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Transcript
5/14/2017
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Superposition and
Op-Amp Circuits
Consider this op-amp circuit, with two input voltages (v1 and v2):
R2
R1
v1
-
v2
ideal
R3
vout
+
R4
The easiest way to analyze this circuit is to apply superposition!
Recall that op-amp circuits are linear, so superposition applies.
Our first step is to set all sources to zero, except v2 —in other
words, set v1 =0 (connect it to ground potential):
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R2
R1
-
v2
ideal
R3
v
vo2
+
i  0
R4
Since the current into the non-inverting input of the op-amp is
zero (i  0 ), it is evident that:
v 
R4
v
R3  R4 2
Likewise, the remainder of the circuit is simply the noninverting amplifier, where:

R 
vo 2   1  2  v 
R1 

Combining these two equations, we get:
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
R   R4 
vo 2   1  2  
 v2
R
R

R

1   3
4 
Now for the second step. Turn off all sources except v1 —in
other words set v2 =0:
R2
R1
v1
i  0
v
ideal
R3
v
vo1
+
i  0
R4
It is evident that the since the current into the non-inverting
terminal of the op-amp is zero, the voltage v+ is likewise zero.
Thus, the circuit above is simply an inverting amplifier, where:
vo 1  
R2
v
R1 1
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There are no more sources in this circuit, so that we can
conclude from superposition that the output voltage is the sum
of our two, single-source solutions:
vout  vo 1  vo 2

 R2 
R   R4 
 1  2  
 v 2    v1
R1   R3  R4 

 R1 
Note this circuit is effectively a weighted difference amplifier.