G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Chapter 8 Instructor Notes
Chapter 8 introduces the notion of integrated circuit electronics through the most common building
block of electronic instrumentation, the operational amplifier. This is, in practice, the area of modern
electronics that is most likely to be encountered by a practicing non-electrical engineer. Thus, the aim of
the chapter is to present a fairly complete functional description of the operational amplifier, including a
discussion of the principal limitations of the op-amp and of the effects of these limitations on the
performance of op-amp circuits employed in measuring instruments and in signal conditioning circuits.
The material presented in this chapter lends itself particularly well to a series of laboratory experiments
(see for example1), which can be tied to the lecture material quite readily.
After a brief introduction, in which ideal amplifier characteristics are discussed, open- and closed- loop
models of the op-amp are presented in section 8.2; the use of these models is illustrated by application of
the basic circuit analysis methods of Chapters 2 and 3. Thus, the Instructor who deems it appropriate can
cover the first two sections in conjunction with the circuit analysis material. A brief, intuitive discussion of
feedback is also presented to explain some of the properties of the op-amp in a closed-loop configuration.
The closed-loop models include a fairly detailed introduction to the inverting, non-inverting and differential
amplifier circuits; however, the ultimate aim of this section is to ensure that the student is capable of
recognizing each of these three configurations, so as to be able to quickly determine the closed loop gain of
practical amplifier circuits, summarized in Table 8.1 (p. 402). The section is sprinkled with various
practical examples, introducing practical op-amp circuits that are actually used in practical instruments,
including the summing amplifier (p. 938), the voltage follower (p. 400)), a differential amplifier (Focus on
Measurements: Electrocardiogram (EKG) Amplifier, pp. 402-404), the instrumentation amplifier (pp. 404405), the level shifter (p.406), and a transducer calibration circuit (Focus on Measurements: Sensor
calibration circuit, pp. 407-409). Two features, new in the third edition, will assist the instructor in
introducing practical design considerations: the box Practical Op-Amp Design Considerations (p. 410)
illustrates some standard design procedures, providing an introduction to a later section on op-amp
limitations; the box Focus on Methodology: Using Op-amp Data Sheets (pp. 410-412) illustrates the use of
device data sheets for two common op-amps. The use of Device Data Sheets is introduced in this chapter
for the first time. In a survey course, the first two sections might be sufficient to introduce the device.
Section 8.3 presents the idea of active filters; this material can also be covered quite effectively
together with the frequency response material of Chapter 6 to reinforce these concepts. Section 8.4
discusses integrator and differentiator circuits, and presents a practical application of the op-amp integrator
in the charge amplifier (Focus on Measurements: Charge Amplifiers, pp. 420-421). The latter example is
of particular relevance to the non-electrical engineer, since charge amplifiers are used to amplify the output
of piezo-electric transducers in the measurement of strain, force, torque and pressure (for additional
material on piezo-electric transducers, see, for example2). A brief section (8.5) is also provided on analog
computers, since these devices are still used in control system design and evaluation.
Coverage of sections 8.4 and 8.5 is not required to complete section 8.6.
The last section of the chapter, 8.6, is devoted to a discussion of the principal performance limits of the
operational amplifier. Since the student will not be prepared to fully comprehend the reason for the
saturation, limited bandwidth, limited slew rate, and other shortcomings of practical op-amps, the section
focuses on describing the effects of these limitations, and on identifying the relevant parameters on the data
sheets of typical op-amps. Thus, the student is trained to recognize these limits, and to include them in the
design of practical amplifier circuits. Since some of these limitations are critical even in low frequency
applications, it is easy (and extremely useful) to supplement this material with laboratory exercises. The
box Focus on Methodology: Using Op-amp Data Sheets – Comparison of LM 741 and LMC 6061, pp. 441446) further reinforces the value of data sheets in realizing viable designs.
The homework problems present a variety of interesting problems at varying levels of difficulty; many
of these problems extend the ideas presented in the text, and present practical extensions of the circuits
discussed in the examples. In a one-semester course, Chapter 8 can serve as a very effective capstone to a
first course in circuit analysis and electronics by stimulating curiosity towards integrated circuit electronics,
1
2
Rizzoni, G., A Practical Introduction to Electronic Instrumentation, 3rd Ed. Kendall-Hunt, 1998
Doebelin E. O., Measurement Systems, McGraw-Hill, Fourth Edition, 1987.
8.1
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
and by motivating the student to pursue further study in electronics or instrumentation. In many respects
this chapter is the centerpiece of the book.
Learning Objectives
1. Understand the properties of ideal amplifiers, and the concepts of gain, input
impedance, and output impedance. Section 1.
2.
Understand the difference between open-loop and closed-loop op-amp configuration,
and compute the gain (or complete the design of) simple inverting, non-inverting,
summing and differential amplifiers using ideal op-amp analysis. Analyze more
advanced op-amp circuits, using ideal op-amp analysis, and identify important
performance parameters in op-amp data sheets. Section 2.
3.
Analyze and design simple active filters. Analyze and design ideal integrator and
differentiator circuits. Sections 3 and 4.
4.
Understand the structure and behavior of analog computers, and design analog
computer circuits to solve simple differential equations. Section 5.
5.
Understand the principal physical limitations of an op-amp. Section 6.
Section 8.1: Ideal Amplifiers
Problem 8.1
Solution:
Known quantities:
For the circuit shown in Figure P8.1:
G = Po
R s = 0.6 kΩ
R L = 0.6 kΩ
PS
Ro1 = 2 kΩ Ro 2 = 2 kΩ
Ri1 = 3 kΩ Ri 2 = 3 kΩ
Avo1 = 100
Find:
The power gain G
-1
G M 2 = 350 mΩ
Vi 2 = Vl 2
= Po/PS, in dB.
Analysis:
Starting from the last stage and going backward, we get
8.2
Vi1 = Vl1
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
V02
Ro 2 R L = 161.5 V
Ÿ
P
=
V
I
=
= 43.49 Vl 22
V O = GM 2 V l 2
l2
0
0 0
+
R
Ro 2 R L
L
Ri 2
Vl 2 = Av 01Vl1
= 60 Vl1 Ÿ P0 = 1.566 ⋅ 10 5 Vl12
Ri 2 + R01
Vl1 = VS
Ri1
= 0.833 VS Ÿ P0 = 1.0875 ⋅ 10 5 VS2
Ri1 + RS
I S = VS
1
= 2.778 ⋅10 −4 VS Ÿ PS = VS I S = 2.778 ⋅ 10 −4 VS2
Ri1 + RS
P0 1.0875 ⋅ 10 5
G=
=
= 3.915 ⋅ 10 8 = 20 dB Log10 [ 3.915 ⋅ 10 8 ] = 171.85 dB
−4
PS 2.778 ⋅10
______________________________________________________________________________________
Problem 8.2
Solution:
Known quantities:
The temperature sensor shown in Figure P8.2 produces a no load (i.e., sensor current = 0) voltage:
vs = V so cos ωt
R s = 400 Ω
V so = 500 mV
ω = 6.28 k
rad
s
The temperature is monitored on a display (the load) with a vertical line of light emitting diodes. Normal
conditions are indicated when a string of the bottommost diodes 2 cm in length are on. This requires that
a
voltage
be
supplied
to
the
display
input
terminals
where:
=
12
kΩ
=
cos
ω
t
=
6
V
.
vo V o
Vo
RL
The signal from the sensor must therefore be amplified. Therefore, a voltage amplifier is connected
between the sensor and CRT with: Ri = 2 kΩ
Ro = 3 kΩ .
Find:
The required no load gain of the amplifier.
Analysis:
The overall loaded voltage gain, using the amplitudes of the sensor voltage and the specified CRT voltage
must be:
V o = 6 V = 12
AV =
500 mV
V so
An expression for the overall voltage gain can also be obtained using two voltage divider relationships:
Ri ] R L
RL =
Avo [ V so
+
+
Ri R s R o + R L
Ro R L
[ 12 ] [ 0.4 + 2 ] [ 3 + 12 ]
[ + ] [ Ro + R L ]
= 18
=
= Av R s Ri
[ 2 ] [ 12 ]
Ri R L
V o = Avo V io
Vo
Avo =
V so
The loss in gain due to the two voltage divisions or "loading" is characteristic of all practical amplifiers.
Ideally, there is no reduction in gain due to loading. This would require an ideal signal source with a
source resistance equal to zero and a load resistance equal to infinity.
______________________________________________________________________________________
8.3
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Problem 8.3
Solution:
Known quantities:
For the circuit shown in Figure P8.3:
Po
Pi
Ri1 = 1.1 kΩ
G=
2
vI1
Pi =
Ri1
Ri 2 = 19 kΩ
R s = 0.7 kΩ
Ro1 = 2.9 kΩ
Avo1 = 65
Find:
The power gain G
R L = 16 Ω
Ro 2 = 22 Ω
G m 2 = 130 mS
= Po/Pi, in dB.
Analysis:
Ro 2 R L Ÿ 1.204 ⋅
vI 2
Ro 2 + R L
[ 65 ] [ 19 ]
vO
Ri 2
[ 1.204 ] = 67.91
=
] [1.204 ] Ÿ
VD : vO = [ Avo1 v I 1
2.9 + 19
vI1
Ro1 + Ri 2
OL : vO = G m 2 v I 2
2
vO
1100
v
= 317.1 ⋅10 3 = 20 dB Log10 [ 317.1 ⋅10 3 ] = 110.02 dB
G = R L2 = [ o ] 2 Ri1 = [ 67.91 ]2
16
vI1 RL
vI1
Ri1
______________________________________________________________________________________
Problem 8.4
Solution:
Known quantities:
For the circuit shown in Figure P8.4:
R s = 0.3 kΩ
R L = 2 kΩ
Ri1 = Ri 2 = 7.7 kΩ
Ro1 = Ro 2 = 1.3 kΩ
vO
= 149.9
vI1
Avo1 = Avo 2 = 17
Find:
a) The power gain in, dB.
b) The overall voltage gain vO/vS.
Analysis:
a)
vO2
7.7
= 86509 = 20 Log10 [ 86509 ] = 98.74 dB
G = PO = R L2 = [ vO ] 2 Ri1 = [ 149.9 ]2
2
vI1 R L
PI
vI1
Ri1
8.4
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
b) The input voltage of the first stage can be determined in terms of the source voltage using voltage
division:
Av =
vO vO1 v I 1 vO
=
=
vS v I 1 v S v I 1
vS
Ri1
Ri1 + R s
vS
= 149.9
7.7
= 144.3= 20 Log10 [ 144.3 ] = 43.18 dB
0.3 + 7.7
______________________________________________________________________________________
Problem 8.5
Solution:
Known quantities:
Figure P8.5.
Find:
What approximations are usually made about the voltages and currents shown for the ideal operational
amplifier [op-amp] model.
Analysis:
iP ≈ 0
iN ≈ 0
vD ≈ 0 .
______________________________________________________________________________________
Problem 8.6
Solution:
Known quantities:
Figure P8.6.
Find:
What approximations are usually made about the circuit components and parameters shown for the ideal
operational amplifier [op-amp] model.
Analysis:
µ ≈ ∞
ri ≈ ∞
ro ≈ 0 .
______________________________________________________________________________________
8.5
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Section 8.2: The Operational Amplifier
Problem 8.7
Solution:
Known quantities:
The circuit shown in Figure P8.7.
