Op-amp's input bias current

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enter image description here



Could you please tell me how do you compute the impedance seen by the inverting input? The result is 100k//1M but according to me the voltage between the 100k and the 1M is not the same. So the resistors are not in parallel. This result is true in small signal analysis. But the input bias current is present at DC. Is there a way to resolve this problem with the thevenin's theorem ?










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  • 1




    $begingroup$
    The result is 100k//1M No it is not unless there is a voltage source to ground connected at the left of the 100k resistor. As drawn there is "nothing" so the 100 k resistor does nothing. How does an opamp's output behave? Does it have a low or high output impedance? You should first think about how this circuit normally operates, what will be connected to it, how is it used?
    $endgroup$
    – Bimpelrekkie
    Jan 2 at 14:30







  • 2




    $begingroup$
    @Bimpelrekkie: In such circuits, it is generally understood that unknown inputs are voltage sources, unless otherwise specified.
    $endgroup$
    – Dave Tweed
    Jan 2 at 14:31






  • 3




    $begingroup$
    @DaveTweed Sure, but I think that is a very bad habit! Nothing is specified at the input here. How much trouble is it to put a "Vin" there and then it would be clear. Very often I see assumptions being made and assumptions are the source of many errors and misunderstandings.
    $endgroup$
    – Bimpelrekkie
    Jan 2 at 14:32







  • 3




    $begingroup$
    @Bimpelrekkie - I'm sure you are too polite to say it, but there is an old saying which goes, "When you 'assume', you make an 'ass' out of 'u' and 'me',"
    $endgroup$
    – WhatRoughBeast
    Jan 2 at 15:50










  • $begingroup$
    The answer must assume that the Thévenin resistance and voltage at Vin- includes the source resistance, Rs.
    $endgroup$
    – Sunnyskyguy EE75
    Jan 2 at 18:36















3












$begingroup$


enter image description here



Could you please tell me how do you compute the impedance seen by the inverting input? The result is 100k//1M but according to me the voltage between the 100k and the 1M is not the same. So the resistors are not in parallel. This result is true in small signal analysis. But the input bias current is present at DC. Is there a way to resolve this problem with the thevenin's theorem ?










share|improve this question











$endgroup$







  • 1




    $begingroup$
    The result is 100k//1M No it is not unless there is a voltage source to ground connected at the left of the 100k resistor. As drawn there is "nothing" so the 100 k resistor does nothing. How does an opamp's output behave? Does it have a low or high output impedance? You should first think about how this circuit normally operates, what will be connected to it, how is it used?
    $endgroup$
    – Bimpelrekkie
    Jan 2 at 14:30







  • 2




    $begingroup$
    @Bimpelrekkie: In such circuits, it is generally understood that unknown inputs are voltage sources, unless otherwise specified.
    $endgroup$
    – Dave Tweed
    Jan 2 at 14:31






  • 3




    $begingroup$
    @DaveTweed Sure, but I think that is a very bad habit! Nothing is specified at the input here. How much trouble is it to put a "Vin" there and then it would be clear. Very often I see assumptions being made and assumptions are the source of many errors and misunderstandings.
    $endgroup$
    – Bimpelrekkie
    Jan 2 at 14:32







  • 3




    $begingroup$
    @Bimpelrekkie - I'm sure you are too polite to say it, but there is an old saying which goes, "When you 'assume', you make an 'ass' out of 'u' and 'me',"
    $endgroup$
    – WhatRoughBeast
    Jan 2 at 15:50










  • $begingroup$
    The answer must assume that the Thévenin resistance and voltage at Vin- includes the source resistance, Rs.
    $endgroup$
    – Sunnyskyguy EE75
    Jan 2 at 18:36













3












3








3


1



$begingroup$


enter image description here



Could you please tell me how do you compute the impedance seen by the inverting input? The result is 100k//1M but according to me the voltage between the 100k and the 1M is not the same. So the resistors are not in parallel. This result is true in small signal analysis. But the input bias current is present at DC. Is there a way to resolve this problem with the thevenin's theorem ?










share|improve this question











$endgroup$




enter image description here



Could you please tell me how do you compute the impedance seen by the inverting input? The result is 100k//1M but according to me the voltage between the 100k and the 1M is not the same. So the resistors are not in parallel. This result is true in small signal analysis. But the input bias current is present at DC. Is there a way to resolve this problem with the thevenin's theorem ?







op-amp






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share|improve this question













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edited Jan 2 at 15:40









Transistor

81.6k778176




81.6k778176










asked Jan 2 at 14:18









LeoLeo

162




162







  • 1




    $begingroup$
    The result is 100k//1M No it is not unless there is a voltage source to ground connected at the left of the 100k resistor. As drawn there is "nothing" so the 100 k resistor does nothing. How does an opamp's output behave? Does it have a low or high output impedance? You should first think about how this circuit normally operates, what will be connected to it, how is it used?
    $endgroup$
    – Bimpelrekkie
    Jan 2 at 14:30







