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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 I_{B+}and I_{B-}works, but when i solve it open up one of the I_{B}to solve the problem.

for I_{B+}, (I_{B-}is opened)

V_{out}= V_{-}( 1 + R_{2}/R_{1}), ..... V_{-}= I_{B+}R_{3}

V_{out}= I_{B+}R_{3}( 1 + R_{2}/R_{1})

for I_{B-}(I_{B+}is opened)

I_{B-}= (V_{-}- 0) / R_{1}+ (V_{-}- V_{out}) / R_{2}

V_{out }= V_{-}(1+ R_{2}/R_{1}) + I_{B-}....can i consider that when I_{B+}is opened, V_{+}= 0 = V_{-}?

if I assume that I am right, i get what the page gives:

V_{out}= -R_{2}(I_{B-})

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

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

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# Need help in understanding input offset curent of Op-Amp

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