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 EE8451 LINEAR INTEGRATED CIRCUITS AND APPLICATIONS

LTPC

3003

UNIT I IC FABRICATION

IC classification, fundamental of monolithic IC technology, epitaxial growth, masking and etching,

diffusion of impurities. Realisation of monolithic ICs and packaging. Fabrication of diodes, capacitance,
resistance, FETs and PV Cell.

UNIT II CHARACTERISTICS OF OPAMP

Ideal OP-AMP characteristics, DC characteristics, AC characteristics, differential amplifier; frequency

response of OP-AMP; Basic applications of op-amp – Inverting and Non-inverting Amplifiers, summer,
differentiator and integrator-V/I & I/V converters.

UNIT III APPLICATIONS OF OPAMP

Instrumentation amplifier and its applications for transducer Bridge, Log and Antilog Amplifiers- Analog
multiplier & Divider, first and second order active filters, comparators, multivibrators, waveform
generators, clippers, clampers, peak detector, S/H circuit, D/A converter (R- 2R ladder and weighted
resistor types), A/D converters using opamps.

UNIT IV SPECIAL ICs

Functional block, characteristics of 555 Timer and its PWM application - IC-566 voltage controlled
oscillator IC; 565-phase locked loop IC, AD633 Analog multiplier ICs.

UNIT V APPLICATION ICs

AD623 Instrumentation Amplifier and its application as load cell weight measurement - IC voltage
regulators –LM78XX, LM79XX; Fixed voltage regulators its application as Linear power supply - LM317,
723 Variability voltage regulators, switching regulator- SMPS - ICL 8038 function generator IC.
TEXT BOOKS:

1. David A. Bell, ‘Op-amp & Linear ICs’, Oxford, 2013.

2. D. Roy Choudhary, Sheil B. Jani, ‘Linear Integrated Circuits’, II edition, New Age, 2003.

3. Ramakant A.Gayakward, ‘Op-amps and Linear Integrated Circuits’, IV edition, Pearson Education, 2003
/ PHI. 2000.

REFERENCES

1. Fiore,”Opamps & Linear Integrated Circuits Concepts & applications”, Cengage, 2010.

2. Floyd ,Buchla,”Fundamentals of Analog Circuits, Pearson, 2013.

3. Jacob Millman, Christos C.Halkias, ‘Integrated Electronics - Analog and Digital circuits system’,
McGraw Hill, 2003.

4. Robert F.Coughlin, Fredrick F. Driscoll, ‘Op-amp and Linear ICs’, Pearson, 6th edition,2012.

5. Sergio Franco, ‘Design with Operational Amplifiers and Analog Integrated Circuits’, Mc Graw Hill,
2016.

6. Muhammad H. Rashid,’ Microelectronic Circuits Analysis and Design’ Cengage Learning, 2011.
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SCAD Group of Institutions

Department of Electrical and Electronics Engineering
Detailed Lesson Plan
Name of the Subject& Code: EE 6303 & LINEAR INTEGRATED CIRCUITS
AND APPLICATIONS
Name of the Faculty:
TEXT BOOKS:
1. David A. Bell, ‘Op-amp & Linear ICs’, Oxford, 2013.
2. D. Roy Choudhary, Sheil B.Jani, ‘Linear Integrated Circuits’, II edition, New
Age, 2003.

ww 3. Ramakant A.Gayakward, ‘Op-amps and Linear Integrated Circuits’, IV

edition, Pearson Education, 2003, PHI. 2000.

w.E
REFERENCES:
1. Fiore,‘Op-amps & Linear Integrated Circuits Concepts &

asy
Applications’,Cengage,2010.

En
2. Floyd , Buchla, ‘Fundamentals of Analog Circuits’, Pearson, 2013.
3. Jacob Millman, Christos C.Halkias, ‘Integrated Electronics - Analog and

gin
Digital circuits system’, Tata McGraw Hill, 2003.
4. Robert F.Coughlin, Fredrick F. Driscoll, ‘Op-amp and Linear ICs’, PHI

Sl.
ee
Learning, 6th edition, 2012.
Hours
rin Cumulative Books
No
Unit Topic / Portions to be Covered Required /
Planned g.n Hrs Referred

1 I IC classification
UNIT I IC FABRICATION
1
et 1 T2
Fundamental of monolithic IC
2 I 1 2 T2
technology
3 I Epitaxial growth 1 3 T2
4 I Masking and etching 1 4 T2
5 I Diffusion of impurities 1 5 T2
Realization of monolithic ICs and
6 I 1 6 T2
packaging
7 I Fabrication of diodes 1 7 T2

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8 I Fabrication of capacitance 1 8 T2
9 I Fabrication of resistance 1 9 T2
10 I Fabrication of FETs 1 10 T2
UNIT II CHARACTERISTICS OF OPAMP
11 II Ideal OP-AMP characteristics 1 11 T2
12 II DC characteristics 1 12 T2
13 II AC characteristics 1 13 T2
14 II Differential amplifier 1 14 T2
15 II Frequency response of OP-AMP 1 15 T2
Basic applications of op-amp – Inverting

ww
16 II
and Non-inverting Amplifiers
2 17 T2

17
18w.EII
II
V/I & I/V converters
Summer, differentiator and integrator
1
2
18
20
T2
T2

19 III asy UNIT III

Instrumentation amplifier
APPLICATIONS OF OPAMP
1 21 T2
20
21
III
III En
Log and Antilog Amplifiers
First and second order active filters
1
1
22
23
T2
T2
22 III Comparators gin 1 24 T2
23
24
III
III
Multivibrators
Waveform generators ee
Clippers, clampers, peak detector, S/H
1
1
rin
25
26
T2
T2

25 III
circuit
2
g.n 28 T2

26

27
III

III
D/A converter (R- 2R ladder and
weighted resistor types)
A/D converters using op-amps
1

1
et 29

30
T2

T2
UNIT IV SPECIAL ICs
28 IV Functional block of 555 Timer 1 31 T2
29 IV Characteristics of 555 Timer 1 32 T2
30 IV Application circuits with 555 Timer 2 34 T2
31 IV IC-566 voltage controlled oscillator IC 1 35 T2
32 IV IC 565-phase lock loop IC 1 36 T2
33 IV Analog multiplier ICs 2 38 T2
UNIT V APPLICATION ICs

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34 V IC voltage regulators –LM78XX,79XX 2 40 T2

35 V Fixed voltage regulators 1 41 T2
36 V LM317, 723 Variable voltage regulators 2 43 T2
37 V Switching regulator 1 44 T2
38 V SMPS 1 45 T2
39 V LM 380 power amplifier 1 46 T2
40 V ICL 8038 function generator IC 1 47 T2

Faculty Incharge HoD

ww
w.E
asy
En
gin
ee rin
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et

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INDEX
Unit No Q.NO Content Page No
Cover page 1
Syllabus 2-3
Lesson plan 4-6
I 1-12 Part A 9-11
I 1-5 Part B 11-25
Steps involved in fabrication of
I 1 11-14
IC
Fabrication of diodes and

ww I 2
capacitors
14-17

w.EI
I
3
4
Fabrication of Resistors & FET
Photolithography
17-21
21-23
I
II
5

asy
1-15
Different IC packages
Part A
24-25
26-28
II
II
1-6
1 EnPart B
Dc characteristics of Op-amp
28-52
28-33
II 2 gin
Ac characteristics of Op-amp 33-37
II
II
3
4 ee
Differentiator & Integrator
Differential amplifier
Inverting and Non-inverting rin
37-41
42-45

II 5
amplifier
g.n
45-48

II
III
III
6
1-10
1-5
Applications of Op-amp
Part A
Part B
et
48-52
53-54
55-68
III 1 Instrumentation amplifier 55-58
III 2 Schmitt trigger 58-60
III 3 R-2R DAC 60-62
III 4 Successive approx. type ADC 62-64
III 5 II order LPF 64-68
IV 1-15 Part A 69-70
IV 1-4 Part B 71-79
IV 1 Monostable multivibrator 71-72

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IV 2 Astable multivibrator 73-74

IV 3 PLL 75-77
IV 4 VCO 77-79
V 1-15 Part A 80-83
V 1-5 Part B 83-98
V 1 LM 380 power amplifier 83-87
IC 723 general purpose
V 2 87-88
regulator
V 3 IC 8038 Function generator 89-91
V 4 LM 317 IC 92-94

wwV 5 Optocoupler IC 94-98

w.E University question papers 99-118

asy
En
gin
ee rin
g.n
et

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UNIT I
IC FABRICATION
Part - A
1. Define an Integrated circuit.
An integrated circuit(IC) is a miniature, low cost electronic circuit consisting of
active and passive components fabricated together on a single crystal of silicon. The
active components are transistors and diodes and passive components are resistors
and
capacitors
2. What are the basic processes involved in fabricating ICs using planar

ww
technology? (April/May 2015)
1. Silicon wafer (substrate) preparation

w.E
2. Epitaxial growth
3. Oxidation
4. Photolithography
asy
5. Diffusion
6. Ion implantation En
7. Isolation technique
8. Metallization
gin
9. Assembly processing & packaging
ee
3. List out the steps used in the preparation of Si – wafers.
rin
1. Crystal growth &doping
2. Ingot trimming & grinding g.n
3. Ingot slicing
4. Wafer policing & etching
et
5. Wafer cleaning
4. Write the basic chemical reaction in the Epitaxial growth process of pure
silicon.
The basic chemical reaction in the Epitaxial growth process of pure silicon is the
Hydrogen reduction of silicon tetrachloride
1200oC
SiCl 4 + 2H2 <-----------> Si + 4 HCl

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5. What are the two important properties of SiO2?

SiO2 is an extremely hard protective coating & is unaffected by almost all
reagents except by hydrochloric acid. Thus it stands against any
contamination.
By selective etching of SiO2, diffusion of impurities through carefully
defined windows in the SiO2 can be accomplished to fabricate various
components.
6. Explain the process of oxidation.
The silicon wafers are stacked up in a quartz boat & then inserted into quartz
furnace tube. The Si wafers are raised to a high temperature in the range of 950 to

ww
1150oC & at the same time, exposed to a gas containing O2 or H2O or both. The
chemical action is

w.E Si + 2H2O -----------> SiO2 + 2H2

7. What is lithography?

asy
Lithography is a process by which the pattern appearing on the mask is

En
transferred to the wafer. It involves two steps: the first step requires applying a few
drops of photo resist to the surface of the wafer & the second step is spinning the

gin
surface to get an even coating of the photo resist across the surface of the wafer.
8. What are the two processes involved in photolithography?
a) Making a photographic mask :
ee rin
The development of photographic mask involves the preparation of initial

g.n
artwork and its reduction, decomposition of initial artwork or layout into several mask
layers.
b) Photo etching :
et
Photo etching is used for the removal of SiO2 from desired regions so that the
desired impurities can be diffused.
9. Define diffusion. (April/May 2015)
The process of introducing impurities into selected regions of a silicon wafer
is called diffusion. The rate at which various impurities diffuse into the silicon will be of
the order of 1µm/hr at the temperature range of 900oC to 1100o C .The impurity atoms
have the tendency to move from regions of higher concentrations to lower
concentrations.

10

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10. What are the advantages

ages of ion implantation technique? (April/Ma
ril/May 2015)
• It is performed at low te
temperature. Therefore, previously diffused
sed regions
r have
a lesser tendency for lateral
late spreading.
• In diffusion process,
s, tem
temperature has to be controlled over a large area inside
the oven, whereas in io
ion implantation process, accelerating poten
potential & beam
content are dielectrically
rically controlled from outside.
11. What are the advantages
tages of IC over discrete components? (April/May
(Apri 2015)
(Nov/Dec 2014) (Nov/Dec 201
c 2016)
• Miniaturization and hence
henc increased equipment density.
• Cost reduction due
e to ba
batch processing.

ww • Increased system reliab

reliability due to elimination of soldered joints.

w.E • Increased functional

al performance.
• Increased operating
per
spe
g speed.

asy
• Reduction in powerr consumption
12. Name the different types
cons
ypes IIC packages?
There are three different
• Metal can En
nt pac
packages available. They are
n pack
package
• Ceramic flat pagin
package

Part -B
• Dual-in-line

1. With the help off nea

ne pa
ee
package

neat sketches completely describe vari rin

various stages
involved in the fabricat
brication of ICs
g.n
(OR)
R)
With neat diagram
m ex
explain the steps involved in the fabrica
circuit shown in figure
igure by using IC technology. et
abrication of the

The various steps in the fabric

fabrication of monolithic IC is described below
1. Wafer preparation 2. Epitaxial
Epi growth 3 .Oxidation
4. Isolation diffusion 5. Bas
Base diffusion 6.Emitter diffusion
7. Aluminum metallization
ation
11

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i. Wafer Preparation:
• The starting material called
ca the substrate is a p-type silicon wafer. The wafers
are usually 10cm diamet
iameter and 0.4mm thickness.
• The resistivity is 10Ω/cm
Ω/cm corresponding to the concentration of ac
acceptor atom
NA=1.4X1015atoms/cm

Step 1

p – Type substrate
su
10 Ω – cm rresistivity NA = 1.4 x 1015 atoms/cm3 400 µm

ww Fig 1.1 Wafer preparation

w.E
ii. Epitaxial growth:
•
asy
An n-type epitaxiall film is grown on the p-type as shown in fig. This
becomes the collector
ctor re
Th ultimately
region of the transistor or an elementt of th
the diode and

En
diffused capacitor assoc
associated with the circuit.

gin
N – epi layer 0.1 – 0.5 Ω - cm 5 – 25 µm

p – Typ
ee
ype substrate 10 Ω - cm

rin
iii. Oxidation
Fig 1.2 Epitaxial growth
g.n
•

Step 3
A SiO2 layer of thickness
layer.
kness of the order 0.02-2µcm is grown on the n
et
he n-epitaxial

N – epi layer
yer 0.1
0. – 0.5 Ω - cm 5 – 25 µm

p – Type substr
substrate 10 Ω - cm

Fig 1.3 Oxidation

12

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iv. Isolation layer:

• In the circuit, four components
compo have to be fabricated, so we require
equired four island
which are isolated(four
(four components
c are resistor, transistor, capacito
pacitor, diode)

ww • For this, SiO2 is removed

Fig 1.4 Isolation layer
moved from five different places using Photolitho
tolithography.

w.E • The wafer is next subjected

subjec
that p-type impurities
to heavy p-type diffusion for a long
ties penetrate
pe
ng time
tim interval so
the n-type epitaxial layer and reach
reac the p-type

v.
substrate.
Base Diffusion: asy
•
En
A new SiO2 layer is grown
gr over entire pattern and a new pattern
attern of opening

•
formed using photolitho
Now p-type impurities gin
tolithography technique.
rities such as Boron is diffused into the region
regio of n-type
epitaxial silicon the
penetrate through n-laye ee
e diffusion
diffu of p-type silicon should be such that it should not
layer to the substrate.
rin
g.n
et
Fig 1.5 Base diffusion
vi. Emitter Diffusion :
• A new type of SiO2 layer
laye is grown over the entire wafer and selec
selectively etched
to open a new set of win
windows and n-type impurity is diffused throug
through them.
• This forms the transisto
nsistor emitter and cathode region.

