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Precision Rectifier

Discussion in 'Electronics Homework Help' started by Syed Waqar Ali, Jan 13, 2016.

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  1. Syed Waqar Ali

    Syed Waqar Ali

    16
    0
    Jan 13, 2016
    Hello Guys..

    while working on a precision rectifier, i found a complication as it's rectified output is not symmetrical,someone has any solution to this problem.How can i fix it while using same circuit ? you can see it in the fig below.
    As far as circuit is concern i am using TL072 (IC's) there... while circuit diagram is also provided..
    Thanking in anticipation
     

    Attached Files:

  2. hevans1944

    hevans1944 Hop - AC8NS

    4,617
    2,156
    Jun 21, 2012
    Looks like something is clipping (flat top) and slew-rate limiting (kink in the waveform) your signal. Why is the input attenuated with a 10 Meg-ohm/150 kΩ divider? Is there a clean sine wave on the top end of the 10 Meg-ohm resistor? Is there a cleanly reduced version of the sine wave on the output of U1A? Why is RV2 variable with no provision for a minimum resistance? RV2 should be a fixed resistor about the same value as R4, R5. R6, and R7. Why don't you use a variable resistance for R2 to control the "gain" of absolute value output from U1B? What is the amplitude of the input sine wave? What frequency is the input sine wave?

    This is a "classic" absolute value or precision rectifier circuit, but it only works well at low frequencies with fast signal diodes (such as 1N4148 shown on schematic) and op-amps with high slew rates. Stray capacitance can lead to poor results. The resistors R4, R5, R6 and R7 associated with feedback for U2A and U2B must be closely matched for good symmetry in the output.
     
  3. Syed Waqar Ali

    Syed Waqar Ali

    16
    0
    Jan 13, 2016
    Input Signal is about 110 volts (rms) having frequency 50Hz,
    we used a divider because we need to reduce the amplitude of the voltage to some optimum level where we can use it (like 5 volts output signal to feed in other circuitry) ...
    Yes we are obtaining a fine sine wave at the output of U1A but it makes problem at the final output..
    is this due to IN4148 (they are not matched ) ???
     
  4. hevans1944

    hevans1944 Hop - AC8NS

    4,617
    2,156
    Jun 21, 2012
    I doubt the 1N4148 diodes are a problem. They do not have to be matched for the circuit to work properly. Fifty hertz (or even 60 Hz) is a low enough frequency that slew rate in the op-amps should not be a problem. Since you get a clean sine wave at the output of U1A, there is no problem with the input voltage divider. The only thing I can see that might be a problem is the variable resistance, RV2, being set at too small a value. Try the obvious things first, such as checking all the connections to make sure they really are good connections, and replacing the dual op-amps, U2A and U2B. Make sure U2 is getting both + and - power on pins 8 and 4 and that the power connections are by-passed with at least 0.1 μF capacitors... the usual "good engineering practices".

    This may seem totally obvious: make sure the power supply is actually on. I once spent a whole morning trying to troubleshoot a prototype analog op-amp circuit before discovering the power supply was off. Back in the day, we always prototyped on the workbench with a lab power supply, separate from the circuits we were working with. It was pretty easy to turn off the power supply before making circuit changes to the prototype circuit and then forget to turn it back on.

    This could also be a problem related to ground faults associated with the oscilloscope measurement. I would use a small 6.3 VAC filament transformer to obtain an isolated, low-level, sine wave to supply input to the absolute-value circuit. You can connect a 10 kΩ potentiometer across the secondary winding, connect one end of that potentiometer to circuit common, and connect the wiper through a 10 kΩ (or larger) resistor to the summing junction (inverting input) of U2A. All this would replace RV2, U1, and the input divider circuitry.

    Make sure the oscilloscope probe has a "ground clip" you can connect to your circuit common. If there is a change in waveform with the "ground clip" removed, you have a ground fault. Operating the oscilloscope from a line isolation transformer could help with that.

    Other than the above suggestions, I don't know what else to suggest that you can do. This circuit has always operated "as advertised" for me. The only anomaly I ever experienced with this circuit was a slight distortion near zero output that occurs at the zero-crossing of the sine wave, where one diode is turning on and the other is turning off. The kinks and flat-topping were never observed with my implementation. They are, however, an indication that the absolute-value circuit is being over-driven at the input.
     
    Arouse1973, dorke and chopnhack like this.
  5. dorke

    dorke

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    Jun 20, 2015
    I would like to add:
    Your simulation diagram shows the power connections to the op. amps to be a dual power supply.
    Is that the same in the circuit you actually build,or are you using a single power supply there?

    What are the actual voltage values of the power supplies measured on the op.amps relative to GND?Are they matched?
    How are the power supplies connected to the circuit ?
    How is the GND connected?

    EDIT:What is the accuracy of the 10K resistors you are using? a pic of one may help.


    Can you post some clear photos of the actual circuit you have build and the connections to it.

    Please fill-in the Voltage levels on the pic below ,and mark the power supplies voltages on the pic as well.

    IMG_20160112_084231.jpg
     
    Last edited: Jan 15, 2016
  6. Syed Waqar Ali

    Syed Waqar Ali

    16
    0
    Jan 13, 2016
    Yes It is dual Supply ....
    The biasing voltages supplied to opamps are exactly same +15 and -15 to both...
    Ground of the whole circuit is common...
    Accuracy of 10K resistance is vary about 5%...
    Circuit diagram (which is implemented) is provided in the pdf below...
     

    Attached Files:

    • Doc1.pdf
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  7. Syed Waqar Ali

    Syed Waqar Ali

    16
    0
    Jan 13, 2016
    Thanks alot !!
    i am trying by bringing all these effects under consideration..
     
  8. dorke

    dorke

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    Jun 20, 2015
    "Accuracy of 10K resistance is vary about 5%..."

    That is not good enough you should go below 1%, if possible even 0.1%.
    In your simulation you can verify the bad distortion effects of 5% resistors.
    Create the worst unmatched situation with R4-R7 values.
    min- 9500 ohm
    max- 10500 ohm
    In this case the distortion can be upto 20% between the 2 half cycles.

    "Ground of the whole circuit is common"

    Yes,but the physical way it is connected is very important.
     
    Last edited: Jan 15, 2016
  9. Syed Waqar Ali

    Syed Waqar Ali

    16
    0
    Jan 13, 2016
    I did it by bringing every important thing into consideration but problem is still there.....
    As u can see in the pic.....
     

    Attached Files:

  10. dorke

    dorke

    2,342
    665
    Jun 20, 2015
    Try this:
    Split R7 into a 7.5K fixed resistor in series with a 5K potentiometer(other combinations are also possible to get at least 10k +/- 2k.)
    That will allow you to balance the gain of the two half cycles.
     
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