Help understanding this rectifier IC

Discussion in 'General Electronics Chat' started by diddy02, Aug 12, 2010.

  1. diddy02

    Thread Starter Member

    Sep 26, 2008
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  2. timrobbins

    Active Member

    Aug 29, 2009
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    Which bit don't you understand, or alternatively which bits are you seeking more details on and what type of detail are you interested in (eg. do you want to know the temperature drift of the gain or something esoteric?
     
  3. diddy02

    Thread Starter Member

    Sep 26, 2008
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    Having trouble understanding the basic principles of operation during the negative half cycle of the AC input... I don't need parameter/characteristic details, just want the working principles.

    * During the positive phase of the input AC voltage, the superdiode(?) thing acts as a short, so you're left with an op-amp, with negative feedback, but with both input terminals connected to the same potential. Shouldn't the output = 0?

    * During the negative phase, the superdiode(?) acts as ground, so you're left with an inverting op-amp. The negative input terminal has a negative voltage applied and the positive terminal is grounded... the feedback loop quickly brings the negative input terminal to ~0V > 0V, and so the output is (+ve). Is this right?
     
    Last edited: Aug 12, 2010
  4. timrobbins

    Active Member

    Aug 29, 2009
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    I suggest you try to interpret the datasheet for the device, especially the section on rail-to-rail input. It indicates that U2 will continue to operately as a gain amp as the neg input goes very slightly below 0V rail. As such, the U2 output will start going high by open-loop gain, and the diode D1 will feedback to give a steady state operating point where the neg input of U2 is driven at pretty much 0V.
     
  5. diddy02

    Thread Starter Member

    Sep 26, 2008
    10
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    Tim,

    Thanks for the reply... I wanted to clarify one further point... why does U1 output something other that 0 during the positive phase... aren't both input terminals connected to the same voltage?
     
  6. timrobbins

    Active Member

    Aug 29, 2009
    318
    16
    You need to review the various amplifier schemes for opamps. There is a good TI tutorial app note - will post link later today.
     
  7. sage.radachowsky

    Member

    May 11, 2010
    241
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    Wow, what an interesting circuit!

    During the positive phase of input voltage, then U2 has no effect on the output, because its output is at ground and the diode prevents any influence of the output of U2 on the input of U1. So then, U1 becomes a noninverting amplifier with a gain of 1, which is actually simply a voltage follower.

    During the negative phase of the input voltage, then U2 acts as a ground clamp on the noninverting input of U1, because the output of U2 is high just as much as it takes to keep the inverting input at ground (because that is equal to the noninverting input). Basically, it just has to cover enough forward voltage drop on the diode to bias the inputs and fight the 75K resistor to keep the inverting input of U2 at ground. In that case, then U1 is an inverting amplifier with gain of 1, so it output the negative of the input voltage, which is positive.

    The result is the absolute value of the input voltage.
     
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