Electrical and Electronic Fundamental

a. During the first-half cycle (positive cycle):

Vin R1 Vout
VBIAS
0V to t1 t2 + RL 0V to t1
_

Figure 6.35: Limiter with voltage biasing which limits the negative
alternation in the first-half cycle

When the input voltage goes positive, the diode reverse-biases and
acts as an open circuit. All currents flow through the load resistor, RL. The
output voltage determined by the voltage divider formed by R1 and RL.

VP(out) = �RL R+LR1� VP(in)

b. During the second-half cycle (negative cycle):

Vin R1 Vout
VBIAS
0V to t1 t2 _ RL 0V to t1 t3
+
-VBIAS - 0.7

Figure 6.36: Limiter with voltage biasing which limits the negative
alternation in the second-half cycle

When the input voltage goes below -0.7V, the diode is still in reverse-
bias condition and acts as an open circuit. Therefore, -0.7V flow through
load resistor, RL .However, when the input goes above -0.7V, the diode

143

forward-biases and acts as a closed circuit. Therefore, -VBIAS can flow
through the RL .

The output voltage at RL, VP(out) = -VBIAS - 0.7

6.9.2.2 Limiter with voltage biasing which limits the positive alternation

We also divide the diode limiter with voltage biasing for negative
alternation limiting basic operation into two cycles; first-half cycle and
second-half cycle. Basic operation for positive alternation limiting:

a. During the first-half cycle (positive cycle):

Vin R1 Vout
VBIAS
0V to t1 t2 + VBIAS + 0.7 t1
_
RL 0V to

Figure 6.37: Limiter with voltage biasing which limits the positive alternation
in the first-half cycle

When the input voltage goes below 0.7V, the diode is still in reverse-
bias condition and acts as an open circuit. Therefore, 0.7V flow through
load resistor, RL . However, when the input goes above 0.7V, the diode
forward-biases and acts as a closed circuit. Therefore, -VBIAS can flow
through the RL .

The output voltage at RL, VP(out) = VBIAS + 0.7

144

b. During the second-half cycle (negative cycle):

Vin R1 Vout
VBIAS
0V to t1 t2 + VBIAS + 0.7 t1 t2
_
RL 0V to

Figure 6.39: Limiter with voltage biasing which limits the positive alternation
in the second-half cycle

When the input voltage goes positive, the diode reverse-biases and
acts as an open circuit. All currents flow through the load resistor, RL. The output
voltage determined by the voltage divider formed by R1 and RL

VP(out) = - �RL R+LR1� VP(in)

Exercise
1. Determine Vo and ID for the network of Figure 6.38

ID Ge ID Si +
10V + Si Si 4.7kΩ
Si Vo 16V
1kΩ (b) Vo
-
(a) 12V

-

Figure 6.38

145

2. Determine the output voltage for the circuit in Figure 6.39(a) for each
input voltage in (b),(c), and (d).

R1 Vin Vin Vin
4.7kΩ +25V +12V +5V

Si t0 t0 t
0 -12V -5V

Vin Vout

R2 -25V
4.7kΩ

(a) (b) (c) (d)

Figure 6.39

3. Determine the output waveform for each circuit in Figure 6.40(a), (b),
(c) and (d).

Vin R1 Vin R1
+5V 1kΩ 56Ω
+10V
0 t Si
-5V Ge RL 0 RL
(a) 1kΩ
Vin t 1MΩ
+200V R1 -10V 3V
100Ω
0 Vin (b)
-200V t
+20V R1
(c) 100Ω
Si
RL Si Ge
t
680Ω 0
10V 15V
50V
(d)
-20V

Figure 6.40

146

Summary

1. There is current through a diode only when it is forward-biased.
Ideally, there is no current when there is no bias nor when there is
reverse bias. Actually, there is a very small current in reverse bias due
to the thermally generated minority carriers, but this can usually be
neglected.

2. A diode conducts current when forward-biased and blocks current
when reversed-biased.

3. The I-V characteristic curve shows the diode current as a function of
voltage across the diode.

4. The ideal model represents the diode as a closed switch in forward
bias and as an open switch inreverse bias.

5. The practical model represents the diode as a switch in series with the
barrier potential.

6. The single diode in a half-wave rectifier is forward-biased and
conducts for of the inputcycle.

7. The output frequency of a half-wave rectifier equals the input
frequency.

8. The two basic types of full-wave rectifier are center-tapped and
bridge.

9. Each diode in a full-wave rectifier is forward-biased and conducts for
of the input cycle.

10. The output frequency of a full-wave rectifier is twice the input
frequency.

11. The peak output voltage of a bridge rectifier equals the total peak
secondary voltage less twodiode drops.
147

12. A dc power supply typically consists of a transformer, a diode rectifier,
a filter, and a regulator.

13. Diode limiters cut off voltage above or below specified levels. Limiters
are also called clippers.

148

CHAPTER 7

TRANSISTOR

Introduction
A transistor is a semiconductor device used to amplify and switch

electronic signals. It is made of P and N type of semiconductor material. A
transistor has three terminals connected to three doped semiconductor
regions internally or can be connected to an external circuit. A voltage or
current applied to one pair of the transistor's terminals changes the current
flowing through another pair of terminals. Because the controlled (output)
power can be much more than the controlling (input) power, the transistor
provides amplification of a signal. Some transistors are packaged
individually but many more are found embedded in integrated circuits.
Transistor can be classified in two general types: the Bipolar Junction
Transistor (BJT) and the Field Effect Transistor (FET). This topic will discuss on
the characteristics of the BJT configuration and explore its use in switching
and amplifier circuits.

