Determining Impedances of an Op-Amp Circuit











up vote
2
down vote

favorite












I'm trying to review some old concepts, and thus far, there was one that I couldn't really understand: impedance of circuits involving op-amps. Assuming an ideal op-amp, it has infinite input impedance and no output impedance, but that's only for the op-amp itself. I'm having trouble understanding how to find the input/output impedance of some op-amp circuits as a whole.



Take the standard non-inverting amplifier for instance:





schematic





simulate this circuit – Schematic created using CircuitLab



Using the ideal op-amp rules, since the non-inverting input is connected to the input voltage through Z1, the input impedance would be infinite as no current can flow through the op-amp terminals. However, when you look at the output impedance, you remove the input source by shorting it to ground, making the non-inverting input equal to ground (as well as the inverting input). My book says that the output impedance would be 0, but I don't understand how this is the case. Replacing the load resistor with a current source, you just see an internal op-amp 'output' resistance of 0 ohms in parallel with Z2 to ground, so is that set of parallel impedances the cause of the 0 ohm output impedance? Is this logic correct?



It feels that because of the nature of the ideal op-amp having 0 output resistance, all op-amp circuits would have 0 output resistance. Is this always the case, or are there some exceptions? I'm trying to develop some methodologies for measuring such impedances in circuit problems as it's hard for me to wrap my head around.










share|improve this question


























    up vote
    2
    down vote

    favorite












    I'm trying to review some old concepts, and thus far, there was one that I couldn't really understand: impedance of circuits involving op-amps. Assuming an ideal op-amp, it has infinite input impedance and no output impedance, but that's only for the op-amp itself. I'm having trouble understanding how to find the input/output impedance of some op-amp circuits as a whole.



    Take the standard non-inverting amplifier for instance:





    schematic





    simulate this circuit – Schematic created using CircuitLab



    Using the ideal op-amp rules, since the non-inverting input is connected to the input voltage through Z1, the input impedance would be infinite as no current can flow through the op-amp terminals. However, when you look at the output impedance, you remove the input source by shorting it to ground, making the non-inverting input equal to ground (as well as the inverting input). My book says that the output impedance would be 0, but I don't understand how this is the case. Replacing the load resistor with a current source, you just see an internal op-amp 'output' resistance of 0 ohms in parallel with Z2 to ground, so is that set of parallel impedances the cause of the 0 ohm output impedance? Is this logic correct?



    It feels that because of the nature of the ideal op-amp having 0 output resistance, all op-amp circuits would have 0 output resistance. Is this always the case, or are there some exceptions? I'm trying to develop some methodologies for measuring such impedances in circuit problems as it's hard for me to wrap my head around.










    share|improve this question
























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I'm trying to review some old concepts, and thus far, there was one that I couldn't really understand: impedance of circuits involving op-amps. Assuming an ideal op-amp, it has infinite input impedance and no output impedance, but that's only for the op-amp itself. I'm having trouble understanding how to find the input/output impedance of some op-amp circuits as a whole.



      Take the standard non-inverting amplifier for instance:





      schematic





      simulate this circuit – Schematic created using CircuitLab



      Using the ideal op-amp rules, since the non-inverting input is connected to the input voltage through Z1, the input impedance would be infinite as no current can flow through the op-amp terminals. However, when you look at the output impedance, you remove the input source by shorting it to ground, making the non-inverting input equal to ground (as well as the inverting input). My book says that the output impedance would be 0, but I don't understand how this is the case. Replacing the load resistor with a current source, you just see an internal op-amp 'output' resistance of 0 ohms in parallel with Z2 to ground, so is that set of parallel impedances the cause of the 0 ohm output impedance? Is this logic correct?



      It feels that because of the nature of the ideal op-amp having 0 output resistance, all op-amp circuits would have 0 output resistance. Is this always the case, or are there some exceptions? I'm trying to develop some methodologies for measuring such impedances in circuit problems as it's hard for me to wrap my head around.










      share|improve this question













      I'm trying to review some old concepts, and thus far, there was one that I couldn't really understand: impedance of circuits involving op-amps. Assuming an ideal op-amp, it has infinite input impedance and no output impedance, but that's only for the op-amp itself. I'm having trouble understanding how to find the input/output impedance of some op-amp circuits as a whole.