Find:
The resistor Rs that will accomplish the nominal gain requirement, and state what the maximum and
minimum values of Rs can be. Will a standard 5 percent tolerance resistor be adequate to satisfy this
requirement?
Analysis:
For a non-inverting amplifier the voltage gain is given by:
Av =
vout R f + Rs R f
=
=
+1 ;
vin
Rs
Rs
Avnom = 16 =
To find the maximum and minimum RS we note that
the minimum Av:
Rs max =
Rf
Rs
Rs ∝
+1 Ÿ
Rs =
Rf
16 − 1
= 1kΩ
1
, so to find the maximum RS we consider
Av
15 kΩ
= 1.02 kΩ .
16(1 − 0.02) − 1
15 kΩ
= 980Ω .
16(1 + 0.02) − 1
Since a standard 5% tolerance 1-kΩ resistor has resistance 950 < R < 1050, a standard resistor will not
Conversely, to find the minimum RS we consider the maximum Av:
Rs min =
suffice in this application.
______________________________________________________________________________________
Problem 8.8
Solution:
Known quantities:
The values of the two 10 percent tolerance resistors used in an inverting amplifier:
RF = 33 kΩ ; RS = 1.2 kΩ .
Find:
a. The nominal gain of the amplifier.
b.
The maximum value of
Av .
c.
The minimum value of
Av .
Analysis:
Rf
− 33
= −27.5 .
Rs
1.2
b. First we note that the gain of the amplifier is proportional to Rf and inversely proportional to RS. This
tells us that to find the maximum gain of the amplifier we consider the maximum Rf and the minimum
RS.
R f max 33 + 0.1(33)
Av max =
= 33.6 .
=
RS min 1.2 − 0.1(1.2)
a.
The gain of the inverting amplifier is:
Av = −
8.6
=
G. Rizzoni, Principles and Applications of Electrical Engineering
c.
To find
Av
we consider the opposite case:
min
Av
min
=
R f min
RS max
Problem solutions, Chapter 8
=
33 − 0.1(33)
= 22.5 .
1.2 + 0.1(1.2)
______________________________________________________________________________________
Problem 8.9
Solution:
Known quantities:
For the circuit shown in Figure P8.9, let
v1 (t ) = 10 + 10 −3 sin (ωt ) V , RF = 10 kΩ , Vbatt = 20 V .
Find:
a. The value of RS such that no DC voltage appears at the output.
b. The corresponding value of vout(t).
Analysis:
a. The circuit may be modeled as shown:
Applying the principle of superposition:
For the 20-V source:
vo
20
=
− 10kΩ
(20)
RS
For the 10-V source:
vo
10
§ 10kΩ ·
= ¨¨
+ 1¸¸(10)
R
¹
© S
The total DC output is:
Solving for RS
b.
vo
DC
= vo
20
+ vo
10
10,000
(10 - 20) = -10 Ÿ
RS
=−
§ 10,000 ·
10,000
(20) + ¨¨
+ 1¸¸(10) = 0
RS
¹
© RS
RS = 10 kΩ .
Since we have already determined RS such that the DC component of the output will be zero, we can
simply treat the amplifier as if the AC source were the only source present. Therefore,
§ Rf
·
+ 1¸¸ = 0.001sin (ωt ) ⋅ (1 + 1) = 2 ⋅10 −3 sin (ωt ) V .
vo (t ) = 0.001sin (ωt ) ⋅ ¨¨
© RS
¹
______________________________________________________________________________________
Problem 8.10
Solution:
Known quantities:
For the circuit shown in Figure P8.10, let:
R = 2 MΩ
RS = 1kΩ
AV (OL ) = 200,000 R0 = 50Ω
R1 = 1kΩ
R2 = 100kΩ
RLOAD = 10kΩ
8.7
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Find:
AV =
The gain
v0
vi
Analysis:
The op-amp has a very large input resistance, a very large "open loop gain" [ΑV(OL)], and a very small
output resistance. Therefore, it can be modeled with small error as an ideal op-amp.
The amplifier shown in Figure P8.10 is a noninverting amplifier, so we have
AV = 1 +
R2
100 ⋅103
= 1+
= 101
R1
1⋅103
______________________________________________________________________________________
Problem 8.11
Solution:
Known quantities:
vout (t ) = −(2 sin ω1t + 4 sin ω 2t + 8 sin ω 3t + 16 sin ω 4t ) V, RF = 5 kΩ .
Find:
Design an inverting summing amplifier to obtain
vout (t ) and determine the require source resistors.
Analysis:
The inverting summing amplifier is shown in the following figure.
By superposition and by selecting
4
vout = −¦
i =1
RSi =
RF
, vSi = sin ω i t , i = 1 4 , the output voltage is
2i
4
4
RF
vSi = −¦ 2i vSi = −¦ 2i sin ω i t
RSi
i =1
i =1
that coincides with the desired output voltage.
So, the required source resistors are
RS1 =
RF
R
R
R
= 2.5kΩ, RS 2 = F = 1.25kΩ, RS 3 = F = 625Ω, RS 4 = F = 312.5Ω .
2
4
8
16
______________________________________________________________________________________
Problem 8.12
Solution:
Known quantities:
For the circuit shown in Figure P8.12:
8.8
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
ri = 2 MΩ µ = 200,000 r0 = 25Ω
RS = 2.2kΩ R1 = 1kΩ RF = 8.7kΩ
RL = 20Ω
Find:
a. An expression for the input resistance vi/ii including the effects of the op-amp.
b. The value of the input resistance in including the effects of the op-amp.
c. The value of the input resistance with ideal op-amp.
Analysis:
a. The circuit in Figure P8.12 can be modeled as in the following figure
where
vi = −vd + R1 ii ,
vd = − ri (ii − iF ) ,
iF = −
vd + v0
,
RF
v0 = µ vd + r0 (iF −
v0
).
RL
Substituting the expression for vd in all the other equations, we obtain
vi = ( R1 + ri ) ii − ri iF ,
iF =
ri ii − v0
,
RF + ri
v0 (1 +
r0
) = − µ ri ii + ( µ ri + r0 )iF .
RL
By solving the second equation above for v0 and substituting in the third equation
vi = ( R1 + ri ) ii − ri iF ,
§
·
r
¨¨1 + 0 + µ ¸¸ri
© RL
¹
ii
iF =
§
r0 ·
¨¨1 +
¸¸( RF + ri ) + µ ri + r0
© RL ¹
and finally,
8.9
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
§
·
r
¨¨1 + 0 + µ ¸¸ ri 2
vi
© RL
¹
= ( R1 + ri ) −
ii
§
r ·
¨¨1 + 0 ¸¸( RF + ri ) + µ ri + r0
© RL ¹
b.
The value of the input resistance is
§
·
r
§ 25
·
¨¨1 + 0 + µ ¸¸ ri 2
+ 2 ⋅105 ¸ 4 ⋅1012
¨1 +
vi
© RL
¹
© 20
¹
= ( R1 + ri ) −
= 2.001 ⋅106 −
=
ii
§ 25 ·
§
r0 ·
6
11
¨1 + ¸ 2.0087 ⋅10 + 4 ⋅10 + 25
¨¨1 +
¸¸( RF + ri ) + µ ri + r0
© 20 ¹
© RL ¹
= 1.0001kΩ
c.
In the ideal case:
vi
= R1= 1kΩ
ii
______________________________________________________________________________________
Problem 8.13
Solution:
Known quantities:
For the circuit shown in Figure P8.13:
vS (t ) = 0.02 + 10 −3 cos(ωt ) V , RF = 220 kΩ , R1 = 47 kΩ , R2 = 1.8kΩ .
Find:
a) expression for the output voltage.
b) The corresponding value of vo(t).
Analysis:
a) the circuit is a noninverting amplifier, then
§ R ·
v0 = ¨¨1 + F ¸¸vS
R2 ¹
©
§ 220 ·
b) v0 = ¨1 +
¸vS = 2.464 + 0.1232 cos(ωt ) V
© 1 .8 ¹
______________________________________________________________________________________
Problem 8.14
Solution:
Known quantities:
For the circuit shown in Figure P8.13:
vS (t ) = 0.05 + 30 ⋅10 −3 cos(ωt ) V , RS = 50Ω , RL = 200Ω .
Find:
The output voltage vo.
8.10
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Analysis:
It is a particular case of a noninverting amplifier where v0=vS.
______________________________________________________________________________________
Problem 8.15
Solution:
Known quantities:
For the circuit shown in Figure P8.15:
vS 1 (t ) = 2.9 ⋅10−3 cos(ωt ) V , R1 = 1kΩ , R2 = 3.3kΩ
vS 2 (t ) = 3.1 ⋅10−3 cos(ωt ) V , R3 = 10kΩ , R4 = 18kΩ
.
Find:
The output voltage vo and a numerical value.
Analysis:
By using superposition,
R3
R4 § R3 ·
§ 10 · 18
¨¨1 + ¸¸vS 2 = ¨1 + ¸
vS 1 +
3.1 ⋅10 −3 cos(ωt ) − 2.9 ⋅10 −2 cos(ωt ) =
R1
R2 + R4 © R1 ¹
1 ¹ 18 + 3.3
©
= −1.83 ⋅10 −3 cos(ωt ) V
v0 = −
______________________________________________________________________________________
Problem 8.16
Solution:
Known quantities:
For the circuit shown in Figure P8.15:
vS1 = 5 mV , R1 = 1kΩ , R2 = 15kΩ
vS 2 = 7 mV , R3 = 72kΩ , R4 = 47 kΩ
Find:
The output voltage vo analytically and numerically.
Analysis:
From the solution of Problem 8.15,
v0 = −
R3
R4 § R3 ·
47
¨¨1 + ¸¸vS 2 = (1 + 72)
vS 1 +
7 ⋅10 −3 − 3.60 ⋅10 −1 = 27.37mV
R1
R2 + R4 © R1 ¹
15 + 47
______________________________________________________________________________________
Problem 8.17
Solution:
Known quantities:
If, in the circuit shown in Figure P8.17:
vS1 = v S 2 = 7 mV R1 = 850 Ω R2 = 1.5 kΩ R F = 2.2 kΩ
Op Amp : Motorola MC1741C
r i = 2 MΩ µ = 200,000 r o = 25 Ω
8.11
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Find:
a) The output voltage.
b) The voltage gain for the two input signals.
Analysis:
a) The op amp has a very large input resistance, a very large "open loop gain" [µ], and a very small
output resistance. Therefore, it can be modeled with small error as an ideal op amp with:
KVL : v D + v N = 0
Ÿ vN ≈ 0
vD ≈ 0
KCL :
v N - v S 2 + v N - v S 1 + v N - vO = 0
R2
R1
RF
≈ 0 iN ≈ 0
iN +
vN
2.2
2.2
] [ 7 mV ] + [ ] [ 7 mV ]
Ÿ vO = - R F v S 1 - R F v S 2 = [ 0.85
1.5
R1
R2
= [ - 2.588 ] [ 7 mV ] + [ - 1.467 ] [ 7 mV ] = - 28.38 mV
b) Using the results above: AV1 = -2.588
AV2 = -1.467.
Note: The output voltage and gain are not dependent on either the op amp parameters or the load
resistance. This result is extremely important in the majority of applications where amplification of a
signal is required.
