Why Put a Resistor in Series with Positive Terminal in Non-Inverting Op-Amp Circuit?
non-invertingoperational-amplifier
I've tried to find the answer, but failed.
Why do we add R1 in series with the positive terminal for a non-inverting amplifier? What is the best value we can assume or choose for it?
Best Answer
It's to compensate the effect of bias current in the non-inverting amplifier. The compensating resistor value equals the parallel combination of R2 and R3. The input current creates a voltage drop across R1 that offsets the voltage across the combination of R2 and R3.
If you do that the amplifier will amplifiy its own output and the ouput will go high or low and stick there until a sufficiently low or high input is presented, which will make it go the other way (low or high).
In essence you will have created a schmitt trigger.
What the formulas don't show is that the solution for this amplifier configuration describes the metastable point,
Another limitation of those equations is that they don't represent the limited output voltage range of real amplifiers.
When they talk about "what range of outputs would you expect", they are talking about the output with V1 == 0V. That is not 100% clear in the problem description.
So given that, you should be able to calculate the output offset voltage from the bias and offset currents (assume different signs of the offset to get a range). The rest of the problem should be self-explanatory.
Best Answer
It's to compensate the effect of bias current in the non-inverting amplifier. The compensating resistor value equals the parallel combination of R2 and R3. The input current creates a voltage drop across R1 that offsets the voltage across the combination of R2 and R3.