  • 2




    $begingroup$
    @Bimpelrekkie: In such circuits, it is generally understood that unknown inputs are voltage sources, unless otherwise specified.
    $endgroup$
    – Dave Tweed
    Jan 2 at 14:31






  • 3




    $begingroup$
    @DaveTweed Sure, but I think that is a very bad habit! Nothing is specified at the input here. How much trouble is it to put a "Vin" there and then it would be clear. Very often I see assumptions being made and assumptions are the source of many errors and misunderstandings.
    $endgroup$
    – Bimpelrekkie
    Jan 2 at 14:32







  • 3




    $begingroup$
    @Bimpelrekkie - I'm sure you are too polite to say it, but there is an old saying which goes, "When you 'assume', you make an 'ass' out of 'u' and 'me',"
    $endgroup$
    – WhatRoughBeast
    Jan 2 at 15:50










  • $begingroup$
    The answer must assume that the Thévenin resistance and voltage at Vin- includes the source resistance, Rs.
    $endgroup$
    – Sunnyskyguy EE75
    Jan 2 at 18:36












  • 1




    $begingroup$
    The result is 100k//1M No it is not unless there is a voltage source to ground connected at the left of the 100k resistor. As drawn there is "nothing" so the 100 k resistor does nothing. How does an opamp's output behave? Does it have a low or high output impedance? You should first think about how this circuit normally operates, what will be connected to it, how is it used?
    $endgroup$
    – Bimpelrekkie
    Jan 2 at 14:30







  • 2




    $begingroup$
    @Bimpelrekkie: In such circuits, it is generally understood that unknown inputs are voltage sources, unless otherwise specified.
    $endgroup$
    – Dave Tweed
    Jan 2 at 14:31






  • 3




    $begingroup$
    @DaveTweed Sure, but I think that is a very bad habit! Nothing is specified at the input here. How much trouble is it to put a "Vin" there and then it would be clear. Very often I see assumptions being made and assumptions are the source of many errors and misunderstandings.
    $endgroup$
    – Bimpelrekkie
    Jan 2 at 14:32







  • 3




    $begingroup$
    @Bimpelrekkie - I'm sure you are too polite to say it, but there is an old saying which goes, "When you 'assume', you make an 'ass' out of 'u' and 'me',"
    $endgroup$
    – WhatRoughBeast
    Jan 2 at 15:50










  • $begingroup$
    The answer must assume that the Thévenin resistance and voltage at Vin- includes the source resistance, Rs.
    $endgroup$
    – Sunnyskyguy EE75
    Jan 2 at 18:36







1




1




$begingroup$
The result is 100k//1M No it is not unless there is a voltage source to ground connected at the left of the 100k resistor. As drawn there is "nothing" so the 100 k resistor does nothing. How does an opamp's output behave? Does it have a low or high output impedance? You should first think about how this circuit normally operates, what will be connected to it, how is it used?
$endgroup$
– Bimpelrekkie
Jan 2 at 14:30





$begingroup$
The result is 100k//1M No it is not unless there is a voltage source to ground connected at the left of the 100k resistor. As drawn there is "nothing" so the 100 k resistor does nothing. How does an opamp's output behave? Does it have a low or high output impedance? You should first think about how this circuit normally operates, what will be connected to it, how is it used?
$endgroup$
– Bimpelrekkie
Jan 2 at 14:30





2




2




$begingroup$
@Bimpelrekkie: In such circuits, it is generally understood that unknown inputs are voltage sources, unless otherwise specified.
$endgroup$
– Dave Tweed
Jan 2 at 14:31




$begingroup$
@Bimpelrekkie: In such circuits, it is generally understood that unknown inputs are voltage sources, unless otherwise specified.
$endgroup$
– Dave Tweed
Jan 2 at 14:31




3




3




$begingroup$
@DaveTweed Sure, but I think that is a very bad habit! Nothing is specified at the input here. How much trouble is it to put a "Vin" there and then it would be clear. Very often I see assumptions being made and assumptions are the source of many errors and misunderstandings.
$endgroup$
– Bimpelrekkie
Jan 2 at 14:32





$begingroup$
@DaveTweed Sure, but I think that is a very bad habit! Nothing is specified at the input here. How much trouble is it to put a "Vin" there and then it would be clear. Very often I see assumptions being made and assumptions are the source of many errors and misunderstandings.
$endgroup$
– Bimpelrekkie
Jan 2 at 14:32





3




3




$begingroup$
@Bimpelrekkie - I'm sure you are too polite to say it, but there is an old saying which goes, "When you 'assume', you make an 'ass' out of 'u' and 'me',"
$endgroup$
– WhatRoughBeast
Jan 2 at 15:50