______________________________________________________________________________________
Problem 8.18
Solution:
Known quantities:
For the circuit shown in Figure P8.15:
vS1 = kT1 , R1 = 11kΩ , R2 = 27 kΩ
vS 2 = kT2
, R3 = 33kΩ , R4 = 68kΩ
T1 = 35 C , T2 = 100 , C , k = 50mV / , C
,
Find:
a) The output voltage.
b) The conditions required for the output voltage to depend only on the difference between the two
temperatures.
Analysis:
a) From the solution of Problem 8.15,
33
R4 § R3 ·
33
68
R3
¨¨1 + ¸¸kT2 = §¨1 + ·¸
5 − 1.75 =
kT1 +
11
R2 + R4 © R1 ¹
R1
© 11 ¹ 68 + 27
= 9.065 V
v0 = −
b) For R3=R4, R1=R2= k R3 , the output voltage is
8.12
G. Rizzoni, Principles and Applications of Electrical Engineering
v0 = −
Problem solutions, Chapter 8
R3
R4 § R3 ·
1
R3 § R1 + R3 ·
¨¨1 + ¸¸ kT2 = − kT1 +
¨
¸ kT2 = T2 − T1
kT1 +
R1
R2 + R4 © R1 ¹
k
R1 + R3 ¨© R1 ¸¹
______________________________________________________________________________________
Problem 8.19
Solution:
Find:
In a differential amplifier, if: Av1 = - 20
Av 2 = + 22 , derive expressions for and then
determine the value of the common and differential mode gains.
Analysis:
There are several ways to do this. Using superposition:
vO-C
v S -C
v
= O- D
vS -D
If : v S - D = 0 : vO = vO-C + vO- D = vO-C
Ÿ
Av-C =
If : v S -C = 0 : vO = vO-C + vO- D = vO- D
Ÿ
Av- D
Assume that signal source #2 is connected to the non-inverting input of the op-amp. The common-mode
output voltage can be obtained using the gains given above but assuming the signal voltages have only a
common-mode component:
vO = v S1 Av1 + v S 2 Av 2
v S1 = v S -C -
1
vS -D
2
v S 2 = v S -C +
1
vS - D
2
Let : v S - D = 0
vO-C = v S -C Av1 + v S -C Av 2 = v S -C [ Av1 + Av 2 ] = v S -C Av-C
Ÿ Av-C = Av1 + Av 2 = [ - 20 ] + [ + 22 ] = 2
The difference-mode output voltage can be obtained using the gains given but assuming the signal voltage
have only a difference-mode component:
Let : v S -C = 0
1
1
1
vO- D = [ - v S - D ] Av1 + [ v S - D ] Av 2 = v S - D [ - Av1 + Av 2 ] = v S - D Av- D
2
2
2
1
1
Av- D = [ Av 2 - Av1 ] = ( [ + 22 ] - [ - 20 ] ) = + 21
2
2
Note: If signal source #1 were connected to the non-inverting input of the op amp, then the difference mode
gain would be the negative of that obtained above.
______________________________________________________________________________________
Problem 8.20
Solution:
Known quantities:
For the circuit shown in Figure P8.15:
vS1 = 1.3V , R1 = R2 = 4.7kΩ
vS 2 = 1.9V , R3 = R4 = 10kΩ
Find:
a) The output voltage.
b) The common-mode component of the output voltage.
8.13
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
c) The differential-mode component of the output voltage.
Analysis:
a) From the solution of Problem 8.15,
v0 = −
R4 § R3 ·
10000
R3
10000 ·
10000
¨¨1 + ¸¸vS 2 = §¨1 +
vS 1 +
1.9 −
1.3 = 1.28V
¸
R1
R2 + R4 © R1 ¹
4700 ¹ 4700 + 10000
4700
©
b) The common-mode component is zero.
c) The differential-mode component is v0.
______________________________________________________________________________________
Problem 8.21
Solution:
Known quantities:
For the circuit shown in Figure P8.15:
vS1, 2 = A + BP1,2
,
A = 0.3 , B = 0.7 V/psi
R1 = R2 = 4.7 kΩ , R3 = R4 = 10kΩ , RL = 1.8kΩ
P1 = 6kPa , P2 = 5kPa
Find:
a) The common-mode input voltage.
b) The differential-mode input voltage.
Analysis:
a) The common-mode input voltage is
vin+
3
vS 2 + vS 1
P1 + P2
-4 11 ⋅ 10
=
= A+ B
= 0.3 + 0.7 ⋅1.4504 ⋅10
= 0.858V
2
2
2
b) The differential-mode input voltage is
vin− = vS 2 − vS 1 = B( P2 − P1 ) = −0.7 ⋅1.4504 ⋅10 -4 ⋅103 = −0.1015V
______________________________________________________________________________________
Problem 8.22
Solution:
Known quantities:
A linear potentiometer [variable resistor] Rp is used to sense and give a signal voltage vY proportional to
the current y position of an x-y plotter. A reference signal vR is supplied by the software controlling the
plotter. The difference between these voltages must be amplified and supplied to a motor. The motor turns
and changes the position of the pen and the position of the "pot" until the signal voltage is equal to the
reference voltage [indicating the pen is in the desired position] and the motor voltage = 0. For proper
operation the motor voltage must be 10 times the difference between the signal and reference voltage. For
rotation in the proper direction, the motor voltage must be negative with respect to the signal voltage for the
polarities shown. An additional requirement is that iP = 0 to avoid "loading" the pot and causing an
erroneous signal voltage.
Find:
a) Design an op amp circuit which will achieve the specifications given. Redraw the circuit shown in
Figure P8.22 replacing the box [drawn with dotted lines] with your circuit. Be sure to show how the
signal voltage and output voltage are connected in your circuit.
b) Determine the value of each component in your circuit. The op amp is an MC1741.
8.14
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Analysis:
a) The output voltage to the motor must be dependent on the difference between two input voltages. A
difference amp is required. The signal voltage must be connected to the inverting input of the
amplifier. However, the feedback path is also connected to the inverting input and this will caused
loading of the input circuit, ie, cause an input current. This can be corrected by adding an isolation
stage [or voltage follower] between the input circuit and the inverting input of the 2nd stage. The
isolation stage must have a gain = 1 and an input current = 0. The difference amplifier, the second
stage, must give an output voltage:
§ 10 kΩ ·
+ 1¸¸(10 )
v0 10 = ¨¨
R
¹
© S
vM = 10 [ v R - vY ]= [ 10 ] v R + [ - 10 ] vY
Ÿ Avr = 10 Avy = - 10
The circuit configuration shown will satisfy these specifications.
b) In this first approximation analysis, assume the op amps can be modeled as ideal op amps:
Ÿ
vD ≈ 0
iP = i N ≈ 0
Consider the first or isolation or voltage follower stage:
Ideal : Ÿ i P ≈ 0 KVL : - vY + v D + vO1 = 0 v D ≈ 0
KVL : - v N - v D + v P = 0
Ÿ vN = vP
vD ≈ 0
KCL :
v N - vO1
R1
+
v N - vM
R2
+ iN = 0
-0
KCL : v P v R + v P + i P = 0
R3
R4
iN ≈ 0
iP ≈ 0
Ÿ
Ÿ
Ÿ
vO1 = vY
vO1 + v M
R1 R2 R1 R2
vN =
1
1 R1 R2
+
R1 R2
vR
R3 R4
R3
vP =
1
1 R3 R4
+
R3 R4
The second stage:
vO1 R2 + v M R1 = v R R4
R2 + R1 R2 + R1
R4 + R3
R2 ] + R4 [ R2 + R1 ] =
vM = vY [ vR
vY Avy + v R Avr
R1
R1 [ R4 + R3 ]
R2 = - 10
Choose : R 2 = 100 kΩ R1 = 10 kΩ
Avy = R1
R4 [ R2 + R1 ] = 10
Choose : R3 = 10 kΩ R4 = 100 kΩ
Avr =
R1 [ R4 + R3 ]
vO1 = vY
vN = vP
Ÿ
The resistances chosen are standard values and there is some commonality in the choices. Moderately
large values were chosen to reduce currents and the resistor power ratings. Cost of the resistors will be
determined primarily by power rating and tolerance.
______________________________________________________________________________________
8.15
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Problem 8.23
Solution:
Known quantities:
For the circuit shown in Figure P8.15:
vS1 = 13mV , R1 = 1kΩ , R2 = 13kΩ
vS 2 = 19 mV , R3 = 81kΩ , R4 = 56kΩ
Find:
The output voltage vo.
Analysis:
From the solution of Problem 8.15,
v0 =
R
56
R4 § R3 ·
¨¨1 + ¸¸vS 2 − 3 vS1 = (1 + 81)
19 ⋅10 −3 − 81⋅13 ⋅10 −3 = 0.211 V
R2 + R4 © R1 ¹
R1
56 + 13
______________________________________________________________________________________
Problem 8.24
Solution:
Known quantities:
The circuit shown in Figure P8.24.
Find:
Show that the current Iout through the light-emitting diode is proportional to the source voltage VS as long
as :
VS > 0.
Analysis:
Assume the op amp is ideal:
V − ≈ V + = VS ½°
¾ Ÿ if
°¿
I− ≈0
VS > 0
I out =
V − VS
=
.
R2 R2
______________________________________________________________________________________
Problem 8.25
Solution:
Known quantities:
The circuit shown in Figure P8.25.
Find:
Show that the voltage Vout is proportional to the current generated by the CdS solar cell. Show that the
transimpedance of the circuit Vout/Is is –R.
Analysis:
v + ≅ v − = 0 Ÿ Vout = − RI S
V
The transimpedance is given by Rtrans = out = − R
IS
Assuming an ideal op-amp :
______________________________________________________________________________________
8.16
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Problem 8.26
Solution:
Known quantities:
The op-amp voltmeter circuit shown in Figure P8.26 is required to measure a maximum input of E = 20
mV. The op-amp input current is IB = 0.2 µA, and the meter circuit has Im = 100 µA full-scale
deflection and
rm = 10 kΩ.
Find:
Determine suitable values for R3 and R4.
Analysis:
Vout = I m rm = (100µA )(10kΩ) = 1V
Vout R4 + R3
R
E E − Vout
From KCL at the inverting input,
+
= 0 or
=
= 1+ 4 .
E
R3
R3
R3
R4
R4 Vout
1
Then,
=
−1 =
− 1 = 49 . Now,
choose
R3
and
R4
such
R3
E
20 × 10 −3
E
IB =
≤ 0.2µ.
R3 (R4 + rm )
At the limit,
(
20 × 10 −3
R3 49 R3 + 10 ×10
3
)
that
= 0.2 × 10 −6 . Solving for R3, we have R3 ≈ 102kΩ .
Therefore, R4 ≈ 5MΩ .
______________________________________________________________________________________
Problem 8.27
Solution:
Known quantities:
Circuit in Figure P8.27.
Find:
The output voltage vo.
Analysis:
The circuit is a cascade of a noninverting op-amp with an inverting op-amp. Assuming ideal op-amps, the
input-output voltage gain is equal to the product of the single gains, therefore
8.17
G. Rizzoni, Principles and Applications of Electrical Engineering
v0 = −
Problem solutions, Chapter 8
RF 2 § RF 1 ·
¸ vS
¨1 +
RS 2 ¨© RS1 ¸¹
______________________________________________________________________________________
Problem 8.28
Solution:
Known quantities:
Circuit in Figure P8.27.
