Basic question regarding op-amp output voltage












5












$begingroup$


When we model an operational amplifier, we say that the output voltage $V_{out}$ is proportional to the difference between the inputs $V_pm$, $V_{out}=A(V_+-V_-)$. I've read that this voltage is always with respect to ground.



enter image description here



However, when we add a double supply to the op-amp, how is ground defined? Does the circuit take it to be $V_-$? Or maybe it is $(V_+-V_-)/2$?










share|improve this question









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$endgroup$








  • 4




    $begingroup$
    One possible (but unsatisfying) answer is: it doesn't matter. Since op amps are always used with negative feedback (unless you're doing something quite odd), the feedback cancels out any possible dependency on the supply voltages.
    $endgroup$
    – Hearth
    5 hours ago








  • 1




    $begingroup$
    I like your suggestion that its $ {(V_+ - V_-)} over {2} $. Another thought: output current goes to zero: an "imperfect" opamp might settle at a strange output voltage, but I'd hope for half-way between the supply voltages.
    $endgroup$
    – glen_geek
    5 hours ago


















5












$begingroup$


When we model an operational amplifier, we say that the output voltage $V_{out}$ is proportional to the difference between the inputs $V_pm$, $V_{out}=A(V_+-V_-)$. I've read that this voltage is always with respect to ground.



enter image description here



However, when we add a double supply to the op-amp, how is ground defined? Does the circuit take it to be $V_-$? Or maybe it is $(V_+-V_-)/2$?










share|improve this question









New contributor




Gabriel Golfetti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 4




    $begingroup$
    One possible (but unsatisfying) answer is: it doesn't matter. Since op amps are always used with negative feedback (unless you're doing something quite odd), the feedback cancels out any possible dependency on the supply voltages.
    $endgroup$
    – Hearth
    5 hours ago








  • 1




    $begingroup$
    I like your suggestion that its $ {(V_+ - V_-)} over {2} $. Another thought: output current goes to zero: an "imperfect" opamp might settle at a strange output voltage, but I'd hope for half-way between the supply voltages.
    $endgroup$
    – glen_geek
    5 hours ago
















5












5








5





$begingroup$


When we model an operational amplifier, we say that the output voltage $V_{out}$ is proportional to the difference between the inputs $V_pm$, $V_{out}=A(V_+-V_-)$. I've read that this voltage is always with respect to ground.



enter image description here



However, when we add a double supply to the op-amp, how is ground defined? Does the circuit take it to be $V_-$? Or maybe it is $(V_+-V_-)/2$?










share|improve this question









New contributor




Gabriel Golfetti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




When we model an operational amplifier, we say that the output voltage $V_{out}$ is proportional to the difference between the inputs $V_pm$, $V_{out}=A(V_+-V_-)$. I've read that this voltage is always with respect to ground.



enter image description here



However, when we add a double supply to the op-amp, how is ground defined? Does the circuit take it to be $V_-$? Or maybe it is $(V_+-V_-)/2$?







op-amp






share|improve this question









New contributor




Gabriel Golfetti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Gabriel Golfetti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 4 hours ago









Michael Karas

44k348103




44k348103






New contributor




Gabriel Golfetti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 5 hours ago









Gabriel GolfettiGabriel Golfetti

1404




1404




New contributor




Gabriel Golfetti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Gabriel Golfetti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Gabriel Golfetti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 4




    $begingroup$
    One possible (but unsatisfying) answer is: it doesn't matter. Since op amps are always used with negative feedback (unless you're doing something quite odd), the feedback cancels out any possible dependency on the supply voltages.
    $endgroup$
    – Hearth
    5 hours ago








  • 1




    $begingroup$
    I like your suggestion that its $ {(V_+ - V_-)} over {2} $. Another thought: output current goes to zero: an "imperfect" opamp might settle at a strange output voltage, but I'd hope for half-way between the supply voltages.
    $endgroup$
    – glen_geek
    5 hours ago
















