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Op-Amp Bandpass Filter

  1. Dec 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Select component values to design for corner frequencies at 500 rad/s and 1500 rad/s.


    2. Relevant equations
    KCL, transfer function


    3. The attempt at a solution
    Gain:

    [1] G = 1 + [itex]\frac{R_{A}}{R_{B}}[/itex]

    Voltage division:

    [2] V2 = V1[itex]\frac{R_{2}}{R_{2}+1/sC_{2}}[/itex]

    KCL:

    [3] 0 = [itex]\frac{V_{1}-V_{i}}{R_{1}}[/itex] + [itex]\frac{V_{1}-V_{2}}{1/sC_{2}}[/itex] + [itex]\frac{V_{1}-V_{o}}{R_{3}}[/itex]

    I am stuck at this point and don't know what to do next. There is a similar problem in the book and it said to substitute equations [1] and [2] into [3].

    I sort of know what to do once I find [itex]\frac{V_{o}}{V_{i}}[/itex], but I just don't know how to get there from where I am.

    Thanks!
     

    Attached Files:

  2. jcsd
  3. Dec 5, 2012 #2
    Equations 2 and 3 are two equations in three unknowns (Vo/Vi is one unknown). The third equation is [1] the gain formula which relates V2 to Vo. The gain formula gives gain from the + terminal of the amp to the output but you could also relate V2 to Vo just by using the resistor divider Ra/Rb. So this equation lets you replace V2 by an expression in Vo and equation [2] will let you replace V1 by an expression in Vo. Then equation [3] will only contain Vo and Vi.

    But honestly it seems like you are throwing equations at it, particularly the gain equation, when it's much simpler not to memorize any equations at all. I'd be looking at removing as many unknowns as possible at the outset. So I would have found V2 in terms of Vo by the resistor divider Ra/Rb. This voltage in Vo would then appear at the + terminal of the op amp. Then KCL at the + terminal would find V1 as a function of Vo and the final KCL at V1 would involve only Vi and Vo.
     
  4. Dec 5, 2012 #3
    I looked up the circuit on Google and found this: http://en.wikipedia.org/wiki/Sallen–Key_topology#Application:_Bandpass_filter

    That is the same circuit and the transfer function is in the image attached. According to the page R2 = 2R1 and C1 = C2.

    Would be a good idea to pick C = 1F or and R1 = 1kΩ, and then use scaling to find the standard values to use?
     

    Attached Files:

  5. Dec 5, 2012 #4

    rude man

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  6. Dec 5, 2012 #5
    No matter what values I try, none work :| (I am always ~30-50Hz away on one or both of them). RA and RB don't have much effect, but I notice R3 changes the bandwidth (or maybe it's something else, but it makes graph wider/smaller depending on value, as well as the center frequency). I am not sure where I am going wrong.
     
  7. Dec 5, 2012 #6

    rude man

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    Go to your link. First, pick your mid-band gain A. That defines Ra and Rb. You might pick A = 1 since nobody told you what it should be.

    Then: what is f0, given your corner frequencies f1 and f2? Then, how is Q expressed as a function of f0, f1 and f2? This part is standard textbook stuff you should have covered by now.

    Once you know f0 and Q, use the formulas in your link to determine all the R's and C's. Note that you have only 2 formulas, so you need to pick some of the R's and/or C's arbitrarily to solve for the remaining two components.

    (Your instructor might have wanted you to determine the transfer function yourself).
     
  8. Dec 5, 2012 #7
    If I set A to 1, then G = 1.5, so I set RB = 10kΩ resulting in RA = 5kΩ.

    f0 = [itex]\sqrt{\frac{500}{2\pi}\frac{1500}{2\pi}}[/itex] = ~138Hz

    Q = [itex]\frac{f_{0}}{f_{2}-f_{1}}[/itex] = ~0.86

    (2[itex]\pi[/itex]f0)2 = [itex]\frac{R_{3}+R_{1}}{C_{1}C_{2}R_{1}R_{2}R_{3}}[/itex]

    R3 = [itex]\frac{R_{1}}{(2\pi f_{0})^{2}(C_{1}C_{2}R_{1}R_{2})}[/itex]

    Now, arbitrary values:
    R1 = 10kΩ
    R2 = 20kΩ (2R1)
    C1 = C2 = 0.1μF

    R3 = ~6.6kΩ

    The simulation is still giving the wrong corner frequencies (this simulation shows ~103Hz and ~310Hz). I am using the universal op-amp 2 in LTSpice IV. Either my circuit is wrong or I calculated something wrong. An image of the circuit I made in LTSpice IV is attached.
     

    Attached Files:

    Last edited: Dec 6, 2012
  9. Dec 6, 2012 #8
    Does 'AC 1 0' mean a 1V amplitude ac signal? If so, try changing V2 and V3 to +/-10 rather than 1. You could be clipping the output.
     
  10. Dec 6, 2012 #9
    I tried a different schematic (attached) and same issue. I am so confused, I have gone over my calculations numerous times and I get the same answers, but it just doesn't give the correct simulation.
     

    Attached Files:

  11. Dec 6, 2012 #10
    You can't set them all to arbitrary values; Q is a function of these values too. You have to find component values that satisfy both equations for Q and fo


    If that's too annoying, I have a simpler design procedure that doesn't let you control the max gain:

    1. G=2 (it must be 2 because this causes a term to zero in the Q equation)
    2. all C the same
    3. R1=R3=1/(woQC)
    4. R2=2Q/(woC)

    Max gain is 2Q2

    (p116 ISBN 9783527407668)
     
    Last edited: Dec 6, 2012
  12. Dec 6, 2012 #11
    EDIT: See my last reply.
     
    Last edited: Dec 6, 2012
  13. Dec 6, 2012 #12

    rude man

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    Check you op amp power supplies. They look backwards to me in polarity, and the voltages (+/-1V) look too low.
     
  14. Dec 6, 2012 #13
    I figured it out while in the lab today:

    Q = ~0.86
    G = ~1.8
    C1 = C2 = 0.47μF
    R1 = R3 = 2.4kΩ
    R2 = 4.7kΩ
    RA = 12kΩ
    RB = 10kΩ

    Thank you for all of your help!
     
  15. Dec 6, 2012 #14

    rude man

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    Also make sure you're not overdriving the input. Does your output look like a nice clean sine wave?
     
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