Learning Objectives
At the end of this topic, students should be able to:
1. Describe the basic structure of the bipolar junction transistor (BJT)
2. Discuss the parameters and characteristics of a transistor and how

they apply to transistor circuits
149

3. Use the BJT as a switch.
4. Identify the BJT terminals and construct the BJT amplifier circuits

7.1 Introduction to Bipolar Junction Transistor (BJT)

7.1.1 BJT Structure and Symbol

The basic structure of the bipolar junction transistor (BJT) determines
its operating characteristics. It is constructed with three doped
semiconductor regions separated by two pn junctions as shown in Figure
7.1: Basic structure below. These junctions are similar to the junctions we saw
in the diodes and thus they may be forward biased or reverse biased. The
BJT can be divided into two types NPN and PNP based on the fabrication.
The NPN type has one p region between two n regions and the PNP type
has one n region between two p regions. The three terminals or leg of the
BJT are known as base, collector and emitter.

C (collector) C (collector)

n Base-Collector p Base-Collector
junction junction

B p B n
(Base) n Base-Emitter (Base)
p Base-Emitter
junction
junction

E (emitter) E (emitter)

NPN PNP

Figure 7.1: Basic structure

The symbols of the NPN and PNP is shown in the figure below. The
different between these two symbols is only on the arrow direction between
base and emitter leg. For the NPN the arrow is point toward the emitter and

150

vise verse for the PNP. By referring to the arrow direction, we can say the
NPN stand for not pointing in.

C (collector) C (collector)

B (base) B (base)

E (emitter) E (emitter)

NPN PNP

Figure 7.2: NPN and PNP symbols of the BJT

7.2 BJT Characteristic

7.2.1 Basic Transistor Operation

In order to used or operate the transistor as an amplifier or switch, the
two pn junctions must be correctly biased with external DC voltages. Figure
7.3 below shows the proper bias arrangement for npn transistors for active
operations as an amplifier (notice that in both cases the base-emitter (BE)
junction is forward-biased and the base-collector (BC) junction is reverse-
biased). The biasing for the PNP is vise verse of the NPN transistor. In this
chapter we only focus on npn BJT.

151

RB B RC IC
C

IB E +
IE
VCC

-

+

VBB

-

Figure 7.3: Biasing for NPN BJT

In general the transistor can be operated in three region known as
active region, saturation region and cutoff region. The biasing for active
operation is illustrated in Figure above. Notice that the BE junction is forward
bias by connecting base terminal (p-type) to the positive terminal of VBB
and emitter terminal (n-type) to the negative terminal of VBB. In contrast,
the collector terminal (n-type) is connected to the positive terminal of VCC.
VCC is set so that collector terminal is more positive than the base. This is to
ensure the reverse bias in BC junction. This condition is required to the
transistor used as the amplifier.

For the saturation region, the BE junction must be in forward biased
condition and the BC junction must also in forward bias condition.
Meanwhile for the cutoff region, the BE junction must be in reverse biased
condition and the BC junction must also in reverse bias condition. This
condition is required to enable the transistor to be used as a switch.

Cutoff region occurred when the transistor is in cutoff region of its
operation, as the base lead open. There is very small amount of collector
leakage current, due to mainly thermally produced carriers. Because is very
small, it will usually be neglected in circuit analysis so that . In cutoff, both
the base-emitter and the base-collector junction are reverse-biased.

152

Saturation region occurred when the base emitter junction becomes

forward-biased and the base current is increased, the collector current also
increases = and decreases as a result of more drop across the
collector resistor = − . When reaches its saturation value,
( ), the base-collector junction becomes forward-biased and is not
increases anymore even with a continued increase in . At the point of
saturation, the relation = is no longer valid. And ( ), for a

transistor occurs somewhere below the knee of the collector curves.

7.2.2 DC Load Line

Cutoff and saturation can be illustrated in relation to the collector
characteristics curves by the use of a load line.
The bottom of load line is at ideal cutoff where = 0 and =
The top of the load line is at saturation where = ( ) and = ( )

In between cutoff and saturation along the load line is the active
region of the transistor’s operation as shown in Figure below.

IC (mA)

saturation

(VCE(sat), IC(sat))

DC Load Line

Cut-off

0 (VCC, 0) VCE (V)

Figure 7.4: DC load line of a BJT consisting of saturation and cutoff points

153

Example 7.1
Draw the DC load line of the circuit below.