Find:
Select appropriate components using standard 5% resistor values to obtain a gain of magnitude
approximately equal to 1,000. How closely can you approximate the gain? Compute the error in the gain
assuming that the resistors have the nominal value.
Analysis:
From the solution of Problem 8.27, the gain is given by
AV = −
RF 2 § RF 1 ·
¸
¨1 +
RS 2 ¨© RS1 ¸¹
From Table 2.2, if we select
RS1 = 1.8kΩ , RS 2 = 1kΩ
RF 1 = 8.2kΩ , RF 2 = 180kΩ
we obtain
AV = −
RF 2 § RF 1 ·
180 § 1.8 + 8.2 ·
¸¸ = −
¨¨1 +
¨
¸ = −1000
RS 2 © RS 1 ¹
1 © 1.8 ¹
So, we can obtain a nominal error equal to zero.
______________________________________________________________________________________
Problem 8.29
Solution:
Known quantities:
Circuit in Figure P8.27.
Find:
Same as in Problem 8.28, but use the ±5% tolerance range to compute the possible range of gains for this
amplifier.
Analysis:
From the solution of Problem 8.27, the gain is given by
AV = −
RF 2 § RF 1 ·
¸
¨1 +
RS 2 ¨© RS1 ¸¹
From Table 2.2, if we select
RS1n = 1.8kΩ , RS 2 n = 1kΩ
RF 1n = 8.2kΩ , RF 2 n = 180kΩ
we obtain
8.18
G. Rizzoni, Principles and Applications of Electrical Engineering
AVn = −
Problem solutions, Chapter 8
RF 2 n § RF 1n ·
180 § 1.8 + 8.2 ·
¸¸ = −
¨¨1 +
¸ = −1000
¨
RS 2 n © RS1n ¹
1 © 1.8 ¹
So, we can obtain a nominal error equal to zero.
The maximum gain is given by
AV + = −
RF 2 § RF 1 ·
R (1 + 0.05) § RF 1n (1 + 0.05) ·
¸¸ = − F 2 n
¸ = −1200
¨¨1 +
¨1 +
RS 2 © RS1 ¹
RS 2 n (1 − 0.05) ¨© RS1n (1 − 0.05) ¸¹
while the minimum gain is
AV − = −
RF 2 § RF 1 ·
R (1 − 0.05) § RF 1n (1 − 0.05) ·
¸¸ = − F 2 n
¸ = −834
¨¨1 +
¨1 +
RS 2 © RS1 ¹
RS 2 n (1 + 0.05) ¨© RS 1n (1 + 0.05) ¸¹
______________________________________________________________________________________
Problem 8.30
Solution:
Known quantities:
For the circuit in Figure P8.30,
RL = 20kΩ , v0 = 0 ÷ 10 V
iinmax = 1mA
Find:
The resistance R such that
iinmax = 1mA .
Analysis:
iin =
v0
v0
10
Ÿ R = max = −3 = 10 kΩ
R
iinmax 10
______________________________________________________________________________________
Problem 8.31
Solution:
Known quantities:
Circuit in Figure P8.13.
Find:
Select appropriate components using standard 5% resistor values to obtain a gain of magnitude
approximately equal to 200. How closely can you approximate the gain? Compute the error in the gain
assuming that the resistors have the nominal value.
Analysis:
From the solution of Problem 8.13, the gain is given by
§ R ·
AV = ¨¨1 + F ¸¸
R2 ¹
©
From Table 2.2, if we select
R2 = 33Ω
RF = 6.8kΩ
we obtain
8.19
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
§ R ·
6800
AVn = ¨¨1 + F ¸¸ = 1 +
= 207
R2 ¹
33
©
The error in the gain is
εA =
V
207 − 200
= 3 .5 %
200
______________________________________________________________________________________
Problem 8.32
Solution:
Known quantities:
Circuit in Figure P8.13.
Find:
Same as in Problem 8.31, but use the ±5% tolerance range to compute the possible range of gains for this
amplifier.
Analysis:
From the solution of Problem 8.31, the gain is given by
§ R ·
AV = ¨¨1 + F ¸¸
R2 ¹
©
From Table 2.2, if we select
R2 n = 33Ω
RFn = 6.8kΩ
we obtain
§ R ·
6800
AVn = ¨¨1 + Fn ¸¸ = 1 +
= 207
33
© R2 n ¹
The maximum gain is given by
§ R (1 + 0.05) ·
6800 ⋅1.05
¸¸ = 1 +
AV + = ¨¨1 + Fn
= 228.75
33 * 0.95
© R2 n (1 − 0.05) ¹
while the minimum gain is
§ R (1 − 0.05) ·
6800 ⋅ 0.95
¸¸ = 1 +
AV − = ¨¨1 + Fn
= 187.44
33 *1.05
© R2 n (1 + 0.05) ¹
______________________________________________________________________________________
Problem 8.33
Solution:
Known quantities:
For the circuit in Figure P8.15,
R1 = R2
, R3 = R4
Find:
Select appropriate components using standard 1% resistor values to obtain a differential amplifier gain of
magnitude approximately equal to 100. How closely can you approximate the gain? Compute the error in
the gain assuming that the resistors the have nominal value.
Analysis:
From the solution of Problem 8.15, the gain is given by
8.20
G. Rizzoni, Principles and Applications of Electrical Engineering
AV =
Problem solutions, Chapter 8
R3
R1
From Table 2.2, if we select
R1 = 1kΩ , R3 = 100kΩ
we obtain
AV =
R3
= 100
R1
and the error for the gain with nominal values is zero.
______________________________________________________________________________________
Problem 8.34
Solution:
Known quantities:
For the circuit in Figure P8.15,
R1 = R2
, R3 = R4
Find:
Same as in Problem 8.33, but use the ±1% tolerance range to compute the possible range of gains for this
amplifier.
Analysis:
From the solution of Problem 8.33
AV =
R3
100(1 + 0.01)
Ÿ AV max =
= 102
R1
1(1 − 0.01)
AV =
R3
100(1 − 0.01)
Ÿ AV min =
= 98
R1
1(1 + 0.01)
______________________________________________________________________________________
8.21
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Section 8.3: Active Filters
Problem 8.35
Solution:
Known quantities:
For the circuit shown in Figure P8.35:
C = 1 µF
R = 10 kΩ
RL = 1kΩ .
Find:
a) The gain [in dB] in the pass band.
b) The cutoff frequency.
c) If this is a low or high pass filter.
Analysis:
a) Assume the op-amp is ideal. Determine the transfer function in the form:
1
V o [j ]
= Ho
1 + j f[ ]
V s [j ]
- Vn - Vd = 0
0
Vd
Vn - Vs +
Vn - Vo = 0
In +
1
R
j C
H v [j ] =
KVL :
KCL :
Vn
0
V o [j ]
= j RC
V s [j ]
This is not in the standard form desired but is the best that can be done. There are no cutoff
frequencies and no clearly defined pass band. The gain [i.e., the magnitude of the transfer function]
and output voltage increases continuously with frequency, at least
until the output voltage tries to exceed the DC supply voltages and
clipping occurs. In a normal high pass filter, the gain will increase
with frequency until the cutoff frequency is reached above which the
gain remains constant.
b) There is no cutoff frequency.
c) This filter is best called a high pass filter; however, since the output
will be clipped and severely distorted above some frequency, it is
not a particularly good high pass filter. It could even be called a
terrible filter with few redeeming graces.
______________________________________________________________________________________
In
0
Vn
0
H v [j ] =
Problem 8.36
Solution:
Known quantities:
For the circuit shown in Figure P8.36:
C = 1 µF
8.22
R1 = 1.8 kΩ
R2 = 8.2kΩ
RL = 333Ω .
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Find:
a) Whether the circuit is a low- or high-pass filter.
b) The gain V0 /VS in decibel in the passband.
c) The cutoff frequency.
Analysis:
a) From Figure 8.24, it results that the amplifier in Figure P8.36 is a high-pass filter.
In fact, the output voltage is
V0 ( jω ) = −
b)
c)
lim
ω →∞
jω CR2
VS ( jω )
1 + jω CR1
V0 ( jω )
R
8.2
= 20 Log 2 = 20 Log
= 13.17 dB
VS ( jω ) dB
R1
1.8
ω0 =
1
1
=
= 5555 rad/s
−6
CR1 1 ⋅10 ⋅1.8 ⋅103
______________________________________________________________________________________
Problem 8.37
Solution:
Known quantities:
For the circuit shown in Figure P8.36:
C = 200 pF
R1 = 10 kΩ
R2 = 220kΩ
RL = 1kΩ .
Find:
a) Whether the circuit is a low- or high-pass filter.
b) The gain V0 /VS in decibel in the passband.
c) The cutoff frequency.
Analysis:
a) From Figure 8.24, it results that the amplifier in Figure P8.36 is a high-pass filter.
In fact, the output voltage is
V0 ( jω ) = −
jω CR2
VS ( jω )
1 + jω CR1
b)
V0 ( jω )
R
220
= 20 Log 2 = 20 Log
= 26.84 dB
ω →∞ V ( jω )
R
10
1
S
dB
c)
ω0 =
lim
1
1
=
= 5 ⋅105 rad/s
3
−12
CR1 200 ⋅10 ⋅10 ⋅10
______________________________________________________________________________________
Problem 8.38
Solution:
Known quantities:
For the circuit shown in Figure P8.38:
C = 100 pF
R1 = 4.7 kΩ
R2 = 68kΩ
RL = 220kΩ .
Find:
Determine the cutoff frequencies and the magnitude of the voltage frequency response function at very low
and at very high frequencies.
Analysis:
The output voltage in the frequency domain is
8.23
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
§
·
¨
¸
R2
¨
¸ V ( jω ) = 1 + jω C ( R1 + R2 ) V ( jω )
V0 ( jω ) = 1 +
i
¨
1
¸ i
1 + jω CR1
+
R
1
¨
¸
jω C
©
¹
The circuit is a high-pass filter. The cutoff frequencies are
ω1 =
1
1
=
= 2.127 ⋅10 6 rad/s
3
−12
CR1 100 ⋅10 ⋅ 4.7 ⋅10
ω2 =
1
1
=
= 1.375 ⋅105 rad/s
−12
C ( R1 + R2 ) 100 ⋅10 ⋅ 72.7 ⋅103
For high frequencies we have
68
V0 ( jω )
R
= 1+ 2 = 1+
= 15.46
ω →∞ V ( jω )
4.7
R1
i
A∞ = lim
At low frequencies
A0 = lim
ω →0
V0 ( jω )
=1
Vi ( jω )
______________________________________________________________________________________
Problem 8.39
Solution:
Known quantities:
For the circuit shown in Figure P8.39:
Find:
a)
An expression for
H υ ( jω ) =
C = 20 nF
R1 = 1kΩ
R2 = 4.7kΩ
R3 = 80kΩ .
V0 ( jω )
.
Vi ( jω )
b) The cutoff frequencies.
c) The passband gain.
d) The Bode plot.
