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Operational
Amplifiers
4
LEARNING GOALS
I n this chapter we discuss a very important commercially
available circuit known as the operational amplifier, or op-amp. Op-amps
are used in literally thousands of applications, including such things as
compact disk (CD) players, random access memories (RAMs), analogto-digital (A/D) and digital-to-analog (D/A) converters, headphone
amplifiers, and electronic instrumentation of all types. Finally, we discuss
the terminal characteristics of this circuit and demonstrate its use in
practical applications as well as circuit design.
●
4.1 Introduction A brief introduction is given to
an extremely important electronic circuit known as
the operational amplifier, or op-amp...Page 134
4.2 Op-Amp Models Models for the op-amp are
developed. This electronic circuit is characterized
by high-input resistance, low-output resistance,
and very high gain...Page 134
4.3 Fundamental Op-Amp Circuits Basic
op-amp configurations, including inverting and
noninverting circuits, are widely used in a host of
practical applications...Page 140
4.4 Comparators A comparator is a variant of
the op-amp circuit that finds wide application in
circuits where comparisons between two quantities
are required...Page 147
4.5 Application Examples...Page 148
4.6 Design Examples...Page 152
Summary...Page 155
Problems...Page 155
Access Problem-Solving Videos P S V
and Circuit Solutions C S at:
http://www.justask4u.com/irwin8/
using the registration code on the inside cover
and see a website with answers and more!
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Introduction
Figure 4.1
A selection of op-amps. On the
left in (a) is a discrete op-amp
assembled on a printed circuit
board (PCB). On the right,
top-down, a LM324 DIP,
LMC6492 DIP, and MAX4240
in a SO-5 package (small
outline/5 pins). The APEX PA03
with its lid removed (b) showing
individual transistors and
resistors.
It can be argued that the operational amplifier, or op-amp as it is commonly known, is the single most important integrated circuit for analog circuit design. It is a versatile interconnection
of transistors and resistors that vastly expands our capabilities in circuit design, from engine
control systems to cellular phones. Early op-amps were built of vacuum tubes, making them
bulky and power hungry. The invention of the transistor at Bell Labs in 1947 allowed engineers to create op-amps that were much smaller and more efficient. Still, the op-amp itself
consisted of individual transistors and resistors interconnected on a printed circuit board
(PCB). When the manufacturing process for integrated circuits (ICs) was developed around
1970, engineers could finally put all of the op-amp transistors and resistors onto a single IC
chip. Today, it is common to find as many as four high-quality op-amps on a single IC for as
little as $0.40. A sample of commercial op-amps is shown in Fig. 4.1.
(a)
(b)
Why are they called operational amplifiers? Originally, the op-amp was designed to
perform mathematical operations such as addition, subtraction, differentiation, and integration. By adding simple networks to the op-amp, we can create these “building blocks” as well
as voltage scaling, current-to-voltage conversion, and a myriad of more complex applications.
4.2
Op-Amp Models
How can we, understanding only sources and resistors, hope to comprehend the performance of
the op-amp? The answer lies in modeling. When the bells and whistles are removed, an op-amp
is just a really good voltage amplifier. In other words, the output voltage is a scaled replica of
the input voltage. Modern op-amps are such good amplifiers that it is easy to create an accurate,
first-order model. As mentioned earlier, the op-amp is very popular and is used extensively in
circuit design at all levels. We should not be surprised to find that op-amps are available for
every application—low voltage, high voltage, micro-power, high speed, high current, and so
forth. Fortunately, the topology of our model is independent of these issues.
We start with the general-purpose LM324 quad (four in a pack) op-amp from National
Semiconductor, shown in the upper right corner of Fig. 4.1a. The pinout for the LM324 is shown
in Fig. 4.2 for a DIP (Dual Inline Pack) style package with dimensions in inches. Recognizing
there are four identical op-amps in the package, we will focus on amplifier 1. Pins 3 and 2 are the
input pins, IN 1+ and IN 1-, and are called the noninverting and inverting inputs, respectively.
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SECTION 4.2
OP-AMP MODELS
0.78
14
13
12
11
10
9
8
OUT 4 IN 4– IN 4+ VEE IN 3– IN 3+ IN 3
0.04
4
–
±
–3
±
0.3
1
±
–
±2
–
OUT 1 IN 1– IN 1+
1
2
3
0.1
135
Figure 4.2
The pinout (a) and
dimensional diagram
(b) of the LM324 quad
op-amp. Note the pin
pitch (distance pin-to-pin)
is 0.1 inches—standard
for DIP packages.
0.06
VCC IN 2– IN 2+ OUT 2
4
5
6
7
(a)
(b)
The output is at pin 1. A relationship exists between the output and input voltages,
4.1
Vo = A o(IN + - IN - )
where all voltages are measured with respect to ground and A o is the gain of the op-amp.
(The location of the ground terminal will be discussed shortly.) From Eq. (4.1), we see that
when IN+ increases, so will Vo . However, if IN- increases, then Vo will decrease—hence the
names noninverting and inverting inputs. We mentioned earlier that op-amps are very good
voltage amplifiers. How good? Typical values for A o are between 10,000 and 1,000,000!
Amplification requires power that is provided by the dc voltage sources connected to pins 4 and
11, called VCC and VEE , respectively. Figure 4.3 shows how the power supplies, or rails, are connected for both dual- and single-supply applications and defines the ground node to which all input
and output voltages are referenced. Traditionally, VCC is a positive dc voltage with respect to
ground, and VEE is either a negative voltage or ground itself. Actual values for these power supplies
can vary widely depending on the application, from as little as one volt up to several hundred.
How can we model the op-amp? A dependent voltage source can produce Vo ! What about
the currents into and out of the op-amp terminals (pins 3, 2, and 1)? Fortunately for us, the
currents are fairly proportional to the pin voltages. That sounds like Ohm’s law. So, we model
the I-V performance with two resistors, one at the input terminals (Ri) and another at the
output (Ro). The circuit in Fig. 4.4 brings everything together.
IN+
IN–
VCC
±
–
VCC
IN+
out
IN–
VEE
VEE
±
–
VCC
VCC
VEE
(a)
(b)
+
Ro
+
vin(t)
IN+(t)
IN–(t)
-
+
Ri
+
-
Figure 4.3
Schematics showing the power supply
connections and ground location for
(a) dual-supply and (b) single-supply
implementations.
Aovin
±
–
vo(t)
-
Figure 4.4
A simple model for the gain characteristics of an op-amp.
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Figure 4.5
A network that depicts an op-amp circuit. VS
and RTh1 model the driving circuit, while the
load is modeled by RL. The circuit in Fig. 4.4 is
the op-amp model.
RTh1
Ro
+
+
VS ±
–
vin(t)
Ri
-
±
Aovin –
RL
vo(t)
-
What values can we expect for A o , Ri , and Ro ? We can reason through this issue with the
help of Fig. 4.5 where we have drawn an equivalent for the circuitry that drives the input
nodes and we have modeled the circuitry connected to the output with a single resistor, RL .
Since the op-amp is supposed to be a great voltage amplifier, let’s write an equation for the
overall gain of the circuit VoVS . Using voltage division at the input and again at the output,
we quickly produce the expression
Ri
Vo
RL
= c
d Ao c
d
VS
R i + R Th1
Ro + RL
To maximize the gain regardless of the values of RTh1 and RL , we make the voltage division ratios as close to unity as possible. The ideal scenario requires that A o be infinity, Ri be
infinity, and Ro be zero, yielding a large overall gain of A o . Table 4.1 shows the actual values
of A o , Ri , and Ro for a sampling of commercial op-amps intended for very different applications. While A o , Ri , and Ro are not ideal, they do have the correct tendencies.
