Precision half-wave rectifier

Discussion in 'General Electronics Chat' started by Vorador, Oct 12, 2013.

  1. Vorador

    Thread Starter Member

    Oct 5, 2012
    87
    1
    Hello everyone,

    Could anyone please explain how a silicon diode with a forward voltage of about 0.7V is turned on in a precision half-wave rectifier, even when the input is so much smaller than the forward voltage? Also, why isn't the output 0.7V less than the input? I've read my book and searched around on the internet, but I couldn't find any explanation clear enough to help me understand this... :confused:

    Thank you!
     
  2. bertus

    Administrator

    Apr 5, 2008
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  3. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    One word: amplification.
     
  4. crutschow

    Expert

    Mar 14, 2008
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    I'd make that two words and add "feedback". ;)
     
  5. #12

    Expert

    Nov 30, 2010
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    The op-amp only cares about getting its inputs to be the same voltage. It will provide the extra .7 volts and you can measure that voltage on the output pin. The other side of the diode will be the same voltage as the input pin of the chip because that is the whole purpose of building a precision rectifier.
     
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  6. Vorador

    Thread Starter Member

    Oct 5, 2012
    87
    1
    I thought there would be no amplification if it's behaving like a voltage follower.

    Thanks everyone. Appreciate it.
     
  7. crutschow

    Expert

    Mar 14, 2008
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    There is no apparent amplification at the output node and it looks like a follower at that point, but there is across the diode due to the negative feedback.
     
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