Help with passive filters

Discussion in 'General Electronics Chat' started by mike., Feb 24, 2013.

  1. mike.

    Thread Starter New Member

    Feb 24, 2013
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    Hi,

    New here and a little stuck on a problem on passive filters, I'm hoping someone may be able to help.

    I need to work out the cut off frequency, and understand the basic formula is 1/2(pi)RC

    However, for example in the low pass filter, I have 2 additional resistors, each individually in parallel with the capacitor.

    How do these additional resistances effect the COF?

    Thanks in advanced
    Mike
     
  2. blah2222

    Well-Known Member

    May 3, 2010
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    So remember, in the equation of f_{cutoff} = \frac{1}{2\pi RC}, the resistance R is the equivalent resistance path that the capacitor C sees.
     
  3. #12

    Expert

    Nov 30, 2010
    16,283
    6,794
    You can calculate a single stage filter much like a resistive voltage divider.
    Figure the impedance of the capacitive part compared to the series resistor.
     
  4. mike.

    Thread Starter New Member

    Feb 24, 2013
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    Thanks for the reply.

    So, the additional resistances would not have an effect on the COF then presumably?
     
  5. #12

    Expert

    Nov 30, 2010
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    They don't change the frequency. They change the Q. The response slope gets more horizontal.
     
  6. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Not much if they are in parallel with the capacitor.
     
  7. crutschow

    Expert

    Mar 14, 2008
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    3,232
    The resistances do affect the COF. You have to calculate the equivalent value of all the resistances and then use that to calculate the COF as blah2222 noted. They will also attenuate the signal in the flat portion of the filter response. They have no affect on the filter Q.
     
  8. mike.

    Thread Starter New Member

    Feb 24, 2013
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    This is what I thought when I first posted, but I am not sure how to do this.

    Any help appreciated :)

    Mike
     
  9. tubeguy

    Well-Known Member

    Nov 3, 2012
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    The capacitor acts like a frequency variable resistor. Its AC resistance continues to drop as frequency increases and it eventually 'swamps out' the resistors in parallel with it.
     
  10. Jaguarjoe

    Active Member

    Apr 7, 2010
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    You have 2 R's in parallel. Calculate their equivalent R.
    Now you only have 1 R across the C.
    Calculate the reactance of the C at your frequency of interest.
    Calc the equivalent resistance of the C&R in parallel.
    Now add the 2nd (or is it 3rd?) R into the mix as the bottom leg of a voltage divider and slide into home.

    Divide and conquer.
     
  11. #12

    Expert

    Nov 30, 2010
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    I guess I'm going to have to go back and study filters again.
    I was under the impression that a resistance in series with an inductor lowered the Q, a resistance in parallel with a capacitor lowers the Q, and a flatter line on graph of the frequency response indicates a lower Q.
     
  12. Jaguarjoe

    Active Member

    Apr 7, 2010
    770
    90
    Take it to the extreme-

    If the series R was very big, there wouldn't be much, if any effective capacitance which would then make the Q very small.
     
  13. crutschow

    Expert

    Mar 14, 2008
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    Q can only vary in a passive circuit with both and L and C. If you have only R and C, or R and L then the Q is fixed.
     
    #12 likes this.
  14. crutschow

    Expert

    Mar 14, 2008
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    You can take it to a bizillion ohms if you like but that still won't affect the Q or effective capacitance. A very large R will just make the RC time-constant very large and the consequent COF go to a very low frequency.
     
  15. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Maybe I am picturing this wrong. The calculation is for a series R, parallel C, and it seems the OP is referring to parallel R and parallel C, with no series R.

    Did I miss something?
     
  16. crutschow

    Expert

    Mar 14, 2008
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    In his original post he stated
    That, to me, implies there is also a resistor in series for the low pass function, but I'm open to other interpretations. ;)
     
  17. blah2222

    Well-Known Member

    May 3, 2010
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    That is what I thought as well.
     
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