The power supplies affect performance in two ways. First, each op-amp has minimum and
maximum supply ranges over which the op-amp is guaranteed to function. Second, for proper operation, the input and output voltages are limited to no more than the supply voltages.*
If the inputs/output can reach within a few dozen millivolts of the supplies, then the inputs/
output are called rail-to-rail. Otherwise, the inputs/output voltage limits are more severe—
usually a volt or so away from the supply values. Combining the model in Fig. 4.4, the values
in Table 4.1, and these I/O limitations, we can produce the graph in Fig. 4.6 showing the
output–input relation for each op-amp in Table 4.1. From the graph we see that LMC6492 and
MAX4240 have rail-to-rail outputs while the LM324 and PA03 do not.
Table 4.1
A list of commercial op-amps and their model values
Manufacturer
Part No.
Ao (V /V )
Ri (M)
Ro ()
100,000
1.0
20
National
LM324
National
LMC6492
50,000
107
150
Maxim
MAX4240
20,000
45
160
Apex
PA03
125,000
105
2
Comments
General purpose, up to ;16 V supplies,
very inexpensive
Low voltage, rail-to-rail inputs and
outputs*
Micro-power (1.8 V supply @ 10 A),
rail-to-rail inputs and outputs
High-voltage, ;75 V, and high-output
current capability, 30 A. That’s 2 kW!
*Rail-to-rail is a trademark of Motorola Corporation. This feature is discussed further in the following paragraphs.
Even though the op-amp can function within the minimum and maximum supply voltages,
because of the circuit configuration, an increase in the input voltage may not yield a corresponding increase in the output voltage. In this case, the op-amp is said to be in saturation.
The following example addresses this issue.
*Op-amps are available that have input and/or output voltage ranges beyond the supply rails. However, these devices
constitute a very small percentage of the op-amp market and will not be discussed here.
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60
Output voltage, vo (V)
10
5
0
–5
40
20
0
–20
–10
LM324@ +/– 15 V
LMC6492@ +/– 5 V
–15
LM324@ +/– 15 V
–40
MAX4240@ +/– 1.5 V
PA03@ +/– 75 V
–60
LMC6492@ +/– 5 V
MAX4240@ +/– 1.5 V
–20
–250 –200 –150 –100 –50
0
50
–80
–1.0 –0.8 –0.6 –0.4 –0.2 0.0
100 150 200 250
Input voltage, vin (V)
0.2
0.4
0.6
Input voltage, vin (mV)
Example 4.1
The input and output signals for an op-amp circuit are shown in Fig. 4.7. We wish to determine (a) if the op-amp circuit is linear and (b) the circuit’s gain.
SOLUTION
a. We know that if the circuit is linear, the output must be linearly related, that is, proportional,
to the input. An examination of the input and output waveforms in Fig. 4.7 clearly indicates
that in the region t = 1.25 to 2.5 and 4 to 6 ms the output is constant while the input is
changing. In this case, the op-amp circuit is in saturation and therefore not linear.
b. In the region where the output is proportional to the input, that is, t = 0 to 1 ms, the input
changes by 1 V and the output changes by 3.3 V. Therefore, the circuit’s gain is 3.3.
Voltage (V)
Figure 4.7
An op-amp input–output
characteristic.
Output
3
2
Input
1
0
1
2
3
4
5
–1
–2
6
7
t (ms)
Input
Output
–3
–4
To introduce the performance of the op-amp in a practical circuit, consider the network in
Fig. 4.8a called a unity gain buffer. Notice that the op-amp schematic symbol includes the
power supplies. Substituting the model in Fig. 4.4 yields the circuit in Fig. 4.8b, containing just
resistors and dependent sources, which we can easily analyze. Writing loop equations, we have
VS = IRi + IRo + A o Vin
Vout = IRo + A o Vin
Vin = IRi
0.8
137
Figure 4.6
Transfer plots for the
op-amps listed in
Table 4.1. The supply
voltages are listed in
the plot legends. Note
that the LMC6492 and
MAX4240 have rail-torail output voltages
(output voltage range
extends to power
supply values), while
the LM324 and PA03
do not.
80
15
Output voltage, vo (V)
OP-AMP MODELS
1.0
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Figure 4.8
Circuit (a) and model (b) for the unity
gain buffer.
+
Vin
VCC
±
–
VS ±
–
Ri
Ro
-
VS ±
–
+
Vo
-
VEE
AoVin
Vo
-
I
(a)
±
–
+
(b)
Solving for the gain, VoVS , we find
Vo
=
VS
1
1 +
Ri
Ro + A o Ri
For Ro V Ri , we have
Vo
L
VS
1
1 +
1
Ao
And, if A o is indeed W 1,
Vo
L 1
VS
The origin of the name unity gain buffer should be apparent. Table 4.2 shows the actual
gain values for VS = 1 V using the op-amps listed in Table 4.1. Notice how close the gain is
to unity and how small the input voltage and current are. These results lead us to simplify the
op-amp in Fig. 4.4 significantly. We introduce the ideal op-amp model, where A o and Ri are
infinite and Ro is zero. This produces two important results for analyzing op-amp circuitry,
listed in Table 4.3.
Table 4.2 Unity gain buffer performance
for the op-amps listed in Table 4.1
Op-Amp
v+
i+
v–
i–
±∞
–
Figure 4.9
Ideal model for an operational
amplifier. Model parameters:
i+ = i - = 0, v+ = v- .
Buffer Gain Vin (V)
I (pA)
Table 4.3
Consequences of the ideal
op-amp model on input terminal IV values
Model Assumption
Terminal Result
LM324
0.999990
9.9999
9.9998
Ao S q
input voltage S 0 V
LMC6492
0.999980
19.999 1.9999 * 10-6
Ri S q
input current S 0 A
MAX4240 0.999950
49.998
PA05
7.9999 7.9999 * 10-5
0.999992
1.1111
From Table 4.3 we find that the ideal model for the op-amp is reduced to that shown in
Fig. 4.9. The important characteristics of the model are as follows: (1) Since Ri is extremely
large, the input currents to the op-amp are approximately zero (i.e., i+ L i- L 0); and (2) if the
output voltage is to remain bounded, then as the gain becomes very large and approaches infinity, the voltage across the input terminals must simultaneously become infinitesimally small so
that as A o S q, v+ - v- S 0 (i.e., v+ - v- = 0 or v+ = v-). The difference between these
input voltages is often called the error signal for the op-amp (i.e., v+ - v- = ve).
The ground terminal
shown on the op-amp is necessary for
signal current return, and it guarantees that Kirchhoff’s current law is satisfied at both the opamp and the ground node in the circuit.
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SECTION 4.2
OP-AMP MODELS
In summary, then, our ideal model for the op-amp is simply stated by the following conditions:
4.2
i+ = i- = 0
v+ = v-
139
0A+
+
0V
-
Vin ±
–
0 A-
+
Vo
These simple conditions are extremely important because they form the basis of our analysis VS ±
+
–
of op-amp circuits.
VS
Let’s use the ideal model to reexamine the unity gain buffer, drawn again in Fig. 4.10,
where the input voltage and currents are shown as zero. Given Vin is zero, the voltage at both
op-amp inputs is VS . Since the inverting input is physically connected to the output, Vo is also
VS —unity gain!