$begingroup$
@Bimpelrekkie - I'm sure you are too polite to say it, but there is an old saying which goes, "When you 'assume', you make an 'ass' out of 'u' and 'me',"
$endgroup$
– WhatRoughBeast
Jan 2 at 15:50












$begingroup$
The answer must assume that the Thévenin resistance and voltage at Vin- includes the source resistance, Rs.
$endgroup$
– Sunnyskyguy EE75
Jan 2 at 18:36




$begingroup$
The answer must assume that the Thévenin resistance and voltage at Vin- includes the source resistance, Rs.
$endgroup$
– Sunnyskyguy EE75
Jan 2 at 18:36










3 Answers
3






active

oldest

votes


















3












$begingroup$

It doesn't matter whether the two voltages are the same. Regardless of what they are, the two voltage sources and their resistors can be replaced with a single source and a single resistor (this is called the "Thévenin equivalent"). The Thévenin resistance is always simply the parallel combination of the original two resistors.






share|improve this answer









$endgroup$








  • 1




    $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Dave Tweed
    Jan 2 at 17:30










  • $begingroup$
    It is also assumed that the Thévenin resistance and voltage at Vin- includes the source resistance.
    $endgroup$
    – Sunnyskyguy EE75
    Jan 2 at 18:34


















2












$begingroup$

DC can still be "small signal", "small signal" simply means that the changes we are looking at are sufficiciently small that nonlinear effects can be ignored.



Still I think the people bringing up impedances and thevinin are not reasoning correctly. The effective impedance at the - input of an non-saturated op-amp in negative feedback is always very low (zero for an ideal op-amp), but nevertheless small currents at that node can have a large impact on output voltage.



Instead I think we need to ask the more fundamental question "what is the affect of input bias currents on output voltage". If we assume that both inputs have the same input bias current then to avoid errors we want the affects of the two input bias currents to be of equal magnitude and opposite polarity.



Lets start with some basic assumptions.



  1. The op-amp has finite input bias currents but is otherwise ideal (in particular this means that all signals are small signals).

  2. The circuit is fed from a voltage source of zero impedance.

  3. We will consider an input bias current flowing in to the op-amps input to be positive and an input bias current flowing out of the op-amps input to be negative.

Assumption 1 means the circuit is linear, which means we can use the principle of superposition to seperately analyse the results of input bias currents.



Lets call the three resistors $R_F$, $R_I$ and $R_+$ for the feedback resistor, input resistor and + terminal resistor.



The inverting input is the simple one. It can't affect the voltage at the op-amps + input and hence can't affect the voltage at the op-amps - input (remember we are assuming the op-amp is otherwise ideal). So the input bias current must be supplied entirely by the op-amps ouput. So for bias currents at the inverting input we can simply calculate the resulting change in output voltage as



$V_OIB- = R_F I_B$



The non-inverting input is more complex. We first need to calculate an input voltage, then calculate a gain. The current flowing into the op-amps input requries a negative voltage at the + terminal.



$V_+ = -R_+ I_B$



We now use the non-inverting op-amp equation to calculate it's affect on output voltage.



$V_OIB+ = (1+fracR_FR_I)V_+ = -(1+fracR_FR_I)R_+ I_B$



Now add them together to get the total contribution of bias currents to output voltage.



$V_OIB = R_F I_B - (1+fracR_FR_I)R_+ I_B$



We want the bias currents to not affect the output voltage.



$V_OIB = 0$



$R_F I_B = (1+fracR_FR_I)R_+ I_B$



$R_F = (1+fracR_FR_I)R_+$



$R_+ = fracR_F(1+fracR_FR_I) = fracR_F R_IR_I + R_F$






share|improve this answer











$endgroup$












  • $begingroup$
    Thank you for this very good explanation :D Nice to you ! :D
    $endgroup$
    – Leo
    Jan 4 at 16:47


















1












$begingroup$



schematic





simulate this circuit – Schematic created using CircuitLab



Let the input bias currents for the inverting and the non-inverting terminals of the op-amp be $I_B^-$ and $I_B^+$ respectively. Thus, the input bias current to the op-amp is their average. That is,
beginequation
I_B = fracI_B^- +I_B^+2
endequation

The input voltage when set to zero that is $ V_i = 0 $, then the the output voltage will be
beginequation
V_o = I_B^- R_f
endequation