Analysis:
a) The frequency response is
H υ ( jω ) =
V0 ( jω )
= 1+
Vi ( jω )
b) The cutoff frequencies are
ω1 =
RR
1
1 + jω C 2 3
R2 + R3
jωC R2 + R3 + jωCR2 R3 § R3 ·
=
= ¨¨1 + ¸¸
R2
R2 + jωCR2 R3
© R2 ¹ 1 + jωCR3
R3 ||
1
1
=
= 625 rad/s
−9
CR3 20 ⋅10 ⋅ 80 ⋅103
( R2 + R3 )
84.7 ⋅103
ω2 =
= 11263 rad/s
=
C R2 R3
20 ⋅10 −9 ⋅ 80 ⋅103 ⋅ 4.7 ⋅103
c)
The passband gain is obtained by evaluating the frequency response at low frequencies,
A0 = lim Hυ ( jω ) = 1 +
ω →0
R3
80
= 1+
= 18
R2
4 .7
d) The magnitude Bode plot for the given amplifier is as shown.
8.24
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
______________________________________________________________________________________
Problem 8.40
Solution:
Known quantities:
For the circuit shown in Figure P8.40:
C = 0.47 µF
R1 = 9.1kΩ
R2 = 22kΩ
RL = 2.2kΩ .
Find:
a) Whether the circuit is a low- or high-pass filter.
b) An expression in standard form for the voltage transfer function.
c) The gain in decibels in the passband, that is, at the frequencies being passed by the filter, and the cutoff
frequency.
Analysis:
a) From Figure 8.21, it results that the amplifier in Figure P8.40 is a low-pass filter.
In fact, the output voltage is
R2
1
VS ( j ω )
R1 1 + jω CR 2
V ( jω )
R
1
H υ ( jω ) = 0
=− 2
Vi ( jω )
R1 1 + jω CR 2
V0 ( jω ) = −
b)
c)
The gain in decibel is obtained by evaluating
Hυ ( j 0) dB = 20 Log
Hυ ( jω ) at ω=0, i.e.
R2
22
= 20 Log
= 7.66 dB .
R1
9 .1
The cutoff frequency is
ω0 =
1
1
=
= 96.71 rad/s
−6
CR2 0.47 ⋅10 ⋅ 22 ⋅103
______________________________________________________________________________________
8.25
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Problem 8.41
Solution:
Known quantities:
For the circuit shown in Figure P8.40:
C = 0.47 nF
R1 = 2.2 kΩ
R2 = 68kΩ
RL = 1kΩ .
Find:
a) An expression in standard form for the voltage frequency response function.
b) The gain in decibels in the passband, that is, at the frequencies being passed by the filter, and the cutoff
frequency.
Analysis:
a) From Figure 8.21, it results that the amplifier in Figure P8.40 is a low-pass filter.
The voltage frequency response function is
H υ ( jω ) =
V0 ( jω )
R
1
=− 2
Vi ( jω )
R1 1 + jω CR 2
b) The gain in decibel is obtained by evaluating
Hυ ( j 0) dB = 20 Log
Hυ ( jω ) at ω=0, i.e.
R2
68
= 20 Log
= 29.8 dB .
R1
2 .2
The cutoff frequency is
ω0 =
1
1
=
= 31289 rad/s
−9
CR2 0.47 ⋅ 10 ⋅ 68 ⋅ 103
______________________________________________________________________________________
Problem 8.42
Solution:
Known quantities:
For the circuit shown in Figure P8.42:
C1 = C2 = 0.1 µF
R1 = R2 = 10 kΩ .
Find:
a) The passband gain.
b) The resonant frequency.
c) The cutoff frequencies.
d) The circuit Q.
e) The Bode plot.
Analysis:
a) From Figure 8.26, we have
ABP ( jω ) = −
Z2
jωC1 R2
jωC1 R1
jωC1 R1
=−
=−
=−
2
2
(1 + jω C1 R1 )(1 + jω C2 R2 ) (1 + jω C1 R1 )
Z1
( jω ) (C1 R1 ) 2 + 2 jω C1 R1 + 1
The magnitude of the frequency response is ABP ( jω ) =
ωC1 R1
2
1 + ω 2 (C1 R1 )
The passband gain is the maximum over ω of the magnitude of the frequency response, i.e.
ABP ( j
1
1
)=
C1 R1
2
b) The resonant frequency is
c)
ωn =
1
= 1000 rad/s
R1C1
The cutoff frequencies are obtained by solving with respect to ω the equation
8.26
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
ωC1 R1
1
=
Ÿ ω 2 − 2 2ω nω + ω n2 = 0 Ÿ ω1, 2 = ω n ( 2 ± 1)
2
2
1 + ω (C1 R1 )
2 2
ω1 = 2414 rad/s
ω 2 = 414 rad/s
ABP ( jω ) =
d) In this case ζ=1, which implies
e)
Q=
1
1
=
2ζ 2
The Bode plot of the frequency response is as shown.
______________________________________________________________________________________
Problem 8.43
Solution:
Known quantities:
For the circuit shown in Figure P8.43:
C = 0.47 nF
R1 = 220Ω
R2 = 68kΩ
RL = 1kΩ .
Find:
a) An expression in standard form for the voltage frequency response function.
b) The gain in decibels in the passband, that is, at the frequencies being passed by the filter, and the cutoff
frequency.
Analysis:
a) The voltage frequency response function is
H υ ( jω ) =
VO ( jω )
R
1
=− 2
R1 1 + jω CR2
Vi ( jω )
b) The gain in decibel is obtained by evaluating
Hυ ( j 0) dB = 20 Log
Hυ ( jω ) at ω=0, i.e.
R2
68000
= 20 Log
= 49.8 dB .
R1
220
The cutoff frequency is
ω0 =
1
1
=
= 31289 rad/s
−9
CR2 0.47 ⋅ 10 ⋅ 68 ⋅ 103
______________________________________________________________________________________
8.27
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Problem 8.44
Solution:
Known quantities:
For the circuit shown in Figure P8.44:
C1 = 2.2 µF C2 = 1 nF
R1 = 2.2kΩ
R2 = 100kΩ .
Find:
Determine the passband gain.
Analysis:
The voltage frequency response is
ABP ( jω ) = −
Z2
ωC1 R2
jωC1 R2
=−
Ÿ ABP ( jω ) =
2
2
2
Z1
(1 + jω C1 R1 )(1 + jω C2 R2 )
1 + ω (C1 R1 ) 1 + ω 2 (C2 R2 )
The cutoff frequencies are
ω1 =
1
1
=
= 206.6 rad/s
−6
C1R1 2.2 ⋅10 ⋅ 2.2 ⋅103
ω2 =
1
1
=
= 10000 rad/s
−9
C2 R2 1 ⋅10 ⋅100 ⋅103
The passband gain can be calculated approximately by evaluating the magnitude of the frequency response
at frequencies greater than ω1 and smaller than ω2, i.e.
ω1 << ω << ω 2 Ÿ 1 + ω 2 (C1 R1 ) ≈ ω 2 (C1 R1 )
ωC1 R2
R 100
= 2 =
= 45.45
ABP ≅
2
R1 2.2
ω 2 (C1 R1 )
2
2
, 1 + ω 2 (C2 R2 ) ≈ 1
2
______________________________________________________________________________________
Problem 8.45
Solution:
Known quantities:
For the circuit shown in Figure P8.45, let
C = C1 = C2 = 220 µF
Find:
Determine the frequency response.
Analysis:
With reference to the Figure shown below, we have
8.28
R = R1 = R2 = RS = 2kΩ .
G. Rizzoni, Principles and Applications of Electrical Engineering
I R 2 ( jω ) =
Problem solutions, Chapter 8
1
1
1
VO ( jω ) Ÿ VC 2 ( jω ) =
I R 2 ( jω ) =
VO ( jω ),
R2
jω C 2
jωC 2 R2
I R 1 ( jω ) = −
1
1
VC 2 ( jω ) = −
VO ( jω ),
R1
jωC 2 R2 R1
§
1 · 1
¨¨ R2 +
¸¸ VO ( jω )
ω
j
C
§
·
1
2 ¹ R2
¸ jωC 1 VO ( jω ),
= ¨¨1 +
I C1 ( jω ) = ©
1
jωC 2 R2 ¸¹
©
jω C 1
§
·
1
1
C
¸ VO ( jω ) =
I in ( jω ) = I C1 ( jω ) + I R 2 ( jω ) − I R1 ( jω ) = ¨¨ jωC 1+ 1 +
+
C 2 R2 R2 jωC 2 R2 R1 ¸¹
©
§2
1 ·
¸ VO ( jω ),
= ¨¨ + jωC +
jωCR 2 ¸¹
©R
Vin ( jω ) = − VC 2 ( jω ) − RS I in ( jω ) = −
§
2 ·
¸VO ( jω ) Ÿ
= −¨¨ 2 + jωCR +
jωCR ¸¹
©
V ( jω )
1
=−
H υ ( jω ) = O
Vin ( jω )
2 + jωCR +
§
1
1 ·
¸VO ( jω ) =
VO ( jω ) − ¨¨ 2 + jωCR +
jωCR
jωCR ¸¹
©
2
jωCR
=−
jωCR
( jω ) (CR )2 + 2 jωCR + 2
2
______________________________________________________________________________________
Problem 8.46
Solution:
Known quantities:
The inverting amplifier shown in Figure P8.46.
Find:
a) The frequency response of the circuit.
b) If R1 = R2 = 100 kΩ and C = 0.1 µF, compute the attenuation in dB at ω = 1,000
c) Compute gain and phase at ω = 2,500 rad/s.
d) Find range of frequencies over which the attenuation is less than 1 dB.
Analysis:
a) Applying KCL at the inverting terminal:
R2 +
1
jω C
vOUT
1 + jωR2 C
=−
=−
v IN
R1
jωR1C
b) Gain = 0.043 dB
o
c) Gain = 0.007 dB ; Phase = 177.71
d) To find the desired frequency range we need to solve the equation:
8.29
rad/s.
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
1 + jωR2 C
< 0.8913 since 20 log10 (0.8913) = −1dB .
jωR1C
This yields a quadratic equation in ω, which can be solved to find ω > 196.5 rad/s .
______________________________________________________________________________________
Problem 8.47
Solution:
Known quantities:
For the circuit shown in Figure P8.47, let
C = C1 = C2 = 100 µF
R1 = 3kΩ
R2 = 2kΩ .
Find:
Determine an expression for the gain.
Analysis:
With reference to the Figure shown above, we have
VC1 ( jω ) = VO ( jω ) Ÿ I C1 ( jω ) = jωC1 VO ( jω ),
VA ( jω ) = R2I C1 ( jω ) + VO ( jω ) = (1 + jωC1R2 )VO ( jω ),
I C 2 ( jω ) = jωC2 (VO ( jω ) − VA ( jω ) ) = −( jω ) C1C2 R2 VO ( jω ),
2
(
)
Vin ( jω ) = VA ( jω ) − R1 (I C 2 ( jω ) − I C1 ( jω ) ) = 1 + jωC1 ( R1 + R2 ) + ( jω ) C1C2 R1 R2 VO ( jω )
And finally, the expression for the gain is
Av ( jω ) =
2
VO ( jω )
1
1
=
=
2 2
Vin ( jω ) 1 + jωC ( R1 + R2 ) + ( jω ) C R1 R2 (1 + jωCR1 )(1 + jωCR2 )
______________________________________________________________________________________
Problem 8.48
Solution:
Known quantities:
The circuit shown in Figure P8.48.
