Basic Electric Circuits Introduction To Operational Amplifiers Lesson 8.
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Transcript of Basic Electric Circuits Introduction To Operational Amplifiers Lesson 8.
Basic Electric Circuits
Introduction To Operational Amplifiers
Lesson 8
Basic Electric Circuits
Operational Amplifiers
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
Basic Electric CircuitsOperational Amplifiers
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
Basic Electric CircuitsOperational Amplifiers
The Philbrick Operational Amplifier.
From “Operational Amplifier”, by Tony van Roon: http://www.uoguelph.ca/~antoon/gadgets/741/741.html
Basic Electric Circuits
Operational Amplifiers
My belief is that “operational” was used as a descriptorearly-on because this form of amplifier can performoperations of
• adding signals
• subtracting signals
• integrating signals, dttx )(
The applications of operational amplifiers ( shortenedto op amp ) have grown beyond those listed above.
3
Basic Electric Circuits
Operational Amplifiers
At this level of study we will be concerned with howto use the op amp as a device.
The internal configuration (design) is beyond basiccircuit theory and will be studied in later electroniccourses. The complexity is illustrated in the followingcircuit.
4
Basic Electric Circuits
Operational Amplifiers
The op amp is built using VLSI techniques. The circuitdiagram of an LM 741 from National Semiconductor isshown below.
5
V+
V-
Vo
Vin(-)
Vin(+)
Figure 8.1: Internal circuitry of LM741.Taken from National Semiconductor
data sheet as shown on the web.
Basic Electric Circuits
Operational Amplifiers
Fortunately, we do not have to sweat a circuit with 22transistors and twelve resistors in order to use the op amp
The circuit in the previous slide is usually encapsulated intoa dual in-line pack (DIP). For a single LM741, the pin connections for the chip are shown below.
Taken from National Semiconductordata sheet as shown on the web.6
Figure 8.2: Pin connection, LM741.
Basic Electric Circuits
Operational Amplifiers
i n ve r t i n g i n p u t
n o n i n ve r t i n g i n p u to u t p u t
V -
V +
The basic op amp with supply voltage included is shownin the diagram below.
7
Figure 8.3: Basic op am diagram with supply voltage.
Basic Electric Circuits
Operational Amplifiers
In most cases only the two inputs and the output areshown for the op amp. However, one should keep inmind that supply voltage is required, and a ground.The basic op am without a ground is shown below.
8Figure 8.4: Outer op am diagram.
Basic Electric Circuits
Operational Amplifiers
A model of the op amp, with respect to the symbol, isshown below.
V 1
V 2
_
+
V d R i
R o
A V d
V o
Figure 8.5: Op Amp Model.9
Basic Electric Circuits
Operational Amplifiers
The previous model is usually shown as follows:
R i
R i
A V d
_
+
V d
V 1
V 2
V o
+
_
Figure 8.6: Working circuit diagram of op amp.
10
Basic Electric Circuits
Operational Amplifiers
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.
+
_
R 2
R 1
+
_
+
_
V i nV o
11 Figure 8.7: Op amp functional circuit.
Basic Electric Circuits
Operational AmplifiersIn terms of the circuit model we have the following:
R i
R i
A V i
_
+
V iV i n V o
+
_
+
_
R 1
R 2
ab
Figure 8.8: Total op amp schematic for voltage gain configuration.12
Basic Electric Circuits
Operational Amplifiers
R i
R i
A V i
_
+
V iV i n V o
+
_
+
_
R 1
R 2
ab
Circuit values are:
R1 = 10 k R2 = 40 kA = 100,000 Ri = 1 meg
13
Basic Electric Circuits
Operational Amplifiers
We can write the following equations for nodes a and b.
Eq 8.1
Eq 8.2
14
ioin
o
oiiiin
AVk
)VV(V
k
VV
meg
V
k
)VV(
4050
40110
Basic Electric Circuits
Operational Amplifiers
Equation 8.1 simplifies to;
inio VVV 10012625 Eq 8.3
Equation 8.2 simplifies to;
010410005.4 95 io VxVx Eq 8.4
15
Basic Electric Circuits
Operational Amplifiers
From Equations 8.3 and 8.4 we find;
ino VV 99.3
This is an expected answer.
Fortunately, we are not required to do elaborate circuitanalysis, as above, to find the relationship between theoutput and input of an op amp. Simplifying the analysisis our next consideration.
16
Eq 8.5
Basic Electric Circuits
Operational AmplifiersFor most all operational amplifiers, Ri is 1 meg orlarger 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
RohmsinputInfinite
RohmsoutputZero
AgainloopopenInfinite
;.3
0;.2
;.1
17
Basic Electric Circuits
Ideal Op Amp:
_
+ ++
+
+
_
__ _
V i
V 1
V 2 = V 1V o
i 1
i 2
= 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.
Basic Electric Circuits
Ideal Op Amp:
Find Vo in terms of Vin for the following configuration.