Due the flow of the current $I_B^+$ the compensation resistor $R_comp$ has a voltage drop of $ V_1$. The difference of voltage between the node $a$ and the non-inverting terminal of the op-amp is zero as mentioned in the schematic.
Applying Kirchoff's voltage law,
beginequation
-V_1+V_2-V_o=0
endequation

beginequation
V_o = V_2-V_1
endequation

By selecting appropriate resistance for the compensation resistor, the output voltage can be made zero. We know that,
beginequation
V_1 = I_B^+ R_comp
endequation

beginequation
I_B^+ = fracV_1R_comp
endequation

Since, non-inverting terminal is at $-V_1$ voltage, the node $a$ is also at the same voltage with the same polarity since they have no potential difference between them.
Thus,
beginequation
I_1 = fracV_1R_1
endequation

beginequation
I_2 = fracV_2R_f
endequation

For compensation as said earlier, the output voltage must be zero since there is no input given. That is, $V_2=V_1$
Thus,
beginequation
I_2=fracV_1R_f
endequation

KCL at node $a$ gives,
beginequation
I_B^-=I_1+I_2=fracV_1R_1+fracV_1R_f
endequation

Assuming $I_B^-=I_B^+$
beginequation
fracV_1R_comp = fracV_1R_1+fracV_1R_f
endequation

Thus,
beginequation
R_comp = fracR_1R_fR_1+R_f = R_1||R_f
endequation






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    3 Answers
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    3 Answers
    3






    active

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    active

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    $begingroup$

    It doesn't matter whether the two voltages are the same. Regardless of what they are, the two voltage sources and their resistors can be replaced with a single source and a single resistor (this is called the "Thévenin equivalent"). The Thévenin resistance is always simply the parallel combination of the original two resistors.






    share|improve this answer









    $endgroup$








    • 1




      $begingroup$
      Comments are not for extended discussion; this conversation has been moved to chat.
      $endgroup$
      – Dave Tweed
      Jan 2 at 17:30










    • $begingroup$
      It is also assumed that the Thévenin resistance and voltage at Vin- includes the source resistance.
      $endgroup$
      – Sunnyskyguy EE75
      Jan 2 at 18:34















    3












    $begingroup$

    It doesn't matter whether the two voltages are the same. Regardless of what they are, the two voltage sources and their resistors can be replaced with a single source and a single resistor (this is called the "Thévenin equivalent"). The Thévenin resistance is always simply the parallel combination of the original two resistors.






    share|improve this answer









    $endgroup$








    • 1




      $begingroup$
      Comments are not for extended discussion; this conversation has been moved to chat.
      $endgroup$
      – Dave Tweed
      Jan 2 at 17:30










    • $begingroup$
      It is also assumed that the Thévenin resistance and voltage at Vin- includes the source resistance.
      $endgroup$
      – Sunnyskyguy EE75
      Jan 2 at 18:34













    3












    3








    3





    $begingroup$

    It doesn't matter whether the two voltages are the same. Regardless of what they are, the two voltage sources and their resistors can be replaced with a single source and a single resistor (this is called the "Thévenin equivalent"). The Thévenin resistance is always simply the parallel combination of the original two resistors.






    share|improve this answer









    $endgroup$



    It doesn't matter whether the two voltages are the same. Regardless of what they are, the two voltage sources and their resistors can be replaced with a single source and a single resistor (this is called the "Thévenin equivalent"). The Thévenin resistance is always simply the parallel combination of the original two resistors.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Jan 2 at 14:29









    Dave TweedDave Tweed

    118k9145256




    118k9145256







    • 1




      $begingroup$
      Comments are not for extended discussion; this conversation has been moved to chat.
      $endgroup$
      – Dave Tweed
      Jan 2 at 17:30










    • $begingroup$
      It is also assumed that the Thévenin resistance and voltage at Vin- includes the source resistance.
      $endgroup$
      – Sunnyskyguy EE75
      Jan 2 at 18:34












    • 1




      $begingroup$
      Comments are not for extended discussion; this conversation has been moved to chat.
      $endgroup$
      – Dave Tweed
      Jan 2 at 17:30










    • $begingroup$
      It is also assumed that the Thévenin resistance and voltage at Vin- includes the source resistance.
      $endgroup$
      – Sunnyskyguy EE75
      Jan 2 at 18:34







    1




    1




    $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Dave Tweed
    Jan 2 at 17:30




    $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Dave Tweed
    Jan 2 at 17:30












    $begingroup$
    It is also assumed that the Thévenin resistance and voltage at Vin- includes the source resistance.
    $endgroup$
    – Sunnyskyguy EE75
    Jan 2 at 18:34




    $begingroup$
    It is also assumed that the Thévenin resistance and voltage at Vin- includes the source resistance.
    $endgroup$
    – Sunnyskyguy EE75
    Jan 2 at 18:34













    2












    $begingroup$

    DC can still be "small signal", "small signal" simply means that the changes we are looking at are sufficiciently small that nonlinear effects can be ignored.



