Understanding Operational Transconductance Amplifiers

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I understand the high-level function of OTAs, that they provide an output current proportional to an input voltage. But what I don't understand, and what I can't seem to find any information on online, is exactly how they're used and what the function of the different connections is.



wikipedia's image of the schematic symbol of an OTA



The most pressing questions I have are thus:



  1. What is the function of $I_bias$? Why are there diodes between this pin and the two inputs? I've seen them called linearizing diodes; how do they linearize the response of the circuit?

  2. What is the function of $I_abc$? As I understand it, it provides a scaling factor; is this correct? How does it do so?

  3. What is the (exact or approximate) formula for $I_out$ as a function of $V_in+$, $V_in-$, $I_bias$, and $I_abc$?

  4. What are the standard application topologies for an OTA? Op amps have the standard inverting and non-inverting amplifier topologies, and slightly more complicated summing amplifier and difference amplifier, integrator and differentiator, but are there any common circuit idioms for the OTA?

  5. Internally, how does an OTA work? (this is perhaps best left to a separate question, though.)









share|improve this question

















  • 1




    Some of your questions are answered in the LM13700 datasheet
    – Scott Seidman
    3 hours ago










  • @ScottSeidman I've looked at that datasheet and unfortunately had trouble understanding much of it. Perhaps I should go through it again.
    – Felthry
    3 hours ago










  • Have you found ti.com/lit/an/sboa117a/sboa117a.pdf?
    – Scott Seidman
    3 hours ago










  • @ScottSeidman Yes, I just read through that again. It did help; perhaps after organizing my thoughts enough to write this question, I also had them organized enough to understand the datasheet better. Still, I think a basic explanation would be worthwhile, as (though they are fairly rare) OTAs can be confusing and beginners may want an accessible explanation of what they do.
    – Felthry
    2 hours ago














up vote
1
down vote

favorite












I understand the high-level function of OTAs, that they provide an output current proportional to an input voltage. But what I don't understand, and what I can't seem to find any information on online, is exactly how they're used and what the function of the different connections is.



wikipedia's image of the schematic symbol of an OTA



The most pressing questions I have are thus:



  1. What is the function of $I_bias$? Why are there diodes between this pin and the two inputs? I've seen them called linearizing diodes; how do they linearize the response of the circuit?

  2. What is the function of $I_abc$? As I understand it, it provides a scaling factor; is this correct? How does it do so?

  3. What is the (exact or approximate) formula for $I_out$ as a function of $V_in+$, $V_in-$, $I_bias$, and $I_abc$?

  4. What are the standard application topologies for an OTA? Op amps have the standard inverting and non-inverting amplifier topologies, and slightly more complicated summing amplifier and difference amplifier, integrator and differentiator, but are there any common circuit idioms for the OTA?

  5. Internally, how does an OTA work? (this is perhaps best left to a separate question, though.)









share|improve this question

















  • 1




    Some of your questions are answered in the LM13700 datasheet
    – Scott Seidman
    3 hours ago










  • @ScottSeidman I've looked at that datasheet and unfortunately had trouble understanding much of it. Perhaps I should go through it again.
    – Felthry
    3 hours ago










  • Have you found ti.com/lit/an/sboa117a/sboa117a.pdf?
    – Scott Seidman
    3 hours ago










  • @ScottSeidman Yes, I just read through that again. It did help; perhaps after organizing my thoughts enough to write this question, I also had them organized enough to understand the datasheet better. Still, I think a basic explanation would be worthwhile, as (though they are fairly rare) OTAs can be confusing and beginners may want an accessible explanation of what they do.
    – Felthry
    2 hours ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I understand the high-level function of OTAs, that they provide an output current proportional to an input voltage. But what I don't understand, and what I can't seem to find any information on online, is exactly how they're used and what the function of the different connections is.



wikipedia's image of the schematic symbol of an OTA



The most pressing questions I have are thus:



  1. What is the function of $I_bias$? Why are there diodes between this pin and the two inputs? I've seen them called linearizing diodes; how do they linearize the response of the circuit?

  2. What is the function of $I_abc$? As I understand it, it provides a scaling factor; is this correct? How does it do so?

