Need help with sallen key bandpass filter

Discussion in 'Homework Help' started by alialiali, Mar 12, 2009.

  1. alialiali

    Thread Starter Member

    Dec 12, 2008
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    Hi every one
    I'm working on bandpass secound order filter and I have to have band pass between 23k and 53k using sallen key filters whaen I use the formola to dsign my high pass which should gave me 23k fc its working and I used the folwing calculation

    R1 = mR, R2= R, C1=C , C2=nc FSF = 1 from the table
    FSF * fc = 1/ 2pi RC squrt mn
    m= 0.229 from the table
    n= 3.3 from the table
    I'm looking for LP for 53k
    1*23k=1 / 2pi * (C=0.01u from th table) * (R which we need 2 find) * squrt 0.229*3.3
    R = 1 / 2pi * (C=0.01u ) * (23k) * squrt 0.229*3.3 = 707 ohms
    Since R = R2 so my R2 value is 707 ohms
    R1 = R*m = 0.229 * 707 = 161.9 = 162 ohms
    C1 = C which is 0.01u so the value of my C1 is 0.01u
    C2 = n * C which is 3.3 * 0.01u = 33n
    so my C2= 33n



    and my LP filter which should gave me the fc=53k as

    R1 = mR, R2= R, C1=C , C2=nc FSF = 1 from the table
    FSF * fc = 1/ 2pi RC squrt mn
    m= 0.229 from the table
    n= 3.3 from the table
    I'm looking for LP for 53k
    1*53k=1 / 2pi * (C=0.01u from th table) * (R which we need 2 find) * squrt 0.229*3.3
    R = 1 / 2pi * (C=0.01u ) * (53k) * squrt 0.229*3.3 = 345
    Since R = R2 so my R2 value is 345 ohms
    R1 = R*m = 0.229 * 345 = 79.005 = 80 ohms
    C1 = C which is 0.01u so the value of my C1 is 0.01u
    C2 = n * C which is 3.3 * 0.01u = 33n
    so my C2= 33n


    the main problem when I contact the output from my HP to the input of my LP filter the over all fc is not maching the fcs I looking for

    can any one advice me how to conact the two filters together this can be using athered filter but my quation is do I need to change the holl calcultion and how?

    many thanks
     
  2. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Your resistors have a very low value.
    What happens when you divide the capacitors by 10 and multiply the resistors by 10 ?
    The frequency will be the same but the impedance will be changed.

    Greetings,
    Bertus
     
  3. alialiali

    Thread Starter Member

    Dec 12, 2008
    26
    0
    I dont know can you explin bit more please
     
  4. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    What I am trying to say is : recalculate the filters with capacitors of 1 nF and n* 1nF in stead of 10 nF and n* 10nF.

    Greetings,
    Bertus
     
  5. alialiali

    Thread Starter Member

    Dec 12, 2008
    26
    0
    Ok I did that divide the capacitors by 10 and multiply the resistors by 10 I can see some difrernt but the problem now is I am getting 57.71kHz for the LP which it should be 53kHz and the HP= 20.84kHz which should be = 23kHz
    please can you help me more
     
  6. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    What kind of filter are you making ?
    Where is the table you are talking about ?
    Most sallen-key filters I know have equal resistors for the low pass and equal capacitors for the high pass.

    Greetings,
    Bertus
     
  7. alialiali

    Thread Starter Member

    Dec 12, 2008
    26
    0
    Many thanks for your king reply

    I enclod copy of the table so you can have look and gave me some advice

    Table 1. Butterworth Filter Table
    FILTER FILTER
    ORDER
    Stage 2
    FSF Q
    1.000 0.7071
    1.000 1.0000 there is full table in the picture below
     
    Last edited: Mar 13, 2009
  8. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    The table does not look familiar to me.
    Here is a link to a PDF from TI about reference designs.

    focus.ti.com/lit/an/slod006b/slod006b.pdf

    From chapter 16 filters are discussed, sallen-key from 16.3.

    Greetings,
    Bertus
     
  9. alialiali

    Thread Starter Member

    Dec 12, 2008
    26
    0
    Thx a lot for the Pdf file but the problem it's very difcult 2 understand it can you please help me bit mor how 2 apply the formlu to my filter as I need band pass betwen 23k and 53k by uding 2 secound order sallen keys filter only
     
    Last edited: Mar 13, 2009
  10. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Here ia a PDF special on sallen key filters.
    Chapter 3 is for low pass
    Chapter 4 is for high pass
    Each chapter has simplifications in it.

    I used the calculator on the web to have the 23 kHz highpass filter
    .http://www.pronine.ca/acthpf.htm

    [​IMG]

    R1 = 4.89 k Ohm, R2 = 9.78 k Ohm and C1 = C2 = 1 nF

    The same I did with the 53 kHz lowpass filter.
    http://www.pronine.ca/actlpf.htm

    [​IMG]

    R1 = R2 = 2.12 k Ohm , C1 = 2 nF , C2 = 1 nF


    Greetings,
    Bertus
     
    Last edited: Mar 14, 2009
  11. alialiali

    Thread Starter Member

    Dec 12, 2008
    26
    0
    Many many thx
     
  12. alialiali

    Thread Starter Member

    Dec 12, 2008
    26
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    This two circuits is great just I’d like to ask how to connect them together so I get my final out put from one port instead to have tow deferent out puts port ?
    Do I need to implement a there’d filter to feed this two in it ?

    Please gave me some ideas
    Many thx
     
  13. alialiali

    Thread Starter Member

    Dec 12, 2008
    26
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    When I contact them together as in the picture it gave me defrnt anwsers
     
  14. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    You can connect the highpass filter at the output of the lowpass filter.
    This will create the bandpass filter.

    [​IMG]

    Greetings,
    Bertus
     
  15. alialiali

    Thread Starter Member

    Dec 12, 2008
    26
    0
    Thats great the high pass signal is ok as it's gaving me 23kHz but when I test the output for the 53k it's gaving me 61KHz so please can you gave more advice.

    I think we need to add another filter such as adder or first order filter
     
    Last edited: Mar 15, 2009
  16. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Perhaps a buffer between the highpass and lowpass will help.

    Greetings,
    Bertus
     
    Last edited: Mar 15, 2009
  17. alialiali

    Thread Starter Member

    Dec 12, 2008
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    Sorry but what do you mean with buffer as I have no idea a bout it at all
     
  18. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    A buffer is an opamp with the + as input and the output conected to the - input.

    [​IMG]


    When you place this between the two filters they will not "see" each other.

    Greetings,
    bertus
     
  19. alialiali

    Thread Starter Member

    Dec 12, 2008
    26
    0
    Do you know what I learned hear from you I wouldn’t learn it in one year
    I greatly appreciate your kind help

    The result of the test now still negative as nothing changed do you think we need to add some resistors to the buffer ?
     
  20. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    What happens when you change the TL072 (wich has a bandwidth of 4 Mhz)
    to a LF357 ( wich has a bandwidth of 20 Mhz) ?
    Perhaps the limited bandwidth is giving some problems.
    (you are using rather high frequencies).

    Greetings,
    Bertus
     
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