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Precision rectifier query

Discussion in 'General Electronics Discussion' started by srxy8, Mar 8, 2013.

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  1. srxy8

    srxy8

    2
    0
    Mar 8, 2013
    Hi referring to my attachment

    I would just like to check if my understanding of this circuit is correct.

    1. When a +ve half cycle of a sine wave goes into first op amp through R1, the expected output of the first op amp is the same as Vin (just negative). Is this correct?

    But at the same time, current flows through R3, and that together with the current through R4 from the output of the first op amp results in a signal with a gain of 1 at output of second op amp.

    Is this correct so far?

    2. For -ve half cycle, the output of first op amp is vout = 0.
    But current is still flowing through R3 and going through second op amp, the output at second op amp is twice the input at first op amp, but positive.

    Again, is this correct?

    Thank you for any help.
    Regards.
     
    Last edited: Mar 8, 2013
  2. Harald Kapp

    Harald Kapp Moderator Moderator

    11,517
    2,652
    Nov 17, 2011
    Mostly correct. It is the voltage at the junction D2|R4 that is controlled to -Vin by U2A. Let's call this intermediate voltage Vx.

    The second amplifier is a summing amplifier. The transfer function is
    Vout=-(2Vin + Vx)

    For Vin >0 D2 is conducting and Vx=-Vin. From that it follows that
    Vout=-(2Vin-Vin)=-Vin This voltage is negative, since Vin>0.

    2) For Vin <0V, Vx=0V (D2 is non-conducting).
    The transfer function of the second OpAmp stays at
    Vout=-(2Vin + Vx)
    which results in
    Vout = -2(Vin+0V) = -2Vin. This voltage is negative, because Vin>0.
     
  3. srxy8

    srxy8

    2
    0
    Mar 8, 2013
    thanks a lot
    :)
     
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