Objectives
Operational amplifiers (op amp) are used extensively in a variety of electronic circuits, from audio systems, filters, and engine control to appliances. In this lab we introduce the operational-amplifier (op-amp), an active circuit that is designed for certain characteristics (high input resistance, low output resistance, and a large differential gain) that make it a nearly ideal amplifier and useful building-block in circuits.
The goal of this experiment is to study operational amplifier (op amp) and its applications. Some basic op ‐ amp circuits, including the four most common types, i.e., the inverting, non ‐ inverting, differencing, and summing amplifiers will be introduced, simulated and applied on a board.
Introduction
An operational amplifier is one of the most important integrated circuit. Many integrated circuits are much larger: a computer's microprocessor can contain several million separate elements. However, in this laboratory, we will show how the functioning of op-amp circuits can be understood without knowing anything about the individual transistors of which op-amps are composed.
In many electronic circuits, the signals (voltage differences) that are generated and manipulated are very small. Therefore, amplification is often essential. When playing a CD for instance, the signals generated in the CD player are quite small and will not adequately drive a speaker system. The signals from the CD player are therefore passed into the stereo amplifier (which often comes as a

tuner/amplifier combination in modern home stereo systems). The heart of the stereo amplifier is the operational amplifier, or op-amp, which takes low level voltage signals as inputs and produces large output voltages that vary linearly with the input voltage.
The op-amp symbol is depicted as in Figure 1. The device has two input terminals: a non-inverting and an inverting terminal, v p and v n
, respectively. An interesting property of the op-amp is that the output voltage is only a function of the difference of the two input terminals, as follows:
V out

( V p

) n v
(1)
+Vcc
V n
Inverting input
V out
+
V p
Non-inverting input
-Vcc
Figure 1. Op-amp symbol
The device is characterized by a very large amplification (gain) A v
, a large input resistance, and small output resistance. A typical value of the gain A v
is
200,000. As a result of the large amplification A v
, the required input voltage difference (v p
- v n
) to obtain a finite output voltage is very small.
Another aspect of the op-amp is that the maximum output voltage is always limited to a certain value determined by the power supplies. This is true for both open-loop, and, as you will find out in lab, closed-loop configurations. The openloop configuration effect is schematically indicated in Figure 2. The corresponding input voltage range to keep the op-amp in the active region is then given by,

V cc
/ A v
   i
V cc
/ A v
(2)

Figure 2. Op-amp open-loop input/output transfer characteristic
Where A v
is the open-loop voltage gain and +V
CC
and –V
CC
are the positive and negative DC power supply voltages, respectively. There is no internal
"ground" or "common" connection; voltages are measured relative to the common connection of the two power supplies.
The numbers on the diagram refer to the pin numbers on the 741 (dual in-line package) integrated circuit (IC) package as shown in Figure 3. Pins 1 and 5 are used for nulling the offset voltage. We will not use these pins in this lab. Pin 8 is not connected (NC) to the internal circuits of the op-amp. One of the more popular op-amps is the UA741, specs.
Figure 3. View of an op-amp DIP package.

It is clear from the Figure 2 that the input should be kept very small in order to stay in the active (or linear region). This makes it difficult to use op-amps in this open-loop configuration. It is for this reason that op-amps are rarely used in open-loop. Two popular feedback configurations are the inverting and noninverting op-amp circuits as shown in Figure 4. circuit components
-
+ circuit components
+
-
Negative feedback--stable Positive feedback --unstable
Figure 4: Feedback structure of op amp devices
A much more useful way to use op-amps is in a negative feedback configuration. This involves connecting the output back to the inverting input of the op-amp. The feedback will keep the differential input voltage (v p
– v n
) close to zero. With feedback, the overall gain is called the closedloop gain, which will be drastically reduced as compared to the open-loop gain.
A major advantage of using feedback is that the gain is now a function of the resistors only and is independent of the op-amp open-loop gain A v
, which can vary from device to device.
Notice that in both cases the output voltage is feedback to the negative input terminal. The only difference is the connection of the input voltage to the inverting or non-inverting input terminal. Feedback can be positive (returned to the non-