      Take the standard non-inverting amplifier for instance:





      schematic





      simulate this circuit – Schematic created using CircuitLab



      Using the ideal op-amp rules, since the non-inverting input is connected to the input voltage through Z1, the input impedance would be infinite as no current can flow through the op-amp terminals. However, when you look at the output impedance, you remove the input source by shorting it to ground, making the non-inverting input equal to ground (as well as the inverting input). My book says that the output impedance would be 0, but I don't understand how this is the case. Replacing the load resistor with a current source, you just see an internal op-amp 'output' resistance of 0 ohms in parallel with Z2 to ground, so is that set of parallel impedances the cause of the 0 ohm output impedance? Is this logic correct?



      It feels that because of the nature of the ideal op-amp having 0 output resistance, all op-amp circuits would have 0 output resistance. Is this always the case, or are there some exceptions? I'm trying to develop some methodologies for measuring such impedances in circuit problems as it's hard for me to wrap my head around.







      op-amp impedance output






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 5 hours ago









      user101402

      15911




      15911






















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          3
          down vote













          The op-amps themselves don't have zero output impedance but when configured with negative feedback they do.




          Using the ideal op-amp rules, since the non-inverting input is connected to the input voltage through Z1, the input impedance would be infinite as no current can flow through the op-amp terminals.




          That's OK.




          However, when you look at the output impedance, you remove the input source by shorting it to ground, making the non-inverting input equal to ground (as well as the inverting input). My book says that the output impedance would be 0, but I don't understand how this is the case.




          I find that the 0 V input case is less helpful than, say a 1 V input case as it is too easy to balance out circuits with 0s everywhere. Let's use 1 V. And lets add some external Rout to make the output obviously non-ideal.





          schematic





          simulate this circuit – Schematic created using CircuitLab



          Figure 1. Rout is the output impedance of the op-amp.



          With Rout in circuit, ask yourself what does the output of OA1 have to do to get 1 V at the inverting input? Answer: it has to get the voltage at OUT = $ frac {Z_3}{Z_2+Z_3} V_{IN} $. That means that the output of the op-amp actually has to go some distance beyond Vout
          to compensate for the voltage drop across the output resistance.




          It feels that because of the nature of the ideal op-amp having 0 output resistance, all op-amp circuits would have 0 output resistance.




          Again, it's not the op-amp itself but rather the feedback that gives the circuit this nature.



          Does that help?






          share|improve this answer




























            up vote
            1
            down vote













            Just to add a little more to Transistor's answer (a mathematical approach).



            You can look at it this way. This is the circuit model of the op amp (with intrinsic output impedance) plus the external resistors:





            schematic





            simulate this circuit – Schematic created using CircuitLab



            Obviously from the circuit, $V^+=0$. You can try to find and expression for $dfrac{V_{text{test}}}{I_{text{test}}}$.



            Also $V^-=V_{text{test}}dfrac{R_1}{R_1+R_2}$, $I_o=dfrac{V_{text{test}}+AV^-}{R_o}$, and $I_f=dfrac{V_{text{test}}}{R_F+R_1}$.



            You combine those, and obtain:



            $$ dfrac{V_{text{test}}}{I_{text{test}}}=dfrac{R_o(R_1+R_F)}{R_1+R_F+AR_1+R_o}$$



            Given the fact that $A$ is a really large number (the op amp open-loop gain), the $AR_1$ term dominates in the denominator.



            So,



            $$ dfrac{V_{text{test}}}{I_{text{test}}} approx dfrac{R_o(R_1+R_F)}{AR_1}$$



            Remember that $G=dfrac{R_1+R_F}{R_1}=1+dfrac{R_F}{R_1}$ is the closed loop gain. Which allows to further re-write this as:



            $$ dfrac{V_{text{test}}}{I_{text{test}}} approx dfrac{R_oG}{A}to 0$$



            And that approaches zero or a very low value. As you can see the output impedance under negative feedback is even smaller than the intrinsic output impedance of the op amp given the fact that $A$ is large and $G$ is no doubt much smaller. If you were to just ignore $R_o$ this would be ideally zero.






            share|improve this answer























              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("schematics", function () {
              StackExchange.schematics.init();
              });
              }, "cicuitlab");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "135"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              convertImagesToLinks: false,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: null,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f412127%2fdetermining-impedances-of-an-op-amp-circuit%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              3
              down vote













              The op-amps themselves don't have zero output impedance but when configured with negative feedback they do.