Armed with the ideal op-amp model, let’s change the circuit in Fig. 4.10 slightly as shown Figure 4.10
in Fig. 4.11 where VS and RS are an equivalent for the circuit driving the buffer and RL models An ideal op-amp configured as
a unity gain buffer.
the circuitry connected to the output. There are three main points here. First, the gain is still
unity. Second, the op-amp requires no current from the driving circuit. Third, the output current (Io = VoRL) comes from the power supplies, through the op-amp and out of the output
pin. In other words, the load current comes from the power supplies, which have plenty of
current output capacity, rather than the driving circuit, which may have very little. This isolation of current is called buffering.
An obvious question at this point is this: If Vo = VS , why not just connect VS to Vo via two
parallel connection wires; why do we need to place an op-amp between them? The answer
to this question is fundamental and provides us with some insight that will aid us in circuit
analysis and design.
Consider the circuit shown in Fig. 4.12a. In this case Vo is not equal to VS because of the
voltage drop across RS :
Vo = VS - IRS
However, in Fig. 4.12b, the input current to the op-amp is zero and, therefore, VS appears at the
op-amp input. Since the gain of the op-amp configuration is 1, Vo = VS . In Fig. 4.12a the resistive
network’s interaction with the source caused the voltage vo to be less than VS . In other words, the
resistive network loads the source voltage. However, in Fig. 4.12b the op-amp isolates the source
from the resistive network; therefore, the voltage follower is referred to as a buffer amplifier
because it can be used to isolate one circuit from another. The energy supplied to the resistive network in the first case must come from the source VS , whereas in the second case it comes from
the power supplies that supply the amplifier, and little or no energy is drawn from VS .
RS
VCC
0A
+
0V
-
VS ±
–
±
–
0A
Figure 4.11
A unity gain buffer with a load resistor.
Io
+
VEE
+
VS
RL
1 k
Vo
-
RS
RS
I
±
–
+
VS ±
–
Vo
-
I
Resistive
network
VS ±
–
+
Vo
-
Resistive
network
Figure 4.12
Illustration of the
isolation capability
of a voltage follower.
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Fundamental Op-Amp Circuits
As a general rule, when analyzing op-amp circuits we write nodal equations at the op-amp
input terminals, using the ideal op-amp model conditions. Thus, the technique is straightforward and simple to implement.
Example 4.2
Let us determine the gain of the basic inverting op-amp configuration shown in Fig. 4.13a
using both the nonideal and ideal op-amp models.
R1
d
–
±A
a
±
–
vS
R2
b
i+
v+
+
c
v–
vo
i–
a
b
d
±A
–
c
-
(a)
(b)
b
a
+
d
d
+
Ro
ve
v+
-
R2
±
–
Ri
b
vo
v–
c
-
v–
Ri
vS ±
–
A(v+-v–)
=Ave
+
-
R1
+
Ro
+
±
– A(v -v )
+
–
vo
v+
c
(c)
-
(d)
v1
vo
R1
R2
+
-
vS ±
–
Ri
Ro
ve
vo
Ave
+
±
–
-
(e)
Figure 4.13 Op-amp circuit.
SOLUTION
Our model for the op-amp is shown generically in Fig. 4.13b and specifically in
terms of the parameters Ri , A, and Ro in Fig. 4.13c. If the model is inserted in the network
in Fig. 4.13a, we obtain the circuit shown in Fig. 4.13d, which can be redrawn as shown in
Fig. 4.13e.
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F U N D A M E N TA L O P - A M P C I R C U I T S
The node equations for the network are
v1 - vS
v1 - vo
v1
+
+
= 0
R1
Ri
R2
vo - v1
vo - Ave
+
= 0
R2
Ro
where ve = -v1 . The equations can be written in matrix form as
-a
1
vS
b
R2
v1
T B R = C R1 S
1
1
vo
0
+
R2
Ro
1
1
1
+
+
R1
Ri
R2
D
1
A
-a
b
R2
Ro
Solving for the node voltages, we obtain
1
1
+
v1
1 R2
Ro
B R =
D
vo
¢
1
A
R2
Ro
1
vS
R2
T C R1 S
1
1
1
0
+
+
R1
Ri
Ro
where
¢ = a
1
1
1
1
1
1
1
A
+
+
ba
+
b - a ba
b
R1
Ri
R2
R2
Ro
R2
R2
Ro
Hence,
vo =
a
a
vS
1
A
ba b
R2
Ro
R1
1
1
1
1
1
1
1
A
+
+
ba
+
b - a ba
b
R1
Ri
R2
R2
Ro
R2
R2
Ro
which can be written as
vo
=
vS
1 - ca
- AR 2R 1 B
1
1
1
1
1
1
1
A
+
+
ba
+
bna b a
bd
R1
Ri
R2
R2
Ro
R2
R2
Ro
If we now employ typical values for the circuit parameters (e.g., A = 105, Ri = 108 ,
Ro = 10 , R1 = 1 k, and R2 = 5 k), the voltage gain of the network is
vo
= -4.9996994 L -5.000
vS
However, the ideal op-amp has infinite gain. Therefore, if we take the limit of the gain equation as A S q, we obtain
vo
R2
lim a b = = -5.000
R1
AS q vS
Note that the ideal op-amp yielded a result accurate to within four significant digits of that
obtained from an exact solution of a typical op-amp model. These results are easily repeated
for the vast array of useful op-amp circuits.
We now analyze the network in Fig. 4.13a using the ideal op-amp model. In this model
i+ = i- = 0
v+ = v-
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As shown in Fig. 4.13a, v+ = 0 and, therefore, v- = 0. If we now write a node equation at
the negative terminal of the op-amp, we obtain
vS - 0
vo - 0
+
= 0
R1
R2
or
vo
R2
= vS
R1
and we have immediately obtained the results derived previously.
Notice that the gain is a simple resistor ratio. This fact makes the amplifier very versatile in
that we can control the gain accurately and alter its value by changing only one resistor. Also,
the gain is essentially independent of op-amp parameters. Since the precise values of A o , Ri ,
and Ro are sensitive to such factors as temperature, radiation, and age, their elimination results
in a gain that is stable regardless of the immediate environment. Since it is much easier to
employ the ideal op-amp model rather than the nonideal model, unless otherwise stated we will
use the ideal op-amp assumptions to analyze circuits that contain operational amplifiers.
P R O B L E M - S O LV I N G S T R AT E G Y
Op-Amp Circuits
Step 1. Use the ideal op-amp model: A o = q, Ri = q, Ro = 0.
● v = v
i+ = i- = 0
+
Step 2. Apply nodal analysis to the resulting circuit.
●
Step 3. Solve nodal equations to express the output voltage in terms of the op-amp
input signals.
Example 4.3
Let us now determine the gain of the basic noninverting op-amp configuration shown in Fig. 4.14.
Figure 4.14
The noninverting
op-amp configuration.
vin
±
–
vo
RF
RI
SOLUTION
Once again we employ the ideal op-amp model conditions, that is, v- = v+ and
i- = i+ . Using the fact that i- = 0 and v- = Vin , the KCL equation at the negative terminal of
the op-amp is
vo - vin
vin
=
RI
RF
or
vin a
vo
1
1
+
b =
RI
RF
RF
Thus
vo
RF
= 1 +
vin
RI
Note the similarity of this case to the inverting op-amp configuration in the previous example. We find that the gain in this configuration is also controlled by a simple resistor ratio, but
is not inverted; that is, the gain ratio is positive.