13

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Fig 1.6 Emitter diffusion

The reason for using
ng hea
heavily doped n-region can be explained as follows:
fo
• Aluminum normal
ormally used for making interconnection, is a p-type
p impurity

ww in silicon and
nd can
ca produce an unwanted rectifying contac
doped n-material.
terial.
ontact with lightly

w.E • avy cconcentration of Phosphorous (n+) layer

However heavy
ohmic contact
ct with
er makes
ma
wit Al-layer. Hence it is desirable to use
a good
e heavily
heav doped n-

vii.
region.
asy
Aluminium Metallization:
tion:
• Now the IC chip
En
ip is complete
c with all active and passive devic
devices and only

•
interconnection betwe
Then a thin coating
ting ogin
between the various components have to be
e made.
ma
of Al is vacuum deposited over the entire
tire su
surface of the

•
wafer.
The interconnection
ee
ction between the components is then formed
rin
forme by photo

•
resistive techniques.
ques.
The undesired Al areas
are are etched away leaving a pattern g.n
rn interconnection
inte
between transistor,
tor, re
et
resistor, diode, and capacitor is shown below.
below

Fig 1.7 Aluminium metallization

14

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2. Explain the various methods of fabricating diodes and capacitors in

monolithic integrated circuits. (April/May 2015)
Fabrication of Diode:
• Diodes find extensive use in Integrated Circuits. The integrated diode is
used as Schottky Barrier Diode. The other name of this diode is metal
semiconductor diode.
• The metal to semiconductor junction can be ohmic as well as rectifying. The
ohmic contact is used when a load is to be attached to a semiconductor
device. This rectifying conduct is called as a metal semiconductor diode.
• Aluminum is a p-type impurity in silicon. The ohmic contact will be formed

ww between Al and n-type silicon and no pn junction is formed.

w.E
•

•
The metal contacts are required to be ohmic and no PN junctions to be
formed between the metal and silicon layers.
The N+ diffusion region serves the purpose of generating ohmic contacts.
• asy
On the other hand, if aluminum is deposited directly on the N-type silicon,

• En
then a metal semiconductor diode can be said to be formed.
Such a metal semiconductor diode junction exhibits the same type of V-I

gin
Characteristics as that of an ordinary PN junction.

ee rin
g.n
et
• The cross sectional view and symbol of a Schottky barrier diode as shown
in figure. Contact 1 shown in figure is a Schottky barrier and the contact 2 is
an ohmic contact.
• The contact potential between the semiconductor and the metal generated
a barrier for the flow of conducting electrons from semiconductor to metal.

15

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• When the junction is forward biased this barrier is lowered and the electron
flow is allowed from semiconductor to metal, where the electrons are in
large quantities.
• The majority carriers carry the conduction current in the Schottky diode
whereas in the PN junction diode, minority carriers carry the conduction
current and it incurs an appreciable time delay from ON state to OFF state.
• This is due to the fact that the minority carriers stored in the junction have
to be totally removed.
Integrated Capacitor:
• Monolithic capacitors are not frequently used in integrated circuits since they

ww are limited in the range of values obtained and their performance.

w.E
• The capacitance is proportional to the area of the junction and inversely
proportional to the depletion thickness.
C α A, where a is the area of the junction and

asy
C α T, where t is the thickness of the depletion layer

i) En
There are, however, two types available,
The junction capacitor (ii)MOS and thin film capacitor
Junction Capacitor:
gin
•
ee
In monolithic ICs junction capacitor is a reverse biased PN junction formed

rin
by the collector-base or emitter-base diffusion of the transistor. Figure
shows the cross sectional view the junction capacitor and the equivalent
circuit.
g.n
et

Fig 1.9 Junction capacitor & its equivalent circuit

• There are two junctions in the diffused capacitor. They are J1 and J2. The
two diodes are idealized diodes.

16

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• The parasitic capacitance C1 is inevitable due to the junction J1 between n-

type epitaxial layer and the substrate.
• The substrate must be held at the most negative point in the circuit to
minimize C1.
• During the reverse biased condition J2 will produce the desired capacitance.
• The value of the capacitance C2 will depends upon the area of the junction,
impurity concentration of the n-type epitaxial layer and the voltage across
the junction.
• The capacitor C2 is polarized and is obtained only when the junction J2 is
reverse biased.

ww MOS and Thin film capacitor:

w.E • Commonly used capacitor is the metal oxide semiconductor capacitor,

the cross sectional view and the equivalent circuit is,

asy
En
gin
ee rin
•
Fig 1.10 MOS and Thin film capacitor
g.n
It is basically a parallel plate capacitor with SiO2 as the dielectric. During

et
emitter diffusion the heavily doped n+ region formed in the lower plate.
The thin film Al metallization is formed in upper plate of the capacitor with
SiO2 as the dielectric.
• In the equivalent circuit, the parasitic effect consist of a small series
resistance R due to n+ region, a collector substrate junction J1 and its
associated capacitance C1.
Thin film Capacitors:
• Thin film capacitors structures used in thin dielectric film layer between two
metal layers.

17

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• It requires additional masking and deposition steps beyond the MOS

structure. In thin film structure aluminum or tantalum is used as capacitor
plates Al2O5 or Ta2O5 as dielectric material.
• Ta2O5 is preferred for large value of capacitors. This is a destructive and
irreversible failure mechanism and may require over voltage protection.
Disadvantage:
• Thin film capacitor fails when the voltage rating exceeds due to breakdown
of the dielectric.
3. Completely describe in detail the various stages involved in the
fabrication of a R and FET in a single chip. (April/May 2015)

ww
Integrated Resistors:
A resistor in a monolithic integrated circuit is obtained by utilizing the bulk

w.E •
resistivity of the diffused volume of semiconductor region. The commonly
used methods for fabricating integrated resistors are
a. Diffused resistor
asy
b. Epitaxial resistor
c. Pinched resistor En
d. Thin film techniques.
gin
a. Diffused Resistor:
• ee
The diffused resistor is formed in any one of the isolated regions of epitaxial
layer during base or emitter diffusion processes. rin
•
the bipolar transistor fabrication.
g.n
This type of resistor fabrication is very economical as it runs in parallel to

•
to realize the monolithic resistor.
et
The N-type emitter diffusion and P-type base diffusion are commonly used

• The diffused resistor has a severe limitation in that, only small valued
resistors can be fabricated.
• The surface geometry such as the length, width and the diffused impurity
profile determine the resistance value.
• The commonly used parameter for defining this resistance is called the
sheet resistance. It is defined as the resistance in ohms/square offered by
the diffused area.

18

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• In the monolithic
ic res
resistor, the resistance value is expressed
ed by R = Rs L/W,
where R= resistanc
istance offered (in ohms), Rs = sheet resistance
resist of the
particular fabrication
ication process involved (in ohms/square), L = length
l of the
diffused area and W = width of the diffused area.
• The sheet resistance
istance of the base and emitter diffusion in 200
200Ω/square and
2.2Ω/square respect
spectively.

b. Epitaxial Resistor:

ww
w.E
asy Fig 1.11 Epitaxial resistor
• The N-epitaxial layer
En
yer ca
figure shows the cross
can be used for realizing large resistance
ance values. The
cross-sectional view of the epitaxial resistor
tor fo
formed in the
epitaxial layer between gin
t two N+ aluminum metal contacts.
een the
c. Pinched resistor:
ee rin
g.n
et
Fig 1.12 Pinched resistor
• The sheet resistance
nce o
offered by the diffusion regions can be increased
in by
narrowing down itss cros
cross-sectional area. This type of resistance
tance is normally
achieved in the base
se reg
region. Figure shows a pinched base diffused
fused resistor.
• It can offer resistance
nce o
of the order of mega ohms in a comparati
paratively smaller
area.

19

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• In the structure shown, no current can flow in the N-type material since the
diode realized at contact 2 is biased in reversed direction.
• Only very small reverse saturation current can flow in conduction path for the
current has been reduced or pinched.
• Therefore, the resistance between the contact 1 and 2 increases as the width
narrows down and hence it acts as a pinched resistor.

d. Thin film resistor:

ww
w.E
asy Fig 1.13 Thin film resistor
•
En
The thin film deposition technique can also be used for the fabrication of

•
monolithic resistors.
gin
A very thin metallic film of thickness less than 1µm is deposited on the silicon

•
ee
dioxide layer by vapour deposition techniques.

rin
Normally, Nichrome (NiCr) is used for this process. Desired geometry is

• g.n
achieved using masked etching processes to obtain suitable value of resistors.
Ohmic contacts are made using aluminum metallization as discussed in earlier

•
sections.
et
The cross-sectional view of a thin film resistor as shown in figure. Sheet
resistances of 40 to 400Ω/ square can be easily obtained in this method and
thus 20kΩ to 50kΩ values are very practical.
Fabrication of FET:
Unipolar monolithic ICs use JFET or MOSFET as active device .The fabrication
techniques of (i). JFET
(ii). MOSFET
(iii) CMOS is discussed below
JFET Fabrication:
The basic process of JFET Fabrication is same as in BJT fabrication

20

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Fig 1.14 JFET fabrication

• The epitaxial layer whic
which formed the collector of the BJT is used
use as the n-

ww channel of the JFET. The P+ gate is formed in the n-channell by th

T. Th
plantation. The n+ regions have been formed
diffusion or Ion implanta
the process of
ed th
the drain and

w.E source contact region

4. Explain the
ion to provide good ohmic contact.
proces
rocess involved in the epitaxiall gr
growth and

asy
photolithography. (April/May
(Apr 2015)

Epitaxial Growth:
En
• Epitaxy is described
d as a
gin
arranged atoms in a single crystal fashion
crystal substrate. The basic
b
hion upon
chemical reaction used for the epitaxi
u a single
pitaxial growth of
pure silicon is the hydrog

SiCl4+2H2 1200
o
C
ee
ydrogen reaction of SiCl4

Si+4HCL
rin
• In an IC fabrication
tion e
required. This is accom
conce
g.n
epitaxial films with specific impurity concentration are
accomplished by introducing phosphine (PH3) for the n-type
Bi-Borane (B2H6) for P
stream.
P-type doping into the silicon-tetrachloride
et
ride hydrogen
h gas

Fig 1.15 Epitaxial growth

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• The process is carried out in a reaction chamber consisting of a long cylindrical

quartz tube encircled by an RF induction coil.
• The silicon wafers are placed on a rectangular graphite rod called a boat. This
boat is then placed in the reaction chamber where the graphite is heated
inductively to a temperature 12000C.
• The various gases required for the growth of desired epitaxial layers are
introduced into the system through a control console.
Photolithography:
It can be used to produce microscopically small circuit and device patterns on

ww
Si-wafers. Photolithography involves two processes, namely
1. Making of a photographic mask

w.E 2. Photo etching

Making of a photographic mask:

asy
It involves the following sequence of operations

En
1. The preparation of artwork
2. Its reduction
•
gin
The initial layout or artwork of an IC is normally done at a scale several hundred

ee
times larger than the final dimension of the finished monolithic circuit. This is
because for a tiny chip more accurate is the final mask. This initial layout is then
decomposed into several mask layers.
rin
•
coordinagraph.
g.n
The artwork is usually produced on a precision drafting machine known as

•
moved along two perpendicular axes.
et
The coordinagraph has a cutting head that can be positioned accurately and

• The coordinagraph out lines pattern cutting through the red mylar without
damaging the clear layer underneath.
• This rubylith pattern of individual mask is photographed and then reduced in
step by a factor of 5 or 10 several times to finally obtain the exact image size.
• The final image size also must be repeated many times in a matrix array, so
that many ICs will be produced in one process.
• The photo repeating is done with a step and repeat camera. This is an imaging
device with a photographic plate on a removable platform.

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Photo etching:
• Photo etching is used for the removal of SiO2 from desired regions so that the
desired impurities can be diffused. The wafer is coated with a film of
photosensitive emulsion (Kodak Photo resist KPR). The thickness of the film in
the range 5000-10000A0

ww Fig 1.16 (a) Photo etching step 1

w.E
• Then the wafer is exposed to ultraviolet light so that KPR becomes polymerized
beneath the transparent regions of the mask.

asy
En
gin
ee
Fig 1.16 (b) Photo etching step 2 rin
•
g.n
Then the mask is removed and the wafer is developed using a chemical
(trichloroethylene) which dissolves the unexposed/un polymerized regions, in
the photo resist and leaves the pattern et
Fig 1.16 (c) Photo etching step 3
• The polymerized photo resist next fixed or curved, so that it becomes immune
to certain chemicals called etchants used in subsequent processing steps. The
chip is immersed in the etching solution of HCl, which removes the SiO2 from
the area which are not protected by KPR

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Fig 1.16 (d) Photo etching step 4

• After diffusion of impurities the photo resist is removed with a chemical solvent
(H2SO4) and mechanical abrasion. This etching process is a wet etching
process and the chemical reagents used are in liquid form.

