Lesson 8 Opetational_Amplifier_Appliacations(1).ppt

Basic Electric Circuits
Introduction To
Operational Amplifiers
Lesson 8

One might ask, why are operational amplifiers
included in Basic Electric Circuits?
The operational amplifier has become so cheap in
price (often less than $1.00 per unit) and it can be
used in so many applications, we present an
introductory study early-on in electric circuits.
1

What is an operational amplifier? This particular
form of amplifier had the name “Operational”
attached to it many years ago.
As early as 1952, Philbrick Operational Amplifiers
(marketed by George A. Philbrick) were constructed
with vacuum tubes and were used in analog
computers.*
Even as late as 1965, vacuum tube
operational amplifiers were still in use and cost in the
range of $75.
* Some reports say that Loebe Julie actually developed the operational amplifier circuitry.
2

The Philbrick Operational Amplifier.
From “Operational Amplifier”, by Tony van Roon: http://www.uoguelph.ca/~antoon/gadgets/741/741.html

My belief is that “operational” was used as a descriptor
early-on because this form of amplifier can perform
operations of
• adding signals
• subtracting signals
• integrating signals,  dt
t
x )
(
The applications of operational amplifiers ( shortened
to op amp ) have grown beyond those listed above.
3

At this level of study we will be concerned with how
to use the op amp as a device.
The internal configuration (design) is beyond basic
circuit theory and will be studied in later electronic
courses. The complexity is illustrated in the following
circuit.
4

The op amp is built using VLSI techniques. The circuit
diagram of an LM 741 from National Semiconductor is
shown below.
5
V+
V-
Vo
Vin(-)
Figure 8.1: Internal circuitry of LM741.
Taken from National Semiconductor
data sheet as shown on the web.
Vin(+)

Fortunately, we do not have to sweat a circuit with 22
transistors and twelve resistors in order to use the op amp
The circuit in the previous slide is usually encapsulated into
a dual in-line pack (DIP). For a single LM741, the pin
connections for the chip are shown below.
Taken from National Semiconductor
data sheet as shown on the web.
6
Figure 8.2: Pin connection, LM741.

inverting input
no ninverting input
o utput
V-
V+
The basic op amp with supply voltage included is shown
in the diagram below.
7
Figure 8.3: Basic op am diagram with supply voltage.

In most cases only the two inputs and the output are
shown for the op amp. However, one should keep in
mind that supply voltage is required, and a ground.
The basic op am without a ground is shown below.
8
Figure 8.4: Outer op am diagram.

A model of the op amp, with respect to the symbol, is
shown below.
V1
V2
_
+
Vd Ri
Ro
AV d
Vo
Figure 8.5: Op Amp Model.
9

The previous model is usually shown as follows:
Ri
Ro
AV d
_
+
Vd
V1
V2
Vo
+
_
Figure 8.6: Working circuit diagram of op amp.
10

Application: As an application of the previous model,
consider the following configuration. Find Vo as a
function of Vin and the resistors R1 and R2.
+
_
R2
R1
+
_
+
_
Vin
Vo
11 Figure 8.7: Op amp functional circuit.

In terms of the circuit model we have the following:
Ri
Ro
AV i
_
+
Vi
Vin Vo
+
_
+
_
R1
R2
a
b
Figure 8.8: Total op amp schematic for voltage
gain configuration.
12

Ri
Ro
AV i
_
+
Vi
Vin Vo
+
_
+
_
R1
R2
a
b
Circuit values are:
R1 = 10 k R2 = 40 k Ro = 50 
A = 100,000 Ri = 1 meg 
13

We can write the following equations for nodes a and b.
Eq 8.1
Eq 8.2
14
 
( )
10 1 40
( )
50
40
i o
in i i
i o
o i
V V
V V V
k meg k
V V
V AV
k

 
 
 
 
 
 
 

Equation 8.1 simplifies to;
in
i
o V
V
V 100
126
25 

 Eq 8.3
Equation 8.2 simplifies to;
0
10
4
10
005
.
4 9
5

 i
o V
x
V
x Eq 8.4
15

From Equations 8.3 and 8.4 we find;
in
o V
V 99
.
3


This is an expected answer.
Fortunately, we are not required to do elaborate circuit
analysis, as above, to find the relationship between the
output and input of an op amp. Simplifying the analysis
is our next consideration.
16
Eq 8.5

For most all operational amplifiers, Ri is 1 meg  or
larger and Ro is around 50  or less. The open-loop gain,
A, is greater than 100,000.
Ideal Op Amp:
The following assumptions are made for the ideal op amp.






i
o
R
ohms
input
Infinite
R
ohms
output
Zero
A
gain
loop
open
Infinite
;
.
3
0
;
.
2
;
.
1
17

Ideal Op Amp:
_
+ +
+
+
+
_
_
_ _
Vi
V1
V2 = V1
Vo
i1
i2
= 0
= 0
(a) i1 = i2 = 0: Due to infinite input resistance.
(b) Vi is negligibly small; V1 = V2.
18
Figure 8.9: Ideal op amp.