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SECTION 4.3
F U N D A M E N TA L O P - A M P C I R C U I T S
The remaining examples, though slightly more complicated, are analyzed in exactly the
same manner as those outlined above.
Example 4.4
Gain error in an amplifier is defined as
GE = c
actual gain - ideal gain
d * 100%
ideal gain
We wish to show that for a standard noninverting configuration with finite gain A o , the gain
error is
GE =
-100%
1 + Ao
where = R1(R1 + R2).
SOLUTION
The standard noninverting configuration and its equivalent circuit are shown in
Fig. 4.15a and b, respectively. The circuit equations for the network in Fig. 4.15b are
vS = vin + v1 ,
vin =
vo
Ao
and
v1 =
R1
vo = vo
R1 + R2
The expression that relates the input and output is
vS = vo c
1 + Ao
1
+ d = vo c
d
Ao
Ao
and thus the actual gain is
Ao
vo
=
vS
1 + Ao
Recall that the ideal gain for this circuit is (R1 + R2)R1 = 1. Therefore, the gain error is
Ao
1
§ 100%
GE = £ 1 + A o 1
which when simplified yields
GE =
-100%
1 + Ao
vS
vo
+
vS
±
–
R2
vin
vo
Aovin
-
v1
R2
R1
R1
(a)
(b)
±
–
Figure 4.15
Circuits used in Example 4.4.
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Example 4.5
Consider the op-amp circuit shown in Fig. 4.16. Let us determine an expression for the output voltage.
R2
Figure 4.16
Differential amplifier
operational
amplifier circuit.
R1
v–
i–
+
+
v1
v2
v+ i+
R4
+
vo
-
-
-
SOLUTION
R3
–
±
The node equation at the inverting terminal is
vo - vv1 - v+
= iR1
R2
At the noninverting terminal KCL yields
v2 - v+
v+
=
+ i+
R3
R4
However, i+ = i- = 0 and v+ = v- . Substituting these values into the two preceding equations
yields
vo - vv1 - v+
= 0
R1
R2
and
v2 - vv=
R3
R4
Solving these two equations for vo results in the expression
vo =
R2
R1
R4
R2
a1 +
b
v2 v
R1
R2 R3 + R4
R1 1
Note that if R 4 = R 2 and R 3 = R 1 , the expression reduces to
vo =
R2
Av - v1 B
R1 2
Therefore, this op-amp can be employed to subtract two input voltages.
Example 4.6
The circuit shown in Fig. 4.17a is a precision differential voltage-gain device. It is used to
provide a single-ended input for an analog-to-digital converter. We wish to derive an expression for the output of the circuit in terms of the two inputs.
SOLUTION
To accomplish this, we draw the equivalent circuit shown in Fig. 4.17b. Recall
that the voltage across the input terminals of the op-amp is approximately zero and the currents into the op-amp input terminals are approximately zero. Note that we can write node
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F U N D A M E N TA L O P - A M P C I R C U I T S
145
equations for node voltages v1 and v2 in terms of vo and va . Since we are interested in
an expression for vo in terms of the voltages v1 and v2 , we simply eliminate the va terms from
the two node equations. The node equations are
v1 - va
v1 - vo
v1 - v2
+
+
= 0
R2
R1
RG
v2 - va
v2 - v1
v2
+
+
= 0
R1
RG
R2
Combining the two equations to eliminate va , and then writing vo in terms of v1 and v2 , yields
vo = Av1 - v2 B a 1 +
v1
±
–
R2
2R 2
+
b
R1
RG
R2
R2
v1
v1
i1=0
R1
v2
±
–
va
Figure 4.17
Instrumentation amplifier circuit.
vo
vo
R1
va
RG
R1
RG
R1
v2
v2
i2=0
R2
R2
(a)
(b)
LEARNING EXTENSIONS
E4.1 Find Io in the network in Fig. E4.1.
Figure E4.1
ANSWER: Io = 8.4 mA.
PSV
±
–
Vo
12 k
12 V
±
–
10 k
2 k
Io
(continues on the next page)
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O P E R AT I O N A L A M P L I F I E R S
E4.2 Determine the gain of the op-amp circuit in Fig. E4.2.
Figure E4.2
PSV
ANSWER:
Vo
R2
= 1 +
.
VS
R1
±
–
+
R2
±
VS –
Vo
R1
-
E4.3 Determine both the gain and the output voltage of the op-amp configuration shown in
Fig. E4.3.
Figure E4.3
ANSWER: Vo = 0.101 V;
gain=101.
±
–
+
1 mV
±
–
100 k
Vo
1 k
-
Example 4.7
The two op-amp circuits shown in Fig. 4.18 produce an output given by the equation
Vo = 8V1 - 4 V2
where
1 V V1 2 V
and
2 V V2 3 V
We wish to determine (a) the range of Vo and (b) if both of the circuits will produce the full
range of Vo given that the dc supplies are ;10 V.
SOLUTION
a. Given that Vo = 8 V1 - 4 V2 and the range for both V1 and V2 as 1 V V1 2 V and
2 V V2 3 V, we find that
Vo max = 8(2) - 4(2) = 8 V
and Vo min = 8(1) - 4(3) = -4 V
and thus the range of Vo is -4 V to +8 V.
b. Consider first the network in Fig. 4.18a. The signal at Vx , which can be derived using the
network in Example 4.5, is given by the equation Vx = 2 V1 - V2 . Vx is a maximum when
V1 = 2 V and V2 = 2 V, that is, Vx max = 2(2) - 2 = 2 V. The minimum value for Vx
occurs when V1 = 1 V and V2 = 3 V, that is Vx min = 2(1) - 3 = -1 V. Since both the max
and min values are within the supply range of ;10 V, the first op-amp in Fig. 4.18a will
not saturate. The output of the second op-amp in this circuit is given by the expression
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C O M PA R AT O R S
Vo = 4Vx . Therefore, the range of Vo is -4 V Vo 8 V. Since this range is also within
the power supply voltages, the second op-amp will not saturate, and this circuit will
produce the full range of Vo .
Next, consider the network in Fig. 4.18b. The signal Vy = -8V1 and so the range of Vy is
-16 V Vy -8 V and the range of Vy is outside the power supply limits. This circuit will
saturate and fail to produce the full range of Vo .
V1
Vx
±
–
±
–
10 k
Figure 4.18
Circuits used in Example 4.7.
Vo
30 k
10 k
10 k
V2
(a)
80 k
V1
10 k
–
±
V2
±
–
Vy
Vz
10 k
10 k
10 k
–
±
Vo
30 k
10 k
(b)
4.4
Comparators
A comparator, a variant of the op-amp, is designed to compare the noninverting and inverting
input voltages. As shown in Fig. 4.19, when the noninverting input voltage is greater, the output goes as high as possible, at or near VCC . On the other hand, if the inverting input voltage
is greater, the output goes as low as possible, at or near VEE . Of course, an ideal op-amp can
do the same thing, that is, swing the output voltage as far as possible. However, op-amps are
not designed to operate with the outputs saturated, whereas comparators are. As a result, comparators are faster and less expensive than op-amps.
We will present two very different quad comparators in this text, National Semiconductor’s
LM339 and Maxim’s MAX917. Note that the LM339 requires a resistor, called a pull-up resistor, connected between the output pin and VCC . The salient features of these products are listed in
Table 4.4. From Table 4.4, it is easy to surmise that the LM339 is a general-purpose comparator,
whereas the MAX917 is intended for low-power applications such as hand-held products.
