To demonstrate the use of operational amplifiers (op-amps) for performing inverting amplifiers, non-inverting amplifiers and some mathematical functions.
For the differential amplifier given in Figure-4.5,

Analyze the circuit by using Eq-4.1

Compare your result with the Eq-4.10 for A=∞.

For the same circuit by using Eq-4.9, find the most proper resistor values (from E12 resistor series) for a=1 and b=3.
Submit all calculations as hard copy before the session.
Please read the laboratory notes carefully before the experiment.
You should study the following topics before the experiment:

What are the operational amplifiers?

Find the operational ranges of the op-amps.

Basic principles of the op-amps.

Mathematical functions that can be obtained by operating an op-amp in linear mode.

Dual output power supply

Multimeter

Oscilloscope

1 piece 10K resistor

1 piece 100K resistor

1 piece 10K variable resistor

1 piece 100K variable resistor

3 pieces resistors (resistor values should be found in pre-lab assignment)

1 piece 10nF capacitor

1 piece 741 op-amp

Wire

1.
BASIC CONCEPTS OF OPERATIONAL AMPLIFIERS
Operational amplifiers (Op-amps) are among the most widely used electronic devices today, being used in a vast array of consumer, industrial, and scientific devices. Operational amplifiers had their origins in analog computers where they were used to do mathematical operations in many linear, non-linear and frequency-dependent circuits.
Operational amplifier is a differential input, single-ended output amplifier, as shown symbolically in Figure 4-1. This device is an amplifier intended for use with external feedback elements, where these elements determine the resultant function, or operation. This gives rise to the name
“operational amplifier,” denoting an amplifier that, by virtue of different feedback hookups, can perform a variety of operations.
Ideal Op-amp Attributes:

Infinite open-loop gain





Infinite input impedance
Zero output impedance
Infinite bandwidth
Infinite common-mode-rejection-ratio (CMRR)
Unconditionally stable
Figure-4.1
The ideal op-amp and its attrributes
Since it is physically impossible to obtain infinite input impedance, infinite open loop-gain and zero output impedance, real op-amps will depart from some of the ideal attributes given in Figure
4.1. However, for real world applications it will be enough to obtain approximate values such as
10.000 open-loop gain, 10Ω input impedance and 1MΩ output impedance. Since in almost all of the linear applications of operational amplifiers negative feedback is used, unconditionally stability must be provided.
As it can be shown from Figure 4.1, op-amp has two inputs namely V p and V n
and ideally the opamp amplifies only the difference voltage of these two terminal, which is called the differential input voltage. The output voltage of the op-amp is then given by,
V o

A ( V p

V n
) (4.1)
1.1
Output Voltage Limits
The output of the operational amplifiers cannot exceed the power supply voltages E sat
+
and E sat
-
(it can be lower because of voltage drop on some active elements). The fact that the output voltage of a practical amplifier cannot exceed certain threshold value is called saturation. So, for an amplifier with a given gain A, output voltage is defined as below;

V o



E
 sat

A ( V




E
 sat p

V n
)
V
E
V p
 sat
A p


V
 n
V n
V
 p

E

E
 sat
A
V
 sat
A n

E
 sat
A
(4.2)
An operational amplifier behaves linearly as long as its output voltage remains below the saturation voltage. As it can be shown from Eq-4.2, if the open-loop voltage gain (A) of an op-amp increases, input differential voltage range (V p
-V n
) decreases and for infinite A, it converges to zero. Thus, if the open-loop voltage gain of an op-amp is high enough, it will have equal input voltages for linear region operation. Practically it is not possible to apply two different voltages to V p and V n
terminals of an op-amp. Because, any voltage difference between two terminals will be multiplied with the high open-loop voltage gain and the device will be saturated. For this reason, op-amps operate with a negative feedback circuit for linear applications.
Operational amplifiers have single pole, which leads a decrease in the gain for high frequency input voltages. When doing your measurements, please note that the op-amp will not operate properly after a certain frequency value.
1.2
Input Offset Voltage
Ideally, if both inputs of an op-amp are at exactly the same voltage, then the output should be at zero volts. However, due to manufacturing process, the differential input transistors of real opamps may not be exactly matched. This causes the output to be non-zero. In practice, a small differential voltage must be applied to the inputs to force the output to be zero. This is known as the input offset voltage, V
OS.
2.
INVERTING OP-AMP
The op-amp inverting gain stage, also known as the negative gain amplifier is shown in Figure-
4.2. Since the input impedance is high enough, the input current of the op-amp can be neglected.
According to the Kirchoff’s current law,
I
R
1