inverting input) or negative (returned to the inverting input), but negative feedback is used primarily in analog circuits because it yields stable, controllable outputs, and we will concentrate on it. For positive feedback, an increasing output
V out
drives the inputs even further positive, resulting in a still more positive V out
.
(A similar argument can be made that once V out
swings negative, it will result in a large negative swing.) As a result, the device will always be in saturation. While this can be useful for some purposes (e.g. for making oscillators and in digital circuits), we shall concentrate on "negative feedback" here.
In the case of negative feedback , we assume that the input terminals are virtually short-circuited . This is an important property that we will make use of extensively to analyze op-amp circuits. Also, because the input resistance is very large the input current will be very small. This gives us the two important assumptions we have to remember when working with negative-feedback op-amp circuits:
1.
The input current is zero. (this is true for open-loop also)
2.
The two input terminals are virtually shorted together (the voltage difference between the inverting and non-inverting inputs is zero) when the amplifier is used with negative feedback.
Both of these rules are idealizations which assume that the op-amp is "ideal"; however, the assumptions are usually very good approximations of "real" opamps. The reasons we say that the two input terminals are "virtually" shorted is because there is no current flowing between them (if it were a real short, there would be current flow).

Using the golden rules, the negative feedback "inverting amplifier" circuit shown in Figure 5a can be analyzed. From golden rule number 1, the voltage at the inverting input must be at ground because V+ is at ground. (The inverting input isn't actually connected to ground, rather the internal circuitry of the op-amp labors to keep it very near ground. We call such a voltage a virtual ground.) From golden rule number 2, all of the current through R
1
must flow through R
2
, because no current flows into the op-amp inputs. We arbitrarily take the direction of the conventional (positive) current to be to the right. Then applying Kirchoff's laws gives
V in
0


IR
1
IR
2


V
0 out

IR
1

V in
(3a)
(3b)
The closed loop gain G is defined as the ratio of the output voltage to the input voltage. Dividing (3b) by (3a) gives
G

V out
V in
 
R
2
R
1
. (4)
This means that you can choose the gain by selecting the values of the resistors!
Also, the gain is not dependent on the details of the particular op-amp (e.g. the exact value of the open loop gain A
0
), but rests only on the open loop gain being large, and on the values of the resistors. (The largeness of A
0
is the underlying justification for Golden Rule 1.) Since the output is negative for positive inputs, the amplifier called an inverting amplifier.

R
1
R
2
-
+
+Vcc
-Vcc
V out
R
1
R
2
-
+
+Vcc
-Vcc
V out
V in V in
Figure 5. (a) Inverting op-amp and (b) non-inverting op-amp circuit.
An op-amp connected in a closed-loop configuration as a non-inverting amplifier is shown in Fig 3b. The input signal is applied to the non-inverting (+) input. The output is applied back to the inverting (-) input through the feedback circuit
(closed loop) formed by the input resistor R
1
and the feedback resistor R f
. This creates –V e
feedback as follows. Resistors R
1
and R
2
form a voltage-divider circuit, which reduces V
O
and connects the reduced voltage V f to the inverting input. The feedback is expressed as
V f

( R
1
R
1

R
2
)
V o
(5)
The difference of the input voltage, V in
and the feedback voltage, V f
is the differential input of the op amp. This differential voltage is amplified by the gain of the op-amp and produces an output voltage expressed as
V o
 
R
(1 2 )
R
1
V in (6)
The closed-loop gain of the non-inverting amplifier is, thus
A
( )
R
2
R
1
(7)

The function of a subtractor is to provide an output proportional to or equal to the difference of two input signals. A basic differential amplifier or a subtractor circuit is shown in Fig. 6.
R
2
R
1
V
1
V
2
R
1
-
+Vcc
+
-
Vcc
V out
R
2
Figure 6. A subtractor or differential amplifier
The output voltage of the differential amplifier can be expressed as
V o

R
2
R
1
( V
2

V
1
)
(8)
Thus it can be seen that the output voltage depends on the difference of the input voltages. (V
2
-V
1
) can be suitably amplified choosing the values of R
2
/R
1
. The circuit also behaves as a subtractor if R
2
=R
1
.

A summing amplifier:
R
F
R
1
V
3
R
2
V
2
V
1 R
3
-
+Vcc
+
-Vcc
V out
Figure 7. A subtractor or differential amplifier
V out
 


R f
R
1
V
1

R f
R
2
V
2

R f
R
3
V
3


  i n 

0
R f
R i
V i (9)
An integrator:
By adding a capacitor in parallel with the feedback resistor R
2
in an inverting amplifier as shown in Figure 8, the op-amp can be used to perform integration.
An ideal or lossless integrator (R
2
= ∞) performs the computation V out


1 
V dt in
.
Thus a square wave input would cause a triangle wave output. However, in a real circuit (R
2
< ∞) there is some decay in the system state at a rate proportional to the state itself. This leads to exponential decay with a time constant of τ = R
2
C.