              Using the ideal op-amp rules, since the non-inverting input is connected to the input voltage through Z1, the input impedance would be infinite as no current can flow through the op-amp terminals.




              That's OK.




              However, when you look at the output impedance, you remove the input source by shorting it to ground, making the non-inverting input equal to ground (as well as the inverting input). My book says that the output impedance would be 0, but I don't understand how this is the case.




              I find that the 0 V input case is less helpful than, say a 1 V input case as it is too easy to balance out circuits with 0s everywhere. Let's use 1 V. And lets add some external Rout to make the output obviously non-ideal.





              schematic





              simulate this circuit – Schematic created using CircuitLab



              Figure 1. Rout is the output impedance of the op-amp.



              With Rout in circuit, ask yourself what does the output of OA1 have to do to get 1 V at the inverting input? Answer: it has to get the voltage at OUT = $ frac {Z_3}{Z_2+Z_3} V_{IN} $. That means that the output of the op-amp actually has to go some distance beyond Vout
              to compensate for the voltage drop across the output resistance.




              It feels that because of the nature of the ideal op-amp having 0 output resistance, all op-amp circuits would have 0 output resistance.




              Again, it's not the op-amp itself but rather the feedback that gives the circuit this nature.



              Does that help?






              share|improve this answer

























                up vote
                3
                down vote













                The op-amps themselves don't have zero output impedance but when configured with negative feedback they do.




                Using the ideal op-amp rules, since the non-inverting input is connected to the input voltage through Z1, the input impedance would be infinite as no current can flow through the op-amp terminals.




                That's OK.




                However, when you look at the output impedance, you remove the input source by shorting it to ground, making the non-inverting input equal to ground (as well as the inverting input). My book says that the output impedance would be 0, but I don't understand how this is the case.




                I find that the 0 V input case is less helpful than, say a 1 V input case as it is too easy to balance out circuits with 0s everywhere. Let's use 1 V. And lets add some external Rout to make the output obviously non-ideal.





                schematic





                simulate this circuit – Schematic created using CircuitLab



                Figure 1. Rout is the output impedance of the op-amp.



                With Rout in circuit, ask yourself what does the output of OA1 have to do to get 1 V at the inverting input? Answer: it has to get the voltage at OUT = $ frac {Z_3}{Z_2+Z_3} V_{IN} $. That means that the output of the op-amp actually has to go some distance beyond Vout
                to compensate for the voltage drop across the output resistance.




                It feels that because of the nature of the ideal op-amp having 0 output resistance, all op-amp circuits would have 0 output resistance.




                Again, it's not the op-amp itself but rather the feedback that gives the circuit this nature.



                Does that help?






                share|improve this answer























                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  The op-amps themselves don't have zero output impedance but when configured with negative feedback they do.




                  Using the ideal op-amp rules, since the non-inverting input is connected to the input voltage through Z1, the input impedance would be infinite as no current can flow through the op-amp terminals.




                  That's OK.




                  However, when you look at the output impedance, you remove the input source by shorting it to ground, making the non-inverting input equal to ground (as well as the inverting input). My book says that the output impedance would be 0, but I don't understand how this is the case.




                  I find that the 0 V input case is less helpful than, say a 1 V input case as it is too easy to balance out circuits with 0s everywhere. Let's use 1 V. And lets add some external Rout to make the output obviously non-ideal.





                  schematic





                  simulate this circuit – Schematic created using CircuitLab



                  Figure 1. Rout is the output impedance of the op-amp.



                  With Rout in circuit, ask yourself what does the output of OA1 have to do to get 1 V at the inverting input? Answer: it has to get the voltage at OUT = $ frac {Z_3}{Z_2+Z_3} V_{IN} $. That means that the output of the op-amp actually has to go some distance beyond Vout
                  to compensate for the voltage drop across the output resistance.




                  It feels that because of the nature of the ideal op-amp having 0 output resistance, all op-amp circuits would have 0 output resistance.




                  Again, it's not the op-amp itself but rather the feedback that gives the circuit this nature.



                  Does that help?






                  share|improve this answer












                  The op-amps themselves don't have zero output impedance but when configured with negative feedback they do.




                  Using the ideal op-amp rules, since the non-inverting input is connected to the input voltage through Z1, the input impedance would be infinite as no current can flow through the op-amp terminals.