  • 4




    $begingroup$
    One possible (but unsatisfying) answer is: it doesn't matter. Since op amps are always used with negative feedback (unless you're doing something quite odd), the feedback cancels out any possible dependency on the supply voltages.
    $endgroup$
    – Hearth
    5 hours ago








  • 1




    $begingroup$
    I like your suggestion that its $ {(V_+ - V_-)} over {2} $. Another thought: output current goes to zero: an "imperfect" opamp might settle at a strange output voltage, but I'd hope for half-way between the supply voltages.
    $endgroup$
    – glen_geek
    5 hours ago










4




4




$begingroup$
One possible (but unsatisfying) answer is: it doesn't matter. Since op amps are always used with negative feedback (unless you're doing something quite odd), the feedback cancels out any possible dependency on the supply voltages.
$endgroup$
– Hearth
5 hours ago






$begingroup$
One possible (but unsatisfying) answer is: it doesn't matter. Since op amps are always used with negative feedback (unless you're doing something quite odd), the feedback cancels out any possible dependency on the supply voltages.
$endgroup$
– Hearth
5 hours ago






1




1




$begingroup$
I like your suggestion that its $ {(V_+ - V_-)} over {2} $. Another thought: output current goes to zero: an "imperfect" opamp might settle at a strange output voltage, but I'd hope for half-way between the supply voltages.
$endgroup$
– glen_geek
5 hours ago






$begingroup$
I like your suggestion that its $ {(V_+ - V_-)} over {2} $. Another thought: output current goes to zero: an "imperfect" opamp might settle at a strange output voltage, but I'd hope for half-way between the supply voltages.
$endgroup$
– glen_geek
5 hours ago












4 Answers
4






active

oldest

votes


















3












$begingroup$

"Ground" in this context is really a misnomer. It's only a reference. For example, take this circuit that you probably know:





schematic





simulate this circuit – Schematic created using CircuitLab



and connect the "ground" to another control instead:





schematic





simulate this circuit



You can think of it as being the exact same circuit with the exact same behavior, provided that the reference control stays constant. It simply centers on this new reference, and there's now no need for an explicit ground.





The actual rules for opamps are:




  1. The output goes as far as it needs to, to keep the two inputs equal.

  2. If the "+" input is higher than the "-" input, it goes up; if lower,
    it goes down.

  3. If the output hits one or the other supply rail, it stops there.

    (actually a volt or two short unless it's explicitly a rail-to-rail design)


You'll notice that there's no mention of "ground" in any of that. Only comparing the two inputs to each other and the output to the supply rails. That's it!






share|improve this answer











$endgroup$





















    2












    $begingroup$

    Look at how input offset voltage is defined for that particular op-amp. Remember that's the input differential voltage (at a particular common mode voltage) that is required to drive the output to exactly zero.



    Use the same ground reference.



    In almost all cases, the maximum offset voltage can drive the output quite a bit more than between the supply rails so it doesn't much matter.






    share|improve this answer









    $endgroup$





















      2












      $begingroup$

      Your question is a good one. Since the differential gain of an opamp is typically very large 100 dB and up, it is often not a detail that is considered.



      Take for example the following simulation of a highly idealized opamp, where each opamp has a different output reference potential:



      enter image description here



      The output is nearly identical for the three cases, and for basic design, the output reference potential can be ignored.



      In cases where you would use the transfer function of the opamp $A(s)$ for stability and frequency analysis its also a mute point. Good design practices ensure all three reference potentials become one single AC ground due to decoupling/bypass capacitors.






      share|improve this answer









      $endgroup$





















        0












        $begingroup$

        It’s like building a house and deciding what is considered ground level (0 meters altitude).You could use a GPS or something to get the actual altitude, but who cares. Just pick a reference that makes sense for the design. Negative altitudes are obviously referring to the basement. Positive altitudes are obviously floors above ground. With your custom coordinate system, your blueprints become easy to read and the house is easier to build. So, if you instead arbitrarily define a certain elevation to be +10 meters, then you have just defined ground level as being 10 meters below the +10 meter mark. (Also, the meter is already defined.)



