Example: the Input Offset Voltage

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Example: The Input Offset Voltage

Consider an inverting amp constructed with an op-amp exhibiting an input offset voltage of Vos: R 2

R 1 v v 1 - i - i ideal 1 vo v Vos + 2 + - +

Applying the concept of a virtual short to the ideal op-amp, we find that:

v1 0  Vos  0

Vos

Thus, v1 v 2 ! For an op-amp with an input offset voltage, the virtual “short” equation turns out to be: 4/5/2018 0257df836ff057d5ac3663aef774ef0c.doc 2/3

v1 Vos  v 2

Recall, however, that the input offset voltage is typically very small (i.e., Vos  5mV ), so that v1 v 2 .

The current into each terminal of the op-amp is still zero, so that:

i1 i 2 where:

vi  v1 i1   R1 and:

v1  vo i2   R2 Combining, we find:

v V V v i os  os o R1 R 2

Performing a little algebra, we can solve this equation for output voltage vo :

Vos R1 V os R 2  vi R 2 vo  R1 and rearranging: 4/5/2018 0257df836ff057d5ac3663aef774ef0c.doc 3/3

     vo   vi    V os    

Hey! We could have easily found this result by applying superposition!

Note that if the input offset voltage is zero (its ideal value), this expression simply reduces to the normal inverting amplifier expression:

R2  vo     vi R1 

Thus, the term: R   2 1 Vos R1  represents an output offset voltage. Note this offset voltage is a constant with respect to vi --its value does not change, even if the input voltage is zero!.