5. Explain the different IC packages.

ww
MONOLITHIC IC PAKAGE:
There are 3 different package configurations are available they are

w.E
1. Metal can package
2. Ceramic flat package

asy
3. Dual-in-line package

• En
METAL CAN or TRANSISTOR PACKAGE:
The chip is encapsulated in a metal or plastic case. The transistor pack is

gin
available with 3, 5, 8, 10 or 12 pins.
•

ee
The metal can package is best suited for power amplifiers because metal is
good heat conductor and consequently has a better dissipation capability than
the flat back or dual-in-line package.
rin
g.n
et

Fig 1.17 Metal can

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• The metal can package permits the use of external heat sink. Most of the
general purpose of Op-Amps comes in 8, 10 or 12 pin packages.
• Voltage regulator ICs such as LM117 has 3-pins. Power Op-Amps and audio
power amplifiers are usually available in 5 pin packages.
CERAMIC FLAT PACKAGE
• The chip is enclosed in a rectangular ceramic case with terminal leads
extending through the sides and ends.
• The flat pack comes with 8, 10, 14 or 16 leads.
• These leads accommodate the power supplies, inputs, outputs and several
special connections required to complete the circuits.

ww
w.E
asy
En
gin
DUAL-IN-LINE PACKAGE
ee
Fig 1.18 Ceramic flat package

rin
•
widely used package type because it can be mounted easily. g.n
The chip is mounted inside a plastic or ceramic case. The DIP is the most

•
•
The 8-pin dual-in-line package is called as mini DIPs.
et
DIPs are also available with 12, 14, 16 and 20 pins. The density of components
integrated on the same chip increases.
• Digital ICs are DIP packages; Metal can packages are also available with dual-
in-line formed leads (DIL-CAN) and with radial formed leads.
• Different outlines exist within each package style to accommodate various die
sizes and number of pins.
• For Example TO-99, TO-100 and TO-101 are some of the outlines available in
a transistor.

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Fig 1.19 Dual-in-line package

ww
w.E
asy
En
gin
ee rin
g.n
et

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UNIT II
CHARACTERISTICS OF OP-AMP
Part - A
1. What are the advantages of ICs over discrete circuits?
Minimization & hence increased equipment density.
Cost reduction due to batch processing.
Increased system reliability
Improved functional performance.
Matched devices.
Increased operating speeds

ww Reduction in power consumption

2. What is OPAMP?

w.E An operational amplifier is a direct coupled high gain amplifier consisting of one
or more differential amplifiers, followed by a level translator and an output stage. It is a

asy
versatile device that can be used to amplify ac as well as dc input signals & designed

En
for computing mathematical functions such as addition, subtraction, multiplication,
integration & differentiation

gin
.3. List out the ideal characteristics of OPAMP?
Characteristics of an ideal operational amplifier:

2. Input impedance
ee
1. Open loop voltage gain AOL = ∞ (infinity)
Ri = ∞ (infinity)
rin
3. Output impedance
4. Zero offset
Ro
Vo
= 0 (zero)
= 0 (zero) g.n
5. Band width BW = ∞ (infinity)
4.what are the different kinds of packages of IC741?
et
a) Metal can (TO) package
b) Dual- in- line package
c) Flat package or flat pack
5. What are the assumptions made from ideal op amp characteristics?
• The current drawn by either of the input terminals(non -inverting/inverting)
is negligible.
• The potential difference between the inverting & non- inverting input terminals is
zero.

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6. Mention some of the linear applications of op – amps:

Adder, subtractor, voltage –to- current converter, current –to- voltage
converters, Instrumentation amplifier, analog computation, power amplifier, etc are
some of the linear op-amp circuits.
7. Mention some of the non – linear applications of op-amps:-
Rectifier, peak detector, clipper, clamper, sample and hold circuit, log amplifier,
anti –log amplifier, multiplier are some of the non – linear op-amp circuits.
8. What are the areas of application of non- linear op- amp circuits?
Industrial instrumentation
Communication

ww Signal processing
9. Define input offset voltage.

w.E A small voltage applied to the input terminals to make the output voltage as
zero when the two input terminals are grounded is called input offset voltage.

asy
10. Define input offset current. State the reasons for the offset currents at the
input of the op-amp.

En
The difference between the bias currents at the input terminals of the op-amp

gin
is called as input offset current. The input terminals conduct a small value of dc current
to bias the input transistors. Since the input transistors cannot be made identical, there
exists a difference in bias currents.
11. Define CMRR of an op-amp.
ee rin
g.n
The relative sensitivity of an op-amp to a difference signal as compared to a
common –mode signal is called the common –mode rejection ratio. It is expressed in
decibels. CMRR= Ad/Ac
12. Define slew rate.
et
The slew rate is defined as the maximum rate of change of output voltage
caused by a step input voltage. An ideal slew rate is infinite which means that op-
amp’s output voltage should change instantaneously in response to input step voltage.
13. Why IC 741 is not used for high frequency applications?
IC741 has a low slew rate because of the predominance of capacitance
present in the circuit at higher frequencies. As frequency increases the output gets
distorted due to limited slew rate.

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14. What causes slew rate?

There is a capacitor with- in or outside of an op-amp to prevent oscillation. It is
this capacitor which prevents the output voltage from responding immediately to a fast
changing input.
15. Define thermal drift.
The bias current, offset current & offset voltage change with temperature. A circuit
carefully nulled at 25oC may not remain so when the temperature rises to 35oC.This is
called thermal drift. Often, offset current drift is expressed in nA/oC and offset voltage
drift in mV/oC.

ww
Part - B
1. Explain in detail about the DC characteristics of the OP-AMP.

w.E
DC Characteristics of op-amp:
DC output voltages are,
1. Input bias current
asy
2. Input offset current
3. Input offset voltage En
4. Thermal drift
1. Input bias current:
gin
ee
The op-amp’s input is differential amplifier, which may be made of BJT or FET.

rin
In an ideal op-amp, we assumed that no current is drawn from the input
terminals.
g.n
The base currents entering into the inverting and non-inverting terminals (IB- &
IB+ respectively).
et
Even though both the transistors are identical, IB- and IB+ are not exactly equal
due to internal imbalance between the two inputs.
Manufacturers specify the input bias current IB
Input bias current IB as the average value of the base currents entering into the
terminals of the op-amp

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Fig 2.1 Input bias current

I B+ + I B−
IB = V0 = ( I B − ) R f
ww So,
2 and
Where a compensation resistor Rcomp has been added between the non-inverting input

w.E
terminal and ground as shown in the figure below.

asy
En
gin
ee rin
g.n
Fig 2.2 Bias current compensation
Current IB+ flowing through the compensating resistor Rcomp, then by KVL we get,
et
-V1 + 0 + V2 -Vo = 0 (or)
Vo = V2 – V1 (1)
By selecting proper value of Rcomp, V2 can be cancelled with V1 and the Vo = 0. The
value of Rcomp is derived a
V1 = IB+ Rcomp (or)
IB+ = V1/Rcomp (2)
The node ‘a’ is at voltage (-V1). Because the voltage at the non-inverting input terminal
is (-V1). So with Vi = 0 we get,
I1 = V 1 / R 1 (3)

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I2 = V2 / Rf (4)
For compensation, Vo should equal to zero (Vo = 0, Vi = 0). i.e. from equation (1) V2 =
V1. So that,
I2 = V1 / Rf (5)
KCL at node ‘a’ gives,
IB- = I2 + I1
f / fa
A= − V1 V1 ( R1 + R f )
1 + ( f / fb ) 2 I B = + = V1
R f R1 R1R f

Assume IB- = IB+ and using equation (4) & (8) we get

ww V1
(R + R ) =
1

R1R f
f V1
Rcomp

w.E Rcomp =
R1 R f
R1 + R f
= R1 R f

Rcomp = R1 || Rf
asy (6)

En
i.e. to compensate for bias current, the compensating resistor, Rcomp should be equal
to the parallel combination of resistor R1 and Rf.
2. Input offset current:
gin
Bias current compensation will work if both bias currents IB+ and IB- are equal.

ee
Since the input transistor cannot be made identical. There will always be some

rin
small difference between IB+ and IB-. This difference is called the offset current
|Ios| = IB+ - IB-
Offset current Ios for BJT op-amp is 200nA and for FET op-amp is 10pA. Even with g.n (1)

bias current compensation, offset current will produce an output voltage when Vi = 0.
V1 = IB+ Rcomp
et (2)
And I1 = V1/R1 (3)
KCL at node ‘a’ gives,
I2 = (IB—I1)
 Rcomp 
I 2 = ( I B − − I1 ) = I B − −  I B +  (Sub the value of I2)
 R1 

Again, V0 = I2 Rf – V1
Vo = I2 Rf - IB+ Rcomp

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 R 
Vo =  I B − − I B+ comp  R f − I B+ Rcomp (4)
 R1 

Substitute the value Rcomp in equation (4) and after algebraic manipulation,
V o = R f  I B− − I B+ 

Vo = R f I 0 s
The offset current can be minimized by keeping feedback resistance small.
Unfortunately to obtain high input impedance, R1 must be kept large.
R1 large, the feedback resistor Rf must also be high, so as to obtain reasonable

ww gain.
The T-feedback network is a good solution. This will allow large feedback

w.E resistance, while keeping the resistance to ground low (in dotted line).
The T-network provides a feedback signal as if the network were a single
feedback resistor.
asy
En
gin
ee rin
g.n
3. Input offset voltage:
Fig 2.3 T-network
et
• Inspite of the use of the above compensating techniques, it is found that the
output voltage may still not be zero with zero input voltage [Vo ≠ 0 with Vi = 0].
• This is due to unavoidable imbalances inside the op-amp and one may have to
apply a small voltage at the input terminal to make output (Vo) = 0.
• This voltage is called input offset voltage Vos. This is the voltage required to be
applied at the input for making output voltage to zero (Vo = 0).

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ww
w.E
asy
En
gin
ee rin
Fig 2.4 Input offset voltage & its equivalent circuit g.n
et
Let us determine the Vos on the output of inverting and non-inverting amplifier. If Vi = 0
(Fig (b) and (c)) become the same as in figure (d). The voltage V2 at the negative input
 R1 
terminal is given by, V2 =  Vo
 R1 + R f
 
 R1 + R f 
Vo =  V2
 R1 

 Rf 
Vo = 1 + V2
 R1 
Vios = Vi − V2 and Vi=0, Vios = 0 − V2 = V2

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 Rf 
Vo = 1 +  Vios
 R1 

This is the output offset voltage of an op-amp in closed loop configuration

4. Thermal drift:
A circuit nulled at 250C may not remain so the temperature rises to 350C.This is called
drift. Bias current, offset current, offset voltage change with temperature.
offset current drift is expressed as nA/oC
offset voltage drift is expressed as mV/oC.
These indicate the change in offset for each degree Celsius change in

ww temperature.

w.E
2. Explain in detail about the AC characteristics of the OP-AMP.
AC Characteristics:
•

asy
For small signal sinusoidal (AC) application one has to know the ac
characteristics such as frequency response and slew-rate.

Frequency Response:
En
•
its phase angle. gin
The variation in operating frequency will cause variations in gain magnitude and

•
ee
The manner in which the gain of the op-amp responds to different frequencies
is called the frequency response.
rin
•
g.n
Op-amp should have an infinite bandwidth Bw = ∞ (i.e) if its open loop gain in
90dB with dc signal its gain should remain the same 90 dB through audio and

•
onto high radio frequency.
et
The op-amp gain decreases (roll-off) at higher frequency what reasons to
decrease gain after a certain frequency reached.
• There must be a capacitive component in the equivalent circuit of the op-amp.
• For an op-amp with only one break (corner) frequency all the capacitors effects
can be represented by a single capacitor C.
• Below fig is a modified variation of the low frequency model with capacitor C at
the o/p.

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Fig 2.5 Frequency response equivalent circuit

There is one pole due to R0 C and one -20dB/decade. The open loop voltage gain of
an op-amp with only one corner frequency is obtained from above fig.

− jXc

ww
Vo =
Ro − jXc
AolVd

w.E
Divide by jXc both numerator and denominator

Vo =
AolVd
 Ro


+ 1
 − jXc  asy (Vin=Vd)

Vo Aol En
Vin
=
1+ j
Ro
gin (Xc=1/2πfc)

A=
Xc
Aol ee rin
(1 + 2πRoC
Aol g.n
A=
1 + j ( f / f 1)
where f1= 1/2πfRoC
et (1)

f1 is the corner frequency or the upper 3 dB frequency of the op-amp. The magnitude
and phase angle of the open loop volt gain are fu of frequency can be written as,
Aol
Magnitude A= (2)
1 + ( f / f 1)
2

Phase angle φ = tan −1 ( f / f 1)

(3)

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The magnitude and phase angle characteristics from equations (2) and (3)
• For frequency f<< f1 the magnitude of the gain is 20 log AOL in dB.
• At frequency f = f1 the gain in 3 dB down from the dc value of AOL in dB. This
frequency f1 is called corner frequency.
• For f > > f1 the fain roll-off at the rate off -20dB/decade or -6dB/decade.

ww
w.E
asy
En
gin
ee rin
g.n
Fig 2.6 Frequency response characteristics
From the phase characteristics
et
the phase angle is zero at frequency f =0.
At the corner frequency f=f1 the phase angle is -450
-900 phase angle occurs at frequency (at f=∞)

The voltage transfer in a S-domain can be written as

Aol Aol
A= =
1 + j ( f / f 1) 1 + j (ω / ω1)

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Aol.ω1 Aol.ω1
A= = (4)
jω + ω1 S + ω1
The transfer f0 of as op-amp with 3 break frequency can be assumed as,

Aol
A= 0 < f1< f 2< f3
(1 + jf / f 1)(1 + jf / f 1)(1 + f / f 3)
Aol.ω1.ω 2.ω 3
A= 0<ω1<ω2< ω3
( S + ω1)(S + ω 2)(S + ω 3)

ww
w.E
asy
En
gin
Slew Rate:
•
ee
Fig 2.7 Frequency response of op-amp

rin
Another important frequency related parameter of an op-amp is the slew rate.

g.n
(Slew rate is the maximum rate of change of output voltage with respect to time.
Specified in V/µs).