        Electrical ground is just what you define to be 0V. That means you implicitly defined ground when you defined V+ and V-. If you have a 9V battery, you know that the positive terminal is 9V higher than the negative terminal, but you could call the positive terminal +6V and the negative terminal -3V, or whatever else you want. What is already defined is 1 volt and positive/negative. Lastly, the reference is chosen so that the circuit is easy to understand.





        share









        $endgroup$













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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          "Ground" in this context is really a misnomer. It's only a reference. For example, take this circuit that you probably know:





          schematic





          simulate this circuit – Schematic created using CircuitLab



          and connect the "ground" to another control instead:





          schematic





          simulate this circuit



          You can think of it as being the exact same circuit with the exact same behavior, provided that the reference control stays constant. It simply centers on this new reference, and there's now no need for an explicit ground.





          The actual rules for opamps are:




          1. The output goes as far as it needs to, to keep the two inputs equal.

          2. If the "+" input is higher than the "-" input, it goes up; if lower,
            it goes down.

          3. If the output hits one or the other supply rail, it stops there.

            (actually a volt or two short unless it's explicitly a rail-to-rail design)


          You'll notice that there's no mention of "ground" in any of that. Only comparing the two inputs to each other and the output to the supply rails. That's it!






          share|improve this answer











          $endgroup$


















            3












            $begingroup$

            "Ground" in this context is really a misnomer. It's only a reference. For example, take this circuit that you probably know:





            schematic





            simulate this circuit – Schematic created using CircuitLab



            and connect the "ground" to another control instead:





            schematic





            simulate this circuit



            You can think of it as being the exact same circuit with the exact same behavior, provided that the reference control stays constant. It simply centers on this new reference, and there's now no need for an explicit ground.





            The actual rules for opamps are:




            1. The output goes as far as it needs to, to keep the two inputs equal.

            2. If the "+" input is higher than the "-" input, it goes up; if lower,
              it goes down.

            3. If the output hits one or the other supply rail, it stops there.

              (actually a volt or two short unless it's explicitly a rail-to-rail design)


            You'll notice that there's no mention of "ground" in any of that. Only comparing the two inputs to each other and the output to the supply rails. That's it!






            share|improve this answer











            $endgroup$
















              3












              3








              3





              $begingroup$

              "Ground" in this context is really a misnomer. It's only a reference. For example, take this circuit that you probably know:





              schematic





              simulate this circuit – Schematic created using CircuitLab



              and connect the "ground" to another control instead:





              schematic





              simulate this circuit



              You can think of it as being the exact same circuit with the exact same behavior, provided that the reference control stays constant. It simply centers on this new reference, and there's now no need for an explicit ground.





              The actual rules for opamps are:




              1. The output goes as far as it needs to, to keep the two inputs equal.

              2. If the "+" input is higher than the "-" input, it goes up; if lower,
                it goes down.

              3. If the output hits one or the other supply rail, it stops there.

                (actually a volt or two short unless it's explicitly a rail-to-rail design)


              You'll notice that there's no mention of "ground" in any of that. Only comparing the two inputs to each other and the output to the supply rails. That's it!






              share|improve this answer











              $endgroup$



              "Ground" in this context is really a misnomer. It's only a reference. For example, take this circuit that you probably know:





              schematic





              simulate this circuit – Schematic created using CircuitLab



              and connect the "ground" to another control instead:





              schematic





              simulate this circuit



              You can think of it as being the exact same circuit with the exact same behavior, provided that the reference control stays constant. It simply centers on this new reference, and there's now no need for an explicit ground.





              The actual rules for opamps are:




              1. The output goes as far as it needs to, to keep the two inputs equal.

              2. If the "+" input is higher than the "-" input, it goes up; if lower,
                it goes down.

              3. If the output hits one or the other supply rail, it stops there.

                (actually a volt or two short unless it's explicitly a rail-to-rail design)


              You'll notice that there's no mention of "ground" in any of that. Only comparing the two inputs to each other and the output to the supply rails. That's it!







