Post
In this case, my question is now a little different. Everywhere else I've looked it said that the natural frequency should be 1/sqrt(LC) so could you please explain to me why f=sqrt(fh*fl) is valid?.
Fo for LC circuit is equal 1/sqrt(LC) becaues only for this frequency
Xc = XL
And Fo is located "halfway" between F2, F1.
But for mother nature "halfway" in not in the "linear" axis (arithmetic mean) but in geometric mean.
And that why Fo = √(F1*F2) it's true.

So for you circuit
Fo = 100Hz and for R = 1KΩ ---> Q = 0.101
L = 0.160762H
C = 15.7563μF
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  #9  
Old 10-04-2010, 06:55 AM
finley finley is offline
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Great. Thanks for being so clear and helpful! Now...to the computer lab!
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