    Still I think the people bringing up impedances and thevinin are not reasoning correctly. The effective impedance at the - input of an non-saturated op-amp in negative feedback is always very low (zero for an ideal op-amp), but nevertheless small currents at that node can have a large impact on output voltage.



    Instead I think we need to ask the more fundamental question "what is the affect of input bias currents on output voltage". If we assume that both inputs have the same input bias current then to avoid errors we want the affects of the two input bias currents to be of equal magnitude and opposite polarity.



    Lets start with some basic assumptions.



    1. The op-amp has finite input bias currents but is otherwise ideal (in particular this means that all signals are small signals).

    2. The circuit is fed from a voltage source of zero impedance.

    3. We will consider an input bias current flowing in to the op-amps input to be positive and an input bias current flowing out of the op-amps input to be negative.

    Assumption 1 means the circuit is linear, which means we can use the principle of superposition to seperately analyse the results of input bias currents.



    Lets call the three resistors $R_F$, $R_I$ and $R_+$ for the feedback resistor, input resistor and + terminal resistor.



    The inverting input is the simple one. It can't affect the voltage at the op-amps + input and hence can't affect the voltage at the op-amps - input (remember we are assuming the op-amp is otherwise ideal). So the input bias current must be supplied entirely by the op-amps ouput. So for bias currents at the inverting input we can simply calculate the resulting change in output voltage as



    $V_OIB- = R_F I_B$



    The non-inverting input is more complex. We first need to calculate an input voltage, then calculate a gain. The current flowing into the op-amps input requries a negative voltage at the + terminal.



    $V_+ = -R_+ I_B$



    We now use the non-inverting op-amp equation to calculate it's affect on output voltage.



    $V_OIB+ = (1+fracR_FR_I)V_+ = -(1+fracR_FR_I)R_+ I_B$



    Now add them together to get the total contribution of bias currents to output voltage.



    $V_OIB = R_F I_B - (1+fracR_FR_I)R_+ I_B$



    We want the bias currents to not affect the output voltage.



    $V_OIB = 0$



    $R_F I_B = (1+fracR_FR_I)R_+ I_B$



    $R_F = (1+fracR_FR_I)R_+$



    $R_+ = fracR_F(1+fracR_FR_I) = fracR_F R_IR_I + R_F$






    share|improve this answer











    $endgroup$












    • $begingroup$
      Thank you for this very good explanation :D Nice to you ! :D
      $endgroup$
      – Leo
      Jan 4 at 16:47















    2












    $begingroup$

    DC can still be "small signal", "small signal" simply means that the changes we are looking at are sufficiciently small that nonlinear effects can be ignored.



    Still I think the people bringing up impedances and thevinin are not reasoning correctly. The effective impedance at the - input of an non-saturated op-amp in negative feedback is always very low (zero for an ideal op-amp), but nevertheless small currents at that node can have a large impact on output voltage.



    Instead I think we need to ask the more fundamental question "what is the affect of input bias currents on output voltage". If we assume that both inputs have the same input bias current then to avoid errors we want the affects of the two input bias currents to be of equal magnitude and opposite polarity.



    Lets start with some basic assumptions.



    1. The op-amp has finite input bias currents but is otherwise ideal (in particular this means that all signals are small signals).

    2. The circuit is fed from a voltage source of zero impedance.

    3. We will consider an input bias current flowing in to the op-amps input to be positive and an input bias current flowing out of the op-amps input to be negative.

    Assumption 1 means the circuit is linear, which means we can use the principle of superposition to seperately analyse the results of input bias currents.



    Lets call the three resistors $R_F$, $R_I$ and $R_+$ for the feedback resistor, input resistor and + terminal resistor.



    The inverting input is the simple one. It can't affect the voltage at the op-amps + input and hence can't affect the voltage at the op-amps - input (remember we are assuming the op-amp is otherwise ideal). So the input bias current must be supplied entirely by the op-amps ouput. So for bias currents at the inverting input we can simply calculate the resulting change in output voltage as



    $V_OIB- = R_F I_B$



    The non-inverting input is more complex. We first need to calculate an input voltage, then calculate a gain. The current flowing into the op-amps input requries a negative voltage at the + terminal.



    $V_+ = -R_+ I_B$



    We now use the non-inverting op-amp equation to calculate it's affect on output voltage.



    $V_OIB+ = (1+fracR_FR_I)V_+ = -(1+fracR_FR_I)R_+ I_B$



    Now add them together to get the total contribution of bias currents to output voltage.



    $V_OIB = R_F I_B - (1+fracR_FR_I)R_+ I_B$



    We want the bias currents to not affect the output voltage.



