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Using capacitors in bridge t-type

Discussion in 'General Electronics Discussion' started by pump, Mar 15, 2012.

  1. pump

    pump

    24
    0
    Nov 4, 2011
    Hope somebody could help me explain this from the circuit below.....What effect do the capacitors have?
     
    Last edited: Mar 16, 2012
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,191
    2,693
    Jan 21, 2010
    You may have to try posting the image or link again...
     
  3. pump

    pump

    24
    0
    Nov 4, 2011
    Thanks will try again
     

    Attached Files:

  4. pump

    pump

    24
    0
    Nov 4, 2011
    Not sure but my first thought is that the configuration will only attenuate under a certian frequency. If over this frequency then te signal will not get attenuated?

    Is that correct?
    :confused:
     
  5. pump

    pump

    24
    0
    Nov 4, 2011
    BUMP :eek:
     
  6. poor mystic

    poor mystic

    1,059
    28
    Apr 8, 2011
    The circuit looks incomplete to me; I think that in order to understand it you need to consider the load, which will be most easily modelled as a pure resistance across the right hand side of the circuit.
    Now we see that the capacitors are in series with the load, so that they have a "high pass" characteristic.
     
  7. pump

    pump

    24
    0
    Nov 4, 2011
    Thanks for your reply poor mystic,

    Yes that circuit is incomplete I removed the load just for the picture. So my thoughts were correct. At high frequency the input signal will not be attenuated?
     
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,191
    2,693
    Jan 21, 2010
    As has been suggested, the load is all-important.

    If the impedance of the load is high enough, the circuit won't attenuate anything at all.

    The lower the load impedance the lower the frequency the 3db point moves.
     
  9. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    My analysis of the circuit is in the attachment. The node equations are on the first line. These are manipulated into a transfer function of the form, 20log(Vout/Vin). For the graph, component values selected have Rp=Rl so for very low frequency or DC input, Vo/Vi=0.5 or 6dB. Very high input frequency passes through the capacitors without attenuation, or 0dB. The depth of the notch seems to be related to the value of Rt.
     

    Attached Files:

  10. pump

    pump

    24
    0
    Nov 4, 2011
    Laplace thank you for the explaination. Thats quite a bit of work you have put in there :)

    Thank you all for helping me understand this circuit. Sometimes reserching by yourslef isn't enough, by speaking to people like yousleves and having it explained in plain English is whats needed. Thanks again for all your help
     
    Last edited: Mar 17, 2012
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