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who can explain this circuits?

Discussion in 'Electronic Design' started by [email protected], Nov 21, 2006.

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  1. Guest

    http://www.analog.com/UploadedFiles/Data_Sheets/AD843.pdf

    On the above datasheet, page 11th, there is a peak detector circuits. I
    will adapt it but I didn't figure out why they put two diodes like this
    and I don't know how the 1k ohm work between the two op-amps either.

    Could you explain it to me, thanks a lot.
     
  2. When the input voltage is positive, the right diode
    conducts, and the pair of opamps act, together to produce a
    follower (100% feedback to the first opamp) that drives the
    storage capacitor to follow the input voltage as it heads
    up. The resistor in series with the capacitor keeps the
    capacitor from causing enough phase shift to destabilize the
    pair of opamps in a feedback loop. The 1 k feedback
    resistor has no current passing through it, except the bias
    current of the first opamp, so it acts essentially like a
    wire during this phase.

    When the signal is negative, the left diode conducts,
    keeping the first opamp from saturating its output all the
    way against the negative rail so that it will recover
    quickly when the signal goes positive, again. During that
    time, the right diode isolates the first opamp form the
    voltage that has been stored across the capacitor.

    The 1 k resistor conducts current from the second opamp
    output back to the first. This current is wasted, but keeps
    from having to disconnect the the feedback path between peaks.
     
  3. Guest

    thanks. then, why not just put a wire instead of a 1 K ohm? and how to
    pick this resistor's value?
     
  4. Tom Bruhns

    Tom Bruhns Guest

    Because if it were zero ohms, A1's more negative output would be
    fighting A2's more positive output. With 1k in between, the current is
    limited to (V(A2out)-V(A1out))*1mA/volt.

    Cheers,
    Tom
     
  5. Guest

    to limit current, a larger value > 1K would be better, apparently
    there are other considerations like phase compensation?
     
  6. You can't tie the two outputs directly together with only a
    diode between them, and have them battle for control of the
    voltage. The value is not critical, but lower resistance
    looks more like a wire when the feedback diode is off, than
    a 10k resistor does (as far as driving the input and stray
    capacitance with little delay or phase shift. But it is
    about as low as you can go to keep the outputs from going
    into current limit when it is being used for isolation (of
    the two outputs) during the lower than peak parts of the
    waveform.
     
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