+
_
R 2
R 1
+
_
+
_
V i nV o
19 Figure 8.10: Gain amplifier op amp set-up.
Basic Electric Circuits
Ideal Op Amp:
+
_
R 2
R 1
+
_
+
_
V i nV o
a
V i
Writing a nodal equation at (a) gives;
21
)(
R
VV
R
VV oiiin
20
Eq 8.6
Basic Electric Circuits
Ideal Op Amp:
21
)(
R
VV
R
VV oiiin
With Vi = 0 we have;
With R2 = 4 k and R1 = 1 k, we have
ino VV 4 Earlierwe got ino VV 99.3
21
Eq 8.7
inVR
RV
1
20
Basic Electric Circuits
Ideal Op Amp:
When Vi = 0 in Eq 8.7 and we apply the Laplace Transform;
1
20
R
R
)s(V
)s(V
in
Eq 8.8
In fact, we can replace R2 with Zfb(s) and R1 with Z1(s) andwe have the important expression;
)s(Z
)s(Z
)s(V
)s(V
in
fb
in
0 Eq 8.9
22
Basic Electric Circuits
Ideal Op Amp:
At this point in circuits we are not able to appreciate theutility of Eq 8.9. We will revisit this at a later point incircuits but for now we point out that judicious selectionsof Zfb(s) and Zin(s) leads to important applications in
• Analog Filters
• Analog Compensators in Control Systems
• Application in Communications
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Basic Electric Circuits
Ideal Op Amp:
Example 8.1: Consider the op amp configuration below.
+
+
+
_
__
3 VV in
6 k
1 k
V 0
a
Figure 8.11: Circuit for Example 8.1.24
Assume Vin = 5 V
Basic Electric Circuits
Operational Amplifiers
+
+
+
_
__
3 VV in
6 k
1 k
V 0
a
At node “a” we can write;
k
V
k
)V( in
6
3
1
3 0
From which; V0 = -51 V25
Eq 8.10
Example 8.1 cont.
Basic Electric CircuitsOperational Amplifiers
Example 8.2: Summing Amplifier. Given the following:R fb
R 1
R 2
V 2
V 1V 0
a
Figure 8.12: Circuit for Example 8.2.
fbR
V
R
V
R
V 0
2
2
1
1 Eq 8.11
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Basic Electric CircuitsOperational Amplifiers
Example 8.2: Summing Amplifier. continued
Equation 8.11 can be expressed as;
2
21
10 V
R
RV
R
RV fbfb Eq 8.12
If R1 = R2 = Rfb then,
210 VVV Eq. 8.13
Therefore, we can add signals with an op amp.
27
Basic Electric CircuitsOperational Amplifiers
Example 8.3: Isolation or Voltage Follower.
Applications arise in which we wish to connect one circuitto another without the first circuit loading the second. This requires that we connect to a “block” that has infinite inputimpedance and zero output impedance. An operational amplifier does a good job of approximating this. Considerthe following:
T he" B lo c k "
C irc u it 1 C irc u it 2+
_
+
_V in V out
Figure 8.13: Illustrating Isolation.28
Basic Electric CircuitsOperational Amplifiers
Example 8.3: Isolation or Voltage Follower. continued
C i r c u i t 1 C i r c u i t 2
Th e B l o c k
+
_
+V i n V 0_
Figure 8.14: Circuit isolation with an op amp.
It is easy to see that: V0 = Vin29
Basic Electric CircuitsOperational Amplifiers
Example 8.4: Isolation with gain.
+
_
__
+
+
2 0 k
V in
V in
V 0
1 0 k
1 0 k
a
+
_
Figure 8.15: Circuit for Example 8.4:
30
Writing a nodal equation at point “a” and simplifying gives;
inVV 20
Basic Electric CircuitsOperational Amplifiers
Example 8.5: The noninverting op amp.
Consider the following:
R 0
R fb
V 0V 2_
+
+
_
a+
_
Figure 8.16: Noninverting op am configuration.
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Basic Electric CircuitsOperational Amplifiers
Example 8.5: The noninverting op amp. Continued
Writing a node equation at “a” gives;
20
0
02
0
02
0
2
1
,
11
0)(
VR
RV
giveswhich
RRV
R
V
so
R
VV
R
V
fb
fbfb
fb
Remember this
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Basic Electric CircuitsOperational Amplifiers
Example 8.6: Noninverting Input.
Find V0 for the following op amp configuration.
+
_
+
+
_
_ 4 V
2 k
6 k
5 k
1 0 k
V 0
a
V x
Figure 8.17: Op amp circuit for example 8.6.
33
Basic Electric CircuitsOperational Amplifiers
Example 8.6: Noninverting Input.
The voltage at Vx is found to be 3 V.
Writing a node equation at “a” gives;
0105
0
k
)VV(
k
V xx
or
VVV x 930
34
End of Lesson 8
CIRCUITS
Operational Amplifiers