Download 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:
R2
i2
R1
vi
v1
-
i1
v2 +
Vos
ideal
+
vo
- +
Applying the concept of a virtual short to the ideal op-amp, we
find that:
v1  0 Vos  0
 Vos
Thus, v1  v2 ! For an op-amp with an input offset voltage, the
virtual “short” equation turns out to be:
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v1  Vos  v2
Recall, however, that the input offset voltage is typically very
small (i.e., Vos  5 mV ), so that v1  v2 .
The current into each terminal of the op-amp is still zero, so
that:
i1  i2
where:
i1 
vi  v1

R1
i2 
v1  vo

R2
and:
Combining, we find:
vi Vos Vos  vo

R1
R2
Performing a little algebra, we can solve this equation for output
voltage vo :
vo 
and rearranging:
Vos R1 Vos R2  vi R2
R1
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
vo   



v

 i 


3/3


Vos

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:
R 
vo    2 vi
 R1 
Thus, the term:
R2 

1


Vos
R

1 
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!.