Ideal Op Amp:
Find Vo in terms of Vin for the following configuration.
+
_
R2
R1
+
_
+
_
Vin
Vo
19 Figure 8.10: Gain amplifier op amp set-up.

Ideal Op Amp:
+
_
R2
R1
+
_
+
_
Vin
Vo
a
Vi
Writing a nodal equation at (a) gives;
 
2
1
)
(
R
V
V
R
V
V o
i
i
in 



20
Eq 8.6

Ideal Op Amp:
 
2
1
)
(
R
V
V
R
V
V o
i
i
in 



With Vi = 0 we have;
With R2 = 4 k and R1 = 1 k, we have
in
o V
V 4

 Earlier
we got
in
o V
V 99
.
3


21
Eq 8.7
in
V
R
R
V
1
2
0



Ideal Op Amp:
When Vi = 0 in Eq 8.7 and we apply the Laplace Transform;
1
2
0
R
R
)
s
(
V
)
s
(
V
in

 Eq 8.8
In fact, we can replace R2 with Zfb(s) and R1 with Z1(s) and
we have the important expression;
)
s
(
Z
)
s
(
Z
)
s
(
V
)
s
(
V
in
fb
in


0
Eq 8.9
22

Ideal Op Amp:
At this point in circuits we are not able to appreciate the
utility of Eq 8.9. We will revisit this at a later point in
circuits but for now we point out that judicious selections
of Zfb(s) and Zin(s) leads to important applications in
• Analog Filters
• Analog Compensators in Control Systems
• Application in Communications
23

Ideal Op Amp:
Example 8.1: Consider the op amp configuration below.
+
+
+
_
_
_
3 V
Vin
6 k
1 k
V0
a
Figure 8.11: Circuit for Example 8.1.
24
Assume Vin = 5 V

+
+
+
_
_
_
3 V
Vin
6 k
1 k
V0
a
At node “a” we can write;
 
k
V
k
)
V
( in
6
3
1
3 0




From which; V0 = -51 V (op amp will saturate)
25
Eq 8.10
Example 8.1 cont.

Example 8.2: Summing Amplifier. Given the following:
R fb
R 1
R 2
V2
V1
V0
a
Figure 8.12: Circuit for Example 8.2.
fb
R
V
R
V
R
V 0
2
2
1
1 

 Eq 8.11
26

Example 8.2: Summing Amplifier. continued
Equation 8.11 can be expressed as;
























 2
2
1
1
0 V
R
R
V
R
R
V
fb
fb
Eq 8.12
If R1 = R2 = Rfb then,
 
2
1
0 V
V
V 

 Eq. 8.13
Therefore, we can add signals with an op amp.
27

Example 8.3: Isolation or Voltage Follower.
Applications arise in which we wish to connect one circuit
to another without the first circuit loading the second. This
requires that we connect to a “block” that has infinite input
impedance and zero output impedance. An operational
amplifier does a good job of approximating this. Consider
the following:
T he
"B lock"
C ircuit 1 C ircuit 2
+
_
+
_
Vin Vout
Figure 8.13: Illustrating Isolation.
28

Example 8.3: Isolation or Voltage Follower. continued
C ircuit 1 C ircuit 2
The B lo ck
+
_
+
Vin V0
_
Figure 8.14: Circuit isolation with an op amp.
It is easy to see that: V0 = Vin
29

Example 8.4: Isolation with gain.
+
_
_
_
+
+
20 k
Vin
Vin
V0
10 k
10 k
a
+
_
Figure 8.15: Circuit for Example 8.4:
30
Writing a nodal equation at point “a” and simplifying gives;
in
V
V 2
0 

Example 8.5: The noninverting op amp.
Consider the following:
R 0
R fb
V0
V2
_
+
+
_
a
Figure 8.16: Noninverting op am configuration.
31

Example 8.5: The noninverting op amp. Continued
Writing a node equation at “a” gives;
2
0
0
0
2
0
0
2
0
2
1
,
1
1
0
)
(
V
R
R
V
gives
which
R
R
V
R
V
so
R
V
V
R
V
fb
fb
fb
fb























Remember this
32

Example 8.6: Noninverting Input.
Find V0 for the following op amp configuration.
+
_
+
+
_
_ 4 V
2 k
6 k
5 k
10 k
V0
a
Vx

Figure 8.17: Op amp circuit for example 8.6.
33

Example 8.6: Noninverting Input.
The voltage at Vx is found to be 3 V.
Writing a node equation at “a” gives;
0
10
5
0



k
)
V
V
(
k
V x
x
or
V
V
V x 9
3
0 

34

End of Lesson 8
CIRCUITS

Lesson 8 Opetational_Amplifier_Appliacations(1).ppt

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