    $V_OIB = 0$



    $R_F I_B = (1+fracR_FR_I)R_+ I_B$



    $R_F = (1+fracR_FR_I)R_+$



    $R_+ = fracR_F(1+fracR_FR_I) = fracR_F R_IR_I + R_F$






    share|improve this answer











    $endgroup$












    • $begingroup$
      Thank you for this very good explanation :D Nice to you ! :D
      $endgroup$
      – Leo
      Jan 4 at 16:47













    2












    2








    2





    $begingroup$

    DC can still be "small signal", "small signal" simply means that the changes we are looking at are sufficiciently small that nonlinear effects can be ignored.



    Still I think the people bringing up impedances and thevinin are not reasoning correctly. The effective impedance at the - input of an non-saturated op-amp in negative feedback is always very low (zero for an ideal op-amp), but nevertheless small currents at that node can have a large impact on output voltage.



    Instead I think we need to ask the more fundamental question "what is the affect of input bias currents on output voltage". If we assume that both inputs have the same input bias current then to avoid errors we want the affects of the two input bias currents to be of equal magnitude and opposite polarity.



    Lets start with some basic assumptions.



    1. The op-amp has finite input bias currents but is otherwise ideal (in particular this means that all signals are small signals).

    2. The circuit is fed from a voltage source of zero impedance.

    3. We will consider an input bias current flowing in to the op-amps input to be positive and an input bias current flowing out of the op-amps input to be negative.

    Assumption 1 means the circuit is linear, which means we can use the principle of superposition to seperately analyse the results of input bias currents.



    Lets call the three resistors $R_F$, $R_I$ and $R_+$ for the feedback resistor, input resistor and + terminal resistor.



    The inverting input is the simple one. It can't affect the voltage at the op-amps + input and hence can't affect the voltage at the op-amps - input (remember we are assuming the op-amp is otherwise ideal). So the input bias current must be supplied entirely by the op-amps ouput. So for bias currents at the inverting input we can simply calculate the resulting change in output voltage as



    $V_OIB- = R_F I_B$



    The non-inverting input is more complex. We first need to calculate an input voltage, then calculate a gain. The current flowing into the op-amps input requries a negative voltage at the + terminal.



    $V_+ = -R_+ I_B$



    We now use the non-inverting op-amp equation to calculate it's affect on output voltage.



    $V_OIB+ = (1+fracR_FR_I)V_+ = -(1+fracR_FR_I)R_+ I_B$



    Now add them together to get the total contribution of bias currents to output voltage.



    $V_OIB = R_F I_B - (1+fracR_FR_I)R_+ I_B$



    We want the bias currents to not affect the output voltage.



    $V_OIB = 0$



    $R_F I_B = (1+fracR_FR_I)R_+ I_B$



    $R_F = (1+fracR_FR_I)R_+$



    $R_+ = fracR_F(1+fracR_FR_I) = fracR_F R_IR_I + R_F$






    share|improve this answer











    $endgroup$



    DC can still be "small signal", "small signal" simply means that the changes we are looking at are sufficiciently small that nonlinear effects can be ignored.



    Still I think the people bringing up impedances and thevinin are not reasoning correctly. The effective impedance at the - input of an non-saturated op-amp in negative feedback is always very low (zero for an ideal op-amp), but nevertheless small currents at that node can have a large impact on output voltage.



    Instead I think we need to ask the more fundamental question "what is the affect of input bias currents on output voltage". If we assume that both inputs have the same input bias current then to avoid errors we want the affects of the two input bias currents to be of equal magnitude and opposite polarity.



    Lets start with some basic assumptions.



    1. The op-amp has finite input bias currents but is otherwise ideal (in particular this means that all signals are small signals).

    2. The circuit is fed from a voltage source of zero impedance.

    3. We will consider an input bias current flowing in to the op-amps input to be positive and an input bias current flowing out of the op-amps input to be negative.

    Assumption 1 means the circuit is linear, which means we can use the principle of superposition to seperately analyse the results of input bias currents.



    Lets call the three resistors $R_F$, $R_I$ and $R_+$ for the feedback resistor, input resistor and + terminal resistor.



    The inverting input is the simple one. It can't affect the voltage at the op-amps + input and hence can't affect the voltage at the op-amps - input (remember we are assuming the op-amp is otherwise ideal). So the input bias current must be supplied entirely by the op-amps ouput. So for bias currents at the inverting input we can simply calculate the resulting change in output voltage as



    $V_OIB- = R_F I_B$



    The non-inverting input is more complex. We first need to calculate an input voltage, then calculate a gain. The current flowing into the op-amps input requries a negative voltage at the + terminal.



    $V_+ = -R_+ I_B$



    We now use the non-inverting op-amp equation to calculate it's affect on output voltage.



    $V_OIB+ = (1+fracR_FR_I)V_+ = -(1+fracR_FR_I)R_+ I_B$



    Now add them together to get the total contribution of bias currents to output voltage.