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Figure 4.19
(a) An ideal comparator and
(b) its transfer curve.
VCC
vin+
-
V+ ±
–
±
–
+
VEE
Vo
V– ±
–
Output voltage
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0
VEE
–1.5
–1
–0.5
0
0.5
1
1.5
Input voltage (V+-V–)
(a)
(b)
A common comparator application is the zero-crossing detector, shown in Fig. 4.20a using
a LM339 with ;5 V supplies. As seen in Fig. 4.20b, when VS is positive, Vo should be near
+5 V and when VS is negative, Vo should be near -5 V. The output changes value on every
zero crossing!
Table 4.4
A listing of some of the features of the LM339 and MAX917 comparators
Product
Min. Supply
Max. Supply
Supply Current
Max. Output Current
Typical Rpull - up
LM339
MAX919
2V
1.8 V
36 V
5.5 V
3 mA
0.8 A
50 mA
8 mA
3 k
NA
Figure 4.20
(a) A zero-crossing
detector and (b) the
corresponding
input/output waveforms.
Input
Output
6
3 k
±
–
–5 V
VS ±
–
+
Vo
-
4
I/O voltages (V)
+5 V
2
0
–2
–4
–6
Time
(a)
4.5
(b)
Application Examples
At this point, we have a new element, the op-amp, which we can effectively employ in both
applications and circuit design. This device is an extremely useful element that vastly expands
our capability in these areas. Because of its ubiquitous nature, the addition of the op-amp to
our repertoire of circuit elements permits us to deal with a wide spectrum of practical circuits.
Thus, we will employ it here, and also use it throughout this text.
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A P P L I C AT I O N E X A M P L E S
Application Example 4.8
In a light meter, a sensor produces a current proportional to the intensity of the incident radiation. We wish to obtain a voltage proportional to the light’s intensity using the circuit in
Fig. 4.21. Thus, we select a value of R that will produce an output voltage of 1 V for each
10 A of sensor current. Assume the sensor has zero resistance.
R
–
±
I
Figure 4.21
Light intensity to voltage
converter.
+
Incident
light
Light
sensor
Vo
-
SOLUTION
Applying KCL at the op-amp input,
I = VoR
Since VoI is 105,
R = 100 k
Application Example 4.9
The circuit in Fig. 4.22 is an electronic ammeter. It operates as follows: The unknown current,
I through RI produces a voltage, VI . VI is amplified by the op-amp to produce a voltage, Vo ,
which is proportional to I. The output voltage is measured with a simple voltmeter. We want
to find the value of R2 such that 10 V appears at Vo for each milliamp of unknown current.
SOLUTION
Since the current into the op-amp + terminal is zero, the relationship between VI
and I is
VI = IRI
Figure 4.22
Electronic ammeter.
±
–
I
Unknown
current
Voltmeter
+
+
VI
RI=1 k
Vo
R2
-
R1=1 k
+
-
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O P E R AT I O N A L A M P L I F I E R S
The relationship between the input and output voltages is
Vo = VI a 1 +
R2
b
R1
or, solving the equation for Vo I, we obtain
Vo
R2
= RI a 1 +
b
I
R1
Using the required ratio Vo I of 104 and resistor values from Fig. 4.22, we can find that
R2 = 9 k
Application Example 4.10
Let us return to the dc motor control example in Chapter 3 (Example 3.22). We want to define
the form of the power amplifier that reads the speed control signal, Vspeed and outputs the dc
motor voltage with sufficient current to drive the motor as shown in Fig. 4.23. Let us make
our selection under the condition that the total power dissipation in the amplifier should not
exceed 100 mW.
5V
Rpot
R1
=1
R2
=0
Power
amp
+
Vspeed
VM /Vspeed=4
+
VM
–
–
dc
motor
Figure 4.23 The dc motor example from Chapter 3.
SOLUTION
From Table 4.1 we find that the only op-amp with sufficient output voltage—
that is, a maximum output voltage of (4)(5) = 20 V—for this application is the PA03 from
APEX. Since the required gain is +4, we can employ the standard noninverting amplifier configuration shown in Fig. 4.24. If the PA03 is assumed to be ideal, then
VM = Vspeed c 1 +
RB
d = 4Vspeed
RA
There are, of course, an infinite number of solutions that will satisfy this equation.
In order to select reasonable values, we should consider the possibility of high currents in
RA and RB when VM is at its peak value of 20 V. Assuming that Rin for the PA03 is much
greater than RA , the currents in RB and RA essentially determine the total power dissipated.
±
–
5V
R1
R2
+
+
Vspeed
–
RB
RA
VM
–
Figure 4.24 The power amplifier configuration using the PA03 op-amp.
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A P P L I C AT I O N E X A M P L E S
The total power dissipated in RA and RB is
Ptotal =
V 2M
202
400
=
RA + RB
RA + RB
RA + RB
Since the total power should not exceed 100 mW, we can use 14 W resistors—an inexpensive
industry standard—with room to spare. With this power specification, we find that
RA + RB =
V 2M
400
= 4000
=
Ptotal
0.1
Also, since
1 +
RB
= 4
RA
then RB = 3 RA . Combining this result with the power specification yields RA = 1 k and
RB = 3 k. Both are standard 5% tolerance values.
Application Example 4.11
An instrumentation amplifier of the form shown in Fig. 4.25 has been suggested. This amplifier
should have high-input resistance, achieve a voltage gain Vo(V1 - V2) of 10, employ the
MAX4240 op-amp listed in Table 4.1, and operate from two 1.5 V AA cell batteries in series. Let
us analyze this circuit, select the resistor values, and explore the validity of this configuration.
Differential amplifier
Figure 4.25
An instrumentation amplifier
using the MAX4240 op-amp.
3V
+
V1
–
Vx
±
–
A
RB=RA
R1
3V
3V
RA
±
–
+
Vo
–
R
1.5 V
R2
B
RA
–
±
1.5 V
Vy
+
V2
–
RB=RA
3V
SOLUTION
As indicated, the op-amp on the right side of the circuit is connected in the
traditional differential amplifier configuration. Example 4.5 indicates that the voltage gain
for this portion of the network is
Vo = (Vx - Vy) c
RB
d
RA
And if RA = RB , the equation reduces to
Vo = Vx - Vy
If we can find a relationship between V1 , V2 , and Vx and Vy , then an expression for the overall
voltage can be written. Applying KCL at node A yields
Vx - V1
V1 - V2
=
R
R1
or
Vx = V1 c 1 +
R1
R1
d - V2 c d
R
R
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In a similar manner, at node B we obtain
V2 - Vy
V1 - V2
=
R
R2
or
Vy = -V1 c
R2
R2
d + V2 c 1 +
d
R
R
By combining these equations, the output voltage can be expressed as
Vo = Vx - Vy = V1 c 1 +
R1
R2
R1
R2
d - V2 c d + V1 c d - V2 c 1 +
d
R
R
R
R
If the resistors are selected such that R1 = R2 , then the voltage gain is
Vo
2R 1
= 1 +
V1 - V2
R
For a gain of +10, we set R1 = 4.5 R. To maintain low power, we will use fairly large values
for these resistors. We somewhat arbitrarily choose R = 100 k and R1 = R2 = 450 k. We
can use 100 k resistors in the differential amplifier stage as well.