Reason for Slew rate:

et
• There is usually a capacitor within 0, outside an op-amp oscillation.
• It is this capacitor which prevents the o/p voltage from fast changing input.
• The rate at which the volt across the capacitor increases is given by

dVc / dt = I/C (1)

• I = Maximum amount furnished by the op-amp to capacitor C. Op-amp should
have the either a higher current or small compensating capacitors.
• For 741 IC, the maximum internal capacitor charging current is limited to about
15µA. So the slew rate of 741 IC is

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SR = dVc / dt |max = Imax/C.

• For a sine wave input,
put, th
the effect of slew rate can be calculated
d as consider
c volt
follower -> The input
ut is la
large amp, high frequency sine wave.
• If Vs = Vm Sinwt then output V0 = Vm sinwt . The rate of change
hange of output is
given by dV0/dt = Vm w coswt.

ww
w.E
asy
Fig
En
ig 2.8 Input and output waveforms
The max rate of change off outp
gin
output across when coswt =1
(i.e) SR = dV0/dt |max = wVm.
wVm
SR = 2∏fVm V/s = 2∏fVm v/m
v/ms.
Thus the maximum frequency
ency fmax
f
of peak value Vm is given by
ee
at which we can obtain an undistorte
rin
storted output volt

fmax (Hz) = Slew rate/6.28

8 * Vm
V .
g.n
called the full power response
ponse. It is maximum frequency of a large
wave with which op-amp can ha
have without distortion.
ge amplitude
am
et
sine

3. Explain the operation

tion of
o differentiator and integrator by using
sing OP-AMP.
O
(a) Differentiator:
• One of the simplest
est o
of the op-amp circuits that contains capacitor
capa in the
differentiating amplifier.
lifier.
• As the name implies,
plies, the circuit performs the mathematical
tical o
operation of
differentiation (i.e) the output
ou waveform is the derivative of the input waveform.
• The differentiator may
ay be constructed from a basic inverting amplifi
mplifier if an input
resistor R1 is replaced
ced by a capacitor C1.

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Fig 2.9 Differentiator circuit

Analysis:

ww Node N is virtual
rtual g
VN = 0
grounded.

w.E( ) Current iC through

ic = C1
d
ugh tthe capacitor,

dV
Vi − VN = C1 i → (1)
asy dt
Current if through
ugh fe
dt
feedback resistor is

En if =
V0

gin Rf -----------
------------>(2)

ee
Apply KCL at node N

ic + i f = 0
rin
dV V
C1 i + 0 = 0
dt R f g.n
V0 = − R f C1
dVi
dt
-----------------
------>(3)
et
Thus the output V0 is equal
al to RF C1 times the negative rate of change of th
the input
voltage Vin with time.
The –sign indicates a 1800 phase
ph shift of the output waveform V0 with
ith respect
re to the
input signal.
Phasor equivalent of output
ut vol
voltage is

V0 ( S ) = − R f C1.SVi ( S )

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V0
A= = R f C1S = jω R f C1 = ω R f C1
Vi

f 1
A= where fa =
fa 2π R f C1
(2)
Input and Output Waveforms:
rms:

ww
w.E
asy
En
gin
ee rin
g.n
et

Fig
ig 2.10 Input and output waveforms
The input signal will be differen
ifferentiated properly, if the time period T of the input
in signal is
larger than or equal to RF C1 (i.e)
(i.e T > RF C1

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(b) Integrator:
• A circuit in which the
e out
output voltage waveform is the integral off the iinput voltage
waveform is the integrato
tegrator or Integration Amplifier.
• Such a circuit is obtained
tained by using a basic inverting amplifier config
configuration if the
feedback resistor RF is re
replaced by a capacitor CF .

ww
w.E Fig 2.11 Integrator circuit
Analysis:
asy
Let Vin is the input voltage appli
applied to the inverting terminal

En
Current through Capacitor Cf= C
Current through Ri
Vin − VB
Ri gind
= C f (VB − Vo )
dt
(1)

ee
(∴VB = VA = 0)
rin
(1)------------->
Vin
Ri
d
= C f Vo
dt g.n
d
dt
Vo = −
1
Ri C f
Vin
et
1
Ri C f ∫
Vo = − Vin dt (2)

Equation (2) indicates thatt the output is directly proportional to the negativ
egative integral of
the input volts and inversely
ely pro
proportional to the time constant R1 CF . In phasor
ph method
the output voltage can be writte
written as

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R f / Ri
A=− 1
(1 + jf / f a ) Vo ( s ) = − Vi ( s )
sR1C f
H ( jω ) = R f / Ri
In steady state, put s = jω and we
w get

1
Vo ( jω ) = − Vi ( jω )
jω R1C f
So the magnitude of the gain
ain or integrator transfer function is

Vo ( jω ) 1 1
A= =− =
ww Vi ( jω ) jω R1C f ω R1C f
Ex: If the input is sine wave ->
> output
o is cosine wave.

w.E
If the input is square wave ->
> output
ou is triangular wave.

asy
En
gin
ee rin
g.n
et

Fig 2.12 Input and output waveforms

• These waveform with
ith as
assumption of R1 Cf = 1, Vout = 0V (i.e) C = 0.
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• When Vin = 0 the integrator works as an open loop amplifier because the
capacitor CF acts an open circuit to the input offset voltage Vio. (Or)
• The Input offset voltage Vio and the part of the input is charging capacitor CF
produce the error voltage at the output of the integrator.

4. Explain in detail about differential amplifier using op-amp.

• The op-amp amplifies the difference between the two input signals is called
differential amplifier. Classify these arrangements according to the number of
op-amps used. i.e

ww 1. Differential amplifier with one op-amp

2. Differential amplifier with two op-amps.

w.E
• Differential amplifiers are used in instrumentation and industrial applications to
amplify differences between 2 input signals such as output of the wheatstone

•
bridge circuit.
asy
Differential amplifier preferred to these application because they are better able

En
to reject common mode (noise) voltages than single input circuit such as

gin
inverting and non-inverting amplifier.

ee
1. Differential Amplifier with one op-amp:
Figure shows the differential amplifier with one op-amp.

rin
g.n
et
Fig 2.13 Differential amplifier circuit with 1 op-amp
To analyze this circuit by deriving voltage gain and input resistance. This circuit is a
combination of inverting and non-inverting amplifier. (i.e)
• When Vx is reduced to zero the circuit is non-inverting amplifier and
• When Vy is reduced to zero the circuit is inverting amplifier.

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Voltage Gain:
The circuit has 2 inputs Vx and Vy . Use superposition theorem, when Vy = 0V,
becomes inverting amplifier. Hence the output due to Vx only is

−R f (Vx )
Vox =
R1
Similarly, when Vx = 0V, becomes Non-inverting amplifier having a voltage divider
network composed of R2 and R3 at the Non – inverting input.

R3 (Vy )
V1 =
R2 + R3
ww
And the output due to Vy then is

w.E  R
Voy = 1 + f
 R1

V1

asy
Note: the gain of the differential amplifier is same as that of inverting amplifier.
Input Resistance:
En
gin
The input resistance Rif of the differential amplifier is resistance determined looking
into either one of the 2 input terminals with the other grounded,
With Vy = 0V,
ee
Inverting amplifier, the input resistance which is,
RiFx ≈ R1 rin (1)
Similarly Vx = 0V,
g.n
Non-inverting amplifier, the input resistance which is,

•
RiFy ≈ (R2 + R3) et
Vx and Vy are not the same. Both the input resistance can be made equal, if we
(2)

modify the basic differential amplifier. R1 and (R2 + R3) can be made much
larger than the source resistances. So that the loading of the signal sources
does not occur.

Note: If we need a variable gain, we can use the differential amplifier. In this circuit
R1 = R2 , RF = R3 and the potentiometer Rp = R4.Depending on the position of the
wiper in R voltage can be varied from the closed loop gain of -2RF /R1 to the open loop
gain of A.

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Fig 2.14 Differential amplifier circuit with 1 op-amp i/p resistance

2. Differential Amplifier with 2 op-amps:

ww We can increase the gain of the differential amplifier and also increase the
input resistance Rif if we use 2 op-amps.

w.E
Voltage gain:
It is composed of 2 stages 1. Non-inverting amplifier 2. Differential amplifier with gain.

asy
En
gin
ee rin
g.n
Fig 2.15 Differential amplifier circuit with 2 op-amps
et
By finding the gain of these 2 stages, we can obtain the overall gain of the circuit, The
o/p
 R 
V2 = 1 + 3 V y − − − − − − − − − −− > (1)
 R2 
By applying superposition theorem to the second stage, we can obtain the output
voltage,
Rf  Rf 
V0 = − V2 + 1 + V x − − − −− > (2)
R1  R1 
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Sub. The value of V2 in eqn. (2)

R f  R3   R 
V0 = − 1 + V y + 1 + f Vx − − > (3)
R1  R2   R1 
Since R1=R3 and Rf=R2

 Rf 
V0 = 1 + (Vx − V y )
 R1 

R3  R1 + R f 
V0 y =  V y − − − − − − − −− > (4)
R2 + R3  R1 
Since R1=R2 and Rf=R3

ww
Voy =
Rf
R1
V y − − − − − − − − − − − − − − − − − −− > (5)

w.E
From eqn.(1) and (5) the net output voltage is,
V0 = V0 x + Voy

V0 = −
Rf
R1
(V x − Vy )
asy
V0 = −
Rf
R1
(V ) xy
En
gin
Or the output voltage gain,

AD =
V0
V xy
=−
Rf
R1
ee rin
The gain of the differential amplifier is same as that of the inverting amplifier.
g.n
5. Explain in detail about inverting and non inverting operational amplifier?
Inverting Amplifier:
et
• An amplifier which provides a phase shift of 1800 between the input and the
output is called as inverting amplifier. The basic circuit diagram of inverting
amplifier is shown below

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ww
Analysis:
Fig 2.16
2. Inverting amplifier circuit

w.E
Current through the resistance

I=
Vin − V A
tance R is,

R1
asy [VA=VB=0 Virtual ground]

I=
Vin
En
R1
Current through RF is,
gin
I=
V A − V0
RF
V
=− 0
RF
ee rin
Vin V
=− 0 g.n
RF

A=
RF
V0 R
=− F
et
Vin R1
RF/R1 is called gain of amplifier
plifiers. Negative sign indicates that polarity of the output is
opposite to that of the input.
ut. So the inverting amplifier is also called ass sign changer.
Sign Changer:
Let K= RF/R1 is called scale factor.
V0=-KVin
Since the output voltage iss chan
changing according to the scale factor K and input
in voltage
Vin the inverting amplifier is cal
called as Scale changer. The input and output
utput waveforms
are shown below
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ww Fig 2.17 Inverting amplifier waveforms

If the value of RF/R1=1, then the output is exactly 1800 phase shift with
en th ith respect
res to the

w.E
input. Hence it is called ass Phas
Non-inverting amplifier:
Phase shift circuit (or) phase inverter.

•
asy
An amplifier which amplifies
ampli
called a non-inverting
ing am
the input without any phase shift betwe
amplifier.
between them is

The basic circuit diagram is shoEn

shown below

gin
ee rin
g.n
et
Fig 2.18 Non-inverting amplifier circuit
Analysis:
I through R1 is,

V − VA
I=
R1
Vin
I= −
R1

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Current through RF is,

V A − V0 Vin − V0
I= =
RF RF

Vin Vin V0
− = −
R1 R F RF
V0 Vin Vin
= −
R1 R1 R F

V0 Vin  R F 
= 1 + 
ww Vin RF 
V R
R1 

w.E A = 0 = 1+ F
Vin R1

asy
The above equation is called
lled as the gain for non-inverting amplifier.
The input and output waveform
eforms are shown below.

En
gin
ee rin
g.n
et
Fig 2.19
.19 N
Non-Inverting amplifier waveforms
Applications of op-amp:
It is classified into 2 types,
• Linear application
• Non-linear Application

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6. Explain the basic appli

applications of op-amp?

Summing Amplifier:
Op-amp may be used
sed tto design a circuit whose output is the
e sum of several
input signals. Such a circuit
uit is ccalled a summing amplifier or a summer.
Adder is classified as
Inverting summer
mmer
Non-inverting
g summer
sum
Inverting Summing Amplifier:
lifier:
In this circuit all the inputt signals
signa to be added are applied to the inverting
verting terminal of

ww
the op-amp. The circuit with two
feedback resistor Rf is shown
tw input signals V1, V2 , input resistors
own in figure
tors R1, R2 and a

w.E
asy
En
gin
Fig 2.20
ee
.20 Inverting
In summing amplifier circuit rin
As point B is grounded
g.n
nded, due to virtual ground concept the node A is also at
virtual ground potential.VA=0
Now from the inputt side
V1 − V A V1
et
I1 = = − − − − − − − − − −− > (1)
R1 R1
V2 − V A V 2
I2 = = − − − − − − − − − −− > (2)
R2 R2
Applying KCL at node
de A and as input op-amp current is zero,
I=I1+I2 ------------------------
------------------------>(3)
From the output side
V A − V0 V
I= = − 0 − − − − − − − − − − > ( 4)
Rf Rf
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Substitute eqn.(1), (2), (4)

(4 in eqn.(3)
− V0 V1 V2
= + − − − − − − − − − − − −− > (5)
Rf R1 R2
Rf Rf 
V0 = −  V1 + V 2  − − − − − − − − > ( 6)
 R1` R2 
If the three resistance
ce ar
are equal R1= R2= Rf

V 0 = − (V1 + V 2 )

By properly selecting Rf , R1 and R2 we can have weighted addition

dition of the input

ww
signals like aV1+aV2 as indicat
ndicated in eqn(6)
Non-Inverting Summing Amplifier:
Amp

w.E Here the input signals

nals to be added and applied to the non-inverting
the op-amp. The circuit diagram
iagram is shown in figure
verting terminal of

asy
En
gin
ee rin
Analysis:
Fig 2.21 Non
Non-Inverting summing amplifier circuit
g.n
Let the voltage of node B is VB . Now the node A is at the same potential
due to virtual ground.
et
tential as that of B

VA=VB------------------------
------------------>(1)
From the input side
V1 − V B
I1 = − − − − − − − − − −− > (2)
R1
V2 − V B
I2 = − − − − − − − − − −− > (3)
R2
Input current of op-amp is zero
I1 + I2 = 0--------------------
--------------------->(4)

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V1 − VB V2 − VB
+ =0
R1 R2
V1 V2 1 1
+ = VB  + 
R1 R2  R1 R2 
R V + R1V2
VB = 2 1 − − − − − − − − > (5)
R + R2
Now at node A
V A VB
I= = − − − − − − − − − −− > (6)
R R

ww and

I=
V0 − V A V0 − V B
= − − − −− > (7)

w.E Rf Rf
Equqting eqn. (6) and (7)

R + Rf
V0 = V B 
 R
asy

 − − − − − − − − > (8)

Substitute eqn.(5) in (8)
R2 (R + R f ) (
En) gi
R1 R + R f
V0 =
R(R1 + R2 )
V1 +
nee
R(R1 + R2 )
V2 − − > (9)

The eqn(9) shows that the output is weighted some of the inputs.
If R1 = R2 = R = Rf we get rin
V0=V1+V2 g.n
Subtractor:
•
et
A basic differential amplifier can be used as a subtractor as shown in the above
figure. If all resistors are equal in value, then the output voltage can be derived
by using superposition principle.