RC IC
2.5kΩ C

RB + VCC
50kΩ B 20V

IB E -
IE
VBB +
5V

-

Solution:

Cutoff point when IC = 0, VCE = VCC = 20V

Saturation point, when VCE = 0 and IC = VCC = 20 = 8mA
RC 2.5k

Base on this point, draw the DC load line across this two points

7.3 DC Biasing

The BJT works as an amplifier when they meet the above
requirement. The BJT must be operating in active region. In general, we only

154

discuss on determination of the quiescent point (Q point) of a given BJT
biasing circuit. Q point is the operating point of the amplifier with IC and VCE
parameter. It will determine the performance of an amplifier. The biasing
circuit is design to have a center Q point to ensure full amplification of the
input signal.

There are a few types of configuration in analyzing DC biasing circuit.
One of them is fixed biased configuration as discuss in above topic. Another
type of biasing cover in this chapter is emitter biased and voltage divider
biased.

7.3.1 Fixed Bias Circuit

The fixed-bias circuit of Figure 7.5 provides a relatively straightforward
and simple introduction to transistor dc bias analysis. Even though the
network employs an npn transistor, the equations and calculations apply
equally well to a pnp transistor configuration merely by changing all current
directions and voltage polarities.

VCC

RB RC IC
C1
ac IB C ac
input B C2 output
signal signal

+

VBE E

- IE

Figure 7.5: Fixed bias circuit
155

The current directions of Figure 7.6 are the actual current directions,
and the voltages are defined by the standard double-subscript notation.
For the dc analysis the network can be isolated from the indicated ac levels
by replacing the capacitors with an open circuit equivalent. In addition,
the dc supply VCC can be separated into two supplies (for analysis purposes
only) as shown in Figure 7.6 to permit a separation of input and output
circuits. It also reduces the linkage between the two to the base current IB.
The separation is certainly valid, as we note in Figure 7.6 that VCC is
connected directly to RB and RC just as in Figure 7.5.

VCC VCC

RB RC IC

IB C
B
+

+ VCE

VBE -

- E
IE

Figure 7.6: DC equivalent of Figure 7.5

156

7.3.1.1 Forward Bias of Base–Emitter
Consider first the base–emitter circuit loop of Figure 7.7.

+

RB

+ _
VCC _
B

+ -

IB VBE _ E

Figure 7.7

Writing Kirchhoff’s voltage equation in the clockwise direction for the loop,
will obtain

-VCC + IBRB + VBE = 0

Note the polarity of the voltage drop across RB as established by the
indicated direction of IB. Solving the equation for the current IB will result in
the following:

IB = VCC - VBE (7.1)
RB

Equation (7.1) is certainly not a difficult one to remember if one simply
keeps in mind that the base current is the current through RB and by Ohm’s

157

law that current is the voltage across RB divided by the resistance RB. The
voltage across RB is the applied voltage VCC at one end less the drop across
the base-to-emitter junction (VBE). In addition, since the supply voltage VCC
and the base–emitter voltage VBE are constants, the selection of a base
resistor, RB, sets the level of base current for the operating point.

7.3.1.2 Collector-Emitter Loop

The collector–emitter section of the network appears in Figure 7.8
with the indicated direction of current IC and the resulting polarity across
RC. The magnitude of the collector current is related directly to IB through

=

+ +
_ VCC
RC IC

_

C

+

VCE

_-

E

Figure 7.8: Collector-Emitter loop

It is interesting to note that since the base current is controlled by the
level of RB and IC is related to IB by a constant, the magnitude of IC is not a
function of the resistance RC. Change RC to any level and it will not affect
the level of IB or IC as long as we remain in the active region of the device.
However, as we shall see, the level of RC will determine the magnitude of
VCE, which is an important parameter.

158

Applying Kirchhoff’s voltage law in the clockwise direction around
the indicated closed loop of Figure 7.4 will result in the following:

and -VCC + ICRC + VCE = 0
VCE = VCC - ICRC
(7.2)

which states in words that the voltage across the collector–emitter region
of a transistor in the fixed-bias configuration is the supply voltage less the
drop across RC. As a brief review of single- and double-subscript notation
recall that

VCE = VC - VE (7.3)

where VCE is the voltage from collector to emitter and VC and VE are
the voltages from collector and emitter to ground respectively. But in this
case, since VE = 0 V, we have

In addition, since VCE = VC (7.4)
and VE = 0 V, then VBE = VB - VE (7.5)
(7.6)
VBE = VB

159

Keep in mind that voltage levels such as VCE are determined by
placing the red (positive) lead of the voltmeter at the collector terminal
with the black (negative) lead at the emitter terminal as shown in Figure 7.9.
VC is the voltage from collector to ground and is measured as shown in the
same figure. In this case the two readings are identical, but in the networks
to follow the two can be quite different. Clearly understanding the
difference between the two measurements can prove to be quite
important in the troubleshooting of transistor networks.

VCC
VC

RC

+_

C VCE

+_

E

Figure 7.9: Measuring VCE dan VC

160

Example 7.2

Determine the following for the fixed-bias configuration of Figure 7.10.