  3. What is the (exact or approximate) formula for $I_out$ as a function of $V_in+$, $V_in-$, $I_bias$, and $I_abc$?

  4. What are the standard application topologies for an OTA? Op amps have the standard inverting and non-inverting amplifier topologies, and slightly more complicated summing amplifier and difference amplifier, integrator and differentiator, but are there any common circuit idioms for the OTA?

  5. Internally, how does an OTA work? (this is perhaps best left to a separate question, though.)









share|improve this question













I understand the high-level function of OTAs, that they provide an output current proportional to an input voltage. But what I don't understand, and what I can't seem to find any information on online, is exactly how they're used and what the function of the different connections is.



wikipedia's image of the schematic symbol of an OTA



The most pressing questions I have are thus:



  1. What is the function of $I_bias$? Why are there diodes between this pin and the two inputs? I've seen them called linearizing diodes; how do they linearize the response of the circuit?

  2. What is the function of $I_abc$? As I understand it, it provides a scaling factor; is this correct? How does it do so?

  3. What is the (exact or approximate) formula for $I_out$ as a function of $V_in+$, $V_in-$, $I_bias$, and $I_abc$?

  4. What are the standard application topologies for an OTA? Op amps have the standard inverting and non-inverting amplifier topologies, and slightly more complicated summing amplifier and difference amplifier, integrator and differentiator, but are there any common circuit idioms for the OTA?

  5. Internally, how does an OTA work? (this is perhaps best left to a separate question, though.)






amplifier transconductance ota






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asked 3 hours ago









Felthry

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  • 1




    Some of your questions are answered in the LM13700 datasheet
    – Scott Seidman
    3 hours ago










  • @ScottSeidman I've looked at that datasheet and unfortunately had trouble understanding much of it. Perhaps I should go through it again.
    – Felthry
    3 hours ago










  • Have you found ti.com/lit/an/sboa117a/sboa117a.pdf?
    – Scott Seidman
    3 hours ago










  • @ScottSeidman Yes, I just read through that again. It did help; perhaps after organizing my thoughts enough to write this question, I also had them organized enough to understand the datasheet better. Still, I think a basic explanation would be worthwhile, as (though they are fairly rare) OTAs can be confusing and beginners may want an accessible explanation of what they do.
    – Felthry
    2 hours ago












  • 1




    Some of your questions are answered in the LM13700 datasheet
    – Scott Seidman
    3 hours ago










  • @ScottSeidman I've looked at that datasheet and unfortunately had trouble understanding much of it. Perhaps I should go through it again.
    – Felthry
    3 hours ago










  • Have you found ti.com/lit/an/sboa117a/sboa117a.pdf?
    – Scott Seidman
    3 hours ago










  • @ScottSeidman Yes, I just read through that again. It did help; perhaps after organizing my thoughts enough to write this question, I also had them organized enough to understand the datasheet better. Still, I think a basic explanation would be worthwhile, as (though they are fairly rare) OTAs can be confusing and beginners may want an accessible explanation of what they do.
    – Felthry
    2 hours ago







1




1




Some of your questions are answered in the LM13700 datasheet
– Scott Seidman
3 hours ago




Some of your questions are answered in the LM13700 datasheet
– Scott Seidman
3 hours ago












@ScottSeidman I've looked at that datasheet and unfortunately had trouble understanding much of it. Perhaps I should go through it again.
– Felthry
3 hours ago




@ScottSeidman I've looked at that datasheet and unfortunately had trouble understanding much of it. Perhaps I should go through it again.
– Felthry
3 hours ago












Have you found ti.com/lit/an/sboa117a/sboa117a.pdf?
– Scott Seidman
3 hours ago




Have you found ti.com/lit/an/sboa117a/sboa117a.pdf?
– Scott Seidman
3 hours ago












@ScottSeidman Yes, I just read through that again. It did help; perhaps after organizing my thoughts enough to write this question, I also had them organized enough to understand the datasheet better. Still, I think a basic explanation would be worthwhile, as (though they are fairly rare) OTAs can be confusing and beginners may want an accessible explanation of what they do.
– Felthry
2 hours ago




@ScottSeidman Yes, I just read through that again. It did help; perhaps after organizing my thoughts enough to write this question, I also had them organized enough to understand the datasheet better. Still, I think a basic explanation would be worthwhile, as (though they are fairly rare) OTAs can be confusing and beginners may want an accessible explanation of what they do.
– Felthry
2 hours ago










3 Answers
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up vote
2
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The most important point to note is that the input diodes and $I_bias$ pin allow you to operate the amplifier open-loop with a current input signal. If you are apply a low-level differential voltage input or operating the amplifier closed-loop, you may ignore the diodes and bias pin all together.