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ww Fig 2.22 Subtractor amplifier circuit

To find the output V01 due to V1 alone, make V2 = 0.

w.E Then the circuit of figure as shown in the above becomes a non-inverting
amplifier having input voltage V1/2 at the non-inverting input terminal and the output
becomes

V  R
asy
En
V01 = 1 1 +  = V1 − − − − − − − − − −− > (1)
2  R
gin
Similarly the output V02 due to V2 alone (with V1 grounded) can be written
simply for an inverting amplifier as
R ee
V02 = − V2 = −V2 − − − − − − − − − − − −− > (2)
R rin
g.n
Thus the output voltage Vo due to both the inputs can be written as

V0 = V01 + V02
V0 = V1 − V2
et
Thus the output voltage is the difference between the two inputs and hence it act as
the subtractor.

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UNIT-III
APPLICATIONS OF OPAMP
Part – A
1. What is the need for an instrumentation amplifier?
In a number of industrial and consumer applications, the measurement of physical
quantities is usually done with the help of transducers. The output of transducer has to
be amplified So that it can drive the indicator or display system. This function is
performed by an instrumentation amplifier.
2. What is a sample and hold circuit? Where it is used?
A sample and hold circuit is one which samples an input signal and holds on to its last

ww
sampled value until the input is sampled again. This circuit is mainly used in digital
interfacing, analog to digital systems, and pulse code modulation systems.

w.E
3. What is a comparator?
A comparator is a circuit which compares a signal voltage applied at one input of an

asy
opamp with a known reference voltage at the other input. It is an open loop op - amp
with output ± Vsat .

En
4. What is a multivibrator?

gin
Multivibrators are a group of regenerative circuits that are used extensively in timing
applications. It is a wave shaping circuit which gives symmetric or asymmetric square

multivibrator.
ee
output. It has two states either stable or quasi- stable depending on the type of

rin
5. Draw the circuit of log amplifier using op-amps g.n
et

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6. Draw the circuit of log amplifier using op-amps

7. What is a zero crossing detector?

Zero crossing detector(ZCD) is a voltage comparator that switches the output between

ww
+Vsat and –Vsat. The output is driven into –Vsat when the input signal passes through
zero to positive direction. Conversely, when input signal passes through zero to

w.E
negative direction, the output switches to +Vsat.
8. What are the applications of comparator?
• Zero crossing detectors
• Window detector asy
• Time marker generator
En
• Phase detector
9. Define conversion time. gin
ee
It is defined as the total time required converting an analog signal into its digital output.

rin
It depends on the conversion technique used & the propagation delay of circuit

g.n
components. The conversion time of a successive approximation type ADC is given by
T (n+1) where T---clock period Tc---conversion time n- no. of bits.
10. What are the different types of filters?
• et
Based on functions: Low pass filter, High pass filter, Band pass filter, Band
reject filter
• Based on order of transfer function : first, second, third higher order filters.
• Based on configuration: Bessel, Chebychev, Butterworth filters.

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Part - B
1. Draw and explain the working of Instrumentation amplifier using Op-Amp
and derive its output voltage equation.
• In a number of industrial and consumer applications, one is required to measure
and control physical quantities.
• Some typical examples are measurement and control of temperature, humidity,
light intensity, water flow etc. these physical quantities are usually measured
with help of transducers.
• The output of transducer has to be amplified so that it can drive the indicator or
display system.

ww • This function is performed by an instrumentation amplifier.

w.E
• The important features of an instrumentation amplifier are
high gain accuracy
high CMRR

asy
high gain stability with low temperature coefficient

• En
low output impedance
There are specially designed op-amps such as µA725 to meet the above stated

gin
requirements of a good instrumentation amplifier.
• Monolithic (single chip)
eeinstrumentation amplifier are

rin
commercially such as AD521, AD524, AD620, AD624 by Analog Devices,
also

LM363.XX (XX -->10,100,500) by National Semiconductor and INA101, 104,

available

3626, 3629 by Burr Brown.

g.n
et

Fig 3.1 Basic differential amplifier circuit

Consider the basic differential amplifier,
The output voltage Vout is given by,

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R2 1  R 
V0 = − V2 + V1 1 + 2 
R1 R R1 
1+ 3 
R4

 + R4 
V = V1 
 R3 + R4 

 
R  1  R1  
V0 = − 2 V2 −  + 1V1 
R1  R  R2  
 1+ 3 
 R4 
R1 R3
=
ww For
R2 R4
we obtain

w.E V0 =
R2
R1
(V1 − V2 )

•
asy
In the circuit of figure 6(a), source V1 sees an input impedance = R3+R4 (=101K)

• En
and the impedance seen by source V2 is only R1 (1K).
This low impedance may load the signal source heavily.
•
gin
Therefore, high resistance buffer is used preceding each input to avoid this
loading effect as shown in figure 6(b).
• ee
The op-amp A1 and A2 have differential input voltage as zero. For V1=V2, that is,
under common mode condition, the voltage across R will be zero.
rin
g.n
et

Fig 3.2 Instrumentation amplifier circuit

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As no current flows through R and R’ the non-inverting amplifier. A1 acts as voltage

follower, so its output
V2’=V2.
Similarly op-amp A2 acts as voltage follower having output
V1’=V1.
However, if V1≠V2, current flows in R and R’, and (V2’-V1’)>(V2-V1). Therefore, this
circuit has differential gain and CMRR more compared to the single op-amp circuit of
figure 6(a).
The output voltage Vo can be calculated as follows
R2V1'

ww
The voltage at the (+) input terminal of op-amp A3 is

theorem, we have,
R1 + R2
. Using superposition

w.E V0 = −
R2 '  R2  R2V1' 
V2 + 1 +  
R1  R1 + R2 

asy
R1 

=− (
R2 '
)
V1 − V2' ---------------->(1)

En R1

gin
Since, no current flows into op-amp, the current I flowing (upwards) in R is
I = (V1-V2)/R and passes through the resistor R’.

V1' = R ' I + V1 =
ee
R'
R
R'
(V1 − V2 ) + V1
rin
And V2' = − R ' I + V2 = −
R
(V1 − V2 ) + V2
g.n
Putting the values of V1’ and V2’ in equation (1), we obtain,

V0 =
R2  2 R '

R1  R

(V1 − V2 ) + (V1 − V2 )

et
R2  2 R '  
V0 = 1 + (V1 − V2 ) ----------------->(2)
R1  R  
In equation (2), if we choose R2 = R1 = 25K (say) and R’ = 25K; R = 50Ω, then a
gain=1001
The difference gain of this instrumentation amplifier R, however should never
be made zero, as this will make the gain infinity. To avoid such a situation, in a
practical circuit, a fixed resistance in series with a potentiometer is used in place of R.

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Fig 3.3 Instrumentation amplifier practical circuit

• Figure 3.3 shows a differential instrumentation amplifier using Transducer

ww Bridge. The circuit uses a resistive transducer whose resistance changes as a

function of the physical quantity to be measured.

w.E
• The bridge is initially balanced by a dc supply voltage Vdc so that V1=V2. As the
physical quantity changes, the resistance RT of the transducer also changes,

asy
causing an unbalance in the bridge (V1≠V2). This differential voltage now gets

En
amplified by the three op-amp differential instrumentation amplifier.

Applications:
• Temperature indicator gin
•
•
Temperature controller
Light intensity meter
ee rin
2. With a neat circuit diagram, explain the working of Schmitt trigger using
Op-Amp. (May 2015) g.n
Schmitt Trigger:
et

Fig 3.4 Schmitt trigger circuit

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ww
w.E
asy
• En
Fig 3.5 Schmitt trigger circuit waveforms
This circuit converts an irregular shaped waveform to a square wave or pulse.

gin
The circuit is known as Schmitt Trigger or squaring circuit.
•

ee
The input voltage Vin triggers (changes the state of) the o/p V0 every time it
exceeds certain voltage levels called the upper threshold Vut and lower
threshold voltage.
rin
•
where the voltage across R1 is feedback to the (+) input. g.n
These threshold voltages are obtained by using theh voltage divider R1 – R2,

•
the value of the output voltage.
et
The voltage across R1 is variable reference threshold voltage that depends on

• When V0 = +Vsat, the voltage across R1 is called “upper threshold” voltage Vut.
The input voltage Vin must be more positive than Vut in order to cause the
output V0 to switch from +Vsat to –Vsat. As long as Vin < Vut , V0 is at +Vsat,
using voltage divider rule.
• Similarly, when V0 = -Vsat, the voltage across R1 is called lower threshold
voltage Vlt . the Vin must be more negative than Vlt in order to cause V0 to switch
from –Vsat to +Vsat.
• In other words, for Vin > Vlt , V0 is at –Vsat.

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• Thus, if the threshold voltages Vut and Vlt are made larger than the input noise
voltages, the positive feedback will eliminate the false o/p transitions.
• Also the positive feedback, because of its regenerative action, will make V0
switch faster between +Vsat and –Vsat. Resistance Rcomp R1 || R2 is used to
minimize the offset problems.
• The comparator with positive feedback is said to exhibit hysteresis, a dead
band condition. (i.e) when the input of the comparator exceeds Vut its output
switches from +Vsat to –Vsat and reverts to its original state, +Vsat when the
input goes below Vlt.
• The hysteresis voltage is equal to the difference between Vut and Vlt.

ww
Therefore

w.E Vref = Vut – Vlt

Vref = R1

asy
R1 + R2 [+Vsat -(-Vsat)]

En
3. With a neat circuit diagram explain the operation of R-2R D/A converter. (May
2015)
• gin
An enhancement of the binary-weighted resistor DAC is the R-2R ladder

output voltages.
ee
network. This type of DAC utilizes Thevenin’s theorem in arriving at the desired

rin
•
• g.n
The R-2R network consists of resistors with only two values - R and 2xR.
If each input is supplied either 0 volts or reference voltage, the output voltage

•
will be an analog equivalent of the binary value of the three bits.
et
VS2 corresponds to the most significant bit (MSB) while VS0 corresponds to the
least significant bit (LSB).

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ww
w.E Fig
ig 3.6 R-2R ladder type D/A circuit
Vout = - (VMSB
SB + Vn + VLSB) = - (VRef + VRef/2 + VRef/
An alternative to the binary-weighted-input
bi
ef/ 4)
DAC is the so-called
called R/2R DAC,
•

asy
which uses fewer unique
nique resistor values.
•
En
A disadvantage off the former DAC design was its requiremen
different precise input
put re
ement of several
resistor values: one unique value per binary
nary in
input bit.
• Manufacture may be si
gin
simplified if there are fewer different resistor
resist values to

•
purchase, stock, and
nd sor
Of course, we could ee
sort prior to assembly.

rin
ld take our last DAC circuit and modify it to use a single input
resistance value, byy con
connecting multiple resistors together in series
eries

g.n
et
Fig 3.7 Ladder type D/A circuit
• Mathematically analyzing
alyzing this ladder network is a bit more complex
mplex than for the
previous circuit, where
ere e
each input resistor provided an easily-calcul
calculated gain for
that bit.
• For those who are inter
interested in pursuing the intricacies of this circuit
ci further,
you may opt to use
se Th
Thevenin's theorem for each binary input
put (remember
(r to

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consider the effects of the virtual ground), and/or use a simulation program like
SPICE to determine circuit response.
• Either way, you should obtain the following table of figures:

| Binary | Output voltage |

---------------------------------
| 000 | 0.00 V |
---------------------------------
| 001 | -1.25 V |
---------------------------------

ww | 010 | -2.50 V |
---------------------------------

w.E | 011 | -3.75 V |

---------------------------------

asy | 100 | -5.00 V |

---------------------------------

En | 101 | -6.25 V |

gin 110 | -7.50 V |

---------------------------------

•
ee| 111 | -8.75 V |
---------------------------------
rin
As was the case with the binary-weighted DAC design, we can modify the value
of the feedback resistor to obtain any "span" desired.
g.n
•
et
For example, if we're using +5 volts for a "high" voltage level and 0 volts for a
"low" voltage level, we can obtain an analog output directly corresponding to the
binary input (011 = -3 volts, 101 = -5 volts, 111 = -7 volts, etc.) by using a
feedback resistance with a value of 1.6R instead of 2R.

4. Explain the working of successive approximation type A/D converter. (May

2016)
• Successive-approximation ADC is a conversion technique based on a
successive-approximation register (SAR).

