Example 11
2 kΩ
8 kΩ
2.4 kΩ
10 V
3 kΩ
_
5V
500 Ω
_
+
+
_
+
1.5 kΩ
500 Ω
+
vo
_
Given the op amp circuit shown in the figure, find the output voltage v o .
The circuit consists of three ideal op amps. The problem is to find the output voltage v o .
As we know, the op amps are building blocks for designing complex circuit. It is very
often that we need connect the op amp circuits in cascade to achieve a large gain. Two
circuits are cascaded when the output of one op amp circuit is connected to the input of
the next circuit. This circuit can be divided into 3 stages. We may label the output of the
first stage as vout1 and that of the second stage as v out 2 . Let us get started from the left side.
Each of the input voltage is passed to the inverting terminal of the op amp through a input
resistor. The feedback resistor connects the inverting terminal with the output terminal.
The circuit in the first stage is known as a summing circuit. The output voltage should be
the weighted sum of the input voltages. Each of the input is weighted by the feedback
resistance divided by resistance of the input resister connected to this input.
That is v out1  
8k
 2k
 10 
8k

 5   (40  13 .33)  53 .33V
3k

. (1)
If we could not recognize that it is a summing amplifier, we have to analyze the circuit to
figure out vout1 . The circuit is connected to two input voltages. According the principle of
superposition, for a linear circuits, we can determine the total response by finding the
response to each independent source with all other independent source set to zero and
then summing the individual responses. In this case, firstly, we turned off the 5V source
and make it zero volt. The circuit becomes an inverting amplifier, the feed back
resistance Rf=8k and the input resistance is 2k. Applying KCL gives us,
v i v out1 '

0
Ri
Rf
Multiply the equation by Rf on both sides and then move the first term to the right side
v out1 '  
Rf
Ri
vi  
8k
 10  40V
2k
Similarly, setting the 10V source to zero, and the output voltage caused by the 5V source
acting alone is
v out1 "  
8k
 5  13 .33V
3k
.
So the output voltage is equal to the sum of the effects of the individual causes acting one
at a time,
v out1  v out1 'v out1 "  40  13.33  53 .33V
We got the same answer as equation (1).
The second stage of the circuit is simply an inverting amplifier, which can be considered
as a special case of the summing amplifier. For summing amplifier, there are multiple
inputs but inverting amplifier takes only 1 input voltage. The output voltage is the input
multiplied by the feedback resistance and then divided by the input resistance. The
polarity of the input voltage is also reversed.
v out 2 
2.4k
 (53 .33)  256V
0.5k
The third stage is a non-inverting amplifier with the input voltage connected to the
noninverting terminal. The voltage gain is 1 plus feedback resistance over the input
resistance, that is 4. The output should be 4 times v out 2
v o  4  256  1024 V
Or by writing the KCL at inverting node,
v out 2 v o  v o

0
0.5
1.5
So vo  4  256  1024 V

.