# Example 6 . ```Example 6
3 kΩ
1.5 kΩ
3 kΩ
_
100 kΩ
1 kΩ
_
+
5V
+
_
+
_
+
2V
+
vo
_
For the op amp circuit shown in the figure, obtain the output voltage v o .
For the circuit which consists of two ideal operational amplifiers, we want to calculate
the output voltage v o . The circuit can be divided into two sub circuits. Let’s look at the
first part of the circuit. The noninverting input of the op amp is grounded. The inverting
input node is connected with two voltage sources through input resistors. There is a
feedback resistor connected between the inverting node and the output node This circuit
is known as summing amplifier. To make it general, we can label the power supplies as
v1 and v2. The two input resistances as R1 and R2, and the feedback resistance as Rf. For
the summing amplifier, the output should be a weighted sum of the inputs.
Rf
 Rf

v out  
v1 
v 2 
R2
 R1

The weight for each input is simply the feedback resistance divided by the input
resistance which is connected to the input.
We can verify it here by analyzing the circuit based on the basic KCL.
Summing the current at node 0,
i1  i 2  i o  0
The two input terminals have the same voltages. That is 0V. i1 should be v1 divided by
R1. Similarly, i 2 is v2 divided by R2. io should be v out minus 0, and then divided by Rf.
We get
v1 v 2 v out


0
R1 R 2 R f
Move the first two items to the right hand side of the equation,
v
v out
v 
  1  2 
Rf
 R1 R2 
Both side of the equation are multiplied by Rf.
Rf
 Rf

v out  
v1 
v 2 
R2
 R1

In this problem v out    5 
3
3
3

 2   9V
1.5 
The second part of the circuit is a inverting amplifier. It can be considered as a special
case the summing amplifier. V2=0. So v o  
100
 (9)  900V
1
```