              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited 1 hour ago

























              answered 1 hour ago









              AaronDAaronD

              4,116528




              4,116528

























                  2












                  $begingroup$

                  Look at how input offset voltage is defined for that particular op-amp. Remember that's the input differential voltage (at a particular common mode voltage) that is required to drive the output to exactly zero.



                  Use the same ground reference.



                  In almost all cases, the maximum offset voltage can drive the output quite a bit more than between the supply rails so it doesn't much matter.






                  share|improve this answer









                  $endgroup$


















                    2












                    $begingroup$

                    Look at how input offset voltage is defined for that particular op-amp. Remember that's the input differential voltage (at a particular common mode voltage) that is required to drive the output to exactly zero.



                    Use the same ground reference.



                    In almost all cases, the maximum offset voltage can drive the output quite a bit more than between the supply rails so it doesn't much matter.






                    share|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      Look at how input offset voltage is defined for that particular op-amp. Remember that's the input differential voltage (at a particular common mode voltage) that is required to drive the output to exactly zero.



                      Use the same ground reference.



                      In almost all cases, the maximum offset voltage can drive the output quite a bit more than between the supply rails so it doesn't much matter.






                      share|improve this answer









                      $endgroup$



                      Look at how input offset voltage is defined for that particular op-amp. Remember that's the input differential voltage (at a particular common mode voltage) that is required to drive the output to exactly zero.



                      Use the same ground reference.



                      In almost all cases, the maximum offset voltage can drive the output quite a bit more than between the supply rails so it doesn't much matter.







                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered 5 hours ago









                      Spehro PefhanySpehro Pefhany

                      208k5158417




                      208k5158417























                          2












                          $begingroup$

                          Your question is a good one. Since the differential gain of an opamp is typically very large 100 dB and up, it is often not a detail that is considered.



                          Take for example the following simulation of a highly idealized opamp, where each opamp has a different output reference potential:



                          enter image description here



                          The output is nearly identical for the three cases, and for basic design, the output reference potential can be ignored.



                          In cases where you would use the transfer function of the opamp $A(s)$ for stability and frequency analysis its also a mute point. Good design practices ensure all three reference potentials become one single AC ground due to decoupling/bypass capacitors.






                          share|improve this answer









                          $endgroup$


















                            2












                            $begingroup$

                            Your question is a good one. Since the differential gain of an opamp is typically very large 100 dB and up, it is often not a detail that is considered.



                            Take for example the following simulation of a highly idealized opamp, where each opamp has a different output reference potential:



                            enter image description here



                            The output is nearly identical for the three cases, and for basic design, the output reference potential can be ignored.



                            In cases where you would use the transfer function of the opamp $A(s)$ for stability and frequency analysis its also a mute point. Good design practices ensure all three reference potentials become one single AC ground due to decoupling/bypass capacitors.






                            share|improve this answer









                            $endgroup$
















                              2












                              2








                              2





                              $begingroup$

                              Your question is a good one. Since the differential gain of an opamp is typically very large 100 dB and up, it is often not a detail that is considered.



                              Take for example the following simulation of a highly idealized opamp, where each opamp has a different output reference potential:



                              enter image description here



                              The output is nearly identical for the three cases, and for basic design, the output reference potential can be ignored.



                              In cases where you would use the transfer function of the opamp $A(s)$ for stability and frequency analysis its also a mute point. Good design practices ensure all three reference potentials become one single AC ground due to decoupling/bypass capacitors.






                              share|improve this answer









                              $endgroup$



                              Your question is a good one. Since the differential gain of an opamp is typically very large 100 dB and up, it is often not a detail that is considered.



                              Take for example the following simulation of a highly idealized opamp, where each opamp has a different output reference potential:



                              enter image description here



                              The output is nearly identical for the three cases, and for basic design, the output reference potential can be ignored.



                              In cases where you would use the transfer function of the opamp $A(s)$ for stability and frequency analysis its also a mute point. Good design practices ensure all three reference potentials become one single AC ground due to decoupling/bypass capacitors.







