Basic question regarding op-amp output voltage












4












$begingroup$


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 $(V_+-V_-)/2$?






op-amp







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Gabriel Golfetti is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 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
    4 hours ago


















4












$begingroup$


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 $(V_+-V_-)/2$?






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.







$endgroup$








  • 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
    4 hours ago
















4












4








4





$begingroup$


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 $(V_+-V_-)/2$?






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.







$endgroup$




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




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 3 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

1354




1354




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
    4 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
    4 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
4 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
4 hours ago












3 Answers
3






active

oldest

votes


















2












$begingroup$

"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











$endgroup$





















    1












    $begingroup$

    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









    $endgroup$





















      1












      $begingroup$

      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









      $endgroup$













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        3 Answers
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        3 Answers
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        2












        $begingroup$

        "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












          $begingroup$

          "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












            2








            2





            $begingroup$

            "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 32 mins ago

























            answered 43 mins ago









            AaronDAaronD

            4,106528




            4,106528

























                1












                $begingroup$

                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









                $endgroup$


















                  1












                  $begingroup$

                  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









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    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 4 hours ago









                    Spehro PefhanySpehro Pefhany

                    208k5158417




                    208k5158417























                        1












                        $begingroup$

                        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









                        $endgroup$


















                          1












                          $begingroup$

                          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









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            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

                            2,14038




                            2,14038






















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                                格利澤436b

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                                格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
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