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Operational Amplifier (Op-Amp)
Microelectronic Circuits - Fifth Edition © Sedra/Smith®
Prodi S1-SK
2017
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.
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.
Operational Amplifiers
The Philbrick Operational Amplifier
From “Operational Amplifier”, by Tony van Roon: http://www.uoguelph.ca/~antoon/gadgets/741/741.html
Operational Amplifiers
“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.
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.
Operational Amplifiers
The op amp is built using VLSI techniques. The circuit
diagram of an LM 741 from National Semiconductor is
shown below. V+
V-
Vo
Vin(-)
Vin(+)
Taken from National Semiconductor
data sheet as shown on the web.
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.
Operational Amplifiers
inverting input
noninverting input
output
V-
V+
The basic op amp with supply voltage included is shown
in the diagram below.
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.
Operational Amplifiers
Inverting input
Non-inverting input
Output
A model of the op amp, with respect to the symbol, is
shown below.
V1
V2
_
+
Vd Ri
Ro
AVd
Vo
Operational Amplifiers
The previous model is usually shown as follows:
Ri
Ri
AVd
_
+
Vd
V1
V2
Vo
+
_
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
+
_
+
_
VinVo
Operational Amplifiers
In terms of the circuit model we have the following total
op amp schematic for voltage gain configuration :
Ri
Ri
AVi
_
+
ViVin Vo
+
_
+
_
R1
R2
ab
Operational Amplifiers
Ri
Ri
AVi
_
+
ViVin Vo
+
_
+
_
R1
R2
ab
Given circuit values :
R1 = 10 k R2 = 40 k
A = 100,000 Ri = 1 M
Ro = 50
Operational Amplifiers
We can write the following equations for nodes a and b.
(Eq 8.1)
(Eq 8.2)
ioin
o
oiiiin
AVk
)VV(V
k
VV
meg
V
k
)VV(
4050
40110
Operational Amplifiers
Equation 8.1 simplifies to:
inio VVV 10012625
Equation 8.2 simplifies to:
010410005.4 95 io VxVx
Operational Amplifiers
(Eq 8.3)
(Eq 8.4)
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.
Operational Amplifiers
(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.
Based on those values, the following assumptions are
made for the ideal op amp:
i
o
RohmsinputInfinite
RohmsoutputZero
AgainloopopenInfinite
;.3
0;.2
;.1
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, so V1 = V2.
Ideal Op-Amp
Find Vo in terms of Vin for the following configuration if
R1 = 1 k and R2 = 4 k.
+
_
R2
R1
+
_
+
_
VinVo
Ideal Op-Amp
+
_
R2
R1
+
_
+
_
VinVo
a
Vi
Writing a nodal equation at (a) gives;
21
)(
R
VV
R
VV oiiin
Ideal Op-Amp
(Eq 8.6)
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
inVR
RV
1
20
(Eq 8.7)
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
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
Ideal Op-Amp
(Eq 8.8)
(Eq 8.9)
Example 8.1: Inverting Amplifier.
Consider the op amp configuration below. Determine
VO and Vin relation if assume Vin = 5 V
+
+
+
_
__
3 V
Vin
6 k
1 k
V0
a
+
+
+
_
__
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 Vin = 5 V, then V0 = -51 V
Example 8.1 continued
(Eq 8.10)
Example 8.2: Summing Amplifier. Given the following
Rfb
R1
R2
V2
V1V0
a
Equation at node “a” is:
fbR
V
R
V
R
V 0
2
2
1
1 (Eq 8.11)
Example 8.2 continued
Equation 8.11 can be expressed as;
2
2
1
1
0 VR
RV
R
RV
fbfb
If R1 = R2 = Rfb then,
210 VVV
Therefore, we can add signals with an op amp.
(Eq 8.13)
(Eq 8.12)
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
Example 8.3 continued
Circuit 1 Circuit 2
The Block
+
_
+Vin V0_
It is easy to see that: V0 = Vin
Example 8.4: Isolation with gain.
+
_
__
+
+
20 k
Vin
Vin
V0
10 k
10 k
a
+
_
Writing a nodal equation at point “a” and simplifying gives:
inVV 20
Example 8.5: Noninverting Amplifier.
Consider the following:
R0
Rfb
V0V2_
+
+
_
a+
_
Example 8.5 continued
Writing a node equation at “a” gives;
2
0
0
0
20
02
0
2
1
,
11
0)(
VR
RV
giveswhich
RRV
R
V
so
R
VV
R
V
fb
fbfb
fb
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
Example 8.6 continued
The voltage at Vx is found to be 3 V.
+
_
+
+
_
_ 4 V
2 k
6 k
5 k
10 k
V0
a
Vxix iy
Write node equation at Vx
3
312
6
0
2
4
x
xx
xx
yx
V
VV
k
V
k
V
ii
Example 8.6 continued
Then, writing a node equation at “a” gives;
0105
0
k
)VV(
k
V xx
or
VVV x 930
Exercise 1
Find IX, VO and IO for the following op amp configuration.
Answer: IX = 0.5 mA; VO = –4 V; IO = –4/3 mA
Exercise 2
Find I and VO for the following op amp configuration.
Answer: I = 0.5 mA; VO = 4 V