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• This is also called bit-we

weighing conversion that employs a comparat
parator to weigh
the applied input voltage
oltage against the output of an N-bit digital-to-ana
analog
converter (DAC).
• The circuit operatess as fo
follows, With the arrival of start command,
nd, th
the SAR sets
MSB d1=1 with the other bits 0.
• Hence the trail bit is 10000000.
100 It is then converted to Vd by DAC
AC an
and
compared with analog
log in
input voltage Va by using the comparator.
If Va > Vd, then change
hange the next LSB is set to 1, by keeping MSB as 1.
If Va < Vd, then MSB
B of SAR shifts to 0 and made the next LSB
B to 1.
1
If Va = Vd EOC signal
ignal will be activated and hence the digital
ital output
ou can be

ww obtained.

w.E
asy
En
gin
The truth table is shown below,
elow,
ee
ccessive approximation type A/D converter
Fig 3.8 Success

rin
Correct digital
SAR Output Vd at
different stages in the g.n
Comparato
arator output
representation

11010100 10000000
conversion
et
1 (initial output)
utput)
11000000 1
11100000 0
11010000 1
11011000 0
11010100 1
11010110 0
11010101 0
11010100

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• A comparison of the
he spe
speed of 8-bit tracking ADC and SAR type ADC
A is given
below, from the figure
igure it is noted that the conversion time of tr
tracking ADC
increases with the incre
increase in the number of bits. But the conver
onversion type of
SAR ADC remains constant
const irrespective of the number of bits used.

ww
w.E
asy
Fig 3.9 Successive
En
ive approximation
a type A/D converter wavefo
aveform

5. Explain the second

nd order
or gin
low pass filter with a neat diagram
gram. Derive its

•
frequency response
se an
A second order LPF
F having
hav ee
and plot the same. (May 2016)

rin
a gain 40dB/decade 1in stop band. First
Firs orders LPF
can be converted into a II order type simply by using an additional
tional RC network.
The gain of the III order
g.n
orde filter is set by R1 and RF, while the high cut off
frequency fH is determin
filter
ermined by R2,C2, R3 and C3.It is also called
et
lled as Sallen-Key

Fig 3.10
3. Second order LPF circuit
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ww Fig 3.11 Second

cond order LPF circuit frequency response

w.E This above fig trans

transferred into S domain.

asy
En
gin
ee rin
Fig 3.12 Second orde
order LPF circuit frequency response S domain
In this circuit all the componen
dom
ponents and the circuit parameters are express g.n
pressed in the S-
domain where S = jω .

 Rf 
V0 = 1 + VB = A0VB
et
 Ri 
Rf
Where A0 = 1 + VB= voltage at node B
Ri
Apply KCL at node A
I1 = I2 + I 3
Let Y1 = 1/R, Y2 = 1 / R, Y3 = Y4 = SC
(Vi-VA) Y1 = (VA - VB) Y2 + (VA - V0) Y3-------------->(1)
Vi Y1 = VA (Y1 + Y2 + Y3) - VB Y2 - V0 Y3--------------->(2)

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The gain of the op-amp is

V0  V0 
= A0 ∴VB =  ---------------->(3)
VB  A0 

Sub (3) in (2)

V0
Vi Y1 = VA(Y1+ Y2+ Y3)- Y2- V0 Y3------------->(4)
A0

Apply KCL at node B

(VA-VB) Y2 = VB Y4
VA Y2= VB (Y4 + Y2)

ww  Y + Y4 
V A = VB  2  --------------------->(5)

w.E
Apply (5) in (4)
 Y2 

ViY1 =
V0
A0

asy
 Y2 + Y4 
 Y2 
VY
(Y1 + Y2 + Y3 ) − V0Y3 − 0 2
A0

Vi Y1 =
V0
[ En
(Y2 + Y4 )(Y1 + Y2 + Y3 ) − Y2Y3 A0 − Y22 ]
V0
Y2 A0

Y1Y2 A0gin
Vo
=

=
Y1Y2 A0
ee
Vi (Y2 + Y4 )(Y1 + Y2 + Y3 ) − Y2Y3 A0 − Y22

rin
Vi Y1Y2 + Y4 (Y1 + Y2 + Y3 ) + Y2Y3 (1 − A0 )
Replace Y1 = Y2 = 1/R and Y3 = Y4 = SC g.n
H (S ) =
V0
= 2 2 2
Ao
+ SCR(3 − A0 ) + 1
Vi S C R ---------------------------->(6)
et
From eqn.(6) it is noted that H(0) = A0 for S = 0 and H(∞) = 0 for S = ∞.
The transfer function of low pass second order hydraulic electrical and mechanical
system can be written as

A0ωh2
H (S ) = 2 ----------------->(7)
S + αωh S + ωh2
Where A0 = gain,
ωh = 1/RC = upper cutoff frequency in radians per seconds.
α = (3- A0) = damping coefficient
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put S = jω in eqn.(7)
A0
H ( jω ) =
( jω / ω h )
2
+ jα (ω / ω h ) + 1
The normalized expression
n for LPF is
A0
H ( jω) = 2
S + αS n + 1
n

Where Sn = jω/ωh = Normalized

alized frequency.
The frequency response for different
diff values of α is shown in figure.

ww
w.E
asy
En
gin
•
Fig 3.13 Second order
For heavily damped
d filte
ee
rder LPF circuit frequency response with
h α values
v
filter (α >1.7) the response is stable. The
he ro
rin
roll off begins

•
very early to the pass
As α is reduced the
ss ba
band.
e response
resp exhibits overshoot and ripple begins g.n
egins to appear at
the early stage of pass band. If α is reduced too much the filter m
oscillatory.
may become
et
• The flattest pass band occurs for damping coefficient is 1.414.
14. This
T is called
butter worth filter. Audio filters are usually butter worth filter.
r. The Chebyshev
filters more likely dampe
amped =1.06.
• A Bessel filter is heavily
eavily damped and has a damping coefficient
cient of
o 1.73. This
gives better pulse respon
response however causes attenuation in the
e upper
uppe end of the
pass band.

Filter Design:
1. Choose a value forr a high
hig cut off freq (fH ).

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2. To simplify the design calculations, set R2 = R3 = R and C2 = C3 = C then choose

a value of c<=1µf.
1
3. Calculate the value of R using eqn. R =
2πf H C

4. Finally, because of the equal resistor (R2 = R3) and capacitor (C2 = C3 ) values,
the pass band volt gain AF = 1 + RF / R1 of the second order had to be = to
1.586. RF = 0.586 R1 . Hence choose a value of R1 < =100kΩ and
5. Calculate the value of RF.

ww
w.E
asy
En
gin
ee rin
g.n
et

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UNIT IV
SPECIAL ICs
Part - A
1. What are the applications of 555 Timer?
• Astable multivibrator
• Monostable multivibrator
• Missing pulse detector
• Linear ramp generator
• Pulse width modulation
• FSK generator

ww • Pulse position modulator

w.E • Schmitt trigger

2. List the applications of 555 timer in monostable mode of operation
•
• asy
Missing pulse detector
Linear ramp generator
• Frequency divider
En
•

gin
Pulse width modulation.
3. List the applications of 555 timer in Astable mode of operation:
*
*
FSK generator
Pulse-position modulator
4. Define 555 IC?
ee rin
g.n
The 555 timer is an integrated circuit specifically designed to perform signal
generation
and timing functions.
5. List the basic blocks of IC 555 timer?
et
• A relaxation oscillator
• RS flip flop
• Two comparator
• Discharge transistor.
6. List the features of 555 Timer?
• It has two basic operating modes: monostable and astble
• It is available in three packages. 8 pin metal can , 8 pin dip, 14 pin dip.
• It has very high temperature stability.

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7. Define duty cycle?

The ratio of high output and low output period is given by a mathematical
parameter called duty cycle. It is defined as the ratio of ON Time to total time.
8. Define VCO.
A voltage controlled oscillator is an oscillator circuit in which the frequency of
oscillations can be controlled by an externally applied voltage.
9. List the features of 566 VCO.
• Wide supply voltage range(10-24V)
• Very linear modulation characteristics
• High temperature stability

ww
10. What does you mean by PLL?
A PLL is a basically a closed loop system designed to lock output frequency

w.E
and phase to the frequency and phase of an input signal.
11. Define lock range.

asy
When PLL is in lock, it can trap freq changes in the incoming signal. The range

En
of frequencies over which the PLL can maintain lock with the incoming signal is called
as lock range.
12. Define capture range.
gin
signal is called as capture range.
13. Define pull-in time.
ee
The range of frequencies over which the PLL can acquire lock with the input

rin
g.n
The total time taken by the PLL to establish lock is called pull-in time.
14. List the applications of 565 PLL.
•
•
Frequency multiplier
Frequency synthesizer
et
• FM detector

15. What are the two types of analog multiplier Ics?

a) IC AD 533
b) IC AD 534

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Part - B
1) Discuss in detail about Mo
Monostable multivibrator using 555 timer
er IC
Block Diagram of 555 Timer
er IC:

ww
w.E
asy
•
Fig
En
g 4.1 Block Diagram of 555 Timer IC
From the above figure
figure, three 5k internal resistors act as
s voltage
vol divider
providing bias voltage
gin
ge of 2/3 Vcc to the upper comparator & 1/3
/3 Vcc to the lower
comparator. It is possib
voltage to the control ee
ossible to vary time electronically by applying
rol voltage
vol input terminal.
ying a modulation

rin
MONOSTABLE OPERATION:
ION:

g.n
et

Fig. 4.2. 555

55 con
connected as a Monostable Multivibrator
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Model Graph:

ww • Initially when the output

Fig. 4.3. Model graph
utput is low, i.e. the circuit is in a stable state,
te, tra
transistor Q1 is

w.E
•
ON & capacitor C is shor
shorted to ground. The output remains low.
During negative going
oing trigger
tr pulse, transistor Q1 is OFF, which
hich releases the

•
asy
short circuit across the e
external capacitor C & drives the outputt high
Now the capacitor C starts
high.
sta charging toward Vcc through RA. When the voltage
across the capacitor
tor eq
En
equals 2/3 Vcc, upper comparator switches
tches from low to

• Since C is unclamped, gin

high. i.e. Q = 0, the trans
transistor Q1 = OFF ; the output is high.
ped, voltage
v across it rises exponentially throug
hrough R towards

•
Vcc with a time constant
nstant RC.
After the time period,
ee rin
d, the upper comparator resets the FF, i.e.. Q = 1, Q1 = ON;
the output is low. i.e
.e disc
voltage across the capacitor
capac as in fig (b) is given by g.n
discharging the capacitor C to ground potentia
tential (Fig.c) The

Vc = Vcc (1-e-t/RC
Therefore At t = T, Vc = 2/3
t/RC

/3 Vcc
) fffff. (1) et
cc(1-e-T/RC)
2/3 Vcc = Vcc(1 or T = RC ln (1/3)
or T = 1.1RC
RC se
seconds ffff. (2)
If the reset is applied Q2 = OFF
OFF, Q1 = ON, timing capacitor C immediately
iately discharged.
The output now will be as in figure (d & e). If the reset is released
ed output
ou will still
remain low until a negative
e goin
going trigger pulse is again applied at pin 2.

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2) Explain astable multivibrat

vibrator using 555 timer with neat sketch
The 555 timer as an Astable
able M
Multivibrator:
• An Astable multivibrato
vibrator, often called a free running multivib
ultivibrator, is a
rectangular wave genera
enerating circuit.
• Unlike the monostable
table multivibrator, this circuit does not require
quire an external
trigger to change the
he sta
state of the output, hence the name free runnin
running.
• However, the time during which the output is either high or low is determined
de by
2 resistors and capacitor
ors, which are externally connected to the
he 555
5 timer.

ww
w.E
asy
En
gin
ee rin
Fig 4.4 Astable Multivibrator
g.n
Model Graph et

Fig. 4.5. Model graph

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The above figures show the 555 timer connected as an astable multivibrator and its
model graph
Initially, when the output is high :
• Capacitor C starts charging toward Vcc through RA & RB.
• However, as soon as voltage across the capacitor equals 2/3 Vcc. Upper
comparator triggers the FF & output switches low.

When the output becomes Low:

• Capacitor C starts discharging through RB and transistor Q1, when the voltage
across C equals 1/3 Vcc, lower comparator output triggers the FF & the output

ww •
goes High.
Then cycle repeats. The capacitor is periodically charged & discharged

w.E
•
between 2/3 Vcc & 1/3 Vcc respectively.
The time during which the capacitor charges from 1/3 Vcc to 2/3 Vcc equal to

asy
the time the output is high & is given by

En
tc = ln 2(RA+RB)C ffffff.(1)
Where [ln 2 = 0.69]

gin
= 0.69 (RA+RB)C

ee
Where, RA & RB are in ohms. And C is in farads.
Similarly, the time during which the capacitors discharges from 2/3 Vcc to 1/3 Vcc is
equal to the time, the output is low and is given by, rin
tc = RB C ln 2
td = 0.69 RB C ffffffff.(2) g.n
where, RB is in ohms and C is in farads.
Thus the total period of the output waveform is
et
T = tc + td = 0.69 (RA+2RB)C ff..(3)
This, in turn, gives the frequency of oscillation as,
f 0 = 1/T = 1.45/(RA+2RB)C fff(4)
Equation 4 indicates that the frequency f 0 is independent of the supply voltage Vcc.
Often the term duty cycle is used in conjunction with the astable multivibrator. The duty
cycle is the ratio of the time tc during which the output is high to the total time period T.
It is generally expressed as a percentage.