Note that the voltage gain of the instrumentation amplifier is essentially the same as that of
a generic differential amplifier. So why add the additional cost of two more op-amps? In this
configuration the inputs V1 and V2 are directly connected to op-amp input terminals; therefore,
the input resistance of the intrumentation amplifier is extremely large. From Table 4.1 we
see that Rin for the MAX4240 is 45 M. This is not the case in the traditional differential
amplifier where the external resistor can significantly decrease the input resistance.
4.6
Design Examples
Design Example 4.12
We are asked to construct an amplifier that will reduce a very large input voltage (i.e., Vin
ranges between ;680 V) to a small output voltage in the range <5 V. Using only two resistors, we wish to design the best possible amplifier.
SOLUTION
Since we must reduce +680 V to -5 V, the use of an inverting amplifier seems
to be appropriate. The input/output relationship for the circuit shown in Fig. 4.26 is
Vo
R2
= Vin
R1
Since the circuit must reduce the voltage, R1 must be much larger than R2 . By trial and error,
one excellent choice for the resistor pair, selected from the standard Table 2.1, is R1 = 27 k
and R2 = 200 . For a Vin = 680 V, the resulting output voltage is 5.037 V, resulting in a
percent error of only 0.74%.
R2
Figure 4.26
A standard inverting
amplifier stage.
I
R1
–
±
Vin ±
–
+
Vo
-
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SECTION 4.6
DESIGN EXAMPLES
Design Example 4.13
There is a requirement to design a noninverting op-amp configuration with two resistors
under the following conditions: the gain must be +10, the input range is ;2 V, and the total
power consumed by the resistors must be less than 100 mW.
SOLUTION
For the standard noninverting configuration in Fig. 4.27a, the gain is
Vo
R2
= 1 +
Vin
R1
For a gain of 10, we find R2R1 = 9. If R1 = 3 k and R2 = 27 k, then the gain requirement
is met exactly. Obviously, a number of other choices can be made, from the standard Table 2.1,
with a 327 ratio. The power limitation can be formalized by referring to Fig. 4.27b where the
maximum input voltage (2 V) is applied. The total power dissipated by the resistors is
PR =
(20 - 2)2
22
4
324
+
=
+
6 0.1
R1
R2
R1
9R 1
The minimum value for R1 is 400 .
±
–
±
–
+
Vin ±
–
Vo
-
R2
±
–
2V
Vo=20 V
2V
R2
R1
+
Figure 4.27
The noninverting op-amp
configuration employed in
Example 4.13.
-
R1
(a)
(b)
Design Example 4.14
We wish to design a weighted-summer circuit that will produce the output
Vo = -0.9V1 - 0.1V2
The design specifications call for use of one op-amp and no more than three resistors.
Furthermore, we wish to minimize power while using resistors no larger than 10 k.
SOLUTION
A standard weighted-summer configuration is shown in Fig. 4.28. Our problem
is reduced to finding values for the three resistors in the network.
Using KCL, we can write
I1 + I2 = -
Vo
R
where
I1 =
V1
R1
and
I2 =
V2
R2
Combining there relationships yields
Vo = - c
R
R
d V1 - c d V2
R1
R2
Therefore, we require
R
= 0.9
R1
and
R
= 0.1
R2
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O P E R AT I O N A L A M P L I F I E R S
R1
I1
R
I2
V1 ±
–
V2 ±
–
R2
–
±
+
Vo
-
Figure 4.28 A standard weighted-summer configuration.
From these requirements, we see that the largest resistor is R2 and that R is the smallest. Also,
note that the RR1 ratio can be expressed as 2730. Finally, to minimize power, we should use
the largest possible resistor values. Based on this information, the best resistor values are
R = 270 , R1 = 300 , and R2 = 2.7 k, which yield the desired performance exactly.
Design Example 4.15
In Example 2.36, a 250- resistor was used to convert a current in the 4- to 20-mA range to
a voltage such that a 20-mA input produced a 5-V output. In this case, the minimum current
(4 mA) produces a resistor voltage of 1 V. Unfortunately, many control systems operate on
a 0- to 5-V range rather than a 1- to 5-V range. Let us design a new converter that will output
0 V at 4 mA and 5 V at 20 mA.
SOLUTION
The simple resistor circuit we designed in Example 2.36 is a good start. However,
the voltage span is only 4 V rather than the required 5 V, and the minimum value is not zero.
These facts imply that a new resistor value is needed and the output voltage should be shifted
down so that the minimum is zero. We begin by computing the necessary resistor value.
R =
Vmax - Vmin
5 - 0
=
= 312.5 Imax - Imin
0.02 - 0.004
The resistor voltage will now range from (0.004)(312.5) to (0.02)(312.5) or 1.25 to 6.25 V.
We must now design a circuit that shifts these voltage levels so that the range is 0 to 5 V. One
possible option for the level shifter circuit is the differential amplifier shown in Fig. 4.29.
Recall that the output voltage of this device is
Vo = (VI - Vshift)
R2
R1
Differential amplifier with shifter
R1
±
–
+
4 –20 mA
R VI
R2
R1
+
R2
± Vshift
–
-
Figure 4.29 A 4–20 mA to 0–5 V converter circuit.
Vo
-
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PROBLEMS
155
Since we have already chosen R for a voltage span of 5 V, the gain of the amplifier should be
1 (i.e., R1 = R2). Clearly, the value of the required shift voltage is 1.25 V. However, we can
verify this value by inserting the minimum values into this last equation
0 = [(312.5)(0.004) - Vshift]
R2
R1
and find
Vshift = (312.5)(0.004) = 1.25 V
There is one caveat to this design. We don’t want the converter resistor, R, to affect the differential amplifier, or vice versa. This means that the vast majority of the 4–20 mA current
should flow entirely through R and not through the differential amplifier resistors. If we
choose R1 and R2 W R, this requirement will be met. Therefore, we might select
R1 = R2 = 100 k so that their resistance values are more than 300 times that of R.
SUMMARY
●
Op-amps are characterized by
High-input resistance
Low-output resistance
Very high gain
●
●
Op-amp problems are typically analyzed by writing node
equations at the op-amp input terminals
●
The output of a comparator is dependent on the difference
in voltage at the input terminals
The ideal op-amp is modeled using
i+ = i- = 0
v+ = v-
PROBLEMS
PSV
CS
both available on the web at: http://www.justask4u.com/irwin8/
SECTION 4.2
4.1 An amplifier has a gain of 15 and the input waveform
shown in Fig. P4.1. Draw the output waveform.
vo (V)
12
vin (mV)
10
150
6
100
5
50
4
0
2
0.5
1
1.5
2.0
t (s)
–50
0
1
–2
–100
–4
–150
Figure P4.1
–6
–8
–10
4.2 An amplifier has a gain of -5 and the output waveform
shown in Fig. P4.2. Sketch the input waveform.
–12
Figure P4.2
2
3
4
5
6
7
8
9
t (ms)
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O P E R AT I O N A L A M P L I F I E R S
4.3 An op-amp base amplifier has supply voltages of ;5 V
and a gain of 20.
(a) Sketch the input waveform from the output waveform
in Fig. P4.3.
(a) Rin = q, Rout = 0, A o Z q.
(b) Rin = q, Rout 7 0, A o = q.
(b) Double the amplitude of your results in (a) and
sketch the new output waveform.
(c) Rin Z q, Rout = 0, A o = q.