A Operational Transconductance Amplifier (OTA), is often just a differential input pair followed by a current-mirror. One branch of the diff-pair is mirrored to the high-side supply. The alternate branch of the diff-pair mirrored to the low-side supply. Both current mirrors are connected to the output pin of the amplifier.



So, if the differential input voltage is 0 V, each branch of the differential pair has the same collector current (equal $V_be$ for $V_diff$ = 0). Hence the output is sourcing and sinking equal values of current resulting in a net output current of 0 A.



The input diodes are not just ordinary diodes, they are geometrically, thermally, and process matched to the input differential pair transistors (also a good chance they are actually diode connected transistors). Such that the current following through each of the diodes is scaled value of the current its respective element in the diff-pair. This allows you to apply an current input-signal and still operate the OTA open-loop.



If you go through the derivation for a current input signal (or voltage signal with high source resistance) the gain of input signal to output current becomes a simple linear function of $I_abc$ and $I_bias$.



You can find full derivations in many of the app-notes such as the one listed by tek,http://www.ti.com/lit/ds/symlink/lm13700.pdf






share|improve this answer



























    up vote
    1
    down vote













    Your questions can be answered by carefully reading the graphs on the datasheet: http://www.ti.com/lit/ds/symlink/lm13700.pdf



    Ibias can help decrease the nonlinearity that occurs when v+ and v- are too large. See figure 13.



    For v+ and v- very small, iout = gm * (v+ - v-). There is a graph of gm vs iabc on the datasheet. It is linear on a log-log plot.




    Internally, how does an OTA work? (this is perhaps best left to a separate question, though.)




    Like an opamp without the output stage. The bias current iabc controls the bias current for the whole amplifier, much like with the LM346. Hooking up the output buffer to the output with nothing between gives you an uncompensated opamp.






    share|improve this answer





























      up vote
      1
      down vote













      The Operational Transconductance Amplifier (OTA) cab be desribed as a two stage amplifier: A differential amplifier stage (providing a differential input) followed by a high-impedance output stage (providing the output current proportional to the input diff. voltage).



      Hence, the transfer function is simply Iout=gm*Vdiff with transconductance gm=f(Iabc). The current Iabc (amplifier bias current) is the quantity which directly controls the "open-loop-gain" gm. It is simply the current in the common leg of the diff. input stage.



      There are many applications for OTA devices, for example in filter and oscillator circuits. In most cases, OTA`s are also operated with negative feedback. Because - in contrast to voltage opamps - the "open-loop" function is finite, the value of gm appears in the formulas for the closed-loop gain - hence, the transconductanve (resp. the controlling current Iabc) can be used to control/tune the gain or other filter/oscillator parameters.



      Simple example: An idealized OTA (infinite output resistance) with a capacitor at the output as a load gives a simple integrator circuit (often the heart of an oscillator).



      There is another important application: An OTA with 100% feedback (direct connection between output and inv. input) can be used as grounded resistor R=1/gm that can be externally controlled with Iabc.



      A very good introductory text (Applications) can be found here:



      https://www.ece.uic.edu/~vahe/spring2016/ece412/OTA-structures2.pdf






      share|improve this answer






















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        3 Answers
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        up vote
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        The most important point to note is that the input diodes and $I_bias$ pin allow you to operate the amplifier open-loop with a current input signal. If you are apply a low-level differential voltage input or operating the amplifier closed-loop, you may ignore the diodes and bias pin all together.



        A Operational Transconductance Amplifier (OTA), is often just a differential input pair followed by a current-mirror. One branch of the diff-pair is mirrored to the high-side supply. The alternate branch of the diff-pair mirrored to the low-side supply. Both current mirrors are connected to the output pin of the amplifier.



        So, if the differential input voltage is 0 V, each branch of the differential pair has the same collector current (equal $V_be$ for $V_diff$ = 0). Hence the output is sourcing and sinking equal values of current resulting in a net output current of 0 A.



        The input diodes are not just ordinary diodes, they are geometrically, thermally, and process matched to the input differential pair transistors (also a good chance they are actually diode connected transistors). Such that the current following through each of the diodes is scaled value of the current its respective element in the diff-pair. This allows you to apply an current input-signal and still operate the OTA open-loop.