Figure 8. An integrator
Differentiator
By adding a capacitor in series with the input resistor R
1
in an inverting amplifier, the op-amp can be used to perform differentiation. An ideal differentiator (R
1
=
0) has no memory and performs the computation V out
 
R C
2 dV in dt
. Thus a triangle wave input would cause a square wave output. However, a real circuit (R
1
> 0) will have some memory of the system state (like an lossy integrator) with exponential decay of time constant τ = R
1
C
Equipment
Breadboard
Cables & Wires and connectors
Operational Amplifier UA741
Digital Oscilloscope,
Multimeter
Signal Generator,
DC Power Supply,
Resistors
Potentiometer

Preliminary work (Calculation and simulation section)
Important NOTE: Please check and read the following documents for grasping the theoretical part of the operational amplifier and its applications.
[1] http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-
002-circuits-and-electronics-spring-2007/video-lectures/
[2] http://www.ti.com/lit/an/sboa092a/sboa092a.pdf
[3] https://groups.yahoo.com/neo/groups/LTspice/info
1) Consider the circuit of Figure 5a with Vin= 1V, R
1
= 1kOhm and R
2
=
4.7kOhm. Simulate the circuit of Fig.5a with these values. Find the value of the amplification A=v o
/v i
. Indicate how you got your result.
2) a) Derive the expression of the amplification for the non-inverting amplifier of Figure 5b, using the rules given above. Simulate the circuit of Fig.5b with these values. b) What value of feedback resistor R
1
is needed to give an amplification equal to 2 when R
2
=10kOhm.
3) Figure 9 shows the circuit schematic of a summing amplifier. Prove that the output voltage is given by
Vo
 
[( R
F
/
1
)
1

( R
F
/ )
2

( R
F
/
3
) ] (10)

R
F
R
1
V
3
R
2
V
2
V
1
R
3
-
+Vcc
+
-Vcc
V out
Figure 9. Summing amplifier
4) Derive the gain (Vout/Vin) expression in terms of R
1
and R
2
for the subtractor or difference amplifiers as given in Fig. 6.
5) Show the relation of output and input voltage of the following circuit
(Integration)
Figure 10. View of an inverting amplifier
6) Simulate the circuit of Fig.11, Fig12, and Figure 14 record all the voltage and current values as described in related tables. (Note: In order to use potentiometer please check the tutorial which is given in [3].)

1) Please construct the op amp circuit which is given in Fig.5a. The following voltage and resistor values should be considered for each experiment. a) Vin=5V, R1=2.2k, R2=2.2k, b) Vin=-5V, R1=2.2k, R2=2.2k
For each conditions, measure and record the desired values to the Table.1.
V o
I
R1
I
R2
Table.1. a b
2) Please construct the op amp circuit which is given in Fig.5b. The following voltage and resistor values should be considered for each experiment. Compare experimental results with your simulations and theoretical results. a) Vin=5V, R1=2.2k, R2=2.2k, b) Vin=-5V, R1=1k, R2=2.2k c) Vin=2V, R1=4.7k, R2=10k a
Table.2 b c
V o
I
R1
I
R2

3) Summing Amplifier 1: Please construct the op amp circuit which is depicted in below. The following voltage and resistor values should be considered for each experiment.
R
1
R
3
Ra
V
1
-
+Vcc
+
V
2
-Vcc
R
2
-
+
+Vcc
-Vcc
V out
Rb
Figure 11. View of an inverting amplifier a) V1=5V, Ra=1k, Rb=1k, R1=2.2k, R2=2.2k, R3=1k measure V2, Vout b) Vin=-5V, Ra=1k, Rb=1k, R1=2.2k, R2=2.2k, R3=2.2k measure V2, Vout c) Vin=5V, Ra=2.2k, Rb=1k, R1=2.2k, R2=2.2k, R3=2.2k measure V2, Vout d) Vin=5V, Ra=2.2k, Rb=1k, R1=2.2k, R2=2.2k, R3=1k measure V2, Vout a
Table.3 b c
V
1
V
2
V out

