Need help in understanding input offset curent of Op-Amp

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Discussion Overview

The discussion revolves around understanding the input offset current of operational amplifiers (op-amps), specifically focusing on the behavior of input bias currents (IB+ and IB-) and their implications in circuit analysis. Participants explore theoretical concepts, mathematical relationships, and practical applications related to op-amp circuits.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant expresses uncertainty about the operation of IB+ and IB- and attempts to derive output voltage equations based on different scenarios of opening these currents.
  • Another participant suggests using superposition to analyze the output voltage, providing equations for both scenarios where IB+ and IB- are considered separately.
  • There is a clarification regarding the definitions of input bias current and input offset current, with one participant stating that these parameters are typically found in op-amp datasheets.
  • Some participants discuss the relationship between the input bias currents and the average input bias current, as well as the offset current, providing formulas for calculating these values.
  • One participant questions the concept of adding current into a resistor, indicating a need for further clarification on this point.

Areas of Agreement / Disagreement

Participants generally agree on the definitions and relationships between input bias currents and input offset current, but there are differing interpretations of how to apply these concepts in circuit analysis. The discussion remains exploratory with no consensus on all aspects presented.

Contextual Notes

Some assumptions and dependencies on circuit configurations are not fully explored, and there are unresolved questions regarding the addition of current into resistors.

Who May Find This Useful

This discussion may be useful for students and practitioners interested in operational amplifier circuits, particularly those looking to deepen their understanding of input bias currents and their effects on circuit behavior.

null void
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image002.gif

This picture is from http://www.ecircuitcenter.com/Circuits/op_ibias/op_ibias.htm

I am not very sure if I am getting the right idea of how the IB+ and IB- works, but when i solve it open up one of the IB to solve the problem.

for IB+, (IB- is opened)
Vout = V-( 1 + R2/R1), ... V- = IB+R3
Vout = IB+R3( 1 + R2/R1)

for IB- (IB+ is opened)
IB- = (V- - 0) / R1 + (V- - Vout) / R2
Vout = V-(1+ R2/R1) + IB-...can i consider that when IB+ is opened, V+ = 0 = V- ?
if I assume that I am right, i get what the page gives:

Vout = -R2(IB-)

one more thing, the IB+ is positive input biase current while the IB- is the negative input bias current? And the Iinput offset = difference of IB- and IB+ ->|IB+ - IB-| ?

These two input bias current and input offset current are the parameters that always appear in the datasheet right?
 
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Try watch this video


Also how can you add current into resistor ?
Vout = IB+R3( 1 + R2/R1)
 
Last edited by a moderator:
Jony130 said:
Try watch this video
Also how can you add current into resistor ?
Vout = IB+R3( 1 + R2/R1)

it is (IB+)R3(1 + R2/R1), sorry for the confusing writing.

I consider that the IB+ current will pass through the R3 and therefore the V+ is equivalent to (IB+)R3 . And i just apply the normal concept for finding the voltage amplification of operational amplifier circuit.
 
For sure we can use a superposition to find Vout.

First we assume IB+ > 0A and IB- = 0A
So have

Vout'1 = IB+*R3 * (1 + R2/R1) and this voltage is positive

Next we open IB+ = 0A and we left with IB-.

Vout'2 = - R2*IB- and for this IB current direction the output voltage is negative.

And finally

Vout = Vout'1 + Vout'2 = IB+•R3•(R2/R1 + 1) - IB-•R2
 
Last edited:
null void said:
These two input bias current and input offset current are the parameters that always appear in the datasheet right?

No, in datasheet you will find the value for the average input bias current.
IB = (IB+ + IB-)/2
and offset current
Ios = IB+ - IB-

so

IB+ = IB + Ios/2

IB- = IB - Ios/2
 
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get it, thanks you very much !
 

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