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Transcript
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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
vin
v-
-
i1
Vos
v+
+
ideal
vout
+
- +
We know that because of the input offset voltage:
v   v  Vos
For the circuit above, the non-inverting terminal of the op-amp
is connected to ground (i.e., v   0 ), and so the virtual “ground”
is now described by:
v  Vos
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The current into each terminal of the op-amp is still zero, so
that from KCL:
i1  i2
where form KCL and Ohm’s Law:
i1 
and:
i2 
vin  v  vin Vos

R1
R1
v   vout Vos  vout

R2
R2
Combining, we find:
vin Vos Vos  vout

R1
R2
Performing a little algebra, we can solve this equation for output
voltage vout :
V R V R  v R
vout  os 1 os 2 in 2
R1
and rearranging:
R 
R 

vout    2 vin   1  2 Vos
R1 
 R1 

Q: Hey! Couldn’t we have easily found this result by applying
superposition?
A: Absolutely!
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Note that if the input offset voltage is zero (its ideal value),
this expression simply reduces to the normal inverting amplifier
expression:
R 
vout    2 vin
 R1 
Thus, the term:
R2 

1

V

R1  os

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!.
R2
i2
R1
vin
v-
-
i1
ideal
+
v+
+
vout
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