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A question about the current detection

Discussion in 'Hobby Electronics' started by [email protected], Jan 30, 2009.

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  1. Guest

    Dear friends,

    I have got an AC current detecting circuit. But I failed to
    understand. Can anyone help me? Thanks a lot in advance!
    http://
    images.elektroda.net/17_1233301094_thumb.jpg


    It seems that the circuit between the input node SI and node 8 is
    useless. Anyone agree with me?

    And what is the capacitors C19 and C24 for?
     
  2. Guest

    By the way the current has been converted to be voltage at the input
    node SI.
     
  3. Ross Herbert

    Ross Herbert Guest

    On Fri, 30 Jan 2009 00:00:06 -0800 (PST), ""

    :Dear friends,
    :
    :I have got an AC current detecting circuit. But I failed to
    :understand. Can anyone help me? Thanks a lot in advance!
    : http://
    :images.elektroda.net/17_1233301094_thumb.jpg

    :
    :It seems that the circuit between the input node SI and node 8 is
    :useless. Anyone agree with me?
    :
    :And what is the capacitors C19 and C24 for?


    You need to learn to copy and paste the correct url link.
     
  4. Franc Zabkar

    Franc Zabkar Guest

    The voltage at U4A-1 = SI/2 since R1 and R2 form a potential divider.

    The voltage at U4C-9 = voltage at U4C-10 (if op-amp is operating in
    linear region).

    Therefore voltage at U4C-8 = SI.

    So it would seem that U4A and U4C are functioning as a unity gain
    buffer with a 10K input impedance, in which case I don't understand
    why they are needed, either.

    They reduce the gain of the high frequency components in the signal.

    - Franc Zabkar
     
  5. Guest

    Thank you very much!

    The Capacitors C19 and C24 are for the stability of the opamp, and
    filter for the higher frequency. They make the opamp a lower unity
    gain frequency, and a phase margin of 90 degree.

     
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