VCC
12V

RB IC RC
240kΩ 2.2kΩ
C1 C2
ac
10μF IB + 10μF output

ac VCE
input

_ ß = 50

Figure 7.10

a. IBQ and ICQ.
b. VCEQ.
c. VB and VC.
d. VBC

Solution:

a. IB = VCC - VBE = 12 - 0.7 = 47.08μA
RB 240k

IC = βIB = (50)(47.08μ) = 2.35mA

161

b. VCE = VCC - ICRC = 12 - (2.35m)(2.2k) = 6.83V

c. VB = VBE = 0.7V
VC = VCE = 6.83V

d. Using double-subscript notation yields
VBC = VB - VC = 0.7 - 6.83 = -6.13V

with the negative sign revealing that the junction is reversed-biased,
as it should be for linear amplification.

7.3.2 Emitter Bias
The emitters biased obtain by adding resistor to the base bias

configuration. The emitter resistor gives some stability to the bias circuit by
providing current series feedback. This circuit will give better performance
and more stable compare to the fixed bias circuit which related to the
stability of the β towards heat effect. Figure below shows the emitter biased
configuration. Figure 7.11 shows the emitter bias configuration.

162

+ RC
VRC IC

_

+ VRB _ C + +
B VC
RB VCC
IB + _
E IE + -

+ VB + VE

VBB VRE RE

-

__ _

Figure 7.11: Emitter biased configuration

From Figure 7.11, the analysis of the emitter biased can be stated with
determine the value of IB and apply similar step in analyzing fixed bias. Using
Kirchhoff’s voltage law to the input loop or BE loop, we get

VBB - IBRB - VBE - IERE = 0

Substituting IE = (1 + β)IB into equation (7.7)

IB = VBB - VBE (7.7)
RB + (1 + β)RE

The collector current can be find by (7.8)
IC = βIB

and from the output loop or CB loop, the value of VCE can easily define by
KVL at the loop,

VCC - ICRC - VCE - IERE = 0

163

and finally

VCE = VCC - ICRC - IERE (7.9)

Example 7.3

Base on the emitter biased circuit, find Q point, VB and VC of the circuit in
Figure 7.12

VCC
20V

RC
2kΩ

RB IC
430kΩ IB B C

+ ß = 50

VBE E IE

-

RE
1kΩ

Figure 7.12

Solution:
The q point is ICQ and VCEQ of the circuit.
From in input loop, we get

VCC - IBRB - VBE - IERE = 0 and IE = (1 + β)IB
164

Rearrange the equation,

IB = VBB - VBE = 20 - 0.7 = 40.1μA
RB + (1 + β)RE 430k + (1 + 50)1k

The value of collector current

IC = βIB = (50)(40.1μ) = 2.01mA

The value of emitter current

IE = (1 + β)IB = (1 + 50)40.1μ = 2.05mA

For the output loop;
VCC - ICRC - VCE - IERE = 0

Rearrange the equation,
VCE = VCC - ICRC - IERE = 20 - (2.01m)(2k) - (2.05m)(1k) = 13.53V

The value of base voltage
VB = VBE + IERE = 0.7 + (2.05m)(1k) = 2.75V

7.3.3 Voltage Divider Bias

Voltage divider biased configuration is the most popular compared
to the previous configuration. This circuit only need one power supply and
voltage-divider bias is more stable (β independent) than other bias types. It
means that in this circuits, ICQ and VCEQ are almost independent of β. The
level of IBQ will change with β so as to maintain the values of ICQ and VCEQ
almost same, thus maintaining the stability of Q point. Refer to the Figure
7.13 below, the circuit has two resistors connected to the base terminal. This
R1 and R2 is the main components in determining the circuit name. These

165

resistors play a rule to provide the voltage terminal by applying voltage
divider rule.

VCC

R1 RC

IC
IB B C

+ E IE

VBE

-

R2 RE

Figure 7.13: Voltage divider bias configuration

There are two method in analyzing the voltage divider biased
configuration; Exact Method and Approximate Method. The Exact method
can be applied to any voltage divider circuit. Meanwhile Approximate
method is direct method, simpler, saves time and energy, can be applied
in most of the circuits. The approximate method is only suitable or
applicable when the βRE > 10R2. By approximate method, IE can be
assumed same as IC.

7.3.3.1 Exact Method Analysis

In this method, the Thevenin equivalent network for the network to
the left of the base terminal to be found. This input part of the circuits is
determined first by using Thevenin theorm. By Thevenin theorm the Thevenin
voltage, VTH and Thevenin resistance, RTH can easily define.

166

VCC

VCC R1 B RC
IC
+
C
R2 VTH
E IE
_ RE

Figure 7.14: Thevenin Voltage view for voltage divider biased

To find the Thevenin Voltage, VTH the voltage divider rules can be
applied

VTH = �R1R+2R2� × VCC

From the thevenin theorm, the circuit can be illustrated as below to
find thevenin resistance, RTH. The power supply is consider short circuit, then
the R1 is parallel with R2 as shown in Figure 7.15.

