Precision Rectifier

Discussion in 'Homework Help' started by vincent19-mas, Jun 10, 2013.

  1. vincent19-mas

    Thread Starter Member

    Dec 27, 2012
    83
    1
    Dear all, I am trying to understand the precision rectifier as in the attached image.

    As you can see from the attached image, when supply voltage is positive, output voltage will be equal to the supply voltage and V1 will be the output voltage + the voltage drop across diode.

    So I would like to ask, how if it is negative input voltage ? I understand that the diode will be reverse biased and the output voltage will be 0V.

    So, I want to ask, what will be the V1 ? Is it 0.6V ? Or we ignore it ?

    Thanks and appreciate !

    Quite urgent. Exam soon !
     
    • 1.jpg
      1.jpg
      File size:
      18.3 KB
      Views:
      46
    • 2.png
      2.png
      File size:
      230.8 KB
      Views:
      35
    • 3.png
      3.png
      File size:
      210.2 KB
      Views:
      38
  2. Shagas

    Active Member

    May 13, 2013
    802
    74
  3. t06afre

    AAC Fanatic!

    May 11, 2009
    5,939
    1,222
    Ouch that was a very bad design idea. Do not use that design in any real world application. That said I understand it is an academic exercise ;)
    For any negative input the diode will block any input and the output will be tied to ground via the resistor R. For positive signal think of the diode as 0.6 volt battery with the negative pole connected to the resistor R. The opamp will try to force the inputs to the same voltage, hence it will always output 0.6 volt more (at the output pin) than the input
     
  4. vincent19-mas

    Thread Starter Member

    Dec 27, 2012
    83
    1
    Do you mean for the positive input, I will get 2.6 as the output after the op-amp ? and get 2.6 - 0.6 = 2.0 volts at Vo ?

    For negative inputs, what will be the voltage V1 after the op-amp ?

    Thanks !
     
    Last edited: Jun 10, 2013
  5. t06afre

    AAC Fanatic!

    May 11, 2009
    5,939
    1,222
    Yes a opamp will always try to make the input voltage (V+)-(V-) equal to zero. In order to do that it has to compensate for the diode voltage drop.
     
  6. crutschow

    Expert

    Mar 14, 2008
    13,022
    3,236
    With a negative input, the feedback loop becomes open and the op amp output voltage goes to its negative rail (as determined by the negative supply voltage).
     
Loading...