4.6 Revisit the exact analysis of the inverting configuration in
Section 4.3.
vo (V)
(a) Find an expression for the gain if Rin = q, Rout = 0,
3
2
1
0
4.5 Revisit your answers in Problem 4.4 under the following
nonideal scenarios.
25
50
75
100
125
t (ms)
–1
A o Z q.
(b) Plot the ratio of the gain in (a) to the ideal gain versus
A o for 1 A o 1000 for an ideal gain of -10.
(c) From your plot, does the actual gain approach the
ideal value as A o increases or decreases?
(d) From your plot, what is the minimum value of A o if
the actual gain is within 5% of the ideal case?
4.7 Revisit the exact analysis of the inverting amplifier in
Section 4.3.
–2
(a) Find an expression for the voltage gain if Rin Z q,
–3
Rout = 0, A o Z q.
(b) For R2 = 27 k and R1 = 3 k, plot the ratio of the
Figure P4.3
4.4 For an ideal op-amp, the voltage gain and input resistance
are infinite while the output resistance is zero. What are the
consequences for
(a) the op-amp’s input voltage?
(b) the op-amp’s input currents?
actual gain to the ideal gain for A o = 1000 and
1 k Rin 100 k.
(c) From your plot, does the ratio approach unity as Rin
increases or decreases?
(d) From your plot in (b), what is the minimum value of
Rin if the gain ratio is to be at least 0.98?
4.8 An op-amp based amplifier has ;18 V supplies and a gain
of -80. Over what input range is the amplifier linear?
(c) the op-amp’s output current?
SECTION 4.3
4.9 Determine the gain of the amplifier in Fig. P4.9. What is
the value of Io ?
Vin
±
–
Io
Vo
R2
R2=20 kΩ
R1=3.3 kΩ
Vin=2 V
4.10 For the amplifier in Fig. P4.10, find the gain and Io .
±
–
Io
Vo
R2
R1
R1
VS
Figure P4.9
Figure P4.10
R2=20 kΩ
R1=3.3 kΩ
VS=2 V
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PROBLEMS
4.11 Using the ideal op-amp assumptions, determine the
values of Vo and I1 in Fig. P4.11.
I1
11 V
±
–
157
4.15 The op-amp in the amplifier in Fig. P4.15 operates with
;15 V supplies and can output no more than 200 mA.
What is the maximum gain allowable for the amplifier if
the maximum value of VS is 1 V?
Vo
VS
Io
±
–
R1+R2=10 kΩ
Vo
10 k
1 k
RL
R2
50 R1
Figure P4.11
4.12 Using the ideal op-amp assumptions, determine I1 , I2 ,
and I3 in Fig. P4.12. P S V
4.16 For the amplifier in Fig. P4.16, the maximum value of VS
is 2 V and the op-amp can deliver no more than 100 mA.
I3
R2
I1
(a) If ;10 V supplies are used, what is the maximum
I2 –
±
R1
1 mA
Figure P4.15
Vo
allowable value of R2 ?
(b) Repeat for ;3 V supplies.
(c) Discuss the impact of the supplies on the maximum
allowable gain.
Figure P4.12
4.13 In a useful application, the amplifier drives a load. The
circuit in Fig. P4.13 models this scenario.
VS
Io
±
–
Vo
RL=10 Ω
R1=100 Ω
(a) Sketch the gain Vo VS for 10 RL q.
(b) Sketch Io for 10 RL q if VS = 0.1 V.
(c) Repeat (b) if VS = 1.0 V.
R2
R1
RL
(d) What is the minimum value of RL if ∑Io∑ must be less
than 100 mA for ∑VS∑ 6 0.5 V?
(e) What is the current IS if RL is 100 ? Repeat for
RL = 10 k.
4.17 For the circuit in Fig. P4.17,
IS
Io
±
–
VS
Figure P4.16
Vo
R2=27 kΩ
R1=3 kΩ
R2
(a) find Vo in terms of V1 and V2 .
(b) If V1 = 2 V and V2 = 6 V, find Vo .
(c) If the op-amp supplies are ;12 V, and V1 = 4 V,
what is the allowable range of V2 ?
RL
V1
R1
±
–
V2
2 k
Figure P4.13
2 k
1 k
4.14 Repeat Problem 4.13 for the circuit in Fig. P4.14.
R2
VS
R2=27 kΩ
R1=3 kΩ
R1
Figure P4.14
+
Io
Vo
RL
Figure P4.17
PSV
Vo
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O P E R AT I O N A L A M P L I F I E R S
4.18 Find Vo in the circuit in Fig. P4.18 assuming the op-amp
is ideal.
1 k
10 5 k
2
+
±
–
4.22 Find Vo in the network in Fig. P4.22 and explain what
effect R1 has on the output.
+
+
1V
Vo
2V
-
±
–
10 +
Vo
R1
-
Figure P4.18
4.19 The network in Fig. P4.19 is a current-to-voltage converter
or transconductance amplifier. Find vo iS for this network.
1
Figure P4.22
4.23 Determine the expression for vo in the network in
Fig. P4.23.
R2
+
+
iS
vo
R1
vA
+
vB
vo
Figure P4.23
Figure P4.19
4.20 Calculate the transfer function io v1 for the network
shown in Fig. P4.20.
v1
4.24 Show that the output of the circuit in Fig. P4.24 is
Vo = c 1 +
R2
R2
d V1 V2
R1
R1
io
±
–
R3
V1
±
–
RF
Vo
RI
R1
Figure P4.20
4.21 Determine the relationship between v1 and io in the
circuit shown in Fig. P4.21.
RF
v1
V2
Figure P4.24
4.25 Find Vo in the network in Fig. P4.25.
RI
+
4
R1
io
1
Figure P4.21
R4
R2
5V
±
–
4V
±
–
+
+
Vo
-
Figure P4.25
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PROBLEMS
4.26 Find the voltage gain of the op-amp circuit shown in
Fig. P4.26.
4.30 Find Vo in the circuit in Fig. P4.30.
40 k
20 k
V1 ±
–
80 k
±
–
+
24 k
Vo
5 k
+
+
5 k
1 k
5V
±
–
4V
-
±
–
Vo
20 k
Figure P4.26
Figure P4.30
4.27 For the circuit in Fig. 4.27 find the value of R1 that
produces a voltage gain of 10.
4.31 Find Vo in the circuit in Fig. P4.31.
10 k
±
–
100 k
+
9V
V1 ±
–
20 k
±
–
+
30 k
Vo
18 k
R1
10 V
-
±
–
- 12 V
+
40 k
±
–
+
20 k
Vo
6V
-
Figure P4.27
Figure P4.31
4.28 Determine the relationship between vo and vin in the
circuit in Fig. P4.28.
4.32 Determine the expression for the output voltage, vo , of
the inverting summer circuit shown in Fig. P4.32.
R1
vin
±
–
vo
R2
RI
v1
v2
RF
v3
RF
R1
R2
+
R3
vo
Figure P4.28
Figure P4.32
4.29 In the network in Fig. P4.29 derive the expression for vo
in terms of the inputs v1 and v2 .
v1
v2
4.33 Determine the output voltage, vo , of the noninverting
averaging circuit shown in Fig. P4.33.
RF
R1
v1
+
R2
RI
vo
v2
v3
R1
R2
±
–
R3
Figure P4.29
RI
Figure P4.33
vo
RF
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O P E R AT I O N A L A M P L I F I E R S
4.34 Find the input/output relationship for the current
amplifier shown in Fig. P4.34.
4.37 Find the expression for vo in the differential amplifier
circuit shown in Fig. P4.37.