167

VCC

R1 B RC
IC
+
C
R2 VTH
E IE
_ RE

Figure 7.15: Thevenin resistance view for voltage divider biased

The Thevenin Resistance, RTH is determining by

RTH = R1 ∥ R2= R1R2
R1 + R2

From the figure above, the voltage divider circuit can be reform like
the emitter biased circuit as Figure 7.16.

RC IC

C

RTH B +

IB E VCC
IE
-
RE
VTH +
-

Figure 7.16: Reformation of voltage divider circuit
168

In analyzing the above circuit is similar with analyzing the emitter
biased circuit.

From the input loop,

IB = VTH - VBE
RTH + (1 + β)RE

Then,

IC = βIB and IE = (1 + β)IB

From the output loop,

VCE = VCC - ICRC - IERE

169

7.3.3.2 Approximate method
The approximate method only applicable when βRE > 10R2 condition

is meet. Then, the collector and emitter current can be assumed similar.
IC = IE.

VCC

R1 RC

IC
IB B C

+ - E IE

VBE

R2 RE

The voltage at terminal base

VB = �R1R+2R2� × VCC

As we can see in the circuit

VB = VBE +VRE

Thus,

VRE = VB - VBE

Finally,

IE = �VRREE� = IC
170

The collector to emitter voltage can be find by

VCE = VCC - IC(RC + RE)

The approximate method offer simpler step and calculation in determining the Q
point of a voltage divider circuit. Thus saving our time and energy.

Example 7.4

Based on the voltage divider circuit, find Q point by using exact and approximate
method. Compare the result.

VCC
18V

R1 IB B RC
39kΩ 4kΩ
+ -
R2 IC
3.9kΩ VBE C

ß = 140

E IE

RE
1.5kΩ

Solution:
Exact method:
1. Find Thevenin voltage, VTH:

VTH = �39k3+.9k3.9k� ×18 = 1.64V

171

2. Find Thevenin resistance, RTH:

RTH = 39k ∥3.9k = (39k)(3.9k) = 3.55kΩ
39k + 3.9k

3. Reforming the circuit like emitter bias

4. Find base current, IB:

1.64 -0.7
IB = 3.55k + (1 + 140)1.5k = 4.37μA
5. Find collector current, IC:
IC = (140)(4.37μ) = 0.612mA
6. Find collector to emitter voltage, VCE:
VCE =1.64 - (0.612m)(4k) -(4.37μ + 0.612m)(1.5k) = 15.63V

Approximate method:

1. Find base voltage, VB:

VB = �R1R+2R2� × VCC = �39k3+.9k3.9k� ×18 = 1.64V

2. Find emitter voltage, VE (same as VRE):

VRE =VRE = VB - VBE = 1.64 - 0.7 = 0.94V

3. Fine emitter current, IE:

IE = VRE = 0.94 = 0.63mA = IC
RE 1.5k

4. Find collector to emitter voltage, VCE:

172

VCE = VCC - IC(RC + RE) = 18 - (0.63m)(4k +1.5k) = 14.54V

Exercise
1. Base on the fixed biased circuit shown in Figure 7.15, calculate the

following:

VCC

IB RC
20μA 2.2kΩ

IC

RB B C
VCE = 7.2V
+

VBE E IE

-

4mA

Figure 7.17

a. Collector current, IC
b. VCC
c. DC beta, β
d. Base resistance, RB

173

2. Given IC = 2mA, VC = 7.6V and VE = 2.4V, determine the following:

VCC
12V

RC

IC

RB IB B C
ß = 80
+

VBE E IE

-

RE

a. Collector resistance, RC
b. Emitter resistance, RE
c. Base resistance, RB
d. Collector to emitter voltage, VCE
e. Base voltage, VB

174

3. Determine Q-point for the voltage divider circuit shown below by
using exact and approximate method. Compare the result.

VCC
18V

R1 IB B RC
82kΩ 5.6kΩ
+ -
R2 IC
22kΩ VBE C

ß = 50

E IE

RE
1.2kΩ

175

Summary
In this topic, we have studies:
1. Basic structures and symbol of NPN and PNP Bipolar Junction

Transistors.
2. The relation between current parameter IB, IE and IC. Beta is the

current gain of a transistor which is the ratio of IC/IB.
3. A transistor can be operated as an electronics switch. When the

transistor is off it is in cutoff condition (no current) flow. When the
transistor is on, it is in saturation condition (maximum current).
4. For the BJT to operate as an amplifier, the base-emitter junction is
forward biased and the collector-base junction is reverse biased. It is
said to be in active region or the linear region of a transistor is the
region of operation.
5. The purpose of biasing is to establish a stable operating point (Q-
point). The Q-point is the best point for operation of a transistor for a
given collector current.

176

CHAPTER 8

OPERATIONAL AMPLIFIER

Introduction
An operational amplifier or op-amp is basically a complete high-gain

voltage amplifier in a small package. It was constructed from cascading of
4 main stages which are two stages of differential amplifier, DC level shifter
and Current driver. The op-amps are now made using integrated circuit
technology, so they come in the typical multi-pin IC packages. Its near-
ideal performance makes it a popular device in electronic circuits. These
little parts are so versatile, useful, handy, easy to use and cheap. With the
proper external components, the operational amplifier can perform a wide
variety of “operations” on the input voltage. The respective circuits are
called comparator, inverting, non-inverting, summers (adders) and
subtractors. Most of the op-amp circuits applied as the main component or
devices in the LED driver, Oscillator, Filter and voltage regulator
applications.