Find:
Sketch the amplitude response of V2
ideal.
/ V1, indicating the half-power frequencies. Assume the op-amp is
8.30
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Analysis:
From KCL at the inverting input:
Vx − V1 Vx
+
=0
ZC
R
§
¨
1
Vx = V1 ¨
¨
1
¨ 1 + jωRC
©
§
1 ·
¸ = V1
Vx ¨¨1 +
jωRC ¸¹
©
1·
§
Vx ¨ jωC + ¸ = jωCV1
R¹
©
Similarly, from KCL at the output of the op-amp:
V2 − Vx V2
+
=0
R
ZC
§1
· V
V2 ¨ + jωC ¸ = x
Vx = V2 (1 + jωRC )
©R
¹ R
V2
jωRC
=
Combining the above results, we find
V1 (1 + jωRC )2
V2
ωRC
=
V1 1 + (ωRC )2
or
This function has the form of a band-pass filter, with maximum value determined as follows:
G=
V2
ωRC
=
V1 1 + (ωRC )2
[
]
[
dG 1 + (ωRC ) 2 RC − ωRC 2ω ( RC ) 2
=
2
dω
1 + (ωRC )2
[
]
]
Setting the derivative equal to zero and solving for the center frequency,
RC + ω 2 R 3C 3 − 2ω 2 R 3C 3 = 0
Then
ω=
1
1
= , and the half-power frequencies are given by:
1+1 2
1
1 1
=
Gmax =
ω 2 R 2 C 2 + 1 = 2 2ωRC
2
22
1
RC
Gmax =
ωRC
1 + (ωRC )2
ω=
1 − ω 2 R 2C 2 = 0
2 2 RC ± 8R 2 C 2 − 4 R 2 C 2
2 ±1
=
2 2
RC
2R C
The curve is sketched below.
8.31
R 2 C 2ω 2 − 2 2 RCω + 1 = 0
·
¸
¸
¸
¸
¹
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
______________________________________________________________________________________
Problem 8.49
Solution:
Known quantities:
The circuit shown in Figure P8.49.
Find:
Determine an analytical expression for the output voltage for the circuit shown in Figure P8.49. What kind
of filter does this circuit implement?
Analysis:
The resistance RF1 does not influence the output voltage, so
1
1
jωC F
VO ( jω ) = −
VS1 ( jω ) = −
VS 1 ( jω )
RS 1
jωC F RS1
The circuit is an integrator (low-pass filter).
______________________________________________________________________________________
Problem 8.50
Solution:
Known quantities:
The circuit shown in Figure P8.50.
Find:
Determine an analytical expression for the output voltage for the circuit shown in Figure P8.50. What kind
of filter does this circuit implement?
Analysis:
Figure P8.50 shows a noninverting amplifier, so the output voltage is given by
1
§
¨ RF +
jωC F
VO ( jω ) = ¨1 +
¨
1
¨
jωCS
©
·
¸
¸V ( jω ) = §¨1 + CS + jωC R ·¸ V ( jω )
S F ¸ S
¨ C
¸ S
F
©
¹
¸
¹
The filter is clearly a high-pass filter.
______________________________________________________________________________________
8.32
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Sections 8.4, 8.5: Integrator and Differentiator Circuits, Analog
Computers
Problem 8.51
Solution:
Known quantities:
Figure P8.50.
Find:
If: C = 1 µ F
R = 10 kΩ
function of time.
Analysis:
RL = 1kΩ , determine an expression for and plot the output voltage as a
KVL : - v N - v D = 0
vD ≈ 0 Ÿ vN ≈ 0
dvC = C d[ v N - v S ]
iC = C
dt
dt
d[ v N - v S ]
KCL : C
+ i N + v N vO = 0
dt
R
dv S
i N ≈ 0 v N ≈ 0 Ÿ vO = - RC
dt
The derivative is the slope of the curve for the source voltage, which is zero for:
t < 2.5 ms, 5 ms < t < 7.5 ms, and t > 15 ms
3 -0
3
−6
For 2.5 ms < t < 5 ms: vO = [ 10 ⋅ 10 ] [ 1 ⋅10 ]
= - 12 V
−3
5 ⋅10 - 2.5 ⋅ 10 −3
[ - 1.5 ] - [ + 3 ]
−3
=+6V
For 7.5 ms < t < 15 ms:
vO = - 10 ⋅10
15 ⋅10 −3 - 7.5 ⋅ 10 −3
______________________________________________________________________________________
Problem 8.52
Solution:
Known quantities:
Figure P8.52(a) and Figure P8.52(b).
Find:
If: C = 1 µ F
R = 10 kΩ RL = 1kΩ
a) An expression for the output voltage.
8.33
G. Rizzoni, Principles and Applications of Electrical Engineering
b) The value of the output voltage at t = 5, 7.5, 12.5, 15, and 20
a function of time.
Analysis:
a) As usual, assume the op amp is ideal so:
Problem solutions, Chapter 8
ms and a plot of the output voltage as
KVL : - v N - v D = 0
Ÿ vN ≈ 0
vD ≈ 0
d[ v N - vO ]
iC = C
dt
d[ v N - vO ]
KCL : v N v S + i N + C
=0
R
dt
1
Ÿ d vO = vN ≈ 0 iN ≈ 0
v S dt
RC
t
1 f
Integrating: vO [ t f ] = vO [ t i ] ³ v S dt
RC ti
b) Integrating gives the area under a curve. Recall area of triangle =
1/2[base x height] and area of rectangle = base x height.
Integrating the source voltage when it is constant gives an output
voltage which is a linear function of time.
Integrating the source voltage when it is a linear function of time
gives an output voltage which is a quadratic function of time.
1
1
rad
= 100
=
3
−6
s
RC
[ 10 ⋅ 10 ] [ 1 ⋅10 ]
1
-3
vO [5ms] = 0 V - [ 100 ] [ 2.5 ⋅10 ] [ 3 V ] = - 375 mV
2
-3
-3
vO [7.5ms] = - 375 ⋅10 - [ 100 ] [ 2.5 ⋅10 ] [ 3 ] = - 1125 mV
1
-3
-3
vO [12.5ms] = - 1124 ⋅10 - [ 100 ] [ 5 ⋅10 ] [ 3 ] = - 1875 mV
2
1
-3
-3
vO [15ms] = - 1875 ⋅10 - [ 100 ] [ 2.5 ⋅10 ] [ - 1.5 ] = - 1687.5 mV
2
-3
-3
vO [20ms] = - 1687.5 ⋅10 - [ 100 ] [ 5 ⋅10 ] [ - 1.5 ] = - 937.5 mV
______________________________________________________________________________________
Problem 8.53
Solution:
Known quantities:
In the circuit shown in Figure P8.53, the capacitor is initially uncharged, and the source voltage is:
vin (t ) = 10 ⋅10 −3 + sin (2,000πt ) V .
Find:
a) At t
= 0, the switch S1 is closed. How long does it take before clipping occurs at the output if
RS = 10 kΩ and C F = 0.008 µ F ?
b) At what times does the integration of the DC input cause the op-amp to saturate fully?
8.34
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Analysis:
vout = −
a)
=−
1
1 τ
vin (t )dt =−
³
³ [0.01 + sin (2000πt )]dt
RS C F
RS C F 0
1 τ
1 τ
0.01dt −
³
³ sin (2000πt )dt
RS C F 0
RS C F 0
The peak amplitude of the AC portion of the output is:
The output will begin to clip when:
vp =
1 § 1
¨
RS C F © 2000π
·
¸ = 1.989 V
¹
v0 (DC ) − v p = −15 V so we need to find at what time the
1 τ
³ 0.01dt = −13 V is satisfied. The answer is found below:
RS C F 0
13RS C F
1
−
0.01τ = −13 Ÿ τ =
= 104 ms
0.01
RS C F
15 RS C F
τ=
= 120 ms
b) Using the results obtained in part a.:
0.01
condition:
−
______________________________________________________________________________________
Problem 8.54
Solution:
Known quantities:
The circuit shown in Figure 8.21.
Find:
a) If RS = 10 kΩ , RF = 2 MΩ ,
C F = 0.008 µ F , and vS (t ) = 10 + sin (2,000πt ) V , find vout
using phasor analysis.
b) Repeat part a if RF = 200 kΩ , and if R F = 20 kΩ .
c) Compare the time constants with the period of the waveform for part a and b. What can you say about
the time constant and the ability of the circuit to integrate?
Analysis:
a) Replacing the circuit elements with the corresponding impedances:
Zf
vin
Zf =−
Rf
1 + jω R f C f
ZS
+
Z S = RS
For the signal component at
-
ω = 2,000π
:
8.35
vout
G. Rizzoni, Principles and Applications of Electrical Engineering
vout
Rf §
1
¨
=−
RS ¨© 1 + jωR f C f
= vin
200
(
(
§
·
·
¸
1
¸vin = −200¨¨
¸vin
¸
¨ 1 + jω
¸
¹
62.5 ¹
©
(
∠ 180 o − arctan ω
)
2
1+ ω
62.5
For the signal component at
Problem solutions, Chapter 8
ω =0
(
62.5
)) = 1.9839∠90.57
o
V
vout = −200vin = −2,000 V . Thus,
(DC):
)
vout (t ) = −2000 + 1.9839 sin 2,000πt + 90.57 o V ≈ −2000 + 2 cos(2,000πt ) V .
b) RF = 200 kΩ
ω = 2,000π :
·
Rf §
1
¨
¸vin = 1.9797∠95.68o V
vout = −
RS ¨© 1 + jωR f C f ¸¹
For the signal component at ω = 0 (DC): vout = −20vin = −200 V .
For the signal component at
(
)
Thus,
vout (t ) = −200 + 1.9797 sin 2,000πt + 95.68 V ≈ −200 + 2 cos(2,000πt ) V
RF = 20 kΩ
o
ω = 2,000π :
·
Rf §
1
¨
¸vin = 1.41∠134.8 o V
vout = −
RS ¨© 1 + jωR f C f ¸¹
For the signal component at ω = 0 (DC): vout = −2vin = −20 V .
For the signal component at
(
)
(
Thus,
)
vout (t ) = −20 + 1.41sin 2,000πt + 135 V ≈ −20 + 1.41cos 2,000πt + 45o V .
o
c)
Rf
τ
T
2 MΩ
16 ms
1 ms
200 kΩ
1.6 ms
1 ms
20 kΩ
0.16 ms
1 ms
In order to have an ideal integrator, it is desirable to have τ >> T.
______________________________________________________________________________________
Problem 8.55
Solution:
Known quantities:
v S (t ) = 10 ⋅10 −3 sin (2,000πt ) V ,
C S = 100 µ F , C F = 0.008 µ F , RF = 2 MΩ , and RS = 10 kΩ .
For the circuit of Figure 8.26, assume an ideal op-amp with
Find:
a)
The frequency response,
v0
( jω ) .
vS
b) Use superposition to find the actual output voltage (remember that DC = 0
8.36
Hz).