I
R
2

0 (4.3) can be written. Eq-4.3 can be restated as follows;
V in

R
1
V n 
V out

R
2
V n 
0
Output voltage and gain of the circuit can be restated by substituting

V out
A
1
V in

V out
A
R
1

V out
 1
R
2
A

0

V
V out in

R
1


R
2
1
A
( R
1

R
2
)
(4.4)
for V n
in Eq-4.1,
(4.5)

can be written. Since the open-loop voltage gain A is high enough, V n restating the Eq-4.4,
V in 
V out
R
2

0
R
1 can be obtained. Thus, the voltage gain of the circuit can be given as follows;
V out
V in
 
R
2
R
1
Note that, Eq-4.5 turns into Eq-4.7 for A = ∞.


V out
A

0 . Then,
(4.6)
(4.7)
Figure-4.2
Inverting op-amp configuration
Exp #4.1

In order to function, the op-amp must be connected to an external power supply .
Since we want to produce both positive and negative output voltages, we need both positive and negative voltages for the power supply. For a 741, the nominal values are +15V for positive supply and -15V for negative supply. To avoid clutter, we won't show the power supply terminals (pins 4 and 7) on any of the subsequent circuit diagrams. However, they must be connected or your amplifier will not operate.

Note that there is no ground terminal on the op-amp. The zero reference point is established by the external circuit and is not important to the op-amp itself.

Connect the R
1
=10K and R
2=
100K resistors as shown in Figure-4.2.

Connect the function generator output to V in
of the circuit given in Figure-4.2. Turn on the function generator and apply a proper sinusoidal signal to V in
. Observe the output from the scope. Note that, the output signal must not be clipped.

Change the R
2
value for 100K, 75K, 50K, 25K and 10K. For each step write down the output voltage and gain. (To change its value disconnect the R
2
from the circuit and then do your measurement)

At the end of your experiment plot the gain-R
2 graph and compare it with the ideal one.
3.
NON-INVERTING OP-AMP
Non-inverting (positive gain) circuit configuration is shown in Figure-4.3. After circuit analysis, the voltage gain of this circuit is obtained as below;
V out
V in

1

R
2
R
1
(4.8)

Figure-4.3
Non-inverting op-amp configuration
Exp #4.2

Connect the circuit under test according to the Figure-4.3

Change the R
2
value as done before in the Exp#4.1 and write down the results.

At the end of your experiment plot the gain-R
2 graph and compare it with the ideal one.
4.
SUMMING AND DIFFERENTIAL AMPLIFIERS
Operational amplifiers are usually used as summing and differential amplifiers in analog computers and control circuits. Therefore, in this section summing and differential amplifiers are studied.
Figure-4.4
Summing amplifier
The summing amplifier given in Figure-4.4 contains two inverting amplifiers. According to the superposition principle, the output of this circuit is the sum of all the individual input signals. So, for V
2
=0, the circuit amplifies the input V
1
with a gain

R
3
R
1
, and for V
1
=0, the circuit amplifies the input V
2
with a gain

R
R
3
2
. Once the superposition principle is applied, the output voltage can be given as in Eq-4.9.
V o
 


