Noob alert: Low pass audio filter help

Discussion in 'The Projects Forum' started by skase, Oct 14, 2008.

  1. skase

    Thread Starter New Member

    Oct 14, 2008
    2
    0
    Hi everyone,

    I have very little electronics experience so be nice :)

    I'm trying to build a simple acoustic stomp box which takes an audio signal from a small piezo transducer, filters out all the high frequencies above say 150Hz and sends the output to a auido jack which is plugged into an amp. It's a simple device to replicate a bass drum sound commonly used in blues. (The transducer sits on the wall inside a hollow wooden box to amplify the sound of a foot taping on it)

    The circuit below seems to be what I'm after however my limited knowledge of this stuff has me stumped on a few (read most) things:
    1. What does the circle represent?
    2. Where exactly I would wire in the transducer and jack output? I'm guesing they sit in paralell with the unknown component but am not 100%.
    3. Would this need to be powered at all?
    4. If I wanted to adjust the cut-off to 150Hz what changes would I make.

    [​IMG]

    This probably seems pretty basic for you guys, sorry. Any diagrams (layman or otherwise) would be helpful. :D

    Thanks in advance!
     
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
  3. skase

    Thread Starter New Member

    Oct 14, 2008
    2
    0
    Thanks for the reply.

    Just so I'm clear then, the ground from my output jack would connect to point 0 and the active to point 2 on the right of the diagram.

    My transducer active wire would connect to point 1 and the ground to point 0 on the left. Then my ground connection can be anywhere along the bottom.

    And according to my calculations I need around a 150 ohm resistor in position R1 to give me my 150Hz cut-off.

    No additional power is required?

    Is this right? Cheers again.
     
  4. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    Yes but i forgot to mention something about the cut-off frequency.

    It is equal to:

    f=1/(2*pi*(R1//Rload)*C1)

    R1//Rload means R1 in parallel with Rload

    so make your calculations again.

    Also, you have to take into account how much current the output of your pick up can source/sink as not to burn it. Do you have its specifications?
     
Loading...