        If you go through the derivation for a current input signal (or voltage signal with high source resistance) the gain of input signal to output current becomes a simple linear function of $I_abc$ and $I_bias$.



        You can find full derivations in many of the app-notes such as the one listed by tek,http://www.ti.com/lit/ds/symlink/lm13700.pdf






        share|improve this answer
























          up vote
          2
          down vote













          The most important point to note is that the input diodes and $I_bias$ pin allow you to operate the amplifier open-loop with a current input signal. If you are apply a low-level differential voltage input or operating the amplifier closed-loop, you may ignore the diodes and bias pin all together.



          A Operational Transconductance Amplifier (OTA), is often just a differential input pair followed by a current-mirror. One branch of the diff-pair is mirrored to the high-side supply. The alternate branch of the diff-pair mirrored to the low-side supply. Both current mirrors are connected to the output pin of the amplifier.



          So, if the differential input voltage is 0 V, each branch of the differential pair has the same collector current (equal $V_be$ for $V_diff$ = 0). Hence the output is sourcing and sinking equal values of current resulting in a net output current of 0 A.



          The input diodes are not just ordinary diodes, they are geometrically, thermally, and process matched to the input differential pair transistors (also a good chance they are actually diode connected transistors). Such that the current following through each of the diodes is scaled value of the current its respective element in the diff-pair. This allows you to apply an current input-signal and still operate the OTA open-loop.



          If you go through the derivation for a current input signal (or voltage signal with high source resistance) the gain of input signal to output current becomes a simple linear function of $I_abc$ and $I_bias$.



          You can find full derivations in many of the app-notes such as the one listed by tek,http://www.ti.com/lit/ds/symlink/lm13700.pdf






          share|improve this answer






















            up vote
            2
            down vote










            up vote
            2
            down vote









            The most important point to note is that the input diodes and $I_bias$ pin allow you to operate the amplifier open-loop with a current input signal. If you are apply a low-level differential voltage input or operating the amplifier closed-loop, you may ignore the diodes and bias pin all together.



            A Operational Transconductance Amplifier (OTA), is often just a differential input pair followed by a current-mirror. One branch of the diff-pair is mirrored to the high-side supply. The alternate branch of the diff-pair mirrored to the low-side supply. Both current mirrors are connected to the output pin of the amplifier.



            So, if the differential input voltage is 0 V, each branch of the differential pair has the same collector current (equal $V_be$ for $V_diff$ = 0). Hence the output is sourcing and sinking equal values of current resulting in a net output current of 0 A.



            The input diodes are not just ordinary diodes, they are geometrically, thermally, and process matched to the input differential pair transistors (also a good chance they are actually diode connected transistors). Such that the current following through each of the diodes is scaled value of the current its respective element in the diff-pair. This allows you to apply an current input-signal and still operate the OTA open-loop.



            If you go through the derivation for a current input signal (or voltage signal with high source resistance) the gain of input signal to output current becomes a simple linear function of $I_abc$ and $I_bias$.



            You can find full derivations in many of the app-notes such as the one listed by tek,http://www.ti.com/lit/ds/symlink/lm13700.pdf






            share|improve this answer












            The most important point to note is that the input diodes and $I_bias$ pin allow you to operate the amplifier open-loop with a current input signal. If you are apply a low-level differential voltage input or operating the amplifier closed-loop, you may ignore the diodes and bias pin all together.



            A Operational Transconductance Amplifier (OTA), is often just a differential input pair followed by a current-mirror. One branch of the diff-pair is mirrored to the high-side supply. The alternate branch of the diff-pair mirrored to the low-side supply. Both current mirrors are connected to the output pin of the amplifier.



            So, if the differential input voltage is 0 V, each branch of the differential pair has the same collector current (equal $V_be$ for $V_diff$ = 0). Hence the output is sourcing and sinking equal values of current resulting in a net output current of 0 A.



            The input diodes are not just ordinary diodes, they are geometrically, thermally, and process matched to the input differential pair transistors (also a good chance they are actually diode connected transistors). Such that the current following through each of the diodes is scaled value of the current its respective element in the diff-pair. This allows you to apply an current input-signal and still operate the OTA open-loop.