Learning Objectives
At the end of this topic, student should be able to:
1. explain about the op-amp devices and its applications
2. explain the operation of the open loop of Op-amp circuit such as

comparators
177

3. calculate the output of the close loop Op-amp circuits such as
Inverting Amplifier, Non-inverting Amplifier, Unity Follower, Summing
Amplifier, Integrator, Differentiator andMultistage Amplifier.

8.1 Introduction to Operational Amplifier

Operational amplifier (op amp) is a very high gain differential
amplifier with a high input impedance (typically a few mega ohms) and
low output impedance (less than 100 ohms). Figure 8.1 shows the symbol of
an op amp with the power supplies and without power supply. The power
supplies are usually not shown in the schematic symbol due to the simplicity
but are always understood to be there.

Input 1 +Vcc Output Input 1 + Output
Input 2 Input 2 -
+
-

-Vcc

Figure 8.1: Op Amp symbol with and without power supply

The Op-amp have several very important characteristics that make
them so useful

1. An op-amp has two inputs and it amplifies the voltage difference
between those two Inputs.These two inputs are known as the non-
inverting input, labeled (+), and the Inverting input,labeled (-), as
shown in Figure 8.1. The output voltage is a function of the non-
Inverting inputvoltage minus the inverting input voltage.

Vo = A(V1 - V2)

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where A = voltage gain of the op amp
2. The op-amp must be connected to external power supplies (+ VCC

and –VCC). The outputvoltage cannot be more positive than the
positive power supply or more negative than thenegative power
supply. The gain (A) is very high, typically more than 100,000.
3. Since the gain (A) is very high, therefore small different between V1
and V2 cause saturation tooutput (VO). Thus most op-amp
application used negative feedback to control the output orgain.
4. Op-amps amplify DC as well as AC.
5. The input currents are almost zero. In more technical terms, the op-
amp has very high input impedance.

Example 8.1
Question 1:
Name of the three common terminals shown in the op-amp symbol.
Answer:
It has three terminals: two input terminals (+ve input and –ve input) and one
output terminal.

Question 2:
The +ve input is called ______ input and the –ve input is called the
____________input.

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Answer:
non-inverting and inverting inputs

Question 3:
What is the +Vcc and –Vcc mean.
Answer:
The op-amp devices require external power supplies (+ve and –ve supplies)
to operate.

8.1.1 Practical symbol for IC op-amp
During the laboratories or practical session, the practical symbols is

used to identify the terminal of the inputs, output and the power supplies of
the op-amp. Normally this symbol can be given in the datasheet. For
instance, the 741 op-amp comes in 8 pins dual inline package (DIP) and
symbols are shown in Figure 8.2 below. The pins are numbered like diagram
below. From the figure, the Pin 8 is not connected (NC) and Pin 1 and 5 are
used to eliminate the offset voltage. The device is normally requires to be
powered by a dual power supplies (V+ and V- with respect to ground). So,
the Pin 4 and 5 are connected to a positive voltage V+ and a negative
voltage V- respectively.

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18V+ null 7 V+
27 3
36-+ 5 2- 7 6 Vout
45 +U1 V+OS2
null 2 -uA741V- OS1 6 3+ 4
1
4 V-
V-

Figure 8.2: The schematic symbols of op-amp and terminals identification.

8.1.2 Ideal and Practical op-amp characteristics

Op-amp is designed to sense the difference between the voltage
signals applied at its two input terminals (that is, the quantity 2 − 1),
multiply this by a gain A, and cause the resulting voltage ( 2 − 1) to
appear at output, = ( 2 − 1). The characteristic of the ideal and
practical op-amp is directly translated from its equivalent circuits as follows:

Ro Ro

Vd Ri AdVd Vo Vd AdVd Vo

(a) (b)

Figure 8.3: (a) Practical and (b) Ideal op-amp equivalent circuits

Based on the equivalents circuits, ideal op-amp has the following
characteristics;

• It has infinite gain (i.e. Open Loop Voltage gain is infinite)
• It has infinite input impedance (i.e. no current demanded at amplifier

input terminals)
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• It has zero output impedance (i.e. Output Voltage unaffected by
load)

• It has infinite bandwidth (i.e. the response extends from dc to ∞)
In practice the ic op-amp falls short of the ideal characteristics, however
the following applies;

• Very HIGH input resistance
• Very LOW output resistance
• Very large Open Loop Voltage Gain

For example, Table 8.1 below shows the characteristics of the most
common op-amp ic available in the market.

