Separate pulsed component from DC and record it?

Discussion in 'Wireless & RF Design' started by jam_27, Aug 15, 2012.

  1. jam_27

    Thread Starter New Member

    Aug 15, 2012
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    0
    Attached is a circuit with a photo-detector on which a pulsed light (from a laser) is falling. In addition a continuous wave light from a laser is also added which is also falling on the photo-detector simultaneously. The photo-detector is a fast detector and I can easily see the pulsed and the dc current in a scope.

    Now what I want to do is the following:
    I want to separate and record the pulsed response without doing anything to the load resistor (R_load) in the attached circuit, i.e the total load in the circuit cannot be altered.

    I am probably wrong, but a capacitor if used in parallel keeping the total impedance same as R_load would work. The pulse signal could be read across the capacitor? If not, please advice.
     
  2. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    488
    Use a non-inverting voltage follower with an Opamp. This gives you the signal without altering the load resistance.

    Then subtract the DC with another Opamp stage.
     
  3. vk6zgo

    Active Member

    Jul 21, 2012
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    85
    A parallel capacitor wouldn't do it,as it would short out the pulsed signal.

    What you can do,is connect a capacitor to the top end of R_load,& connect the other end into an amplifier with an input impedance Zin>>>R_load.

    The effect of Zin on the load appearing to the photo-detector will be negligible,& the pulsed signal will have the DC removed,& be fed into the amplifier.
    Obviously if you remove the DC,the pulsed waveform will now be balanced around zero volts,but if you want to clamp one level,you can do this later.
     
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