            If you go through the derivation for a current input signal (or voltage signal with high source resistance) the gain of input signal to output current becomes a simple linear function of $I_abc$ and $I_bias$.



            You can find full derivations in many of the app-notes such as the one listed by tek,http://www.ti.com/lit/ds/symlink/lm13700.pdf







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 1 hour ago









            sstobbe

            1,65628




            1,65628






















                up vote
                1
                down vote













                Your questions can be answered by carefully reading the graphs on the datasheet: http://www.ti.com/lit/ds/symlink/lm13700.pdf



                Ibias can help decrease the nonlinearity that occurs when v+ and v- are too large. See figure 13.



                For v+ and v- very small, iout = gm * (v+ - v-). There is a graph of gm vs iabc on the datasheet. It is linear on a log-log plot.




                Internally, how does an OTA work? (this is perhaps best left to a separate question, though.)




                Like an opamp without the output stage. The bias current iabc controls the bias current for the whole amplifier, much like with the LM346. Hooking up the output buffer to the output with nothing between gives you an uncompensated opamp.






                share|improve this answer


























                  up vote
                  1
                  down vote













                  Your questions can be answered by carefully reading the graphs on the datasheet: http://www.ti.com/lit/ds/symlink/lm13700.pdf



                  Ibias can help decrease the nonlinearity that occurs when v+ and v- are too large. See figure 13.



                  For v+ and v- very small, iout = gm * (v+ - v-). There is a graph of gm vs iabc on the datasheet. It is linear on a log-log plot.




                  Internally, how does an OTA work? (this is perhaps best left to a separate question, though.)




                  Like an opamp without the output stage. The bias current iabc controls the bias current for the whole amplifier, much like with the LM346. Hooking up the output buffer to the output with nothing between gives you an uncompensated opamp.






                  share|improve this answer
























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    Your questions can be answered by carefully reading the graphs on the datasheet: http://www.ti.com/lit/ds/symlink/lm13700.pdf



                    Ibias can help decrease the nonlinearity that occurs when v+ and v- are too large. See figure 13.



                    For v+ and v- very small, iout = gm * (v+ - v-). There is a graph of gm vs iabc on the datasheet. It is linear on a log-log plot.




                    Internally, how does an OTA work? (this is perhaps best left to a separate question, though.)




                    Like an opamp without the output stage. The bias current iabc controls the bias current for the whole amplifier, much like with the LM346. Hooking up the output buffer to the output with nothing between gives you an uncompensated opamp.






                    share|improve this answer














                    Your questions can be answered by carefully reading the graphs on the datasheet: http://www.ti.com/lit/ds/symlink/lm13700.pdf



                    Ibias can help decrease the nonlinearity that occurs when v+ and v- are too large. See figure 13.



                    For v+ and v- very small, iout = gm * (v+ - v-). There is a graph of gm vs iabc on the datasheet. It is linear on a log-log plot.




                    Internally, how does an OTA work? (this is perhaps best left to a separate question, though.)




                    Like an opamp without the output stage. The bias current iabc controls the bias current for the whole amplifier, much like with the LM346. Hooking up the output buffer to the output with nothing between gives you an uncompensated opamp.







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited 3 hours ago

























                    answered 3 hours ago









                    τεκ

                    3,17511116




                    3,17511116




















                        up vote
                        1
                        down vote













                        The Operational Transconductance Amplifier (OTA) cab be desribed as a two stage amplifier: A differential amplifier stage (providing a differential input) followed by a high-impedance output stage (providing the output current proportional to the input diff. voltage).



                        Hence, the transfer function is simply Iout=gm*Vdiff with transconductance gm=f(Iabc). The current Iabc (amplifier bias current) is the quantity which directly controls the "open-loop-gain" gm. It is simply the current in the common leg of the diff. input stage.



                        There are many applications for OTA devices, for example in filter and oscillator circuits. In most cases, OTA`s are also operated with negative feedback. Because - in contrast to voltage opamps - the "open-loop" function is finite, the value of gm appears in the formulas for the closed-loop gain - hence, the transconductanve (resp. the controlling current Iabc) can be used to control/tune the gain or other filter/oscillator parameters.



                        Simple example: An idealized OTA (infinite output resistance) with a capacitor at the output as a load gives a simple integrator circuit (often the heart of an oscillator).