RF
RF
iin
+
io
±
–
R1
v1
RL
–
±
R1
v2
RI
vo
R1
–
±
R1
Figure P4.37
Figure P4.34
4.38 Find vo in the circuit in Fig. P4.38.
4.35 Find Vo in the circuit in Fig. P4.35.
PSV
PSV
R2
R3
80 k
R4
40 k
10 k
±
–
+
R1
v1
+
5V
–
±
vo
Figure P4.38
+
Vo
20 k
40 k
4.39 Find the output voltage, vo , in the circuit in Fig. P4.39.
Figure P4.35
R4
v1
4.36 Find vo in the circuit in Fig. P4.36.
+
-
R2
v1
R1
R2
R1
R1
±
–
–
±
R2
vo
v2
v2
–
±
R2
Figure P4.36
–
±
R3
R3
R4
Figure P4.39
vo
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PROBLEMS
4.40 The electronic ammeter in Example 4.9 has been modified
and is shown in Fig. P4.40. The selector switch allows the
user to change the range of the meter. Using values for R1
and R2 from Example 4.9, find the values of RA and RB
that will yield a 10-V output when the current being
measured is 100 mA and 10 mA, respectively.
4.47 Design an op-amp-based circuit to produce the function
Vo = 5 V1 - 4 V2
4.48 Design an op-amp-based circuit to produce the function
Vo = 5 V1 - 7 V2
±
–
I
Unknown
current
161
RA
RB
RC=1 k
Voltmeter
+
Vo
R2=9 k
Selector
switch
-
+
R1=1 k
-
Figure P4.40
4.41 Given a box of 10-k resistors and an op-amp, design a
circuit that will have an output voltage of
Vo = -2V1 - 4V2
4.42 Design an op-amp circuit that has a gain of -50 using
resistors no smaller than 1 k.
4.49 Show that the circuit in Fig. P4.49 can produce the output
Vo = K1 V1 - K2 V2
only for 0 K1 K2 + 1.
V1
R1
4.43 Design a two-stage op-amp network that has a gain
of -50,000 while drawing no current into its input
terminal. Use no resistors smaller than 1 k.
±
–
R2
R4
4.44 Design an op-amp circuit that has the following
input/output relationship:
R3
Vo = -5 V1 + 0.5 V2
4.45 A voltage waveform with a maximum value of
200 mV must be amplified to a maximum of 10 V
and inverted. However, the circuit that produces the
waveform can provide no more than 100 A. Design
the required amplifier.
4.46 An amplifier with a gain of ; 1% is needed. Using
resistor values from Table 2.1, design the amplifier.
Use as few resistors as possible.
Vo
V2
Figure P4.49
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4.50 A 170°C maximum temperature digester is used in a
paper mill to process wood chips that will eventually
become paper. As shown in Fig. P4.50a, three
electronic thermometers are placed along its length.
Each thermometer outputs 0 V at 0°C, and the voltage
changes 25 mV °C. We will use the average of the three
thermometer voltages to find an aggregate digester
temperature. Furthermore, 1 volt should appear at Vo for
every 10°C of average temperature. Design such an
averaging circuit using the op-amp configuration shown
in Fig. 4.50b if the final output voltage must be positive.
4.51 A 0.1- shunt resistor is used to measure current in a
fuel-cell circuit. The voltage drop across the shunt
resistor is to be used to measure the current in the circuit.
The maximum current is 20 A. Design the network
shown in Fig. P4.51 so that a voltmeter attached to the
output will read 0 volts when the current is 0 A and 20 V
when the current is 20 A. Be careful not to load the shunt
resistor, since loading will cause an inaccurate reading.
Fuel
cell
Paper mill digester
Thermometer #1
Thermometer #2
Load
0.1 Thermometer #3
Design
circuit
Voltmeter
(a)
Figure P4.51
R1
Rf
RB
R2
±
V1 –
R3
RA
–
±
+
±
V2 –
±
V3 –
–
±
+
Vx
Vo
-
-
(b)
Figure P4.50
4.52 Wood pulp is used to make paper in a paper mill.
The amount of lignin present in pulp is called the kappa
number. A very sophisticated instrument is used to
measure kappa, and the output of this instrument ranges
from 1 to 5 volts, where 1 volt represents a kappa
number of 12 and 5 volts represents a kappa number of
20. The pulp mill operator has asked to have a kappa
meter installed on his console. Design a circuit that will
employ as input the 1- to 5-volt signal and output the
kappa number. An electronics engineer in the plant has
suggested the circuit shown in Fig. P4.52.
24 V
±
–
24 V
±
–
24 V
±
–
Figure P4.52
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PROBLEMS
according to the plot in Fig. P4.54a, is available. The
problem then is to use this RTD to design a circuit that
employs this device as an input and produces a 0- to
12-V signal at the output, where 0 V corresponds to 0°C
and 12 V corresponds to 500°C. An engineer familiar
with this problem suggests the use of the circuit shown
in Fig. P4.54b in which the RTD bridge circuit provides
the input to a standard instrumentation amplifier.
Determine the component values in this network needed
to satisfy the design requirements.
4.53 An operator in a chemical plant would like to have a set
of indicator lights that indicate when a certain chemical
flow is between certain specific values. The operator
wants a RED light to indicate a flow of at least 10 GPM
(gallons per minute), RED and YELLOW lights to
indicate a flow of 60 GPM, and RED, YELLOW, and
GREEN lights to indicate a flow rate of 80 GPM. The
4–20 mA flow meter instrument outputs 4 mA when the
flow is zero and 20 mA when the flow rate is 100 GPM.
An experienced engineer has suggested the
circuit shown in Fig. P4.53. The 4–20 mA flow meter
and 250 resistor provide a 1–5 V signal, which serves
as one input for the three comparators. The light bulbs
will turn on when the negative input to a comparator is
higher than the positive input. Using this network, design
a circuit that will satisfy the operator’s requirements.
4–20 mA
600
Resistance in ohms
12 V
12 V
–
±
250 Green
R3
400
300
200
–
±
100
Yellow
R5
200
R4
400
500
Figure P4.54a
–
±
Red
R2
300
Temperature in °C
12 V
R1
500
100
12 V
12 V
163
v1
R6
Figure P4.53
4.54 An industrial plant has a requirement for a circuit that
uses as input the temperature of a vessel and outputs a
voltage proportional to the vessel’s temperature. The
vessel’s temperature ranges from 0°C to 500°C, and the
corresponding output of the circuit should range from 0
to 12 V. A RTD (resistive thermal device), which is a
linear device whose resistance changes with temperature
R1
±
–
RTD
±
–
R2
12 V
v2
R3
24 V
24 V
±
–
Vo (0–12V)
R5
R4
RG
R4
R5
Figure P4.54b
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O P E R AT I O N A L A M P L I F I E R S
TYPICAL PROBLEMS FOUND ON THE FE EXAM
4FE-1 Given the summing amplifier shown in Fig. 4PFE-1,
select the values of R 2 that will produce an output
voltage of –3 V.
R2
4 k
–
±
4V
±
–
+
12 k
–
±
Vo
2V
-
Figure 4PFE-1
4FE-2 Determine the output voltage V o of the summing
op-amp circuit shown in Fig. 4PFE-2.
18 k
36 k
6 k
6 k
–
±
2V
±
–
–
±
12 k
+
12 k
–
±
Figure 4PFE-2
1V
±
–
Vo
3V
-