Table 8.1: Typical characteristics of op-amp IC’s

Parameters LM741 LF347 LM318 Ideal
2 x105 1 x105 2 x105 ∞
Open Loop Gain
(AOL) 2MΩ 1012MΩ 3MΩ ∞
Input Resistance (Rin) 75Ω 75Ω 75Ω 0

Output resistance 1MHz 4MHz 15MHz ∞
(RO) 90dB 100dB 100dB ∞
Gain Bandwidth

CMRR

Example 8.2

Question 1:

Describe some of the characteristics of an ideal op-amp

Answer:

A practical op amp has a infinite input impedance, zero output
impedance, infinite voltage gain and infinite band width.

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Question 2:

Describe some of the characteristics of a practical op-amp

Answer:

A practical op amp has a very high input impedance, very low output
impedance and very high voltage gain

8.1.3 Open Loop and Close Loop Configurations

In general, the configurations using op-amp can be divided into
open loop configuration or close loop configuration. For the open loop
configuration, there is no feedback from output back to the input. The
open-loop voltage gain, AOL, of an op-amp is the internal voltage gain of
the device and represents the ratio of output voltage to input voltage when
there are no external components. The open-loop voltage gain is set
entirely by the internal design. Open-loop voltage gain can range up to
10000 and is not a well-controlled parameter. Data sheets often refer to the
open-loop voltage gain as the large-signal voltage gain. For the open loop
configuration, comparator is one of the best examples of the applications.

As the Open-Loop Voltage Gain of the Op-Amp is very large, an
extremely small difference in the two input voltages drives the Op-Amp into
its saturated output states, i.e. it will cause the output voltage to go all the
way to its extreme positive or negative voltage limit. As this is seldom
desirable the full gain of the Op-Amp is not usually applied to an input,
instead negative feedback (using external resistors) is applied to reduce
the overall gain through signal feedback. When the negative feedback is
applied, the configuration is known as the close loop configuration as
shown in

Figure 8.4 below. The negative feedback or close loop configuration
will reduce the gain and improve many characteristics of the op-amp.

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Negative
feedback

circuit

Vf - Vo = Acl VIN
VIN +

Figure 8.4: Close loop configuration circuits

Where:

VIN is the input signal
Vf is a portion of the output signal fed back to the inverting input
Acl the closed loop gain is the voltage gain with negative feedback

The Op-Amp responds to the voltage VIN at its non-inverting input,
which moves the output towards saturation. However, a fraction of this
output is returned to the inverting terminal through the feedback path. As
the feedback signal approaches the value of VIN, there is nothing left for
the Op-Amp to amplify as Vf – VIN = 0. Thus the feedback signal tries (but
never quite succeeds) in matching the input signal and thus the gain is
controlled by the amount of feedback used.

Example 8.3
Question 1:
What is the different between close loop and open loop configuration?
Answer:
Close loop configuration have a feedback connected from the output
terminal to the input legs.

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Question 2:
What is the purpose of the feedback circuit?
Answer:
Typically, it used to control the voltage gain of the op-amp circuit.

8.2 Comparator

The comparator configuration does not used feedback component in the
circuit. The purpose of the comparator is to compare two voltages and
produce a signal that indicates which voltage is greater. Figure 8-5 shows
the comparator circuit configuration.

- Vo
V2 +

V1

Figure 8.5: Open loop configuration: Comparator circuit

The principle of the comparator is based in Equation = ( 1 − 2).
Since the voltage gain, A, of an op-amp is very large, any difference will be
magnified to the power supply rails ±Vcc. If V1 is greater than V2 then the
difference V+ -V- will be positive and the result will be amplified to +Vcc. If,
however, V2 is greater, then the difference is negative and the result will be
amplified to -Vcc. Finally, if the two voltages are exactly equal, then the
difference will be zero and the output will also be zero.

The nonzero voltage level detectors are the commonly as the basic
principle of the comparator is applied as shown in Figure 8.6. The reference

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battery voltage or voltage divider or zener diode sets the reference voltage
at which the op-amp turns goes to the maximum voltage level.

+V +V

_ R1 _ R _
+ VREF VREF +
VIN
VOUT + VOUT VOUT
R2
VREF
VIN
VIN

(a) Battery reference (b) Voltage divider reference (b) Zener diode reference voltage

Figure 8.6: Comparator circuit: nonzero level detectors configuration

Figure 8.7 shows the inputs and output waveform of the nonzero level
detector. When the Vin greater than the Vref, the output of the op amp will
be +Vcc (+Vsat), meanwhile if the Vin less than the Vref, the result will be
amplified to –Vcc (-Vsat).

VREF t
VIN 0

+VOUT(max) t

VOUT 0

-
VOUT(max)

Figure 8.7: output waveform of the nonzero level detector

186

Example 8.4

In the comparator circuit, the inverting leg of the op amp is
connected to the 1.63V reference voltage, Vref and the non-inverting leg
is connected to the 5VP AC voltage. Sketch the output waveform of the
comparator circuit.

VIN

5V

1.63
V

t

VOUT Answer
+12V
t
-12V

One of the application examples of comparator is an over-
temperature sensing circuit as shown in Figure 8.8. The comparator circuit
will trigger the transistor Q1 to on in order to energize the relay. The
energized relay coil will pull the contactor to perform some designed task.

