2. 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
3. Basic Electric Circuits
Operational 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
4. Basic Electric Circuits
Operational Amplifiers
The Philbrick Operational Amplifier.
From “Operational Amplifier”, by Tony van Roon: http://www.uoguelph.ca/~antoon/gadgets/741/741.html
5. Basic Electric Circuits
Operational Amplifiers
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, dttx )(
The applications of operational amplifiers ( shortened
to op amp ) have grown beyond those listed above.
3
6. Basic Electric Circuits
Operational Amplifiers
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
7. Basic Electric Circuits
Operational Amplifiers
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(+)
8. Basic Electric Circuits
Operational Amplifiers
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.
9. Basic Electric Circuits
Operational Amplifiers
inverting input
noninverting input
output
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.
10. Basic Electric Circuits
Operational Amplifiers
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.
11. Basic Electric Circuits
Operational Amplifiers
A model of the op amp, with respect to the symbol, is
shown below.
V1
V2
_
+
Vd Ri
Ro
AVd
Vo
Figure 8.5: Op Amp Model.
9
12. Basic Electric Circuits
Operational Amplifiers
The previous model is usually shown as follows:
Ri
Ro
AVd
_
+
Vd
V1
V2
Vo
+
_
Figure 8.6: Working circuit diagram of op amp.
10
13. 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.
+
_
R2
R1
+
_
+
_
Vin
Vo
11 Figure 8.7: Op amp functional circuit.
14. Basic Electric Circuits
Operational Amplifiers
In terms of the circuit model we have the following:
Ri
Ro
AVi
_
+
ViVin Vo
+
_
+
_
R1
R2
a
b
Figure 8.8: Total op amp schematic for voltage
gain configuration.12
15. Basic Electric Circuits
Operational Amplifiers
Ri
Ro
AVi
_
+
ViVin Vo
+
_
+
_
R1
R2
a
b
Circuit values are:
R1 = 10 k R2 = 40 k Ro = 50
A = 100,000 Ri = 1 meg
13
16. Basic Electric Circuits
Operational Amplifiers
We can write the following equations for nodes a and b.
Eq 8.1
Eq 8.2
14
( )
10 1 40
( )
50
40
i oin i i
i o
o i
V VV V V
k meg k
V V
V AV
k
18. 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 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
19. Basic Electric Circuits
Operational Amplifiers
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
RohmsinputInfinite
RohmsoutputZero
AgainloopopenInfinite
;.3
0;.2
;.1
17
20. Basic Electric Circuits
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.
21. Basic Electric Circuits
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.
22. Basic Electric Circuits
Ideal Op Amp:
+
_
R2
R1
+
_
+
_
Vin
Vo
a
Vi
Writing a nodal equation at (a) gives;
21
)(
R
VV
R
VV oiiin
20
Eq 8.6
23. 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 Earlier
we got
ino VV 99.3
21
Eq 8.7
inV
R
R
V
1
2
0
24. 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) and
we have the important expression;
)s(Z
)s(Z
)s(V
)s(V
in
fb
in
0
Eq 8.9
22
25. Basic Electric Circuits
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
26. Basic Electric Circuits
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
27. Basic Electric Circuits
Operational Amplifiers
+
+
+
_
_
_
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.
28. Basic Electric Circuits
Operational Amplifiers
Example 8.2: Summing Amplifier. Given the following:
Rfb
R1
R2
V2
V1
V0
a
Figure 8.12: Circuit for Example 8.2.
fbR
V
R
V
R
V 0
2
2
1
1
Eq 8.11
26
29. Basic Electric Circuits
Operational Amplifiers
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
fbfb
Eq 8.12
If R1 = R2 = Rfb then,
210 VVV Eq. 8.13
Therefore, we can add signals with an op amp.
27
30. Basic Electric Circuits
Operational Amplifiers
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:
The
"Block"
Circuit 1 Circuit 2
+
_
+
_
Vin Vout
Figure 8.13: Illustrating Isolation.28
31. Basic Electric Circuits
Operational Amplifiers
Example 8.3: Isolation or Voltage Follower. continued
Circuit 1 Circuit 2
The Block
+
_
+
Vin V0_
Figure 8.14: Circuit isolation with an op amp.
It is easy to see that: V0 = Vin
29
32. Basic Electric Circuits
Operational Amplifiers
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;
inVV 20
33. Basic Electric Circuits
Operational Amplifiers
Example 8.5: The noninverting op amp.
Consider the following:
R0
Rfb
V0V2
_
+
+
_
a
Figure 8.16: Noninverting op am configuration.
31
34. Basic Electric Circuits
Operational Amplifiers
Example 8.5: The noninverting op amp. Continued
Writing a node equation at “a” gives;
2
0
0
0
2
0
02
0
2
1
,
11
0
)(
V
R
R
V
giveswhich
RR
V
R
V
so
R
VV
R
V
fb
fbfb
fb
Remember this
32
35. Basic Electric Circuits
Operational Amplifiers
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
36. Basic Electric Circuits
Operational Amplifiers
Example 8.6: Noninverting Input.
The voltage at Vx is found to be 3 V.
Writing a node equation at “a” gives;
0
105
0
k
)VV(
k
V xx
or
VVV x 930
34