How does this opamp full-wave rectifier work?

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I want to understand how the following circuit works. I understand OpAmps, diodes, and voltage dividers pretty well; but put this circuit together, and I don't know how it affects positive or negative voltage input.



enter image description here






op-amp rectifier







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  • D1 is in the middle of a feedback loop, so it becomes an 'ideal' diode, up to the limits of what the op-amp can handle. The topology still has a gain of -1.
    – Sparky256
    4 hours ago






  • 1




    @TonyEErocketscientist. Look again Tony. It is a full wave rectifier. Think about the path for both positive and negative inputs.
    – Sparky256
    4 hours ago










  • @TonyEErocketscientist. I agree with you Tony and put that in my answer. It is NOT an ideal FW rectifier because of R1+R2 add 2K of impedance to positive inputs. As you commented and extra op-amp and diode is needed. This is real old-school stuff...
    – Sparky256
    4 hours ago










  • @TonyEErocketscientist - what 0V? With a diode in the feedback loop, the -ve input to the opamp is only 0V for a -ve input voltage. +ve input is simply fed straight through the 1k + 1k. So as long as you can tolerate a 2k impedance, it's a full-wave rectifier.
    – brhans
    4 hours ago











  • @TonyEErocketscientist. My "off your meds" comment was not meant to be an insult, but a wake-up call. No more than that.
    – Sparky256
    4 hours ago














up vote
1
down vote

favorite












I want to understand how the following circuit works. I understand OpAmps, diodes, and voltage dividers pretty well; but put this circuit together, and I don't know how it affects positive or negative voltage input.



enter image description here






op-amp rectifier







share|improve this question























  • D1 is in the middle of a feedback loop, so it becomes an 'ideal' diode, up to the limits of what the op-amp can handle. The topology still has a gain of -1.
    – Sparky256
    4 hours ago






  • 1




    @TonyEErocketscientist. Look again Tony. It is a full wave rectifier. Think about the path for both positive and negative inputs.
    – Sparky256
    4 hours ago










  • @TonyEErocketscientist. I agree with you Tony and put that in my answer. It is NOT an ideal FW rectifier because of R1+R2 add 2K of impedance to positive inputs. As you commented and extra op-amp and diode is needed. This is real old-school stuff...
    – Sparky256
    4 hours ago










  • @TonyEErocketscientist - what 0V? With a diode in the feedback loop, the -ve input to the opamp is only 0V for a -ve input voltage. +ve input is simply fed straight through the 1k + 1k. So as long as you can tolerate a 2k impedance, it's a full-wave rectifier.
    – brhans
    4 hours ago











  • @TonyEErocketscientist. My "off your meds" comment was not meant to be an insult, but a wake-up call. No more than that.
    – Sparky256
    4 hours ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I want to understand how the following circuit works. I understand OpAmps, diodes, and voltage dividers pretty well; but put this circuit together, and I don't know how it affects positive or negative voltage input.



enter image description here






op-amp rectifier







share|improve this question















I want to understand how the following circuit works. I understand OpAmps, diodes, and voltage dividers pretty well; but put this circuit together, and I don't know how it affects positive or negative voltage input.



enter image description here






op-amp rectifier




op-amp rectifier






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 4 hours ago









brhans

8,39421826




8,39421826










asked 4 hours ago









Bee

1259




1259











  • D1 is in the middle of a feedback loop, so it becomes an 'ideal' diode, up to the limits of what the op-amp can handle. The topology still has a gain of -1.
    – Sparky256
    4 hours ago






  • 1




    @TonyEErocketscientist. Look again Tony. It is a full wave rectifier. Think about the path for both positive and negative inputs.
    – Sparky256
    4 hours ago










  • @TonyEErocketscientist. I agree with you Tony and put that in my answer. It is NOT an ideal FW rectifier because of R1+R2 add 2K of impedance to positive inputs. As you commented and extra op-amp and diode is needed. This is real old-school stuff...
    – Sparky256
    4 hours ago










  • @TonyEErocketscientist - what 0V? With a diode in the feedback loop, the -ve input to the opamp is only 0V for a -ve input voltage. +ve input is simply fed straight through the 1k + 1k. So as long as you can tolerate a 2k impedance, it's a full-wave rectifier.
    – brhans
    4 hours ago











  • @TonyEErocketscientist. My "off your meds" comment was not meant to be an insult, but a wake-up call. No more than that.
    – Sparky256
    4 hours ago
















  • D1 is in the middle of a feedback loop, so it becomes an 'ideal' diode, up to the limits of what the op-amp can handle. The topology still has a gain of -1.
    – Sparky256
    4 hours ago






  • 1




    @TonyEErocketscientist. Look again Tony. It is a full wave rectifier. Think about the path for both positive and negative inputs.
    – Sparky256
    4 hours ago