                              share|improve this answer












                              share|improve this answer



                              share|improve this answer










                              answered 4 hours ago









                              sstobbesstobbe

                              2,15038




                              2,15038























                                  0












                                  $begingroup$

                                  It’s like building a house and deciding what is considered ground level (0 meters altitude).You could use a GPS or something to get the actual altitude, but who cares. Just pick a reference that makes sense for the design. Negative altitudes are obviously referring to the basement. Positive altitudes are obviously floors above ground. With your custom coordinate system, your blueprints become easy to read and the house is easier to build. So, if you instead arbitrarily define a certain elevation to be +10 meters, then you have just defined ground level as being 10 meters below the +10 meter mark. (Also, the meter is already defined.)



                                  Electrical ground is just what you define to be 0V. That means you implicitly defined ground when you defined V+ and V-. If you have a 9V battery, you know that the positive terminal is 9V higher than the negative terminal, but you could call the positive terminal +6V and the negative terminal -3V, or whatever else you want. What is already defined is 1 volt and positive/negative. Lastly, the reference is chosen so that the circuit is easy to understand.





                                  share









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    It’s like building a house and deciding what is considered ground level (0 meters altitude).You could use a GPS or something to get the actual altitude, but who cares. Just pick a reference that makes sense for the design. Negative altitudes are obviously referring to the basement. Positive altitudes are obviously floors above ground. With your custom coordinate system, your blueprints become easy to read and the house is easier to build. So, if you instead arbitrarily define a certain elevation to be +10 meters, then you have just defined ground level as being 10 meters below the +10 meter mark. (Also, the meter is already defined.)



                                    Electrical ground is just what you define to be 0V. That means you implicitly defined ground when you defined V+ and V-. If you have a 9V battery, you know that the positive terminal is 9V higher than the negative terminal, but you could call the positive terminal +6V and the negative terminal -3V, or whatever else you want. What is already defined is 1 volt and positive/negative. Lastly, the reference is chosen so that the circuit is easy to understand.





                                    share









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      It’s like building a house and deciding what is considered ground level (0 meters altitude).You could use a GPS or something to get the actual altitude, but who cares. Just pick a reference that makes sense for the design. Negative altitudes are obviously referring to the basement. Positive altitudes are obviously floors above ground. With your custom coordinate system, your blueprints become easy to read and the house is easier to build. So, if you instead arbitrarily define a certain elevation to be +10 meters, then you have just defined ground level as being 10 meters below the +10 meter mark. (Also, the meter is already defined.)



                                      Electrical ground is just what you define to be 0V. That means you implicitly defined ground when you defined V+ and V-. If you have a 9V battery, you know that the positive terminal is 9V higher than the negative terminal, but you could call the positive terminal +6V and the negative terminal -3V, or whatever else you want. What is already defined is 1 volt and positive/negative. Lastly, the reference is chosen so that the circuit is easy to understand.





                                      share









                                      $endgroup$



                                      It’s like building a house and deciding what is considered ground level (0 meters altitude).You could use a GPS or something to get the actual altitude, but who cares. Just pick a reference that makes sense for the design. Negative altitudes are obviously referring to the basement. Positive altitudes are obviously floors above ground. With your custom coordinate system, your blueprints become easy to read and the house is easier to build. So, if you instead arbitrarily define a certain elevation to be +10 meters, then you have just defined ground level as being 10 meters below the +10 meter mark. (Also, the meter is already defined.)



                                      Electrical ground is just what you define to be 0V. That means you implicitly defined ground when you defined V+ and V-. If you have a 9V battery, you know that the positive terminal is 9V higher than the negative terminal, but you could call the positive terminal +6V and the negative terminal -3V, or whatever else you want. What is already defined is 1 volt and positive/negative. Lastly, the reference is chosen so that the circuit is easy to understand.






                                      share











                                      share


                                      share










                                      answered 4 mins ago









                                      Joe MacJoe Mac

                                      1666




                                      1666






















                                          Gabriel Golfetti is a new contributor. Be nice, and check out our Code of Conduct.










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