                        There is another important application: An OTA with 100% feedback (direct connection between output and inv. input) can be used as grounded resistor R=1/gm that can be externally controlled with Iabc.



                        A very good introductory text (Applications) can be found here:



                        https://www.ece.uic.edu/~vahe/spring2016/ece412/OTA-structures2.pdf






                        share|improve this answer


























                          up vote
                          1
                          down vote













                          The Operational Transconductance Amplifier (OTA) cab be desribed as a two stage amplifier: A differential amplifier stage (providing a differential input) followed by a high-impedance output stage (providing the output current proportional to the input diff. voltage).



                          Hence, the transfer function is simply Iout=gm*Vdiff with transconductance gm=f(Iabc). The current Iabc (amplifier bias current) is the quantity which directly controls the "open-loop-gain" gm. It is simply the current in the common leg of the diff. input stage.



                          There are many applications for OTA devices, for example in filter and oscillator circuits. In most cases, OTA`s are also operated with negative feedback. Because - in contrast to voltage opamps - the "open-loop" function is finite, the value of gm appears in the formulas for the closed-loop gain - hence, the transconductanve (resp. the controlling current Iabc) can be used to control/tune the gain or other filter/oscillator parameters.



                          Simple example: An idealized OTA (infinite output resistance) with a capacitor at the output as a load gives a simple integrator circuit (often the heart of an oscillator).



                          There is another important application: An OTA with 100% feedback (direct connection between output and inv. input) can be used as grounded resistor R=1/gm that can be externally controlled with Iabc.



                          A very good introductory text (Applications) can be found here:



                          https://www.ece.uic.edu/~vahe/spring2016/ece412/OTA-structures2.pdf






                          share|improve this answer
























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            The Operational Transconductance Amplifier (OTA) cab be desribed as a two stage amplifier: A differential amplifier stage (providing a differential input) followed by a high-impedance output stage (providing the output current proportional to the input diff. voltage).



                            Hence, the transfer function is simply Iout=gm*Vdiff with transconductance gm=f(Iabc). The current Iabc (amplifier bias current) is the quantity which directly controls the "open-loop-gain" gm. It is simply the current in the common leg of the diff. input stage.



                            There are many applications for OTA devices, for example in filter and oscillator circuits. In most cases, OTA`s are also operated with negative feedback. Because - in contrast to voltage opamps - the "open-loop" function is finite, the value of gm appears in the formulas for the closed-loop gain - hence, the transconductanve (resp. the controlling current Iabc) can be used to control/tune the gain or other filter/oscillator parameters.



                            Simple example: An idealized OTA (infinite output resistance) with a capacitor at the output as a load gives a simple integrator circuit (often the heart of an oscillator).



                            There is another important application: An OTA with 100% feedback (direct connection between output and inv. input) can be used as grounded resistor R=1/gm that can be externally controlled with Iabc.



                            A very good introductory text (Applications) can be found here:



                            https://www.ece.uic.edu/~vahe/spring2016/ece412/OTA-structures2.pdf






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                            The Operational Transconductance Amplifier (OTA) cab be desribed as a two stage amplifier: A differential amplifier stage (providing a differential input) followed by a high-impedance output stage (providing the output current proportional to the input diff. voltage).



                            Hence, the transfer function is simply Iout=gm*Vdiff with transconductance gm=f(Iabc). The current Iabc (amplifier bias current) is the quantity which directly controls the "open-loop-gain" gm. It is simply the current in the common leg of the diff. input stage.



                            There are many applications for OTA devices, for example in filter and oscillator circuits. In most cases, OTA`s are also operated with negative feedback. Because - in contrast to voltage opamps - the "open-loop" function is finite, the value of gm appears in the formulas for the closed-loop gain - hence, the transconductanve (resp. the controlling current Iabc) can be used to control/tune the gain or other filter/oscillator parameters.



                            Simple example: An idealized OTA (infinite output resistance) with a capacitor at the output as a load gives a simple integrator circuit (often the heart of an oscillator).



                            There is another important application: An OTA with 100% feedback (direct connection between output and inv. input) can be used as grounded resistor R=1/gm that can be externally controlled with Iabc.



                            A very good introductory text (Applications) can be found here:



                            https://www.ece.uic.edu/~vahe/spring2016/ece412/OTA-structures2.pdf







                            share|improve this answer














                            share|improve this answer



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