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  • opamp
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Related

Experimenting with Op Amps: active full-wave rectifier

Jan Cumps
Jan Cumps over 2 years ago

The Art Of Electronics features a set of practical op-amp circuits. They have several full-wave rectifiers. I'm trying the simplest one. You don't need fancy devices. A pair of 741s will do.

The circuit:

image
image source: my copy of the book. fair use.

Looking at the node where the arrow points to:

  • for positive values, the first op-amp is an inverting buffer.
  • for negative values, the op-amp output has a  0.6V dc bias, the right side of  the feedback resistor will be at 0 V (ground). 
    hint: the 0.6 V output is generated by the op-amp to drive the inverting input to the same value as the + input (ground), effectively clamping the output to 1 diode drop above ground.

image

Looking at the second op-amp:

  • it is an op-amp with a summing node. Two signals at the inverting input are summed and inverted. The amplification factor is defined by the input resistor and the feedback resistor.
  • for the input signal, the amplification = -1 * R/R = -1 (the input resistance equals the feedback resistance, resulting in the -1 amplification factor, doubling and inverting the signal)
  • for the signal coming from op-amp 1, the amplification = -1 * R / (R/2) = -2 (the input resistance is half the feedback resistance, resulting in the -2 amplification factor)
  • it is an inverting amplifier for the input. Both positive and negative signals are amplified by -1.
  • for the signal coming from the first op-amp, it acts as a -2 amplifier.
  • for positive inputs, this results in  Vin * -1 + Vin * -1 * -2 = -Vin + 2Vin = Vin ->  result: Vin, positive Vin comes out as positive Vin (see comment)
  • for negative inputs, this results in Vin * -1 (making Vin positive)  + Vin * 0 * -1 -> result -Vin: negative Vin comes out as positive Vin

image

Here is the result: The blue line is the output. Underneath it, you see the input. The output overlaps the input for the positive part. It inverses the negative part. No diode drop.

image

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  • Jan Cumps
    Jan Cumps over 2 years ago +5
    If you are puzzled by the circuit, there is a way to understand it: Build the first op-amp circuit and feed a sinus into it. Try it with and without the vertical diode. Try to understand why it is…
  • anniel747
    anniel747 over 2 years ago in reply to Jan Cumps +4
    Deleted
  • javagoza
    javagoza over 2 years ago in reply to anniel747 +4
  • Jan Cumps
    Jan Cumps over 2 years ago in reply to michaelkellett
    michaelkellett said:
    Just for fun try scoping the output of A1 and high frequencies

    40 kHz:

    image

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  • Jan Cumps
    Jan Cumps over 2 years ago in reply to Jan Cumps
    michaelkellett said:
    Just for fun try scoping the output of A1 and [...] with a low amplitude signal

    500 Hz, 70 mVpp. I had to turn on averaging because the whole scope screen was smeared with noise:

    image

    (in this capture, channel 3 is in different vertical range than the 3 other channels)

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  • michaelkellett
    michaelkellett over 2 years ago in reply to Jan Cumps

    The first op amp in this design has a hard time because it has to swing the diode drop and more in a small fraction of a cycle time. The smaller the signal the worse the effect.

    Some of the true RMS chips have a similar problem, good frequency response with large signals but much less good with small. I can remember struggling to get the required accuracy over amplitude and frequency in a meter design for BT in about 1978.

    MK

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  • Jan Cumps
    Jan Cumps over 2 years ago in reply to Jan Cumps

    Measurements when the input is  +3.23 V dc.

    image

    You can see here that for positive inputs, the horizontal diode does not conduct, and the 10K feedback resistor comes into play.
    The 3.24 V input is inverted, and its inverse is passed to circuit 2.

    The second op-amp, the inverting summing amp, has different gain for both inputs. The gain = -(Rfeedback / Rin)
    For the original +3.23 V input, with gain -1 (-10K/10K), we'd get -3.23 V output.
    For the -3.23 V V signal coming from the first circuit, with gain -2 (-10K/5K), we'd get +6.46V output
    Summed up, this is +3.23 V output.

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  • Jan Cumps
    Jan Cumps over 2 years ago in reply to michaelkellett
    michaelkellett said:
    The first op amp in this design has a hard time because it has to swing the diode drop and more in a small fraction of a cycle time.

    The struggle is visible on this 11 kHz capture (input: 2Vpp)

    image

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  • DAB
    DAB over 2 years ago

    Very cool experiment Jan.

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  • Jan Cumps
    Jan Cumps over 2 years ago in reply to Jan Cumps

    (I burned half the weekend reading that book)

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  • flyingbean
    flyingbean over 2 years ago
    Jan Cumps said:
    for positive inputs, this results in  Vin * -1 + Vin * -2 = Vin ->  result: Vin, positive Vin comes out as Vin

    I think for positive input, Vin*(-1) +(-1)*Vin *(-2) = Vin -> result: Vin

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  • jc2048
    jc2048 over 2 years ago

    This is another one you see sometimes. I got this from Analog Signal Processing by R Pallas-Areny and J G Webster. One less resistor and high impedance input. Works ok in a simulator.

    image

    image
    Has same problems as all of these when the input transitions through zero and it switches between the two modes of operation.

    If you want to practice your circuit-analysing skills (in-head simulation), there is a potential problem with it that makes it less flexible than the one you showed.

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  • anniel747
    anniel747 over 2 years ago in reply to Jan Cumps
    [deleted]
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