    $V_OIB = R_F I_B - (1+fracR_FR_I)R_+ I_B$



    We want the bias currents to not affect the output voltage.



    $V_OIB = 0$



    $R_F I_B = (1+fracR_FR_I)R_+ I_B$



    $R_F = (1+fracR_FR_I)R_+$



    $R_+ = fracR_F(1+fracR_FR_I) = fracR_F R_IR_I + R_F$







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Jan 2 at 19:30

























    answered Jan 2 at 18:20









    Peter GreenPeter Green

    11.6k11939




    11.6k11939











    • $begingroup$
      Thank you for this very good explanation :D Nice to you ! :D
      $endgroup$
      – Leo
      Jan 4 at 16:47
















    • $begingroup$
      Thank you for this very good explanation :D Nice to you ! :D
      $endgroup$
      – Leo
      Jan 4 at 16:47















    $begingroup$
    Thank you for this very good explanation :D Nice to you ! :D
    $endgroup$
    – Leo
    Jan 4 at 16:47




    $begingroup$
    Thank you for this very good explanation :D Nice to you ! :D
    $endgroup$
    – Leo
    Jan 4 at 16:47











    1












    $begingroup$



    schematic





    simulate this circuit – Schematic created using CircuitLab



    Let the input bias currents for the inverting and the non-inverting terminals of the op-amp be $I_B^-$ and $I_B^+$ respectively. Thus, the input bias current to the op-amp is their average. That is,
    beginequation
    I_B = fracI_B^- +I_B^+2
    endequation

    The input voltage when set to zero that is $ V_i = 0 $, then the the output voltage will be
    beginequation
    V_o = I_B^- R_f
    endequation

    Due the flow of the current $I_B^+$ the compensation resistor $R_comp$ has a voltage drop of $ V_1$. The difference of voltage between the node $a$ and the non-inverting terminal of the op-amp is zero as mentioned in the schematic.
    Applying Kirchoff's voltage law,
    beginequation
    -V_1+V_2-V_o=0
    endequation

    beginequation
    V_o = V_2-V_1
    endequation

    By selecting appropriate resistance for the compensation resistor, the output voltage can be made zero. We know that,
    beginequation
    V_1 = I_B^+ R_comp
    endequation

    beginequation
    I_B^+ = fracV_1R_comp
    endequation

    Since, non-inverting terminal is at $-V_1$ voltage, the node $a$ is also at the same voltage with the same polarity since they have no potential difference between them.
    Thus,
    beginequation
    I_1 = fracV_1R_1
    endequation

    beginequation
    I_2 = fracV_2R_f
    endequation

    For compensation as said earlier, the output voltage must be zero since there is no input given. That is, $V_2=V_1$
    Thus,
    beginequation
    I_2=fracV_1R_f
    endequation

    KCL at node $a$ gives,
    beginequation
    I_B^-=I_1+I_2=fracV_1R_1+fracV_1R_f
    endequation

    Assuming $I_B^-=I_B^+$
    beginequation
    fracV_1R_comp = fracV_1R_1+fracV_1R_f
    endequation

    Thus,
    beginequation
    R_comp = fracR_1R_fR_1+R_f = R_1||R_f
    endequation






    share|improve this answer











    $endgroup$

















      1












      $begingroup$



      schematic





      simulate this circuit – Schematic created using CircuitLab



      Let the input bias currents for the inverting and the non-inverting terminals of the op-amp be $I_B^-$ and $I_B^+$ respectively. Thus, the input bias current to the op-amp is their average. That is,
      beginequation
      I_B = fracI_B^- +I_B^+2
      endequation

      The input voltage when set to zero that is $ V_i = 0 $, then the the output voltage will be
      beginequation
      V_o = I_B^- R_f
      endequation

      Due the flow of the current $I_B^+$ the compensation resistor $R_comp$ has a voltage drop of $ V_1$. The difference of voltage between the node $a$ and the non-inverting terminal of the op-amp is zero as mentioned in the schematic.
      Applying Kirchoff's voltage law,
      beginequation
      -V_1+V_2-V_o=0
      endequation

      beginequation
      V_o = V_2-V_1
      endequation

      By selecting appropriate resistance for the compensation resistor, the output voltage can be made zero. We know that,
      beginequation
      V_1 = I_B^+ R_comp
      endequation

      beginequation
      I_B^+ = fracV_1R_comp
      endequation

      Since, non-inverting terminal is at $-V_1$ voltage, the node $a$ is also at the same voltage with the same polarity since they have no potential difference between them.
      Thus,
      beginequation
      I_1 = fracV_1R_1
      endequation

      beginequation
      I_2 = fracV_2R_f
      endequation

      For compensation as said earlier, the output voltage must be zero since there is no input given. That is, $V_2=V_1$
      Thus,
      beginequation
      I_2=fracV_1R_f
      endequation