4) Difference Amplifier:
V
1
Ra
Rb
R
3
R
1
-
+Vcc
+
V
2
-Vcc
R
2
R
4
-
+
+Vcc
-Vcc
V out
Figure 12. View of an difference amplifier
Wire the above circuit, using 10 kΩ for all four resistors. if we let R3=R1 and R4=R2, then we have Vout=(R2/R1)(v2-v1) i.e.
Using the DC power supply for V1 and the 0-6 V power supply for V2 (or vice versa), verify that Vout is indeed the difference of v1 and v2 with the expected sign. Substitute 100 kΩ resistors for R2 and R4 and verify that the gain changes as expected. a) Ra=1k, Rb=2.2k, V1=4V Measure V2, R1=2.2k, R2=2.2k, R1=R3 and
R2=R4 b) Ra=1k, Rb=1k, for V1=5V measure V2, R1=2.2k, R2=2.2k c) Ra=2.2k, Rb=1k, for V1=6V measure V1 and V2, R1=2.2k, R2=2.2k

V
1
A
Table.2 b c
V
2
Vo
5) Non-inverting Operational Amplifier: a) Build the circuit of Figure 13. Notice that this is the same non-inverting amplifier as the one in Figure 5b. The input voltage is derived from the power supply by a potentiometer, used as a voltage divider. Measure the actual values of the resistors R1 and R2 and the potentiometer and record them in your lab notebook. Compute these values and record them.
Figure 13. View of an noninverting amplifier b) When you finish building the circuit, set the power supply to 10 and −10V
(from the +/− 25V). It is important that the Vcc and −Vcc are exactly equal in magnitude. Set the current limiter to 100 mA. Double check the connection of the board before connecting and switching on the power supply to your circuit.

c) Measure the input-output characteristic of the op-amp circuit: vary the input voltage V in
between –6V and +6V in steps of about 1V by adjusting the potentiometer. Measure the actual value of V in
and the corresponding output voltage V o
with the multimeter. Make a table with entries V in
, V o
(measured), and V o
(calculated). Note that the output voltage saturates above a certain input voltage. You can fill in the measured values now and do the calculations later for your report. d) For your report, m ake a plot of the measured and calculated output voltage versus the input voltage. You can use your favorite plotting tool (e.g. a
LTspice or Matlab) to make the plot. Indicate on the graph the transition between the active and saturated regions of the circuit. Find the slope of the graph (i.e. the amplification) and compare it with the calculated one, based on measured resistance values. Note also the maximum and minimum output voltage (i.e. the saturation levels).
6) Summing Amplifiers: a) Build the summing amplifier of Figure 14. First, sketch how you will layout the circuit on your board. For voltage source V
2
use the 6V power supply.
Make sure you connect all the grounds together to get a common ground.
This is a crucial step in any circuit.
Explain why common ground is important in your report.

Figure 14. View of a summing amplifier b) Measure and record the actual values of the resistors. c) Write the expression of the output voltage as a function of the input voltages
V
1
and V
2
. d) Make the voltage V
1
=1V and V
2
=2V and measure the output voltage. Next, change the voltage V
1
=3V and measure the output voltage. How do the measured values compare to the calculated ones? e) For V
1
=1V and V
2
=2V, measure v n
to verify that the inverting input terminal is virtually shorted to ground.
Besides the general guidelines, report the following for this lab:

The lab report should be segmented into 4 parts: o Inverting amplifier o Non-inverting amplifier o Differencing amplifier o Summing amplifier


Each part should be comprised of the circuit design with LSpice Capture and the simulated voltages at each node as well as the closed-loop gain,
 the analytical equation for each amplifier and the calculated output voltage and gain,
 the experimental voltage readout with a description of the experimental procedure ,
 a table of all three values for the op-amp terminal voltages: from simulation, theory, and experiment,
 a table of all three values for the closed-loop op-amp gain seen by V out
: from simulation, theory, and experiment,
 and your explanation of the variations between the three values of the last two steps.

Answer the questions in the analysis section
http://eecs.oregonstate.edu/education/docs/tutorials/Simulating%20an%20op%2
0amp.pdf http://www.mathworks.com/help/physmod/simscape/examples/op-amp-circuitinverting-amplifier.html
http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002circuits-and-electronics-spring-2007/video-lectures/ http://www.ti.com/lit/an/sboa092a/sboa092a.pdf