When we analize the effects of the offset voltage at the outlet of an amplifier configuration, we consider that this voltage is applied to the non-inverting input. At least it appears so in all the books I have reviewed so far. Is there any particual reason for this? Thanks for your help
There is only one simple reason: The calculation of the resulting output DC offset is easier in case of non-inverting configuration. In this case, we just have to multiply the input offset with the "noise gain" which is the inverse of the feedback factor (and identical with the non-inv. gain factor).
However, you also can add the input DC offset in form of a DC voltage source at the inverting input. In this case, the calculation is a bit more involved - however, the result is the same (except the sign).
Why is the calculation more involved? The offset voltage is measured (specified) as a DC voltage directly at the opamp input against ground. In the non-inv. configuration we, therefore, can replace the "normal" signal input voltage by the offset voltage Voff - and the result is known (noise gain). However, in the inverting configuration this is not possible (the inv. input terminal does not receive the signal input). First, we have to recalculate the influence of the known DC offset Voff and place the resulting voltage source (taking the feedback network into account) in series to the input resistor.
However, please note that the input offset voltage Voff is specified for the special case Vout=0. That means: Voff is identical to an externally applied DC Voltage at the non-inv. input that brings the DC output to zero! In this context, the sign of Voff is fixed.