  • @TonyEErocketscientist. I agree with you Tony and put that in my answer. It is NOT an ideal FW rectifier because of R1+R2 add 2K of impedance to positive inputs. As you commented and extra op-amp and diode is needed. This is real old-school stuff...
    – Sparky256
    4 hours ago










  • @TonyEErocketscientist - what 0V? With a diode in the feedback loop, the -ve input to the opamp is only 0V for a -ve input voltage. +ve input is simply fed straight through the 1k + 1k. So as long as you can tolerate a 2k impedance, it's a full-wave rectifier.
    – brhans
    4 hours ago











  • @TonyEErocketscientist. My "off your meds" comment was not meant to be an insult, but a wake-up call. No more than that.
    – Sparky256
    4 hours ago















D1 is in the middle of a feedback loop, so it becomes an 'ideal' diode, up to the limits of what the op-amp can handle. The topology still has a gain of -1.
– Sparky256
4 hours ago




D1 is in the middle of a feedback loop, so it becomes an 'ideal' diode, up to the limits of what the op-amp can handle. The topology still has a gain of -1.
– Sparky256
4 hours ago




1




1




@TonyEErocketscientist. Look again Tony. It is a full wave rectifier. Think about the path for both positive and negative inputs.
– Sparky256
4 hours ago




@TonyEErocketscientist. Look again Tony. It is a full wave rectifier. Think about the path for both positive and negative inputs.
– Sparky256
4 hours ago












@TonyEErocketscientist. I agree with you Tony and put that in my answer. It is NOT an ideal FW rectifier because of R1+R2 add 2K of impedance to positive inputs. As you commented and extra op-amp and diode is needed. This is real old-school stuff...
– Sparky256
4 hours ago




@TonyEErocketscientist. I agree with you Tony and put that in my answer. It is NOT an ideal FW rectifier because of R1+R2 add 2K of impedance to positive inputs. As you commented and extra op-amp and diode is needed. This is real old-school stuff...
– Sparky256
4 hours ago












@TonyEErocketscientist - what 0V? With a diode in the feedback loop, the -ve input to the opamp is only 0V for a -ve input voltage. +ve input is simply fed straight through the 1k + 1k. So as long as you can tolerate a 2k impedance, it's a full-wave rectifier.
– brhans
4 hours ago





@TonyEErocketscientist - what 0V? With a diode in the feedback loop, the -ve input to the opamp is only 0V for a -ve input voltage. +ve input is simply fed straight through the 1k + 1k. So as long as you can tolerate a 2k impedance, it's a full-wave rectifier.
– brhans
4 hours ago













@TonyEErocketscientist. My "off your meds" comment was not meant to be an insult, but a wake-up call. No more than that.
– Sparky256
4 hours ago




@TonyEErocketscientist. My "off your meds" comment was not meant to be an insult, but a wake-up call. No more than that.
– Sparky256
4 hours ago










2 Answers
2






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up vote
4
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Though not ideal it is a full wave rectifier.



For a negative input it is inverted by the op-amp thus becomes a positive output of the opposite polarity, passing through the diode. R1 and R2 set the gain at -1 so a negative input (compared to ground) becomes a positive value at the output.



For a positive input it is inverted but the diode will not let a negative value pass, thus disabling the op-amp and R1 + R2 become bypass resistors so a positive value appears at the outputs, but with 2K of resistance as they have no buffer. If the load is very light or buffered it will behave more like an ideal FW rectifier.



For a true full wave rectifier you need 2 op-amps and 2 diodes, so each polarity can be buffered.






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    up vote
    0
    down vote













    enter image description here
    See the 1st trace for Diode V is only < 0.25V not 0.7 because diode current with 10k is only 10uA here with 100mV input. Place with drag and drop components. Falstad is awesome to learn new things but here uses an "Ideal OpAmp" that I defined only by right mouse properties.



    Rather than explain every detail, try to examine diode voltage and current and load effects and use your training and logic to figure it out for each polarity.



    My simulation for you. You cannot put a cap. on this output to get an envelope voltage due to a mismatched output impedance of the diode switch in series with the output. Subtle errors arise from your choice of FET input OP Amp. So choose wisely and go for a buffered dual Op Amp FW rectifier design (search web)






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      2 Answers
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      2 Answers
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      up vote
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      Though not ideal it is a full wave rectifier.



      For a negative input it is inverted by the op-amp thus becomes a positive output of the opposite polarity, passing through the diode. R1 and R2 set the gain at -1 so a negative input (compared to ground) becomes a positive value at the output.



      For a positive input it is inverted but the diode will not let a negative value pass, thus disabling the op-amp and R1 + R2 become bypass resistors so a positive value appears at the outputs, but with 2K of resistance as they have no buffer. If the load is very light or buffered it will behave more like an ideal FW rectifier.



      For a true full wave rectifier you need 2 op-amps and 2 diodes, so each polarity can be buffered.






      share|improve this answer


























        up vote
        4
        down vote













        Though not ideal it is a full wave rectifier.