% duty cycle = (tc / T )* 100

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% DC = [(RA+RB)/ /(RA+2RB)] * 100

3) Explain phase locked loop with neat sketch:

ww
w.E Fig 4.6 Basic Block Diagram of a PLL
Phase locked loop construction and operation:
•
asy
The PLL consists of i) Phase detector ii) LPF iii) VCO. The phase detector or
comparator compares the input frequency fIN with feedback frequency fOUT.
•
En
The output of the phase detector is proportional to the phase difference

•
between fIN & fOUT.
gin
The output of the phase detector is a dc voltage & therefore is often referred

•
to as the error voltage.
ee rin
The output of the phase detector is then applied to the LPF, which removes

g.n
the high frequency noise and produces a dc level. This dc level in turn, is
input to the VCO.
•
•
et
The output frequency of VCO is directly proportional to the dc level.
The VCO frequency is compared with input frequency and adjusted until it is
equal to the input frequencies.
• PLL goes through 3 states, i) free running ii) Capture iii) Phase lock.
• Before the input is applied, the PLL is in free running state. Once the input
frequency is applied the VCO frequency starts to change and PLL is said to
be in the capture mode.
• The VCO frequency continuous to change until it equals the input frequency
and the PLL is in phase lock mode.
• When Phase locked, the loop tracks any change in the input frequency
through its repetitive action.
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• If an input signal vs of frequency fs is applied to the PLL, the phase detector

compares the phase and frequency of the incoming signal to that of the
output vo of the VCO.
• If the two signals differ in frequency of the incoming signal to that of the
output vo of the VCO. If the two signals differ in frequency and/or phase, an
error voltage ve is generated.
• The phase detector is basically a multiplier and produces the sum (fs + fo)
and difference (fs - fo) components at its output.
• The high frequency component (fs + fo) is removed by the low pass filter
and the difference frequency component is amplified then applied as control

ww voltage vc to VCO.

w.E • The signal vc shifts the VCO frequency in a direction to reduce the
frequency difference between fs and fo.
•

asy
Once this action starts, we say that the signal is in the capture range. The
VCO continues to change frequency till its output frequency is exactly the

• En
same as the input signal frequency. The circuit is then said to be locked.
Once locked, the output frequency fo of VCO is identical to fs except for a

gin
finite phase difference φ.
•
ee
This phase difference φ generates a corrective control voltage vc to shift the

rin
VCO frequency from f0 to fs and thereby maintain the lock. Once locked,
PLL tracks the frequency changes of the input signal. Thus, a PLL goes

g.n
through three stages (i) free running, (ii) capture and (iii) locked or tracking.
Capture range:
• et
The range of frequencies over which the PLL can acquire lock with an input
signal is called the capture range. This parameter is also expressed as
percentage of fo.

Pull-in time:
• The total time taken by the PLL to establish lock is called pull-in time. This
depends on the initial phase and frequency difference between the two signals
as well as on the overall loop gain and loop filter characteristics.

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(a) Phase Detector:

• Phase detector compares the input frequency and VCO frequency and
generates DC voltage i.e., proportional to the phase difference between the two
frequencies.
• Depending on whether the analog/digital phase detector is used, the PLL is
called either an analog/digital type respectively.
• Even though most monolithic PLL integrated circuits use analog phase
detectors.
• Example for Analog: Double-balanced mixer , Example for Digital: Ex-OR, Edge
trigger, monolithic Phase detector.

ww
(b) Low – Pass filter:

w.E
• The function of the LPF is to remove the high frequency components in the
output of the phase detector and to remove the high frequency noise.
•
asy
LPF controls the characteristics of the phase locked loop. i.e, capture range,
lock ranges, bandwidth
•
En
Lock range (Tracking range): The lock range is defined as the range of

gin
frequencies over which the PLL system follows the changes in the input
frequency fIN.
•
ee
Capture range: Capture range is the frequency range in which the PLL acquires

rin
phase lock. Capture range is always smaller than the lock range.
•
g.n
Filter Bandwidth: Filter Bandwidth is reduced, its response time increases.
However reduced Bandwidth reduces the capture range of the PLL. Reduced

et
Bandwidth helps to keep the loop in lock through momentary losses of signal
and also minimizes noise.

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4.Explain Voltage Controlled Oscillator (VCO) with neat diagram

• A common type of VCO available in IC form is Signetics NE/SE566. The pin
configuration and basic block diagram of 566 VCO are shown in figures below.

ww
w.E
asy
En
gin
• ee
Fig. 4.7. VCO pin and block diagram

rin
Referring to the circuit in the above figure, the capacitor c1 is linearly charged
or discharged by a constant current source/sink.
•
g.n
The amount of current can be controlled by changing the voltage vc applied at

•
IC chip. The voltage at pin 6 is held at the same voltage as pin 5. et
the modulating input (pin 5) or by changing the timing resistor R1 external to the

Thus, if the modulating voltage at pin 5 is increased, the voltage at pin 6 also
increases, resulting in less voltage across R1 and thereby decreasing the
charging current.
• The voltage across the capacitor C1 is applied to the inverting input terminal of
Schmitt trigger via buffer amplifier.
• The output voltage swing of the Schmitt trigger is designed to Vcc and 0.5 Vcc.
If Ra = Rb in the positive feedback loop, the voltage at the non-inverting input
terminal of Schmitt trigger swings from 0.5 Vcc to 0.25 Vcc.

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• When the voltage on the

th capacitor c1 exceeds 0.5 Vcc during
ring ccharging, the
output of the Schmitt
itt trig
trigger goes LOW (0.5 Vcc).
• The capacitor now disch
ischarges and when it is at 0.25 Vcc, the outpu
output of Schmitt
trigger goes HIGH
H (Vc
(Vcc). Since the source and sink currents
rrents are equal,
capacitor charges and di
discharges for the same amount of time.
• This gives a triangular
ular vo
voltage waveform across c1 which is also
lso available
av at pin
4. The square wave
e output
out of the Schmitt trigger is inverted by bu
buffer amplifier
at pin 3. The outputt wave
waveforms are shown near the pins 4 and 3.

The output frequency of the

he VC
VCO can be given as follows:

ww
w.E
asy
where
re V+ is Vcc.
• The output frequency
the voltage vc at the En
cy of the VCO can be changed either by (i)
e mod
modulating input terminal pin 5.
i) R1, (ii) c1 or (iii)

• The voltage vc can

gin
n be varied by connecting a R1R2 circuitt as shown
s in the

•
figure below.
The components R1and
in the centre of the op
ee
1and c1 are first selected so that VCO output

rin
tput frequency
fr
operating frequency range. Now the modulating
modu input
lies

voltage is usually varied from 0.75 Vcc to Vcc which can produce
variation of about 10
0 to 1.
1
g.n
duce a frequency

et

Fig. 4.8. SE/NE 566 pin diagram

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UNIT – V
APPLICATION ICs
Part - A

1. What are the limitations of three terminal regulators? (April/May 2015)

• No short circuit protection
• Output voltage (+ve or –ve) is fixed
2. How current boosting is achieved in a 723 IC? (April/May 2015)
The current boosting is achieved in a 723 IC by adding a boost transistor Q1 to
the voltage regulator

ww 3. How power amplifies is classified? Mention any one power amplifier IC?
(April/May 2015)

w.E A variety of monolithic as well as hybrid power amplifiers are commercially

available. LM 380 is a popular power audio amplifier

asy
4. What is an opto-coupler? (April/May 2015), (April/May 2012)

En
Opto-coupler IC is a combined package of a photo-emitting device and a photo-
sensing device.
•
gin
Better isolation between the two stages.
•
•
•
Wide frequency response ee
Impedance problem between the stages is eliminated.

Easily interfaced with digital circuit rin

• Compact and light weight
g.n
• Problems such as noise, transients, contact bounce, are eliminated
5. What is meant by thermal shut down applied to voltage regulators?
(Nov/Dec 2014)
et
The IC has a temperature sensor (built-in) which turns off the IC when it
becomes too hot (usually 125°C to 150°C). The output current will drop and
remains there until the IC has cooled significantly.

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6. Draw the internal block diagram of a function generator IC (Nov/Dec 2014)

ww
w.E
asy
7. What is a switching regulator? (Nov/Dec 2014)
Switching regulators are those which operate the power transistor as a high

En
frequency on/off switch, so that the power transistor does not conduct current

gin
continuously. This give improved efficiency over series regulators.
8. Draw the pin diagram of IC 8038 (Nov/Dec 2014)

Sine adjust 1 ee 14 NC

rin
Sine out

Triangle out
2

3
13

12
NC

Sine adjust g.n

Duty cycle

Frequency adjust
4 IC
8038
11 -VEE Or GND et
5 10 Timing Capacitor

+Vcc 6 9 Square out

F.M bias FM sweep out

7 8

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9. State the need for protection diodes in voltage regulators based on LM

317 regulators (April/May 2014), (Nov/Dec 2012)
When external capacitors are used with LM 317, it is sometimes necessary to
add protection diodes to prevent the capacitors from discharging through low
current points into the regulators. Protection diodes are included for use with
outputs greater than 25V and high value of output capacitance.
10. What is an isolation amplifier? (Nov/Dec 2012), (Nov/Dec 2011)
An isolation amplifier is an amplifier that offers electrical isolation between its
input and output terminals. Easy to use, ultra low leakage 18 pin DIP package
• Better isolation between the two stages.

ww
•
•
Impedance problem between the stages is eliminated.
Wide frequency response
•
•
w.EEasily interfaced with digital circuit
Compact and light weight
• asy
Problems such as noise, transients, contact bounce, are eliminated

En
11. Define load regulation (Nov/Dec 2013)
It is defined as the change in output voltage for a change in load current and of

gin
V0. Typical value of load regulation for 7805 is 15mV for 5mA < I0 < 1.5A

•
2012) ee
12. What are the applications of switch mode power supplies? (April/May

Adjustable high voltage constant current sources rin

• Battery powered systems
g.n
•
•
•
Telecommunication circuits
Personal computers
Printers
et
• Video games
• Motor and industrial control systems
• Automotive applications
13. Why do switching regulators have better efficiency than series regulator?
(April/May 2012)
Switching regulators have better efficiency than series regulator because the
switching regulator have applied a very high frequency signal (40 kHz and
above), the transistors used are acting as the switches and become alternately

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ON and OFF at a frequency of 20 kHz. The time duration, power consumption,

size and cost is very small when compared with series regulator.
14. What are the disadvantages of linear voltage regulator? (Nov/Dec 2011)
• Low efficiency
• Use of step down transformer is bulky and expensive
• Weight is high
• Response to load variation is fast
15. What are the advantages of switched capacitor filter over active filters?
(April/May 2010)
The switched capacitor filter have very small size, low power consumption and

ww reliability and price which are more favorable than those of passive LC and RC
active filters.

w.E
Part -B

asy
En
1. With a neat diagram explain the operation of LM 380 power amplifier
(Nov/Dec 2014), (Nov/Dec 2013)
Features of LM380:
gin
ee
1. Internally fixed gain of 50 (34dB)
2. Output is automatically self centering to one half of the supply voltage.
3. Output is short circuit proof with internal thermal limiting.
rin
4. Input stage allows the input to be ground referenced or ac coupled.
5. Wide supply voltage range (5 to 22V). g.n
6. High peak current capability.
7. High impedance
et
8. Low total harmonic distortion
9. Bandwidth of 100 KHz at Pout = 2W & RL = 8Ω
Introduction:
• Small signal amplifiers are essentially voltage amplifiers that supply their loads
with larger amplifier signal voltage.
• On the other hand, large signal or power amplifier supply a large signal current
to current operated loads such as speakers & motors.
• In audio applications, however, the amplifier called upon to deliver much higher
current than that supplied by general purpose op-amps.
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• This means that loads

ads such
su as speakers & motors requiring substan
ubstantial currents
cannot be driven directly
irectly by the output of general purpose op-amp
amps. However
there are two possible
ible so
solutions,
• To use discrete orr mon
monolithic power transistors called powerr boosters
boo at the
output of the op-amp
• To use specialized ICs d
designed as power amplifiers.

ww
w.E
asy
En
gin
Fig 5.1 Functional
ee rin
tional block diagram of Audio Power Amplifier
plifier

g.n
et

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ww
w.E
asy Fig 5.2 Pin diagram

En
gin
ee rin
g.n
et
Fig 5.3 Block diagram
LM380 circuit description:
It is connected of 4 stages,
i. PNP emitter follower
ii. Different amplifier
iii. Common emitter
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iv. Emitter follower

(i) PNP Emitter follower:
• The input stage is emitter follower composed of PNP transistors Q1 & Q2 which
drives the PNP Q3-Q4 differential pair.
• The choice of PNP input transistors Q1 & Q2 allows the input to be referenced to
ground
• i.e., the input can be direct coupled to either the inverting & non-inverting
terminals of the amplifier.
(ii) Differential Amplifier:
• The current in the PNP differential pair Q3-Q4 is established by Q7, R3 & +V.

ww• The current mirror formed by transistor Q7, Q8 & associated resistors then

w.E
•
establishes the collector current of Q9.
Transistor Q5 & Q6 constitute of collector loads for the PNP differential pair.
•

asy
The output of the differential amplifier is taken at the junction of Q4 & Q6
transistors & is applied as an input to the common emitter voltage gain.
(iii)
•
Common Emitter:
En
Common Emitter amplifier stage is formed by transistor Q9 with D1, D2 & Q8 as
a current source load.
gin
•

•
ee
The capacitor C between the base & collector of Q9 provides internal

rin
compensation & helps to establish the upper cutoff frequency of 100 KHz.
Since Q7 & Q8 form a current mirror, the current through D1 & D2 is
approximately the same as the current through R3.
g.n
•
et
D1 & D2 are temperature compensating diodes for transistors Q10 & Q11 in that
D1 & D2 have the same characteristics as the base-emitter junctions of Q11.
Therefore the current through Q10 & (Q11-Q12) is approximately equal to the
current through diodes D1 & D2.
(iv) (Output stage) - Emitter follower:
• Emitter follower formed by NPN transistor Q10 & Q11. The combination of PNP
transistor Q11 & NPN transistor Q12 has the power capability of NPN transistors
but the characteristics of a PNP transistor.
• The negative dc feedback applied through R5 balances the differential amplifier
so that the dc output voltage is stabilized at +V/2

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• To decouple the input stage from the supply voltage +V, by pass capacitor in
order of micro farad should be connected between the by-pass terminal (pin 1)
& ground (pin 7).
• The overall internal gain of the amplifier is fixed at 50. However gain can be
increased by using positive feedback.

