input bias current of opamp

[electroboy]

hello... please tell me how v0 is obtained in the following picture..particularly the R2*In term...though it is said superposition theorem , i cant understand.. also i am confused whether R2*In term will be positive or negative...

 

[electroboy]

is my question wrong ?? please help me..

 

[DXNewcastle]

In most applications which use any popular Op-Amp IC, the currents In and Ip will be very small currents. These are input currents and the designer might not even bother to calculate any terms which include these currents. They would probably use a low or zero value for Rp and maybe a low value for R1 so that these currents become insignificant.

To answer your question, the term R2*In is introducing a small voltage at the output (Vo) which will be a result of the small current flowing into the inverting input, In.
R2*In will be a very small positive value.
Have you built this circuit? Have you been able to measure any voltage at the output, Vo?

 

[electroboy]

In most applications which use any popular Op-Amp IC, the currents In and Ip will be very small currents. These are input currents and the designer might not even bother to calculate any terms which include these currents. They would probably use a low or zero value for Rp and maybe a low value for R1 so that these currents become insignificant.

To answer your question, the term R2*In is introducing a small voltage at the output (Vo) which will be a result of the small current flowing into the inverting input, In.
R2*In will be a very small positive value.
Have you built this circuit? Have you been able to measure any voltage at the output, Vo?

thank you sir.. i have not yet built the circuit.. i will do the circuit and update my level....also my doubt is "In is flowing inside opamp.. how can it produce voltage with feedback resistor??? " please dont ignore my doubt even it is too elementary...

 

[DXNewcastle]

.. how can it produce voltage with feedback resistor???

because that is one of the few things that Op-Amps are designed to do!
They amplify voltage and they amplify current.

It is the designer's job to choose the circuit's components which will provide the required voltage amplification. In this case, the amount of voltage amplification is defined by the values of the two resistors R1 and R2.
The feedback resistor only presents a proportion of the output voltage to the invertng input, which is going to allow for some gain. I'm sure you have other diagrams which include calculations of voltage amplification. These should help you to understand how output voltages are calculated.

 

[Audioguru]

If you are worried that the opamp amplifies its input bias current in the resistors then seleect an opamp with Jfet inputs because it has no input bias current.

 

[FvM]

In my words, the above suggestion says:
"If you don't understand how to calculate bias current induced errors, use only FET OPs". I mean to remember bad jokes with a similar payoff. I think, it's still good to have these problems in text books, and of course, the question can be answered. By the way, in some applications (e.g. high speed) we have OPs with uA input currents that can't be neglected.

Regarding the R2*In term in particular, I want to suggest a simple explanation. The equation is based on an ideal OP, except for the input currents, it has infinite gain. So the feedback will zero the input voltage by setting a respective output voltage. With zero input voltage, no current flows through R1, all must go to R2. So the part of the ouput voltage caused by In will be R2*In. And about polarity, for the In direction drawn in the circuit R2*In will be positive. I think it's obvious, that a positive output voltage will drive a positive current into the input.

 

[Audioguru]

If you worry about the input bias current for an opamp then you must look to see if it has NPN input transistors or PNP. Many opamps have PNP like the LM324, LM358, MC3317x, MC3407x and TLEx134.

 

[electroboy]

thank you very much..this helped me a lot...