        For a negative input it is inverted by the op-amp thus becomes a positive output of the opposite polarity, passing through the diode. R1 and R2 set the gain at -1 so a negative input (compared to ground) becomes a positive value at the output.



        For a positive input it is inverted but the diode will not let a negative value pass, thus disabling the op-amp and R1 + R2 become bypass resistors so a positive value appears at the outputs, but with 2K of resistance as they have no buffer. If the load is very light or buffered it will behave more like an ideal FW rectifier.



        For a true full wave rectifier you need 2 op-amps and 2 diodes, so each polarity can be buffered.






        share|improve this answer
























          up vote
          4
          down vote










          up vote
          4
          down vote









          Though not ideal it is a full wave rectifier.



          For a negative input it is inverted by the op-amp thus becomes a positive output of the opposite polarity, passing through the diode. R1 and R2 set the gain at -1 so a negative input (compared to ground) becomes a positive value at the output.



          For a positive input it is inverted but the diode will not let a negative value pass, thus disabling the op-amp and R1 + R2 become bypass resistors so a positive value appears at the outputs, but with 2K of resistance as they have no buffer. If the load is very light or buffered it will behave more like an ideal FW rectifier.



          For a true full wave rectifier you need 2 op-amps and 2 diodes, so each polarity can be buffered.






          share|improve this answer














          Though not ideal it is a full wave rectifier.



          For a negative input it is inverted by the op-amp thus becomes a positive output of the opposite polarity, passing through the diode. R1 and R2 set the gain at -1 so a negative input (compared to ground) becomes a positive value at the output.



          For a positive input it is inverted but the diode will not let a negative value pass, thus disabling the op-amp and R1 + R2 become bypass resistors so a positive value appears at the outputs, but with 2K of resistance as they have no buffer. If the load is very light or buffered it will behave more like an ideal FW rectifier.



          For a true full wave rectifier you need 2 op-amps and 2 diodes, so each polarity can be buffered.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 4 hours ago

























          answered 4 hours ago









          Sparky256

          10.1k21434




          10.1k21434






















              up vote
              0
              down vote













              enter image description here
              See the 1st trace for Diode V is only < 0.25V not 0.7 because diode current with 10k is only 10uA here with 100mV input. Place with drag and drop components. Falstad is awesome to learn new things but here uses an "Ideal OpAmp" that I defined only by right mouse properties.



              Rather than explain every detail, try to examine diode voltage and current and load effects and use your training and logic to figure it out for each polarity.



              My simulation for you. You cannot put a cap. on this output to get an envelope voltage due to a mismatched output impedance of the diode switch in series with the output. Subtle errors arise from your choice of FET input OP Amp. So choose wisely and go for a buffered dual Op Amp FW rectifier design (search web)






              share|improve this answer


























                up vote
                0
                down vote













                enter image description here
                See the 1st trace for Diode V is only < 0.25V not 0.7 because diode current with 10k is only 10uA here with 100mV input. Place with drag and drop components. Falstad is awesome to learn new things but here uses an "Ideal OpAmp" that I defined only by right mouse properties.



                Rather than explain every detail, try to examine diode voltage and current and load effects and use your training and logic to figure it out for each polarity.



                My simulation for you. You cannot put a cap. on this output to get an envelope voltage due to a mismatched output impedance of the diode switch in series with the output. Subtle errors arise from your choice of FET input OP Amp. So choose wisely and go for a buffered dual Op Amp FW rectifier design (search web)






                share|improve this answer
























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  enter image description here
                  See the 1st trace for Diode V is only < 0.25V not 0.7 because diode current with 10k is only 10uA here with 100mV input. Place with drag and drop components. Falstad is awesome to learn new things but here uses an "Ideal OpAmp" that I defined only by right mouse properties.



                  Rather than explain every detail, try to examine diode voltage and current and load effects and use your training and logic to figure it out for each polarity.



                  My simulation for you. You cannot put a cap. on this output to get an envelope voltage due to a mismatched output impedance of the diode switch in series with the output. Subtle errors arise from your choice of FET input OP Amp. So choose wisely and go for a buffered dual Op Amp FW rectifier design (search web)






                  share|improve this answer














                  enter image description here
                  See the 1st trace for Diode V is only < 0.25V not 0.7 because diode current with 10k is only 10uA here with 100mV input. Place with drag and drop components. Falstad is awesome to learn new things but here uses an "Ideal OpAmp" that I defined only by right mouse properties.



                  Rather than explain every detail, try to examine diode voltage and current and load effects and use your training and logic to figure it out for each polarity.



                  My simulation for you. You cannot put a cap. on this output to get an envelope voltage due to a mismatched output impedance of the diode switch in series with the output. Subtle errors arise from your choice of FET input OP Amp. So choose wisely and go for a buffered dual Op Amp FW rectifier design (search web)







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 3 hours ago

























                  answered 3 hours ago









                  Tony EE rocketscientist

                  59.2k22088




                  59.2k22088



























                       

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