Need help in understanding input offset curent of Op-Amp

[null void]

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 + R₂/R₁), ... V_- = I_B+R₃
V_out = I_B+R₃( 1 + R₂/R₁)

for I_B- (I_B+ is opened)
I_B- = (V_- - 0) / R₁ + (V_- - V_out) / R₂
V_out = V_-(1+ R₂/R₁) + 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 = -R₂(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?

 

[Jony130]

Try watch this video


Also how can you add current into resistor ?
Vout = IB+R3( 1 + R2/R1)

 

[null void]

Try watch this video
Also how can you add current into resistor ?
Vout = IB+R3( 1 + R2/R1)

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

I consider that the I_B+ current will pass through the R3 and therefore the V₊ is equivalent to (I_B+)R₃ . And i just apply the normal concept for finding the voltage amplification of operational amplifier circuit.

 

[Jony130]

For sure we can use a superposition to find Vout.

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

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

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

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

And finally

Vout = Vout'1 + Vout'2 = I_B+•R3•(R2/R1 + 1) - I_B-•R2

 

[Jony130]

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 = (I_B+ + I_B-)/2
and offset current
Ios = I_B+ - I_B-

so

I_B+ = IB + Ios/2

I_B- = IB - Ios/2

 

[null void]

get it, thanks you very much !