                  That's OK.




                  However, when you look at the output impedance, you remove the input source by shorting it to ground, making the non-inverting input equal to ground (as well as the inverting input). My book says that the output impedance would be 0, but I don't understand how this is the case.




                  I find that the 0 V input case is less helpful than, say a 1 V input case as it is too easy to balance out circuits with 0s everywhere. Let's use 1 V. And lets add some external Rout to make the output obviously non-ideal.





                  schematic





                  simulate this circuit – Schematic created using CircuitLab



                  Figure 1. Rout is the output impedance of the op-amp.



                  With Rout in circuit, ask yourself what does the output of OA1 have to do to get 1 V at the inverting input? Answer: it has to get the voltage at OUT = $ frac {Z_3}{Z_2+Z_3} V_{IN} $. That means that the output of the op-amp actually has to go some distance beyond Vout
                  to compensate for the voltage drop across the output resistance.




                  It feels that because of the nature of the ideal op-amp having 0 output resistance, all op-amp circuits would have 0 output resistance.




                  Again, it's not the op-amp itself but rather the feedback that gives the circuit this nature.



                  Does that help?







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 4 hours ago









                  Transistor

                  79.5k777172




                  79.5k777172
























                      up vote
                      1
                      down vote













                      Just to add a little more to Transistor's answer (a mathematical approach).



                      You can look at it this way. This is the circuit model of the op amp (with intrinsic output impedance) plus the external resistors:





                      schematic





                      simulate this circuit – Schematic created using CircuitLab



                      Obviously from the circuit, $V^+=0$. You can try to find and expression for $dfrac{V_{text{test}}}{I_{text{test}}}$.



                      Also $V^-=V_{text{test}}dfrac{R_1}{R_1+R_2}$, $I_o=dfrac{V_{text{test}}+AV^-}{R_o}$, and $I_f=dfrac{V_{text{test}}}{R_F+R_1}$.



                      You combine those, and obtain:



                      $$ dfrac{V_{text{test}}}{I_{text{test}}}=dfrac{R_o(R_1+R_F)}{R_1+R_F+AR_1+R_o}$$



                      Given the fact that $A$ is a really large number (the op amp open-loop gain), the $AR_1$ term dominates in the denominator.



                      So,



                      $$ dfrac{V_{text{test}}}{I_{text{test}}} approx dfrac{R_o(R_1+R_F)}{AR_1}$$



                      Remember that $G=dfrac{R_1+R_F}{R_1}=1+dfrac{R_F}{R_1}$ is the closed loop gain. Which allows to further re-write this as:



                      $$ dfrac{V_{text{test}}}{I_{text{test}}} approx dfrac{R_oG}{A}to 0$$



                      And that approaches zero or a very low value. As you can see the output impedance under negative feedback is even smaller than the intrinsic output impedance of the op amp given the fact that $A$ is large and $G$ is no doubt much smaller. If you were to just ignore $R_o$ this would be ideally zero.






                      share|improve this answer



























                        up vote
                        1
                        down vote













                        Just to add a little more to Transistor's answer (a mathematical approach).



                        You can look at it this way. This is the circuit model of the op amp (with intrinsic output impedance) plus the external resistors:





                        schematic





                        simulate this circuit – Schematic created using CircuitLab



                        Obviously from the circuit, $V^+=0$. You can try to find and expression for $dfrac{V_{text{test}}}{I_{text{test}}}$.



                        Also $V^-=V_{text{test}}dfrac{R_1}{R_1+R_2}$, $I_o=dfrac{V_{text{test}}+AV^-}{R_o}$, and $I_f=dfrac{V_{text{test}}}{R_F+R_1}$.



                        You combine those, and obtain:



                        $$ dfrac{V_{text{test}}}{I_{text{test}}}=dfrac{R_o(R_1+R_F)}{R_1+R_F+AR_1+R_o}$$



                        Given the fact that $A$ is a really large number (the op amp open-loop gain), the $AR_1$ term dominates in the denominator.