2. Draw and explain the functional diagram of 723 general purpose regulator
(April/May 2012), (Nov/Dec 2012), (Nov/Dec 2011)
Features of IC723:
i. Unregulated dc supply voltage at the input between 9.5V & 40V

wwii.
iii.
Adjustable regulated output voltage between 2 to 3V
Maximum load current of 150 mA (ILmax = 150mA)

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iv.
v.
With the additional transistor used, ILmax upto 10A is obtainable
Positive or Negative supply operation
vi.
asy
Internal Power dissipation of 800mW
vii.
viii. En
Built in short circuit protection
Very low temperature drift
ix. High ripple rejection
gin
ii.
Reference generating block
Error Amplifier
ee
The simplified functional block diagram can be divided in to 4 blocks.
i.

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iii.
iv.
Series Pass transistor
Circuitry to limit the current g.n
(i) Reference Generating block:
The temperature compensated Zener diode, constant current source & voltage
et
reference amplifier together from the reference generating block. The Zener diode is
used to generate a fixed reference voltage internally. Constant current source will
make the Zener diode to operate at affixed point & it is applied to the Non – inverting
terminal of error amplifier. The Unregulated input voltage} Vcc is applied to the voltage
reference amplifier as well as error amplifier.
(ii) Error Amplifier:
Error amplifier is a high gain differential amplifier with 2 inputs (inverting & non
inverting). The Non-inverting terminal is connected to the internally generated

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reference voltage. The Inverti

nverting terminal is connected to the fulll regu
regulated output
voltage.

ww
w.E
asy Fig 5.4 Pin diagram of IC723

(iii) Series Pass Transis

En
ansistor:
Q1 is the internal series
es pas
gin
pass transistor which is driven by the error
rror am
transistor actually acts ass a vvariable resistor & regulates the output
amplifier. This
tput voltage.
v The
collector of transistor Q1
maximum collector voltage
ee
1 is connected to the Un-regulated power
wer supply. The
e of Q1 is limited to 36Volts. The maximum
rin
um current
c which
can be supplied by Q1 is 150m
(iv)
150mA.
Circuitry to limit
it the current: g.n
• The internal transistor
OFF transistor. It turns
urns ON
O when the IL exceeds a predetermined
et
tor Q2 is used for current sensing & limiting.. Q2 is normally
ed limit.
lim Low
voltage, Low current
nt is capable
c of supplying load voltage which
h is equal
eq to or
between 2 to 7Volts.

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3. Explain the working

king of IC 8038 function generator (April/
April/May 2015),
(Nov/Dec 2013), (Nov/D
Nov/Dec 2011), (Nov/Dec 2010)

ww
w.E
asy
En
Fig 5.5 Function
gin
ctional block diagram of Function generator
ator

ee rin
g.n
et
Fig. 5.6. Output waveform

It consists of two current sources,

source two comparators, two buffers, one FF an
and a sine
wave converter.

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Pin description:
Pin 1 & Pin 12: Sine wave adjusts:
• The distortion in the sine wave output can be reduced by adjusting the 100KΩ
pots connected between pin12 & pin11 and between pin 1 & 6.

Pin 2 Sine Wave Output:

• Sine wave output is available at this pin. The amplitude of this sine wave is 0.22
Vcc. Where} 5V ≤ Vcc ≤ } 15 V.

Pin 3 Triangular Wave output:

• Triangular wave is available at this pin. The amplitude of the triangular wave is

ww 0.33Vcc. Where} 5V ≤ Vcc ≤ } 15 V.

w.E
Pin 4 & Pin 5 Duty cycle / Frequency adjust:
• The symmetry of all the output wave forms & 50% duty cycle for the square

asy
wave output is adjusted by the external resistors connected from Vcc to pin 4.
These external resistors & capacitors at pin 10 will decide the frequency of the
output wave forms.
En
Pin 6 + Vcc:
gin
•
pin. ee
Positive supply voltage the value of which is between 10 & 30V is applied to this

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Pin 7: FM Bias:
• This pin along with pin no8 is used to TEST the IC 8038. g.n
Pin 9: Square Wave Output:
•
et
A square wave output is available at this pin. It is an open collector output so
that this pin can be connected through the load to different power supply
voltages. This arrangement is very useful in making the square wave output.

Pin 10: Timing Capacitors:

• The external capacitor C connected to this pin will decide the output frequency
along with the resistors connected to pin 4 & 5.

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Pin 11: -VEE or Ground:

• If a single polarity supply is to be used then this pin is connected to supply
ground & if supply voltages are to be used then (-) supply is connected to this
pin.

Pin 13 & Pin 14: NC (No Connection)

Important features of IC 8038:
• All the outputs are simultaneously available.
• Frequency range: 0.001Hz to 500 kHz
• Low distortion in the output wave forms.

ww •
•
Low frequency drifts due to change in temperature.
Easy to use

w.E
Parameters:
(i) Frequency of the output wave form:
•
asy
The output frequency dependent on the values of resistors R1 & R2 along with
the external capacitor C connected at pin 10.
•
En
If RA= RB = R & if RC is adjusted for 50% duty cycle then fo = RC 0.3; RA = R1,

(ii)
RB = R3, RC = R2
gin
Duty cycle / Frequency Adjust: (Pin 4 & 5):
•
ee
Duty cycle as well as the frequency of the output wave form can be adjusted by
controlling the values of external resistors at pin 4 & 5.
rin
•
g.n
The values of resistors RA & RB connected between Vcc * pin 4 & 5 respectively
along with the capacitor connected at pin 10 decide the frequency of the wave

•
form.
The values of RA & RB should be in the range of 1kΩ to 1MΩ.
et
(iii) FM Bias:
• The FM Bias input (pin7) corresponds to the junction of resistors R1 & R2.
• The voltage Vin is the voltage between Vcc & pin8 and it decides the output
frequency.
• The output frequency is proportional to Vin as given by the following expression
For RA = RB (50% duty cycle).
fo = CRAVcc
1.5Vin ; where C is the timing capacitor

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With pin 7 & 8 connected to each other the output frequency is given by fo = RC 0.3
where R = RA = RB for 50% duty cycle.
(iv) FM Sweep input (pin 8):
• This input should be connected to pin 7, if we want a constant output frequency.
But if the output frequency is supposed to vary, then a variable dc voltage
should be applied to this pin. The voltage between Vcc & pin 8 is called Vin and it
decides the output frequency as, 1.5 Vin fo = C RA Vcc
• A potentiometer can be connected to this pin to obtain the required variable
voltage required to change the output frequency.

ww
4. What are IC voltage regulators? Explain the principle of IC LM 317 as a voltage
regulator (Nov/Dec 2010)

w.E
Classifications of IC voltage regulators:
IC Voltage Regulator:
•
• Positive/negative asy
Fixed Volt Reg. Adjustable O/P Volt Reg Switching Reg

•
En
Fixed & Adjustable output Voltage Regulators are known as Linear Regulator.
•
Switching Regulator: gin
A series pass transistor is used and it operates always in its active region.

•
•
ee
Series Pass Transistor acts as a switch.
The amount of power dissipation in it decreases considerably.
rin
•
Adjustable Voltage Regulator: g.n
Power saving result is higher efficiency compared to that of linear.

Advantages of Adjustable Voltage Regulator over fixed voltage regulator are,

• Adjustable output voltage from 1.2v to 57 v
et
• Output current 0.10 to 1.5 A
• Better load & line regulation
• Improved overload protection
• Improved reliability under the 100% thermal overloading

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Adjustable Positive Voltage

tage R
Regulator (LM317):

Fig. 5.7. LM317 circuit

• LM317 series adjustab
justable 3 terminal positive voltage regulator
ulator, the three

ww •
terminals are Vin, Vout & a
LM317 requires only
ly 2 e
adjustment (ADJ).
external resistors to set the output voltage.

w.E
• LM317 produces a volta
voltage of 1.25v between its output & adjustme
This voltage is called
ed as Vref.
ustment terminals.

•
asy
Vref (Reference Voltage)
ltage) is a constant; hence current I1 flows
s through
thro R1 will
also be constant. Becau
“program resistor”. En
ecause resistor R1 sets current I1. It is called
led “current
“cu set” or

• Resistor R2 is called
lled as
gin
a “Output set” resistors, hence current
rrent through this

•
resistor is the sum of I1 & Iadj
LM317 is designed in su ee
such as that Iadj is very small & constant
line voltage & load curre
current. rin
nt with changes in

The output voltage Vo is, Vo=R1I1+ (I1+Iadj) R2 ------------- (1) g.n

Where I1= Vref/R1
Vo = (Vref/R1) R1 + Vref/R1 + Iadj R2
= Vref + (Vref/R1) R2 + Iadj R2
et
Vo = Vref [1 + R2/R1] + Iadj R2 ------------- (2)
R1 = Curre
Current (I1) set resistor
R2 = outpu
output (Vo) set resistor
Vref = 1.25vv whi
which is a constant voltage between outpu
output and ADJ
terminals.
Current Iadj is very small. Theref
Therefore the second term in (2) can be neglecte
lected.
Thus the final expression for the output voltage is given by
Vo= 1.25v [1 + R2/R1] -------------
--------- (3)

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Eqn (3) indicates that we can va

vary the output voltage by varying the resista
esistance R2.
The value of R1 is normally
ly kep
kept constant at 240 ohms for all practicall appl
applications.
Practical Regulator using
g LM
LM317:

ww
w.E
asy
Fig. 5.8. Practical Regulator using LM317
• If LM317 is far away
En
ay from
fro the input power supply, then 0.1µff disc type or 1µf

•
tantalum capacitor shoul
The output capacitor gin
should be used at the input of LM317.
or Co is optional. Co should be in the range of 1 tto 1000µf.
• The adjustment termina
ripple rejection ratio
o as h
ee
rminal is bypassed with a capacitor C2 this
is will improve the

rin
high as 80 dB is obtainable at any output
ut lev
level.
• When the filter capacitor
acitor is used, it is necessary to use the protectiv
tective diodes.
• These diodes do not a
allow the capacitor C2 to discharge through
throu g.n
the low

•
current point of the regulator.
These diodes are requ
regul
required only for high output voltages (above
higher values of output
tput ccapacitance 25µf and above.
et
above 25v) & for

5. Explain in detail about

ut optocouplers
opt (April/May 2012), (Nov/Dec
ec 2013)
20
• Optocouplers or Optoisolato
isolators is a combination of light source & ligh
light detector in
the same package.
• They are used to couple
ple ssignal from one point to other optically,
lly, by providing a
complete electric isolation
tion b
between them.
• This kind of isolation is prov
provided between a low power control circuit
rcuit & high power
output circuit, to protect
ct the control circuit.

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• Depending on the type of light source & detector used we can get a variety of
optocouplers. They are as follows,
(i) LED – LDR optocoupler
(ii) LED – Photodiode optocoupler
(iii) LED – Phototransistor optocoupler
Characteristics of optocoupler:
(i) Current Transfer Ratio (CTR)
(ii) Isolation Voltage
(iii) Response Time
(iv) Common Mode Rejection

ww
Types of optocoupler:
(i) LED – Photodiode optocoupler:

w.E
asy
En
gin
ee rin
g.n
et
Fig. 5.9. Optocoupler circuit and its waveform

• LED photodiode shown in figure, here the infrared LED acts as a light source &
photodiode is used as a detector.
• The advantage of using the photodiode is its high linearity. When the pulse at the
input goes high, the LED turns ON. It emits light. This light is focused on the
photodiode.

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• In response to this light the photocurrent will start flowing though the photodiode.
As soon as the input pulse reduces to zero, the LED turns OFF & the photocurrent
through the photodiode reduces to zero. Thus the pulse at the input is coupled to
the output side.

(ii) LED – Phototransistor Optocoupler:

ww
w.E
asy
En
gin
ee rin
g.n
•
Fig. 5.10. LED – Phototransistor Optocoupler circuit & its waveform
et
The LED phototransistor optocoupler shown in figure. An infrared LED acts as a
light source and the phototransistor acts as a photo detector.
• This is the most popularly used optocoupler, because it does not need any
additional amplification.
• When the pulse at the input goes high, the LED turns ON. The light emitted by the
LED is focused on the CB junction of the phototransistor.
• In response to this light photocurrent starts flowing which acts as a base current for
the phototransistor.

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• The collector current of phototransistor starts flowing. As soon as the input pulse
reduces to zero, the LED turns OFF & the collector current of phototransistor
reduces to zero. Thus the pulse at the input is optically coupled to the output side.
Advantages of Optocoupler:
• Control circuits are well protected due to electrical isolation.
• Wideband signal transmission is possible.
• Due to unidirectional signal transfer, noise from the output side does not get
coupled to the input side.
• Interfacing with logic circuits is easily possible.
• It is small size & light weight device.

ww
Disadvantages:
•
• w.E
Slow speed.
Possibility of signal coupling for high power signals.
Applications:
• asy
Optocouplers are used basically to isolate low power circuits from high power

•
circuits.
En
At the same time the control signals are coupled from the control circuits to the
high power circuits. gin
Optocoupler IC:
ee rin
The optocouplers are available in the IC form MCT2E is the standard
optocoupler IC which is used popularly in many electronic application.
• g.n
This input is applied between pin 1& pin 2. An infrared light emitting diode is

•
connected between these pins.
et
The infrared radiation from the LED gets focused on the internal phototransistor.
• The base of the phototransistor is generally left open. But sometimes a high value
pull down resistance is connected from the Base to ground to improve the
sensitivity.
• The block diagram shows the opto-electronic-integrated ciruit (OEIC) and the major
components of a fiber-optic communication facility.

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ww
w.E
Fig. 5.11 Block
k dia
diagram of opto-electronic-integrated circ
cuit

asy
En
gin
ee rin
g.n
et

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gin
ee rin
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et

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et

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et

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et

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ee rin
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et

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et

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ee rin
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et

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ee rin
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et

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ee rin
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et

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ee rin
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et

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ee rin
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et

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ee rin
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et

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ee rin
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