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Wheatstone bridge +V

R1 R3 _ Relay
+
Thermistor R5
Q1
R2 R4

Figure 8.8: Over-temperature sensing circuit

8.3 Inverting and Non-Inverting Operational Amplifier

Let’s analyze the result of connecting a resistor in feedback between
the output terminal and the inverting input terminal as shown in Figure 8.9.
The input is applied to the inverting (-) input and the non-inverting input (+)
is grounded. The resistor Rf is the feedback resistor is connected from the
output to the negative (inverting) input.

- Vo = - Rf × Vi
+ Ri

Figure 8.9: Inverting amplifier

In order to analyze the circuit, the virtual equivalent circuit of the op
amp can be used. When applying this concept, we should obey with these
two rules. First, the current at both of the op-amp inputs is zero. I+ = I- = 0.

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This is due to the very high input resistance in the op-amp. Rules number two
is the op-amp attempts to make the voltages at the two inputs equal. V+ =
V-. Thus all the analysis of the op amp amplifier will abide by this rule as
illustrate in Figure 8.10.

V1 Ri V- Rf Vo

Virtual
V+ short

Figure 8.10: Virtual equivalent circuit for op amp

Applying first rule, where V+ is connected to ground, thus V- = 0V.
Applying the second rule, I1 - I2 = 0. Thus I1 = I2. By applying KCL;

At V- node, I1 - I2=0

So, Vi - V- = V- - Vo
Ri Rf

Thus, Vi - 0 = 0 - Vo
Ri Rf

Finally the voltage gain for inverting amplifier is Vo = Av = - Rf
Vi Ri

The minus sign means that Vo will be inverted with respect to Vi,
hence the name of this amplifier is inverting amplifier. When Vi is positive,
Vo is negative, and when Vi is negative, Vo is positive. The gain of the
inverting amplifier, is completely dependent on our choices of Rf and Ri. This
is an inverting amplifier, where the gain can be greater than, less than, or
equal to unity.

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Example 8.5
Determine the voltage gain of the inverting amplifier below:

360kΩ Vo

12kΩ

Vi -
+

Answer:

Av = - Rf 360k = -30
Ri = - 12k

The basic configuration for a non-inverting amplifier and its virtual
circuit is shown in Figure 8-11. Now suppose we feedback only a fraction of
the output voltage rather than all of it. The noninverting input pin is at V1
and the inverting input pin is at a voltage constitute from R1 and Rf voltage
divider. Remember, the current flowing into an op-amp input is virtually nil,
so we can neglect its effect on the voltage divider. This is one of the very
nice features of an op-amp.

Vi + Vo= �1+ RRfi� ×Vi Vi V
- V

Rf
Ri

Figure 8.11: Non-inverting amplifier and its virtual equivalent circuit
190

In this circuit, as in the inverting amplifier, V+ is equal with V- = V1.
Applying the second rule, I1 - I2 = 0. Thus I1 = I2. By applying KCL;

0 - Vi = Vi - Vo
Ri Rf

Vo - Vi = Vi
Rf Ri

Thus, Vo = �Rf × Vi� + Vi = �1 + Rf� × Vi

Ri Ri

Finally the voltage gain for inverting amplifier,

Vo = Av = �1 + RRfi�
Vi

Notice that by adjusting the ratio of Rf and R1, we can control the gain
of the op-amp circuit almost anything we want. The circuit in is called a
non-inverting amplifier because the output voltage is in phase with the
input voltage. It is not inverted. When the input voltage increases, the
output voltages will also increase and vice versa.

Example 8.6

Determine the voltage gain of the non-inverting op-amp with Rf=100kΩ and
R1=4.7kΩ. Calculate the output voltage if the input voltage is 200mV.

Answer:

Av = �1 + Rf � = 1 + 100k = 22.3
Ri 4.7k

Vo = AvVi=(122.3)(200m) = 4.46V

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8.4 Voltage Follower

The voltage follower or unity gain buffer shown in Figure 8.12 is
probably the simplest linear op-amp circuit. The V+ is connected to the
input voltage V1. Notice the feedback connected from the output to the
inverting (-) input. This is the concept of negative feedback. A fraction (in
this case all) of the output voltage is "fed back" to the input in order to
control the gain of the op-amp. V- try to equal with the V+ thus also to the
Vo. Actually, the op-amp works very hard to maintain a very small
difference between the voltages on its inputs. This circuit is known as a
voltage follower because the output “follows” the input.

- Vo Vi V+ Vo
Vi +
Virtual

V - short

Figure 8.12: Voltage follower and its virtual equivalent circuit

Since; V+ = V- = Vi, thus also Vo, The voltage gain is Vo = Av = 1. The
Vi

unity-follower circuit provides a gain of unity (1) with no polarity or phase

reversal between the input and the output.

8.5 Summing Amplifier
The inverting amplifier can also be used as a summing amplifier. That is, it
can be made to add the effects of several input voltages together. Look
at the circuit in Figure 8.13. Notice that we can input multiple currents and
effectively sum them, as shown in the equivalent circuit.

192