Need help in understanding input offset curent of Op-Amp

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The discussion focuses on understanding the input bias currents (IB+ and IB-) of operational amplifiers and their impact on output voltage calculations. Users explore how to derive output voltage equations based on the assumptions of opening one bias current at a time, leading to equations that incorporate resistor values. Clarifications are made regarding the definitions of input bias current and input offset current, emphasizing their presence in datasheets. The average input bias current is calculated as the mean of IB+ and IB-, while the offset current is the difference between them. Overall, the thread provides insights into operational amplifier behavior and the mathematical relationships involved.
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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)
 
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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 throught 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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