      KCL at node $a$ gives,
      beginequation
      I_B^-=I_1+I_2=fracV_1R_1+fracV_1R_f
      endequation

      Assuming $I_B^-=I_B^+$
      beginequation
      fracV_1R_comp = fracV_1R_1+fracV_1R_f
      endequation

      Thus,
      beginequation
      R_comp = fracR_1R_fR_1+R_f = R_1||R_f
      endequation






      share|improve this answer











      $endgroup$















        1












        1








        1





        $begingroup$



        schematic





        simulate this circuit – Schematic created using CircuitLab



        Let the input bias currents for the inverting and the non-inverting terminals of the op-amp be $I_B^-$ and $I_B^+$ respectively. Thus, the input bias current to the op-amp is their average. That is,
        beginequation
        I_B = fracI_B^- +I_B^+2
        endequation

        The input voltage when set to zero that is $ V_i = 0 $, then the the output voltage will be
        beginequation
        V_o = I_B^- R_f
        endequation

        Due the flow of the current $I_B^+$ the compensation resistor $R_comp$ has a voltage drop of $ V_1$. The difference of voltage between the node $a$ and the non-inverting terminal of the op-amp is zero as mentioned in the schematic.
        Applying Kirchoff's voltage law,
        beginequation
        -V_1+V_2-V_o=0
        endequation

        beginequation
        V_o = V_2-V_1
        endequation

        By selecting appropriate resistance for the compensation resistor, the output voltage can be made zero. We know that,
        beginequation
        V_1 = I_B^+ R_comp
        endequation

        beginequation
        I_B^+ = fracV_1R_comp
        endequation

        Since, non-inverting terminal is at $-V_1$ voltage, the node $a$ is also at the same voltage with the same polarity since they have no potential difference between them.
        Thus,
        beginequation
        I_1 = fracV_1R_1
        endequation

        beginequation
        I_2 = fracV_2R_f
        endequation

        For compensation as said earlier, the output voltage must be zero since there is no input given. That is, $V_2=V_1$
        Thus,
        beginequation
        I_2=fracV_1R_f
        endequation

        KCL at node $a$ gives,
        beginequation
        I_B^-=I_1+I_2=fracV_1R_1+fracV_1R_f
        endequation

        Assuming $I_B^-=I_B^+$
        beginequation
        fracV_1R_comp = fracV_1R_1+fracV_1R_f
        endequation

        Thus,
        beginequation
        R_comp = fracR_1R_fR_1+R_f = R_1||R_f
        endequation






        share|improve this answer











        $endgroup$





        schematic





        simulate this circuit – Schematic created using CircuitLab



        Let the input bias currents for the inverting and the non-inverting terminals of the op-amp be $I_B^-$ and $I_B^+$ respectively. Thus, the input bias current to the op-amp is their average. That is,
        beginequation
        I_B = fracI_B^- +I_B^+2
        endequation

        The input voltage when set to zero that is $ V_i = 0 $, then the the output voltage will be
        beginequation
        V_o = I_B^- R_f
        endequation

        Due the flow of the current $I_B^+$ the compensation resistor $R_comp$ has a voltage drop of $ V_1$. The difference of voltage between the node $a$ and the non-inverting terminal of the op-amp is zero as mentioned in the schematic.
        Applying Kirchoff's voltage law,
        beginequation
        -V_1+V_2-V_o=0
        endequation

        beginequation
        V_o = V_2-V_1
        endequation

        By selecting appropriate resistance for the compensation resistor, the output voltage can be made zero. We know that,
        beginequation
        V_1 = I_B^+ R_comp
        endequation

        beginequation
        I_B^+ = fracV_1R_comp
        endequation

        Since, non-inverting terminal is at $-V_1$ voltage, the node $a$ is also at the same voltage with the same polarity since they have no potential difference between them.
        Thus,
        beginequation
        I_1 = fracV_1R_1
        endequation

        beginequation
        I_2 = fracV_2R_f
        endequation

        For compensation as said earlier, the output voltage must be zero since there is no input given. That is, $V_2=V_1$
        Thus,
        beginequation
        I_2=fracV_1R_f
        endequation

        KCL at node $a$ gives,
        beginequation
        I_B^-=I_1+I_2=fracV_1R_1+fracV_1R_f
        endequation

        Assuming $I_B^-=I_B^+$
        beginequation
        fracV_1R_comp = fracV_1R_1+fracV_1R_f
        endequation

        Thus,
        beginequation
        R_comp = fracR_1R_fR_1+R_f = R_1||R_f
        endequation







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Jan 2 at 18:27

























        answered Jan 2 at 17:56









        RAMASUBRAMANIYAN GRAMASUBRAMANIYAN G

        415




        415



























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