Overthinking Filter Design?

Discussion in 'The Projects Forum' started by Mawangs1, Apr 20, 2013.

  1. Mawangs1

    Thread Starter New Member

    Mar 26, 2012
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    With the Sallen Key filter pictured below, I'm designing a 3-4th order LP filter. This filter is supposed to have variable adjustment so that I can block frequencies higher than 250Hz all the way up to ones higher than 1000Hz. I'm getting frustrated though because I can't figure out how to select my component values.

    [​IMG]
    I assume that:

    - cascading this circuit 2 times should make this a 4th order LP filter

    -Picking 2 random values for C1 and C2 and utilizing a variable resistor in place of R1 or R2 in each of the 2 levels will allow me to adjust to the values I mentioned earlier

    -The equation for the cutoff frequency of this circuit is: [​IMG]
    - Since I know the above equation, I can place in the frequency I want the limits at (250-1000Hz) for f0 in the equation and then solve for R1 or R2

    What I can't figure out is this. How do I know what values to select for the 2 capacitors and one of the resistors such that as I rotate my variable resistor, the frequency ranges I want to block are attained?

    I've been mixing and matching values for awhile and I'm getting nowhere doing it this way. There's gotta be a smarter way to go about doing this....
     
    Last edited: Apr 20, 2013
  2. blah2222

    Well-Known Member

    May 3, 2010
    553
    33
    So what you are saying is you want to pass all frequencies lower than 250Hz and attenuate everything higher or do you want to be able to move the 3db frequency between 250Hz and 1000Hz?

    Most often Sallen-Key filters are used where C1 = C2 = C and R1 = R2 = R to create a simpler equation of:

    w_{0} = \frac{1}{RC}

    Using a dual-ganged potentiometer would be useful to control both R1 and R2 simulataneously.

    http://www.onlinetps.com/shop/images/item_images/Dual_Potentiometer.jpg
     
  3. Mawangs1

    Thread Starter New Member

    Mar 26, 2012
    23
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    I would like to pass all frequencies lower than the value I adjust to. So, if I get everything setup right and I turn my pot up , I should be able to pass frequencies lower than what I turn to. Whether this is 250, 600, or even 1000Hz.

    Your reply explains the diagrams at this link to me now to.
     
  4. #12

    Expert

    Nov 30, 2010
    16,248
    6,745
    For R1 = R2 and C1 = C2
    For F = 1000
    if R = 1000, C = .159uf
    for F = 250 and C = .159uf
    R = 4k

    Check my math. I do make mistakes.
     
  5. tubeguy

    Well-Known Member

    Nov 3, 2012
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    Last edited: Apr 20, 2013
    #12 likes this.
  6. Mawangs1

    Thread Starter New Member

    Mar 26, 2012
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    Your math is fine, but I'm still overthinking things.


    If I know I can achieve my lower limit at 250Hz using 4kΩ and my higher limit at 1000Hz using 1kΩ, how can set my variable resistor such that these values are reached by turning the pot all the way down or all the way up?


    My first thought is to place a resistor in parallel with with pot, but if I was to turn the pot all the way down I'd have 0 resistance right?
     
  7. #12

    Expert

    Nov 30, 2010
    16,248
    6,745
    Put 1k in series with the pot, then parallel the pot to make it a 3k pot.
     
  8. blah2222

    Well-Known Member

    May 3, 2010
    553
    33
    Just remember that in order for your filter to operate as expected, you will need to maintain R1 = R2 at all times.
     
  9. #12

    Expert

    Nov 30, 2010
    16,248
    6,745
    Ps, I don't think anyone mentioned, your schematic in the first post is a second order filter.

    Assuming you stack 2 of them,
    Keeping R1 = R2 = R3 = R4 "the same" will have some small ability to tolerate mismatch, but it can be difficult considering the quality of potentiometers you can buy.
     
  10. crutschow

    Expert

    Mar 14, 2008
    12,977
    3,220
    Cascading two 2nd order filters does not make an optimum 4th order filter, as the roll-off at the corner frequency will not be as sharp. You can use a tool, such as the free FilterPro software from Texas Instrument, to easily determine the best values for an active filter of any order and type.
     
    aws505 and #12 like this.
  11. Mawangs1

    Thread Starter New Member

    Mar 26, 2012
    23
    0
    Ugh, I keep getting confused.


    For my situation, I'm going to be using a 100K dual ganged pot. Thus, I've found the following:


    at F = 1000Hz, if R= 100kΩ, then C will be 0.00159uF


    at F = 250Hz, if C = 0.00159uF, then R ≈ 400.4kΩ



    But then this is where I get myself confused again.

    I then assume that I want to put 100kΩ in series with a parallel total resistance of 300.4kΩ, but how can I get that total resistance if I have to use a potentiometer that's less than my total resistance?


    Also how would I go about utilizing this FilterPro program? I Dl'ed it and I entered some parameters, but I can't tell what to change and what to leave the same..


    *EDIT* The attached picture is what I think you guys want me to do.
     
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  12. #12

    Expert

    Nov 30, 2010
    16,248
    6,745
    You will either make the capacitors larger or buy 1 megohm pots.
     
  13. tubeguy

    Well-Known Member

    Nov 3, 2012
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    Here' a hint.
    Put 33,000 ohms in series with the 100 k pot and calculate caps with that.
     
  14. tubeguy

    Well-Known Member

    Nov 3, 2012
    1,157
    197
    Looked at your bmp. No parallel resistor needed, just 33,000 ohm in series with the 100k pot and recalculate. A very common value works.
     
  15. Mawangs1

    Thread Starter New Member

    Mar 26, 2012
    23
    0
    I've added a new circuit up. With this new one, it appears f = 1000Hz at R= 33kΩ , C = 5nF and then f =250Hz at R = 133kΩ, C = 5nF. How did you know to select 33kΩ?
     
    • kkg2.JPG
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  16. #12

    Expert

    Nov 30, 2010
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    Apparently he can do arithmetic. (Very old word for math.)
     
  17. tubeguy

    Well-Known Member

    Nov 3, 2012
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    ;)
    #12 got us started with the ratio of 1k for the minimum resistor value and 4k for the max value, so the variable part was 3k. Then scale the resistors by subbing 100k for 3k.
     
    Last edited: Apr 21, 2013
  18. Mawangs1

    Thread Starter New Member

    Mar 26, 2012
    23
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    Sorry, could you clarify this a little more? In my mind, you're telling me that you divided 1k by 3k which yields 1/3 than you multiplied this times 100k. Am I off here?
     
  19. #12

    Expert

    Nov 30, 2010
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    He said he scaled the resistors by substituting 100 k for 3 k. He did not say he divided 1k by 3k.

    Are you trying to follow this through a computerized translator because you don't speak English?
    We can be careful to use simple sentences.
     
    Last edited: Apr 22, 2013
  20. Mawangs1

    Thread Starter New Member

    Mar 26, 2012
    23
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    No, I just don't clearly see what he's saying. I get what he says about the ratio of the 1kΩ and 4kΩ resistors, but then he says "...so the variable part was 3k. Then scale the resistors by subbing 100k for 3k."


    I don't understand what he does after he finds the 1/4 ratio. Forgive me but I've been thinking about this so long today, that I can't think (if that makes sense).
     
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