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Analysis:
a)
v (t)
S
Zf
:
ZS
Z S = RS +
+
vo (t)
Rf
1
Zf =
jω C S
1 + R f C f jω
Rf
Zf
1 + jω R f C f
jω R f C S
vo
=−
=−
=−
1
( jωRS C S + 1) jωR f C f + 1
vS
ZS
RS +
jω C S
jω
vo
5 ⋅10 −3
( jω ) = −
vS
( jω + 1)§¨ jω + 1·¸
© 62.5 ¹
v
j 0v S
= 0V
b) vo = v S o ( jω ) . By superposition,
vo 10 mV = vS
vS
1+1
(
vo ω =2000π =
)
j1.257 ⋅10 6 v S
= 20 ⋅10 −3 ∠ − 89.43o V
(1 + j 6283)(1 + j100)
(
)
vo (t ) = 20 ⋅10 −3 sin 2000πt − 89.43o V
We can say that the practical differentiator is a good approximation of the ideal differentiator.
______________________________________________________________________________________
Problem 8.56
Solution:
Find:
Derive the differential equation corresponding to the analog computer simulation circuit of Figure P8.56.
Analysis:
x(t ) = −200³ z dt or z = −
Therefore,
Therefore
1 dx
.Also, z = −20 y .
200 dt
dy
1 dx
.
Also, y = ³ (4 f (t ) + x (t )) dt or
= 4 f (t ) + x(t ) .
dt
4000 dt
1 d 2x
d 2x
= 4 f (t ) + x(t ) or
− 4000 x(t ) − 16000 f (t ) = 0 .
4000 dt 2
dt 2
y=
______________________________________________________________________________________
8.37
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Problem 8.57
Solution:
Find:
Construct the analog computer simulation corresponding to the following differential equation:
dx
d 2x
+ 100 + 10 x = −5 f (t ) .
2
dt
dt
Analysis:
d 2x
dt 2
-5 f(t)
-100
- dx
dt
x(t)
dx
dt
-100
-10
______________________________________________________________________________________
8.38
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Section 8.6: Physical Limitations of Operational Amplifiers
Problem 8.58
Solution:
Known quantities:
For the circuit shown in Figure 8.8:
RS = RF = 2.2kΩ .
Find:
Find the error introduced in the output voltage if the op-amp has an input offset voltage of 2 mV.
Assume zero bias currents and that the offset voltage appears as in Figure 8.48.
Analysis:
By superposition, the error is given by
§ R ·
∆vO = ¨¨1 + F ¸¸VOffset = 4mV
© RS ¹
______________________________________________________________________________________
Problem 8.59
Solution:
Known quantities:
For the circuit shown in Figure 8.8:
RS = RF = 2.2kΩ .
Find:
Repeat Problem 8.58 assuming that in addition to the input offset voltage, the op-amp has an input bias
current of 1 µA. Assume that the bias currents appear as in Figure 8.49.
Analysis:
By superposition, the effect of the bias currents on the output is
v+ = − RI B+ = v−
§
§ 1
1 · ·¸
¸¸ I B+
Ÿ ∆vO ,IB = RF ¨¨ I B− − R¨¨
+
¸
© RF RS ¹ ¹
©
−6
If R = RF || RS then ∆vO ,IB = RF (I B− − I B+ ) = − RF I OS = −2200 ⋅1 ⋅10 = −2.2mV
∆vO ,IB − v− v−
−
= I B−
RF
RS
The effect of the bias voltage is
§
R ·
∆ v O ,VB = ¨¨ 1 + F ¸¸V Offset = 4 mV
RS ¹
©
So, the total error on the output voltage is
∆vO = ∆vO ,VB + ∆vO ,IB = 1.8mV
______________________________________________________________________________________
Problem 8.60
Solution:
Known quantities:
The circuit shown in Figure P8.60.
The input bias currents are equal and the input bias voltage is zero.
8.39
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Find:
The value of Rx that eliminates the effect of the bias currents in the output voltage.
Analysis:
We can write
v+ = − Rx I B+ = v−
, I B+ = I B−
∆vO ,IB − v− v−
− = I B−
RF
R1
By selecting
§
§ 1
1· ·
Ÿ ∆vO ,IB = RF ¨¨ I B− − Rx ¨¨
+ ¸¸ I B+ ¸¸
© RF R1 ¹ ¹
©
Rx = RF || R1 , a zero output voltage error is obtained.
______________________________________________________________________________________
Problem 8.61
Solution:
Known quantities:
For the circuit shown in Figure P8.60:
RF = 3.3kΩ
R1 = 1kΩ vS = 1.5 sin(ωt ) V
Find:
The highest-frequency input that can be used without exceeding the slew rate limit of 1V/µs.
Analysis:
The maximum slope of a sinusoidal signal at the output of the amplifier is
S0 = A ⋅ ω =
V
1 6
RF
ω = 10 6 Ÿ ω max =
10 = 3.03 ⋅105 rad/s
s
3 .3
R1
______________________________________________________________________________________
Problem 8.62
Solution:
Known quantities:
The Bode plot shown in Figure 8.45:
A0 = 10 6 ω 0 = 10π rad/s
Find:
The approximate bandwidth of a circuit that uses the op-amp with a closed loop gain of A1 =75 and
A2 =350.
Analysis:
The product of gain and bandwidth in any given op-amp is constant, so
ω1 =
A0ω 0 107 π
=
= 4.186 ⋅105 rad/s
A1
75
ω2 =
A0ω 0 107 π
=
= 8.971⋅10 4 rad/s
A2
350
______________________________________________________________________________________
8.40
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Problem 8.63
Solution:
Known quantities:
For the practical charge amplifier circuit shown in Figure P8.63, the user is provided with a choice of three
time constants - τ long = RL C F , τ medium = RM C F , τ short = RS C F , which can be selected by means of
a switch. Assume that
RL = 10 MΩ , RM = 1MΩ , RS = 0.1MΩ , and C F = 0.1 µF .
Find:
Analyze the frequency response of the practical charge amplifier for each case, and determine the lowest
input frequency that can be amplified without excessive distortion for each case. Can this circuit amplify a
DC signal?
Analysis:
Applying KCL at the inverting terminal:
i+
or,
Vout
=0
1
R+
jω C
Vout
jωC
=−
i
1 + jωRC
This response is clearly that of a high-pass filter, therefore the charge amplifier will never be able to
amplify a DC signal.
The low end of the (magnitude) frequency response is plotted below for the three time constants. The figure
illustrates how as the time constant decreases the cut-off frequency moves to the right (solid line: R = 10
MΩ; dashed line: R =1 MΩ; dotted line: R = 0.1 MΩ).
-100
response of practical charge amplifier
-110
dB
-120
-130
-140
-150
-160
10-1
100
101
102
frequency, rad/s
From the frequency response plot one can approximate the minimum useful frequency for distortionless
response to be (nominally) 1
Hz for the 10 MΩ case, 10 Hz for the 1 MΩ case, and 100 Hz for the
0.1 MΩ case.
______________________________________________________________________________________
8.41
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Problem 8.64
Solution:
Find:
Consider a differential amplifier. We would desire the common-mode output to be less than 1% of the
differential-mode output. Find the minimum dB common-mode rejection ratio (CMRR) that fulfills this
requirement if the differential mode gain Adm = 1,000. Let
v1 = sin (2,000πt ) + 0.1sin (120π ) V
(
)
v2 = sin 2,000πt + 180 o + 0.1sin (120π ) V
§v +v ·
v0 = Adm (v1 − v2 ) + Acm ¨ 1 2 ¸
© 2 ¹
Analysis:
We first determine which is the common mode and which is the differential mode signal:
v1 − v2 = 2 sin (2,000πt )
v1 + v2
= 0.1sin (120πt )
2
Therefore, vout = Adm 2 sin (2000πt ) + Acm 2 sin (120πt )
Since we desire the common mode output to be less than 1% of the differential mode output, we require:
Acm (0.1) ≤ 0.01(2) or Acm ≤ 0.2 .
CMRR =
Adif
Acm
So
CMRRmin =
1000
= 5000 = 74 dB .
0 .2
______________________________________________________________________________________
Problem 8.65
Solution:
Known quantities:
As indicated in Figure P8.65, the rise time, tr, of the output waveform is defined as the time it takes for the
waveform to increase from 10% to 90% of its final value, i.e.,
t r ≡ t b − t a = −τ (ln 0.1 − ln 0.9) = 2.2τ , where τ is the circuit time constant.
Find:
Estimate the slew rate for the op-amp.
Analysis:
dv out
Vm (0.9 − 0.1)
15 × 0.8 V
V . Therefore, the slew rate is approximately
=
=
≈ 2.73
−6
4.4 µs
µs
dt max (14.5 − 10.1) × 10
2.73
V
.
µs
______________________________________________________________________________________
Problem 8.66
Solution:
Find:
5
Consider an inverting amplifier with open-loop gain 10 . With reference to Equation 8.18,
8.42
G. Rizzoni, Principles and Applications of Electrical Engineering
a)
If
Problem solutions, Chapter 8
RS = 10 kΩ and RF = 1 MΩ , find the voltage gain AV (CL ) .
RS = 10 kΩ and RF = 10 MΩ .
c) Repeat part a if RS = 10 kΩ and RF = 100 MΩ .
d) Using the resistors values of part c, find AV (CL ) if AV (OL ) → ∞ .
b) Repeat part a if
Analysis:
RF
RS
ACL = −
1
1+
R F + RS
RS AOL
ACL = - 99.899
b) ACL = - 990
c) ACL = -9091
d) As AOL → ∞ , ACL = −10,000 .
a)
______________________________________________________________________________________
Problem 8.67
Solution:
Known quantities:
Figure P8.67.
Find:
a) If the op-amp shown in Figure P8.67 has an open-loop gain of 45 X
105, find the closed-loop gain for
RS = RF = 7.5 kΩ .
b) Repeat part a if RF = 5 RS = 37.5 kΩ .
Analysis:
a)
Rf
RS
vv+
+
-
vin
vin = v +
and
(
v0 = AOL vin − v −
-
Writing KCL at v :
)
+
+ A (v+-v - )
OL
v out
-
·
§ v
Ÿ v − = −¨¨ 0 − vin ¸¸ .
¹
© AOL
v − − 0 v − − v0
+
=0.
RS
RF
Substituting,
− v0
− v0
+ vin
+ vin
AOL
AOL
v
+
= 0
RS
RF
RF
§
§ 1
1
1
1 ·
1 ·
¸¸ = −vin ¨¨
¸¸
Ÿ v0 ¨¨ −
−
−
+
© AOL RS AOL RF RF ¹
© RS RF ¹
8.43
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
§ 1
1 ·
¸
− ¨¨
+
RS RF ¸¹
§ 1
v0
1 ·
©
¸¸
=
= AOL RS RF ¨¨
+
1
1
1
vin
K
K
2 ¹
© 1
−
−
−
AOL RS AOL RF RF
2
where:
K1 = RF RS + RS + RS AOL RF
K 2 = RF RS + RF 2 + RS AOL RF
For the conditions of part a we obtain:
ACL =
b)
1
2
ACL = 5
45
5
×10 + 1
+
1
6
45
5
× 10 + 1
1
2
+
45
×105 + 1
= 1.999
1
6
45
§
· §
·
¨
¸ ¨
¸
v0 RF ¨
1
1
¸
¨
¸
Æ
=
+
¨
¸
¨
¸
+
+
R
R
R
R
vin RS
S
F
F
+1¸ ¨ S
+1¸
¨
© AOL RS
¹ © AOL RS
¹
× 10 5 + 1
= 5.999 .