R
R
3
1
V
1

R
R
3
2
V
2


 
 aV
1
 bV
2

(4.9)

Figure-4.5
Differential amplifier
Different from the summing amplifier, the differential amplifier given in Figure-4.5 contains both inverting and non-inverting amplifiers. Again by using the superposition principle the output of the circuit can be obtained as done before. So, for V
2
=0, the circuit behaves as an inverting amplifier and amplifies the input V
1
with a gain

R
4
R
1
, and for V
1
=0, the circuit behaves as a noninverting amplifier and amplifies the input V p
with a gain 1

R
4
R
1
. Here, V p
is the voltage of the positive terminal and given as, V
2
R
2
R
3

R
3 voltage can be given as in Eq-4.10.
V o

V
2
R
2
R
3

R
3


1

R
R
4
1



R
R
4
1
V
1
 aV
2
. Once the superposition principle is applied, the output
 bV
1
(4.10)
Exp #4.3

Set up the circuit by using the resistor values calculated for a=1 and b=3.

Apply 5V DC voltage to input V

Plot the output signal.
1
and a sine wave with proper amplitude to input V
2
.
5.
DIFFERENTIATOR AND INTEGRATOR CIRCUITS
By introducing electrical reactance into the feedback loops of op-amp amplifier circuits, we can cause the output to respond to changes in the input voltage over time. Deriving their names from their respective calculus functions, the integrator produces a voltage output proportional to the product (multiplication) of the input voltage and time; and the differentiator (not to be confused with differential) produces a voltage output proportional to the input voltage's rate of change. In
Figure-4.6 and 4.7 integrator and differentiator circuits are given respectively.

Figure-4.6
Op-amp integrator
Figure-4.7
Op-amp differentiator
At proper operating frequencies the current flowing through the R
2
resistor in the integrator circuit and the voltage drop over the R
1
resistor in the differentiator circuit can be neglected. Thus, by assuming R
2
=∞ for the integrator circuit,
V o
 
1 
V
1 dt

V o ( t

0 )
RC and R
1
=0 for the differentiator circuit,
(4.11)
V o
 
R
2
C dV
1 dt the voltage transfer functions for both circuit can be obtained.
(4.12)
Suppose that, a sine function sin(
 t ) is applied to the input of a differentiator. In this case, the output will be the derivative of this function, which is
  cos(
 t ) . So, when the input frequency is increased, the gain of this circuit also increases. But please note that, the op-amp will not operate properly after a certain frequency value as mentioned before in Seciton-1.
Exp #4.4
Integrator

Set up the integrator circuit for R
1
=10K, R
2
=100K, C=10nF, R eq
=10K.

Apply an asymmetric square wave with proper amplitude and frequency to the input of the circuit.

Plot the output signal.
Differentiator

Set up the differentiator circuit with the circuit elements used for the integrator.

Apply an asymmetric triangle wave with proper amplitude and frequency to the input of the circuit.

Plot the output signal


Simulate all experiment by a computer aided design tool such as PSpice, Cadence etc.
 Compare the gain-R
2
graphs with ideal ones for the Exp#4.1 and Exp#4.2

In Exp#4.2, find the voltage gain of the circuit for R
1
=∞ and R
2
=0. How the circuit behaves for this case?

Explain the meaning of following terms:
1.
Common mode rejection ratio (CMRR)
2.
Unity gain bandwidth
3.
Slew rate (SW)
4.
Rail to Rail
1.
A.B. Williams, “Designer Handbook of Integrated Circuits” (1984)
2.
A.B. Grebene, “Bipolar and MOS Analog Integrated Circuit Design” (1984)
3.
U. Tietze, Ch. Schenk, “Electronic Circuits: Design and Applications” (1991)
4.
M.S. Türköz, “Elektronik Devreleri”, Sistem Yayıncılık (1995)
5.
H.Kuntman, “Analog Tümdevre Tasarımı”, Sistem Yayıncılık (1992)