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Precision active half-wave rectifier

Discussion in 'Electronic Design' started by MRW, Oct 25, 2006.

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  1. MRW

    MRW Guest

    Circuit link:

    In Figure 1 of the above link (Basic Precision Halfwave Rectifier), how
    would I analyse this circuit? At first, I thought R1 and R2 didn't
    really do anything because it seemed like they weren't part of any
    feedback networks, but when I ran a simulation without them, the
    circuit did not rectify.

    Any tips?

    I initially analyzed the circuit by replacing the diode with a voltage
    drop. I assumed 0.7V, so I wrote my output as Vout = Immediate voltage
    output of opamp - 0.7 V. Then, set Vout equal to my Vin. That's why I
    had the impression that R1 and R2 did not do anything.

  2. john jardine

    john jardine Guest

    R1 shouldn't really be needed. It's just defining the input voltage as being
    wrt 0V.
    R2 in reality wouldn't normally be explicity shown as part of the circuit.
    Some kind of lowish value is needed though in order to hold the output
    voltage near 0V when the opamp and it's reverse biased diode has flopped
    down to near the negative rail (any negative input voltage).

    Without the resistor your sim will be measuring the output voltage with some
    kind of infinite resistance voltmeter and the multi megohms due to a
    reversed biased diode, would look essentially like a short circuit.
  3. Guest

    Download LTSpice from Then you can see the waveforms
    everywhere in the circuit.

    - Tom Gootee
  4. Tam/WB2TT

    Tam/WB2TT Guest

    Do you actually want the half wave rectified sine wave, or a DC voltage
    equal to the peak?

  5. Jamie

    Jamie Guest

    R1 is there for capacitive/inductive coupling., the input of the OP-amp
    could be bipolar, thus you need to rid that diode effect and give some
    in range impedance for the AC source.
    the R2 is needed to create load on the OP-amp since, at the zero state,
    you will not have any load coupling for the next stage. A Oscope with
    AC input set for example, would not work very well because of the
    diode in line.
    there are other more in death reasons for the state of the op-amp,
    but i won't get into that.

    That is about as simple as i can explain it.
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