                        So,



                        $$ dfrac{V_{text{test}}}{I_{text{test}}} approx dfrac{R_o(R_1+R_F)}{AR_1}$$



                        Remember that $G=dfrac{R_1+R_F}{R_1}=1+dfrac{R_F}{R_1}$ is the closed loop gain. Which allows to further re-write this as:



                        $$ dfrac{V_{text{test}}}{I_{text{test}}} approx dfrac{R_oG}{A}to 0$$



                        And that approaches zero or a very low value. As you can see the output impedance under negative feedback is even smaller than the intrinsic output impedance of the op amp given the fact that $A$ is large and $G$ is no doubt much smaller. If you were to just ignore $R_o$ this would be ideally zero.






                        share|improve this answer

























                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          Just to add a little more to Transistor's answer (a mathematical approach).



                          You can look at it this way. This is the circuit model of the op amp (with intrinsic output impedance) plus the external resistors:





                          schematic





                          simulate this circuit – Schematic created using CircuitLab



                          Obviously from the circuit, $V^+=0$. You can try to find and expression for $dfrac{V_{text{test}}}{I_{text{test}}}$.



                          Also $V^-=V_{text{test}}dfrac{R_1}{R_1+R_2}$, $I_o=dfrac{V_{text{test}}+AV^-}{R_o}$, and $I_f=dfrac{V_{text{test}}}{R_F+R_1}$.



                          You combine those, and obtain:



                          $$ dfrac{V_{text{test}}}{I_{text{test}}}=dfrac{R_o(R_1+R_F)}{R_1+R_F+AR_1+R_o}$$



                          Given the fact that $A$ is a really large number (the op amp open-loop gain), the $AR_1$ term dominates in the denominator.



                          So,



                          $$ dfrac{V_{text{test}}}{I_{text{test}}} approx dfrac{R_o(R_1+R_F)}{AR_1}$$



                          Remember that $G=dfrac{R_1+R_F}{R_1}=1+dfrac{R_F}{R_1}$ is the closed loop gain. Which allows to further re-write this as:



                          $$ dfrac{V_{text{test}}}{I_{text{test}}} approx dfrac{R_oG}{A}to 0$$



                          And that approaches zero or a very low value. As you can see the output impedance under negative feedback is even smaller than the intrinsic output impedance of the op amp given the fact that $A$ is large and $G$ is no doubt much smaller. If you were to just ignore $R_o$ this would be ideally zero.






                          share|improve this answer














                          Just to add a little more to Transistor's answer (a mathematical approach).



                          You can look at it this way. This is the circuit model of the op amp (with intrinsic output impedance) plus the external resistors:





                          schematic





                          simulate this circuit – Schematic created using CircuitLab



                          Obviously from the circuit, $V^+=0$. You can try to find and expression for $dfrac{V_{text{test}}}{I_{text{test}}}$.



                          Also $V^-=V_{text{test}}dfrac{R_1}{R_1+R_2}$, $I_o=dfrac{V_{text{test}}+AV^-}{R_o}$, and $I_f=dfrac{V_{text{test}}}{R_F+R_1}$.



                          You combine those, and obtain:



                          $$ dfrac{V_{text{test}}}{I_{text{test}}}=dfrac{R_o(R_1+R_F)}{R_1+R_F+AR_1+R_o}$$



                          Given the fact that $A$ is a really large number (the op amp open-loop gain), the $AR_1$ term dominates in the denominator.



                          So,



                          $$ dfrac{V_{text{test}}}{I_{text{test}}} approx dfrac{R_o(R_1+R_F)}{AR_1}$$



                          Remember that $G=dfrac{R_1+R_F}{R_1}=1+dfrac{R_F}{R_1}$ is the closed loop gain. Which allows to further re-write this as:



                          $$ dfrac{V_{text{test}}}{I_{text{test}}} approx dfrac{R_oG}{A}to 0$$



                          And that approaches zero or a very low value. As you can see the output impedance under negative feedback is even smaller than the intrinsic output impedance of the op amp given the fact that $A$ is large and $G$ is no doubt much smaller. If you were to just ignore $R_o$ this would be ideally zero.







                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited 30 mins ago









                          Null

                          4,843102233




                          4,843102233










                          answered 2 hours ago









                          Big6

                          2,8371615




                          2,8371615






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Electrical Engineering Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.





                              Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                              Please pay close attention to the following guidance:


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f412127%2fdetermining-impedances-of-an-op-amp-circuit%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              Popular posts from this blog

                              404 Error Contact Form 7 ajax form submitting

                              How to know if a Active Directory user can login interactively

                              TypeError: fit_transform() missing 1 required positional argument: 'X'