______________________________________________________________________________________
Problem 8.68
Solution:
Find:
Given the unity-gain bandwidth for an ideal op-amp equal to 5.0 MHz, find the voltage gain at frequency
of
f = 500 kHz.
Analysis:
A0ω 0 = K = A1ω1
Ÿ K = 1× 2π × 5.0MHz = 10π × 10 6
A1 =
K 10π × 10 6
=
= 10,000
ω1 2π × 500
______________________________________________________________________________________
Problem 8.69
Solution:
Find:
Determine the relationship between a finite and frequency-dependent open-loop gain
closed-loop gain
AV (OL ) (ω ) and the
AV (CL ) (ω ) of an inverting amplifier as a function of frequency. Plot AV (CL ) versus ω.
Analysis:
As shown in Equation 8.84, if we consider a real op-amp:
8.44
G. Rizzoni, Principles and Applications of Electrical Engineering
A0
AV (OL ) (ω )
AV (CL ) (ω )
=
AC
1 + jω
1 + jω
ω0
, where AC
= - RF / RS, and A0 is the low-frequency open-loop gain.
ω1
If we choose A0
ω1 =
Problem solutions, Chapter 8
= 106, and ω0 = 10 π, and since A0ω 0 = ACω1 :
10 6 ⋅10π
.
RF
RS
Therefore:
20Log10 AVCL
20Log10 AC
ω1
Log10 ω
______________________________________________________________________________________
Problem 8.70
Solution:
Find:
A sinusoidal sound (pressure) wave impinges upon a condenser microphone of sensitivity S (mV / Pa).
The voltage output of the microphone vS is amplified by two cascaded inverting amplifiers to produce an
amplified signal v0. Determine the peak amplitude of the sound wave (in dB) if v0 = 5 VRMS. Estimate
the maximum peak magnitude of the sound wave in order that v0 not contain any saturation effects of the
op-amps.
Analysis:
p (t ) = P0 sin (ωt ) Pa ; v S (t ) = S ⋅ p (t ) mV ½
°
A1 A2
A1 A2
°
v0 (t ) =
v S (t ) V =
S ⋅ P0 sin (ωt ) V ¾ Ÿ A1 A2 ⋅ S ⋅ P0 = 5000 mV Ÿ
1000
1000
°
v0 = 5 VRMS
°¿
5000
dB
Ÿ P0 = 20 log10
A1 A2 ⋅ S
A1 A2 ⋅ S ⋅ P0
≤ 12 V Ÿ P0 ≤ 12,000 ⋅ A1 A2 ⋅ S Pa .
1000
______________________________________________________________________________________
8.45
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Problem 8.71
Solution:
Known quantities:
If, in the circuit shown in Figure P8.71
vS1 = 2.8 + 0.01cos(ωt ) V ; vS 2 = 3.5 − 0.007 cos(ωt ) V
Av1 = −13 ;
Av 2 = 10 ; ω = 4
krad
s
Find:
a) The common and difference mode input signals.
b) The common and differential mode gains.
c) The common and difference mode components of the output voltage.
d) The total output voltage.
e) The common mode rejection ratio.
Analysis:
a)
1
1
[ v S1 + v S 2 ] = ( [ 2.8 V + 10 mV cos ωt ] + [ 3.5 V - 7 mV cos ωt ] )=
2
2
= 3.15 V + 1.5 mV cos ωt
v S -C =
vS - D = vS1 - v S 2 = [ 2.8 V + 10 mV cos ωt ] - [ 3.5 V - 7 mV cos ωt ]=
= - 0.7 V + 17 mV cos ωt
Note that the expression for the differential input voltage (and the differential gain below) depends on
which of the two sources (in this case, vS1) is connected to the non-inverting input.
1
[ Av1 - Av 2 ] = - 11.5
2
vO-C = v S -C Avc = [ 3.15 V + 1.5 mV cos ωt ] [ - 3 ] = - 9.450 V - 4.5 mV cos ωt
c)
vO- D = v S - D Avd = [ - 0.7 V + 17 mV cos ωt ] [ - 11.5 ] = 8.050 V - 195.5 mV cos ωt
d) vO = vO-C + vO - D = - 1.4 V - 200.0 mV cos ωt
| Avc |
3
] = 20 dB Log10 [
] = - 11.67 dB .
e) CMRR = 20 dB Log10 [
| Avd |
11.5
b)
Avc = Av1 + Av 2 = - 3
Avd =
______________________________________________________________________________________
Problem 8.72
Solution:
Known quantities:
If, in the circuit shown in Figure P8.71:
vS1 = 3.5 + 0.01 cos ωt V
Avc = 10 dB Avd = 20 dB
v S 2 = 3.5 - 0.01 cos ωt
krad
ω = 4
s
Find:
a) The common and differential mode input voltages.
b) The voltage gains for vS1 and vS2.
c) The common mode component and differential mode components of the output voltage.
d) The common mode rejection ratio [CMRR] in dB.
8.46
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 8
Analysis:
a)
1
1
[ v S1 + v S 2 ] = ( [ 3.5 V + 10 mV cos ωt ] + [ 3.5 V - 10 mV cos ωt ] )= 3.5 V
2
2
= v S1 - v S 2 = [ 3.5 V + 10 mV cos ωt ] - [ 3.5 V - 10 mV cos ωt ] = 20 mV cos ωt
vS -C =
vS - D
Note the expression for the difference mode voltage depends on which signal [in this case, vS1] is
connected to the non-inverting input. This is also true for the expression for the differential gain
below.
b)
Avc =
10 dB
10 20 dB
20 dB
10 20 dB
= 3.162
= 10
Avd =
1
1
Ÿ
Avc = Av1 + Av 2
Av1 = Avc + Avd = [ 3.162 ] + 10 = 11.581
2
2
1
1
1
Ÿ
Avd = [ Av1 - Av 2 ]
Av 2 = Avc - Avd = [ 3.162 ] - 10 = - 8.419
2
2
2
vO-C = vS-C Avc = [ 3.5 V ] [ 3.162 ] = 11.07 V
c)
vO- D = v S - D Avd = [ 20 mV cosωt ] [ 10 ] = 200 mV cosωt
3.162
Avc | ] = 20 dB
d) CMRR = 20 dB Log10 [ |
] = - 10 dB .
Log10 [
10
Avd
______________________________________________________________________________________
Problem 8.73
Solution:
Known quantities:
If, in the circuit shown in Figure P8.73, the two voltage sources are temperature sensors with T
Temperature [Kelvin] and:
vS1 = k T 1
vS 2 = k T 2
V
Where : k = 120 µ
K
R1 = R3 = R4 = 5 kΩ R 2 = 3 kΩ
R L = 600 Ω
Find:
a) The voltage gains for the two input voltages.
b) The common mode and differential mode input voltage.
c) The common mode and difference mode gains.
d) The common mode component and the differential mode component of the output voltage.
e) The common mode rejection ratio [CMRR] in dB.
Analysis:
a)
8.47
=
G. Rizzoni, Principles and Applications of Electrical Engineering
Assume the op amp is ideal.:
v N - v S1
KVL : - v N - v D + v P = 0
Ÿ vN = vP
vD ≈ 0
v N - vO
= 0
iN ≈ 0 : Ÿ
KCL : i P + v P vS 2 + v P vO = 0
R4
R2
iP ≈ 0 : Ÿ
KCL : i N +
R1
+
Problem solutions, Chapter 8
R3
v S 1 vO
+
R
R3
1
vN =
1
1
+
R1 R3
v S 2 vO
+
R4
R
2
vP =
1
1
+
R 2 R4
R1 R3
R1 R3
R 2 R4
R 2 R4
v S1 R3 + vO R1 = v S 2 R4 + vO R2
R3 + R1
R4 + R 2
R4 + [ - R3 ]
vS 2
v S1
R
R3 + R1 [ R4 + R2 ] [ R3 + R1 ]
4 + R2
vO =
[ R4 + R2 ] [ R3 + R1 ]
R1 - R2
R3 + R1 R4 + R2
R4 [ R3 + R1 ]
R3 [ R4 + R 2 ]
= vS 2
+ v S1 [ ]
R1 [ R4 + R2 ] - R2 [ R3 + R1 ]
R1 [ R4 + R2 ] - R2 [ R3 + R1 ]
[ 5 ] [ 10 ]
R4 [ R3 + R1 ]
= 5
=
Av 2 =
[ 5 ] [ 8 ] - [ 3 ] [ 10 ]
R1 [ R4 + R2 ] - R2 [ R3 + R1 ]
- R3 [ R4 + R 2 ]
-[ 5 ][ 8 ]
=
=-4
Av1 =
[ 5 ] [ 8 ] - [ 3 ] [ 10 ]
R1 [ R4 + R2 ] - R2 [ R3 + R1 ]
Equating:
b)
V
] [ 310 K ] = 37.20 mV
K
V
vS 2 = k T 2 = [ 120 µ ] [ 335 K ] = 40.20 mV
K
1
1
vS -C = [ v S1 + v S 2 ] = [ 37.20 mV + 40.20 mV ] = 38.70 mV
2
2
vS - D = vS 2 - v S1 = 40.20 mV - 37.2 mV = + 3 mV
vS1 = k T 1 = [ 120 µ
Note that the expression for the differential mode voltage (and the differential mode gain below)
depends on Source #2 being connected to the non-inverting input.
c)
vO = v S 2 Av 2 + v S1 AS1 = [ v S -C +
1
1
v S - D ] Av 2 + [ v S -C - v S - D ] Av1
2
2
1
[ Av 2 - Av1 ] = v S -C Av-c + v S - D Avd
2
­ Avc = Av 2 + Av1 = 5 + [- 4] = 1
°
Ÿ®
1
1
°̄ Avd = 2 [ Av 2 - Av1 ] = 2 [ 5 - (- 4) ] = 4.5
= v S -C [ Av 2 + Av1 ] + v S - D
8.48
G. Rizzoni, Principles and Applications of Electrical Engineering
d)
e)
Problem solutions, Chapter 8
vO-C = v S -C Avc = [ 38.70 mV ] [ 1 ] = 38.70 mV
vO- D = v S - D Avd = [ 3 mV ] [ 4.5 ] = 13.5 mV
An ideal difference amplifier would eliminate all common mode output. This did not happen here. A
figure of merit for a differential amplifier is the Common Mode Rejection Ratio [CMRR]:
1
CMRR = 20 dB Log10 [ Avc ] = 20 dB Log10 [
] = - 13.06 dB .
4.5
Avd
______________________________________________________________________________________
Problem 8.74
Solution:
Known quantities:
If, for the differential amplifier shown in Figure P8.73:
vS1 = 13 mV v S 2 = 9 mV
v0C = 33 mV
v0 D = 18 V
v 0 = v0 C + v 0 D
Find:
a) The common mode gain.
b) The differential mode gain.
c) The common mode rejection ratio in dB.
Analysis:
a)
1
1
[ v S1 + vS 2 ] = [ 13 mV + 9 mV ] = 11 mV
2
2
33
mV
Ÿ AVC = vOC =
= 3
11 mV
v SC
vSC =
v SD = v S 2 - vS1 = 9 mV - 13 mV = - 4 mV
b)
18 V
v
Ÿ AVD = OD =
= - 4500
- 4 mV
v SD
| AVC |
c) CMRR = 20 dB Log10 [
] = - 63.52 dB .
| AVD |
______________________________________________________________________________________
8.49