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OpAmp Precision Rectifier

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  1. Apr 2, 2015 #1
    Hi all,

    I am currently studying OpAmps and one of the topics that came up was the applications of OpAmp circuits. In particular, I was learning about precision rectifiers (using non-inverting version of it, thus the diode is pointing away from the "out" terminal of the OpAmp). The thing I cannot figure out is how does the OpAmp bypass the Vf limitations that regular full bridge rectifiers face. In other words, say for V_in = 0, the voltage at the inverting terminal is close to 0 as we assume an ideal device. Therefore, the voltage at the output terminal is 0.7 volts higher than that of the inverting input. My assumption is that if V_in remains at 0, the voltage at the output terminal will stay at 0.7 volts.

    To summarize the question, what properties of the OpAmp cause it to create and sustain that voltage difference that keeps the diode "on". Isn't it just easier for the OpAmp to establish 0 volts at the output and thus 0 volts at the inverting input node?
     
  2. jcsd
  3. Apr 2, 2015 #2

    NascentOxygen

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    Staff: Mentor

  4. Apr 2, 2015 #3
    My wording of the question could be confusing as I am trying to learn rather than stating something I know. But I came across this concept multiple times - that the voltage required to turn on the diode is supplied by the OpAmp rather than by the small signal. That is the reason why the precision rectifier is so useful for small signals that barely or don't at all exceed 0.7 volts
     
  5. Apr 3, 2015 #4

    NascentOxygen

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    And that's true. When you examine the circuit you can see that the output of the OP-AMP is not the point from which we take the output of the circuit. http://en.m.wikipedia.org/wiki/Precision_rectifier

    The OP-AMP output will be 0.7V more than the voltage we desire, but we don't take our rectified output from the OP-AMP output pin.
     
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